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Proposition 2.2 Suppose \( f \in {L}^{\infty } \) is supported on a set of finite measure. Then \( f \in {L}^{p} \) for all \( p < \infty \), and\n\n\[ \parallel f{\parallel }_{{L}^{p}} \rightarrow \parallel f{\parallel }_{{L}^{\infty }}\;\text{ as }p \rightarrow \infty . \] | Proof. Let \( E \) be a measurable subset of \( X \) with \( \mu \left( E\right) < \infty \), and so that \( f \) vanishes in the complement of \( E \) . If \( \mu \left( E\right) = 0 \), then \( \parallel f{\parallel }_{{L}^{\infty }} = \) \( \parallel f{\parallel }_{{L}^{p}} = 0 \) and there is nothing to prove. Otherwise\n\n\[ \parallel f{\parallel }_{{L}^{p}} = {\left( {\int }_{E}{\left| f\left( x\right) \right| }^{p}d\mu \right) }^{1/p} \leq {\left( {\int }_{E}\parallel f{\parallel }_{{L}^{\infty }}^{p}d\mu \right) }^{1/p} \leq \parallel f{\parallel }_{{L}^{\infty }}\mu {\left( E\right) }^{1/p}. \]\n\nSince \( \mu {\left( E\right) }^{1/p} \rightarrow 1 \) as \( p \rightarrow \infty \), we find that \( \lim \mathop{\sup }\limits_{{p \rightarrow \infty }}\parallel f{\parallel }_{{L}^{p}} \leq \parallel f{\parallel }_{{L}^{\infty }} \) .\n\nOn the other hand, given \( \epsilon > 0 \), we have\n\n\[ \mu \left( \left\{ {x : \left| {f\left( x\right) }\right| \geq \parallel f{\parallel }_{{L}^{\infty }} - \epsilon }\right\} \right) \geq \delta \;\text{ for some }\delta > 0, \]\n\nhence\n\n\[ {\int }_{X}{\left| f\right| }^{p}{d\mu } \geq \delta {\left( \parallel f{\parallel }_{{L}^{\infty }} - \epsilon \right) }^{p} \]\n\nTherefore \( \mathop{\liminf }\limits_{{p \rightarrow \infty }}\parallel f{\parallel }_{{L}^{p}} \geq \parallel f{\parallel }_{{L}^{\infty }} - \epsilon \), and since \( \epsilon \) is arbitrary, we have \( \lim \mathop{\inf }\limits_{{p \rightarrow \infty }}\parallel f{\parallel }_{{L}^{p}} \geq \parallel f{\parallel }_{{L}^{\infty }} \) . Hence the limit \( \mathop{\lim }\limits_{{p \rightarrow \infty }}\parallel f{\parallel }_{{L}^{p}} \) exists, and equals \( \parallel f{\parallel }_{{L}^{\infty }} \) . | Yes |
Proposition 3.1 A linear functional on a Banach space is continuous, if and only if it is bounded. | Proof. The key is to observe that \( \ell \) is continuous if and only if \( \ell \) is continuous at the origin.\n\nIndeed, if \( \ell \) is continuous, we choose \( \epsilon = 1 \) and \( g = 0 \) in the above definition so that \( \left| {\ell \left( f\right) }\right| \leq 1 \) whenever \( \parallel f\parallel \leq \delta \), for some \( \delta > 0 \) . Hence, given any non-zero \( h \), an element of \( \mathcal{B} \), we see that \( {\delta h}/\parallel h\parallel \) has norm equal to \( \delta \), and hence \( \left| {\ell \left( {{\delta h}/\parallel h\parallel }\right) }\right| \leq 1 \) . Thus \( \left| {\ell \left( h\right) }\right| \leq M\parallel h\parallel \) with \( M = 1/\delta \) .\n\nConversely, if \( \ell \) is bounded it is clearly continuous at the origin, hence continuous.\n\nThe significance of continuous linear functionals in terms of closed hyperplanes in \( \mathcal{B} \) is a noteworthy geometric point to which we return later on. Now we take up analytic aspects of linear functionals. | Yes |
Theorem 3.2 The vector space \( {\mathcal{B}}^{ * } \) is a Banach space. | Proof. It is clear that \( \parallel \cdot \parallel \) defines a norm, so we only check that \( {\mathcal{B}}^{ * } \) is complete. Suppose that \( \left\{ {\ell }_{n}\right\} \) is a Cauchy sequence in \( {\mathcal{B}}^{ * } \) . Then, for each \( f \in \mathcal{B} \), the sequence \( \left\{ {{\ell }_{n}\left( f\right) }\right\} \) is Cauchy, hence converges to a limit, which we denote by \( \ell \left( f\right) \) . Clearly, the mapping \( \ell : f \mapsto \ell \left( f\right) \) is linear. If \( M \) is so that \( \begin{Vmatrix}{\ell }_{n}\end{Vmatrix} \leq M \) for all \( n \), we see that\n\n\[ \left| {\ell \left( f\right) }\right| \leq \left| {\left( {\ell - {\ell }_{n}}\right) \left( f\right) }\right| + \left| {{\ell }_{n}\left( f\right) }\right| \leq \left| {\left( {\ell - {\ell }_{n}}\right) \left( f\right) }\right| + M\parallel f\parallel \]\n\nso that in the limit as \( n \rightarrow \infty \), we find \( \left| {\ell \left( f\right) }\right| \leq M\parallel f\parallel \) for all \( f \in \mathcal{B} \) . Thus \( \ell \) is bounded. Finally, we must show that \( {\ell }_{n} \) converges to \( \ell \) in \( {\mathcal{B}}^{ * } \) . Given \( \epsilon > 0 \) choose \( N \) so that \( \begin{Vmatrix}{{\ell }_{n} - {\ell }_{m}}\end{Vmatrix} < \epsilon /2 \) for all \( n, m > N \) . Then, if \( n > N \), we see that for all \( m > N \) and any \( f \)\n\n\[ \left| {\left( {\ell - {\ell }_{n}}\right) \left( f\right) }\right| \leq \left| {\left( {\ell - {\ell }_{m}}\right) \left( f\right) }\right| + \left| {\left( {{\ell }_{m} - {\ell }_{n}}\right) \left( f\right) }\right| \leq \left| {\left( {\ell - {\ell }_{m}}\right) \left( f\right) }\right| + \frac{\epsilon }{2}\parallel f\parallel . \]\n\nWe can also choose \( m \) so large (and dependent on \( f \) ) so that we also have \( \left| {\left( {\ell - {\ell }_{m}}\right) \left( f\right) }\right| \leq \epsilon \parallel f\parallel /2 \) . In the end, we find that for \( n > N \),\n\n\[ \left| {\left( {\ell - {\ell }_{n}}\right) \left( f\right) }\right| \leq \epsilon \parallel f\parallel \]\n\nThis proves that \( \begin{Vmatrix}{\ell - {\ell }_{n}}\end{Vmatrix} \rightarrow 0 \), as desired. | Yes |
Theorem 4.1 Suppose \( 1 \leq p < \infty \), and \( 1/p + 1/q = 1 \) . Then, with \( \mathcal{B} = \) \( {L}^{p} \) we have\n\n\[{\mathcal{B}}^{ * } = {L}^{q}\]\n\nin the following sense: For every bounded linear functional \( \ell \) on \( {L}^{p} \) there is a unique \( g \in {L}^{q} \) so that\n\n\[ \ell \left( f\right) = {\int }_{X}f\left( x\right) g\left( x\right) {d\mu }\left( x\right) ,\;\text{ for all }f \in {L}^{p}. \]\n\nMoreover, \( \parallel \ell {\parallel }_{{\mathcal{B}}^{ * }} = \parallel g{\parallel }_{{L}^{q}} \) . | The proof of the theorem is based on two ideas. The first, as already seen, is Hölder's inequality; to which a converse is also needed. The second is the fact that a linear functional \( \ell \) on \( {L}^{p},1 \leq p < \infty \), leads naturally to a (signed) measure \( \nu \) . Because of the continuity of \( \ell \) the measure \( \nu \) is absolutely continuous with respect to the underlying measure \( \mu \), and our desired function \( g \) is then the density function of \( \nu \) in terms of \( \mu \) . | Yes |
Lemma 4.2 Suppose \( 1 \leq p, q \leq \infty \), are conjugate exponents.\n\n(i) If \( g \in {L}^{q} \), then \( \parallel g{\parallel }_{{L}^{q}} = \mathop{\sup }\limits_{{\parallel f{\parallel }_{{L}^{p}} \leq 1}}\left| {\int {fg}}\right| \) . | Proof. We start with (i). If \( g = 0 \), there is nothing to prove, so we may assume that \( g \) is not 0 a.e., and hence \( \parallel g{\parallel }_{{L}^{q}} \neq 0 \) . By Hölder’s inequality, we have that\n\n\[ \parallel g{\parallel }_{{L}^{q}} \geq \mathop{\sup }\limits_{{\parallel f{\parallel }_{{L}^{p}} \leq 1}}\left| {\int {fg}}\right| .\n\]\n\nTo prove the reverse inequality we consider several cases.\n\n- First, if \( q = 1 \) and \( p = \infty \), we may take \( f\left( x\right) = \operatorname{sign}g\left( x\right) \) . Then, we have \( \parallel f{\parallel }_{{L}^{\infty }} = 1 \), and clearly, \( \int {fg} = \parallel g{\parallel }_{{L}^{1}} \).\n\n- If \( 1 < p, q < \infty \), then we set \( f\left( x\right) = {\left| g\left( x\right) \right| }^{q - 1}\operatorname{sign}g\left( x\right) /\parallel g{\parallel }_{{L}^{q}}^{q - 1} \) . We observe that \( \parallel f{\parallel }_{{L}^{p}}^{p} = \int {\left| g\left( x\right) \right| }^{p\left( {q - 1}\right) }{d\mu }/\parallel g{\parallel }_{{L}^{q}}^{p\left( {q - 1}\right) } = 1 \) since \( p(q - 1) = q \), and that \( \int {fg} = \parallel g{\parallel }_{{L}^{q}} \).\n\n- Finally, if \( q = \infty \) and \( p = 1 \), let \( \epsilon > 0 \), and \( E \) a set of finite positive measure, where \( \left| {g\left( x\right) }\right| \geq \parallel g{\parallel }_{{L}^{\infty }} - \epsilon \) . (Such a set exists by the definition of \( \parallel g{\parallel }_{{L}^{\infty }} \) and the fact that the measure \( \mu \) is \( \sigma \) -finite.) Then, if we take \( f\left( x\right) = {\chi }_{E}\left( x\right) \operatorname{sign}g\left( x\right) /\mu \left( E\right) \), where \( {\chi }_{E} \) denotes the characteristic function of the set \( E \), we see that \( \parallel f{\parallel }_{{L}^{1}} = 1 \), and also\n\n\[ \left| {\int {fg}}\right| = \frac{1}{\mu \left( E\right) }{\int }_{E}\left| g\right| \geq \parallel g{\parallel }_{\infty } - \epsilon .\n\]\nThis completes the proof of part (i). | Yes |
Theorem 5.2 Suppose \( {V}_{0} \) is a linear subspace of \( V \), and that we are given a linear functional \( {\ell }_{0} \) on \( {V}_{0} \) that satisfies\n\n\[{\ell }_{0}\left( v\right) \leq p\left( v\right) ,\;\text{ for all }v \in {V}_{0}.\n\]\nThen \( {\ell }_{0} \) can be extended to a linear functional \( \ell \) on \( V \) that satisfies\n\n\[ \ell \left( v\right) \leq p\left( v\right) ,\;\text{ for all }v \in V.\n\] | Proof. Suppose \( {V}_{0} \neq V \), and pick \( {v}_{1} \) a vector not in \( {V}_{0} \) . We will first extend \( {\ell }_{0} \) to the subspace \( {V}_{1} \) spanned by \( {V}_{0} \) and \( {v}_{1} \), as we did before. We can do this by defining a putative extension \( {\ell }_{1} \) of \( {\ell }_{0} \), defined on \( {V}_{1} \) by \( {\ell }_{1}\left( {\alpha {v}_{1} + w}\right) = \alpha {\ell }_{1}\left( {v}_{1}\right) + {\ell }_{0}\left( w\right) \), whenever \( w \in {V}_{0} \) and \( \alpha \in \mathbb{R} \), if \( {\ell }_{1}\left( {v}_{1}\right) \) is chosen so that\n\n\[{\ell }_{1}\left( v\right) \leq p\left( v\right) ,\;\text{ for all }v \in {V}_{1}.\n\]\nHowever, exactly as above, this happens when\n\n\[- p\left( {-{v}_{1} + {w}^{\prime }}\right) + {\ell }_{0}\left( {w}^{\prime }\right) \leq {\ell }_{1}\left( {v}_{1}\right) \leq p\left( {{v}_{1} + w}\right) - {\ell }_{0}\left( w\right)\n\]\nfor all \( w,{w}^{\prime } \in {V}_{0} \) .\n\nThe right-hand side exceeds the left-hand side because of \( {\ell }_{0}\left( {w}^{\prime }\right) + \) \( {\ell }_{0}\left( w\right) \leq p\left( {{w}^{\prime } + w}\right) \) and the sub-linearity of \( p \) . Thus an appropriate choice of \( {\ell }_{1}\left( {v}_{1}\right) \) is possible, giving the desired extension of \( {\ell }_{0} \) from \( {V}_{0} \) to \( {V}_{1} \) .\n\nWe can think of the extension we have constructed as the key step in an inductive procedure. This induction, which in general is necessarily trans-finite, proceeds as follows. We well-order all vectors in \( V \) that do not belong to \( {V}_{0} \), and denote this ordering by \( < \) . Among these vectors we call a vector \( v \) \ | No |
Proposition 5.3 Suppose \( {f}_{0} \) is a given element of \( \mathcal{B} \) with \( \begin{Vmatrix}{f}_{0}\end{Vmatrix} = M \) . Then there exists a continuous linear functional \( \ell \) on \( \mathcal{B} \) so that \( \ell \left( {f}_{0}\right) = M \) and \( \parallel \ell {\parallel }_{{\mathcal{B}}^{ * }} = 1 \) . | Proof. Define \( {\ell }_{0} \) on the one-dimensional subspace \( {\left\{ \alpha {f}_{0}\right\} }_{\alpha \in \mathbb{R}} \) by \( {\ell }_{0}\left( {\alpha {f}_{0}}\right) = {\alpha M} \), for each \( \alpha \in \mathbb{R} \) . Note that if we set \( p\left( f\right) = \parallel f\parallel \) for every \( f \in \mathcal{B} \), the function \( p \) satisfies the basic sub-linear property (10). We also observe that\n\n\[ \left| {{\ell }_{0}\left( {\alpha {f}_{0}}\right) }\right| = \left| \alpha \right| M = \left| \alpha \right| \begin{Vmatrix}{f}_{0}\end{Vmatrix} = p\left( {\alpha {f}_{0}}\right) ,\]\n\nso \( {\ell }_{0}\left( f\right) \leq p\left( f\right) \) on this subspace. By the extension theorem \( {\ell }_{0} \) extends to an \( \ell \) defined on \( \mathcal{B} \) with \( \ell \left( f\right) \leq p\left( f\right) = \parallel f\parallel \), for all \( f \in \mathcal{B} \) . Since this inequality also holds for \( - f \) in place of \( f \) we get \( \left| {\ell \left( f\right) }\right| \leq \parallel f\parallel \), and thus \( \parallel \ell {\parallel }_{{\mathcal{B}}^{ * }} \leq 1 \) . The fact that \( \parallel \ell {\parallel }_{{\mathcal{B}}^{ * }} \geq 1 \) is implied by the defining property \( \ell \left( {f}_{0}\right) = \begin{Vmatrix}{f}_{0}\end{Vmatrix} \), thereby proving the proposition. | Yes |
Proposition 5.4 Let \( {\mathcal{B}}_{1},{\mathcal{B}}_{2} \) be a pair of Banach spaces and \( \mathcal{S} \subset {\mathcal{B}}_{1} \) a dense linear subspace of \( {\mathcal{B}}_{1} \). Suppose \( {T}_{0} \) is a linear transformation from \( \mathcal{S} \) to \( {\mathcal{B}}_{2} \) that satisfies \( {\begin{Vmatrix}{T}_{0}\left( f\right) \end{Vmatrix}}_{{\mathcal{B}}_{2}} \leq M\parallel f{\parallel }_{{\mathcal{B}}_{1}} \) for all \( f \in \mathcal{S} \). Then \( {T}_{0} \) has a unique extension \( T \) to all of \( {\mathcal{B}}_{1} \) so that \( \parallel T\left( f\right) {\parallel }_{{\mathcal{B}}_{2}} \leq M\parallel f{\parallel }_{{\mathcal{B}}_{1}} \) for all \( f \in {\mathcal{B}}_{1} \). | Proof. If \( f \in {\mathcal{B}}_{1} \), let \( \left\{ {f}_{n}\right\} \) be a sequence in \( \mathcal{S} \) which converges to \( f \). Then since \( {\begin{Vmatrix}{T}_{0}\left( {f}_{n}\right) - {T}_{0}\left( {f}_{m}\right) \end{Vmatrix}}_{{\mathcal{B}}_{2}} \leq M{\begin{Vmatrix}{f}_{n} - {f}_{m}\end{Vmatrix}}_{{\mathcal{B}}_{1}} \) it follows that \( \left\{ {{T}_{0}\left( {f}_{n}\right) }\right\} \) is a Cauchy sequence in \( {\mathcal{B}}_{2} \), and hence converges to a limit, which we define to be \( T\left( f\right) \). Note that the definition of \( T\left( f\right) \) is independent of the chosen sequence \( \left\{ {f}_{n}\right\} \), and that the resulting transformation \( T \) has all the required properties. | Yes |
Theorem 5.5 The operator \( {T}^{ * } \) defined by (13) is a bounded linear transformation from \( {\mathcal{B}}_{2}^{ * } \) to \( {\mathcal{B}}_{1}^{ * } \) . Its norm \( \begin{Vmatrix}{T}^{ * }\end{Vmatrix} \) satisfies \( \parallel T\parallel = \begin{Vmatrix}{T}^{ * }\end{Vmatrix} \) . | Proof. First, if \( {\begin{Vmatrix}{f}_{1}\end{Vmatrix}}_{{\mathcal{B}}_{1}} \leq 1 \), we have that\n\n\[ \left| {{\ell }_{1}\left( {f}_{1}\right) }\right| = \left| {{\ell }_{2}\left( {T\left( {f}_{1}\right) }\right) }\right| \leq \begin{Vmatrix}{\ell }_{2}\end{Vmatrix}{\begin{Vmatrix}T\left( {f}_{1}\right) \end{Vmatrix}}_{{\mathcal{B}}_{2}} \leq \begin{Vmatrix}{\ell }_{2}\end{Vmatrix}\parallel T\parallel .\n\]\n\nThus taking the supremum over all \( {f}_{1} \in {\mathcal{B}}_{1} \) with \( {\begin{Vmatrix}{f}_{1}\end{Vmatrix}}_{{\mathcal{B}}_{1}} \leq 1 \), we see that the mapping \( {\ell }_{2} \mapsto {T}^{ * }\left( {\ell }_{2}\right) = {\ell }_{1} \) has norm \( \leq \parallel T\parallel \) .\n\nTo prove the reverse inequality we can find for any \( \epsilon > 0 \) an \( {f}_{1} \in {\mathcal{B}}_{1} \) with \( \parallel {f}_{1}{\parallel }_{{\mathcal{B}}_{1}} = 1 \) and \( \parallel T\left( {f}_{1}\right) {\parallel }_{{\mathcal{B}}_{2}} \geq \parallel T\parallel - \epsilon \) . Next, with \( {f}_{2} = T\left( {f}_{1}\right) \in {\mathcal{B}}_{2} \) , by Proposition 5.3 (with \( \mathcal{B} = {\mathcal{B}}_{2} \) ) there is an \( {\ell }_{2} \) in \( {\mathcal{B}}_{2}^{ * } \) so that \( {\begin{Vmatrix}{\ell }_{2}\end{Vmatrix}}_{{\mathcal{B}}_{2}^{ * }} = 1 \) but \( {\ell }_{2}\left( {f}_{2}\right) \geq \parallel T\parallel - \epsilon \) . Thus by (13) one has \( {T}^{ * }\left( {\ell }_{2}\right) \left( {f}_{1}\right) \geq \parallel T\parallel - \epsilon \), and since \( {\begin{Vmatrix}{f}_{1}\end{Vmatrix}}_{{\mathcal{B}}_{1}} = 1 \), we conclude \( {\begin{Vmatrix}{T}^{ * }\left( {\ell }_{2}\right) \end{Vmatrix}}_{{\mathcal{B}}_{1}^{ * }} \geq \parallel T\parallel - \epsilon \) . This gives \( \begin{Vmatrix}{T}^{ * }\end{Vmatrix} \geq \) \( \parallel T\parallel - \epsilon \) for any \( \epsilon > 0 \), which proves the theorem. | Yes |
Theorem 5.6 There is an extended-valued non-negative function \( \widehat{m} \), defined on all subsets of \( \mathbb{R} \) with the following properties:\n\n(i) \( \widehat{m}\left( {{E}_{1} \cup {E}_{2}}\right) = \widehat{m}\left( {E}_{1}\right) + \widehat{m}\left( {E}_{2}\right) \) whenever \( {E}_{1} \) and \( {E}_{2} \) are disjoint subsets of \( \mathbb{R} \) .\n\n(ii) \( \widehat{m}\left( E\right) = m\left( E\right) \) if \( E \) is a measurable set and \( m \) denotes the Lebesgue measure.\n\ni) \( \widehat{m}\left( {E + h}\right) = \widehat{m}\left( E\right) \) for every set \( E \) and real number \( h \) . | From (i) we see that \( \widehat{m} \) is finitely additive; however it cannot be countably additive as the proof of the existence of non-measurable sets shows. (See Section 3, Chapter 1 in Book III.) | No |
Corollary 5.8 There is a non-negative function \( \widehat{m} \) defined on all subsets of \( \mathbb{R}/\mathbb{Z} \) so that:\n\n(i) \( \widehat{m}\left( {{E}_{1} \cup {E}_{2}}\right) = \widehat{m}\left( {E}_{1}\right) + \widehat{m}\left( {E}_{2}\right) \) for all disjoint subsets \( {E}_{1} \) and \( {E}_{2} \) .\n\n(ii) \( \widehat{m}\left( E\right) = m\left( E\right) \) if \( E \) is measurable.\n\n(iii) \( \widehat{m}\left( {E + h}\right) = \widehat{m}\left( E\right) \) for every \( h \) in \( \mathbb{R} \) . | We need only take \( \widehat{m}\left( E\right) = I\left( {\chi }_{E}\right) \), with \( I \) as in Theorem 5.7, where \( {\chi }_{E} \) denotes the characteristic function of \( E \) . | Yes |
Theorem 7.3 Let \( X \) be a compact metric space and \( C\left( X\right) \) the Banach space of continuous real-valued functions on \( X \) . Then, given any bounded linear functional \( \ell \) on \( C\left( X\right) \), there exists a unique finite signed Borel measure \( \mu \) on \( X \) so that\n\n\[ \ell \left( f\right) = {\int }_{X}f\left( x\right) {d\mu }\left( x\right) \;\text{ for all }f \in C\left( X\right) . \]\n\nMoreover, \( \parallel \ell \parallel = \parallel \mu \parallel = \left| \mu \right| \left( X\right) \) . In other words \( C{\left( X\right) }^{ * } \) is isometric to \( M\left( X\right) \) . | Proof. By the proposition, there exist two positive linear functionals \( {\ell }^{ + } \) and \( {\ell }^{ - } \) so that \( \ell = {\ell }^{ + } - {\ell }^{ - } \) . Applying Theorem 7.1 to each of these positive linear functionals yields two finite Borel measures \( {\mu }_{1} \) and \( {\mu }_{2} \) . If we define \( \mu = {\mu }_{1} - {\mu }_{2} \), then \( \mu \) is a finite signed Borel measure and \( \ell \left( f\right) = \int {fd\mu } \) .\n\nNow we have\n\n\[ \left| {\ell \left( f\right) }\right| \leq \int \left| f\right| d\left| \mu \right| \leq \parallel f\parallel \left| \mu \right| \left( X\right) \]\n\nand thus \( \parallel \ell \parallel \leq \left| \mu \right| \left( X\right) \) . Since we also have \( \left| \mu \right| \left( X\right) \leq {\mu }_{1}\left( X\right) + {\mu }_{2}\left( X\right) = {\ell }^{ + }\left( 1\right) + {\ell }^{ - }\left( 1\right) = \parallel \ell \parallel \), we conclude that \( \parallel \ell \parallel = \left| \mu \right| \left( X\right) \) as desired.\n\nTo prove uniqueness, suppose \( \int {fd\mu } = \int {fd}{\mu }^{\prime } \) for some finite signed Borel measures \( \mu \) and \( {\mu }^{\prime } \), and all \( f \in C\left( X\right) \) . Then if \( \nu = \mu - {\mu }^{\prime } \), one has \( \int {fd\nu } = 0 \), and consequently, if \( {\nu }^{ + } \) and \( {\nu }^{ - } \) are the positive and negative variations of \( f \), one finds that the two positive linear functionals defined on \( C\left( X\right) \) by \( {\ell }^{ + }\left( f\right) = \int {fd}{\nu }^{ + } \) and \( {\ell }^{ - }\left( f\right) = \int {fd}{\nu }^{ - } \) are identical. By the uniqueness in Theorem 7.1, we conclude that \( {\nu }^{ + } = {\nu }^{ - } \), hence \( \nu = 0 \) and \( \mu = {\mu }^{\prime } \), as desired. | Yes |
