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Lemma 7.2 If \( A \) is an \( n \times n \) complex matrix with constant coefficients and \( \parallel A\parallel < 1 \) then for every \( \epsilon > 0 \)\n\n\[ \frac{1}{{\epsilon }^{n/2}}\det \left( {I + A}\right) {\int }_{{\mathbb{R}}^{n}}{e}^{-\frac{\pi }{\epsilon }{\left( \left( I + A\right) v\right) }^{2}}{dv} = 1. \] | To prove the lemma note that \( \operatorname{Re}\left( {\left( \left( I + A\right) v\right) }^{2}\right) \geq {\left| v\right| }^{2} - \parallel A\parallel {\left| v\right| }^{2} \geq c{\left| v\right| }^{2} \), with \( c > 0 \), so that the integral in (27) converges. A change of scale reduces the identity to the case \( \epsilon = 1 \) . Now if \( A \) is real, a further change of variables \( {v}^{\prime } = \left( {I + A}\right) v \) (which is invertible since \( \parallel A\parallel < 1 \) ) reduces this case to the standard Gaussian integral. Finally, we pass to the general situation by analytic continuation, noting that the left-hand side of (27) is holomorphic in the entries of \( A \), whenever \( \parallel A\parallel < 1 \) . | Yes |
Corollary 7.3 If \( f \) is a continuous function of compact support, then\n\n\[ \frac{\det \left( {I + A}\right) }{{\epsilon }^{n/2}}{\int }_{{\mathbb{R}}^{n}}{e}^{-\frac{\pi }{\epsilon }{\left( \left( I + A\right) v\right) }^{2}}f\left( {\xi + v}\right) {dv} \rightarrow f\left( \xi \right) \]\n\nuniformly in \( \xi \) as \( \epsilon \rightarrow 0 \) . | To prove the lemma note that \( \operatorname{Re}\left( {\left( \left( I + A\right) v\right) }^{2}\right) \geq {\left| v\right| }^{2} - \parallel A\parallel {\left| v\right| }^{2} \geq c{\left| v\right| }^{2} \), with \( c > 0 \), so that the integral in (27) converges. A change of scale reduces the identity to the case \( \epsilon = 1 \) . Now if \( A \) is real, a further change of variables \( {v}^{\prime } = \left( {I + A}\right) v \) (which is invertible since \( \parallel A\parallel < 1 \) ) reduces this case to the standard Gaussian integral. Finally, we pass to the general situation by analytic continuation, noting that the left-hand side of (27) is holomorphic in the entries of \( A \), whenever \( \parallel A\parallel < 1 \) . The corollary then follows from the usual arguments about approximations of the identity as in Section 4, Chapter 2 in Book I and Section 2 in Chapter 3 of Book III. | Yes |
Theorem 7.5 Suppose that the Levi form (18) has at least one strictly positive eigenvalue for each \( z \in M \) . Then for each \( {z}^{0} \in M \), there is a ball \( {B}^{\prime } \) centered at \( {z}^{0} \) so that whenever \( {F}_{0} \) is a continuous function on \( M \) that satisfies the tangential Cauchy-Riemann equations in the weak sense, there exists an \( F \) which is holomorphic in \( {\Omega }_{ - } \cap {B}^{\prime } \), continuous in \( {\bar{\Omega }}_{ - } \cap {B}^{\prime } \) and so that \( F\left( z\right) = {F}_{0}\left( z\right) \) for \( z \in M \cap {B}^{\prime } \) . | To prove the theorem we first use Theorem 7.1 to find a ball \( {B}_{1} \) centered at \( {z}_{0} \) so that \( {F}_{0} \) can be uniformly approximated (on \( M \cap {B}_{1} \) ) by polynomials \( \left\{ {{p}_{n}\left( z\right) }\right\} \) . Then we invoke the corollary to Theorem 6.1 to find a ball \( {B}^{\prime } \) so that (22) holds (with \( {B}_{1} \) in place of \( B \) ). Therefore the \( {p}_{n} \) also converge uniformly in \( {\Omega }_{ - } \cap {B}^{\prime } \) . The limit of this sequence, \( F \) , is then holomorphic there, continuous in \( {\bar{\Omega }}_{ - } \cap {B}^{\prime } \), and gives the desired extension of \( {F}_{0} \) . | Yes |
Theorem 8.1 Suppose \( F \in {H}^{2}\left( \mathcal{U}\right) \) . Then, when restricted to \( z \in \partial \mathcal{U} \), the limit\n\n\[ \mathop{\lim }\limits_{{\epsilon \rightarrow 0}}{F}_{\epsilon } = {F}_{0} \]\n\nexists in the \( {L}^{2}\left( {\partial \mathcal{U},{d\beta }}\right) \) norm. Also\n\n\[ \parallel F{\parallel }_{{H}^{2}\left( \mathcal{U}\right) } = {\begin{Vmatrix}{F}_{0}\end{Vmatrix}}_{{L}^{2}\left( {\partial \mathcal{U}}\right) }.\n\] | The proof of the theorem can be given by the Fourier transform representation of each \( F \in {H}^{2}\left( \mathcal{U}\right) \) in analogy with the case \( n = 1 \) treated in Chapter 5 of Book III. | No |
Lemma 8.2 Suppose \( {B}_{1} \) and \( {B}_{2} \) are two open balls in \( {\mathbb{C}}^{n - 1} \), with \( {\bar{B}}_{1} \subset {B}_{2} \). Then, whenever \( f \) is holomorphic in \( {\mathbb{C}}^{n - 1} \n\n\[ \mathop{\sup }\limits_{{{z}^{\prime } \in {B}_{1}}}{\left| f\left( {z}^{\prime }\right) \right| }^{2} \leq c{\int }_{{B}_{2}}{\left| f\left( {w}^{\prime }\right) \right| }^{2}{dm}\left( {w}^{\prime }\right) . \]\n | Indeed for sufficiently small \( \delta \), whenever \( {z}^{\prime } \in {B}_{1} \) then \( {B}_{\delta }\left( {z}^{\prime }\right) \subset {B}_{2} \), so since \( f \) is harmonic in \( {\mathbb{R}}^{{2n} - 2} \), the mean-value property and the Cauchy-Schwarz inequality gives\n\n\[ {\left| f\left( {z}^{\prime }\right) \right| }^{2} \leq \frac{1}{m\left( {B}_{\delta }\right) }{\int }_{{B}_{\delta }\left( {z}^{\prime }\right) }{\left| f\left( {w}^{\prime }\right) \right| }^{2}{dm}\left( {w}^{\prime }\right) ,\]\n\nproving the claim. | No |
Theorem 8.5 Suppose \( F \in {H}^{2}\left( \mathcal{U}\right) \), and let \( {F}_{0} = \mathop{\lim }\limits_{{\epsilon \rightarrow 0}}{F}_{\epsilon } \) as in Theorem 8.1. Then \[ C\left( {F}_{0}\right) \left( z\right) = F\left( z\right) \] | Turning to the proof of the theorem, we observe that \[ S\left( {z, w}\right) = {\int }_{0}^{\infty }{\lambda }^{n - 1}{e}^{-{4\pi \lambda r}\left( {z, w}\right) }{d\lambda } \] since \( {\int }_{0}^{\infty }{\lambda }^{n - 1}{e}^{-{A\lambda }}{d\lambda } = \left( {n - 1}\right) !{A}^{-n} \), whenever \( \operatorname{Re}\left( A\right) > 0 \) . So, at least formally, \[ {\int }_{\partial \mathcal{U}}S\left( {z, w}\right) {F}_{0}\left( w\right) {d\beta } = \] \[ = {\int }_{0}^{\infty }{\int }_{\partial \mathcal{U}}{F}_{0}\left( {{w}^{\prime },{u}_{n} + i{\left| {w}^{\prime }\right| }^{2}}\right) {\lambda }^{n - 1}{e}^{-{4\pi \lambda r}\left( {z, w}\right) }{dm}\left( {{w}^{\prime },{u}_{n}}\right) {d\lambda }. \] But as we have seen \[ {\int }_{\mathbb{R}}{F}_{0}\left( {{w}^{\prime },{u}_{n} + i{v}_{n}}\right) {e}^{-{2\pi i\lambda }\left( {{u}_{n} + i{v}_{n}}\right) }d{u}_{n} = f\left( {{w}^{\prime },\lambda }\right) . \] Now insert this in the above, recalling that \( r\left( {z, w}\right) = - \frac{{\bar{w}}_{n} - {z}_{n}}{2i} - {z}^{\prime } \cdot {\bar{w}}^{\prime } \), and that \( {\left( 4\lambda \right) }^{n - 1}{\int }_{{\mathbb{C}}^{n - 1}}f\left( {{w}^{\prime },\lambda }\right) {e}^{-{4\pi \lambda }{\left| {w}^{\prime }\right| }^{2}}{dm}\left( {w}^{\prime }\right) = f\left( {{z}^{\prime },\lambda }\right) \) . The result is that \[ {\int }_{\partial \mathcal{U}}S\left( {z, w}\right) {F}_{0}\left( w\right) {d\beta }\left( w\right) = {\int }_{0}^{\infty }f\left( {{z}^{\prime },\lambda }\right) {e}^{{2\pi i\lambda }{z}_{n}}{d\lambda } \] which by (33) is what we want to obtain. To make this argument rigorous, we proceed as in the proof of Theorem 8.1, with the improved function \( {F}_{\epsilon }^{\delta } \) in place of \( F \) . Then all the integrals in question converge absolutely, and therefore the interchanges of integration are justified. This gives the reproducing property (38) for \( {F}_{\epsilon }^{\delta } \) instead of \( F \) | Yes |
Lemma 8.6 For \( f \) as above, we have\n\n(39)\n\n\[ f\left( {z}^{\prime }\right) = {\int }_{{\mathbb{C}}^{n - 1}}{K}_{\lambda }\left( {{z}^{\prime },{w}^{\prime }}\right) f\left( {w}^{\prime }\right) {e}^{-{4\pi \lambda }{\left| {w}^{\prime }\right| }^{2}}{dm}\left( {w}^{\prime }\right) \]\n\nwith \( {K}_{\lambda }\left( {{z}^{\prime },{w}^{\prime }}\right) = {\left( 4\lambda \right) }^{n - 1}{e}^{{4\pi \lambda }{z}^{\prime } \cdot {\bar{w}}^{\prime }} \) . | Proof. In fact, consider first the case when \( {4\lambda } = 1 \), and \( {z}^{\prime } = 0 \) . Then (39), which states \( f\left( 0\right) = {\int }_{{\mathbb{C}}^{n - 1}}f\left( {w}^{\prime }\right) {e}^{-\pi {\left| {w}^{\prime }\right| }^{2}}{dm}\left( {w}^{\prime }\right) \), is a simple consequence of the mean-value property of \( f \) (taken on spheres in \( {\mathbb{C}}^{n - 1} \) centered at the origin) and the fact that \( {\int }_{{\mathbb{C}}^{n - 1}}{e}^{-\pi {\left| {z}^{\prime }\right| }^{2}}{dm}\left( {z}^{\prime }\right) = 1 \) .\n\nWe now apply this identity to \( {w}^{\prime } \mapsto f\left( {{z}^{\prime } + {w}^{\prime }}\right) {e}^{-\pi {\bar{z}}^{\prime } \cdot {w}^{\prime }} \) for fixed \( {z}^{\prime } \) . The result is then (39) when \( {4\lambda } = 1 \) . A simple rescaling argument then gives (39) in general. | Yes |
Theorem 8.7 Suppose \( U \) is a distribution defined on \( \mathbb{C} \times \mathbb{R} \), so that \( \bar{L}\left( U\right) = f \) in a neighborhood of the origin. Then (41) must hold. | Proof. Assume first that \( U \) has compact support, and \( \bar{L}\left( U\right) = f \) everywhere. Then\n\n\[ C\left( f\right) \left( z\right) = \left\langle {f, S\left( {z,{u}_{2} + i{\left| {w}_{1}\right| }^{2}}\right) }\right\rangle = \left\langle {\bar{L}\left( U\right), S\left( {z,{u}_{2} + i{\left| {w}_{1}\right| }^{2}}\right) }\right\rangle \]\n\n\[ = - \left\langle {U,\bar{L}\left( {S\left( {z,{u}_{2} + i{\left| {w}_{1}\right| }^{2}}\right) }\right) }\right\rangle \]\n\n\[ = 0\text{,} \]\n\nsince \( \bar{L}\left( {S\left( {z,{u}_{2} + i{\left| {w}_{1}\right| }^{2}}\right) }\right) = 0 \), because \( w \mapsto S\left( {z, w}\right) \) is conjugate holomorphic. Thus trivially \( C\left( f\right) \left( z\right) \) is holomorphic everywhere.\n\nIf \( U \) does not have compact support and \( \bar{L}\left( U\right) = f \) only in a neighborhood of the origin, then replace \( U \) by \( {\eta U} \), with \( \eta \) a \( {C}^{\infty } \) cut-off function that is 1 near the origin. With \( {U}^{\prime } = {\eta U} \), then \( \bar{L}\left( {U}^{\prime }\right) = {f}^{\prime } \) everywhere, so \( C\left( {f}^{\prime }\right) = 0 \) but \( C\left( {f - {f}^{\prime }}\right) \) is analytic near the origin because \( f - {f}^{\prime } \) vanishes near the origin of \( \mathbb{C} \times \mathbb{R} \). Therefore (41) holds. | Yes |
Proposition 1.1 The mapping \( f \mapsto A\left( f\right) \) is bounded from \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) to \( {L}_{k}^{2}\left( {\mathbb{R}}^{d}\right) \), with \( k = \frac{d - 1}{2} \) . | Proof. The proposition is a consequence of the identity\n\n(3)\n\n\[ \widehat{d\sigma }\left( \xi \right) = {2\pi }{\left| \xi \right| }^{-d/2 + 1}{J}_{d/2 - 1}\left( {{2\pi }\left| \xi \right| }\right) \]\n\nwhere \( \widehat{d\sigma }\left( \xi \right) = {\int }_{{S}^{d - 1}}{e}^{-{2\pi ix} \cdot \xi }{d\sigma }\left( x\right) \), and \( {J}_{m} \) is the Bessel function of order \( m \) . In turn this is just another version of the formula for the Fourier transform of a radial function \( f\left( x\right) = {f}_{0}\left( \left| x\right| \right) \), given by \( \widehat{f}\left( \xi \right) = F\left( \left| \xi \right| \right) \) , with\n\n(4)\n\n\[ F\left( \rho \right) = {2\pi }{\rho }^{-d/2 + 1}{\int }_{0}^{\infty }{J}_{d/2 - 1}\left( {2\pi \rho r}\right) {f}_{0}\left( r\right) {r}^{d/2}{dr} \]\n\nfrom which (3) follows by a simple limiting argument. From (3) we obtain the key decay estimate\n\n(5)\n\n\[ \left| {\widehat{d\sigma }\left( \xi \right) }\right| \leq O\left( {\left| \xi \right| }^{-\frac{d - 1}{2}}\right) \;\text{ as }\left| \xi \right| \rightarrow \infty . \]\n\nIndeed, (5) is deducible from (3) and the asymptotic behavior of the Bessel functions that guarantees that \( {J}_{m}\left( r\right) = O\left( {r}^{-1/2}\right) \) as \( r \rightarrow \infty \) .\n\nOnce (5) is established the proof of the proposition is finished via Plancherel’s theorem as in the case \( d = 3 \) . | Yes |
Proposition 2.1 Suppose \( \\left| {\\nabla \\Phi \\left( x\\right) }\\right| \\geq c > 0 \) for all \( x \) in the support of \( \\psi \) . Then for every \( N \\geq 0 \n\\[ \n\\left| {I\\left( \\lambda \\right) }\\right| \\leq {c}_{N}{\\lambda }^{-N},\\;\\text{ whenever }\\lambda > 0.\n\\] | Proof. We consider the following vector field\n\n\\[ \nL = \\frac{1}{i\\lambda }\\mathop{\\sum }\\limits_{{k = 1}}^{d}{a}_{k}\\frac{\\partial }{\\partial {x}_{k}} = \\frac{1}{i\\lambda }\\left( {a \\cdot \\nabla }\\right) ,\n\\]\n\nwith \( a = \\left( {{a}_{1},\\ldots ,{a}_{d}}\\right) = \\frac{\\nabla \\Phi }{{\\left| \\nabla \\Phi \\right| }^{2}} \) . Then the transpose \( {L}^{t} \) of \( L \) is given by\n\n\\[ \n{L}^{t}\\left( f\\right) = - \\frac{1}{i\\lambda }\\mathop{\\sum }\\limits_{{k = 1}}^{d}\\frac{\\partial }{\\partial {x}_{k}}\\left( {{a}_{k}f}\\right) = - \\frac{1}{i\\lambda }\\nabla \\cdot \\left( {af}\\right) .\n\\]\n\nBecause of our assumption on \( \\nabla \\Phi \), the \( {a}_{j} \) and all their partial derivatives are each bounded on the support of \( \\psi \).\n\nNow observe that \( L\\left( {e}^{i\\lambda \\Phi }\\right) = {e}^{i\\lambda \\Phi } \), therefore \( {L}^{N}\\left( {e}^{i\\lambda \\Phi }\\right) = {e}^{i\\lambda \\Phi } \) for every positive integer \( N \) . Thus\n\n\\[ \nI\\left( \\lambda \\right) = {\\int }_{{\\mathbb{R}}^{d}}{L}^{N}\\left( {e}^{i\\lambda \\Phi }\\right) {\\psi dx} = {\\int }_{{\\mathbb{R}}^{d}}{e}^{i\\lambda \\Phi }{\\left( {L}^{t}\\right) }^{N}\\left( \\psi \\right) {dx}.\n\\]\n\nTaking absolute values in the last integral gives \( \\left| {I\\left( \\lambda \\right) }\\right| \\leq {c}_{N}{\\lambda }^{-N} \) for positive \( \\lambda \), thus proving the proposition. | Yes |