Theorem 7.4 Suppose \( X \) is a metric space and \( \ell \) a positive linear functional on \( {C}_{b}\left( X\right) \) . For simplicity assume that \( \ell \) is normalized so that \( \ell \left( 1\right) = 1 \) . Assume also that for each \( \epsilon > 0 \), there is a compact set \( {K}_{\epsilon } \subset X \) so that\n\n(21)\n\n\[ \left| {\ell \left( f\right) }\right| \leq \mathop{\sup }\limits_{{x \in {K}_{\epsilon }}}\left| {f\left( x\right) }\right| + \epsilon \parallel f\parallel ,\;\text{ for all }f \in {C}_{b}\left( X\right) .\n\]\n\nThen there exists a unique finite (positive) Borel measure \( \mu \) so that\n\n\[ \ell \left( f\right) = {\int }_{X}f\left( x\right) {d\mu }\left( x\right) ,\;\text{ for all }f \in {C}_{b}\left( X\right) . \] | The proof of this theorem proceeds as that of Theorem 7.1, save for one key aspect. First we define\n\n\[ \rho \left( \mathcal{O}\right) = \sup \left\{ {\ell \left( f\right) ,\text{ where }f \in {C}_{b}\left( X\right) ,\operatorname{supp}\left( f\right) \subset \mathcal{O}\text{, and }0 \leq f \leq 1}\right\} .\n\]\n\nThe change that is required is in the proof of the countable sub-additivity of \( \rho \), in that the support of \( f \) ’s (in the definition of \( \rho \left( \mathcal{O}\right) \) ) are now not necessarily compact. In fact, suppose \( \mathcal{O} = \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{\mathcal{O}}_{k} \) is a countable union of open sets. Let \( C \) be the support of \( f \), and given a fixed \( \epsilon > 0 \), set \( K = C \cap {K}_{\epsilon } \), with \( {K}_{\epsilon } \) the compact set arising in (21). Then \( K \) is compact and \( \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{\mathcal{O}}_{k} \) covers \( K \) . Proceeding as before, we obtain a partition of unity \( {\left\{ {\eta }_{k}\right\} }_{k = 1}^{N} \), with \( {\eta }_{k} \) supported in \( {\mathcal{O}}_{k} \) and \( \mathop{\sum }\limits_{{k = 1}}^{N}{\eta }_{k}\left( x\right) = 1 \), for \( x \in K \) . Now \( f - \mathop{\sum }\limits_{{k = 1}}^{N}f{\eta }_{k} \) vanishes on \( {K}_{\epsilon } \) . Thus by (21)\n\n\[ \left| {\ell \left( f\right) - \mathop{\sum }\limits_{{k = 1}}^{N}\ell \left( {f{\eta }_{k}}\right) }\right| \leq \epsilon \]\n\nand hence\n\n\[ \ell \left( f\right) \leq \mathop{\sum }\limits_{{k = 1}}^{\infty }\rho \left( {\mathcal{O}}_{k}\right) + \epsilon \]\n\nSince this holds for each \( \epsilon \), we obtain the required sub-additivity of \( \rho \) and thus of \( {\mu }_{ * } \) . The proof of the theorem can then be concluded as before. | Yes |
Theorem 2.1 Suppose \( T \) is a linear mapping from \( {L}^{{p}_{0}} + {L}^{{p}_{1}} \) to \( {L}^{{q}_{0}} + \) \( {L}^{{q}_{1}} \) . Assume that \( T \) is bounded from \( {L}^{{p}_{0}} \) to \( {L}^{{q}_{0}} \) and from \( {L}^{{p}_{1}} \) to \( {L}^{{q}_{1}} \n\n\[ \left\{ \begin{array}{l} \parallel T\left( f\right) {\parallel }_{{L}^{{q}_{0}}} \leq {M}_{0}\parallel f{\parallel }_{{L}^{{p}_{0}}}, \\ \parallel T\left( f\right) {\parallel }_{{L}^{{q}_{1}}} \leq {M}_{1}\parallel f{\parallel }_{{L}^{{p}_{1}}}. \end{array}\right. \]\n\nThen \( T \) is bounded from \( {L}^{p} \) to \( {L}^{q} \) ,\n\n\[ \parallel T\left( f\right) {\parallel }_{{L}^{q}} \leq M\parallel f{\parallel }_{{L}^{p}} \]\n\nwhenever the pair \( \left( {p, q}\right) \) can be written as\n\n\[ \frac{1}{p} = \frac{1 - t}{{p}_{0}} + \frac{t}{{p}_{1}}\;\text{ and }\;\frac{1}{q} = \frac{1 - t}{{q}_{0}} + \frac{t}{{q}_{1}} \]\n\nfor some \( t \) with \( 0 \leq t \leq 1 \) . Moreover, the bound \( M \) satisfies \( M \leq {M}_{0}^{1 - t}{M}_{1}^{t} \) . | Null | No |
Lemma 2.2 (Three-lines lemma) Suppose \( \Phi \left( z\right) \) is a holomorphic function in the strip \( S = \{ z \in \mathbb{C} : 0 < \operatorname{Re}\left( z\right) < 1\} \), that is also continuous and bounded on the closure of \( S \) . If\n\n\[ \n{M}_{0} = \mathop{\sup }\limits_{{y \in \mathbb{R}}}\left| {\Phi \left( {iy}\right) }\right| \;\text{ and }\;{M}_{1} = \mathop{\sup }\limits_{{y \in \mathbb{R}}}\left| {\Phi \left( {1 + {iy}}\right) }\right| \n\]\n\nthen\n\n\[ \n\mathop{\sup }\limits_{{y \in \mathbb{R}}}\left| {\Phi \left( {t + {iy}}\right) }\right| \leq {M}_{0}^{1 - t}{M}_{1}^{t},\;\text{ for all }0 \leq t \leq 1. \n\] | Proof. We begin by proving the lemma under the assumption that \( {M}_{0} = {M}_{1} = 1 \) and \( \mathop{\sup }\limits_{{0 \leq x \leq 1}}\left| {\Phi \left( {x + {iy}}\right) }\right| \rightarrow 0 \) as \( \left| y\right| \rightarrow \infty \) . In this case, let \( M = \sup \left| {\Phi \left( z\right) }\right| \) where the sup is taken over all \( z \) in the closure of the strip \( S \) . We may clearly assume that \( M > 0 \), and let \( {z}_{1},{z}_{2},\ldots \) be a sequence of points in the strip with \( \left| {\Phi \left( {z}_{n}\right) }\right| \rightarrow M \) as \( n \rightarrow \infty \) . By the decay condition imposed on \( \Phi \), the points \( {z}_{n} \) cannot go to infinity, hence there exists \( {z}_{0} \) in the closure of the strip, so that a subsequence of \( \left\{ {z}_{n}\right\} \) converges to \( {z}_{0} \) . By the maximum modulus principle, \( {z}_{0} \) cannot be in the interior of the strip,(unless \( \Phi \) is constant, in which case the conclusion is trivial) hence \( {z}_{0} \) must be on its boundary, where \( \left| \Phi \right| \leq 1 \) . Thus \( M \leq 1 \) , and the result is proved in this special case.\n\nIf we only assume now that \( {M}_{0} = {M}_{1} = 1 \), we define\n\n\[ \n{\Phi }_{\epsilon }\left( z\right) = \Phi \left( z\right) {e}^{\epsilon \left( {{z}^{2} - 1}\right) },\;\text{ for each }\epsilon > 0. \n\]\n\nSince \( {e}^{\epsilon \left\lbrack {{\left( x + iy\right) }^{2} - 1}\right\rbrack } = {e}^{\epsilon \left( {{x}^{2} - 1 - {y}^{2} + {2ixy}}\right) } \), we find that \( \left| {{\Phi }_{\epsilon }\left( z\right) }\right| \leq 1 \) on the lines \( \operatorname{Re}\left( z\right) = 0 \) and \( \operatorname{Re}\left( z\right) = 1 \) . Moreover,\n\n\[ \n\mathop{\sup }\limits_{{0 \leq x \leq 1}}\left| {{\Phi }_{\epsilon }\left( {x + {iy}}\right) }\right| \rightarrow 0\;\text{ as }\left| y\right| \rightarrow \infty , \n\]\n\nsince \( \Phi \) is bounded. Therefore, by the first case, we know that \( \left| {{\Phi }_{\epsilon }\left( z\right) }\right| \leq 1 \) in the closure of the strip. Letting \( \epsilon \rightarrow 0 \), we see that \( \left| \Phi \right| \leq 1 \) as desired.\n\nFinally, for arbitrary positive values of \( {M}_{0} \) and \( {M}_{1} \), we let \( \widetilde{\Phi }\left( z\right) = \) \( {M}_{0}^{z - 1}{M}_{1}^{-z}\Phi \left( z\right) \), and note that \( \widetilde{\Phi } \) satisfies the condition of the previous case, that is, \( \widetilde{\Phi } \) is bounded by 1 on the lines \( \operatorname{Re}\left( z\right) = 0 \) and \( \operatorname{Re}\left( z\right) = 1 \) . | Yes |
Corollary 2.3 With \( T \) as before:\n\n(a) The Riesz diagram of \( T \) is a convex set.\n\n(b) \( \log {M}_{x, y} \) is a convex function on this set. | Conclusion (a) means that if \( \left( {{x}_{0},{y}_{0}}\right) = \left( {1/{p}_{0},1/{q}_{0}}\right) \) and \( \left( {{x}_{1},{y}_{1}}\right) = \) \( \left( {1/{p}_{1},1/{q}_{1}}\right) \) are points in the Riesz diagram of \( T \), then so is the line segment joining them. This is an immediate consequence of Theorem 2.1. Similarly the convexity of the function \( \log {M}_{x, y} \) is its convexity on each line segment, and this follows from the conclusion \( M \leq {M}_{0}^{1 - t}{M}_{1}^{t} \) guaranteed also by Theorem 2.1. | Yes |
Corollary 2.4 If \( 1 \leq p \leq 2 \) and \( 1/p + 1/q = 1 \), then\n\n\[ \parallel T\left( f\right) {\parallel }_{{L}^{q}\left( \mathbb{Z}\right) } \leq \parallel f{\parallel }_{{L}^{p}\left( \left\lbrack {0,{2\pi }}\right\rbrack \right) }.\] | Note that since \( {L}^{2}\left( \left\lbrack {0,{2\pi }}\right\rbrack \right) \subset {L}^{1}\left( \left\lbrack {0,{2\pi }}\right\rbrack \right) \) and \( {L}^{2}\left( \mathbb{Z}\right) \subset {L}^{\infty }\left( \mathbb{Z}\right) \) we have \( {L}^{2}\left( \left\lbrack {0,{2\pi }}\right\rbrack \right) + {L}^{1}\left( \left\lbrack {0,{2\pi }}\right\rbrack \right) = {L}^{1}\left( \left\lbrack {0,{2\pi }}\right\rbrack \right) \), and also \( {L}^{2}\left( \mathbb{Z}\right) + {L}^{\infty }\left( \mathbb{Z}\right) = {L}^{\infty }\left( \mathbb{Z}\right) \).\n\nThe inequality for \( {p}_{0} = {q}_{0} = 2 \) is a consequence of Parseval’s identity, while the one for \( {p}_{1} = 1,{q}_{1} = \infty \) follows from the observation that for all \( n \),\n\n\[ \left| {a}_{n}\right| \leq \frac{1}{2\pi }{\int }_{0}^{2\pi }\left| {f\left( \theta \right) }\right| {d\theta }\n\nThus Riesz’s theorem guarantees the conclusion when \( 1/p = \frac{\left( 1 - t\right) }{2} + t \) , \( 1/q = \frac{\left( 1 - t\right) }{2} \) for any \( t \) with \( 0 \leq t \leq 1 \) . This gives all \( p \) with \( 1 \leq p \leq 2 \) , and \( q \) related to \( p \) by \( 1/p + 1/q = 1 \) . | Yes |
Corollary 2.5 If \( 1 \leq p \leq 2 \) and \( 1/p + 1/q = 1 \), then\n\n\[{\begin{Vmatrix}{T}^{\prime }\left( \left\{ {a}_{n}\right\} \right) \end{Vmatrix}}_{{L}^{q}\left( \left\lbrack {0,{2\pi }}\right\rbrack \right) } \leq {\begin{Vmatrix}\left\{ {a}_{n}\right\} \end{Vmatrix}}_{{L}^{p}\left( \mathbb{Z}\right) }.\] | The proof is parallel to that of the previous corollary. The case \( {p}_{0} = \) \( {q}_{0} = 2 \) is, as has already been mentioned, a consequence of Parseval’s identity, while the case \( {p}_{1} = 1 \) and \( {q}_{1} = \infty \) follows directly from the fact that\n\n\[ \left| {\mathop{\sum }\limits_{{n = - \infty }}^{\infty }{a}_{n}{e}^{in\theta }}\right| \leq \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\left| {a}_{n}\right| \] | Yes |
Corollary 2.6 If \( 1 \leq p \leq 2 \) and \( 1/p + 1/q = 1 \), then the Fourier transform \( T \) has a unique extension to a bounded map from \( {L}^{p} \) to \( {L}^{q} \), with \( \parallel T\left( f\right) {\parallel }_{{L}^{q}} \leq \parallel f{\parallel }_{{L}^{p}} \) | Null | No |
Theorem 3.2 Suppose \( 1 < p < \infty \) . Then the Hilbert transform \( H \), initially defined on \( {L}^{2} \cap {L}^{p} \) by (13) or (14), satisfies the inequality\n\n\[ \parallel H\left( f\right) {\parallel }_{{L}^{p}} \leq {A}_{p}\parallel f{\parallel }_{{L}^{p}},\;\text{ whenever }f \in {L}^{2} \cap {L}^{p}, \] \n\nwith a bound \( {A}_{p} \) independent of \( f \) . The Hilbert transform then has a unique extension to all of \( {L}^{p} \) satisfying the same bound. | ## 3.3 Proof of Theorem 3.2\n\nThe main idea of the proof was already outlined at the end of Section 1 in the context of Fourier series and the corresponding theorem for the conjugate function. While this proof, which depends on complex analysis, is elegant, its approach is essentially limited to this operator and cannot deal with the generalizations of the Hilbert transform in the setting of \( {\mathbb{R}}^{d} \) . The \ | No |
Theorem 4.1 Suppose \( f \in {L}^{p}\left( {\mathbb{R}}^{d}\right) \) with \( 1 < p \leq \infty \) . Then \( {f}^{ * } \in {L}^{p}\left( {\mathbb{R}}^{d}\right) \) , and (26) holds, namely\n\n\[ \n{\begin{Vmatrix}{f}^{ * }\end{Vmatrix}}_{{L}^{p}} \leq {A}_{p}\parallel f{\parallel }_{{L}^{p}}\n\]\n\nThe bound \( {A}_{p} \) depends on \( p \) but is independent of \( f \) . | Let us first see why \( {f}^{ * }\left( x\right) < \infty \), for a.e. \( x \), whenever \( f \in {L}^{p} \) . Observe that we can decompose \( f = {f}_{1} + {f}_{\infty } \), where \( {f}_{1}\left( x\right) = f\left( x\right) \) if \( \left| {f\left( x\right) }\right| > 1 \) , and \( {f}_{1}\left( x\right) = 0 \) elsewhere; also \( {f}_{\infty }\left( x\right) = f\left( x\right) \) if \( \left| {f\left( x\right) }\right| \leq 1 \) and \( {f}_{\infty }\left( x\right) = 0 \) elsewhere. Then \( {f}_{1} \in {L}^{1} \) and \( {f}_{\infty } \in {L}^{\infty } \) . But clearly \( {f}^{ * } \leq {f}_{1}^{ * } + {f}_{\infty }^{ * } \leq \) \( {f}_{1}^{ * } + 1 \), since \( \left| {{f}_{\infty }\left( x\right) }\right| \leq 1 \) everywhere. Now from (27) (with \( {f}_{1} \) in place of \( f \) ), we see that \( {f}_{1}^{ * } \) is finite almost everywhere. Thus the same is true for \( {f}^{ * } \).\n\nThe proof that \( {f}^{ * } \in {L}^{p} \) relies on a more quantitative version of the argument just given. We strengthen the weak-type inequality (27) by incorporating in it the \( {L}^{\infty } \) boundedness of the mapping \( f \mapsto {f}^{ * } \) . The stronger version states\n\n(28)\n\n\[ \nm\left( \left\{ {x : {f}^{ * }\left( x\right) > \alpha }\right\} \right) \leq \frac{{A}^{\prime }}{\alpha }{\int }_{\left| f\right| > \alpha /2}\left| f\right| {dx},\;\text{ for all }\alpha > 0.\n\]\n\nHere \( {A}^{\prime } \) is a different constant; it can be taken to be \( {2A} \) . The improvement of (27), (except for a different constant, which is inessential), is that here we only integrate over the set where \( \left| {f\left( x\right) }\right| > \alpha /2 \), instead of the whole of \( {\mathbb{R}}^{d} \).\n\nTo prove (28) we write \( f = {f}_{1} + {f}_{\infty } \), where now \( {f}_{1}\left( x\right) = f\left( x\right) \), if \( \left| {f\left( x\right) }\right| > \) \( \alpha /2 \), and \( {f}_{\infty }\left( x\right) = f\left( x\right) \) if \( \left| {f\left( x\right) }\right| \leq \alpha /2 \) . Then \( {f}^{ * } \leq {f}_{1}^{ * } + {f}_{\infty }^{ * } \leq {f}_{1}^{ * } + \alpha /2 \) , since \( \left| {{f}_{\infty }\left( x\right) }\right| \leq \alpha /2 \) for all \( x \) . Therefore \( \left\{ {x : {f}^{ * }\left( x\right) > \alpha }\right\} \subset \left\{ {x : {f}_{1}^{ * } > }\right. \) \( \alpha /2\} \), and applying the weak-type inequality (27) to \( {f}_{1} \) in place of \( f \) (and \( \alpha /2 \) in place of \( \alpha \) ) then immediately yields (28), with \( {A}^{\prime } = {2A} \) . | Yes |
Proposition 5.1 Suppose \( f \in {L}^{p}\left( {\mathbb{R}}^{d}\right), p > 1 \), and \( f \) has bounded support. Then \( f \) belongs to \( {\mathbf{H}}_{r}^{1}\left( {\mathbb{R}}^{d}\right) \) if and only if \( {\int }_{{\mathbb{R}}^{d}}f\left( x\right) {dx} = 0 \) . | Note that \( f \) is automatically in \( {L}^{1} \), by Hölder’s inequality (see Proposition 1.4 in Chapter 1), and the cancelation condition is necessary as has been pointed out.\n\nTo prove the sufficiency we assume that \( f \) is supported in a ball \( {B}_{1} \) of unit radius, and that \( {\int }_{{B}_{1}}\left| {f\left( x\right) }\right| {dx} \leq 1 \) . These normalizations can be achieved by a simple change of scale and multiplication of \( f \) by an appropriate constant. We next consider a truncated version of the maximal function \( {f}^{ * } \) . We define \( {f}^{ \dagger } \) by\n\n\[ \n{f}^{ \dagger }\left( x\right) = \sup \frac{1}{m\left( B\right) }{\int }_{B}\left| {f\left( y\right) }\right| {dy} \n\]\n\nwhere the supremum is taken over all balls \( B \) of radius \( \leq 1 \) that contain \( x \) . We note that under our assumptions we have\n\n(32)\n\n\[ \n{\int }_{{\mathbb{R}}^{d}}{f}^{ \dagger }\left( x\right) {dx} < \infty \n\]\n\nIndeed, \( {f}^{ \dagger }\left( x\right) = 0 \) if \( x \notin {B}_{3} \), where \( {B}_{3} \) is the ball with same center as \( {B}_{1} \) , but with radius 3 . This is because \( x \notin {B}_{3} \) and if \( x \in B \) with the radius of \( B \) less than or equal to 1, then \( B \) must be disjoint from \( {B}_{1} \), the support of \( f \) . Thus\n\n\[ \n{\int }_{{\mathbb{R}}^{d}}{f}^{ \dagger }\left( x\right) {dx} = {\int }_{{B}_{3}}{f}^{ \dagger }\left( x\right) {dx} \leq c{\left( {\int }_{{B}_{3}}{\left( {f}^{ \dagger }\left( x\right) \right) }^{p}dx\right) }^{1/p} \n\]\n\nby Hölder's inequality. However the last integral is finite by Theorem 4.1, since clearly \( {f}^{ \dagger }\left( x\right) \leq {f}^{ * }\left( x\right) \) .\n\nNow for each \( \alpha \geq 1 \), we consider a basic decomposition of \( f \) at \ | No |
Lemma 5.2 Suppose \( \Omega \subset {\mathbb{R}}^{d} \) is a non-trivial open set. Then there is a collection \( \left\{ {Q}_{j}\right\} \) of dyadic cubes with disjoint interiors so that \( \Omega = \) \( \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{Q}_{j} \), and\n\n\[ \operatorname{diam}\left( {Q}_{j}\right) \leq d\left( {{Q}_{j},{\Omega }^{c}}\right) \leq 4\operatorname{diam}\left( {Q}_{j}\right) . \] | Proof. We claim first that every point \( \bar{x} \in \Omega \) belongs to some dyadic cube \( {Q}_{\bar{x}} \) for which (37) holds (with \( {Q}_{\bar{x}} \) in place of \( {Q}_{j} \) ).\n\nLet \( \delta = d\left( {\bar{x},{\Omega }^{c}}\right) > 0 \) . Now the dyadic cubes containing \( \bar{x} \) have diameters varying over \( \left\{ {\sqrt{d}{2}^{-k}}\right\}, k \in \mathbb{Z} \) . Hence we can find a dyadic cube \( {Q}_{\bar{x}} \) which contains \( \bar{x} \), with \( \delta /4 \leq \operatorname{diam}\left( {Q}_{\bar{x}}\right) \leq \delta /2 \) . Now \( d\left( {{Q}_{\bar{x}},{\Omega }^{c}}\right) \leq \delta \leq \) 4 \( \operatorname{diam}\left( {Q}_{\bar{x}}\right) \), since \( \bar{x} \in {Q}_{\bar{x}} \) . Also\n\n\[ d\left( {{Q}_{\bar{x}},{\Omega }^{c}}\right) \geq \delta - \operatorname{diam}\left( {Q}_{\bar{x}}\right) \geq \delta /2 \geq \operatorname{diam}\left( {Q}_{\bar{x}}\right) ,\]\n\n\( {}^{11} \) This kind of decomposition already arose in Chapter 1 of Book III. thus (37) is proved for \( {Q}_{\bar{x}} \) . Now let \( \widetilde{\mathcal{Q}} \) be the collection of all cubes \( {Q}_{\bar{x}} \) obtained as \( \bar{x} \) ranges over \( \Omega \) . Their union clearly covers \( \Omega \) but their interiors are far from disjoint. To achieve the desired disjointness select from \( \widetilde{\mathcal{Q}} \) the maximal cubes, that is, those cubes in \( \widetilde{\mathcal{Q}} \) not contained in larger cubes of \( \widetilde{\mathcal{Q}} \) . Clearly, by what has been said above, each \( Q \) is contained in a maximal cube and these maximal cubes necessarily have disjoint interiors. The lemma is therefore proved. | Yes |
Corollary 5.3 Fix \( p > 1 \) . Then any p-atom \( \mathfrak{a} \) is in \( {\mathbf{H}}_{r}^{1} \) . Moreover there is a bound \( {c}_{p} \), independent of the atom \( \mathfrak{a} \), so that\n\n\[ \parallel \mathfrak{a}{\parallel }_{{\mathbf{H}}_{r}^{1}} \leq {c}_{p} \] | Proof. One can rescale a \( p \) -atom \( \mathfrak{a} \), associated to a ball \( B \) of radius \( r \), by replacing \( \mathfrak{a} \) by \( {\mathfrak{a}}_{r} \), with \( {\mathfrak{a}}_{r}\left( x\right) = {r}^{d}\mathfrak{a}\left( {rx}\right) \) . Then clearly \( {\mathfrak{a}}_{r}\left( x\right) \) is supported where \( {rx} \in B \), that is, \( x \in \frac{1}{r}B = {B}^{r} \) and the latter ball has radius one. Also since \( m\left( {B}^{r}\right) = {r}^{-d}m\left( B\right) \) and \( {\begin{Vmatrix}{\mathfrak{a}}_{r}\end{Vmatrix}}_{{L}^{p}} = {r}^{d - d/p}\parallel \mathfrak{a}{\parallel }_{{L}^{p}} \), we have \( {\begin{Vmatrix}{\mathfrak{a}}_{r}\end{Vmatrix}}_{{L}^{p}} \leq m{\left( {B}^{r}\right) }^{-1 + 1/p} \) . Thus \( {\mathfrak{a}}_{r} \) is a \( p \) -atom for the (unit) ball \( {B}^{r} \) . Moreover, as has already been observed \( {\begin{Vmatrix}{r}^{d}f\left( rx\right) \end{Vmatrix}}_{{\mathbf{H}}_{r}^{1}} = \parallel f{\parallel }_{{\mathbf{H}}_{r}^{1}} \), for every \( r > 0 \) . Thus (38) has been reduced to the case of \( p \) -atoms associated to balls of unit radius. Observe that automatically for such \( p \) -atoms one has \( \int \left| {\mathfrak{a}\left( x\right) }\right| {dx} \leq 1 \), therefore we see that we find ourselves exactly in the setting of the proof of Proposition 5.1 with \( f\left( x\right) = \mathfrak{a}\left( x\right) \) . In fact one notes that what is proved there amounts to (38), with the constant \( {c}_{p} \) incorporating the bound \( {A}_{p} \) in (26) for the maximal function, since the calculation for \( {\int }_{{\mathbb{R}}^{d}}{f}^{ \dagger }\left( x\right) {dx} \) used to establish (32) shows that this quantity is bounded by \( c{A}_{p}\parallel f{\parallel }_{{L}^{p}} \) . We have already noted that \( {A}_{p} = O\left( {1/\left( {p - 1}\right) }\right) \) as \( p \rightarrow 1 \) . Because \( f = \mathfrak{a} \), the proof of (38) is complete. | Yes |