Proposition 2.2 In the above situation, \( \\left| {{I}_{1}\\left( \\lambda \\right) }\\right| \\leq c{\\lambda }^{-1} \), all \( \\lambda > 0 \), with \( c = 3 \) . | Proof. The proof uses the operator \( L \) that occurred in the previous proposition. We may assume \( {\\Phi }^{\\prime } > 0 \) on \( \\left\\lbrack {a, b}\\right\\rbrack \), because the case when \( {\\Phi }^{\\prime } < 0 \) follows by taking complex conjugates. So \( L = \\frac{1}{{i\\lambda }{\\Phi }^{\\prime }\\left( x\\right) }\\frac{d}{dx} \), and \( {L}^{t}\\left( f\\right) = - \\frac{1}{i\\lambda }\\frac{d}{dx}\\left( {f/{\\Phi }^{\\prime }}\\right) \), hence\n\n\\[ \n{I}_{1}\\left( \\lambda \\right) = {\\int }_{a}^{b}L\\left( {e}^{i\\lambda \\Phi }\\right) {dx} = {\\int }_{a}^{b}{e}^{i\\lambda \\Phi }{L}^{t}\\left( 1\\right) {dx} + {\\left\\lbrack {e}^{i\\lambda \\Phi }\\frac{1}{{i\\lambda }{\\Phi }^{\\prime }}\\right\\rbrack }_{a}^{b}, \n\\]\n\nand now (because we do not have an amplitude \( \\psi \) that vanishes at the end-points) there are boundary terms. Since \( \\left| {{\\Phi }^{\\prime }\\left( x\\right) }\\right| \\geq 1 \), these two terms contribute a total majorized by \( 2/\\lambda \) . But the integral on the right-hand side is clearly bounded by\n\n\\[ \n{\\int }_{a}^{b}\\left| {{L}^{t}\\left( 1\\right) }\\right| {dx} = \\frac{1}{\\lambda }{\\int }_{a}^{b}\\left| {\\frac{d}{dx}\\left( \\frac{1}{{\\Phi }^{\\prime }}\\right) }\\right| {dx}. \n\\]\n\nHowever \( {\\Phi }^{\\prime } \) is monotonic and continuous while \( \\left| {{\\Phi }^{\\prime }\\left( x\\right) }\\right| \\geq 1 \), so \( \\frac{d}{dx}\\left( {1/{\\Phi }^{\\prime }}\\right) \n\ndoes not change sign in the interval \( \\left\\lbrack {a, b}\\right\\rbrack \) . Therefore\n\n\\[ \n{\\int }_{a}^{b}\\left| {\\frac{d}{dx}\\left( \\frac{1}{{\\Phi }^{\\prime }}\\right) }\\right| {dx} = \\left| {{\\int }_{a}^{b}\\frac{d}{dx}\\left( \\frac{1}{{\\Phi }^{\\prime }}\\right) {dx}}\\right| = \\left| {\\frac{1}{{\\Phi }^{\\prime }\\left( b\\right) } - \\frac{1}{{\\Phi }^{\\prime }\\left( a\\right) }}\\right| . \n\\]\n\nAltogether then \( \\left| {{I}_{1}\\left( \\lambda \\right) }\\right| \\leq 3/\\lambda \) and the proposition is proved. | Yes |
Proposition 2.3 Under the above assumptions, and with \( {I}_{1}\left( \lambda \right) \) given by (7) we have\n\n\[ \left| {{I}_{1}\left( \lambda \right) }\right| \leq {c}^{\prime }{\lambda }^{-1/2}\;\text{ for all }\lambda > 0,\text{ with }{c}^{\prime } = 8. \] | Proof. We may assume that \( {\Phi }^{\prime \prime }\left( x\right) \geq 1 \) throughout the interval, because the case \( {\Phi }^{\prime \prime }\left( x\right) \leq - 1 \) follows from this by taking complex conjugates. Now \( {\Phi }^{\prime \prime }\left( x\right) \geq 1 \) implies that \( {\Phi }^{\prime }\left( x\right) \) is strictly increasing, so if \( \Phi \) has a critical point in \( \left\lbrack {a, b}\right\rbrack \), it can have only one. Assume \( {x}_{0} \) is such a critical point and break the interval \( \left\lbrack {a, b}\right\rbrack \) in three sub-intervals: the first is centered at \( {x}_{0} \) and is \( \left\lbrack {{x}_{0} - \delta ,{x}_{0} + \delta }\right\rbrack \) with \( \delta \) chosen momentarily. The other two make up the complement and are \( \left\lbrack {a,{x}_{0} - \delta }\right\rbrack \) and \( \left\lbrack {{x}_{0} + \delta, b}\right\rbrack \) . Now the first interval has length \( {2\delta } \), so trivially the integral taken over that interval contributes at most \( {2\delta } \) . On the interval \( \left\lbrack {{x}_{0} + \delta, b}\right\rbrack \) we observe that \( {\Phi }^{\prime }\left( x\right) \geq \delta \) (because \( {\Phi }^{\prime \prime } \geq 1 \) ) and so by Proposition 2.2 and the remark that follows it, the integral contributes at most \( 3/\left( {\delta \lambda }\right) \) ; similarly for the interval \( \left\lbrack {a,{x}_{0} - \delta }\right\rbrack \) . Thus altogether \( {I}_{1}\left( \lambda \right) \) is majorized by \( {2\delta } + 6/\left( {\delta \lambda }\right) \) , and upon choosing \( \delta = {\lambda }^{-1/2} \) we get (9). Note that if \( \Phi \) has no critical points in \( \left\lbrack {a, b}\right\rbrack \) and/or one of the three intervals is smaller than indicated, then each of the estimates holds a fortiori, and hence also the conclusion. | Yes |
Corollary 2.4 Assume \( \Phi \) satisfies the hypotheses of Proposition 2.3. Then\n\n\[ \left| {{\int }_{a}^{b}{e}^{{i\lambda \Phi }\left( x\right) }\psi \left( x\right) {dx}}\right| \leq {c}_{\psi }{\lambda }^{-1/2} \]\n\nwhere \( {c}_{\psi } = 8\left( {{\int }_{a}^{b}\left| {{\psi }^{\prime }\left( x\right) }\right| {dx} + \left| {\psi \left( b\right) }\right| }\right) \) . | Proof. Let \( J\left( x\right) = {\int }_{a}^{x}{e}^{{i\lambda \Phi }\left( u\right) }{du} \) . We integrate by parts, using \( J\left( a\right) = \) 0 . Then\n\n\[ {\int }_{a}^{b}{e}^{{i\lambda \Phi }\left( x\right) }\psi \left( x\right) {dx} = - {\int }_{a}^{b}J\left( x\right) \frac{d\psi }{dx}{dx} + J\left( b\right) \psi \left( b\right) ,\]\n\nand the result follows, because \( \left| {J\left( x\right) }\right| \leq 8{\lambda }^{-1/2} \) for each \( x \), by the proposition. | Yes |
Theorem 3.1 Suppose the hypersurface \( M \) has non-vanishing Gauss curvature at each point of the support of \( {d\mu } \) . Then\n\n\[ \left| {\widehat{d\mu }\left( \xi \right) }\right| = O\left( {\left| \xi \right| }^{-\left( {d - 1}\right) /2}\right) \;\text{ as }\left| \xi \right| \rightarrow \infty . \] | First some preliminary remarks. We can assume that the support of \( \psi \) is centered in a sufficiently small ball (so that in particular the representation (18) of \( M \) holds in it), because we can always write a given \( \psi \) as a finite sum of \( {\psi }_{j} \) of that type. Next, all our estimates can be made in the coordinate system used in (18) since the transformations of the \( x \) -space \( {\mathbb{R}}^{d} \) used in that change of coordinates involves only a translation and a rotation. The Fourier transform \( \widehat{d\mu }\left( \xi \right) \) then undergoes a multiplication by a factor of absolute value 1 (a character) and the same rotation in the \( \xi \) variable. Thus the estimate (21) is unchanged.\n\nNow because of (20) we have\n\n\[ \widehat{d\mu }\left( \xi \right) = {\int }_{{\mathbb{R}}^{d - 1}}{e}^{-{2\pi i}\left( {{x}^{\prime } \cdot {\xi }^{\prime } + \varphi \left( {x}^{\prime }\right) {\xi }_{d}}\right) }\widetilde{\psi }\left( {x}^{\prime }\right) d{x}^{\prime }, \]\n\nwith \( \xi = \left( {{\xi }^{\prime },{\xi }_{d}}\right) \in {\mathbb{R}}^{d} \) and \( \widetilde{\psi } \) the \( {C}^{\infty } \) function with compact support given by\n\n\[ \widetilde{\psi }\left( {x}^{\prime }\right) = \psi \left( {{x}^{\prime },\varphi \left( {x}^{\prime }\right) }\right) {\left( 1 + {\left| {\nabla }_{{x}^{\prime }}\varphi \right| }^{2}\right) }^{1/2}. \]\n\nWe divide the \( \xi \) space into two parts: the \ | No |
Corollary 3.2 If \( M \) has at least \( m \) non-vanishing principal curvatures at each point of the support of \( {d\mu } \), then\n\n\[ \left| {\widehat{d\mu }\left( \xi \right) }\right| = O\left( {\left| \xi \right| }^{-m/2}\right) \;\text{ as }\left| \xi \right| \rightarrow \infty . \] | First some preliminary remarks. We can assume that the support of \( \psi \) is centered in a sufficiently small ball (so that in particular the representation (18) of \( M \) holds in it), because we can always write a given \( \psi \) as a finite sum of \( {\psi }_{j} \) of that type. Next, all our estimates can be made in the coordinate system used in (18) since the transformations of the \( x \) -space \( {\mathbb{R}}^{d} \) used in that change of coordinates involves only a translation and a rotation. The Fourier transform \( \widehat{d\mu }\left( \xi \right) \) then undergoes a multiplication by a factor of absolute value 1 (a character) and the same rotation in the \( \xi \) variable. Thus the estimate (21) is unchanged.\n\nNow because of (20) we have\n\n(22)\n\n\[ \widehat{d\mu }\left( \xi \right) = {\int }_{{\mathbb{R}}^{d - 1}}{e}^{-{2\pi i}\left( {{x}^{\prime } \cdot {\xi }^{\prime } + \varphi \left( {x}^{\prime }\right) {\xi }_{d}}\right) }\widetilde{\psi }\left( {x}^{\prime }\right) d{x}^{\prime }, \]\n\nwith \( \xi = \left( {{\xi }^{\prime },{\xi }_{d}}\right) \in {\mathbb{R}}^{d} \) and \( \widetilde{\psi } \) the \( {C}^{\infty } \) function with compact support given by\n\n\[ \widetilde{\psi }\left( {x}^{\prime }\right) = \psi \left( {{x}^{\prime },\varphi \left( {x}^{\prime }\right) }\right) {\left( 1 + {\left| {\nabla }_{{x}^{\prime }}\varphi \right| }^{2}\right) }^{1/2}. \]\n\nWe divide the \( \xi \) space into two parts: the \ | No |
Corollary 3.3 If \( M = \partial \Omega \) has non-vanishing Gauss curvature at each point, then\n\n\[ \n{\widehat{\chi }}_{\Omega }\left( \xi \right) = O\left( {\left| \xi \right| }^{-\frac{d + 1}{2}}\right) ,\;\text{ as }\left| \xi \right| \rightarrow \infty .\n\] | Proof. Using an appropriate partition of unity we can write\n\n\[ \n{\chi }_{\Omega } = \mathop{\sum }\limits_{{j = 0}}^{N}{\psi }_{j}{\chi }_{\Omega }\n\]\n\nwith each \( {\psi }_{j} \) a \( {C}^{\infty } \) function of compact support; \( {\psi }_{0} \) is supported in the interior of \( \Omega \), while each \( {\psi }_{j},1 \leq j \leq N \), is supported in a small neighborhood of the boundary in which the boundary is given as (18). Now since \( {\psi }_{0}{\chi }_{\Omega } = {\psi }_{0} \), it is clear that \( {\left( {\psi }_{0}{\chi }_{\Omega }\right) }^{ \land } \) is rapidly decreasing. Next consider any \( {\left( {\psi }_{j}{\chi }_{\Omega }\right) }^{ \land } \) for \( 1 \leq j \leq N \) . In analogy with (22), this has the form\n\n\[ \n{\int }_{{x}_{d} > \varphi \left( {x}^{\prime }\right) }{e}^{-{2\pi i}\left( {{x}^{\prime } \cdot {\xi }^{\prime } + {x}_{d}{\xi }_{d}}\right) }{\psi }_{j}\left( {{x}^{\prime },{\xi }_{d}}\right) d{x}^{\prime }d{x}_{d},\n\]\n\nwhich is, after changing variables so that \( {x}_{d} = u + \varphi \left( {x}^{\prime }\right) \), \n\n(23)\n\n\[ \n{\int }_{{\mathbb{R}}^{d - 1}}{e}^{-{2\pi i}\left( {{x}^{\prime } \cdot {\xi }^{\prime } + \varphi \left( {x}^{\prime }\right) {\xi }_{d}}\right) }\Psi \left( {{x}^{\prime },{\xi }_{d}}\right) d{x}^{\prime }\n\]\n\nwhere \( \Psi \left( {{x}^{\prime },{\xi }_{d}}\right) = {\int }_{0}^{\infty }{e}^{-{2\pi iu}{\xi }_{d}}{\psi }_{j}\left( {{x}^{\prime }, u + \varphi \left( {x}^{\prime }\right) }\right) {du} \) . Note that \( \Psi \left( {{x}^{\prime },{\xi }_{d}}\right) \) is a \( {C}^{\infty } \) function in \( {x}^{\prime } \) of compact support, uniformly in \( {\xi }_{d} \) . When \( \left| {\xi }_{d}\right| < c\left| {\xi }^{\prime }\right| \), the argument proceeds as before, giving an estimate \( O\left( {\left| \xi \right| }^{-N}\right) \) for each \( N \geq 0 \) . To deal with the situation when \( \left| {\xi }_{d}\right| \geq c\left| {\xi }^{\prime }\right| \) write\n\n\[ \n\Psi \left( {{x}^{\prime },{\xi }_{d}}\right) = - \frac{1}{{2\pi i}{\xi }_{d}}{\int }_{0}^{\infty }\frac{d}{du}\left( {e}^{-{2\pi iu}{\xi }_{d}}\right) \psi \left( {{x}^{\prime }, u + \varphi \left( x\right) }\right) {du}.\n\]\n\nand integrate by parts, giving us an additional decay of \( O\left( {1/\left| {\xi }_{d}\right| }\right) = \) \( O\left( {1/\left| \xi \right| }\right) \) in (23). This proves the corollary. | Yes |
Theorem 4.1 Suppose the Gauss curvature is non-vanishing at each point \( x \in M \) in the support of \( {d\mu } \) . Then\n\n(a) The map \( A \) given by (24) takes \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) to \( {L}_{k}^{2}\left( {\mathbb{R}}^{d}\right) \), with \( k = \frac{d - 1}{2} \) . | The proof of part (a) in the theorem is the same as that for the sphere once we invoke the decay (21), which implies that \( {\left( 1 + {\left| \xi \right| }^{2}\right) }^{k/2}\widehat{d\mu }\left( \xi \right) \) is bounded. Hence\n\n\[ \parallel A\left( f\right) {\parallel }_{{L}_{k}^{2}} = {\begin{Vmatrix}{\left( 1 + {\left| \xi \right| }^{2}\right) }^{k/2}\widehat{Af}\left( \xi \right) \end{Vmatrix}}_{{L}^{2}} \]\n\n\[ = \parallel {\left( 1 + {\left| \xi \right| }^{2}\right) }^{k/2}\widehat{f}\left( \xi \right) \widehat{d\mu }\left( \xi \right) {\parallel }_{{L}^{2}} \]\n\n\[ \leq c\parallel \widehat{f}{\parallel }_{{L}^{2}} = c\parallel f{\parallel }_{{L}^{2}} \] | Yes |
Corollary 4.2 The Riesz diagram (see Section 2 in Chapter 2) of the map \( A \) is the closed triangle in the \( \left( {1/p,1/q}\right) \) plane whose vertices are \( \left( {0,0}\right) ,\left( {1,1}\right) \) and \( \left( {\frac{d}{d + 1},\frac{1}{d + 1}}\right) \) . | Null | No |
Corollary 4.3 If we only assume that \( M \) has at least \( m \) non-vanishing principal curvatures, then the same conclusions hold with \( k = m/2 \), and \( p = \frac{m + 2}{m + 1}, q = m + 2 \) . | The proof of part (a) in the theorem is the same as that for the sphere once we invoke the decay (21), which implies that \( {\left( 1 + {\left| \xi \right| }^{2}\right) }^{k/2}\widehat{d\mu }\left( \xi \right) \) is bounded. Hence\n\n\[ \parallel A\left( f\right) {\parallel }_{{L}_{k}^{2}} = {\begin{Vmatrix}{\left( 1 + {\left| \xi \right| }^{2}\right) }^{k/2}\widehat{Af}\left( \xi \right) \end{Vmatrix}}_{{L}^{2}} \]\n\n\[ = \parallel {\left( 1 + {\left| \xi \right| }^{2}\right) }^{k/2}\widehat{f}\left( \xi \right) \widehat{d\mu }\left( \xi \right) {\parallel }_{{L}^{2}} \]\n\n\[ \leq c\parallel \widehat{f}{\parallel }_{{L}^{2}} = c\parallel f{\parallel }_{{L}^{2}} \] | Yes |