Theorem 5.4 If \( f \) belongs to the Hardy space \( {\mathbf{H}}_{r}^{1}\left( \mathbb{R}\right) \), then \( {H}_{\epsilon }\left( f\right) \in \) \( {L}^{1}\left( \mathbb{R}\right) \), for every \( \epsilon > 0 \) . Moreover \( {H}_{\epsilon }\left( f\right) \) (see (14)) converges in the \( {L}^{1} \) norm, as \( \epsilon \rightarrow 0 \) . Its limit, defined as \( H\left( f\right) \), satisfies | Proof. The argument below illustrates a nice feature of \( {\mathbf{H}}_{r}^{1}\left( \mathbb{R}\right) \) : to show the boundedness of an operator on \( {\mathbf{H}}_{r}^{1} \) it often suffices merely to verify it for atoms, and this is usually a simple task.\n\nLet us first see that for all atoms \( \mathfrak{a} \), we have\n\n(39)\n\n\[{\begin{Vmatrix}{H}_{\epsilon }\left( \mathfrak{a}\right) \end{Vmatrix}}_{{L}^{1}\left( \mathbb{R}\right) } \leq A\]\nwith \( A \) independent of the atom \( \mathfrak{a} \) and \( \epsilon \) . Indeed, we can avail ourselves of the translation-invariance and scale-invariance of the Hilbert transform to simplify matters even further by restricting ourselves in proving (39) for the case of atoms associated to the (unit) interval \( I = \left\lbrack {-1/2,1/2}\right\rbrack \) . This reduction proceeds, on the one hand, by recalling that if \( {\mathfrak{a}}_{r}\left( x\right) = \) \( r\mathfrak{a}\left( {rx}\right) \), then \( H\left( {\mathfrak{a}}_{r}\right) \left( x\right) = {rH}\left( \mathfrak{a}\right) \left( {rx}\right) \) ; that \( {\mathfrak{a}}_{r} \) is an atom associated to the interval \( {I}_{r} = \frac{1}{r}I \) whenever \( \mathfrak{a} \) is supported in \( I \) ; and that \( \parallel {rF}\left( {rx}\right) {\parallel }_{{L}^{1}\left( \mathbb{R}\right) } = \) \( \parallel F\left( x\right) {\parallel }_{{L}^{1}\left( \mathbb{R}\right) } \), whenever \( F \in {L}^{1} \) . On the other hand, the translations \( f\left( x\right) \mapsto f\left( {x + h}\right), h \in \mathbb{R} \), commute with the operator \( H \), as is evident from (14); also translation clearly preserves atoms and the radii of their associated balls.\n\nThus in proving (39) we may assume that \( \mathfrak{a} \) is an atom associated to the interval \( \left| x\right| \leq 1/2 \) . We will estimate \( {H}_{\epsilon }\left( \mathfrak{a}\right) \left( x\right) \) differently, according to whether \( \left| x\right| \leq 1 \) ,( \( x \) belongs to the \ | Yes |
Theorem 6.1 Suppose \( \Phi \) is a \( {C}^{1} \) function with compact support on \( {\mathbb{R}}^{d} \) . With \( M \) defined by (41) we have that \( M\left( f\right) \in {L}^{1}\left( {\mathbb{R}}^{d}\right) \), whenever \( f \in \) \( {\mathbf{H}}_{r}^{1}\left( {\mathbb{R}}^{d}\right) \) . Moreover\n\n\[ \parallel M\left( f\right) {\parallel }_{{L}^{1}\left( \mathbb{R}\right) } \leq A\parallel f{\parallel }_{{\mathbf{H}}_{r}^{1}\left( {\mathbb{R}}^{d}\right) }.\] | Proof. Suppose \( f \) is in \( {\mathbf{H}}_{r}^{1}\left( {\mathbb{R}}^{d}\right) \) and \( f = \sum {\lambda }_{k}{\mathfrak{a}}_{k} \) is an atomic decomposition. Then clearly \( M\left( f\right) \leq \sum \left| {\lambda }_{k}\right| M\left( {\mathfrak{a}}_{k}\right) \), and thus it suffices to prove (42) when \( f \) is an atom \( \mathfrak{a} \) .\n\nIn fact, note that with \( {\mathfrak{a}}_{r} \) defined as \( {\mathfrak{a}}_{r}\left( x\right) = {r}^{d}\mathfrak{a}\left( {rx}\right), r > 0 \), we have \( \left( {{\mathfrak{a}}_{r} * {\Phi }_{\epsilon }}\right) \left( x\right) = {r}^{d}\left( {\mathfrak{a} * {\Phi }_{\epsilon r}}\right) \left( {rx}\right) \), and hence \( M\left( {\mathfrak{a}}_{r}\right) \left( x\right) = {r}^{d}M\left( \mathfrak{a}\right) \left( {rx}\right) \) . Also the mapping \( \mathfrak{a} \mapsto M\left( \mathfrak{a}\right) \) clearly commutes with translations. Therefore in proving (42) we may assume that the atom \( \mathfrak{a} \) is associated to the unit ball (centered at the origin).\n\nNow we consider two cases; when \( \left| x\right| \leq 2 \) and when \( \left| x\right| > 2 \) . In the first case clearly \( M\left( \mathfrak{a}\right) \left( x\right) \leq c \) and hence \( {\int }_{\left| x\right| \leq 2}M\left( \mathfrak{a}\right) \left( x\right) {dx} \leq {c}^{\prime } \) . In the second case, we write\n\n\[ \left( {\mathfrak{a} * {\Phi }_{\epsilon }}\right) \left( x\right) = {\epsilon }^{-d}{\int }_{{\mathbb{R}}^{d}}\mathfrak{a}\left( y\right) \Phi \left( \frac{x - y}{\epsilon }\right) {dy} \]\n\n\[ = {\epsilon }^{-d}{\int }_{{\mathbb{R}}^{d}}\mathfrak{a}\left( y\right) \left\lbrack {\Phi \left( \frac{x - y}{\epsilon }\right) - \Phi \left( \frac{x}{\epsilon }\right) }\right\rbrack {dy} \]\n\nsince \( \int \mathfrak{a}\left( y\right) {dy} = 0 \) . However since \( \left| x\right| \geq 2 \) and \( \left| y\right| \leq 1 \), we have that \( \mid x - \) \( y\left| \geq \right| x \mid /2 \) . Moreover since \( \Phi \in {C}^{1} \) we have that \( \left| {\Phi \left( \frac{x - y}{\epsilon }\right) - \Phi \left( \frac{x}{\epsilon }\right) }\right| \leq \) \( c\left| y\right| /\epsilon \leq c/\epsilon \) . In addition the fact that \( \Phi \) has compact support implies that \( \left( {\mathfrak{a} * {\Phi }_{\epsilon }}\right) \left( x\right) \) vanishes unless \( \left| \frac{x - y}{\epsilon }\right| \leq A \) for some bound \( A \), which in turn means that \( \epsilon > \left| x\right| /\left( {2A}\right) \) . Altogether then\n\n\[ {\epsilon }^{-d}\left| {\Phi \left( \frac{x - y}{\epsilon }\right) - \Phi \left( \frac{x}{\epsilon }\right) }\right| \leq c{\epsilon }^{-d - 1} \leq {c}^{\prime }{\left| x\right| }^{-d - 1} \]\n\nfor those \( x \) . As a result \( {\int }_{\left| x\right| > 2}M\left( \mathfrak{a}\right) \left( x\right) {dx} \leq c \) . Therefore (42) is established and the theorem is proved. | Yes |
Theorem 6.2 Suppose \( g \in \mathrm{{BMO}} \) . Then the linear functional \( \ell \) defined by (44), initially considered for \( f \in {H}_{0}^{1} \), has a unique extension to \( {\mathbf{H}}_{r}^{1} \) that satisfies\n\n\[ \parallel \ell \parallel \leq c\parallel g{\parallel }_{\mathrm{{BMO}}} \]\n\nConversely, every bounded linear functional \( \ell \) on \( {\mathbf{H}}_{r}^{1} \) can be written as (44) with \( g \in \mathrm{{BMO}} \) and\n\n\[ \parallel g{\parallel }_{\mathrm{{BMO}}} \leq {c}^{\prime }\parallel \ell \parallel \]\n\nHere \( \parallel \ell \parallel \) stands for \( \parallel \ell {\parallel }_{{\left( {\mathbf{H}}_{r}^{1}\right) }^{ * }} \), the norm of \( \ell \) as a linear functional on \( {\mathbf{H}}_{r}^{1} \) . | Proof. Let us first assume that \( g \in \) BMO is bounded. Start with a general \( f \in {\mathbf{H}}_{r}^{1} \), and let \( f = \mathop{\sum }\limits_{{k = 1}}^{\infty }{\lambda }_{k}{\mathfrak{a}}_{k} \) be an atomic decomposition. Then by the convergence of the sum in the \( {L}^{1} \) norm we get \( \ell \left( f\right) = \) \( \sum {\lambda }_{k}\int {\mathfrak{a}}_{k}g \) . But\n\n\[ \int {\mathfrak{a}}_{k}\left( x\right) g\left( x\right) {dx} = \int {\mathfrak{a}}_{k}\left( x\right) \left\lbrack {g\left( x\right) - {g}_{{B}_{k}}}\right\rbrack {dx} \]\n\nwhere \( {\mathfrak{a}}_{k} \) is supported in the ball \( {B}_{k} \) . However \( \left| {{\mathfrak{a}}_{k}\left( x\right) }\right| \leq \frac{1}{m\left( {B}_{k}\right) } \) and thus\n\n\[ \left| {\ell \left( f\right) }\right| \leq \mathop{\sum }\limits_{k}\left| {\lambda }_{k}\right| \frac{1}{m\left( {B}_{k}\right) }{\int }_{{B}_{k}}\left| {g\left( x\right) - {g}_{{B}_{k}}}\right| {dx}. \]\n\nTherefore considering all possible decompositions of \( f \) then gives\n\n\[ \left| {\int f\left( x\right) g\left( x\right) {dx}}\right| \leq \parallel f{\parallel }_{{\mathbf{H}}_{r}^{1}}\parallel g{\parallel }_{\mathrm{{BMO}}} \]\n\nunder the assumption that \( g \) is bounded. Next, if we restrict ourselves to \( f \in {H}_{0}^{1} \) (in particular to \( f \) that are bounded) with a general \( g \) in BMO, and let \( {g}^{\left( k\right) } \) be the truncation of \( g \) (as defined above), then the fact that\n\n\[ \left| {\int f\left( x\right) {g}^{\left( k\right) }\left( x\right) {dx}}\right| \leq \parallel f{\parallel }_{{\mathbf{H}}_{r}^{1}}{\begin{Vmatrix}{g}^{\left( k\right) }\end{Vmatrix}}_{\mathrm{{BMO}}} \]\n\njust proved, together with a passage to the limit as \( k \rightarrow \infty \) using the dominated convergence theorem, shows that\n\n\[ \left| {\ell \left( f\right) }\right| = \left| {\int f\left( x\right) g\left( x\right) {dx}}\right| \leq c\parallel f{\parallel }_{{\mathbf{H}}_{r}^{1}}\parallel g{\parallel }_{\mathrm{{BMO}}}, \]\n\nwhenever \( f \in {H}_{0}^{1} \) and \( g \in \mathrm{{BMO}} \) . Thus the direct conclusion of the theorem is established. | Yes |
Proposition 1.1 Suppose \( F \) is a distribution and \( \psi \in \mathcal{D} \). Then\n\n(a) The two definitions of \( F * \psi \) given above coincide.\n\n(b) The distribution \( F * \psi \) is a \( {C}^{\infty } \) function. | Proof. Let us observe first that \( F\left( {\psi }_{x}^{ \sim }\right) \) is continuous in \( x \) and in fact indefinitely differentiable. Note that if \( {x}_{n} \rightarrow {x}_{0} \) as \( n \rightarrow \infty \), then \( {\psi }_{{x}_{n}}^{ \sim }\left( y\right) = \psi \left( {{x}_{n} - y}\right) \rightarrow \psi \left( {{x}_{0} - y}\right) = {\psi }_{{x}_{0}}^{ \sim }\left( y\right) \) uniformly in \( y \), and the same is true for all partial derivatives. Therefore \( {\psi }_{{x}_{n}}^{ \sim } \rightarrow {\psi }_{{x}_{0}}^{ \sim } \) in \( \mathcal{D} \) (as functions of \( y \) ) as \( n \rightarrow \infty \), and thus by the assumed continuity of \( F \) on \( \mathcal{D} \) we have that \( F\left( {\psi }_{x}^{ \sim }\right) \) is continuous in \( x \). Similarly, all corresponding difference quotients converge and the result is that \( F\left( {\psi }_{x}^{ \sim }\right) \) is indefinitely differentiable, with \( {\partial }_{x}^{\alpha }F\left( {\psi }_{x}^{ \sim }\right) = F\left( {{\partial }_{x}^{\alpha }{\psi }_{x}^{ \sim }}\right) \).\n\nIt remains to prove conclusion (a), and for this it suffices to show that\n\n(2)\n\n\[ \int F\left( {\psi }_{x}^{ \sim }\right) \varphi \left( x\right) {dx} = F\left( {{\psi }^{ \sim } * \varphi }\right) ,\;\text{ for each }\varphi \in \mathcal{D}. \]\n\nHowever since \( \psi \in \mathcal{D} \), and of course \( \varphi \) is continuous with compact support, then it is easily seen that\n\n\[ \left( {{\psi }^{ \sim } * \varphi }\right) \left( x\right) = \int {\psi }^{ \sim }\left( {x - y}\right) \varphi \left( y\right) {dy} = \mathop{\lim }\limits_{{\epsilon \rightarrow 0}}S\left( \epsilon \right) \]\n\nwhere \( S\left( \epsilon \right) = {\epsilon }^{d}\mathop{\sum }\limits_{{n \in {\mathbb{Z}}^{d}}}{\psi }^{ \sim }\left( {x - {n\epsilon }}\right) \varphi \left( {n\epsilon }\right) \). Here the convergence of the Riemann sums \( S\left( \epsilon \right) \) to \( {\psi }^{ \sim } * \varphi \) is in \( \mathcal{D} \). Clearly, \( S\left( \epsilon \right) \) is finite for each \( \epsilon > 0 \), and thus \( F\left( {S}_{\epsilon }\right) = {\epsilon }^{d}\mathop{\sum }\limits_{{n \in {\mathbb{Z}}^{d}}}F\left( {\psi }_{n\epsilon }^{ \sim }\right) \varphi \left( {n\epsilon }\right) \). Hence by the continuity of \( x \mapsto F\left( {\psi }_{x}^{ \sim }\right) \), a passage to the limit \( \epsilon \rightarrow 0 \) yields (2), proving the proposition. | Yes |
Corollary 1.2 Suppose \( F \) is a distribution on \( {\mathbb{R}}^{d} \) . Then there exists a sequence \( \left\{ {f}_{n}\right\} \), with \( {f}_{n} \in {C}^{\infty } \), and \( {f}_{n} \rightarrow F \) in the weak sense. | Proof. Let \( \left\{ {\psi }_{n}\right\} \) be an approximation to the identity constructed as follows. Fix a \( \psi \in \mathcal{D} \) with \( {\int }_{{\mathbb{R}}^{d}}\psi \left( x\right) {dx} = 1 \) and set \( {\psi }_{n}\left( x\right) = {n}^{d}\psi \left( {nx}\right) \) . Form \( {F}_{n} = F * {\psi }_{n} \) . Then by the second conclusion of the proposition, each \( {F}_{n} \) is a \( {C}^{\infty } \) function. However by the first conclusion \[ {F}_{n}\left( \varphi \right) = F\left( {{\psi }_{n}^{ \sim } * \varphi }\right) \;\text{ for every }\varphi \in \mathcal{D}. \] Moreover, as is easily verified, \( {\psi }_{n}^{ \sim } * \varphi \rightarrow \varphi \) in \( \mathcal{D} \) . Thus \( {F}_{n}\left( \varphi \right) \rightarrow F\left( \varphi \right) \) , for each \( \varphi \in \mathcal{D} \), and the corollary is established. | Yes |
Proposition 1.3 Suppose \( F \) is a distribution whose support is \( {C}_{1} \), and \( \psi \) is in \( \mathcal{D} \) and has support \( {C}_{2} \) . Then the support of \( F * \psi \) is contained in \( {C}_{1} + {C}_{2} \) | Indeed for each \( x \) for which \( F\left( {\psi }_{x}^{ \sim }\right) \neq 0 \), we must have that the support of \( F \) intersects the support of \( {\psi }_{x}^{ \sim } \) . Since the support of \( {\psi }_{x}^{ \sim } \) is the set \( x - \) \( {C}_{2} \) this means that the set \( {C}_{1} \) and \( x - {C}_{2} \) have a point, say \( y \), in common. Because \( x = y + x - y \), while \( y \in {C}_{1} \) and \( x - y \in {C}_{2} \) (since \( y \in x - {C}_{2} \) ) we have that \( x \in {C}_{1} + {C}_{2} \), and thus our assertion is established. Note that the set \( {C}_{1} + {C}_{2} \) is closed because \( {C}_{1} \) is closed and \( {C}_{2} \) is compact. | Yes |
Proposition 1.4 Suppose \( F \) is a tempered distribution. Then there is a positive integer \( N \) and a constant \( c > 0 \), so that\n\n\[ \left| {F\left( \varphi \right) }\right| \leq c\parallel \varphi {\parallel }_{N},\;\text{ for all }\varphi \in \mathcal{S}. \]\n | Proof. Assume otherwise. Then the conclusion fails and for each positive integer \( n \) there is a \( {\psi }_{n} \in \mathcal{S} \) with \( {\begin{Vmatrix}{\psi }_{n}\end{Vmatrix}}_{n} = 1 \), while \( \left| {F\left( {\psi }_{n}\right) }\right| \geq n \) . Take \( {\varphi }_{n} = {\psi }_{n}/{n}^{1/2} \) . Then \( {\begin{Vmatrix}{\varphi }_{n}\end{Vmatrix}}_{N} \leq {\begin{Vmatrix}{\varphi }_{n}\end{Vmatrix}}_{n} \) as soon as \( n \geq N \), and thus \( {\begin{Vmatrix}{\varphi }_{n}\end{Vmatrix}}_{N} \leq {n}^{-1/2} \rightarrow 0 \) as \( n \rightarrow \infty \), while \( \left| {F\left( {\varphi }_{n}\right) }\right| \geq {n}^{1/2} \rightarrow \infty \), contradicting the continuity of \( F \) . | Yes |
Proposition 1.5 Suppose \( F \) is a tempered distribution and \( \psi \in \mathcal{S} \) . Then \( F * \psi \) is a slowly increasing \( {C}^{\infty } \) function, which when considered as a tempered distribution satisfies \( {\left( F * \psi \right) }^{ \land } = {\psi }^{ \land }{F}^{ \land } \) . | Proof. The fact that \( F\left( {\psi }_{x}^{ \sim }\right) \) is slowly increasing follows from the proposition in Section 1.4 together with the observation that for any function \( \psi \in \mathcal{D} \) and \( N,{\begin{Vmatrix}{\psi }_{x}^{ \sim }\end{Vmatrix}}_{N} \leq c{\left( 1 + \left| x\right| \right) }^{N}\parallel \psi {\parallel }_{N} \), and more generally,\n\n\[{\begin{Vmatrix}{\partial }_{x}^{\alpha }{\psi }_{x}^{ \sim }\end{Vmatrix}}_{N} \leq c{\left( 1 + \left| x\right| \right) }^{N}\parallel \psi {\parallel }_{N + \left| \alpha \right| }.\]\n\nSince \( \left( {F * \psi }\right) \left( \varphi \right) = F\left( {{\psi }^{ \sim } * \varphi }\right) \), it follows that \( {\left( F * \psi \right) }^{ \land }\left( \varphi \right) = F\left( {{\psi }^{ \sim } * {\varphi }^{ \land }}\right) \) . On the other hand, \( {\psi }^{ \land }{F}^{ \land }\left( \varphi \right) = {F}^{ \land }\left( {{\psi }^{ \land }\varphi }\right) = F\left( {\left( {\psi }^{ \land }\varphi \right) }^{ \land }\right) \) . Thus the desired identity, \( {\left( F * \psi \right) }^{ \land }\left( \varphi \right) = \left( {{\psi }^{ \land }{F}^{ \land }}\right) \left( \varphi \right) \) is proved because, as is easily verified, \( {\left( {\psi }^{ \land }\varphi \right) }^{ \land } = {\psi }^{ \sim } * {\varphi }^{ \land } \) . | Yes |
Proposition 1.6 If \( F \) is a distribution of compact support then its Fourier transform \( {F}^{ \land } \) is a slowly increasing \( {C}^{\infty } \) function. In fact, as a function of \( \xi \), one has \( {F}^{ \land }\left( \xi \right) = F\left( {e}_{\xi }\right) \) where \( {e}_{\xi } \) is the element of \( \mathcal{D} \) given by \( {e}_{\xi }\left( x\right) = \eta \left( x\right) {e}^{-{2\pi ix\xi }} \), with \( \eta \) a function in \( \mathcal{D} \) that equals 1 in a neighborhood of the support of \( F \) . | Proof. If we invoke Proposition 1.4, we see immediately that \( \left| {F\left( {e}_{\xi }\right) }\right| \leq \) \( C{\begin{Vmatrix}{e}_{\xi }\end{Vmatrix}}_{N} \leq {c}^{\prime }{\left( 1 + \left| \xi \right| \right) }^{N} \) . By the same estimate, every difference quotient of \( F\left( {e}_{\xi }\right) \) converges and \( \left| {{\partial }_{\xi }^{\alpha }F\left( {e}_{\xi }\right) }\right| \leq {c}_{\alpha }{\left( 1 + \left| \xi \right| \right) }^{N + \left| \alpha \right| } \) . Therefore \( F\left( {e}_{\xi }\right) \) is \( {C}^{\infty } \) and slowly increasing. To prove that the function \( F\left( {e}_{\xi }\right) \) is the Fourier transform of \( F \) it suffices to see that\n\n(3)\n\n\[{\int }_{{\mathbb{R}}^{d}}F\left( {e}_{\xi }\right) \varphi \left( \xi \right) {d\xi } = F\left( \widehat{\varphi }\right) \;\text{ for every }\varphi \in \mathcal{S}.\n\]\n\nWe prove this first when \( \varphi \in \mathcal{D} \) .\n\nNow by what we have already seen, the function \( g\left( \xi \right) = F\left( {e}_{\xi }\right) \varphi \left( \xi \right) \) is continuous and certainly has compact support. Thus\n\n\[{\int }_{{\mathbb{R}}^{d}}F\left( {e}_{\xi }\right) \varphi \left( \xi \right) {d\xi } = {\int }_{{\mathbb{R}}^{d}}g\left( \xi \right) {d\xi } = \mathop{\lim }\limits_{{\epsilon \rightarrow 0}}{S}_{\epsilon }\n\]\n\nwhere for each \( \epsilon > 0,{S}_{\epsilon } \) is the (finite) sum \( {\epsilon }^{d}\mathop{\sum }\limits_{{n \in {\mathbb{Z}}^{d}}}g\left( {n\epsilon }\right) \) . However \( {S}_{\epsilon } = F\left( {s}_{\epsilon }\right) \), with \( {s}_{\epsilon } = {\epsilon }^{d}\mathop{\sum }\limits_{{n \in {\mathbb{Z}}^{d}}}{e}_{n\epsilon }\left( x\right) \varphi \left( {n\epsilon }\right) \) . Clearly as \( \epsilon \rightarrow 0 \), we have\n\n\[{s}_{\epsilon }\left( x\right) \rightarrow \eta \left( x\right) {\int }_{{\mathbb{R}}^{d}}{e}^{-{2\pi ix} \cdot \xi }\varphi \left( \xi \right) {d\xi } = \eta \left( x\right) \widehat{\varphi }\left( x\right)\n\]\n\n in the \( \parallel \cdot {\parallel }_{N} \) norm. Thus, using Proposition 1.4 again, we get that \( {S}_{\epsilon } \rightarrow \) \( F\left( {\eta \widehat{\varphi }}\right) \) . Now since \( \eta = 1 \) in a neighborhood of the support of \( F \), then \( F\left( {\eta \widehat{\varphi }}\right) = F\left( \widehat{\varphi }\right) \) . Altogether we have (3) when \( \varphi \in \mathcal{D} \), and to extend this result to \( \varphi \in \mathcal{S} \) it suffices to recall that \( \mathcal{D} \) is dense in \( \mathcal{S} \). | Yes |
Theorem 1.7 Suppose \( F \) is a distribution supported at the origin. Then \( F \) is a finite sum\n\n\[ F = \mathop{\sum }\limits_{{\left| \alpha \right| \leq N}}{a}_{\alpha }{\partial }_{x}^{\alpha }\delta \]\n\nThat is,\n\n\[ F\left( \varphi \right) = \mathop{\sum }\limits_{{\left| \alpha \right| \leq N}}{\left( -1\right) }^{\left| \alpha \right| }{a}_{\alpha }\left( {{\partial }_{x}^{\alpha }\varphi }\right) \left( 0\right) ,\;\text{ for }\varphi \in \mathcal{D}. \] | Proceeding with the proof of the theorem, we now apply the above lemma to \( {F}_{1} = F - \mathop{\sum }\limits_{{\left| \alpha \right| < N}}{a}_{\alpha }{\partial }_{x}^{\alpha }\delta \) where \( N \) is the index that guarantees the conclusion of Proposition 1.4, while the \( {a}_{\alpha } \) are chosen so that \( {a}_{\alpha } = \frac{{\left( -1\right) }^{\left| \alpha \right| }}{\alpha !}F\left( {x}^{\alpha }\right) \) . Then since \( {\partial }_{x}^{\alpha }\left( \delta \right) \left( {x}^{\beta }\right) = {\left( -1\right) }^{\left| \alpha \right| }\alpha ! \), if \( \alpha = \beta \), and zero otherwise, we see that \( {F}_{1} = 0 \), which proves the theorem. | Yes |
Lemma 1.8 Suppose \( {F}_{1} \) is a distribution supported at the origin that satisfies for some \( N \) the following two conditions:\n\n(a) \( \left| {{F}_{1}\left( \varphi \right) }\right| \leq c\parallel \varphi {\parallel }_{N} \), for all \( \varphi \in \mathcal{D} \).\n\n(b) \( {F}_{1}\left( {x}^{\alpha }\right) = 0 \), for all \( \left| \alpha \right| \leq N \).\n\nThen \( {F}_{1} = 0 \). | In fact, let \( \eta \in \mathcal{D} \), with \( \eta \left( x\right) = 0 \) for \( \left| x\right| \geq 1 \), and \( \eta \left( x\right) = 1 \) when \( \left| x\right| \leq 1/2 \) , and write \( {\eta }_{\epsilon }\left( x\right) = \eta \left( {x/\epsilon }\right) \). Then since \( {F}_{1} \) is supported at the origin, \( {F}_{1}\left( {{\eta }_{\epsilon }\varphi }\right) = {F}_{1}\left( \varphi \right) \). Moreover, by the same token \( {F}_{1}\left( {{\eta }_{\epsilon }{x}^{\alpha }}\right) = {F}_{1}\left( {x}^{\alpha }\right) = 0 \) for all \( \left| \alpha \right| \leq N \), and hence\n\n\[ {F}_{1}\left( \varphi \right) = {F}_{1}\left( {{\eta }_{\epsilon }\left( {\varphi \left( x\right) - \mathop{\sum }\limits_{{\left| \alpha \right| \leq N}}\frac{{\varphi }^{\left( \alpha \right) }\left( 0\right) }{\alpha !}{x}^{\alpha }}\right) }\right) ,\]\n\nwith \( {\varphi }^{\left( \alpha \right) } = {\partial }_{x}^{\alpha }\varphi \left( 0\right) \). If \( R\left( x\right) = \varphi \left( x\right) - \mathop{\sum }\limits_{{\left| \alpha \right| \leq N}}\frac{{\varphi }^{\left( \alpha \right) }\left( 0\right) }{\alpha !}{x}^{\alpha } \) is the remainder, then \( \left| {R\left( x\right) }\right| \leq c{\left| x\right| }^{N + 1} \) and \( \left| {{\partial }_{x}^{\beta }R\left( x\right) }\right| \leq {c}_{\beta }{\left| x\right| }^{N + 1 - \left| \beta \right| } \), when \( \left| \beta \right| \leq N \). However \( \left| {{\partial }_{x}^{\beta }{\eta }_{\epsilon }\left( x\right) }\right| \leq {c}_{\beta }{\epsilon }^{-\left| \beta \right| } \) and \( {\partial }_{x}^{\beta }{\eta }_{\epsilon }\left( x\right) = 0 \) if \( \left| x\right| \geq \epsilon \). Thus by Leibnitz’s rule, \( {\begin{Vmatrix}{\eta }_{\epsilon }R\end{Vmatrix}}_{N} \leq {c\epsilon } \), and our assumption (a) gives \( \left| {{F}_{1}\left( \varphi \right) }\right| \leq {c}^{\prime }\epsilon \) , which yields the desired conclusion upon letting \( \epsilon \rightarrow 0 \). | Yes |
Theorem 2.1 The distribution \( \operatorname{pv}\left( \frac{1}{x}\right) \) equals:\n\n(a) \( \frac{d}{dx}\left( {\log \left| x\right| }\right) \).\n\n(b) \( \frac{1}{2}\left( {\frac{1}{x - {i0}} + \frac{1}{x + {i0}}}\right) \).\n\nAlso, its Fourier transform equals \( \frac{\pi }{i}\operatorname{sign}\left( x\right) \) . | Regarding (a), note that \( \log \left| x\right| \) is a locally integrable function. Here \( \frac{d}{dx}\left( {\log \left| x\right| }\right) \) is its derivative taken as a distribution. Now in that sense\n\n\[ \left( {\frac{d}{dx}\log \left| x\right| }\right) \left( \varphi \right) = - {\int }_{-\infty }^{\infty }\left( {\log \left| x\right| }\right) \frac{d\varphi }{dx}{dx},\;\text{ for every }\varphi \in \mathcal{S}. \]\n\nHowever the integral is the limit as \( \epsilon \rightarrow 0 \) of \( - {\int }_{\left| x\right| \geq \epsilon }\left( {\log \left| x\right| }\right) \frac{d\varphi }{dx}{dx} \), and an integration by parts shows that this equals\n\n\[ {\int }_{\left| x\right| \geq \epsilon }\frac{\varphi \left( x\right) }{x}{dx} + \log \left( \epsilon \right) \left\lbrack {\varphi \left( \epsilon \right) - \varphi \left( {-\epsilon }\right) }\right\rbrack \]\n\nMoreover, \( \varphi \left( \epsilon \right) - \varphi \left( {-\epsilon }\right) = O\left( \epsilon \right) \) since in particular \( \varphi \) is of class \( {C}^{1} \). Therefore \( \log \left( \epsilon \right) \left\lbrack {\varphi \left( \epsilon \right) - \varphi \left( {-\epsilon }\right) }\right\rbrack \rightarrow 0 \) as \( \epsilon \rightarrow 0 \), and we have established (a). | Yes |
Proposition 2.2 Suppose \( F \) is a tempered distribution on \( {\mathbb{R}}^{d} \) that is homogeneous of degree \( \lambda \) . Then its Fourier transform \( {F}^{ \land } \) is homogeneous of degree \( - d - \lambda \) . | To deal with \( {\left( {F}^{ \land }\right) }_{a} \) we write successively,\n\n\[ \n{\left( {F}^{ \land }\right) }_{a}\left( \varphi \right) = {F}^{ \land }\left( {\varphi }^{a}\right) = F\left( {\left( {\varphi }^{a}\right) }^{ \land }\right) = F\left( {\left( {\varphi }^{ \land }\right) }_{a}\right) \n\] \n\n\[ \n= {F}^{a}\left( {\varphi }^{ \land }\right) = {a}^{-d}{F}_{{a}^{-1}}\left( {\varphi }^{ \land }\right) = {a}^{-d - \lambda }F\left( {\varphi }^{ \land }\right) = {a}^{-d - \lambda }{F}^{ \land }\left( \varphi \right) . \n\] \n\nThus \( {\left( {F}^{ \land }\right) }_{a} = {a}^{-d - \lambda }{F}^{ \land } \), as was to be proved. | Yes |
Theorem 2.3 If \( - d < \lambda < 0 \), then\n\n\[{\left( {H}_{\lambda }\right) }^{ \land } = {c}_{\lambda }{H}_{-d - \lambda },\;\text{with}\;{c}_{\lambda } = \frac{\Gamma \left( \frac{d + \lambda }{2}\right) }{\Gamma \left( \frac{-\lambda }{2}\right) }{\pi }^{-d/2 - \lambda }.\] | To prove the theorem we start with the fact that \( \psi \left( x\right) = {e}^{-\pi {\left| x\right| }^{2}} \) is its own Fourier transform. Then since \( {\left( {\psi }_{a}\right) }^{ \land } = {\left( {\psi }^{ \land }\right) }^{a} \) we get (with \( a = {t}^{1/2} \) )\n\n\[{\int }_{{\mathbb{R}}^{d}}{e}^{-{\pi t}{\left| x\right| }^{2}}\widehat{\varphi }\left( x\right) {dx} = {t}^{-d/2}{\int }_{{\mathbb{R}}^{d}}{e}^{-\pi {\left| x\right| }^{2}/t}\varphi \left( x\right) {dx}.\n\nWe now multiply both sides by \( {t}^{-\lambda /2 - 1} \) and integrate over \( \left( {0,\infty }\right) \), and then interchange the order of integration. We note that\n\n\[{\int }_{0}^{\infty }{e}^{-{tA}}{t}^{-\lambda /2 - 1}{dt} = {A}^{\lambda /2}\Gamma \left( {-\lambda /2}\right) ,\n\nif \( A > 0 \) and \( \lambda > 0 \), by making the indicated change of variables that reduces the identity to the case \( A = 1 \) . Thus using the above identity with \( A = \pi {\left| x\right| }^{2} \), we get\n\n\[{\int }_{{\mathbb{R}}^{d}}{\int }_{0}^{\infty }{e}^{-{\pi t}{\left| x\right| }^{2}}\widehat{\varphi }\left( x\right) {t}^{-\lambda /2 - 1}{dtdx} = {\pi }^{\lambda /2}\Gamma \left( {-\lambda /2}\right) {\int }_{{\mathbb{R}}^{d}}{\left| x\right| }^{\lambda }\widehat{\varphi }\left( x\right) {dx}.\n\nSimilarly, we deal with \( {\int }_{0}^{\infty }{t}^{-d/2}{t}^{-\lambda /2 - 1}{e}^{-A/t}{dt} \) by making the change of variables \( t \rightarrow 1/t \) which shows that this integral equals\n\n\[{\int }_{0}^{\infty }{t}^{d/2 + \lambda /2 - 1}{e}^{-{At}}{dt} = {A}^{-d/2 - \lambda /2}\Gamma \left( {\frac{d}{2} + \frac{\lambda }{2}}\right) .\n\nInserting this in \( {\int }_{{\mathbb{R}}^{d}}{\int }_{0}^{\infty }{t}^{-d/2}{e}^{-\pi {\left| x\right| }^{2}/t}\varphi \left( x\right) {dtdx} \) yields\n\n\[{\pi }^{\lambda /2}\Gamma \left( {-\lambda /2}\right) {\int }_{{\mathbb{R}}^{d}}{\left| x\right| }^{\lambda }\widehat{\varphi }\left( x\right) {dx} =\n\n\[{\pi }^{-d/2 - \lambda /2}\Gamma \left( {d/2 + \lambda /2}\right) {\int }_{{\mathbb{R}}^{d}}{\left| x\right| }^{-d - \lambda }\varphi \left( x\right) {dx}\n\nand this is our theorem. | Yes |
Theorem 2.4 The Fourier transform of a regular homogeneous distribution \( K \) of degree \( \lambda \) is a regular homogeneous distribution of degree \( - d - \lambda \), and conversely. | Proof. We already know from Proposition 2.2 that \( {K}^{ \land } \) is homogeneous of degree \( - d - \lambda \) . To prove that \( {K}^{ \land } \) agrees with a \( {C}^{\infty } \) function away from the origin, we decompose \( K = {K}_{0} + {K}_{1} \), with \( {K}_{0} \) supported near the origin and \( {K}_{1} \) supported away from the origin. To do this, fix a cut-off function \( \eta \) that is \( {C}^{\infty } \), is supported in \( \left| x\right| \leq 1 \), and that equals 1 on \( \left| x\right| \leq 1/2 \) . Write \( {K}_{0} = {\eta K},{K}_{1} = \left( {1 - \eta }\right) K \) . In particular \( {K}_{1} \) is the function \( \left( {1 - \eta }\right) k \), since \( 1 - \eta \) vanishes near the origin. Also \( {K}^{ \land } = {K}_{0}^{ \land } + {K}_{1}^{ \land } \n\nNow by Proposition 1.6, \( {K}_{0}^{ \land } \) is an (everywhere) \( {C}^{\infty } \) function. To prove that \( {K}_{1}^{ \land } \) is \( {C}^{\infty } \) away from the origin we observe that by the usual manipulations of the Fourier transform valid for tempered distributions,\n\n(7)\n\n\[ \n{\left( -4{\pi }^{2}{\left| \xi \right| }^{2}\right) }^{N}{\partial }_{\xi }^{\alpha }\left( {K}_{1}^{ \land }\right) = {\left( {\bigtriangleup }^{N}\left\lbrack {\left( -2\pi ix\right) }^{\alpha }{K}_{1}\right\rbrack \right) }^{ \land }. \n\] \n\nRecall that \( \bigtriangleup \) denotes the Laplacian, \( \bigtriangleup = {\partial }^{2}/\partial {x}_{1}^{2} + \cdots + {\partial }^{2}/\partial {x}_{d}^{2} \) .\n\nNow when \( \left| x\right| \geq 1,{K}_{1} = k \), so there \( {\partial }_{x}^{\beta }\left( {K}_{1}\right) \) is a bounded homogeneous function of degree \( \lambda - \left| \beta \right| \) and thus is \( O\left( {\left| x\right| }^{\lambda - \left| \beta \right| }\right) \), for \( \left| x\right| \geq 1 \) . Therefore \( {\bigtriangleup }^{N}\left\lbrack {{x}^{\alpha }{K}_{1}}\right\rbrack \) is \( O\left( {\left| x\right| }^{\lambda + \left| \alpha \right| - {2N}}\right) \) for \( \left| x\right| \geq 1 \) while it is certainly a bounded function for \( \left| x\right| \leq 1 \) . Hence for \( N \) sufficiently large \( \left( {{2N} > \lambda + \left| \alpha \right| + d}\right) \) this function belongs to \( {L}^{1}\left( {\mathbb{R}}^{d}\right) \) . As a result its Fourier transform is continuous. (See Chapter 2 in Book III.) This shows by (7) that \( {\partial }_{x}^{\alpha }\left( {K}_{1}^{ \land }\right) \) agrees with a continuous function away from the origin. Since this holds for every \( \alpha \), it follows from Exercise 2 that \( {K}_{1}^{ \land } \) is a \( {C}^{\infty } \) function away from the origin, as desired.\n\nNote that since the inverse Fourier transform is the Fourier transform followed by reflection, that is, \( {K}^{ \vee } = {\left( {K}^{ \land }\right) }^{ \sim } \), the converse is a consequence of the direction we have just proved. | Yes |
Lemma 2.6 Suppose \( {\lambda }_{1},{\lambda }_{2},\ldots ,{\lambda }_{n} \), are distinct real numbers and that for constants \( {a}_{j} \) and \( {b}_{j},1 \leq j \leq n \), we have\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{n}\left( {{a}_{j}{x}^{{\lambda }_{j}} + {b}_{j}{x}^{{\lambda }_{j}}\log x}\right) = 0\;\text{ for all }x > 0. \]\n\nThen \( {a}_{j} = {b}_{j} = 0 \) for all \( 1 \leq j \leq n \) . | To prove the lemma we assume, as one may, that \( {\lambda }_{n} \) is the largest of the \( {\lambda }_{j} \) ’s. Then multiplying the identity by \( {x}^{-{\lambda }_{n}} \) and letting \( x \) tend to infinity we see that \( {b}_{n} \) as well as \( {a}_{n} \) must vanish. Thus we are reduced to the case when \( n \) is replaced by \( n - 1 \), and this induction gives the lemma. | No |
Corollary 2.7 There is no distribution \( {K}_{0} \) that is homogeneous of degree \( - d \) and that agrees with the function \( 1/{\left| x\right| }^{d} \) away from the origin. | If such a \( {K}_{0} \) existed, then \( {K}_{0} - \left\lbrack \frac{1}{{\left| x\right| }^{d}}\right\rbrack \) would be supported at the origin, and hence equal to \( \mathop{\sum }\limits_{{\left| \alpha \right| \leq M}}{c}_{\alpha }{\partial }_{x}^{\alpha }\delta \) . Applying this difference to \( {\varphi }^{a} \) would yield that\n\n\[ \n{a}^{-d}{K}_{0}\left( \varphi \right) - {a}^{-d}\left\lbrack \frac{1}{{\left| x\right| }^{d}}\right\rbrack \left( \varphi \right) - {a}^{-d}\log \left( a\right) {A}_{d}\varphi \left( 0\right) -\n\]\n\n\[ \n- \mathop{\sum }\limits_{{\left| \alpha \right| \leq M}}{c}_{\alpha }{a}^{-d - \left| \alpha \right| }{\partial }_{x}^{\alpha }\delta \left( \varphi \right) = 0\n\]\n\nfor all \( a > 0 \) . This leads to a contradiction with Lemma 2.6 if we take \( \varphi \) so that \( \varphi \left( 0\right) \neq 0 \) . | Yes |
Theorem 2.8 For \( d \geq 3 \), the locally integrable function \( F \) defined by \( F\left( x\right) = {C}_{d}{\left| x\right| }^{-d + 2} \) is a fundamental solution for the operator \( \bigtriangleup \), with \( {C}_{d} = - \frac{\Gamma \left( {\frac{d}{2} - 1}\right) }{4{\pi }^{\frac{d}{2}}}. \) | This follows by taking \( \lambda = - d + 2 \) (in Theorem 2.3), then \( \Gamma \left( \frac{d + \lambda }{2}\right) = \) \( \Gamma \left( 1\right) = 1 \), while \( \Gamma \left( {d/2}\right) = \left( {d/2 - 1}\right) \Gamma \left( {d/2 - 1}\right) \) . Therefore \( \widehat{F}\left( \xi \right) \) equals \( 1/\left( {-4{\pi }^{2}{\left| \xi \right| }^{2}}\right) \), and hence\n\n\[ \n{\left( \bigtriangleup F\right) }^{ \land } = 1,\;\text{ which means }\;\bigtriangleup F = \delta .\n\] | No |
Theorem 2.10 \( F \) is a fundamental solution of \( L = \frac{\partial }{\partial t} - {\bigtriangleup }_{x} \) . | Proof. Since \( {LF}\left( \varphi \right) = F\left( {{L}^{\prime }\varphi }\right) \) with \( {L}^{\prime } = - \frac{\partial }{\partial t} - {\bigtriangleup }_{x} \), it suffices to see that \( F\left( {{L}^{\prime }\varphi }\right) \), which equals\n\n\[ \mathop{\lim }\limits_{{\epsilon \rightarrow 0}}{\int }_{t \geq \epsilon }{\int }_{{\mathbb{R}}^{d}}F\left( {x, t}\right) \left( {-\frac{\partial }{\partial t} - {\bigtriangleup }_{x}}\right) \varphi \left( {x, t}\right) {dxdt} \]\n\nis \( \delta \left( \varphi \right) = \varphi \left( {0,0}\right) \) .\n\nNow \( F\left( {x, t}\right) = {\mathcal{H}}_{t}\left( x\right) \) when \( t > 0 \), so an integration by parts in the \( x \) -variables gives\n\n\[ {\int }_{t \geq \epsilon }{\int }_{{\mathbb{R}}^{d}}F\left( {x, t}\right) \left( {-\frac{\partial }{\partial t} - {\bigtriangleup }_{x}}\right) \varphi \left( {x, t}\right) {dxdt} = \]\n\n\[ = - {\int }_{{\mathbb{R}}^{d}}\left( {{\int }_{t \geq \epsilon }{\mathcal{H}}_{t}\frac{\partial \varphi }{\partial t} + \left( {{\bigtriangleup }_{x}{\mathcal{H}}_{t}}\right) {\varphi dt}}\right) {dx} \]\n\n\[ = - {\int }_{{\mathbb{R}}^{d}}\left( {{\int }_{t \geq \epsilon }{\mathcal{H}}_{t}\frac{\partial \varphi }{\partial t} + \frac{\partial {\mathcal{H}}_{t}}{\partial t}{\varphi dt}}\right) {dx} \]\n\n\[ = {\int }_{{\mathbb{R}}^{d}}{\mathcal{H}}_{\epsilon }\left( x\right) \varphi \left( {x,\epsilon }\right) {dx} \]\n\nHowever, because \( \varphi \in \mathcal{S} \), one has \( \left| {\varphi \left( {x,\epsilon }\right) - \varphi \left( {x,0}\right) }\right| \leq O\left( \epsilon \right) \) uniformly in \( x \) . Therefore\n\n\[ {\int }_{{\mathbb{R}}^{d}}{\mathcal{H}}_{\epsilon }\left( x\right) \varphi \left( {x,\epsilon }\right) {dx} = {\int }_{{\mathbb{R}}^{d}}{\mathcal{H}}_{\epsilon }\left( x\right) \left( {\varphi \left( {x,0}\right) + O\left( \epsilon \right) }\right) {dx} \]\n\nand this tends to \( \varphi \left( {0,0}\right) \), since \( {\mathcal{H}}_{t} \) is an approximation to the identity. | Yes |
Theorem 2.11 Every constant coefficient (linear) partial differential equation \( L \) on \( {\mathbb{R}}^{d} \) has a fundamental solution. | Proof. After a possible change of coordinates consisting of a rotation and multiplication by a constant, we may assume that the characteristic polynomial of \( L \) will be of the form\n\n\[ P\left( \xi \right) = P\left( {{\xi }_{1},{\xi }^{\prime }}\right) = {\xi }_{1}^{m} + \mathop{\sum }\limits_{{j = 0}}^{{m - 1}}{\xi }_{1}^{j}{Q}_{j}\left( {\xi }^{\prime }\right) ,\]\n\nwhere each \( {Q}_{j}^{\prime } \) is a polynomial of degree at most \( m - j \) . A proof that a general polynomial \( P \) can be written in the above form, can be found for instance in Section 3, Chapter 5, Book III, where an earlier version of the \ | No |
Theorem 2.12 Every elliptic operator has a regular parametrix. | Proof. Observe first by a straightforward inductive argument in \( k \) , that whenever \( \left| \alpha \right| = k \) and \( P \) is any polynomial\n\n\[{\left( \frac{\partial }{\partial \xi }\right) }^{\alpha }\left( \frac{1}{P\left( \xi \right) }\right) = \mathop{\sum }\limits_{{0 \leq \ell \leq k}}\frac{{q}_{\ell }\left( \xi \right) }{P{\left( \xi \right) }^{\ell + 1}}\]\n\nwhere each \( {q}_{\ell } \) is a polynomial of degree \( \leq \ell m - k \) .\n\nNow suppose \( \left| {P\left( \xi \right) }\right| \geq c{\left| \xi \right| }^{m} \), whenever \( \left| \xi \right| \geq {c}_{1} \), and let \( \gamma \) be a \( {C}^{\infty } \) function which is equal to 1 for all large values of \( \xi \) and is supported in \( \left| \xi \right| \geq {c}_{1} \) . Then observe from the above identity that\n\n(17)\n\n\[ \left| {{\partial }_{\xi }^{\alpha }\left( \frac{\gamma \left( \xi \right) }{P\left( \xi \right) }\right) }\right| \leq {A}_{\alpha }{\left| \xi \right| }^{-m - \left| \alpha \right| }\]\n\nNow let \( Q \) be the tempered distribution whose Fourier transform is the (bounded) function \( \gamma \left( \xi \right) /P\left( \xi \right) \) . Taking up the same argument as in the proof of Theorem 2.4, we have\n\n\[{\left( {\left( -4{\pi }^{2}{\left| x\right| }^{2}\right) }^{N}{\partial }_{x}^{\beta }Q\right) }^{ \land } = {\bigtriangleup }_{\xi }^{N}\left\lbrack {{\left( 2\pi i\xi \right) }^{\beta }\left( {\gamma /P}\right) }\right\rbrack .\n\nBecause of (17) and Leibnitz's rule, the right-hand side above is clearly dominated by \( {A}_{\alpha }^{\prime }{\left| \xi \right| }^{-m - {2N} + \left| \beta \right| } \) for \( \left| \xi \right| \geq 1 \) ; it is also bounded when \( \left| \xi \right| \leq \) 1. Thus as soon as \( {2N} + m - \left| \beta \right| > d \), this function is integrable, and therefore \( {\left| x\right| }^{2N}{\partial }_{x}^{\beta }Q \), being its inverse Fourier transform up to a multiplicative constant, is continuous. Since this is true for each \( \beta \), we see that \( Q \) agrees with a \( {C}^{\infty } \) function away from the origin.\n\nNote moreover that \( {\left( LQ\right) }^{ \land } = P\left( \xi \right) \left\lbrack {\gamma \left( \xi \right) )/P\left( \xi \right) }\right\rbrack = \gamma \left( \xi \right) = 1 + (\gamma \left( \xi \right) - 1). By its definition, \( \gamma \left( \xi \right) - 1 \) is in \( \mathcal{D} \), and hence \( \gamma \left( \xi \right) - 1 = \widehat{r} \), for some \( r \in \mathcal{S} \) . Finally, \( {\left( LQ\right) }^{ \land } = 1 + \widehat{r} \), which means \( {LQ} = \delta + r \), as was to be shown. | Yes |
Corollary 2.13 Given any \( \epsilon > 0 \), the elliptic operator \( L \) has a regular parametrix \( {Q}_{\epsilon } \) that is supported in the ball \( \{ x : \left| x\right| \leq \epsilon \} \) . | In fact, let \( {\eta }_{\epsilon } \) be a cut-off function in \( \mathcal{D} \), that is 1 when \( \left| x\right| \leq \epsilon /2 \) , and that is supported where \( \left| x\right| \leq \epsilon \) . Set \( {Q}_{\epsilon } = {\eta }_{\epsilon }Q \), and observe that \( L\left( {{\eta }_{\epsilon }Q}\right) - {\eta }_{\epsilon }L\left( Q\right) \) involves only terms that are derivatives of \( {\eta }_{\epsilon } \) of positive order, and these vanish when \( \left| x\right| < \epsilon /2 \) . The difference is therefore a \( {C}^{\infty } \) function. However, \( {\eta }_{\epsilon }L\left( Q\right) = {\eta }_{\epsilon }\left( {\delta + r}\right) = \delta + {\eta }_{\epsilon }r \) . Altogether, this gives \( L\left( {Q}_{\epsilon }\right) = \delta + {r}_{\epsilon } \), where \( {r}_{\epsilon } \) is a \( {C}^{\infty } \) function. Notice that \( {r}_{\epsilon } \) is automatically also supported in \( \left| x\right| \leq \epsilon \) . | Yes |
Theorem 2.14 Suppose the partial differential operator \( L \) has a regular parametrix. Assume \( U \) is a distribution given in an open set \( \Omega \subset {\mathbb{R}}^{d} \) and \( L\left( U\right) = f \), with \( f \) a \( {C}^{\infty } \) function in \( \Omega \) . Then \( U \) is also a \( {C}^{\infty } \) function on \( \Omega \) . In particular, this holds whenever \( L \) is elliptic. | Proof of the theorem. It suffices to show that \( U \) agrees with a \( {C}^{\infty } \) function on any ball \( B \) with \( \bar{B} \subset \Omega \) . Fix such a ball (say of radius \( \rho \) ), and let \( {B}_{1} \) be the concentric ball having radius \( \rho + \epsilon \), with \( \epsilon > 0 \) so small that \( \overline{{B}_{1}} \subset \Omega \) . Next, choose a cut-off function \( \eta \) in \( \mathcal{D} \), supported in \( \Omega \) , with \( \eta \left( x\right) = 1 \) in a neighborhood of \( \overline{{B}_{1}} \) . Define \( {U}_{1} = {\eta U} \) . Then \( {U}_{1} \) and \( L\left( {U}_{1}\right) = {F}_{1} \) are distributions of compact support in \( {\mathbb{R}}^{d} \) and moreover \( {F}_{1} \) agrees with a \( {C}^{\infty } \) function (that is, \( f \) ) in a neighborhood of \( \overline{{B}_{1}} \) . Thus \( {F}_{1} \) agrees in a smaller neighborhood of \( \overline{{B}_{1}} \) with a \( {C}^{\infty } \) function \( {f}_{1} \) that has compact support.\n\nWe now apply the parametrix \( {Q}_{\epsilon } \) supported in \( \{ \left| x\right| \leq \epsilon \} \) whose existence is guaranteed by Corollary 2.13. On the one hand,\n\n\[ \n{Q}_{\epsilon } * L\left( {U}_{1}\right) = L\left( {Q}_{\epsilon }\right) * {U}_{1} = \left( {\delta + {r}_{\epsilon }}\right) * {U}_{1} = {U}_{1} + {r}_{\epsilon } * {U}_{1}, \n\] \n\nand since \( {r}_{\epsilon } * {U}_{1} = {U}_{1} * {r}_{\epsilon } \) by Proposition 1.1, we have that \( {r}_{\epsilon } * {U}_{1} \) is a \( {C}^{\infty } \) function in \( {\mathbb{R}}^{d} \) . On the other hand, \n\n\[ \n{Q}_{\epsilon } * L\left( {U}_{1}\right) = {Q}_{\epsilon } * {F}_{1} = {Q}_{\epsilon } * {f}_{1} + {Q}_{\epsilon } * \left( {{F}_{1} - {f}_{1}}\right) . \n\] \n\nNow again, \( {Q}_{\epsilon } * {f}_{1} \) is a \( {C}^{\infty } \) function, while by Proposition \( {1.3},{Q}_{\epsilon } * \left( {{F}_{1} - }\right. \) \( \left. {f}_{1}\right) \) is supported in the closure of the \( \epsilon \) -neighborhood of the support of \( {F}_{1} - {f}_{1} \) . Since \( {F}_{1} - {f}_{1} \) vanishes in a neighborhood of \( \overline{{B}_{1}} \) it follows that \( {Q}_{\epsilon } * \left( {{F}_{1} - {f}_{1}}\right) \) vanishes in \( B \) . Altogether then \( {U}_{1} \) is a \( {C}^{\infty } \) function on \( B \) . Since \( {U}_{1} = {\eta U} \) and \( \eta \) equals 1 in \( B \), then \( U \) is a \( {C}^{\infty } \) function in \( B \) , and the theorem is therefore proved. | Yes |