Proposition 4.4 With the above assumptions,\n\n\[ \n{\begin{Vmatrix}{T}_{c}\end{Vmatrix}}_{{L}^{q}} \leq M\parallel f{\parallel }_{{L}^{p}} \n\]\n\nfor any \( c \) with \( a \leq c \leq b \), where \( c = \left( {1 - \theta }\right) a + {\theta b} \) and \( 0 \leq \theta \leq 1 \) ; and\n\n\[ \n\frac{1}{p} = \frac{1 - \theta }{{p}_{0}} + \frac{\theta }{{p}_{1}}\;\text{ and }\;\frac{1}{q} = \frac{1 - \theta }{{q}_{0}} + \frac{\theta }{{q}_{1}}.\n\] | Once we have formulated this result, we in fact observe that we can prove it by essentially the same argument as in Section 2 in Chapter 2.\n\nWe write \( s = a\left( {1 - z}\right) + {bz} \), so \( z = \frac{s - a}{b - a} \), and the strip \( S \) is thereby transformed into the strip \( 0 \leq \operatorname{Re}\left( z\right) \leq 1 \) . For \( f \) and \( g \) given simple functions, we write \( {f}_{s} = {\left| f\right| }^{\gamma \left( s\right) }\frac{f}{\left| f\right| } \) and \( {g}_{s} = {\left| g\right| }^{\delta \left( s\right) }\frac{g}{\left| g\right| } \) where we define \( \gamma \left( s\right) = p\left( {\frac{1 - s}{{p}_{0}} + \frac{s}{{p}_{1}}}\right) \), and \( \delta \left( s\right) = q\left( {\frac{1 - s}{{q}_{0}^{\prime }} + \frac{s}{{q}_{1}^{\prime }}}\right) \) . We then check that\n\n\[ \n\Phi \left( s\right) = {\int }_{{\mathbb{R}}^{d}}{T}_{s}\left( {f}_{s}\right) {g}_{s}{dx}\n\]\n\n is continuous and bounded in the strip \( S \) and analytic in the interior. We then apply the three-lines lemma to \( \Phi \left( s\right) \) and obtain the desired conclusion as in the proof of Theorem 2.1 in Chapter 2. | Yes |
Proposition 4.5 The Fourier transform \( {\widehat{K}}_{s}\left( \xi \right) \) is analytically continuable into the half-plane \( - \frac{d - 1}{2} \leq \operatorname{Re}\left( s\right) \) and satisfies \[ \mathop{\sup }\limits_{{\xi \in {\mathbb{R}}^{d}}}\left| {{\widehat{K}}_{s}\left( \xi \right) }\right| \leq M\;\text{ in the strip } - \frac{d - 1}{2} \leq \operatorname{Re}\left( s\right) \leq 1. \] | Proof. Write \( s\left( {s + 1}\right) \cdots \left( {s + N}\right) {u}^{s - 1} = {\left( \frac{d}{du}\right) }^{N + 1}{u}^{s + N} \) . Then an \( \left( {N + 1}\right) \) -fold integration by parts yields \[ {I}_{s}\left( \rho \right) = {\left( -1\right) }^{N + 1}{\int }_{0}^{\infty }{u}^{s + N}{\left( \frac{d}{du}\right) }^{N + 1}\left( {F\left( u\right) {e}^{-{2\pi iu\rho }}}\right) {du}, \] from which the analytic continuation of \( {I}_{s} \) to the half-space \( \operatorname{Re}\left( s\right) > \) \( - N - 1 \) is evident. It also proves the estimate (a) when \( \rho \) is bounded, for example when \( \left| \rho \right| \leq 1 \) . The proof of the size estimate (a) when \( \left| \rho \right| > 1 \) is similar but requires a little more care. We break the range of integration in (27) into two parts, essentially according to \( u\left| \rho \right| \leq 1 \) or \( u\left| \rho \right| > 1 \) . We suppose \( \eta \) is a \( {C}^{\infty } \) cut-off function on \( \mathbb{R} \) with \( \eta \left( u\right) = 1 \) if \( \left| u\right| \leq 1/2 \), and \( \eta \left( u\right) = 0 \) if \( \left| u\right| \geq 1 \) , and insert \( \eta \left( {u\rho }\right) \) or \( 1 - \eta \left( {u\rho }\right) \) in the integral (27). When we insert \( \eta \left( {u\rho }\right) \) we write the resulting integral as \[ {\left( -1\right) }^{N + 1}{\int }_{0}^{\infty }{u}^{s + N}{\left( \frac{d}{du}\right) }^{N + 1}\left( {\eta \left( {u\rho }\right) {e}^{-{2\pi iu\rho }}F\left( u\right) }\right) {du} \] and so it is dominated by a constant multiple of \[ {\left( 1 + \left| \rho \right| \right) }^{N + 1}{\int }_{0 \leq u \leq 1/\left| \rho \right| }{u}^{\sigma + N}{du},\;\text{ with }\sigma = \operatorname{Re}\left( s\right) . \] Since \( \sigma + N > - 1 \) this quantity is itself dominated by the product \( (1 + \) \( \left| \rho \right| {)}^{N + 1}{\left| \rho \right| }^{-\sigma - N - 1} \), which is \( \lesssim {\left( 1 + \left| \rho \right| \right) }^{-\sigma } \), since we have assumed \( \left| \rho \right| \geq 1 \) . When we insert \( 1 - \eta \left( {u\rho }\right) \) we write the resulting integral as \[ s\left( {s + 1}\right) \cdots \left( {s + N}\right) \frac{1}{{\left( -2\pi i\rho \right) }^{k}}{\int }_{0}^{\infty }{u}^{s - 1}F\left( u\right) \left( | Yes |
Lemma 4.6 \( {I}_{s}\left( \rho \right) \) initially given above for \( \operatorname{Re}\left( s\right) > 0 \), has an analytic continuation into the half-space \( \operatorname{Re}\left( s\right) > - N - 1 \) . | Proof. Write \( s\left( {s + 1}\right) \cdots \left( {s + N}\right) {u}^{s - 1} = {\left( \frac{d}{du}\right) }^{N + 1}{u}^{s + N} \) . Then an \( \left( {N + 1}\right) \) -fold integration by parts yields\n\n\[ \n{I}_{s}\left( \rho \right) = {\left( -1\right) }^{N + 1}{\int }_{0}^{\infty }{u}^{s + N}{\left( \frac{d}{du}\right) }^{N + 1}\left( {F\left( u\right) {e}^{-{2\pi iu\rho }}}\right) {du}, \n\]\n\nfrom which the analytic continuation of \( {I}_{s} \) to the half-space \( \operatorname{Re}\left( s\right) > \) \( - N - 1 \) is evident. It also proves the estimate (a) when \( \rho \) is bounded, for example when \( \left| \rho \right| \leq 1 \) . | Yes |
Proposition 5.1 Suppose \( f \in {L}^{p}\left( {\mathbb{R}}^{d}\right) \) is a radial function. Then \( \widehat{f} \) is continuous for \( \xi \neq 0 \) whenever \( 1 \leq p < {2d}/\left( {d + 1}\right) \) . Note the sequence of exponents \( \frac{2d}{\left( d + 1\right) } : 1,\frac{4}{3},\frac{3}{2},\frac{8}{5},\ldots \) that tends to 2 as \( d \rightarrow \infty \) . | Proof. Suppose \( f\left( x\right) = {f}_{0}\left( \left| x\right| \right) \) . Then \( \widehat{f}\left( \xi \right) = F\left( \left| \xi \right| \right) \) with \( F \) defined by (4), namely,\n\n(29)\n\n\[ F\left( \rho \right) = {2\pi }{\rho }^{-d/2 + 1}{\int }_{0}^{\infty }{J}_{d/2 - 1}\left( {2\pi \rho r}\right) {f}_{0}\left( r\right) {r}^{d/2}{dr}. \]\n\nWe can make the simplifying assumption that \( f \) vanishes in the unit ball (thus the integral above is taken for \( r \geq 1 \) ) because an \( {L}^{p} \) function supported in a ball is automatically in \( {L}^{1} \) and its Fourier transform is then continuous.\n\nWe also restrict \( \rho = \left| \xi \right| \) to a bounded interval excluding the origin, and note that then the integral in (29) converges absolutely and uniformly in \( \rho \) . In fact the integral is dominated by a constant multiple of\n\n(30)\n\n\[ {\int }_{1}^{\infty }\left| {{f}_{0}\left( r\right) }\right| {r}^{d/2 - 1/2}{dr} \]\n\nsince \( \left| {{J}_{d/2 - 1}\left( u\right) }\right| \leq A{u}^{-1/2} \) if \( u > 0 \), as we have already seen. Now let \( q \) be the exponent dual to \( p,\left( {1/p + 1/q = 1}\right) \), and write\n\n\[ {r}^{d/2 - 1/2} = {r}^{\frac{d - 1}{p}}{r}^{\frac{d - 1}{q}}{r}^{-\frac{d - 1}{2}}. \]\n\nThen by Hölder's inequality the integral (30) is majorized by the product of an \( {L}^{p} \) and an \( {L}^{q} \) norm. The \( {L}^{p} \) factor is\n\n\[ {\left( {\int }_{1}^{\infty }{\left| {f}_{0}\left( r\right) \right| }^{p}{r}^{d - 1}dr\right) }^{1/p} = c\parallel f{\parallel }_{{L}^{p}\left( {\mathbb{R}}^{d}\right) } \]\n\nwhile the second factor is\n\n\[ {\left( {\int }_{1}^{\infty }{r}^{d - 1 - q\left( \frac{d - 1}{2}\right) }dr\right) }^{1/q} \]\n\nand this is finite if \( d - 1 - q\left( \frac{d - 1}{2}\right) < - 1 \), which means \( q > {2d}/\left( {d - 1}\right) \) , and thus \( p < {2d}/\left( {d + 1}\right) \) . The asserted convergence of (30) therefore proves the continuity in \( \rho \) of \( F \) in (29) and establishes the proposition.\n\nAn examination of the proof shows the range \( 1 \leq p < {2d}/\left( {d + 1}\right) \) cannot be extended. | Yes |
Theorem 5.2 Suppose \( M \) has non-zero Gauss curvature at each point of the support of \( {d\mu } \) . Then the restriction inequality (31) holds for \( q = 2 \) and \( p = \frac{{2d} + 2}{d + 3} \) . | The proof starts with several quick observations. Let \( \mathcal{R} \) denote the restriction operator\n\n\[ \mathcal{R}\left( f\right) = {\left. \widehat{f}\left( \xi \right) \right| }_{M} = {\left. {\int }_{{\mathbb{R}}^{d}}{e}^{-{2\pi ix} \cdot \xi }f\left( x\right) dx\right| }_{M}, \]\n\nwhich is initially defined to map continuous functions \( f \) of compact support on \( {\mathbb{R}}^{d} \) to continuous functions on \( M \) . Consider also the \ | No |
Corollary 5.4 Under the assumptions of the theorem, the restriction inequality (31) holds for \( 1 \leq p \leq \frac{{2d} + 2}{d + 3} \) and \( q \leq \left( \frac{d - 1}{d + 1}\right) {p}^{\prime } \) . | This follows by combining the critical case \( p = \frac{{2d} + 2}{d + 3}, q \leq 2 \) (a consequence of the theorem and Hölder's inequality) with the trivial case \( p = 1, q = \infty \) via the Riesz interpolation theorem. | Yes |
Corollary 5.5 Suppose that for some \( \delta > 0 \), we have \( \left| {\widehat{d\mu }\left( \xi \right) }\right| = O\left( {\left| \xi \right| }^{-\delta }\right) ,\; \) as \( \left| \xi \right| \rightarrow \infty , \) for all measures of the above form. Then the restriction property (31) holds for \( p = \frac{{2\delta } + 2}{\delta + 2}, q = 2 \). | Null | No |
Proposition 6.1 For each \( t \) :\n\n(i) \( {e}^{{it}\bigtriangleup } \) maps \( \mathcal{S} \) to \( \mathcal{S} \) . | Proof. That \( {e}^{{it}\bigtriangleup } \) maps \( \mathcal{S} \) to \( \mathcal{S} \) is clear because the multiplier \( {e}^{-{it4}{\pi }^{2}{\left| \xi \right| }^{2}} \) has the property that each derivative in \( \xi \) is of at most polynomial increase. | Yes |
Proposition 6.2 For each \( t \) :\n\n(i) The operator \( {e}^{{it}\bigtriangleup } \) is unitary on \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) .\n\n(ii) For every \( f \), the mapping \( t \mapsto {e}^{{it}\bigtriangleup }\left( f\right) \) is continuous in the \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) norm.\n\n(iii) If \( f \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \), then \( u\left( {x, t}\right) = {e}^{{it}\bigtriangleup }\left( f\right) \left( x\right) \) satisfies (34) in the sense of distributions. | Proof. Conclusion (i) is immediate from Plancherel's theorem, since the multiplier \( {e}^{-{it4}{\pi }^{2}{\left| \dot{\xi }\right| }^{2}} \) has absolute value one. Now if \( \widehat{f} \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \) , then clearly \( {e}^{-{it4}{\pi }^{2}{\left| \xi \right| }^{2}}\widehat{f}\left( \xi \right) \rightarrow {e}^{-i{t}_{0}4{\pi }^{2}{\left| \xi \right| }^{2}}\widehat{f}\left( \xi \right) \) in the \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) norm when \( t \rightarrow {t}_{0} \), so (ii) follows again from Plancherel’s theorem.\n\nTo prove the third conclusion we use the short-hand \( \mathcal{L} = \frac{1}{i}\frac{\partial }{\partial t} - \bigtriangleup \) , and \( {\mathcal{L}}^{\prime } = - \frac{1}{i}\frac{\partial }{\partial t} - \bigtriangleup \) for its transpose. Conclusion (iii) asserts that whenever \( \varphi \) is a \( {C}^{\infty } \) function on \( {\mathbb{R}}^{d} \times \mathbb{R} \) of compact support, then\n\n(38)\n\n\[{\iint }_{{\mathbb{R}}^{d} \times \mathbb{R}}{\mathcal{L}}^{\prime }\left( \varphi \right) \left( {x, t}\right) \left( {{e}^{{it}\bigtriangleup }f}\right) \left( x\right) {dxdt} = 0.\]\n\nNow if \( f \in \mathcal{S} \), then (38) holds for such \( f \), because then \( u\left( {x, t}\right) = {e}^{{it}\bigtriangleup }\left( f\right) \left( x\right) \) satisfies \( \mathcal{L}\left( u\right) = 0 \) in the usual sense, as we have seen. For general \( f \in {L}^{2} \) , approximate \( f \) in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) by a sequence \( \left\{ {f}_{n}\right\} \) with \( {f}_{n} \in \mathcal{S} \) . Then because of conclusion (i) we may pass to the limit and obtain (38) for any \( f \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \), finishing the proof of the proposition. | Yes |
Theorem 6.4 The solution \( {e}^{t{\left( \frac{d}{dx}\right) }^{3}}\left( f\right) \) satisfies | The proof of this is result is parallel with that of the previous theorem and reduces to a restriction theorem on \( {\mathbb{R}}^{2} \) for the cubic curve\n\n\[ \Gamma = \left\{ {\left( {{\xi }_{1},{\xi }_{2}}\right) : {\xi }_{2} = - 4{\pi }^{2}{\xi }_{1}^{3}}\right\} \]\n\nAccording to Corollary 5.5, what is needed is an estimate for \( \widehat{d\mu }\left( \xi \right) \) , where \( {d\mu } \) is a smooth measure carried on the cubic curve \( \Gamma \) . The desired estimate can be rephrased as follows.\n\nLemma 6.5 Let \( I\left( \xi \right) = {\int }_{\mathbb{R}}{e}^{{2\pi i}\left( {{\xi }_{1}t + {\xi }_{2}{t}^{3}}\right) }\psi \left( t\right) {dt} \), where \( \psi \) is a \( {C}^{\infty } \) function of compact support. Then\n\n\[ I\left( \xi \right) = O\left( {\left| \xi \right| }^{-1/3}\right) ,\;\text{ as }\left| \xi \right| \rightarrow \infty . \]\n\nProof. First note that \( I\left( \xi \right) = O\left( {\left| {\xi }_{2}\right| }^{-1/3}\right) \) . In fact\n\n\[ I\left( \xi \right) = {\int }_{\left| t\right| \leq {\left| {\xi }_{2}\right| }^{-1/3}} + {\int }_{\left| t\right| > {\left| {\xi }_{2}\right| }^{-1/3}}. \]\n\nThe first integral is obviously \( O\left( {\left| {\xi }_{2}\right| }^{-1/3}\right) \) . For the second term we use the second derivative test (Proposition 2.3 and Corollary 2.4) noting that the second derivative of the phase exceeds \( c\left| {\xi }_{2}\right| {\left| {\xi }_{2}\right| }^{-1/3} = c{\left| {\xi }_{2}\right| }^{2/3} \), so this term is also \( O\left( {\left| {\xi }_{2}\right| }^{-1/3}\right) \), which proves that \( I\left( \xi \right) = O\left( {\left| {\xi }_{2}\right| }^{-1/3}\right) \) . We apply this result when \( \left| {\xi }_{2}\right| \geq {c}^{\prime }\left| {\xi }_{1}\right| \), where \( {c}^{\prime } \) is a suitably small constant, giving \( I\left( \xi \right) = O\left( {\left| \xi \right| }^{-1/3}\right) \) in this case.\n\nIn the case when \( \left| {\xi }_{1}\right| > \left( {1/{c}^{\prime }}\right) \left| {\xi }_{2}\right| \), we apply the first derivative test (Proposition 2.1) noting that there the first derivative of the phase exceeds a multiple of \( \left| {\xi }_{1}\right| \) . Thus \( I\left( \xi \right) = O\left( {\left| {\xi }_{1}\right| }^{-1}\right) = O\left( {\left| \xi \right| }^{-1/3}\right) \) . A combination of these two cases yields the lemma.\n\nWe can now invoke Corollary 5.5 with \( \delta = 1/3 \) and obtain\n\n\[ \parallel \mathcal{R}\left( f\right) {\parallel }_{{L}^{2}\left( \Gamma \right) } \leq c\parallel f{\parallel }_{{L}^{p}\left( {\mathbb{R}}^{2}\right) } \]\n\nand\n\n\[ {\begin{Vmatrix}{\mathcal{R}}^{ * }\left( F\right) \end{Vmatrix}}_{{L}^{q}\left( {\mathbb{R}}^{2}\right) } \leq c\parallel F{\parallel }_{{L}^{2}\left( \Gamma \right) } \]\n\nfor \( p = \frac{{2\delta } + 2}{\delta + 2} = \frac{8}{7} \), and \( 1/p + 1/q = 1 \), so \( q = 8 \) . The estimate for \( {\mathcal{R}}^{ * } \) then proves our theorem. | Yes |
Lemma 6.5 Let \( I\left( \xi \right) = {\int }_{\mathbb{R}}{e}^{{2\pi i}\left( {{\xi }_{1}t + {\xi }_{2}{t}^{3}}\right) }\psi \left( t\right) {dt} \), where \( \psi \) is a \( {C}^{\infty } \) function of compact support. Then | \[ I\left( \xi \right) = O\left( {\left| \xi \right| }^{-1/3}\right) ,\;\text{ as }\left| \xi \right| \rightarrow \infty . \] Proof. First note that \( I\left( \xi \right) = O\left( {\left| {\xi }_{2}\right| }^{-1/3}\right) \) . In fact \[ I\left( \xi \right) = {\int }_{\left| t\right| \leq {\left| {\xi }_{2}\right| }^{-1/3}} + {\int }_{\left| t\right| > {\left| {\xi }_{2}\right| }^{-1/3}}. \] The first integral is obviously \( O\left( {\left| {\xi }_{2}\right| }^{-1/3}\right) \) . For the second term we use the second derivative test (Proposition 2.3 and Corollary 2.4) noting that the second derivative of the phase exceeds \( c\left| {\xi }_{2}\right| {\left| {\xi }_{2}\right| }^{-1/3} = c{\left| {\xi }_{2}\right| }^{2/3} \), so this term is also \( O\left( {\left| {\xi }_{2}\right| }^{-1/3}\right) \), which proves that \( I\left( \xi \right) = O\left( {\left| {\xi }_{2}\right| }^{-1/3}\right) \) . We apply this result when \( \left| {\xi }_{2}\right| \geq {c}^{\prime }\left| {\xi }_{1}\right| \), where \( {c}^{\prime } \) is a suitably small constant, giving \( I\left( \xi \right) = O\left( {\left| \xi \right| }^{-1/3}\right) \) in this case. In the case when \( \left| {\xi }_{1}\right| > \left( {1/{c}^{\prime }}\right) \left| {\xi }_{2}\right| \), we apply the first derivative test (Proposition 2.1) noting that there the first derivative of the phase exceeds a multiple of \( \left| {\xi }_{1}\right| \) . Thus \( I\left( \xi \right) = O\left( {\left| {\xi }_{1}\right| }^{-1}\right) = O\left( {\left| \xi \right| }^{-1/3}\right) \) . A combination of these two cases yields the lemma. | Yes |
Proposition 6.6 Suppose \( F \) is a \( {C}^{\infty } \) function on \( {\mathbb{R}}^{d} \times \mathbb{R} \) of compact support. Then \( S\left( F\right) \) is a \( {C}^{\infty } \) function that satisfies (43) and (44). | Proof. Write \( F = {e}^{{it}\bigtriangleup }G\left( {\cdot, t}\right) \) with \( G\left( {x, t}\right) = i{\int }_{0}^{t}{e}^{-{is}\bigtriangleup }F\left( {\cdot, s}\right) {ds} \) . Now \( F\left( {\cdot, s}\right) \) is in the Schwartz space \( \mathcal{S}\left( {\mathbb{R}}^{d}\right) \) for each \( s \) and depends smoothly on \( s \) . Thus the same is true for \( G\left( {\cdot, s}\right) \) and then for \( S\left( F\right) \left( {\cdot, s}\right) \), so this function is \( {C}^{\infty } \) . Now differentiate both sides of the identity\n\n\[ \n{e}^{-{it}\bigtriangleup }\left( {S\left( F\right) }\right) \left( {\cdot, t}\right) = i{\int }_{0}^{t}{e}^{-{is}\bigtriangleup }F\left( {\cdot, s}\right) {ds}, \n\]\n\nwith respect to \( t \) .\n\nThe left-hand side gives \( {e}^{-{it}\bigtriangleup }\left( {-i\bigtriangleup + \frac{\partial }{\partial t}}\right) S\left( F\right) \left( {\cdot, t}\right) \) . The right-hand side yields \( i{e}^{-{it}\bigtriangleup }F\left( {\cdot, t}\right) \) . After composing with \( {e}^{{it}\bigtriangleup } \), we see that\n\n\[ \n\left( {-i\bigtriangleup + \frac{\partial }{\partial t}}\right) S\left( F\right) \left( {\cdot, t}\right) = {iF}\left( {\cdot, t}\right) \n\]\n\nas was to be proved. Note that it is obvious that \( S\left( F\right) \left( {\cdot ,0}\right) = 0 \) . | Yes |
Proposition 6.8 If \( F \in {L}^{p}\left( {{\mathbb{R}}^{d} \times \mathbb{R}}\right) \) then \( S\left( F\right) \) can be corrected (that is, redefined on a set of measure zero) so that for each \( t, S\left( F\right) \left( {\cdot, t}\right) \) belongs to \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) and, moreover, the map \( t \mapsto S\left( F\right) \left( {\cdot, t}\right) \) is continuous in the \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) norm. | This is based on the inequality\n\n(52)\n\n\[ \n{\begin{Vmatrix}{\int }_{\alpha }^{\beta }{e}^{-{is}\bigtriangleup }F\left( \cdot, s\right) ds\end{Vmatrix}}_{{L}^{2}\left( {\mathbb{R}}^{d}\right) } \leq c\parallel F{\parallel }_{{L}^{p}\left( {{\mathbb{R}}^{d} \times \mathbb{R}}\right) }, \]\n\nwith \( c \) independent of the finite numbers \( \alpha \) and \( \beta \) .\n\nIn fact, (52) is essentially the dual statement of (40) in Theorem 6.3. We let \( g \) be any element of \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) with \( \parallel g{\parallel }_{{L}^{2}\left( {\mathbb{R}}^{d}\right) } \leq 1 \) . Then by the unitarity of \( {e}^{-{is}\bigtriangleup } \) we have\n\n\[ \n{\int }_{\alpha }^{\beta }\left( {{\int }_{{\mathbb{R}}^{d}}{e}^{-{is}\bigtriangleup }F\left( {x, s}\right) \overline{g\left( x\right) }{dx}}\right) {ds} = {\int }_{\alpha }^{\beta }\left( {{\int }_{{\mathbb{R}}^{d}}F\left( {x, s}\right) \overline{v\left( {x, s}\right) }{dx}}\right) {ds} \]\n\nwhere \( v\left( {x, s}\right) = \left( {{e}^{{is}\bigtriangleup }g}\right) \left( x\right) \) . So by (40), \( \parallel v{\parallel }_{{L}^{q}\left( {{\mathbb{R}}^{d} \times \mathbb{R}}\right) } \leq c \) and Hölder’s inequality gives\n\n\[ \n\left| {{\int }_{{\mathbb{R}}^{d}}\left( {{\int }_{\alpha }^{\beta }{e}^{-{is}\bigtriangleup }F\left( {\cdot, s}\right) {ds}}\right) \overline{g\left( x\right) }{dx}}\right| \leq c\parallel F{\parallel }_{{L}^{p}\left( {{\mathbb{R}}^{d} \times \mathbb{R}}\right) }, \]\n\nand since \( g \) was arbitrary, this suffices to establish (52).\n\nNext, since \( S\left( F\right) \left( {x, t}\right) = i{e}^{{it}\bigtriangleup }{\int }_{0}^{t}{e}^{-{is}\bigtriangleup }F\left( {\cdot, s}\right) {ds} \), taking \( \alpha = 0 \) and \( \beta = \) \( t \) in (52), we see that for each \( t \) the function \( S\left( F\right) \left( {\cdot, s}\right) \) belongs to \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) , and\n\n(53)\n\n\[ \n\mathop{\sup }\limits_{t}\parallel S\left( F\right) \left( {\cdot, t}\right) {\parallel }_{{L}^{2}\left( {\mathbb{R}}^{d}\right) } \leq c\parallel F{\parallel }_{{L}^{p}\left( {{\mathbb{R}}^{d} \times \mathbb{R}}\right) }. \]\n\nFinally, approximate \( F \) in the \( {L}^{p} \) norm by a sequence \( \left\{ {F}_{n}\right\} \) of \( {C}^{\infty } \) functions of compact support. Then for each \( n, S\left( {F}_{n}\right) \left( {\cdot, t}\right) \) is clearly continuous in \( t \) in the \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) norm. Since by (53)\n\n\[ \n\mathop{\sup }\limits_{t}{\begin{Vmatrix}S\left( F\right) \left( \cdot, t\right) - S\left( {F}_{n}\right) \left( \cdot, t\right) \end{Vmatrix}}_{{L}^{2}} \leq c{\begin{Vmatrix}F - {F}_{n}\end{Vmatrix}}_{{L}^{p}} \rightarrow 0, \]\n\nthe continuity in \( t \) carries over to \( S\left( F\right) \left( {\cdot, t}\right) \) and the proposition is proved. | Yes |
Theorem 7.1 The operator \( \mathcal{A} \) extends to a bounded linear map of \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) to \( {L}_{k}^{2}\left( {\mathbb{R}}^{d}\right) \), with \( k = \frac{d - 1}{2} \). | Null | No |
Proposition 7.2 Under the above assumptions we have \( \\begin{Vmatrix}{T}_{\\lambda }\\end{Vmatrix} \\leq c{\\lambda }^{-d/2} \) , \( \\lambda > 0 \), with \( \\parallel \\cdot \\parallel \) denoting the norm of the operator acting on \( {L}^{2}\\left( {\\mathbb{R}}^{d}\\right) \) . | The proof of the proposition is in many ways like that of the scalar version, Proposition 2.5 in Section 2, so we will be brief. As before, we begin by taking the precaution that \( \\psi \) is supported in a small ball. Now if \( T \) is an operator on \( {L}^{2} \), then \( \\begin{Vmatrix}{{T}^{ * }T}\\end{Vmatrix} = \\parallel T{\\parallel }^{2} \), where \( {T}^{ * } \) denotes the adjoint of \( T \) . \( {}^{9} \n\nHowever \( {T}_{\\lambda } \) is given by the kernel \( K\\left( {x, y}\\right) = {e}^{{i\\lambda \\Phi }\\left( {x, y}\\right) }\\psi \\left( {x, y}\\right) \), that is, \( {T}_{\\lambda }\\left( f\\right) \\left( x\\right) = \\int K\\left( {x, y}\\right) f\\left( y\\right) {dy} \), so \( {T}_{\\lambda }^{ * } \) is given by the kernel \( \\bar{K}\\left( {x, y}\\right) \) and \( {T}_{\\lambda }^{ * }{T}_{\\lambda } \) is given by the kernel\n\n\[ M\\left( {x, y}\\right) = {\\int }_{{\\mathbb{R}}^{d}}\\bar{K}\\left( {z, x}\\right) K\\left( {z, y}\\right) {dz} = {\\int }_{{\\mathbb{R}}^{d}}{e}^{{i\\lambda }\\left\\lbrack {\\Phi \\left( {z, y}\\right) - \\Phi \\left( {z, x}\\right) }\\right\\rbrack }\\psi \\left( {x, y, z}\\right) {dz}, \]\n\nwith \( \\psi \\left( {x, y, z}\\right) = \\bar{\\psi }\\left( {z, x}\\right) \\psi \\left( {z, y}\\right) \). The crucial point will be like (14), namely,\n\n\[ \\left| {M\\left( {x, y}\\right) }\\right| \\leq {c}_{N}{\\left( \\lambda \\left| x - y\\right| \\right) }^{-N}\\;\\text{ for every }N \\geq 0. \]\n\nHere, with \( z = \\left( {{z}_{1},\\ldots ,{z}_{d}}\\right) \\in {\\mathbb{R}}^{d} \), we use the vector field\n\n\[ L = \\frac{1}{i\\lambda }\\mathop{\\sum }\\limits_{{j = 1}}^{d}{a}_{j}\\frac{\\partial }{\\partial {z}_{j}} = a \\cdot {\\nabla }_{z} \]\n\nand its transpose, \( {L}^{t}\\left( f\\right) = - \\frac{1}{i\\lambda }\\mathop{\\sum }\\limits_{{j = 1}}^{d}\\frac{\\partial \\left( {{a}_{j}f}\\right) }{\\partial {z}_{j}} \), where\n\n\[ \\left( {a}_{j}\\right) = a = \\frac{{\\nabla }_{z}\\left( {\\Phi \\left( {z, x}\\right) - \\Phi \\left( {z, y}\\right) }\\right) }{{\\left| {\\nabla }_{z}\\left( \\Phi \\left( z, x\\right) - \\Phi \\left( z, y\\right) \\right) \\right| }^{2}}. \]\n\nNow because \( u = x - y \) is sufficiently small in view of the support assumptions made on \( \\psi \), we see as before that \( \\left| a\\right| \\approx {\\left| x - y\\right| }^{-1} \) and \( \\left| {{\\partial }_{x}^{\\alpha }a}\\right| \\lesssim \) \( {\\left| x - y\\right| }^{-1} \), for all \( \\alpha \). Thus\n\n\[ \\left| {M\\left( {x, y}\\right) }\\right| \\leq \\left| {\\int {L}^{N}\\left( {e}^{{i\\lambda }\\left\\lbrack {\\Phi \\left( {z, y}\\right) - \\Phi \\left( {z, x}\\right) }\\right\\rbrack }\\right) \\psi \\left( {x, y, z}\\right) {dz}}\\right| \]\n\n\[ \\leq \\int \\left| {{\\left( {L}^{t}\\right) }^{N}\\psi \\left( {x, y, z}\\right) } | Yes |
Proposition 7.4 Assume that\n\n\[ \n\begin{Vmatrix}{{T}_{k}{T}_{j}^{ * }}\end{Vmatrix} \leq {a}^{2}\left( {k - j}\right) \;\text{ and }\;\begin{Vmatrix}{{T}_{k}^{ * }{T}_{j}}\end{Vmatrix} \leq {a}^{2}\left( {k - j}\right) .\n\]\n\nThen for every \( r \) ,\n\n(72)\n\n\[ \n\begin{Vmatrix}{\mathop{\sum }\limits_{{k = 0}}^{r}{T}_{k}}\end{Vmatrix} \leq A\n\]\n\nThe thrust of this proposition is of course that the bound \( A \) is independent of \( r \) . | Proof. We write \( T = \mathop{\sum }\limits_{{k = 0}}^{r}{T}_{k} \) and recall that \( \parallel T{\parallel }^{2} = \begin{Vmatrix}{T{T}^{ * }}\end{Vmatrix} \) . Since \( T{T}^{ * } \) is self-adjoint we may use this identity repeatedly to obtain \( \parallel T{\parallel }^{2n} = \) \( \begin{Vmatrix}{\left( T{T}^{ * }\right) }^{n}\end{Vmatrix} \) ,(at least when \( n \) is of the form \( n = {2}^{s} \) for some integer \( s \) ). Now\n\n\[ \n{\left( T{T}^{ * }\right) }^{n} = \mathop{\sum }\limits_{{{i}_{1},{i}_{2},\ldots ,{i}_{2n}}}{T}_{{i}_{1}}{T}_{{i}_{2}}^{ * }\cdots {T}_{{i}_{{2n} - 1}}{T}_{{i}_{2n}}^{ * }.\n\]\n\nWe make two estimates for the norm of each term in the above sum. First\n\n\[ \n\begin{Vmatrix}{{T}_{{i}_{1}}{T}_{{i}_{2}}^{ * }\cdots {T}_{{i}_{{2n} - 1}}{T}_{{i}_{2n}}^{ * }}\end{Vmatrix} \leq {a}^{2}\left( {{i}_{1} - {i}_{2}}\right) {a}^{2}\left( {{i}_{3} - {i}_{4}}\right) \cdots {a}^{2}\left( {{i}_{{2n} - 1} - {i}_{2n}}\right) ,\n\]\n\nwhich is obtained by associating the product as \( \left( {{T}_{{i}_{1}}{T}_{{i}_{2}}^{ * }}\right) \cdots \left( {{T}_{{i}_{{2n} - 1}}{T}_{{i}_{2n}}^{ * }}\right) \) . Next\n\n\[ \n\begin{Vmatrix}{{T}_{{i}_{1}}{T}_{{i}_{2}}^{ * }\cdots {T}_{{i}_{{2n} - 1}}{T}_{{i}_{2n}}^{ * }}\end{Vmatrix} \leq {A}^{2}{a}^{2}\left( {{i}_{2} - {i}_{3}}\right) {a}^{2}\left( {{i}_{4} - {i}_{5}}\right) \cdots {a}^{2}\left( {{i}_{{2n} - 2} - {i}_{{2n} - 1}}\right) ,\n\]\n\nwhich is obtained by associating the product as \( {T}_{{i}_{1}}\left( {{T}_{{i}_{2}}^{ * }{T}_{{i}_{3}}}\right) \cdots \) \( \left( {{T}_{{i}_{{2n} - 2}}^{ * }{T}_{{i}_{{2n} - 1}}}\right) {T}_{{i}_{2n}}^{ * } \), and using the fact that \( {T}_{{i}_{1}} \) and \( {T}_{{i}_{2n}}^{ * } \) are both bounded by \( A \) . Taking the geometric mean of these estimates yields\n\n\[ \n\begin{Vmatrix}{{T}_{{i}_{1}}{T}_{{i}_{2}}^{ * }\cdots {T}_{{i}_{{2n} - 1}}{T}_{{i}_{2n}}^{ * }}\end{Vmatrix} \leq {Aa}\left( {{i}_{1} - {i}_{2}}\right) a\left( {{i}_{2} - {i}_{3}}\right) \cdots a\left( {{i}_{{2n} - 1} - {i}_{2n}}\right) .\n\]\n\nNow we sum this first in \( {i}_{1} \), then \( {i}_{2} \), and so on, until \( {i}_{{2n} - 1} \), obtaining a further factor of \( A \) each time, because \( A = \sum a\left( k\right) \) . When we sum in \( {i}_{2n} \) we use the fact that there are \( r + 1 \) terms in the sum. The result is then \( \parallel T{\parallel }^{2n} \leq {A}^{2n}\left( {r + 1}\right) \) . Taking the \( {\left( 2n\right) }^{\text{th }} \) root and letting \( n \rightarrow \infty \) gives (72) and the proposition. | Yes |