Theorem 3.2 Let \( T \) be the operator \( T\left( f\right) = f * K \), with \( K \) as in Proposition 3.1. Then \( T \) initially defined for \( f \) in \( \mathcal{S} \) extends to a bounded operator on \( {L}^{p}\left( {\mathbb{R}}^{d}\right) \), for \( 1 < p < \infty \) . | This means that for each \( p,1 < p < \infty \), there is a bound \( {A}_{p} \) so that\n\n(23)\n\n\[ \parallel {Tf}{\parallel }_{{L}^{p}\left( {\mathbb{R}}^{d}\right) } \leq {A}_{p}\parallel f{\parallel }_{{L}^{p}\left( {\mathbb{R}}^{d}\right) } \]\n\nfor \( f \in \mathcal{S} \) . Thus by Proposition 5.4 in Chapter 1 we see that \( T \) has a (unique) extension to all of \( {L}^{p} \) that satisfies the bound (23) for \( f \in {L}^{p} \) . We break the proof into five steps.\n\nStep 1: \( {L}^{2} \) estimate. The case \( p = 2 \) follows directly from the fact that \( {\left( Tf\right) }^{ \land } = {f}^{ \land }{K}^{ \land } \) ,(see Proposition 1.5) and that\n\n\[ \parallel {Tf}{\parallel }_{{L}^{2}} = {\begin{Vmatrix}{\left( Tf\right) }^{ \land }\end{Vmatrix}}_{{L}^{2}} \leq \left( {\mathop{\sup }\limits_{\xi }\left| {{K}^{ \land }\left( \xi \right) }\right| }\right) \parallel \widehat{f}{\parallel }_{{L}^{2}} \leq A\parallel f{\parallel }_{{L}^{2}}, \]\n\nby Plancherel’s theorem. The inequality \( \mathop{\sup }\limits_{\xi }\left| {{K}^{ \land }\left( \xi \right) }\right| \leq A \) is of course a consequence of Proposition 3.1. | Yes |
Lemma 3.3 For each \( f \) in \( {L}^{1}\left( {\mathbb{R}}^{d}\right) \) and \( \alpha > 0 \), we can find an open set \( {E}_{\alpha } \) and a decomposition \( f = g + b \) so that:\n\n(a) \( m\left( {E}_{\alpha }\right) \leq \frac{c}{\alpha }\parallel f{\parallel }_{{L}^{1}\left( {\mathbb{R}}^{d}\right) } \) .\n\n(b) \( \left| {g\left( x\right) }\right| \leq {c\alpha } \), for all \( x \) .\n\n(c) \( {E}_{\alpha } \) is a union \( \bigcup {Q}_{k} \) of cubes \( {Q}_{k} \) whose interiors are disjoint. Moreover \( b = \mathop{\sum }\limits_{k}{b}_{k} \), with each function \( {b}_{k} \) supported in \( {Q}_{k} \) and\n\n\[ \int \left| {{b}_{k}\left( x\right) }\right| {dx} \leq \operatorname{cam}\left( {Q}_{k}\right) ,\;\text{ while }{\int }_{{Q}_{k}}{b}_{k}\left( x\right) {dx} = 0. \] | The proof of the lemma is a simplified version of the argument used to prove Proposition 5.1 in the previous chapter; in particular, here we use the full maximal function \( {f}^{ * } \) instead of the truncated version \( {f}^{ \dagger } \) . The guiding idea is to try to cut the domain of \( f \) into the set when \( \left| {f\left( x\right) }\right| > \alpha \) and its complement. However, as before, we must be more subtle and in the present situation cut \( f \) according to where \( {f}^{ * }\left( x\right) > \alpha \) . Thus we take \( {E}_{\alpha } = \left\{ {x : {f}^{ * }\left( x\right) > \alpha }\right\} \) . The conclusion (a) is therefore the weak-type estimate for \( {f}^{ * } \) given in (27) of the previous chapter.\n\nNext, since \( {E}_{\alpha } \) is open we can write it as \( \mathop{\bigcup }\limits_{k}{Q}_{k} \), where the \( {Q}_{k} \) are closed cubes with disjoint interiors, with the distance of \( {Q}_{k} \) from \( {E}_{\alpha }^{c} \) comparable to the diameter of \( {Q}_{k} \) . (This is Lemma 5.2 of the previous chapter.) Now set\n\n\[ {m}_{k} = \frac{1}{m\left( {Q}_{k}\right) }{\int }_{{Q}_{k}}{fdx} \]\n\nThus if \( {\bar{x}}_{k} \) is a point of \( {E}_{\alpha }^{c} \) closest to \( {Q}_{k} \), one has \( \left| {m}_{k}\right| \leq c{f}^{ * }\left( {\bar{x}}_{k}\right) \leq \) \( {c\alpha } \) . We define \( g\left( x\right) = f\left( x\right) \) for \( x \notin {E}_{\alpha }^{c} \) and \( g\left( x\right) = {m}_{k} \) for \( x \in {Q}_{k} \) . As a result \( \left| {f\left( x\right) }\right| \leq \alpha \) for \( x \in {E}_{\alpha }^{c} \), because \( {f}^{ * }\left( x\right) \leq \alpha \) there. Altogether then \( \left| {g\left( x\right) }\right| \leq {c\alpha } \), proving conclusion (b).\n\nFinally, \( b\left( x\right) = f\left( x\right) - g\left( x\right) \) is supported in \( {E}_{\alpha } = \mathop{\bigcup }\limits_{k}{Q}_{k} \) and hence \( b = \) \( \mathop{\sum }\limits_{k}{b}_{k} \), where each \( {b}_{k} \) is supported in \( {Q}_{k} \) and equals \( f\left( x\right) - {m}_{k} \) there. Thus\n\n\[ \int \left| {{b}_{k}\left( x\right) }\right| {dx} = {\int }_{{Q}_{k}}\left| {f\left( x\right) - {m}_{k}}\right| {dx} \leq {\int }_{{Q}_{k}}\left| {f\left( x\right) }\right| {dx} + \left| {m}_{k}\right| m\left( {Q}_{k}\right) . \]\n\nAlso as before\n\n\[ {\int }_{{Q}_{k}}\left| {f\left( x\right) }\right| {dx} \leq {cm}\left( {Q}_{k}\right) {f}^{ * }\left( {\bar{x}}_{k}\right) \leq {c\alpha m}\left( {Q}_{k}\right) ,\]\n\nhence\n\n\[ \int \left| {{b}_{k}\left( x\right) }\right| {dx} \leq \operatorname{cam}\left( {Q}_{k}\right) \]\n\nsince \( \left| {m}_{k}\right| \leq {c\alpha } \) . Clearly, \( \int {b}_{k}\left( x\right) {dx} = {\int }_{{Q}_{k}}\left( {f\left( x\right) - {m}_{k}}\right) {dx} = 0 \), and so the decomposition lemma is proved. | Yes |
Theorem 1.1 Every complete metric space \( X \) is of the second category in itself, that is, \( X \) cannot be written as the countable union of nowhere dense sets. | Proof of the theorem. We argue by contradiction, and assume that \( X \) is a countable union of nowhere dense sets \( {F}_{n} \) ,\n\n\[ X = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{n} \]\n\nBy replacing each \( {F}_{n} \) by its closure, we may assume that each \( {F}_{n} \) is closed. It now suffices to find a point \( x \in X \) with \( x \notin \bigcup {F}_{n} \) .\n\nSince \( {F}_{1} \) is closed and nowhere dense, hence not all of \( X \), there exists an open ball \( {B}_{1} \) of some radius \( {r}_{1} > 0 \) whose closure \( \overline{{B}_{1}} \) is entirely contained in \( {F}_{1}^{c} \) .\n\nSince \( {F}_{2} \) is closed and nowhere dense, the ball \( {B}_{1} \) cannot be entirely contained in \( {F}_{2} \), otherwise \( {F}_{2} \) would have a non-empty interior. Since \( {F}_{2} \) is also closed, there exists a ball \( {B}_{2} \) of some radius \( {r}_{2} > 0 \) whose closure \( \overline{{B}_{2}} \) is contained in \( {B}_{1} \) and also in \( {F}_{2}^{c} \) . Clearly, we may choose \( {r}_{2} \) so that \( {r}_{2} < {r}_{1}/2 \) .\n\nContinuing in this fashion, we obtain a sequence of balls \( \left\{ {B}_{n}\right\} \) with the following properties:\n\n(i) The radius of \( {B}_{n} \) tends to 0 as \( n \rightarrow \infty \) .\n\n(ii) \( {B}_{n + 1} \subset {B}_{n} \) .\n\n) \( {F}_{n} \cap \overline{{B}_{n}} \) is empty.\n\nChoose any point \( {x}_{n} \) in \( {B}_{n} \) . Then, \( {\left\{ {x}_{n}\right\} }_{n = 1}^{\infty } \) is a Cauchy sequence because of properties (i) and (ii) above. Since \( X \) is complete, this sequence converges to a limit which we denote by \( x \) . By (ii) we see that \( x \in \overline{{B}_{n}} \) for each \( n \), and hence \( x \notin {F}_{n} \) for all \( n \) by (iii). This contradicts (1), and the proof of the Baire category theorem is complete. | Yes |
Corollary 1.2 In a complete metric space, a generic set is dense. | To prove the corollary, we argue by contradiction and assume that \( E \subset X \) is generic but not dense. Then there exists a closed ball \( \bar{B} \) entirely contained in \( {E}^{c} \) . Since \( E \) is generic we can write \( {E}^{c} = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{n} \) where each \( {F}_{n} \) is nowhere dense, hence\n\n\[ \bar{B} = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }\left( {{F}_{n} \cap \bar{B}}\right) \]\n\nIt is clear that \( {F}_{n} \cap \bar{B} \) is nowhere dense, hence the above contradicts Theorem 1.1 applied to the complete metric space \( \bar{B} \), and the corollary is proved. | Yes |
Lemma 1.4 Suppose \( \left\{ {f}_{n}\right\} \) is a sequence of continuous functions on a complete metric space \( X \), and \( {f}_{n}\left( x\right) \rightarrow f\left( x\right) \) for each \( x \) as \( n \rightarrow \infty \) . Then, given an open ball \( B \subset X \) and \( \epsilon > 0 \), there exists an open ball \( {B}_{0} \subset B \) and an integer \( m \geq 1 \) so that \( \left| {{f}_{m}\left( x\right) - f\left( x\right) }\right| \leq \epsilon \) for all \( x \in {B}_{0} \) . | Proof. Let \( Y \) denote a closed ball contained in \( B \) . Note that \( Y \) is itself a complete metric space. Define\n\n\[ \n{E}_{\ell } = \left\{ {x \in Y : \mathop{\sup }\limits_{{j, k \geq \ell }}\left| {{f}_{j}\left( x\right) - {f}_{k}\left( x\right) }\right| \leq \epsilon }\right\} \n\]\n\nThen, since \( {f}_{n}\left( x\right) \) converges for every \( x \in X \), we must have\n\n(2)\n\n\[ \nY = \mathop{\bigcup }\limits_{{\ell = 1}}^{\infty }{E}_{\ell } \n\]\n\nMoreover, each \( {E}_{\ell } \) is closed since it is the intersection of sets of the type \( \left\{ {x \in Y : \left| {{f}_{j}\left( x\right) - {f}_{k}\left( x\right) }\right| \leq \epsilon }\right\} \) which are closed by the continuity of \( {f}_{j} \) and \( {f}_{k} \) . Therefore, by Theorem 1.1 applied to the complete metric space \( Y \), some set in the union (2), say \( {E}_{m} \), must contain an open ball \( {B}_{0} \) . By construction,\n\n\[ \n\mathop{\sup }\limits_{{j, k \geq m}}\left| {{f}_{j}\left( x\right) - {f}_{k}\left( x\right) }\right| \leq \epsilon \;\text{ whenever }x \in {B}_{0}, \n\]\n\nand letting \( k \) tend to infinity we find that \( \left| {{f}_{m}\left( x\right) - f\left( x\right) }\right| \leq \epsilon \) for all \( x \in \) \( {B}_{0} \) . This proves the lemma. | Yes |
Lemma 1.6 For every \( M > 0 \), the set \( {\mathcal{P}}_{M} \) of zig-zag functions is dense in \( C\left( \left\lbrack {0,1}\right\rbrack \right) \) . | Proof. It is plain that given \( \epsilon > 0 \) and a continuous function \( f \) , there exists a function \( g \in \mathcal{P} \) so that \( \parallel f - g\parallel \leq \epsilon \) . Indeed, since \( f \) is continuous on the compact set \( \left\lbrack {0,1}\right\rbrack \) it must be uniformly continuous, and there exists \( \delta > 0 \) so that \( \left| {f\left( x\right) - f\left( y\right) }\right| \leq \epsilon \) whenever \( \left| {x - y}\right| < \delta \) . If we choose \( n \) so large that \( 1/n < \delta \), and define \( g \) as a linear function on each interval \( \left\lbrack {k/n,\left( {k + 1}\right) /n}\right\rbrack \) for \( k = 0,\ldots, n - 1 \) with \( g\left( {k/n}\right) = f\left( {k/n}\right) \) , \( g\left( {\left( {k + 1}\right) /n}\right) = f\left( {\left( {k + 1}\right) /n}\right) \), we see at once that \( \parallel f - g\parallel \leq \epsilon \) .\n\nIt now suffices to see how to approximate \( g \) on \( \left\lbrack {0,1}\right\rbrack \) by zig-zag functions in \( {\mathcal{P}}_{M} \) . Indeed, if \( g \) is given by \( g\left( x\right) = {ax} + b \) for \( 0 \leq x \leq 1/n \), consider the two segments\n\n\[ \n{\varphi }_{\epsilon }\left( x\right) = g\left( x\right) + \epsilon \;\text{ and }\;{\psi }_{\epsilon }\left( x\right) = g\left( x\right) - \epsilon .\n\]\n\nThen, beginning at \( g\left( 0\right) \), we travel on a line segment of slope \( + M \) until we intersect \( {\varphi }_{\epsilon } \) . Then, we reverse direction and travel on a line segment of slope \( - M \) until we intersect \( {\psi }_{\epsilon } \) (see Figure 1).\n\nWe obtain \( h \in {\mathcal{P}}_{M} \) so that\n\n\[ \n{\psi }_{\epsilon }\left( x\right) \leq h\left( x\right) \leq {\varphi }_{\epsilon }\left( x\right) ,\;\text{ for all }0 \leq x \leq 1/n,\n\]\n\nand therefore \( \left| {h\left( x\right) - g\left( x\right) }\right| \leq \epsilon \) in \( \left\lbrack {0,1/n}\right\rbrack \) .\n\nThen, we begin at \( h\left( {1/n}\right) \) and repeat this argument on the interval \( \left\lbrack {1/n,2/n}\right\rbrack \) . Continuing in this fashion, we obtain a function \( h \in {\mathcal{P}}_{M} \) with \( \parallel h - g\parallel \leq \epsilon \) . Hence \( \parallel f - h\parallel \leq {2\epsilon } \), and the lemma is proved. | Yes |
Theorem 2.1 Suppose that \( \mathcal{B} \) is a Banach space, and \( \mathcal{L} \) is a collection of continuous linear functionals on \( \mathcal{B} \) .\n\n(i) If \( \mathop{\sup }\limits_{{\ell \in \mathcal{L}}}\left| {\ell \left( f\right) }\right| < \infty \) for each \( f \in \mathcal{B} \), then\n\n\[ \mathop{\sup }\limits_{{\ell \in \mathcal{L}}}\parallel \ell \parallel < \infty \]\n\n(ii) This conclusion also holds if we only assume that \( \mathop{\sup }\limits_{{\ell \in \mathcal{L}}}\left| {\ell \left( f\right) }\right| < \infty \) for all \( f \) in some set of the second category. | Proof. It suffices to show (ii) since by Baire’s theorem, \( \mathcal{B} \) is of the second category. So suppose that \( \mathop{\sup }\limits_{{\ell \in \mathcal{L}}}\left| {\ell \left( f\right) }\right| < \infty \) for all \( f \in E \), where \( E \) is of the second category.\n\nFor each positive integer \( M \), define\n\n\[ {E}_{M} = \{ f \in \mathcal{B} : \mathop{\sup }\limits_{{\ell \in \mathcal{L}}}\left| {\ell \left( f\right) }\right| \leq M\} \]\n\nThen, the hypothesis in the theorem guarantees that\n\n\[ E = \mathop{\bigcup }\limits_{{M = 1}}^{\infty }{E}_{M} \]\n\nMoreover, each \( {E}_{M} \) is closed, since it can be written as an intersection \( {E}_{M} = \mathop{\bigcap }\limits_{{\ell \in \mathcal{L}}}{E}_{M,\ell } \), where \( {E}_{M,\ell } = \{ f : \left| {\ell \left( f\right) }\right| \leq M\} \) is closed by the continuity of \( \ell \) . Since \( E \) is of the second category, some \( {E}_{M} \) must have non-empty interior, say when \( M = {M}_{0} \) . In other words, there exists \( {f}_{0} \in \mathcal{B} \), and \( r > 0 \) so that \( {B}_{r}\left( {f}_{0}\right) \subset {E}_{{M}_{0}} \) . Hence for all \( \ell \in \mathcal{L} \) we have\n\n\[ \left| {\ell \left( f\right) }\right| \leq {M}_{0}\;\text{ whenever }\begin{Vmatrix}{f - {f}_{0}}\end{Vmatrix} < r. \]\n\nAs a result, for all \( \parallel g\parallel < r \), and all \( \ell \in \mathcal{L} \) we have\n\n\[ \parallel \ell \left( g\right) \parallel \leq \begin{Vmatrix}{\ell \left( {g + {f}_{0}}\right) }\end{Vmatrix} + \begin{Vmatrix}{\ell \left( {-{f}_{0}}\right) }\end{Vmatrix} \leq 2{M}_{0}, \]\n\nand this implies the conclusion (ii) in the theorem. | Yes |
Lemma 2.3 \( \begin{Vmatrix}{\ell }_{N}\end{Vmatrix} = {L}_{N} \) for all \( N \geq 0 \) . | Proof. We already know from the above that \( \begin{Vmatrix}{\ell }_{N}\end{Vmatrix} \leq {L}_{N} \) . To prove the reverse inequality, it suffices to find a sequence of continuous functions \( \left\{ {f}_{k}\right\} \) so that \( \begin{Vmatrix}{f}_{k}\end{Vmatrix} \leq 1 \), and \( {\ell }_{N}\left( {f}_{k}\right) \rightarrow {L}_{N} \) as \( k \rightarrow \infty \) . To do so, first let \( g \) denote the function equal to 1 when \( {D}_{N} \) is positive and -1 when \( {D}_{N} \) is negative. Then \( g \) is measurable, \( \parallel g\parallel \leq 1 \), and\n\n\[ \n{L}_{N} = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }g\left( {-y}\right) {D}_{N}\left( y\right) {dy} \n\]\n\nwhere we used the fact that \( {D}_{N} \) is even, hence \( g\left( y\right) = g\left( {-y}\right) \) . Clearly, there exists a sequence of continuous functions \( \left\{ {f}_{k}\right\} \) with \( - 1 \leq {f}_{k}\left( x\right) \leq 1 \) for all \( - \pi \leq x \leq \pi \), and so that\n\n\[ \n{\int }_{-\pi }^{\pi }\left| {{f}_{k}\left( y\right) - g\left( y\right) }\right| {dy} \rightarrow 0\;\text{ as }k \rightarrow \infty .\n\]\n\nAs a result, we find that \( {\ell }_{N}\left( {f}_{k}\right) \rightarrow {L}_{N} \) as \( k \rightarrow \infty \), while \( \begin{Vmatrix}{f}_{k}\end{Vmatrix} \leq 1 \), hence \( \begin{Vmatrix}{\ell }_{N}\end{Vmatrix} \geq {L}_{N} \), as desired. | Yes |
Lemma 2.4 There is a constant \( c > 0 \) so that \( {L}_{N} \geq c\log N \) . | Proof. Since \( \left| {\sin y}\right| /\left| y\right| \leq 1 \) for all \( y \), and \( \sin y \) is an odd function, we see that \( {}^{1} \)\n\n\[ \n{L}_{N} \geq c{\int }_{0}^{\pi }\frac{\left| \sin \left( N + 1/2\right) y\right| }{\left| y\right| }{dy} \]\n\n\[ \n\geq c{\int }_{0}^{\left( {N + 1/2}\right) \pi }\frac{\left| \sin x\right| }{x}{dx} \]\n\n\[ \n\geq c\mathop{\sum }\limits_{{k = 0}}^{{N - 1}}{\int }_{k\pi }^{\left( {k + 1}\right) \pi }\frac{\left| \sin x\right| }{x}{dx} \]\n\n\[ \n\geq c\mathop{\sum }\limits_{{k = 0}}^{{N - 1}}\frac{1}{\left( {k + 1}\right) \pi }{\int }_{k\pi }^{\left( {k + 1}\right) \pi }\left| {\sin x}\right| {dx}. \]\n\nHowever, for all \( k \) we have \( {\int }_{k\pi }^{\left( {k + 1}\right) \pi }\left| {\sin x}\right| {dx} = {\int }_{0}^{\pi }\left| {\sin x}\right| {dx} \), so that\n\n\[ \n{L}_{N} \geq c\mathop{\sum }\limits_{{k = 0}}^{{N - 1}}\frac{1}{k + 1} \geq c\log N \]\n\nas was to be shown. | Yes |
Corollary 3.2 If \( X \) and \( Y \) are Banach spaces, and \( T : X \rightarrow Y \) is a continuous bijective linear transformation, then the inverse \( {T}^{-1} : Y \rightarrow X \) of \( T \) is also continuous. Hence there are constants \( c, C > 0 \) with\n\n\[ c\parallel f{\parallel }_{X} \leq \parallel T\left( f\right) {\parallel }_{Y} \leq C\parallel f{\parallel }_{X}\;\text{ for all }f \in X. \] | This follows immediately from the discussion preceding Theorem 3.1. | No |
Corollary 3.3 Suppose the vector space \( V \) is equipped with two norms \( \parallel \cdot {\parallel }_{1} \) and \( \parallel \cdot {\parallel }_{2} \) . If\n\n\[ \parallel v{\parallel }_{1} \leq C\parallel v{\parallel }_{2}\;\text{ for all }v \in V, \]\n\nand \( V \) is complete with respect to both norms, then \( \parallel \cdot {\parallel }_{1} \) and \( \parallel \cdot {\parallel }_{2} \) are equivalent. | Indeed, the hypothesis implies that the identity mapping \( I : \left( {V,\parallel \cdot {\parallel }_{2}}\right) \rightarrow \) \( \left( {V,\parallel \cdot {\parallel }_{1}}\right) \) is continuous, and since it is clearly bijective, its inverse \( I \) : \( \left( {V,\parallel \cdot {\parallel }_{1}}\right) \rightarrow \left( {V,\parallel \cdot {\parallel }_{2}}\right) \) is also continuous. Hence \( c\parallel v{\parallel }_{2} \leq \parallel v{\parallel }_{1} \) for some \( c > 0 \) and all \( v \in V \) . | Yes |
Theorem 3.4 The mapping \( T : {\mathcal{B}}_{1} \rightarrow {\mathcal{B}}_{2} \) given by \( T\left( f\right) = \{ \widehat{f}\left( n\right) \} \) is linear, continuous and injective, but not surjective. | Proof. We first note that \( T \) is clearly linear, and also continuous with \( \parallel T\left( f\right) {\parallel }_{\infty } \leq \parallel f{\parallel }_{{L}^{1}} \) . Moreover, \( T \) is injective since \( T\left( f\right) = 0 \) implies that \( \widehat{f}\left( n\right) = 0 \) for all \( n \), which then implies \( {}^{3} \) that \( f = 0 \) in \( {L}^{1} \) . If \( T \) were surjective, then Corollary 3.2 would imply that there is a constant \( c > 0 \) that satisfies\n\n(9)\n\n\[ c\parallel f{\parallel }_{{L}^{1}} \leq \parallel T\left( f\right) {\parallel }_{\infty },\;\text{ for all }f \in {\mathcal{B}}_{1}. \]\n\nHowever, if we set \( f = {D}_{N} \) the \( {N}^{\text{th }} \) Dirichlet kernel given by \( {D}_{N} = \) \( \mathop{\sum }\limits_{{\left| n\right| \leq N}}{e}^{inx} \), and recall from Lemma 2.4 that \( {\begin{Vmatrix}{D}_{N}\end{Vmatrix}}_{{L}^{1}} = {L}_{N} \rightarrow \infty \) as \( N \rightarrow \infty \), we find that (9) is violated as \( N \) tends to infinity, which is our desired contradiction. | Yes |
Theorem 4.1 Suppose \( X \) and \( Y \) are two Banach spaces. If \( T : X \rightarrow Y \) is a closed linear map, then \( T \) is continuous. | Proof. Since the graph of \( T \) is a closed subspace of the Banach space \( X \times Y \) with the norm \( \parallel \left( {x, y}\right) {\parallel }_{X \times Y} = \parallel x{\parallel }_{X} + \parallel y{\parallel }_{Y} \), the graph \( {G}_{T} \) is itself a Banach space. Consider the two projections \( {P}_{X} : G\left( T\right) \rightarrow X \) and \( {P}_{Y} : G\left( T\right) \rightarrow Y \) defined by\n\n\[ \n{P}_{X}\left( {x, T\left( x\right) }\right) = x\;\text{ and }\;{P}_{Y}\left( {x, T\left( x\right) }\right) = T\left( x\right) .\n\]\n\nThe mappings \( {P}_{X} \) and \( {P}_{Y} \) are continuous and linear. Moreover, \( {P}_{X} \) is bijective, hence its inverse \( {P}_{X}^{-1} \) is continuous by Corollary 3.2. Since \( T = {P}_{Y} \circ {P}_{X}^{-1} \), we conclude that \( T \) is continuous, as was to be shown. | Yes |