Proposition 8.1 \( \mathop{\sum }\limits_{{k = 1}}^{\mu }{r}_{2}\left( k\right) = {\pi \mu } + O\left( {\mu }^{1/2}\right) \), as \( \mu \rightarrow \infty \) . | The proof depends on the realization that \( \mathop{\sum }\limits_{{k = 0}}^{\mu }{r}_{2}\left( k\right) \) represents the number of lattice points in the disc of radius \( R \) with \( {R}^{2} = \mu \) . In fact, with \( {\mathbb{Z}}^{2} \) denoting the lattice points in \( {\mathbb{R}}^{2} \), that is, the points in \( {\mathbb{R}}^{2} \) with integral coordinates, then \( {r}_{2}\left( k\right) = \# \left\{ {\left( {{n}_{1},{n}_{2}}\right) \in {\mathbb{Z}}^{2} : k = {n}_{1}^{2} + {n}_{2}^{2}}\right\} \), and hence\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{\mu }{r}_{2}\left( k\right) = \# \left\{ {\left( {{n}_{1},{n}_{2}}\right) \in {\mathbb{Z}}^{2} : {n}_{1}^{2} + {n}_{2}^{2} \leq {R}^{2}}\right\} .\n\nSo if \( N\left( R\right) \) is the quantity above, then the proposition is equivalent to\n\n(79)\n\n\[ N\left( R\right) = \pi {R}^{2} + O\left( R\right) ,\;\text{ as }R \rightarrow \infty .\n\nTo prove this we write \( {D}_{R} \) for the closed disc \( \left\{ {x \in {\mathbb{R}}^{2} : \left| x\right| \leq R}\right\} \), and let \( {\widetilde{D}}_{R} \) be the rectangular region that is the union of unit squares centered at points \( n \in {\mathbb{Z}}^{2} \) with \( n \in {D}_{R} \), that is,\n\n\[ {\widetilde{D}}_{R} = \mathop{\bigcup }\limits_{{\left| n\right| \leq R, n \in {\mathbb{Z}}^{2}}}\left( {S + n}\right)\n\nwith \( S = \left\{ {x = \left( {{x}_{1},{x}_{2}}\right) : - 1/2 \leq {x}_{i} < 1/2,\;i = 1,2}\right\} .\n\nSince the squares \( S + N \) are mutually disjoint and each has area 1, we see that \( m\left( {\widetilde{D}}_{R}\right) = N\left( R\right) \) . However\n\n(80)\n\n\[ {D}_{R - {2}^{-1/2}} \subset {\widetilde{D}}_{R} \subset {D}_{R + {2}^{-1/2}}.\n\nIn fact if \( x \in S + n \) with \( \left| n\right| \leq R \), then \( \left| x\right| \leq {2}^{-1/2} + \left| n\right| \leq R + {2}^{-1/2} \), so \( {\widetilde{D}}_{R} \subset {D}_{R + {2}^{-1/2}} \) . The reverse inclusion can be proved the same way. It follows from (80) that\n\n\[ m\left( {D}_{R - {2}^{-1/2}}\right) \leq m\left( {\widetilde{D}}_{R}\right) \leq m\left( {D}_{R + {2}^{-1/2}}\right) ,\n\nand hence\n\n\[ \pi {\left( R - {2}^{-1/2}\right) }^{2} \leq N\left( R\right) \leq \pi {\left( R + {2}^{-1/2}\right) }^{2},\n\nproving that \( N\left( R\right) = \pi {R}^{2} + O\left( R\right) \). | Yes |
Proposition 8.2 Suppose \( f \) belongs to the Schwartz space \( \mathcal{S}\left( {\mathbb{R}}^{d}\right) \). Then\n\n\[ \mathop{\sum }\limits_{{n \in {\mathbb{Z}}^{d}}}f\left( n\right) = \mathop{\sum }\limits_{{n \in {\mathbb{Z}}^{d}}}\widehat{f}\left( n\right) \]\n\nHere \( {\mathbb{Z}}^{d} \) denotes the collection of lattice points in \( {\mathbb{R}}^{d} \), the points with integral coordinates, and \( \widehat{f} \) is the Fourier transform of \( f \). | For the proof consider two sums\n\n\[ \mathop{\sum }\limits_{{n \in {\mathbb{Z}}^{d}}}f\left( {x + n}\right) \;\text{ and }\;\mathop{\sum }\limits_{{n \in {\mathbb{Z}}^{d}}}\widehat{f}\left( n\right) {e}^{{2\pi in} \cdot x}. \]\n\nBoth are rapidly converging series (since \( f \) and \( \widehat{f} \) are in \( \mathcal{S}\left( {\mathbb{R}}^{d}\right) \) ), and hence both these sums are continuous functions. Moreover each is periodic, that is, each is unchanged when \( x \) is replaced by \( x + m \), for any \( m \in {\mathbb{Z}}^{d} \). For the sum \( \mathop{\sum }\limits_{{n \in {\mathbb{Z}}^{d}}}f\left( {x + n}\right) \) this is clear, because replacing \( x \) by \( x + m \) merely reshuffles the sum. Also the second sum is unchanged, because of the periodicity of \( {e}^{{2\pi in} \cdot x} \) for each \( n \in {\mathbb{Z}}^{d} \). Moreover both sums have the same Fourier coefficients. To see this, let \( Q \) be the fundamental cube\n\n\[ Q = \left\{ {x \in {\mathbb{R}}^{d} : 0 < {x}_{j} \leq 1, j = 1,\ldots, d}\right\}, \]\n\nand fix any \( m \in {\mathbb{Z}}^{d} \). Then\n\n\[ {\int }_{Q}\left( {\mathop{\sum }\limits_{n}f\left( {x + n}\right) }\right) {e}^{-{2\pi im} \cdot x}{dx} = \mathop{\sum }\limits_{n}{\int }_{Q + n}f\left( x\right) {e}^{-{2\pi im} \cdot x}{dx} \]\n\n\[ = {\int }_{{\mathbb{R}}^{d}}f\left( x\right) {e}^{-{2\pi im} \cdot x}{dx} \]\n\n\[ = \widehat{f}\left( m\right) \]\n\nsince \( \mathop{\bigcup }\limits_{{n \in {\mathbb{Z}}^{d}}}\left( {Q + n}\right) \) is a partition of \( {\mathbb{R}}^{d} \) into cubes \( \{ Q + n{\} }_{n \in {\mathbb{Z}}^{d}} \). Moreover\n\n\[ {\int }_{Q}\left( {\mathop{\sum }\limits_{n}\widehat{f}\left( n\right) {e}^{{2\pi in} \cdot x}}\right) {e}^{-{2\pi im} \cdot x}{dx} = \widehat{f}\left( m\right) ,\]\n\nbecause \( {\int }_{Q}{e}^{{2\pi in} \cdot x}{e}^{-{2\pi im} \cdot x}{dx} = 1 \) if \( n = m \), and is 0 otherwise. Since \( \mathop{\sum }\limits_{n}f\left( {x + n}\right) = \mathop{\sum }\limits_{n}\widehat{f}\left( n\right) {e}^{2\pi inx} \) have the same Fourier coefficients, these functions must be equal, and setting \( x = 0 \) gives us (82). | Yes |
Theorem 8.3 \( N\left( R\right) = \pi {R}^{2} + O\left( {R}^{2/3}\right) \), as \( R \rightarrow \infty \) . | Proof. We replace the characteristic function \( {\chi }_{R} \) by a regularized version as follows. We fix a non-negative \ | No |
Theorem 8.5\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\mu }d\left( k\right) = \mu \log \mu + \left( {{2\gamma } - 1}\right) \mu + O\left( {{\mu }^{1/3}\log \mu }\right) \;\text{ as }\mu \rightarrow \infty .{}^{16} \] | Now as much as we might wish to follow the lines of the proof of Theorem 8.3, there are serious obstacles that seem to stand in the way. In fact, if \( {\chi }_{\mu } \) is the characteristic function of the region\n\n(90)\n\n\[ \left\{ {\left( {{x}_{1},{x}_{2}}\right) \in {\mathbb{R}}^{2} : {x}_{1}{x}_{2} \leq \mu ,{x}_{1} > 0,\text{ and }{x}_{2} > 0}\right\} \]\n\nwhich consists of the area on or below the hyperbola \( {x}_{1}{x}_{2} = \mu \), then indeed\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\mu }d\left( k\right) = \mathop{\sum }\limits_{{n \in {\mathbb{Z}}^{2}}}{\chi }_{\mu }\left( n\right) \]\n\nHowever the other side of the Poisson summation formula (82) for \( f = {\chi }_{\mu } \) is problematic as it stands. In fact, \( {\widehat{\chi }}_{\mu }\left( 0\right) = {\int }_{{\mathbb{R}}^{2}}{\chi }}_{\mu }{dx} = \infty \), and for the same reason the integral giving each term \( {\widehat{\chi }}_{\mu }\left( n\right) \) is not well-defined.\n\nA further issue is that the main term in (89) is \( \mu \log \mu \), while a simple scaling of the region (90) would suggest rather a term linear in \( \mu \) . Connected with this is the mysterious occurrence of Euler’s constant \( \gamma \) in the subsidiary term. | No |
Corollary 8.7 The conclusions for \( {\mathfrak{J}}_{a, b}^{ - } \) are the same as those for \( {\mathfrak{J}}_{a, b}^{ + } \) stated in Proposition 8.6, except that (i) should be modified to read that uniformly in \( a, b \) , \[ \left( {\mathrm{i}}^{\prime }\right) {\mathfrak{J}}_{a, b}^{ - } = O\left( {\left| \lambda \right| }^{-N}\right) \text{for}\left| \lambda \right| \geq 1/2\text{, for every}N \geq 0\text{.} \] | The only change occurs in the treatment of \( {II} \), namely \( \int {e}^{{i\lambda \Phi }\left( u\right) }\alpha \left( u\right) \frac{du}{u} \) , where now \( \Phi \left( u\right) = u - 1/u \) . In this case \( {\Phi }^{\prime }\left( u\right) = 1 + 1/{u}^{2} > 1 \), and there is no critical point. So Proposition 2.1 implies that \( {II} = O\left( {\left| \lambda \right| }^{-N}\right) \) for every \( N \geq 0 \), and then conclusion \( \left( {\mathrm{i}}^{\prime }\right) \) follows by the arguments we have used for \( I \) and \( {III} \) previously. | Yes |
Theorem 8.9 Let \( \widehat{f} \) be the Fourier transform of \( f\left( {x, y}\right) = {f}_{0}\left( {xy}\right) \) . Then \( \widehat{f} \) is a continuous function where \( {\xi \eta } \neq 0 \) . It is given by\n\n\[ \widehat{f}\left( {\xi ,\eta }\right) = 2{\int }_{0}^{\infty }{\mathfrak{J}}^{ + }\left( {-{2\pi }{\left| \xi \eta \right| }^{1/2}\rho }\right) {f}_{0}\left( {\rho }^{2}\right) {\rho d\rho } \]\n\nfor \( \left( {\xi ,\eta }\right) \in {Q}_{1} \) . In \( {Q}_{2},{Q}_{3} \), and \( {Q}_{4} \) it is given by the analogous formulas, with \( {\mathfrak{J}}^{ + }\left( {- \cdot }\right) \) replaced by \( {\mathfrak{J}}^{ - }\left( {- \cdot }\right) ,{\mathfrak{J}}^{ + }\left( {+ \cdot }\right) \) and \( {\mathfrak{J}}^{ - }\left( {+ \cdot }\right) \), respectively. | Proof. We approximate \( f \) by \( {f}_{\epsilon } \), with \( {f}_{\epsilon }\left( {x, y}\right) = {f}_{0}\left( {xy}\right) {\eta }_{\epsilon }\left( x\right) {\eta }_{\epsilon }\left( y\right) \) . Then each \( {f}_{\epsilon } \) is a \( {C}^{\infty } \) function of compact support, and clearly \( {f}_{\epsilon } \rightarrow f \) in the sense of tempered distributions.\n\nNow\n\n\[ {\widehat{f}}_{\epsilon }\left( {\xi ,\eta }\right) = \int {e}^{-{2\pi i}\left( {{\xi x} + {\eta y}}\right) }{f}_{0}\left( {xy}\right) {\eta }_{\epsilon }\left( x\right) {\eta }_{\epsilon }\left( y\right) {dxdy}. \]\n\nWe introduce new variables \( \left( {u,\rho }\right) \) in the first quadrant with \( x = {u\rho } \) , \( y = \frac{\rho }{u} \), and observe that\n\n\[ \frac{\partial \left( {x, y}\right) }{\partial \left( {u,\rho }\right) } = \left( \begin{matrix} \rho & u \\ - \frac{\rho }{{u}^{2}} & \frac{1}{u} \end{matrix}\right) \]\n\nwhich has a determinant equal to \( {2\rho }/u \) . Therefore \( {dxdy} = {2\rho }\frac{du}{u}{d\rho } \) and\n\n\[ {\widehat{f}}_{\epsilon }\left( {\xi ,\eta }\right) = 2{\int }_{0}^{\infty }{\int }_{0}^{\infty }{e}^{-{2\pi i}\left( {{\xi u\rho } + {\eta \rho }/u}\right) }{f}_{0}\left( {\rho }^{2}\right) {\eta }_{\epsilon }\left( {\rho u}\right) {\eta }_{\epsilon }\left( {\rho /u}\right) \rho \frac{du}{u}{d\rho }. \]\n\nAgain, if \( \left( {\xi ,\eta }\right) \) is in the first quadrant and if we make the change of variables \( u \mapsto {\left( \eta /\xi \right) }^{1/2}u \), then we have\n\n\[ {\widehat{f}}_{\epsilon }\left( {\xi ,\eta }\right) = 2{\int }_{0}^{\infty }{\mathfrak{J}}_{a, b}^{ + }\left( {-{2\pi }{\left| \xi \eta \right| }^{1/2}\rho }\right) {f}_{0}\left( {\rho }^{2}\right) {\rho d\rho } \]\n\nwith now \( a = \frac{\epsilon }{\rho }{\left( \frac{\xi }{\eta }\right) }^{1/2} \) and \( b = \frac{\epsilon }{\rho }{\left( \frac{\eta }{\xi }\right) }^{1/2} \).\n\nThe analogous formulas for \( {\widehat{f}}_{\epsilon }\left( {\xi ,\eta }\right) \) hold when \( \left( {\xi ,\eta }\right) \) are in the second, third and fourth quadrants. So the fact that \( {\widehat{f}}_{\epsilon } \) converges in the sense of tempered distributions to the limit \( \widehat{f} \) given by (92) then follows by the same reasoning used in the proof of Proposition 8.8. | Yes |
Corollary 8.10 The Fourier transforms \( {\widehat{f}}_{\epsilon } \) and \( \widehat{f} \) satisfy the following estimate, uniformly in \( \epsilon \) :\n\n\[ \left| {{\widehat{f}}_{\epsilon }\left( {\xi ,\eta }\right) }\right| \leq {A}_{N}{\left| \xi \eta \right| }^{-N}\;\text{ when }\left| {\xi \eta }\right| \geq 1/2, \] \n\nfor every \( N \geq 0 \) . | This is a consequence of the asymptotic behavior of \( {\mathfrak{J}}^{ \pm }\left( \lambda \right) \) for \( \lambda \) as given in Proposition 8.6 and its corollary together with the fact that \( {\int }_{0}^{\infty }{e}^{-{4\pi i\rho }{\left| \xi \eta \right| }^{1/2}}{f}_{0}\left( {\rho }^{2}\right) {\rho d\rho } \) is \( O\left( {\left| \xi \eta \right| }^{-N}\right) \) for every \( N \geq 0 \), since \( {f}_{0}\left( {\rho }^{2}\right) \rho \) is a \( {C}^{\infty } \) function with compact support in \( \left( {0,\infty }\right) \) . | Yes |
Theorem 8.11\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }{f}_{0}\left( k\right) d\left( k\right) = {\int }_{0}^{\infty }\left( {\log \rho + {2\gamma }}\right) {f}_{0}\left( \rho \right) {d\rho } + \mathop{\sum }\limits_{{k = 1}}^{\infty }{F}_{0}\left( k\right) d\left( k\right) ,\] | Proof. We apply the Poisson summation formula\n\n\[ \mathop{\sum }\limits_{{\mathbb{Z}}^{2}}{f}_{\epsilon }\left( {m, n}\right) = \mathop{\sum }\limits_{{\mathbb{Z}}^{2}}{\widehat{f}}_{\epsilon }\left( {m, n}\right) \]\n\nto the approximating functions \( {f}_{\epsilon } \), and then pass to the limit as \( \epsilon \rightarrow 0 \) . Now the sum on the left-hand side is clearly taken over a bounded set of\n\n--- | No |
Theorem 2.6.1 Any collection of subsets of \( I \) that has the finite intersection property can be extended to an ultrafilter on \( I \) . | Proof. If \( \mathcal{H} \) has the fip, then the filter \( {\mathcal{F}}^{\mathcal{H}} \) generated by \( \mathcal{F} \) is proper \( \left( {{2.5}\left( 7\right) }\right) \) . Let \( P \) be the collection of all proper filters on \( I \) that include \( {\mathcal{F}}^{\mathcal{H}} \) , partially ordered by set inclusion \( \subseteq \) . Then every linearly ordered subset of \( P \) has an upper bound in \( P \), since by \( {2.4}\left( 4\right) \) the union of this chain is in \( P \) . Hence by Zorn’s lemma \( P \) has a maximal element, which is thereby a maximal proper filter on \( I \) and thus an ultrafilter by \( {2.5}\left( 6\right) \) . | Yes |
Corollary 2.6.2 Any infinite set has a nonprincipal ultrafilter on it. | Proof. If \( I \) is infinite, the cofinite filter \( {\mathcal{F}}^{co} \) is proper and has the finite intersection property, and so is included in an ultrafilter \( \mathcal{F} \) . But for any \( i \in I \) we have \( I - \{ i\} \in {\mathcal{F}}^{co} \subseteq \mathcal{F} \), so \( \{ i\} \notin \mathcal{F} \), whereas \( \{ i\} \in {\mathcal{F}}^{i} \) . Hence \( \mathcal{F} \neq {\mathcal{F}}^{i} \) . Thus \( \mathcal{F} \) is nonprincipal. | Yes |
Theorem 3.6.1 The structure \( \langle * \mathbb{R}, + , \cdot , < \rangle \) is an ordered field with zero [0] and unity \( \left\lbrack \mathbf{1}\right\rbrack \) . | Proof. (Sketch) As a quotient ring of \( {\mathbb{R}}^{\mathbb{N}} \) ,* \( \mathbb{R} \) is readily shown to be a commutative ring with zero [0] and unity [1], and additive inverses given by\n\n\[ - \left\lbrack \left\langle {{r}_{n} : n \in \mathbb{N}}\right\rangle \right\rbrack = \left\lbrack \left\langle {-{r}_{n} : n \in \mathbb{N}}\right\rangle \right\rbrack ,\]\n\nor more briefly, \( - \left\lbrack {r}_{n}\right\rbrack = \left\lbrack {-{r}_{n}}\right\rbrack \) . To show that it has multiplicative inverses, suppose \( \left\lbrack r\right\rbrack \neq \left\lbrack \mathbf{0}\right\rbrack \) . Then \( r ≢ \mathbf{0} \), i.e., \( \left\{ {n \in \mathbb{N} : {r}_{n} = 0}\right\} \notin \mathcal{F} \), so as \( \mathcal{F} \) is an ultrafilter, \( J = \left\{ {n \in \mathbb{N} : {r}_{n} \neq 0}\right\} \in \mathcal{F} \) . Define a sequence \( s \) by putting\n\n\[ {s}_{n} = \left\{ \begin{array}{ll} \frac{1}{{r}_{n}} & \text{if }n \in J \\ 0 & \text{otherwise.} \end{array}\right.\]\n\nThen \( \llbracket r \odot s = 1\rrbracket \) is equal to \( J \), so \( \llbracket r \odot s = 1\rrbracket \in \mathcal{F} \), giving \( r \odot s \equiv 1 \) and hence\n\n\[ \left\lbrack r\right\rbrack \cdot \left\lbrack s\right\rbrack = \left\lbrack {r \odot s}\right\rbrack = \left\lbrack 1\right\rbrack \]\n\nin * \( \mathbb{R} \) . But this means that \( \left\lbrack s\right\rbrack \) is the multiplicative inverse \( {\left\lbrack r\right\rbrack }^{-1} \) of \( \left\lbrack r\right\rbrack \) .\n\nTo see that the ordering \( < \) on \( {}^{ * }\mathbb{R} \) is linear, observe that \( \mathbb{N} \) is the disjoint union of the three sets\n\n\[ \llbracket r < s\rrbracket ,\;\llbracket r = s\rrbracket ,\;\llbracket s < r\rrbracket ,\]\n\nso exactly one of the three belongs to \( \mathcal{F} \) (by \( {2.5}\left( 4\right) \) ), and so exactly one of\n\n\[ \left\lbrack r\right\rbrack < \left\lbrack s\right\rbrack ,\;\left\lbrack r\right\rbrack = \left\lbrack s\right\rbrack ,\;\left\lbrack s\right\rbrack < \left\lbrack r\right\rbrack \]\n\nis true. It remains to show that the set \( \left\{ {\left\lbrack r\right\rbrack : \left\lbrack 0\right\rbrack < \left\lbrack r\right\rbrack }\right\} \) of \ | No |
Theorem 3.7.1 The map \( r \mapsto {}^{ * }r \) is an order-preserving field isomorphism from \( \mathbb{R} \) into * \( \mathbb{R} \) . | Null | No |
Theorem 3.9.1 Any infinite subset of \( \mathbb{R} \) has nonstandard members. | Proof. Note first that this result must depend on \( \mathcal{F} \) being nonprincipal, because if \( \mathcal{F} \) were principal, there would be no nonstandard elements of \( {}^{ * }\mathbb{R} \) at all.\n\nNow, if \( A \subseteq \mathbb{R} \) is infinite, then there is a sequence \( r \) of elements of \( A \) whose terms are all distinct. Then \( \llbracket r \in A\rrbracket = \mathbb{N} \in \mathcal{F} \), so \( \left\lbrack r\right\rbrack \in {}^{ * }A \) . But for each \( s \in A,\left\{ {n : {r}_{n} = s}\right\} \) is either \( \varnothing \) or a singleton, neither of which can belong to \( \mathcal{F}\left( {{2.5}\left( 5\right) }\right) \), so \( \left\lbrack r\right\rbrack \neq {}^{ * }s \) . Hence \( \left\lbrack r\right\rbrack \in {}^{ * }A - A \) . | Yes |
Theorem 5.6.1 Every limited hyperreal \( b \) is infinitely close to exactly one real number, called the shadow of \( b \), denoted by \( \operatorname{sh}\left( b\right) \) . | Proof. Let \( A = \{ r \in \mathbb{R} : r < b\} \) . Since \( b \) is limited, there exist real \( r, s \) with \( r < b < s \), so \( A \) is nonempty and bounded above in \( \mathbb{R} \) by \( s \) . By the completeness of \( \mathbb{R} \), it follows that \( A \) has a least upper bound \( c \in \mathbb{R} \) .\n\nTo show \( b \simeq c \), take any positive real \( \varepsilon \in \mathbb{R} \) . Since \( c \) is an upper bound of \( A \), we cannot have \( c + \varepsilon \in A \) ; hence \( b \leq c + \varepsilon \) . Also, if \( b \leq c - \varepsilon \), then \( c - \varepsilon \) would be an upper bound of \( A \), contrary to the fact that \( c \) is the smallest such upper bound. Hence \( b \nleqslant c - \varepsilon \) . Altogether then, \( c - \varepsilon < b \leq c + \varepsilon \), so \( \left| {b - c}\right| \leq \varepsilon \) . Since this holds for all positive real \( \varepsilon, b \) is infinitely close to \( c \) .\n\nFinally, for uniqueness, if \( b \simeq {c}^{\prime } \in \mathbb{R} \), then as \( b \simeq c \), we get \( c \simeq {c}^{\prime } \), and so \( c = {c}^{\prime } \), since both are real. | Yes |
Theorem 5.6.2 If \( b \) and \( c \) are limited and \( n \in \mathbb{N} \), then\n\n(1) \( \operatorname{sh}\left( {b \pm c}\right) = \operatorname{sh}\left( b\right) \pm \operatorname{sh}\left( c\right) \) ,\n\n(2) \( \operatorname{sh}\left( {b \cdot c}\right) = \operatorname{sh}\left( b\right) \cdot \operatorname{sh}\left( c\right) \) ,\n\n(3) \( \operatorname{sh}\left( {b/c}\right) = \operatorname{sh}\left( b\right) /\operatorname{sh}\left( c\right) \) if \( \operatorname{sh}\left( c\right) \neq 0 \) (i.e., if \( c \) is appreciable),\n\n(4) \( \operatorname{sh}\left( {b}^{n}\right) = \operatorname{sh}{\left( b\right) }^{n} \) ,\n\n(5) \( \operatorname{sh}\left( \left| b\right| \right) = \left| {\operatorname{sh}\left( b\right) }\right| \) ,\n\n(6) \( \operatorname{sh}\left( \sqrt[n]{b}\right) = \sqrt[n]{\operatorname{sh}\left( b\right) } \) if \( b \geq 0 \) ,\n\n(7) if \( b \leq c \) then \( \operatorname{sh}\left( b\right) \leq \operatorname{sh}\left( c\right) \) . | Proof. Exercise. | No |
Theorem 5.6.3 The quotient ring \( \mathbb{L}/\mathbb{I} \) is isomorphic to the real number field \( \mathbb{R} \) by the correspondence \( \operatorname{hal}\left( b\right) \mapsto \operatorname{sh}\left( b\right) \) . Hence \( \mathbb{I} \) is a maximal ideal of the ring \( \mathbb{L} \) . | Null | No |
Theorem 6.1.1 A real-valued sequence \( \left\langle {{s}_{n} : n \in \mathbb{N}}\right\rangle \) converges to \( L \in \mathbb{R} \) if and only if \( {s}_{n} \simeq L \) for all unlimited \( n \) . | Proof. Suppose \( \left\langle {{s}_{n} : n \in \mathbb{N}}\right\rangle \) converges to \( L \), and fix an \( N \in {}^{ * }{\mathbb{N}}_{\infty } \) . In order to show that \( {s}_{N} \simeq L \) we have to show that \( \left| {{s}_{N} - L}\right| < \varepsilon \) for any positive real \( \varepsilon \) . But given such an \( \varepsilon \), the standard convergence condition implies that there is an \( {m}_{\varepsilon } \in \mathbb{N} \) such that the standard tail beyond \( {s}_{{m}_{\varepsilon }} \) is within \( \varepsilon \) of \( L \) :\n\n\[ \left( {\forall n \in \mathbb{N}}\right) \left( {n > {m}_{\varepsilon } \rightarrow \left| {{s}_{n} - L}\right| < \varepsilon }\right) .\n\]\n\nThen by (universal) transfer this holds for the extended tail as well:\n\n\[ \left( {\forall n \in {}^{ * }\mathbb{N}}\right) \left( {n > {m}_{\varepsilon } \rightarrow \left| {{s}_{n} - L}\right| < \varepsilon }\right) .\n\]\n\nBut in fact, \( N > {m}_{\varepsilon } \) because \( N \) is unlimited and \( {m}_{\varepsilon } \) is limited, and so this last sentence implies \( \left| {{s}_{N} - L}\right| < \varepsilon \) as desired.\n\nFor the converse, suppose \( {s}_{n} \simeq L \) for all unlimited \( n \) . We have to show that any given interval \( \left( {L - \varepsilon, L + \varepsilon }\right) \) in \( \mathbb{R} \) contains some standard tail of the sequence. The essence of the argument is to invoke the fact that the extended tail is infinitely close to \( L \), hence contained in \( {}^{ * }\left( {L - \varepsilon, L + \varepsilon }\right) \), and then apply transfer.\n\nTo spell this out, fix an unlimited \( N \in {}^{ * }{\mathbb{N}}_{\infty } \) . Then for any \( n \in {}^{ * }\mathbb{N} \) , if \( n > N \), it follows that \( n \) is also unlimited, so \( {s}_{n} \simeq L \) and therefore \( \left| {{s}_{n} - L}\right| < \varepsilon \) . This shows that\n\n\[ \left( {\forall n \in {}^{ * }\mathbb{N}}\right) \left( {n > N \rightarrow \left| {{s}_{n} - L}\right| < \varepsilon }\right) .\n\]\n\nHence the sentence\n\n\[ \left( {\exists z \in {}^{ * }\mathbb{N}}\right) \left( {\forall n \in {}^{ * }\mathbb{N}}\right) \left( {n > z \rightarrow \left| {{s}_{n} - L}\right| < \varepsilon }\right)\n\]\n\nis true. But this is the \( * \) -transform of\n\n\[ \left( {\exists z \in \mathbb{N}}\right) \left( {\forall n \in \mathbb{N}}\right) \left( {n > z \rightarrow \left| {{s}_{n} - L}\right| < \varepsilon }\right) ,\n\]\n\nso by (existential) transfer the latter holds true, giving the desired conclusion.\n\nThus convergence to \( L \) amounts to the requirement that the extended tail of the sequence is contained in the halo of \( L \) . In this characterisation the role of the standard tails is taken over by the extended tail, while the standard open neighbourhoods \( \left( {L - \varepsilon, L + \varepsilon }\right) \) are replaced by the \ | Yes |
Theorem 6.2.1 A real-valued sequence \( \left\langle {{s}_{n} : n \in \mathbb{N}}\right\rangle \) converges in \( \mathbb{R} \) if either\n\n(1) it is bounded above in \( \mathbb{R} \) and nondecreasing: \( {s}_{1} \leq {s}_{2} \leq \cdots \) ; or\n\n(2) it is bounded below in \( \mathbb{R} \) and nonincreasing: \( {s}_{1} \geq {s}_{2} \geq \cdots \) . | Proof. Consider case (1). Let \( {s}_{N} \) be an extended term. We will show that \( {s}_{N} \) has a shadow, and that this shadow is a least upper bound of the set \( \left\{ {{s}_{n} : n \in \mathbb{N}}\right\} \) in \( \mathbb{R} \) . Since a set can have only one least upper bound, this implies that all extended terms have the same shadow, and so by Theorem 6.1.1 the original sequence converges to this shadow in \( \mathbb{R} \) .\n\nNow, by hypothesis there is a real number \( b \) that is an upper bound for \( \left\{ {{s}_{n} : n \in \mathbb{N}}\right\} \) . Then the statement \( {s}_{1} \leq {s}_{n} \leq b \) holds for all \( n \in \mathbb{N} \), so it holds for all \( n \in {}^{ * }\mathbb{N} \) by universal transfer. In particular, \( {s}_{1} \leq {s}_{N} \leq b \) , showing that \( {s}_{N} \) is limited, so indeed has a shadow \( L \) .\n\nNext we show that \( L \) is an upper bound of the real sequence. Since this sequence is nondecreasing, by universal transfer we have\n\n\[ n \leq m \rightarrow {s}_{n} \leq {s}_{m} \]\n\nfor all \( n, m \in * \mathbb{N} \) . In particular, if \( n \in \mathbb{N} \), then \( n \leq N \), so \( {s}_{n} \leq {s}_{N} \simeq L \) , giving \( {s}_{n} \leq L \), as both numbers are real.\n\nFinally, we show that \( L \) is the least upper bound in \( \mathbb{R} \) . For if \( r \) is any real upper bound of \( \left\{ {{s}_{n} : n \in \mathbb{N}}\right\} \), then by transfer, \( {s}_{n} \leq r \) for all \( n \in * \mathbb{N} \) , so \( L \simeq {s}_{N} \leq r \), giving \( L \leq r \), as both are real. | Yes |
Theorem 6.3.1 If \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} = L \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{t}_{n} = M \) in \( \mathbb{R} \), then\n\n(1) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{s}_{n} + {t}_{n}}\right) = L + M \) ,\n\n(2) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {c{s}_{n}}\right) = {cL} \), for any \( c \in \mathbb{R} \) ,\n\n(3) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{s}_{n}{t}_{n}}\right) = {LM} \) ,\n\n(4) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{s}_{n}/{t}_{n}}\right) = L/M \), if \( M \neq 0 \) . | Proof. Use Exercise 5.7(1). | No |
Theorem 6.4.1 A real-valued sequence \( \left\langle {s}_{n}\right\rangle \) is bounded in \( \mathbb{R} \) if and only if its extended terms are all limited. | Proof. To say that \( \left\langle {{s}_{n} : n \in \mathbb{N}}\right\rangle \) is bounded in \( \mathbb{R} \) means that it is contained within some real interval \( \left\lbrack {-b, b}\right\rbrack \), or equivalently that its absolute values \( \left| {s}_{n}\right| \) have some real upper bound \( b \) :\n\n\[ \left( {\forall n \in \mathbb{N}}\right) \left| {s}_{n}\right| < b. \]\n\nThen by universal transfer the extended sequence is contained in \( {}^{ * }\left\lbrack {-b, b}\right\rbrack \) , i.e., \( \left| {s}_{n}\right| < b \) for all \( n \in * \mathbb{N} \) ; hence \( {s}_{n} \) is limited in general.\n\nFor the converse, if \( {s}_{n} \) is limited for all unlimited \( n \in {}^{ * }{\mathbb{N}}_{\infty } \), then it is limited for all \( n \in \mathbb{N} \) . Hence if \( r \in {\mathbb{R}}_{\infty }^{ + } \) is any positive unlimited hyperreal, we observe that the entire extended sequence lies in the interval \( \{ x \in \mathbb{{R}^{n}} : \) \( - r < x < r\} \) and apply transfer. More formally, we have \( \left| {s}_{n}\right| < r \) for all \( n \in * \mathbb{N} \), so the sentence\n\n\[ \left( {\exists y \in {}^{ * }\mathbb{R}}\right) \left( {\forall n \in {}^{ * }\mathbb{N}}\right) \left| {s}_{n}\right| < y \]\n\nis true. But then by existential transfer it follows that there is some real number that is an upper bound to \( \left| {s}_{n}\right| \) for all \( n \in \mathbb{N} \) . | Yes |
Theorem 6.4.2 A real-valued sequence\n\n(1) diverges to infinity if and only if all of its extended terms are positive unlimited; and\n\n(2) diverges to minus infinity if and only if all of its extended terms are negative unlimited. | Proof. Exercise. | No |