Lemma 4.3 Under the assumptions of the theorem, there exists \( A > 0 \) so that\n\n\[ \parallel f{\parallel }_{{L}^{\infty }} \leq A\parallel f{\parallel }_{{L}^{2}}\;\text{ for all }f \in E. \] | Proof. If \( 1 \leq p \leq 2 \), then Hölder’s inequality with the conjugate exponents \( r = 2/p \) and \( {r}^{ * } = 2/\left( {2 - p}\right) \) yields\n\n\[ \int {\left| f\right| }^{p} \leq {\left( \int {\left| f\right| }^{2}\right) }^{p/2}{\left( \int 1\right) }^{\frac{2 - p}{2}}.\]\n\nSince \( X \) has finite measure, we see after taking \( {p}^{\text{th }} \) roots in the above, that there is some \( B > 0 \) so that \( \parallel f{\parallel }_{{L}^{p}} \leq B\parallel f{\parallel }_{{L}^{2}} \) for all \( f \in E \) . Together with (10), this proves the lemma when \( 1 \leq p \leq 2 \) .\n\nWhen \( 2 < p < \infty \), we note first that \( {\left| f\left( x\right) \right| }^{p} \leq \parallel f{\parallel }_{{L}^{\infty }}^{p - 2}{\left| f\left( x\right) \right| }^{2} \), and integrating this inequality gives\n\n\[ \parallel f{\parallel }_{{L}^{p}}^{p} \leq \parallel f{\parallel }_{{L}^{\infty }}^{p - 2}\parallel f{\parallel }_{{L}^{2}}^{2} \]\n\nIf we now use (10), and assume that \( \parallel f{\parallel }_{{L}^{\infty }} \neq 0 \), we find that for some \( A > 0 \), we have \( \parallel f{\parallel }_{{L}^{\infty }} \leq A\parallel f{\parallel }_{{L}^{2}} \) whenever \( f \in E \), and the proof of the lemma is complete. | Yes |
Theorem 5.1 The collection of sets in \( \mathcal{K} \) of two-dimensional Lebesgue measure zero is generic. | In particular, this collection is non-empty, and in fact dense. Loosely stated, the key to the argument is to show that sets \( K \) in \( \mathcal{K} \) whose horizontal slices \( \{ x : \left( {x, y}\right) \in K\} \) have \ | No |
Lemma 5.2 For each fixed \( {y}_{0} \) and \( \epsilon \), the collection of sets \( \mathcal{K}\left( {{y}_{0},\epsilon }\right) \) is open and dense in \( \mathcal{K} \) . | To prove that \( \mathcal{K}\left( {{y}_{0},\epsilon }\right) \) is open, suppose \( K \in \mathcal{K}\left( {{y}_{0},\epsilon }\right) \) and pick \( \eta \) so that \( {K}^{\eta } \) satisfies the condition above. Suppose \( {K}^{\prime } \in \mathcal{K} \) with \( \operatorname{dist}\left( {K,{K}^{\prime }}\right) < \) \( \eta /2 \) . This means in particular that \( {K}^{\prime } \subset {K}^{\eta /2} \), and the triangle inequality then shows that \( {\left( {K}^{\prime }\right) }^{\eta /2} \subset {K}^{\eta } \) . Therefore\n\n\[ \n{m}_{1}\left( \left\{ {x : \left( {x, y}\right) \in {\left( {K}^{\prime }\right) }^{\eta /2}}\right\} \right) \leq {m}_{1}\left( \left\{ {x : \left( {x, y}\right) \in {K}^{\eta }}\right\} \right) < {10\epsilon }, \n\] \n\nand as a result \( {K}^{\prime } \in \mathcal{K}\left( {{y}_{0},\epsilon }\right) \), as was to be shown. | Yes |
Proposition 1.1 For each integer \( N \geq 1 \) , \[ {\begin{Vmatrix}{S}_{N}\end{Vmatrix}}_{{L}^{2}} = {N}^{1/2}. \] | This proposition follows from the fact that \( \left\{ {{r}_{n}\left( t\right) }\right\} \) is an orthonormal system on \( {L}^{2}\left( \left\lbrack {0,1}\right\rbrack \right) \) . Indeed, we have that \( {\int }_{0}^{1}{r}_{n}\left( t\right) {dt} = 0 \) because each \( {r}_{n} \) is equal to 1 on a set of measure \( 1/2 \), and equal to -1 on a set of measure also \( 1/2 \) . Moreover, by their mutual independence and (4), we have \[ {\int }_{0}^{1}{r}_{n}\left( t\right) {r}_{m}\left( t\right) {dt} = 0\;\text{ if }n \neq m. \] In addition, we obviously have \( {\int }_{0}^{1}{r}_{n}^{2}\left( t\right) {dt} = 1 \) . Therefore \[ {\begin{Vmatrix}\mathop{\sum }\limits_{{n = 1}}^{N}{a}_{n}{r}_{n}\end{Vmatrix}}_{{L}^{2}}^{2} = \mathop{\sum }\limits_{{n = 1}}^{N}{\left| {a}_{n}\right| }^{2} \] and the assertion follows by taking \( {a}_{n} = 1 \) for \( 1 \leq n \leq N \) . | Yes |
Corollary 1.2 \( {S}_{N}/N \) converges to 0 in probability. | In fact,\n\n\[ m\left( \left\{ {\left| {{S}_{N}\left( x\right) /N}\right| > \epsilon }\right\} \right) = m\left( \left\{ {\left| {{S}_{N}\left( x\right) }\right| > {\epsilon N}}\right\} \right) \leq \frac{1}{{\epsilon }^{2}{N}^{2}}\int {\left| {S}_{N}\left( x\right) \right| }^{2}{dm}, \]\n\nby Tchebychev’s inequality. Hence \( m\left( \left\{ {x : \left| {{S}_{N}\left( x\right) /N}\right| > \epsilon }\right\} \right) \leq 1/\left( {{\epsilon }^{2}N}\right) \) , and the corollary is proved. | Yes |
Corollary 1.5 Let \( {S}_{N}\left( t\right) = \mathop{\sum }\limits_{{n = 1}}^{N}{r}_{n}\left( t\right) \) . Then \( {S}_{N}\left( t\right) /N \rightarrow 0 \), as \( N \rightarrow \) \( \infty \) for almost every \( t \) . In fact, if \( \alpha > 1/2 \), then \( {S}_{N}\left( t\right) /{N}^{\alpha } \rightarrow 0 \) for almost every \( t \) . | Proof. Fix \( 1/2 < \beta < \alpha \), and let \( {a}_{n} = {n}^{-\beta } \) and \( {b}_{n} = {n}^{\beta } \) . Clearly \( \sum {a}_{n}^{2} < \infty \) . Set \( {\widetilde{S}}_{N}\left( t\right) = \mathop{\sum }\limits_{{n = 1}}^{N}{a}_{n}{r}_{n}\left( t\right) \) . Then, by summation by parts, setting \( {\widetilde{S}}_{0} = 0 \), we get\n\n\[ \n{S}_{N}\left( t\right) = \mathop{\sum }\limits_{{n = 1}}^{N}{r}_{n} = \mathop{\sum }\limits_{{n = 1}}^{N}{a}_{n}{r}_{n}{b}_{n} \n\]\n\n\[ \n= \mathop{\sum }\limits_{{n = 1}}^{N}\left( {{\widetilde{S}}_{n} - {\widetilde{S}}_{n - 1}}\right) {b}_{n} \n\]\n\n\[ \n= {\widetilde{S}}_{N}{b}_{N} + \mathop{\sum }\limits_{{n = 1}}^{{N - 1}}{\widetilde{S}}_{n}\left( {{b}_{n} - {b}_{n + 1}}\right) . \n\]\n\nHowever \( \left| {{b}_{n} - {b}_{n + 1}}\right| = {b}_{n + 1} - {b}_{n} \), and \( \mathop{\sum }\limits_{{n = 1}}^{{N - 1}}\left( {{b}_{n + 1} - {b}_{n}}\right) = {b}_{N} - 1 = \) \( O\left( {N}^{\beta }\right) \) while the convergence of the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{r}_{n}\left( t\right) \) for almost all \( t \) guarantees that \( \left| {{\widetilde{S}}_{n}\left( t\right) }\right| = O\left( 1\right) \) for almost every \( t \) . As a result, for those \( t \) , \( {S}_{N}\left( t\right) = O\left( {N}^{\beta }\right) \) and this implies \( {S}_{N}\left( t\right) /{N}^{\alpha } \rightarrow 0 \) for almost all \( t \), proving the corollary. | Yes |
Lemma 1.8 For each \( p < \infty \) there is a bound \( {A}_{p} \) so that\n\n\[ \parallel F{\parallel }_{{L}^{p}} \leq {A}_{p}\parallel F{\parallel }_{{L}^{2}} \]\n\nfor all \( F \in {L}^{p}\left( \left\lbrack {0,1}\right\rbrack \right) \) of the form \( F\left( t\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{a}_{n}{\rho }_{n}\left( t\right) \) . | It clearly suffices to prove the corresponding statement when the \( {a}_{n} \) are assumed real and have been normalized so that \( \parallel F{\parallel }_{{L}^{2}}^{2} = \mathop{\sum }\limits_{{-\infty }}^{\infty }{a}_{n}^{2} = 1 \) .\n\nNow observe that the defining property (3) shows that whenever \( \left\{ {f}_{n}\right\} \) is a sequence of mutually independent (real-valued) functions, so is the sequence \( \left\{ {{\Phi }_{n}\left( {f}_{n}\right) }\right\} \), with \( \left\{ {\Phi }_{n}\right\} \) any sequence of continuous functions from \( \mathbb{R} \) to \( \mathbb{R} \) . As a result the functions \( \left\{ {e}^{{a}_{n}{\rho }_{n}\left( t\right) }\right\} \) are mutually independent. Thus if \( {F}_{N}\left( t\right) = \mathop{\sum }\limits_{{\left| n\right| \leq N}}{a}_{n}{\rho }_{n}\left( t\right) \), then (11)\n\n\[ {\int }_{0}^{1}{e}^{{F}_{N}\left( t\right) }{dt} = {\int }_{0}^{1}\left( {\mathop{\prod }\limits_{{n = - N}}^{N}{e}^{{a}_{n}{\rho }_{n}\left( t\right) }}\right) {dt} = \mathop{\prod }\limits_{{n = - N}}^{N}\left( {{\int }_{0}^{1}{e}^{{a}_{n}{\rho }_{n}\left( t\right) }{dt}}\right) .\n\nHowever, \( {\int }_{0}^{1}{e}^{{a}_{n}{\rho }_{n}\left( t\right) }{dt} = \cosh \left( {a}_{n}\right) \), since each \( {\rho }_{n} \) takes values +1 or -1 on sets of measure \( 1/2 \) respectively. Also, \( \cosh \left( x\right) \leq {e}^{{x}^{2}} \) for real \( x \), as a comparison of their power series clearly shows. Hence\n\n\[ {\int }_{0}^{1}{e}^{{F}_{N}\left( t\right) }{dt} \leq \mathop{\prod }\limits_{{n = - N}}^{N}{e}^{{a}_{n}^{2}} \leq {e}^{\sum {a}_{n}^{2}} \leq e.\n\nA similar inequality holds with the \( {a}_{n} \) replaced by \( - {a}_{n} \) . Altogether then\n\n\[ {\int }_{0}^{1}{e}^{\left| {F}_{N}\left( t\right) \right| }{dt} \leq {2e} \]\n\nA simple passage to the limit, as \( N \rightarrow \infty \), then gives that \( {e}^{\left| F\left( t\right) \right| } \) is integrable over \( \left\lbrack {0,1}\right\rbrack \), and \( {\int }_{0}^{1}{e}^{\left| F\left( t\right) \right| }{dt} \leq {2e} \) . However for each \( p \) there is a constant \( {c}_{p} \) so that \( {u}^{p} \leq {c}_{p}{e}^{u} \) for all \( u \geq 0 \) . Thus \( \parallel F{\parallel }_{{L}^{p}}^{p} \leq {2e}{c}_{p} \), and the lemma is proved with \( {A}_{p} = {\left( 2e{c}_{p}\right) }^{1/p} \). | Yes |
Theorem 2.1 Suppose \( \\left\\{ {f}_{n}\\right\\} \) is a sequence of functions that are mutually independent, are identically distributed, and have mean \( {m}_{0} \) . Then\n\n\[ \n\\frac{1}{N}\\mathop{\\sum }\\limits_{{n = 0}}^{{N - 1}}{f}_{n}\\left( x\\right) \\rightarrow {m}_{0}\\;\\text{ for almost every }x \\in X,\\text{ as }N \\rightarrow \\infty .\n\] | The possibility of reducing this theorem to the ergodic theorem depends on the device of replacing the sequence \( \\left\\{ {f}_{n}\\right\\} \) by another sequence that is \ | No |
Lemma 2.2 If \( \left\{ {f}_{N}\right\} \) and \( \left\{ {g}_{N}\right\} \) have the same joint distribution, then so do the sequences \( \left\{ {{\Phi }_{N}\left( f\right) }\right\} \) and \( \left\{ {{\Phi }_{N}\left( g\right) }\right\} \) . Here \( {\Phi }_{N}\left( f\right) = {\Phi }_{N}\left( {{f}_{1},\ldots ,{f}_{N}}\right) \) , \( {\Phi }_{N}\left( g\right) = {\Phi }_{N}\left( {{g}_{1},\ldots ,{g}_{N}}\right) \), and each \( {\Phi }_{N} \) is a continuous function from \( {\mathbb{R}}^{N} \) to \( \mathbb{R} \) . | To see this, note that if \( B \subset {\mathbb{R}}^{N} \) is a Borel set, and \( \Phi = \left( {{\Phi }_{1},\ldots ,{\Phi }_{N}}\right) \) , then \( {B}^{\prime } = {\Phi }^{-1}\left( B\right) \) is also a Borel set in \( {\mathbb{R}}^{N} \), so if \( f = \left( {{f}_{1},\ldots ,{f}_{N}}\right) \) and \( g = \left( {{g}_{1},\ldots ,{g}_{N}}\right) \), then \( {\mu }_{\Phi \left( f\right) }\left( B\right) = {\mu }_{f}\left( {B}^{\prime }\right) \) and \( {\mu }_{\Phi \left( g\right) }\left( B\right) = {\mu }_{g}\left( {B}^{\prime }\right) \) . Since \( f \) and \( g \) have the same joint distribution we must have \( {\mu }_{f}\left( {B}^{\prime }\right) = {\mu }_{g}\left( {B}^{\prime }\right) \) , and the lemma is proved. | Yes |
Lemma 2.3 If \( \left\{ {F}_{N}\right\} \) and \( \left\{ {G}_{N}\right\} \) have the same joint distribution, then \( {F}_{N}\left( x\right) \rightarrow {m}_{0} \) almost everywhere as \( N \rightarrow \infty \) if and only if \( {G}_{N}\left( y\right) \rightarrow {m}_{0} \) almost everywhere as \( N \rightarrow \infty \) . | To prove this lemma, note that if we define \( {E}_{N, k} = \left\{ {x : \mathop{\sup }\limits_{{r \geq N}} \mid {F}_{r}\left( x\right) - }\right. \) \( \left. {{m}_{0} \mid \leq 1/k}\right\} \), then \( {F}_{N} \rightarrow {m}_{0} \) almost everywhere if and only if \( m\left( {E}_{N, k}\right) \rightarrow \) 1, as \( N \rightarrow \infty \), for each \( k \) . If \( {E}_{N, k}^{\prime } = \left\{ {y : \mathop{\sup }\limits_{{r \geq N}}\left| {{G}_{r}\left( x\right) - {m}_{0}}\right| \leq 1/k}\right\} \) , then \( m\left( {E}_{N, k}\right) = {m}^{ * }\left( {E}_{N, k}^{\prime }\right) \), and this leads to our desired result. | Yes |
Proposition 2.4 Given an integrable function \( f \) and a sub-algebra \( \mathcal{A} \) of \( \mathcal{M} \), there is a unique \( {}^{9} \) function \( F \) so that:\n\n(i) \( F \) is \( \mathcal{A} \) -measurable.\n\n(ii) \( {\int }_{A}{Fdm} = {\int }_{A}{fdm} \) for any set \( A \in \mathcal{A} \) . | Proof. We denote by \( {m}^{\prime } \) the restriction of the measure \( m \) to \( \mathcal{A} \) . Define a ( \( \sigma \) -finite) signed measure \( \nu \) on \( \mathcal{A} \) by \( \nu \left( A\right) = {\int }_{A}f\;{dm} \), for \( A \in \mathcal{A} \) . Then since \( \nu \) is clearly absolutely continuous with respect to \( {m}^{\prime } \), the Lebesgue-Radon-Nikodym theorem \( {}^{10} \) guarantees that there is a function \( F \) that is \( \mathcal{A} \) -measurable so that \( \nu \left( A\right) = {\int }_{A}{Fd}{m}^{\prime } = {\int }_{A}{Fdm} \) . Given the definition of \( \nu \), the existence of the required \( F \) is therefore established. Its uniqueness is clear because if \( G \) is \( \mathcal{A} \) -measurable and \( {\int }_{A}{Gdm} = 0 \) for every \( A \in \mathcal{A} \), then necessarily \( G = 0 \) . | Yes |
(a) If \( f \in {L}^{2} \), then \( \mathbb{E}\left( f\right) \in {L}^{2} \) and \( \parallel \mathbb{E}\left( f\right) {\parallel }_{{L}^{2}} \leq \parallel f{\parallel }_{{L}^{2}} \) . | Proof. To establish (a) observe that if \( g \) is bounded and \( \mathcal{A} \) -measurable, then by the proposition above, \( {\int }_{X}{gfdm} = {\int }_{X}\mathbb{E}\left( {gf}\right) {dm} = {\int }_{X}g\mathbb{E}\left( f\right) {dm} \) . But\n\n\[ \parallel \mathbb{E}\left( f\right) {\parallel }_{{L}^{2}} = \mathop{\sup }\limits_{g}\left| {{\int }_{X}g\mathbb{E}\left( f\right) {dm}}\right| \]\n\nwhere \( g \) ranges over bounded \( \mathcal{A} \) -measurable functions with \( \parallel g{\parallel }_{{L}^{2}} \leq 1 \) (see Lemma 4.2 in Chapter 1), because of the fact that \( \mathbb{E}\left( f\right) \) is \( \mathcal{A} \) -measurable. Moreover \( \left| {{\int }_{X}{gfdm}}\right| \leq \parallel f{\parallel }_{{L}^{2}} \) for such \( g \), gives conclusion (a). | Yes |
Lemma 2.7 Suppose \( {\mathcal{B}}_{0},\ldots ,{\mathcal{B}}_{n} \) are mutually independent algebras. Then for each \( k < n \), the algebras \( \mathop{\bigvee }\limits_{{j = 0}}^{k}{\mathcal{B}}_{j} \) and \( {\mathcal{B}}_{n} \) are mutually independent. | Null | No |
Theorem 2.8 Suppose \( {f}_{0},\ldots ,{f}_{n},\ldots \) are independent functions that are square integrable, and that each has mean zero, and variance \( {\sigma }_{n}^{2} = {\begin{Vmatrix}{f}_{n}\end{Vmatrix}}_{{L}^{2}}^{2} \) . Assume that\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{\sigma }_{n}^{2} < \infty \]\n\nThen \( {s}_{n} = \mathop{\sum }\limits_{{k = 0}}^{n}{f}_{k} \) converges (as \( n \rightarrow \infty \) ) almost everywhere. | We begin the proof of the theorem by noting that under its assumptions the sequence \( {s}_{n} = \mathop{\sum }\limits_{{k = 0}}^{n}{f}_{k} \) converges in the \( {L}^{2} \) norm, as \( n \rightarrow \infty \) . Indeed, since the \( {f}_{n} \) are mutually independent and \( {\int }_{X}{f}_{n}{dm} = 0 \), then by (4) they are mutually orthogonal. Hence by Pythagoras' theorem, if \( m < n,{\begin{Vmatrix}{s}_{n} - {s}_{m}\end{Vmatrix}}_{{L}^{2}}^{2} = \mathop{\sum }\limits_{{k = m + 1}}^{n}{\begin{Vmatrix}{f}_{k}\end{Vmatrix}}^{2} = \mathop{\sum }\limits_{{k = m + 1}}^{n}{\sigma }_{k}^{2} \rightarrow 0 \), as \( n, m \rightarrow \) \( \infty \) . Thus \( {s}_{n} \) converges to a limit (call it \( {s}_{\infty } \) ) in the \( {L}^{2} \) norm. Using (14) and the fact that each \( {\mathbb{E}}_{n} \) is continuous in the \( {L}^{2} \) norm by Lemma 2.5, we arrive at\n\n\[ {s}_{n} = {\mathbb{E}}_{n}\left( {s}_{\infty }\right) ,\;\text{ for all }n. \]\n\nOur desired result now follows from a basic maximal theorem for martingales and its corollary, which gives convergence almost everywhere. | Yes |
Corollary 2.9 If \( \mathop{\sup }\limits_{n}{\sigma }_{n} < \infty \), then for each \( \alpha > 1/2 \)\n\n\[ \frac{{s}_{n}}{{n}^{\alpha }} \rightarrow 0\;\text{ almost everywhere as }n \rightarrow \infty . \]\n\nNote that here, unlike in Theorem 2.1, we have not assumed that the \( {f}_{n} \) are identically distributed. On the other hand, we have made a more restrictive assumption in requiring square integrability. | We begin the proof of the theorem by noting that under its assumptions the sequence \( {s}_{n} = \mathop{\sum }\limits_{{k = 0}}^{n}{f}_{k} \) converges in the \( {L}^{2} \) norm, as \( n \rightarrow \infty \) . Indeed, since the \( {f}_{n} \) are mutually independent and \( {\int }_{X}{f}_{n}{dm} = 0 \), then by (4) they are mutually orthogonal. Hence by Pythagoras' theorem, if \( m < n,{\begin{Vmatrix}{s}_{n} - {s}_{m}\end{Vmatrix}}_{{L}^{2}}^{2} = \mathop{\sum }\limits_{{k = m + 1}}^{n}{\begin{Vmatrix}{f}_{k}\end{Vmatrix}}^{2} = \mathop{\sum }\limits_{{k = m + 1}}^{n}{\sigma }_{k}^{2} \rightarrow 0 \), as \( n, m \rightarrow \) \( \infty \) . Thus \( {s}_{n} \) converges to a limit (call it \( {s}_{\infty } \) ) in the \( {L}^{2} \) norm. Using (14) and the fact that each \( {\mathbb{E}}_{n} \) is continuous in the \( {L}^{2} \) norm by Lemma 2.5, we arrive at\n\n\[ {s}_{n} = {\mathbb{E}}_{n}\left( {s}_{\infty }\right) ,\;\text{ for all }n. \]\n\nOur desired result now follows from a basic maximal theorem for martingales and its corollary, which gives convergence almost everywhere. | Yes |
Theorem 2.10 Suppose \( {s}_{\infty } \) is an integrable function, and \( {s}_{n} = {\mathbb{E}}_{n}\left( {s}_{\infty }\right) \) , where the \( {\mathbb{E}}_{n} \) are conditional expectations for an increasing family \( \left\{ {\mathcal{A}}_{n}\right\} \) of sub-algebras of \( \mathcal{M} \) . Then:\n\n(a) \( m\left( \left\{ {x : \mathop{\sup }\limits_{n}\left| {{s}_{n}\left( x\right) }\right| > \alpha }\right\} \right) \leq \frac{1}{\alpha }{\begin{Vmatrix}{s}_{\infty }\end{Vmatrix}}_{{L}^{1}} \) for every \( \alpha > 0 \) . | For the proof of part (a) we may assume that \( {s}_{\infty } \) is non-negative, for otherwise we may proceed with \( \left| {s}_{\infty }\right| \) instead of \( {s}_{\infty } \) and then obtain the result once we observe that \( \left| {{\mathbb{E}}_{n}\left( {s}_{\infty }\right) }\right| \leq {\mathbb{E}}_{n}\left( \left| {s}_{\infty }\right| \right) \) . For fixed \( \alpha \), let \( A = \left\{ {x : \mathop{\sup }\limits_{n}{s}_{n}\left( x\right) > \alpha }\right\} \) . Then we can partition \( A = \mathop{\bigcup }\limits_{{n = 0}}^{\infty }{A}_{n} \), where \( {A}_{n} \) is the set where \( n \) is the first time that \( {s}_{n}\left( x\right) > \alpha \) . That is \( {A}_{n} = \{ x \) : \( \left. {{s}_{n}\left( x\right) > \alpha \text{, but}{s}_{k}\left( x\right) \leq \alpha \text{, for}k < n}\right\} \) . Note that \( {A}_{n} \in {\mathcal{A}}_{n} \) . Also,\n\n\[ \n{\int }_{A}{s}_{\infty }{dm} = \mathop{\sum }\limits_{{n = 0}}^{\infty }{\int }_{{A}_{n}}{s}_{\infty }{dm} = \mathop{\sum }\limits_{{n = 0}}^{\infty }{\int }_{{A}_{n}}{\mathbb{E}}_{n}\left( {s}_{\infty }\right) {dm} = \mathop{\sum }\limits_{{n = 0}}^{\infty }{\int }_{{A}_{n}}{s}_{n}{dm} \n\]\n\n\[ \n> \alpha \mathop{\sum }\limits_{n}{\int }_{{A}_{n}}{dm} \n\]\n\n\[ \n= {\alpha m}\left( A\right) \text{.} \n\]\n\nThe identity \( {\int }_{{A}_{n}}{\mathbb{E}}_{n}\left( {s}_{\infty }\right) {dm} = {\int }_{{A}_{n}}{s}_{\infty }{dm} \) follows from the definition of the conditional expectation \( {\mathbb{E}}_{n}\left( {s}_{\infty }\right) \) . Thus\n\n(15)\n\n\[ \nm\left( A\right) \leq \frac{1}{\alpha }{\int }_{A}{s}_{\infty }{dm},\;\text{ with }A = \left\{ {x : \mathop{\sup }\limits_{n}{s}_{n}\left( x\right) > \alpha }\right\} , \n\]\n\nand part (a) is proved. (The reader might find it instructive to compare (15) with a corresponding estimate for the Hardy-Littlewood maximal function in equation (28) of Chapter 2.) | Yes |
Theorem 2.11 If the algebras \( {\mathcal{A}}_{0},{\mathcal{A}}_{1},\ldots ,{\mathcal{A}}_{n},\ldots \) are mutually independent then every element of the tail algebra has either measure zero or one. | Proof. Let \( \mathcal{B} \) denote the tail algebra. Note that \( {\mathcal{A}}_{r} \) is automatically independent from \( \mathop{\bigvee }\limits_{{k = r + 1}}^{\infty }{\mathcal{A}}_{k} \), by Lemma 2.7. Hence each \( {\mathcal{A}}_{r} \) is independent of \( \mathcal{B} \), and thus the algebras \( \mathcal{B} \) and \( \mathcal{B} \) are mutually independent! Therefore as observed above, every element of \( \mathcal{B} \) has measure zero or one. | No |
Corollary 2.12 Suppose \( {f}_{0},{f}_{1},\ldots ,{f}_{n},\ldots \) are mutually independent functions. The set where \( \mathop{\sum }\limits_{{k = 0}}^{\infty }{f}_{k} \) converges has measure zero or one. | Proof. Set \( {\mathcal{A}}_{n} = {\mathcal{A}}_{{f}_{n}} \) . Then these algebras are independent. Now with \( {s}_{n} = \mathop{\sum }\limits_{{k = 0}}^{n}{f}_{k} \), and a fixed positive integer \( {n}_{0} \), we have by the Cauchy criterion that\n\n\[ \left\{ {x : \lim {s}_{n}\left( x\right) \text{ exists }}\right\} = \mathop{\bigcap }\limits_{{\ell = 1}}^{\infty }\mathop{\bigcup }\limits_{{r = {n}_{0}}}^{\infty }\left\{ {x : \left| {{s}_{n}\left( x\right) - {s}_{m}\left( x\right) }\right| < \frac{1}{\ell },\text{ all }n, m \geq r}\right\} .\n\]\n\nSince \( \left\{ {x : \left| {{s}_{n}\left( x\right) - {s}_{m}\left( x\right) }\right| < 1/\ell \text{, all}n, m \geq r}\right\} \in \mathop{\bigvee }\limits_{{k = {n}_{0}}}^{\infty }{\mathcal{A}}_{k} \) whenever \( r \geq {n}_{0} \) we conclude that the set of convergence is a tail set, as desired. | Yes |