Theorem 6.5.1 A real-valued sequence \( \left\langle {s}_{n}\right\rangle \) is Cauchy in \( \mathbb{R} \) if and only if all its extended terms are infinitely close to each other, i.e., iff \( {s}_{m} \simeq {s}_{n} \) for all \( m, n \in {}^{ * }{\mathbb{N}}_{\infty } \) . | Proof. Exercise. | No |
Theorem 6.5.2 (Cauchy’s Convergence Criterion). A real-valued sequence converges in \( \mathbb{R} \) if and only if it is Cauchy. | Proof. If \( \left\langle {{s}_{n} : n \in \mathbb{N}}\right\rangle \) is Cauchy, then it is bounded (standard result-why is it true?). Thus taking an unlimited number \( m \in {}^{ * }{\mathbb{N}}_{\infty } \), we have that \( {s}_{m} \) is limited (Theorem 6.4.1) and so it has a shadow \( L \in \mathbb{R} \) . But all extended terms of the sequence are infinitely close to each other (Theorem 6.5.1), hence are infinitely close to \( {s}_{m} \), and therefore are infinitely close to \( L \) as \( {s}_{m} \simeq L \) . This shows that the extended tail of the sequence is contained in the halo of \( L \), implying by Theorem 6.1.1 that \( \left\langle {s}_{n}\right\rangle \) converges to \( L \in \mathbb{R} \) . | No |
Theorem 6.6.1 \( L \in \mathbb{R} \) is a cluster point of the real-valued sequence \( \left\langle {s}_{n}\right. \) : \( n \in \mathbb{N}\rangle \) if and only if the sequence has an extended term infinitely close to \( L \), i.e., iff \( {s}_{N} \simeq L \) for some unlimited \( N \) . | Proof. Assume that (i) holds. Let \( \varepsilon \) be a positive infinitesimal and \( m \in \) \( {}^{ * }{\mathbb{N}}_{\infty } \) . Then by transfer of (i), there is some \( n \in {}^{ * }\mathbb{N} \) with \( n > m \), and hence \( n \) is unlimited, and\n\n\[ \left| {{s}_{n} - L}\right| < \varepsilon \simeq 0. \]\n\nThus \( {s}_{n} \) is an extended term infinitely close to \( L \) . (Indeed, the argument shows that any interval of infinitesimal width around \( L \) contains terms arbitrarily far along the extended tail.)\n\nConversely, suppose there is an unlimited \( N \) with \( {s}_{N} \simeq L \) . To prove (i), take any positive \( \varepsilon \in \mathbb{R} \) and \( m \in \mathbb{N} \) . Then \( N > m \) and \( \left| {{s}_{N} - L}\right| < \varepsilon \) . This shows that\n\n\[ \left( {\exists n \in * \mathbb{N}}\right) \left( {n > m \land \left| {{s}_{n} - L}\right| < \varepsilon }\right) . \]\n\nThus by existential transfer, \( \left| {{s}_{n} - L}\right| < \varepsilon \) for some \( n \in \mathbb{N} \), with \( n > m \) . | Yes |
Theorem 6.8.2 A real number \( L \) is equal to \( \overline{\lim }s \) if and only if\n\n(1) \( {s}_{n} < L \) or \( {s}_{n} \simeq L \) for all unlimited \( n \) ; and\n\n(2) \( {s}_{n} \simeq L \) for at least one unlimited \( n \) . | Proof. The condition \ | No |
Theorem 6.8.3 A bounded real-valued sequence \( s \) converges to \( L \in \mathbb{R} \) if and only if\n\n\[ \mathop{\limsup }\limits_{{n \rightarrow \infty }}{s}_{n} = \mathop{\liminf }\limits_{{n \rightarrow \infty }}{s}_{n} = L. \] | Proof. Since \( \overline{\lim }s \) and \( \underline{\lim }s \) are the maximum and minimum elements of \( {C}_{s} \), requiring that they both be equal to \( L \) amounts to requiring that \( {C}_{s} = \{ L\} \) . But that just means that the shadow of every extended term is equal to \( L \), which is equivalent to having \( s \) converge to \( L \) by Theorem 6.1.1. | Yes |
Theorem 6.8.4 If \( s \) is a bounded real-valued sequence with limit superior \( \overline{\lim } \), then for any positive real \( \varepsilon \) :\n\n(1) some standard tail of \( s \) has all its terms smaller than \( \overline{\lim } + \varepsilon \), i.e., \( {s}_{n} < \overline{\lim } + \varepsilon \) for all but finitely many \( n \in \mathbb{N} \) ;\n\n(2) \( \overline{\lim } - \varepsilon < {s}_{n} \) for infinitely many \( n \in \mathbb{N} \) . | Proof.\n\n(1) If \( m \in * \mathbb{N} \) is unlimited, then \( \operatorname{sh}\left( {s}_{m}\right) \leq \overline{\lim } \), so\n\n\[ \n{s}_{m} \simeq \operatorname{sh}\left( {s}_{m}\right) < \overline{\lim } + \varepsilon \n\]\n\nshowing that \( {s}_{m} < \overline{\lim } + \varepsilon \) because \( \operatorname{sh}\left( {s}_{m}\right) \) and \( \overline{\lim } + \varepsilon \) are both real. Thus all extended terms are smaller than \( \overline{\lim } + \varepsilon \), and in particular, this holds for all terms after \( {s}_{N} \) for any fixed unlimited \( N \) :\n\n\[ \n\left( {\forall m \in {}^{ * }\mathbb{N}}\right) \left( {m \geq N \rightarrow {s}_{m} < \overline{\lim } + \varepsilon }\right) .\n\]\n\nExistential transfer then provides an \( n \in \mathbb{N} \) such that all of\n\n\[ \n{s}_{n},{s}_{n + 1},{s}_{n + 2},\ldots \n\]\n\nare smaller than \( \overline{\lim } + \varepsilon \) .\n\n(2) \( \overline{\lim } \) is the shadow of some extended term \( {s}_{N} \) with \( N \) unlimited. Then \( \overline{\lim } - \varepsilon < \overline{\lim } \simeq {s}_{N} \), so \( \overline{\lim } - \varepsilon < {s}_{N} \) . But now for any limited \( m \in \mathbb{N} \) we have \( m < N \) and \( \overline{\lim } - \varepsilon < {s}_{N} \) . Existential transfer then ensures that there is a limited \( n \) with \( m < n \) and \( \overline{\lim } - \varepsilon < {s}_{n} \) . This shows that \( \overline{\lim } - \varepsilon < {s}_{n} \) for arbitrarily large \( n \in \mathbb{N} \), giving the desired conclusion. | Yes |
Theorem 6.8.5 For any bounded real-valued sequence \( s \) , \n\n\[ \n\mathop{\limsup }\limits_{{n \rightarrow \infty }}{s}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{S}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {\mathop{\sup }\limits_{{m \geq n}}{s}_{m}}\right) .\n\] | Proof. First we show that \n\n\[ \n\overline{\lim } \leq {S}_{m}\;\text{ for all }m \in \mathbb{N}.\n\] \n\n(ii) \n\nTo see this, take an extended term \( {s}_{N} \) whose shadow is infinitely close to the cluster point \( \overline{\lim } \) . Then if \( m \in \mathbb{N} \), we have \( {s}_{n} \leq {S}_{m} \) for all limited \( n \geq m \), and hence for all hypernatural \( n \geq m \) by transfer. In particular, \( {s}_{N} \leq {S}_{m} \), so as \( \overline{\lim } \simeq {s}_{N} \), this forces \( \overline{\lim } \leq {S}_{m} \) as required for (ii). \n\nNow let \( L = \mathop{\lim }\limits_{{n \rightarrow \infty }}{S}_{n} \) . Then \( L \) is the greatest lower bound of the sequence \( S \), and by (ii), \( \overline{\lim } \) is a lower bound for this sequence, so \( \overline{\lim } \leq L \) . But if \( \overline{\lim } < L \), we can choose a positive real \( \varepsilon \) with \( \overline{\lim } + \varepsilon < L \), and then by Theorem 6.8.4(1) there is some \( n \in \mathbb{N} \) such that the standard tail \( {s}_{n},{s}_{n + 1},{s}_{n + 2},\ldots \) is bounded above by \( \overline{\lim } + \varepsilon \) . This implies that the least upper bound \( {S}_{n} \) of this tail is no bigger than \( \overline{\lim } + \varepsilon \) . However, that gives \n\n\[ \n{S}_{n} \leq \overline{\lim } + \varepsilon < L \n\] \n\nwhich contradicts the fact that \( L \) is a lower bound of \( S \), and so \( L \leq {S}_{n} \) . \n\nWe are left with the conclusion that \( \overline{\lim } = L \), as desired. | Yes |
Theorem 7.1.1 \( f \) is continuous at the real point \( c \) if and only if \( f\left( x\right) \simeq \) \( f\left( c\right) \) for all \( x \in {}^{ * }\mathbb{R} \) such that \( x \simeq c \), i.e., iff\n\n\[ f\left( {\operatorname{hal}\left( c\right) }\right) \subseteq \operatorname{hal}\left( {f\left( c\right) }\right) . \] | Proof. The standard definition is that \( f \) is continuous at \( c \) iff for each open interval \( \left( {f\left( c\right) - \varepsilon, f\left( c\right) + \varepsilon }\right) \) around \( f\left( c\right) \) in \( \mathbb{R} \) there is a corresponding open interval \( \left( {c - \delta, c + \delta }\right) \) around \( c \) that is mapped into \( \left( {f\left( c\right) - \varepsilon, f\left( c\right) + \varepsilon }\right) \) by \( f \) . Since \( a < c < b \), the number \( \delta \) can be chosen small enough so that the interval \( \left( {c - \delta, c + \delta }\right) \) is contained with \( \left( {a, b}\right) \), ensuring that \( f \) is indeed defined at all points that are within \( \delta \) of \( c \) .\n\nContinuity at \( c \) is thus formally expressed by the sentence\n\n\[ \left( {\forall \varepsilon \in {\mathbb{R}}^{ + }}\right) \left( {\exists \delta \in {\mathbb{R}}^{ + }}\right) \left( {\forall x \in \mathbb{R}}\right) \left( {\left| {x - c}\right| < \delta \rightarrow \left| {f\left( x\right) - f\left( c\right) }\right| < \varepsilon }\right) .\n\n(i)\n\nNow suppose \( x \simeq c \) implies \( f\left( x\right) \simeq f\left( c\right) \) . To show that (i) holds, let \( \varepsilon \) be a positive real number. Then we have to find a real \( \delta \) small enough to fulfill (i). First we show that this can be achieved if \ | No |
Corollary 7.1.2 The following are equivalent.\n\n(1) \( f \) is continuous at \( c \in \mathbb{R} \) .\n\n(2) \( f\left( x\right) \simeq f\left( c\right) \) whenever \( x \simeq c \) .\n\n(3) There is some positive \( d \simeq 0 \) such that \( f\left( x\right) \simeq f\left( c\right) \) whenever \( \left| {x - c}\right| < d \) . | Null | No |
Theorem 7.1.3 The following are equivalent.\n\n(1) \( f \) is continuous at \( c \) in \( A \) .\n\n(2) \( f\\left( x\\right) \\simeq f\\left( c\\right) \) for all \( x \\in * A \) with \( x \\simeq c \) .\n\n(3) There is some positive \( d \\simeq 0 \) such that \( f\\left( x\\right) \\simeq f\\left( c\\right) \) for all \( x \\in {}^{ * }A \) with \( \\left| {x - c}\\right| < d \) . | Null | No |
Theorem 7.7.1 \( f \) is uniformly continuous on \( A \) if and only if \( x \simeq y \) implies \( f\left( x\right) \simeq f\left( y\right) \) for all hyperreals \( x, y \in {}^{ * }A \) . | Proof. Exercise. | No |
Theorem 7.7.2 If the real function \( f \) is continuous on the closed interval \( \left\lbrack {a, b}\right\rbrack \) in \( \mathbb{R} \), then \( f \) is uniformly continuous on \( \left\lbrack {a, b}\right\rbrack \) . | Proof. Take hyperreals \( x, y \in * \left\lbrack {a, b}\right\rbrack \) with \( x \simeq y \) . Let \( c = \operatorname{sh}\left( x\right) \) . Then since \( a \leq x \leq b \) and \( x \simeq c \), we have \( c \in \left\lbrack {a, b}\right\rbrack \), and so \( f \) is continuous at \( c \) . Applying Theorem 7.1.1, we get \( f\left( x\right) \simeq f\left( c\right) \) and \( f\left( y\right) \simeq f\left( c\right) \), whence \( f\left( x\right) \simeq f\left( y\right) \) . Hence \( f \) is uniformly continuous by Theorem 7.7.1. | Yes |
Theorem 7.12.1 The sequence \( \left\langle {{f}_{n} : n \in \mathbb{N}}\right\rangle \) of real-valued functions defined on \( A \subseteq \mathbb{R} \) converges pointwise to the function \( f : A \rightarrow \mathbb{R} \) if and only if for each \( x \in A \) and each unlimited \( n \in {}^{ * }\mathbb{N},{f}_{n}\left( x\right) \simeq f\left( x\right) \) . | Null | No |
Theorem 7.12.2 \( \left\langle {{f}_{n} : n \in \mathbb{N}}\right\rangle \) converges uniformly to the function \( f \) : \( A \rightarrow \mathbb{R} \) if and only if for each \( x \in {}^{ * }A \) and each unlimited \( n \in {}^{ * }\mathbb{N},{f}_{n}\left( x\right) \simeq \) \( f\left( x\right) \) . | Proof. Exercise. | No |
Theorem 7.13.1 If the functions \( \left\langle {{f}_{n} : n \in \mathbb{N}}\right\rangle \) are all continuous on \( A \subseteq \) \( \mathbb{R} \), and the sequence converges uniformly to the function \( f : A \rightarrow \mathbb{R} \), then \( f \) is continuous on \( A \) . | Proof. Let \( c \) belong to \( A \) . To prove that \( f \) is continuous at \( c \), we invoke Theorem 7.1.3(2). If \( x \in * A \) with \( x \simeq c \), we want \( f\left( x\right) \simeq f\left( c\right) \), i.e., \( \mid f\left( x\right) - \) \( f\left( c\right) \mid < \varepsilon \) for any positive real \( \varepsilon \) . The key to this is to analyse the inequality\n\n\[ \left| {f\left( x\right) - f\left( c\right) }\right| \leq \left| {f\left( x\right) - {f}_{n}\left( x\right) }\right| + \left| {{f}_{n}\left( x\right) - {f}_{n}\left( c\right) }\right| + \left| {{f}_{n}\left( c\right) - f\left( c\right) }\right| .\n\]\n\n(x)\n\nOn the right side, the middle term \( \left| {{f}_{n}\left( x\right) - {f}_{n}\left( c\right) }\right| \) will be infinitesimal for any \( n \in \mathbb{N} \) because \( x \simeq c \) and \( {f}_{n} \) is continuous at \( c \) . By taking a large enough \( n \), the first and last terms on the right can be made small enough that the sum of the three terms is less than \( \varepsilon \) .\n\nTo see how this works in detail, for a given \( \varepsilon \in {\mathbb{R}}^{ + } \) we apply the definition of uniform convergence to the number \( \varepsilon /4 \) to get that there is some integer \( m \in \mathbb{N} \) such that\n\n\[ n > m\text{ implies }\left| {{f}_{n}\left( x\right) - f\left( x\right) }\right| < \varepsilon /4 \]\n\nfor all \( n \in \mathbb{N} \) and all \( x \in A \), and hence for all \( n \in {}^{ * }\mathbb{N} \) and all \( x \in {}^{ * }A \) by universal transfer.\n\nNow fix \( n \) as a standard integer, say by putting \( n = m + 1 \) . Then for any \( x \in {}^{ * }A \) with \( x \simeq c \) it follows, since \( x, c \in {}^{ * }A \), that\n\n\[ \left| {{f}_{n}\left( x\right) - f\left( x\right) }\right| ,\left| {{f}_{n}\left( c\right) - f\left( c\right) }\right| < \varepsilon /4 \]\n\nand so in \( \left( \mathrm{x}\right) \) we get\n\n\[ \left| {f\left( x\right) - f\left( c\right) }\right| < \varepsilon /4 + \text{ infinitesimal } + \varepsilon /4 < \varepsilon \]\n\nas desired.\n\nNote that this proof is a mixture of standard and nonstandard arguments: it uses the hyperreal characterisation of continuity of \( {f}_{n} \) and \( f \), but the standard definition of uniform convergence of \( \left\langle {{f}_{n} : n \in \mathbb{N}}\right\rangle \) rather than the characterisation given by Theorem 7.12.2. | Yes |