Lemma 2.14 \( \widehat{\mu }\left( {\xi /{N}^{1/2}}\right) = 1 - 2{\sigma }^{2}{\pi }^{2}{\xi }^{2}/N + o\left( {1/N}\right) \), as \( N \rightarrow \infty \) . | Proof. Indeed, when \( \xi \) is fixed\n\n\[ \n{e}^{-{2\pi i\xi t}/{N}^{1/2}} = 1 - {2\pi i\xi t}/{N}^{1/2} - 2{\pi }^{2}{\xi }^{2}{t}^{2}/N + {E}_{N}\left( t\right) \n\] \n\nwith \( {E}_{N}\left( t\right) = O\left( {{t}^{2}/N}\right) \), but also \( {E}_{N}\left( t\right) = O\left( {{t}^{3}/{N}^{3/2}}\right) \). Integrating this in \( t \), we get\n\n\[ \n\widehat{\mu }\left( {\xi /{N}^{1/2}}\right) = 1 - \frac{2{\pi }^{2}{\xi }^{2}}{N}{\sigma }^{2} + {\int }_{-\infty }^{\infty }{E}_{N}\left( t\right) {d\mu }\left( t\right) , \n\] \n\nbecause \( {m}_{0} = {\int }_{-\infty }^{\infty }{td\mu }\left( t\right) = 0 \), and \( {\sigma }^{2} = {\int }_{-\infty }^{\infty }{t}^{2}{d\mu }\left( t\right) \). The lemma will be proved as soon as we see that \( {\int }_{-\infty }^{\infty }{E}_{N}\left( t\right) {d\mu }\left( t\right) = o\left( {1/N}\right) \). However, the integral in question can be divided into a part where \( {t}^{2} < {\epsilon }_{N}N \), and a complementary part where \( {t}^{2} \geq {\epsilon }_{N}N \). Here we choose \( {\epsilon }_{N} \) to tend to zero as \( N \rightarrow \infty \), while \( {\epsilon }_{N}N \rightarrow \infty \) as \( N \rightarrow \infty \) ; (for example, the choice \( {\epsilon }_{N} = {N}^{-1/2} \) will do.) Now for the first part\n\n\[ \n{\int }_{{t}^{2} < {\epsilon }_{N}N}{E}_{N}\left( t\right) {dt} = O\left( {{\int }_{{t}^{2} < {\epsilon }_{N}N}{t}^{3}/{N}^{3/2}{d\mu }\left( t\right) }\right) \n\] \n\n\[ \n= O\left( {\frac{{\epsilon }_{N}^{1/2}}{N}{\int }_{-\infty }^{\infty }{t}^{2}{d\mu }\left( t\right) }\right) \n\] \n\n\[ \n= o\left( {1/N}\right) \text{.} \n\] \n\nIn addition, for the second part we can estimate\n\n\[ \n{\int }_{{t}^{2} \geq {\epsilon }_{N}N}{E}_{N}\left( t\right) {dt} = O\left( {\frac{1}{N}{\int }_{{t}^{2} \geq {\epsilon }_{N}N}{t}^{2}{d\mu }\left( t\right) }\right) = o\left( {1/N}\right) . \n\] \n\nHaving thus proved the lemma we see that\n\n\[ \n{\widehat{\mu }}_{N}\left( \xi \right) = \widehat{\mu }{\left( \xi /{N}^{1/2}\right) }^{N} = {\left( 1 - 2{\sigma }^{2}{\pi }^{2}{\xi }^{2}/N + o\left( 1/N\right) \right) }^{N}, \n\] \n\nand this converges to \( {e}^{-2{\sigma }^{2}{\pi }^{2}{\xi }^{2}} \), completing step (ii). | Yes |
Lemma 2.15 Suppose \( \left\{ {\mu }_{N}\right\}, N = 1,2,\ldots \), and \( \nu \) are non-negative \( {fi} \) - nite Borel measures on \( \mathbb{R} \), and that \( \nu \) is continuous. Assume that \( {\widehat{\mu }}_{N}\left( \xi \right) \rightarrow \) \( \widehat{\nu }\left( \xi \right) \), as \( N \rightarrow \infty \), for each \( \xi \in \mathbb{R} \) . Then \( {\mu }_{N}\left( \left( {a, b}\right) \right) \rightarrow \nu \left( {a, b}\right) \) for all \( a < b \) . | Proof. We prove first that\n\n(16)\n\n\[ \n{\mu }_{N}\left( \varphi \right) \rightarrow \nu \left( \varphi \right) \;\text{ as }N \rightarrow \infty \n\] \n\nfor any \( \varphi \) that is \( {C}^{\infty } \) and has compact support, where we have used the notation \( {\mu }_{N}\left( \varphi \right) = {\int }_{-\infty }^{\infty }\varphi \left( t\right) d{\mu }_{N}\left( t\right) \) and \( \nu \left( \varphi \right) = {\int }_{-\infty }^{\infty }\varphi \left( t\right) {d\nu }\left( t\right) \) .\n\nNotice that since \( {\widehat{\mu }}_{N}\left( 0\right) = {\int }_{-\infty }^{\infty }d{\mu }_{N}\left( t\right) \), then the convergent sequence \( {\int }_{-\infty }^{\infty }d{\mu }_{N}\left( t\right) \) must be bounded. As a result, for some \( M \) we have \( \left| {{\widehat{\mu }}_{N}\left( \xi \right) }\right| \leq \) \( M \) for all \( N \) and also \( \left| {\widehat{\nu }\left( \xi \right) }\right| \leq M \) .\n\nNext, the function \( \varphi \) can be represented by its inverse Fourier transform \( \varphi \left( t\right) = {\int }_{-\infty }^{\infty }{e}^{-{2\pi it}}{\varphi }^{ \vee }\left( \xi \right) {d\xi } \), where \( {\varphi }^{ \vee }\left( \xi \right) = \widehat{\varphi }\left( {-\xi }\right) \) is necessarily in the Schwartz space \( \mathcal{S} \) . This shows that\n\n\[ \n{\int }_{\mathbb{R}}\varphi \left( t\right) d{\mu }_{N}\left( t\right) = {\int }_{\mathbb{R}}{\varphi }^{ \vee }\left( \xi \right) {\widehat{\mu }}_{N}\left( \xi \right) {d\xi } \n\] \n\nby applying Fubini’s theorem to \( \int \int {e}^{-{2\pi it\xi }}{\varphi }^{ \vee }\left( \xi \right) \;d{\mu }_{N}\left( t\right) \;{d\xi } \), which is justified by the rapid decrease of \( {\varphi }^{ \vee } \) . Similarly, \( \int {\varphi d\nu } = \int {\varphi }^{ \vee }\left( \xi \right) \widehat{\nu }\left( \xi \right) {d\xi } \) . Then since \( {\widehat{\mu }}_{N}\left( \xi \right) \rightarrow \widehat{\nu }\left( \xi \right) \) pointwise and boundedly we obtain (16).\n\nNow for \( \left( {a, b}\right) \) fixed, let \( {\varphi }_{\epsilon } \) be a sequence of positive \( {C}^{\infty } \) functions with \( {\varphi }_{\epsilon } \leq {\chi }_{\left( a, b\right) } \), and \( {\varphi }_{\epsilon }\left( t\right) \rightarrow {\chi }_{\left( a, b\right) }\left( t\right) \) for every \( t \) as \( \epsilon \rightarrow 0 \) . Then\n\n\[ \n{\mu }_{N}\left( \left( {a, b}\right) \right) \geq {\mu }_{N}\left( {\varphi }_{\epsilon }\right) \rightarrow \nu \left( {\varphi }_{\epsilon }\right) \;\text{ as }N \rightarrow \infty .\n\] \n\nAs a result, \( \mathop{\liminf }\limits_{{N \rightarrow \infty }}{\mu }_{N}\left( {a, b}\right) \geq \nu \left( {\varphi }_{\epsilon }\right) \), and letting \( \epsilon \rightarrow 0 \) gives\n\n\[ \n\mathop{\liminf }\limits_{{N \rightarrow \infty }}{\mu }_{N}\left( \left( {a, b}\right) \right) \geq \nu \left( \left( {a, b}\right) \right) .\n\] | Yes |
Theorem 2.17 Under the above conditions on \( \left\{ {f}_{n}\right\} \), the measures \( {\mu }_{N} \) converge weakly to \( {\nu }_{{\sigma }^{2}} \) as \( N \rightarrow \infty \) . | The proof proceeds essentially as in the case of real-valued functions, showing first the analog of (16) for smooth functions with compact support, and then proceeding as in Corollary 2.16 for continuous functions. The calculation of the characteristic function of the Gaussian is given in Exercise 32. | No |
Lemma 2.1 The \( \sigma \) -algebra \( \mathcal{C} \) is the same as the \( \sigma \) -algebra \( \mathcal{B} \) of Borel sets. | Proof. If \( \mathcal{O} \) is an open set in \( {\mathbb{R}}^{dk} \), then clearly\n\n\[ \left\{ {\mathrm{p} \in \mathcal{P} : \left( {\mathrm{p}\left( {t}_{1}\right) ,\mathrm{p}\left( {t}_{2}\right) ,\ldots ,\mathrm{p}\left( {t}_{k}\right) }\right) \in \mathcal{O}}\right\} \]\n\nis open in \( \mathcal{P} \), and hence this set belongs to \( \mathcal{B} \) . As a result, cylindrical sets are in \( \mathcal{B} \), thus \( \mathcal{C} \subset \mathcal{B} \) .\n\nTo see the reverse inclusion, note that for any fixed \( n \) and \( a \) and a given path \( {\mathfrak{p}}_{0} \), the set \( \left\{ {\mathfrak{p} \in \mathcal{P} : \mathop{\sup }\limits_{{0 \leq t \leq n}}\left| {\mathfrak{p}\left( t\right) - {\mathfrak{p}}_{0}\left( t\right) }\right| \leq a}\right\} \) is the same as the corresponding set where the supremum is restricted to the \( t \) in \( \left\lbrack {0, n}\right\rbrack \) that are rational, and hence this set is in \( \mathcal{C} \) . It is then not too difficult to see that for any \( \delta > 0 \), the ball \( \left\{ {\mathrm{p} \in \mathcal{P} : d\left( {\mathrm{p},{\mathrm{p}}_{0}}\right) < \delta }\right\} \) is in \( \mathcal{C} \) . Since open balls generate \( \mathcal{B} \) we have \( \mathcal{B} \subset \mathcal{C} \), and the lemma is established. | Yes |
Corollary 2.3 Suppose the sequence of probability measures \( \left\{ {\mu }_{N}\right\} \) is tight, and for each \( 0 \leq {t}_{1} \leq {t}_{2} \leq \cdots \leq {t}_{k} \) the measures \( {\mu }_{N}^{\left( {t}_{1},\ldots ,{t}_{k}\right) } \) converge weakly to a measure \( {\mu }_{{t}_{1},\ldots ,{t}_{k}} \), as \( N \rightarrow \infty \) . Then the sequence \( \left\{ {\mu }_{N}\right\} \) converges weakly to a measure \( \mu \), and moreover \( {\mu }^{\left( {t}_{1},\ldots ,{t}_{k}\right) } = {\mu }_{{t}_{1},\ldots ,{t}_{k}} \). | Proof. First, by Lemma 2.2, there is a subsequence \( \left\{ {\mu }_{{N}_{m}}\right\} \) that converges weakly to a measure \( \mu \) . Next, \( {\mu }_{{N}_{m}}^{\left( {t}_{1},\ldots ,{t}_{k}\right) } \rightarrow {\mu }^{\left( {t}_{1},\ldots ,{t}_{k}\right) } \) weakly. In fact, if \( {\pi }^{{t}_{1},{t}_{2},\ldots ,{t}_{k}} \) is the continuous mapping from \( \mathcal{P} \) to \( {\mathbb{R}}^{kd} \) that assigns to \( \mathrm{p} \in \mathcal{P} \) the point \( \left( {\mathrm{p}\left( {t}_{1}\right) ,\mathrm{p}\left( {t}_{2}\right) ,\ldots ,\mathrm{p}\left( {t}_{k}\right) }\right) \in {\mathbb{R}}^{kd} \), then, by definition, \( {\mu }^{\left( {t}_{1},\ldots ,{t}_{k}\right) }\left( A\right) = \mu \left( {{\left( {\pi }^{{t}_{1},\ldots ,{t}_{k}}\right) }^{-1}A}\right) \) for any Borel set \( A \subset {\mathbb{R}}^{dk} \). As a result\n\n\[{\int }_{{\mathbb{R}}^{dk}}{fd}{\mu }^{\left( {t}_{1},\ldots ,{t}_{k}\right) } = {\int }_{\mathcal{P}}\left( {f \circ {\pi }^{{t}_{1},\ldots ,{t}_{k}}}\right) {d\mu }\]\n\nfor any \( f \in {C}_{b}\left( {\mathbb{R}}^{dk}\right) \), with a similar identity with \( \mu \) replaced by \( {\mu }_{{N}_{m}} \). From this, and the weak convergence of \( {\mu }_{{N}_{m}} \) to \( \mu \), it follows that \( {\mu }^{\left( {t}_{1},\ldots ,{t}_{k}\right) } = \) \( {\mu }_{{t}_{1},\ldots ,{t}_{k}} \).\n\nWe now observe that the full sequence \( \left\{ {\mu }_{N}\right\} \) must converge weakly to \( \mu \). Suppose the contrary. Then there is another sequence \( {\mu }_{{N}_{n}^{\prime }} \) and a bounded continuous function \( f \) on \( \mathcal{P} \), so that \( \int {fd}{\mu }_{{N}_{n}^{\prime }} \) converges to a limit that is not equal to \( \int {fd\mu } \). Now using Lemma 2.2 again, there is a further subsequence \( \left\{ {\mu }_{{N}_{n}^{\prime \prime }}\right\} \) and a measure \( \nu \), so that \( {\mu }_{{N}_{n}^{\prime \prime }} \) converges weakly to \( \nu \), while \( \nu \neq \mu \). However by the previous argument we have \( {\nu }^{\left( {t}_{1},\ldots ,{t}_{k}\right) } = {\mu }^{\left( {t}_{1},\ldots ,{t}_{k}\right) } \) for all choices of \( 0 \leq {t}_{1} \leq {t}_{2} \leq \cdots \leq {t}_{k} \). Therefore \( \nu = \mu \), and \( \int {fd\mu } = \int {fd\nu } \). This contradiction completes the proof of the corollary. | Yes |
Lemma 2.4 A closed set \( K \subset \mathcal{P} \) is compact if for each positive \( T \) there is a positive bounded function \( h \mapsto {w}_{T}\left( h\right) \), defined for \( h \in (0,1\rbrack \) with \( {w}_{T}\left( h\right) \rightarrow 0 \) as \( h \rightarrow 0 \), and so that\n\n(7)\n\n\[ \mathop{\sup }\limits_{{\mathsf{p} \in K}}\mathop{\sup }\limits_{{0 \leq t \leq T}}\left| {\mathsf{p}\left( {t + h}\right) - \mathsf{p}\left( t\right) }\right| \leq {w}_{T}\left( h\right) ,\;\text{ for }h \in (0,1\rbrack . \]\n\n | The condition (7) implies that the functions on \( K \) are equicontinuous on each interval \( \left\lbrack {0, T}\right\rbrack \) . The lemma then essentially follows from the Arzela-Ascoli criterion. (Recall, this criterion was used in a special setting in Section 3, Chapter 8 of Book II.) | No |
Lemma 3.2 We have as \( \lambda \rightarrow \infty \) ,\n\n\[ \mathop{\sup }\limits_{{n \geq 1}}m\left( \left\{ {x : \mathop{\sup }\limits_{{k \leq n}}\left| {{s}_{k}\left( x\right) }\right| > \lambda {n}^{1/2}}\right\} \right) = O\left( {\lambda }^{-p}\right) \]\n\nfor every \( p \geq 2 \) . | To prove the lemma we apply the martingale maximal theorem of the previous chapter (Theorem 2.10, in the form that it takes in Exercise 29, part (b)) to the stopped sequence \( \left\{ {s}_{k}^{\prime }\right\} \) defined as \( {s}_{k}^{\prime } = {s}_{k} \) if \( k \leq n,{s}_{k}^{\prime } = \) \( {s}_{n} \) if \( k \geq n \), and \( {s}_{\infty }^{\prime } = {s}_{n} \) . With \( {s}_{n}^{ * } = \mathop{\sup }\limits_{{k \leq n}}\left| {s}_{k}\right| = \mathop{\sup }\limits_{k}\left| {s}_{k}^{\prime }\right| \) we then have\n\n\[ m\left( \left\{ {x : {s}_{n}^{ * } > \alpha }\right\} \right) \leq \frac{1}{\alpha }{\int }_{\left| {s}_{n}\right| > \alpha }\left| {s}_{n}\right| {dm}. \]\n\nMultiplying both sides by \( p{\alpha }^{p - 1} \) and integrating, using an argument similar to the one used in Section 4.1 of Chapter 2 yields\n\n\[ \int {\left( {s}_{n}^{ * }\right) }^{p}{dm} \leq \frac{p}{p - 1}\int {\left| {s}_{n}\right| }^{p}{dm}. \]\n\nNow, the Khinchin inequality of Lemma 1.8 in the previous chapter, applied to the more general setting described in Exercise 10 gives\n\n\[ \int {\left| {s}_{n}\right| }^{p}{dm} \leq A{\left( \int {\left| {s}_{n}\right| }^{2}dm\right) }^{p/2}. \]\nThus\n\n\[ m\left( \left\{ {{s}_{n}^{ * } > \alpha }\right\} \right) \leq \frac{1}{{\alpha }^{p}}{\begin{Vmatrix}{s}_{n}^{ * }\end{Vmatrix}}_{{L}^{p}}^{p} \leq \frac{{A}^{\prime }}{{\alpha }^{p}}{\begin{Vmatrix}{s}_{n}\end{Vmatrix}}_{{L}^{2}}^{p}. \]\n\nSetting \( \alpha = \lambda {n}^{1/2} \) and recalling that \( {\begin{Vmatrix}{s}_{n}\end{Vmatrix}}_{{L}^{2}} = {n}^{1/2} \) completes the proof of the lemma. | Yes |
Theorem 4.1 The following are also Brownian motion processes:\n\n(a) \( {\delta }^{-1/2}{B}_{t\delta } \) for every fixed \( \delta > 0 \) .\n\n(b) \( \mathfrak{o}\left( {B}_{t}\right) \) whenever \( \mathfrak{o} \) is an orthogonal linear transformation on \( {\mathbb{R}}^{d} \) .\n\n(c) \( {B}_{t + {\sigma }_{0}} - {B}_{{\sigma }_{0}} \) whenever \( {\sigma }_{0} \geq 0 \) is a constant. | We need only check that these new processes satisfy the conditions B-1, B-2, and B-3 defining Brownian motion. Thus the assertion (a) of the theorem is clear once we observe that for any function \( f \), the covariance matrix of \( {\delta }^{-1/2}f \) is \( {\delta }^{-1} \) times the covariance matrix of \( f \) . The assertion (b) is also obvious once we note that the covariance matrix of \( \mathfrak{o}\left( f\right) \) is the same as that of \( f \) ; and that if \( {f}_{1},\ldots ,{f}_{n},\ldots \) are mutually independent so are \( \mathfrak{o}\left( {f}_{1}\right) ,\mathfrak{o}\left( {f}_{2}\right) ,\ldots ,\mathfrak{o}\left( {f}_{n}\right) ,\ldots \) Finally (c) is immediate from the definition of Brownian motion. | Yes |
Theorem 4.2 With \( W \) the Wiener measure on \( \mathcal{P} \) we have:\n\n(a) If \( 0 < a < 1/2 \) and \( T > 0 \), then, with respect to \( W \) almost every path \( \mathrm{p} \) satisfies\n\n\[ \mathop{\sup }\limits_{{0 \leq t \leq T,0 < h \leq 1}}\frac{\left| \mathrm{p}\left( t + h\right) - \mathrm{p}\left( t\right) \right| }{{h}^{a}} < \infty . \]\n\n--- | The first conclusion is implicit in our construction of Brownian motion. Indeed, suppose \( {K}^{\left( T\right) } \) is the set arising in the proof of Theorem 3.1. Then we have seen that \( {\mu }_{N}\left( {K}^{\left( T\right) }\right) \geq 1 - \epsilon \) for every \( N \) . Thus the same holds for the weak limit of the \( \left\{ {\mu }_{N}\right\} \) . Hence \( W\left( {K}^{\left( T\right) }\right) \geq 1 - \epsilon \) . But by the definition of \( {K}^{\left( T\right) } \) we have the inequality in (a) for every \( \mathrm{p} \in {K}^{\left( T\right) } \) . Since \( \epsilon \) is arbitrary, the first conclusion holds. | Yes |
Lemma 5.2 \( {\mathcal{A}}_{0 + } = {\mathcal{A}}_{0} \) . | Proof of the lemma. Fix a bounded continuous function \( f \) on \( {\mathbb{R}}^{kd} \), and a sequence \( 0 \leq {t}_{1} < {t}_{2} < \cdots < {t}_{k} \) . For any \( \delta > 0 \), set\n\n\[ \n{f}_{\delta } = f\left( {{B}_{{t}_{1} + \delta } - {B}_{\delta },{B}_{{t}_{2} + \delta } - {B}_{{t}_{1} + \delta },\ldots ,{B}_{{t}_{k} + \delta } - {B}_{{t}_{k - 1} + \delta }}\right) .\n\]\n\nIf \( A \) is any set in \( {\mathcal{A}}_{0 + } \), then \( A \in {\mathcal{A}}_{\delta } \), for \( \delta > 0 \) . Then by the independence of the above increments from \( {B}_{\delta } \), we see that\n\n\[ \n{\int }_{A}{f}_{\delta }{dP} = P\left( A\right) {\int }_{\Omega }{f}_{\delta }{dP}.\n\]\n\nThus by continuity of the paths we can let \( \delta \rightarrow 0 \) and obtain\n\n\[ \n{\int }_{A}{f}_{0}{dP} = P\left( A\right) {\int }_{\Omega }{f}_{0}{dP}.\n\]\n\nNow any bounded continuous function \( g \) on \( {\mathbb{R}}^{kd} \) can be written in the form \( g\left( {{x}_{1},\ldots ,{x}_{k}}\right) = f\left( {{x}_{1},{x}_{2} - {x}_{1},\ldots ,{x}_{k} - {x}_{k - 1}}\right) \) where \( f \) is another such function. As a result\n\n\[ \n{\int }_{A}g\left( {{B}_{{t}_{1}},\ldots ,{B}_{{t}_{k}}}\right) {dP} = P\left( A\right) {\int }_{\Omega }g\left( {{B}_{{t}_{1}},\ldots ,{B}_{{t}_{k}}}\right) {dP}.\n\]\n\nHence by a passage to the limit, this holds if \( g \) is the characteristic function of a Borel set of \( {\mathbb{R}}^{kd} \) . Thus \( P\left( {A \cap E}\right) = P\left( A\right) P\left( E\right) \) whenever \( E \) is a cylindrical set. From this, we deduce the same equality for any Borel set \( E \) by using Exercise 4 in the previous chapter. Therefore \( P\left( A\right) = \) \( P{\left( A\right) }^{2} \), which implies \( P\left( A\right) = 0 \) or \( P\left( A\right) = 1 \) . Since \( A \) was an arbitrary subset of \( {\mathcal{A}}_{0 + } \), the lemma, and also the proposition, are proved. | Yes |
Theorem 5.3 Suppose \( {B}_{t} \) is a Brownian motion and \( \sigma \) is a stopping time. Then the process \( {B}_{t}^{ * } \), defined by\n\n\[ \n{B}_{t}^{ * }\left( \omega \right) = {B}_{t + \sigma \left( \omega \right) }\left( \omega \right) - {B}_{\sigma \left( \omega \right) }\left( \omega \right) \n\]\n\nis also a Brownian motion. Moreover \( {B}_{t}^{ * } \) is independent of \( {\mathcal{A}}_{\sigma } \) . | Proof. We have already noted that if \( \sigma \left( \omega \right) \) is a constant, \( \sigma \left( \omega \right) = \) \( {\sigma }_{0} \), then \( {B}_{t + {\sigma }_{0}} - {B}_{{\sigma }_{0}} \) is a Brownian motion (see Theorem 4.1), so the assertion in the theorem holds in this case.\n\nNext assume that \( \sigma \) is discrete, that is, it takes on only a countable set of values \( {\sigma }_{1} < {\sigma }_{2} < \cdots < {\sigma }_{\ell } < \ldots \) . Also suppose \( 0 \leq {t}_{1} < {t}_{2} < \cdots < {t}_{k} \) are fixed. Let us use the temporary notation\n\n\[ \n\mathbf{B} = \left( {{B}_{{t}_{1}},{B}_{{t}_{2}},\ldots ,{B}_{{t}_{k}}}\right) \n\]\n\n\[ \n{B}^{ * } = \left( {{B}_{{t}_{1}}^{ * },{B}_{{t}_{2}}^{ * },\ldots ,{B}_{{t}_{k}}^{ * }}\right) \n\]\n\n\[ \n{\mathbf{B}}_{\ell }^{ * } = \left( {{B}_{{t}_{1} + {\sigma }_{\ell }} - {B}_{{\sigma }_{\ell }},{B}_{{t}_{2} + {\sigma }_{\ell }} - {B}_{{\sigma }_{\ell }},\ldots ,{B}_{{t}_{k} + {\sigma }_{\ell }} - {B}_{{\sigma }_{\ell }}}\right) , \n\]\n\nwith all these bold-face vectors taking values in \( {\mathbb{R}}^{kd} \) . Now if \( E \) is a Borel set in \( {\mathbb{R}}^{kd} \), then\n\n\[ \n\left\{ {\omega : {\mathbf{B}}^{ * } \in E}\right\} = \mathop{\bigcup }\limits_{\ell }\left\{ {\omega : {\mathbf{B}}_{\ell }^{ * } \in E,\text{ and }\sigma = {\sigma }_{\ell }}\right\} . \n\]\n\n---\n\n\( {}^{8} \) A corresponding independence when \( \sigma \) is an arbitrary positive constant is characteristic of a \ | No |
Proposition 6.2 Suppose \( x \in \partial \mathcal{R} \) and \( x + \Gamma \) is disjoint from \( \mathcal{R} \), for some truncated cone \( \Gamma \) . Then \( x \) is a regular point. | Proof. We assume \( x = 0 \), and consider the set \( A \) of Brownian paths starting at the origin that enter \( \Gamma \) for an infinite sequence of times tending to zero. Let \( {A}_{n} = \mathop{\bigcup }\limits_{{{r}_{k} < 1/n}}\left\{ {\omega : {B}_{{r}_{k}}\left( \omega \right) \in \Gamma }\right\} \) where \( {r}_{k} \) is an enumeration of the positive rationals. Then \( A = \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{A}_{n} \) . However \( {A}_{n} \in {\mathcal{A}}_{n} \) for each \( n \), and hence \( A \in {\mathcal{A}}_{0 + } = {\mathcal{A}}_{0} \), by the zero-one law. So \( m\left( A\right) = 0 \) or \( m\left( A\right) = 1 \), and we show that in fact \( m\left( A\right) = 1 \) . Assume the contrary, that is \( m\left( A\right) = 0 \) . By the rotation invariance of Brownian motion, the same result would hold for any rotation of our truncated cone, and finitely many such rotations cover the ball of radius \( \delta \), with the origin excluded, while every path enters that ball at arbitrarily small times. This is a contradiction. Now returning to our boundary point \( x \), if \( x + \Gamma \) is disjoint from \( \mathcal{R} \) , then there are, for each \( \omega \), arbitrarily small times for which \( {B}_{t}\left( \omega \right) \in \Gamma \) , and hence \( {B}_{t}^{x}\left( \omega \right) \notin \mathcal{R} \) . Thus \( x \) is regular. | Yes |
Corollary 6.3 Suppose the bounded open set \( \mathcal{R} \) satisfies the outside cone condition. Assume \( f \) is a given continuous function on \( \partial \mathcal{R} \). Then there is a unique function \( u \) that is continuous in \( \overline{\mathcal{R}} \), harmonic in \( \mathcal{R} \), and such that \( {\left. u\right| }_{\partial \mathcal{R}} = f \). | Proof. Theorem 6.1 and Proposition 6.2 show that \( u \) is continuous in \( \overline{\mathcal{R}} \) and \( {\left. u\right| }_{\partial \mathcal{R}} = f \). The uniqueness is a consequence of the well-known maximum principle. \( {}^{9} \) | No |
Proposition 1.2 Suppose \( f \) and \( g \) are a pair of holomorphic functions in a region \( {}^{1}\Omega \), and \( f \) and \( g \) agree in a neighborhood of a point \( {z}^{0} \in \Omega \) . Then \( f \) and \( g \) agree throughout \( \Omega \) . | Proof. We may assume that \( g = 0 \) . If we fix any point \( {z}^{\prime } \in \Omega \) , it suffices to prove that \( f\left( {z}^{\prime }\right) = 0 \) . Using the pathwise connectedness of \( \Omega \) we can find a sequence of points \( {z}^{1},\ldots ,{z}^{N} = {z}^{\prime } \) in \( \Omega \) and polydiscs \( {\mathbb{P}}_{{r}_{k}}\left( {z}^{k}\right) \), for \( 0 \leq k \leq N \), so that\n\n\( {}^{1} \) Recall that a region is defined to be an open and connected set.\n\n(a) \( {\mathbb{P}}_{{r}_{k}}\left( {z}^{k}\right) \subset \Omega \) ,\n\n(b) \( {z}^{k + 1} \in {\mathbb{P}}_{{r}_{k}}\left( {z}^{k}\right) \), for \( 0 \leq k \leq N - 1 \) .\n\nNow if \( f \) vanishes in a neighborhood of \( {z}^{k} \), it must necessarily vanish in all of \( {\mathbb{P}}_{{r}_{k}}\left( {z}^{k}\right) \) . (This little fact is established in Exercise 1.) Thus \( f \) vanishes in \( {\mathbb{P}}_{{r}_{0}}\left( {z}^{0}\right) \), and by (b), it vanishes in \( {\mathbb{P}}_{{r}_{k + 1}}\left( {z}^{k + 1}\right) \) if it vanishes in \( {\mathbb{P}}_{{r}_{k}}\left( {z}^{k}\right) \) . Hence, by an induction on \( k \), we arrive at the conclusion that the function \( f \) vanishes on \( {\mathbb{P}}_{{r}_{N}}\left( {z}^{N}\right) \), and therefore \( f\left( {z}^{\prime }\right) = 0 \), and the proposition is proved. | Yes |
Theorem 2.1 Suppose \( F \) is holomorphic in \( \Omega = \left\{ {z \in {\mathbb{C}}^{n},\rho < \left| z\right| < 1}\right\} \) , for some fixed \( \rho ,0 < \rho < 1 \) . Then \( F \) can be analytically continued into the ball \( \left\{ {z \in {\mathbb{C}}^{n} : \left| z\right| < 1}\right\} \) . | Proof. Consider the integral\n\n\[ I\left( {{z}_{1},{z}_{2}}\right) = \frac{1}{2\pi i}{\int }_{\left| {\zeta }_{1}\right| = a + \epsilon }\frac{F\left( {{\zeta }_{1},{z}_{2}}\right) }{{\zeta }_{1} - {z}_{1}}d{\zeta }_{1} \]\n\nwhich is well-defined for small positive \( \epsilon \), when \( \left( {{z}_{1},{z}_{2}}\right) \) is in a neighborhood \( \widetilde{\mathcal{O}} \) of the product set (3). In fact then the variable of integration ranges over a neighborhood of \( {K}_{2} \), where \( F \) is analytic and hence continuous. Moreover \( I\left( {{z}_{1},{z}_{2}}\right) \) is analytic in \( \widetilde{\mathcal{O}} \), since it is visibly analytic in \( {z}_{1} \) for fixed \( {z}_{2} \) when \( \left| {z}_{1}\right| < a + \epsilon \), and \( {z}_{2} \) is near the set \( {b}_{2} \leq \left| {z}_{2}\right| \leq {b}_{1} \) ; also it is analytic in \( {z}_{2} \) (for fixed \( {z}_{1} \) ) in that set, by virtue of the analyticity of \( F \) . Finally when \( \left( {{z}_{1},{z}_{2}}\right) \) is near the set \( {K}_{1} \), then \( I\left( {{z}_{1},{z}_{2}}\right) = F\left( {{z}_{1},{z}_{2}}\right) \) by the Cauchy integral formula, and thus \( I \) provides the desired continuation of \( F \) . | Yes |
Lemma 2.2 If the function \( F \) is holomorphic in a region \( \mathcal{O} \) that contains the union \( {K}_{1} \cup {K}_{2} \) then \( F \) extends analytically to an open set \( \widetilde{\mathcal{O}} \) containing the product set\n\n\[ \left\{ {\left( {{z}_{1},{z}_{2}}\right) : \left| {z}_{1}\right| \leq a,{b}_{2} \leq \left| {z}_{2}\right| \leq {b}_{1}}\right\} . \] | Proof. Consider the integral\n\n\[ I\left( {{z}_{1},{z}_{2}}\right) = \frac{1}{2\pi i}{\int }_{\left| {\zeta }_{1}\right| = a + \epsilon }\frac{F\left( {{\zeta }_{1},{z}_{2}}\right) }{{\zeta }_{1} - {z}_{1}}d{\zeta }_{1} \]\n\nwhich is well-defined for small positive \( \epsilon \), when \( \left( {{z}_{1},{z}_{2}}\right) \) is in a neighborhood \( \widetilde{\mathcal{O}} \) of the product set (3). In fact then the variable of integration ranges over a neighborhood of \( {K}_{2} \), where \( F \) is analytic and hence continuous. Moreover \( I\left( {{z}_{1},{z}_{2}}\right) \) is analytic in \( \widetilde{\mathcal{O}} \), since it is visibly analytic in \( {z}_{1} \) for fixed \( {z}_{2} \) when \( \left| {z}_{1}\right| < a + \epsilon \), and \( {z}_{2} \) is near the set \( {b}_{2} \leq \left| {z}_{2}\right| \leq {b}_{1} \) ; also it is analytic in \( {z}_{2} \) (for fixed \( {z}_{1} \) ) in that set, by virtue of the analyticity of \( F \) . Finally when \( \left( {{z}_{1},{z}_{2}}\right) \) is near the set \( {K}_{1} \), then \( I\left( {{z}_{1},{z}_{2}}\right) = F\left( {{z}_{1},{z}_{2}}\right) \) by the Cauchy integral formula, and thus \( I \) provides the desired continuation of \( F \) . | Yes |
Proposition 3.1 Suppose \( f \) is continuous and has compact support on \( \mathbb{C} \) . Then:\n\n(a) \( {ugivenby}\left( 6\right) \) is also continuous and satisfies (5) in the sense of distributions.\n\n(b) If \( f \) is in the class \( {C}^{k}, k \geq 1 \), then so is \( u \), and \( u \) satisfies (5) in the usual sense.\n\n(c) If \( u \) is any \( {C}^{1} \) function of compact support, then \( u \) is already of the form (6); in fact\n\n\[ u = \frac{\partial u}{\partial \bar{z}} * \Phi \] | Proof. Note first that\n\n\[ u\left( {z + h}\right) - u\left( z\right) = \frac{1}{\pi }{\int }_{{\mathbb{C}}^{1}}f\left( {z + h - \zeta }\right) - f\left( {z - \zeta }\right) \frac{d\zeta }{\zeta }, \]\n\nand that this tends to zero as \( h \rightarrow 0 \), by the uniform continuity of \( f \) and the fact that the function \( 1/\zeta \) is integrable over compact sets in \( {\mathbb{C}}^{1} \) . If \( f \) is in the class \( {C}^{k}, k \geq 1 \), an easy elaboration of this shows that we can differentiate under the integral sign in (6) and find that any partial derivative of \( u \) of order \( \leq k \) is represented in the same way in terms of partial derivatives of \( f \) .\n\nNext we use the fact that \( \Phi \left( z\right) = 1/\left( {\pi z}\right) \) is a fundamental solution of the operator \( \partial /\partial \bar{z} \) . This means that in the sense of distributions \( \frac{\partial }{\partial \bar{z}}\Phi = {\delta }_{0} \), with \( {\delta }_{0} \) the Dirac delta function at the origin. (See Exercise 16 in Chapter 3.) So using the formalism of distributions, as in Chapter 3, we have\n\n\[ \frac{\partial }{\partial \bar{z}}\left( {f * \Phi }\right) = f * \left( \frac{\partial \Phi }{\partial \bar{z}}\right) = \left( \frac{\partial f}{\partial \bar{z}}\right) * \Phi . \]\n\nThe first set of equalities means that \( \partial u/\partial \bar{z} = f \), since \( f * {\delta }_{0} = f \), and so assertions (a) and (b) are now proved. Using the equality of the second and third members above (with \( u \) in place of \( f \) ) gives \( u = u * {\delta }_{0} = \frac{\partial u}{\partial \bar{z}} * \Phi \) , and this is assertion (c). | Yes |
Proposition 3.2 Suppose \( n \geq 2 \) . If \( {f}_{j},1 \leq j \leq n \), are functions of class \( {C}^{k} \) of compact support that satisfy (7), then there exists a function \( u \) of class \( {C}^{k} \) and of compact support that satisfies the inhomogeneous Cauchy-Riemann equations (4). \( {}^{2} \) | Proof. Write \( z = \left( {{z}^{\prime },{z}_{n}}\right) \), where \( {z}^{\prime } = \left( {{z}_{1},\ldots ,{z}_{n - 1}}\right) \in {\mathbb{C}}^{n - 1} \) and set\n\n(8)\n\n\[ u\left( z\right) = \frac{1}{\pi }{\int }_{{\mathbb{C}}^{1}}{f}_{n}\left( {{z}^{\prime },{z}_{n} - \zeta }\right) \frac{{dm}\left( \zeta \right) }{\zeta }.\]\n\nThen by the previous proposition \( \partial u/\partial {\bar{z}}_{n} = {f}_{n} \) . However by differentiating under the integral sign (which is easily justified) we see that for \( 1 \leq j \leq n - 1 \)\n\n\[ \frac{\partial u}{\partial {\bar{z}}_{j}} = \frac{1}{\pi }{\int }_{{\mathbb{C}}^{1}}\frac{\partial {f}_{n}}{\partial {\bar{z}}_{j}}\left( {{z}^{\prime },{z}_{n} - \zeta }\right) \frac{{dm}\left( \zeta \right) }{\zeta }\]\n\n\[ = \frac{1}{\pi }{\int }_{{\mathbb{C}}^{1}}\frac{\partial {f}_{j}}{\partial {\bar{z}}_{n}}\left( {{z}^{\prime },{z}_{n} - \zeta }\right) \frac{{dm}\left( \zeta \right) }{\zeta }\]\n\n\[ = {f}_{j}\left( {{z}^{\prime },{z}_{n}}\right) \text{.} \]\n\nThe next-to-last step results from the consistency condition (7), and the last step is a consequence of part (c) of Proposition 3.1. Therefore \( u \) solves (4).\n\nNext, since the \( {f}_{j} \) have compact support, there is a fixed \( R \), so that the \( {f}_{j} \) vanish when \( \left| z\right| > R \) for all \( j \) . Thus by Proposition 1.1, \( u \) is holomorphic in \( \left| {z}^{\prime }\right| > R \), so by (8), \( u \) also vanishes there. Since the latter is an open subset of the connected set \( \left| z\right| > R \), Proposition 1.2 implies that \( u \) vanishes when \( \left| z\right| > R \), and all our assertions are proved. | Yes |
Theorem 4.1 Assume \( \Omega \) is a bounded region in \( {\mathbb{C}}^{n} \), whose boundary is of class \( {C}^{3} \), and suppose the complement of \( \bar{\Omega } \) is connected. If \( {F}_{0} \) is a function of class \( {C}^{3} \) on \( \partial \Omega \) that satisfies the tangential Cauchy-Riemann equations, then there is a holomorphic function \( F \) in \( \Omega \) that is continuous in \( \bar{\Omega } \), so that \( {\left. F\right| }_{\partial \Omega } = {F}_{0} \) . | The proof of this theorem is in the same spirit as the previous one, but the details are different. The function \( {F}_{0} \) of class \( {C}^{3}\left( {\partial \Omega }\right) \) can, by definition, be thought of as a function of class \( {C}^{3} \) on the whole space. Now \( {F}_{0} \) satisfies the tangential Cauchy-Riemann equations, and we can modify it (without changing its restriction to \( \partial \Omega \) ), so that the modified function \( {F}_{1} \) is of class \( {C}^{2} \) and\n\n(15)\n\n\[ \n{\left. \bar{\partial }{F}_{1}\right| }_{\partial \Omega } = 0 \n\]\n\nThis modification is achieved by taking \( {F}_{1} = {F}_{0} - {a\rho } \), where \( a \) is a suitable \( {C}^{2} \) function. Indeed, \( {F}_{1} \) already satisfies the tangential Cauchy-Riemann equations. An independent Cauchy-Riemann vector field (that is not tangential) is given by \( N \), with\n\n\[ \nN\left( f\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{\bar{\rho }}_{j}\frac{\partial f}{\partial {\bar{z}}_{j}}. \n\]\n\nIn fact, we note that\n\n\[ \nN\left( \rho \right) = \mathop{\sum }\limits_{{j = 1}}^{n}{\left| \frac{\partial \rho }{\partial {\bar{z}}_{j}}\right| }^{2} = \frac{1}{4}{\left| \nabla \rho \right| }^{2} > 0. \n\]\nThus if we set \( a = N\left( {F}_{0}\right) /N\left( \rho \right) \) near the boundary of \( \Omega \) and extend \( a \) strictly away from the boundary to be zero, then (15) is achieved because of (14).\n\nWe now define the one-form \( f \) in \( \Omega \) by \( f = \bar{\partial }{F}_{1} \) . Then \( f \) is continuous on \( \bar{\Omega } \), is of class \( {C}^{1}\left( \bar{\Omega }\right) \), vanishes on \( \partial \Omega \), and satisfies \( \bar{\partial }f = 0 \) in the interior of \( \Omega \) . We can now extend \( f \) to \( {\mathbb{C}}^{n} \) (keeping the same name) so that \( f = 0 \) outside of \( \Omega \) . Then \( f \) satisfies \( \bar{\partial }f = 0 \) in \( {\mathbb{C}}^{n} \) (at least in the sense of distributions). This would be evident if we supposed that \( {F}_{0} \) and \( \partial \Omega \) were of class \( {C}^{4} \) instead of class \( {C}^{3} \) . In the latter case an additional argument is needed (see Exercise 6 in Chapter 3). We can now invoke Proposition 3.2 to obtain a continuous function \( u \) so that \( \bar{\partial }u = f \) and moreover \( u \) has compact support. Since \( u \) is holomorphic on \( {\bar{\Omega }}^{c} \) and this set is connected, it follows that \( u \) vanishes throughout \( {\bar{\Omega }}^{c} \) and by continuity it vanishes on \( \partial \Omega \) . Finally, take \( F = {F}_{1} - u \), then \( F \) is holomorphic in \( \Omega \), continuous in \( \bar{\Omega } \) and \( {\left. F\right| }_{\partial \Omega } = {\left. {F}_{1}\right| }_{\partial \Omega } = {\left. {F}_{0}\right| }_{\partial \Omega } \), completing the proof of the theorem. | Yes |
Proposition 5.1 Near any point \( {z}^{0} \in \partial \Omega \) we can introduce holomorphic coordinates \( \left( {{z}_{1},\ldots ,{z}_{n}}\right) \) centered at \( {z}^{0} \) so that\n\n\[ \Omega = \left\{ {\operatorname{Im}\left( {z}_{n}\right) > \mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{\lambda }_{j}{\left| {z}_{j}\right| }^{2} + E\left( z\right) }\right\} \]\n\nHere the \( {\lambda }_{j} \) are real numbers, and \( E\left( z\right) = {x}_{n}\ell \left( {z}^{\prime }\right) + D{x}_{n}^{2} + o\left( {\left| z\right| }^{2}\right) \), as \( z \rightarrow 0;{}^{5} \) also \( \ell \left( {z}^{\prime }\right) \) is a linear function of \( {x}_{1},\ldots ,{x}_{n - 1},{y}_{1},\ldots ,{y}_{n - 1} \), and \( D \) is a real number. | Proof of the proposition. As in (10), we see that we can introduce complex coordinates (with an affine complex linear change of variables) so that near \( {z}^{0} \) the set \( \Omega \) is given by\n\n\[ \operatorname{Im}\left( {z}_{n}\right) > \varphi \left( {{z}^{\prime },{x}_{n}}\right) \]\n\nwith \( z = \left( {{z}^{\prime },{z}_{n}}\right) ,{z}^{\prime } = \left( {{z}_{1},\ldots ,{z}_{n - 1}}\right) \), and \( {z}_{j} = {x}_{j} + i{y}_{j} \) . We can also arrange matters so that \( \varphi \left( {0,0}\right) = 0 \) and\n\n\[ {\left. \frac{\partial }{\partial {x}_{j}}\varphi \right| }_{\left( 0,0\right) } = {\left. \frac{\partial }{\partial {y}_{j}}\varphi \right| }_{\left( 0,0\right) } = {\left. \frac{\partial }{\partial {x}_{n}}\varphi \right| }_{\left( 0,0\right) },\;1 \leq j \leq n - 1. \]\n\nUsing Taylor’s expansion of \( \varphi \) at the origin up to order 2 we see that\n\n\[ \varphi = \mathop{\sum }\limits_{{1 \leq j, k \leq n - 1}}\left( {{\alpha }_{jk}{z}_{j}{z}_{k} + {\bar{\alpha }}_{jk}{\bar{z}}_{j}{\bar{z}}_{k}}\right) + \]\n\n\[ + \mathop{\sum }\limits_{{1 \leq j, k \leq n - 1}}{\beta }_{jk}{z}_{j}{\bar{z}}_{k} + {x}_{n}{\ell }^{\prime }\left( {z}^{\prime }\right) + \]\n\n\[ + D{x}_{n}^{2} + o\left( {\left| z\right| }^{2}\right) ,\;\text{ as }z \rightarrow 0. \]\n\nHere \( {\beta }_{jk} = {\bar{\beta }}_{kj} \) and \( {\ell }^{\prime } \) is a (real) linear function of the variables \( {x}_{1},\ldots \) , \( {x}_{n - 1} \) and \( {y}_{1},\ldots ,{y}_{n - 1} \), with \( D \) a real number.\n\nNext we introduce the (global) holomorphic change of coordinates \( {\zeta }_{n} = {z}_{n} - {2i}\mathop{\sum }\limits_{{1 < j, k < n - 1}}{\alpha }_{jk}{z}_{j}{z}_{k} \), and \( {\zeta }_{k} = {z}_{k} \), for \( 1 \leq k \leq n - 1 \) . Then \( \operatorname{Im}\left( {\zeta }_{n}\right) = \operatorname{Im}\left( {z}_{n}\right) - \mathop{\sum }\limits_{{1 < j, k < n - 1}}\left( {{\alpha }_{jk}{z}_{j}{z}_{k} + {\bar{\alpha }}_{jk}{\bar{z}}_{j}{\bar{z}}_{k}}\right) \), and thus in t | Yes |
Corollary 6.2 Suppose the Levi form, as given by (18), has at least one strictly positive eigenvalue for each \( z \in M \) . Under these circumstances, for every \( {z}^{0} \in M \) there is a ball \( {B}^{\prime } \) centered at \( {z}^{0} \) so that whenever \( F \) is holomorphic in \( {\Omega }_{ - } \) and continuous in \( {\Omega }_{ - } \cup M \) we have\n\n(22)\n\n\[ \mathop{\sup }\limits_{{z \in {\Omega }_{ - } \cap {B}^{\prime }}}\left| {F\left( z\right) }\right| \leq \mathop{\sup }\limits_{{z \in M}}\left| {F\left( z\right) }\right| \] | The theorem we have just proved tells us that when an eigenvalue of the Levi form is positive, the control of the restriction of a holomorphic function to a small piece of the boundary gives us a corresponding control of the function in an interior region. This is a strong hint that for such boundaries a local version of Bochner's theorem (Theorem 4.1) should be valid. Our proof of this will be based on a remarkable extension of the Weierstrass approximation theorem, to which we now turn. | No |
Theorem 7.1 Suppose \( M \subset {\mathbb{C}}^{n} \) is a hypersurface of class \( {C}^{2} \) as above. Given a point \( {z}^{0} \in M \), there are open balls \( {B}^{\prime } \) and \( B \), centered at \( {z}^{0} \), with \( {\bar{B}}^{\prime } \subset B \), so that: if \( F \) is a continuous function in \( M \cap B \) that satisfies the tangential Cauchy-Riemann equations in the weak sense, then \( F \) can be uniformly approximated on \( M \cap {\bar{B}}^{\prime } \) by polynomials in \( {z}_{1},{z}_{2},\ldots ,{z}_{n} \) . | Proof. We shall first take \( B \) small enough so that in \( B \), the hypersurface \( M \) has been represented by \( M = \left\{ {{y}_{n} = \varphi \left( {{z}^{\prime },{x}_{n}}\right) }\right\} \) where \( {z}^{0} \) corresponds to the origin. Besides \( \varphi \left( {0,0}\right) = 0 \), we can also suppose that the partial derivatives \( \frac{\partial \varphi }{\partial {x}_{j}},1 \leq j \leq n \), and \( \frac{\partial \varphi }{\partial {y}_{j}},1 \leq j \leq n - 1 \), vanish at the origin.\n\nNow for each \( u \in {\mathbb{R}}^{n - 1} \), sufficiently close to the origin we define the slice \( {M}_{u} \) of \( M \) to be the \( n \) -dimensional sub-manifold given by\n\n\[ \n{M}_{u} = \left\{ {z : {y}_{n} = \varphi \left( {{z}^{\prime },{x}_{n}}\right) ,\text{ with }{z}^{\prime } = {x}^{\prime } + {iu}}\right\} .\n\]\n\nWe let \( \Phi = {\Phi }^{u} \) be the mapping identifying the neighborhood of the origin \( {\mathbb{R}}^{n} \) with \( {M}_{u} \) given by \( \Phi \left( x\right) = \left( {{x}^{\prime } + {iu},{x}_{n} + {i\varphi }\left( {{x}^{\prime } + {iu},{x}_{n}}\right) }\right) \) with \( x = \left( {{x}^{\prime },{x}_{n}}\right) \in {\mathbb{R}}^{n - 1} \times \mathbb{R} = {\mathbb{R}}^{n} \) . Observe that \( M \) is fibered by the collection \( {\left\{ {M}_{u}\right\} }_{u} \) . Now for fixed \( u \), the Jacobian of the mapping \( x \mapsto \Phi \left( x\right) \) , that is, \( \frac{\partial \Phi }{\partial x} \), is the complex \( n \times n \) matrix given by \( I + A\left( x\right) \), where the entries of \( A\left( x\right) \) are zero, except in the last row, and in that row we have the vector \( \left( {i\frac{\partial \varphi }{\partial {x}_{1}}, i\frac{\partial \varphi }{\partial {x}_{2}},\ldots, i\frac{\partial \varphi }{\partial {x}_{n}}}\right) \) . So \( A\left( 0\right) = 0 \), and \( \det \left( \frac{\partial \Phi }{\partial x}\right) = 1 + i\frac{\partial \varphi }{\partial {x}_{n}} \) . We shall need to shrink the ball \( B \) further so that \( \parallel A\left( x\right) \parallel \leq 1/2 \), on this ball, where \( \parallel \cdot \parallel \) denotes the matrix-norm.\n\nNow with \( u \) fixed, the map \( \Phi \) carries the Lebesgue measure on \( {\mathbb{R}}^{n} \) to a measure (with complex density) \( d{m}_{u}\left( z\right) = \mathcal{J}\left( x\right) {dx} \) on \( {M}_{u} \) defined by\n\n\[ \n{\int }_{{M}_{u}}f\left( z\right) d{m}_{u}\left( z\right) = {\int }_{{\mathbb{R}}^{n}}f\left( {\Phi \left( x\right) }\right) \mathcal{J}\left( x\right) {dx},\;\text{ where }\mathcal{J}\left( x\right) = \det \left( \frac{\partial \Phi \left( x\right) }{\partial x}\right) \n\]\nfor every continuous function \( f \) with sufficiently small support.\n\nNext take \( {B}^{\prime } \) any ball with the same center as \( B \) but strictly interior to it. Define \( \chi \) to be a smooth (say \( {C}^{1} \) ) cut-off fun | Yes |
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