Theorem 7.14.1 If the functions \( \left\langle {{f}_{n} : n \in \mathbb{N}}\right\rangle \) are all continuous on \( A \subseteq \) \( \mathbb{R} \), then for any \( n \in * \mathbb{N} \) and any \( y \in * A \) there is a positive infinitesimal \( d \) such that \( {f}_{n}\left( x\right) \simeq {f}_{n}\left( y\right) \) for all \( x \in {}^{ * }A \) with \( \left| {x - y}\right| < d \) . | Proof. The fact that \( {f}_{n} \) is continuous on \( A \) for all \( n \in \mathbb{N} \) is expressed by the sentence\n\n\[ \left( {\forall n \in \mathbb{N}}\right) \left( {\forall y \in A}\right) \]\n\n\[ \left( {\forall \varepsilon \in {\mathbb{R}}^{ + }}\right) \left( {\exists \delta \in {\mathbb{R}}^{ + }}\right) \left( {\forall x \in A}\right) \left( {\left| {x - y}\right| < \delta \rightarrow \left| {{f}_{n}\left( x\right) - {f}_{n}\left( y\right) }\right| < \varepsilon }\right) ,\]\n\nwhich states that \ | Yes |
Theorem 8.1.1 If \( f \) is defined at \( x \in \mathbb{R} \), then the real number \( L \in \mathbb{R} \) is the derivative of \( f \) at \( x \) if and only if for every nonzero infinitesimal \( \varepsilon \) , \( f\left( {x + \varepsilon }\right) \) is defined and\n\n\[ \frac{f\left( {x + \varepsilon }\right) - f\left( x\right) }{\varepsilon } \simeq L \] | Proof. Let \( g\left( h\right) = \frac{f\left( {x + h}\right) - f\left( x\right) }{h} \) and apply the characterisation of\n\n\[ \text{“}\mathop{\lim }\limits_{{h \rightarrow 0}}g\left( h\right) = L\text{”} \]\n\ngiven in Section 7.3.\n\nThus when \( f \) is differentiable (i.e., has a derivative) at \( x \), we have\n\n\[ {f}^{\prime }\left( x\right) = \operatorname{sh}\left( \frac{f\left( {x + \varepsilon }\right) - f\left( x\right) }{\varepsilon }\right) \]\n\nfor all infinitesimal \( \varepsilon \neq 0 \) . | No |
Theorem 8.2.1 If \( f \) is differentiable at \( x \in \mathbb{R} \), then \( f \) is continuous at \( x \) . | The differential of \( f \) at \( x \) corresponding to \( {\Delta x} \) is defined to be\n\n\[ \n{df} = {f}^{\prime }\left( x\right) {\Delta x}.\n\]\n\nThus whereas \( {\Delta f} \) represents the increment of the \ | No |
Theorem 8.2.2 (Incremental Equation) If \( {f}^{\prime }\left( x\right) \) exists at real \( x \) and \( {\Delta x} = {dx} \) is infinitesimal, then \( {\Delta f} \) and \( {df} \) are infinitesimal, and there is an infinitesimal \( \varepsilon \), dependent on \( x \) and \( {\Delta x} \), such that\n\n\[ \n{\Delta f} = {f}^{\prime }\left( x\right) {\Delta x} + {\varepsilon \Delta x} = {df} + {\varepsilon dx},\n\]\n\nand so\n\n\[ \nf\left( {x + {\Delta x}}\right) = f\left( x\right) + {f}^{\prime }\left( x\right) {\Delta x} + {\varepsilon \Delta x}.\n\] | This last equation elucidates the role of the derivative function \( {f}^{\prime } \) as the best linear approximation to the function \( f \) at \( x \) . For the graph of the linear function\n\n\[ \nl\left( {\Delta x}\right) = f\left( x\right) + {f}^{\prime }\left( x\right) {\Delta x}\n\]\n\ngives the tangent to \( f \) at \( x \) when the origin is translated to the point \( \left( {x,0}\right) \) , and \( l\left( {\Delta x}\right) \) differs from \( f\left( {x + {\Delta x}}\right) \) by the amount \( {\varepsilon \Delta x} \), which we saw above is itself infinitely smaller than \( {\Delta x} \) when \( {\Delta x} \) is infinitesimal, and in that sense is \ | No |
Theorem 8.7.1 (Incremental Equation for Two Variables) If \( f \) is smooth at the real point \( \left( {a, b}\right) \) and \( {\Delta x} \) and \( {\Delta y} \) are infinitesimal, then\n\n\[{\Delta f} = {df} + {\varepsilon \Delta x} + {\delta \Delta y}\]\n\nfor some infinitesimals \( \varepsilon \) and \( \delta \) . | Proof. The increment of \( f \) at \( \left( {a, b}\right) \) corresponding to \( {\Delta x},{\Delta y} \) can be written as\n\n\[{\Delta f} = \left\lbrack {f\left( {a + {\Delta x}, b + {\Delta y}}\right) - f\left( {a + {\Delta x}, b}\right) }\right\rbrack + \left\lbrack {f\left( {a + {\Delta x}, b}\right) - f\left( {a, b}\right) }\right\rbrack .\n\]\n\n(ii)\n\nThe second main summand of (ii) is the increment at \( a \) corresponding to \( {\Delta x} \) of the one-variable function \( x \mapsto f\left( {x, b}\right) \), whose derivative \( {f}_{x}\left( {a, b}\right) \) is assumed to exist. Applying the one-variable incremental equation (Theorem 8.2.2) thus gives\n\n\[f\left( {a + {\Delta x}, b}\right) - f\left( {a, b}\right) = {f}_{x}\left( {a, b}\right) {\Delta x} + {\varepsilon \Delta x}\]\n\n(iii)\n\nfor some infinitesimal \( \varepsilon \) .\n\nSimilarly, for the first summand we need to show that\n\n\[f\left( {a + {\Delta x}, b + {\Delta y}}\right) - f\left( {a + {\Delta x}, b}\right) = {f}_{y}\left( {a, b}\right) {\Delta y} + {\delta \Delta y}\]\n\n(iv)\n\nfor some infinitesimal \( \delta \) . Then combining (ii)-(iv) will give\n\n\[{\Delta f} = {f}_{x}\left( {a, b}\right) {\Delta x} + {f}_{y}\left( {a, b}\right) {\Delta y} + {\varepsilon \Delta x} + {\delta \Delta y},\]\n\nwhich is the desired result. | Yes |
Theorem 8.10.1 If the nth derivative \( {f}^{\left( n\right) } \) exists on an open interval containing the real number \( x \), and \( {f}^{\left( n\right) } \) is continuous at \( x \), then for any infinitesimal \( {\Delta x} \) ,\n\n\[ f\left( {x + {\Delta x}}\right) = f\left( x\right) + {f}^{\prime }\left( x\right) {\Delta x} + \frac{{f}^{\prime \prime }\left( x\right) }{2!}\Delta {x}^{2} + \cdots + \frac{{f}^{\left( n\right) }\left( x\right) }{n!}\Delta {x}^{n} + {\varepsilon \Delta }{x}^{n} \]\n\nfor some infinitesimal \( \varepsilon \) . | Null | No |
Theorem 8.11.1 Let \( f \) be differentiable on an interval \( \left( {a, b}\right) \) in \( \mathbb{R} \) . Then the derivative \( {f}^{\prime } \) is continuous on \( \left( {a, b}\right) \) if and only if for each hyperreal \( x \) that is well inside \( {}^{ * }\left( {a, b}\right) \) and each infinitesimal \( {\Delta x} \) , \n\n\[ \n f\left( {x + {\Delta x}}\right) = f\left( x\right) + {f}^{\prime }\left( x\right) {\Delta x} + {\varepsilon \Delta x} \n\] \n\nfor some infinitesimal \( \varepsilon \) . | Proof. Assume that the incremental equation holds at points well inside \( {}^{ * }\left( {a, b}\right) \) . To prove continuity of \( {f}^{\prime } \), let \( c \) be a real point in \( \left( {a, b}\right) \) and suppose \( x \simeq c \) . We want \( {f}^{\prime }\left( x\right) \simeq {f}^{\prime }\left( c\right) \) .\n\nNow, if \( \Delta = \left( {x - c}\right) \simeq 0 \), then using Theorem 8.2.2 we get \n\n\[ \n f\left( x\right) = f\left( {c + \Delta }\right) = f\left( c\right) + {f}^{\prime }\left( c\right) \Delta + {\varepsilon \Delta } \n\] \nfor some \( \varepsilon \simeq 0 \) . But \( x \) is well inside \( {}^{ * }\left( {a, b}\right) \), since \( a < c < b \) and \( x \simeq c \), so by the assumed incremental equation at \( x \), applied to the infinitesimal \( - \Delta \) , we have \n\n\[ \n f\left( c\right) = f\left( {x + \left( {-\Delta }\right) }\right) = f\left( x\right) + {f}^{\prime }\left( x\right) \left( {-\Delta }\right) + {\varepsilon }^{\prime }\left( {-\Delta }\right) \n\] \n\nfor some \( {\varepsilon }^{\prime } \simeq 0 \) . Combining these equations leads to \n\n\[ \n {f}^{\prime }\left( x\right) - {f}^{\prime }\left( c\right) = \varepsilon - {\varepsilon }^{\prime } \simeq 0 \n\] \n\ngiving our desired conclusion \( {f}^{\prime }\left( x\right) \simeq {f}^{\prime }\left( c\right) \) .\n\nThe proof that continuity of \( {f}^{\prime } \) implies the incremental equation at points well inside \( {}^{ * }\left( {a, b}\right) \) is indicated in the following exercises, which also give an example to show what can happen when continuity fails. | No |
Theorem 9.4.1 The function \( F\left( x\right) = {\int }_{a}^{x}f\left( t\right) {dt} \) is differentiable on \( \left\lbrack {a, b}\right\rbrack \) , and its derivative is \( f \) . | There is a very intuitive explanation of why this relationship should hold. The increment\n\n\[ \n{\Delta F} = F\left( {x + {\Delta x}}\right) - F\left( x\right) \n\]\n\nof \( F \) at \( x \) corresponding to a positive infinitesimal \( {\Delta x} \) is closely approximated by the area of the rectangle of height \( f\left( x\right) \) and width \( {\Delta x} \), i.e, by \( f\left( x\right) {\Delta x} \) . Thus the quotient \( \frac{\Delta F}{\Delta x} \) should be closely approximated by \( f\left( x\right) \) itself.\n\nDoes \ | No |
Theorem 9.4.2 Fundamental Theorem of Calculus. If a function \( G \) has a continuous derivative \( f \) on \( \left\lbrack {a, b}\right\rbrack \), then \( {\int }_{a}^{b}f\left( x\right) {dx} = G\left( b\right) - G\left( a\right) \) . | Proof. This follows from Theorem 9.4.1 by standard arguments that require no ideas of limits or infinitesimals. For if \( F\left( x\right) = {\int }_{a}^{x}f\left( t\right) {dt} \), then on \( \left\lbrack {a, b}\right\rbrack \) we have \( {\left( G\left( x\right) - F\left( x\right) \right) }^{\prime } = f\left( x\right) - f\left( x\right) = 0 \), so there is a constant \( c \) with \( G\left( x\right) - F\left( x\right) = c \) . This implies \( G\left( b\right) - G\left( a\right) = F\left( b\right) - F\left( a\right) \) . But \( F\left( b\right) - F\left( a\right) = {\int }_{a}^{b}f\left( t\right) {dt}. \) | Yes |
Theorem 10.1.1 If \( A \subseteq \mathbb{R} \) and \( r \in \mathbb{R} \), (1) \( r \) is interior to \( A \) if and only if \( r \simeq x \) implies \( x \in * A \), i.e., iff \( \operatorname{hal}\left( r\right) \subseteq \) *A. | (1) Let \( r \in {A}^{ \circ } \). Then \( \left( {r - \varepsilon, r + \varepsilon }\right) \subseteq A \) for some real \( \varepsilon > 0 \). Then the sentence \[ \left( {\forall x \in \mathbb{R}}\right) \left( {\left| {r - x}\right| < \varepsilon \rightarrow x \in A}\right) \] (i) is true. But now if \( r \simeq x \) in \( {}^{ * }\mathbb{R} \), then \( \left| {r - x}\right| < \varepsilon \), so by universal transfer of (i), \( x \in {}^{ * }A \). This shows that \( \operatorname{hal}\left( r\right) \subseteq {}^{ * }A \). (An alternative way of putting this is to observe that \( \operatorname{hal}\left( r\right) \subseteq {}^{ * }\left( {r - \varepsilon, r + \varepsilon }\right) \subseteq {}^{ * }A \).) Conversely, if \( \operatorname{hal}\left( r\right) \subseteq {}^{ * }A \), then the sentence \[ \left( {\exists \varepsilon \in {}^{ * }{\mathbb{R}}^{ + }}\right) \left( {\forall x \in {}^{ * }\mathbb{R}}\right) \left( {\left| {r - x}\right| < \varepsilon \rightarrow x \in {}^{ * }A}\right) \] is seen to be true by interpreting \( \varepsilon \) as any positive infinitesimal. But then by existential transfer there is some real \( \varepsilon > 0 \) for which (i) holds, so \( \left( {r - \varepsilon, r + \varepsilon }\right) \subseteq A \) and hence \( r \in {A}^{ \circ } \). | Yes |
Theorem 10.2.2 For any real number \( r \) ,\n\n\[\n\operatorname{hal}\left( r\right) = \bigcap \{ {}^{ * }A : r \in A\\text{ and }A\\text{ is open }\} .\n\] | Proof. We have already observed that if \( r \in A \subseteq \mathbb{R} \) and \( A \) is open, then \( \operatorname{hal}\left( r\right) \subseteq {}^{ * }A \) . On the other hand, if \( x \notin \operatorname{hal}\left( r\right) \), then \( x ≄ r \), so there must exist some real \( \varepsilon > 0 \) such that \( \left| {r - x}\right| > \varepsilon \) . Put \( A = \left( {r - \varepsilon, r + \varepsilon }\right) \subseteq \mathbb{R} \) . Then \( r \in A \) and \( A \) is open, but \( x \notin {}^{ * }A = \{ y \in {}^{ * }\mathbb{R} : \left| {r - y}\right| < \varepsilon \} \) . | Yes |
Theorem 10.3.1 (Heine-Borel) A set \( B \subseteq \mathbb{R} \) is compact if and only if it is closed and bounded. | Proof. We have already seen that if \( B \) satisfies Robinson’s criterion, then it is closed and bounded (above and below).\n\nConversely, if \( B \) is closed and bounded, then there is some real \( b \) such that\n\n\[ \left( {\forall x \in B}\right) \left( {\left| x\right| \leq b}\right) \text{.} \]\n\nNow, to prove Robinson’s criterion, suppose \( x \in {}^{ * }B \) . Then by transfer, \( \left| x\right| \leq b \in \mathbb{R} \) . Hence \( x \) is limited, and so has a shadow \( r \in \mathbb{R} \) . Then \( r \simeq \) \( x \in {}^{ * }B \), and so \( r \in B \) because \( B \) is closed. Thus we have shown that \( x \) is infinitely close to the member \( r \) of \( B \), proving that \( B \) is compact. | Yes |
Theorem 10.4.1 The continuous image of a compact set is compact. | Proof. Let \( f \) be a continuous real function, and \( B \) a compact subset of \( \mathbb{R} \) included in the domain of \( f \) . Now, it is true, by definition of \( f\left( B\right) \), that\n\n\[ \left( {\forall y \in f\left( B\right) }\right) \left( {\exists x \in B}\right) \left( {y = f\left( x\right) }\right) .\n\nThus by transfer, if \( y \in \mathcal{N}\left( {f\left( B\right) }\right) \), then \( y = f\left( x\right) \) for some \( x \in \mathcal{B} \) . Since \( B \) is compact, \( x \simeq r \) for some \( r \in B \) . Then by continuity of \( f, f\left( x\right) \simeq f\left( r\right) \) , i.e., \( y \) is infinitely close to \( f\left( r\right) \in f\left( B\right) \) . This shows by Robinson’s criterion that \( f\left( B\right) \) is compact. | Yes |
Theorem 10.4.2 If \( f \) is continuous on a compact set \( B \subseteq \mathbb{R} \), then \( f \) is uniformly continuous on \( B \) . | Proof. By Theorem 7.7.1 we have to show that for all \( x, y \in {}^{ * }B \) ,\n\n\[ x \simeq y\;\text{ implies }\;f\left( x\right) \simeq f\left( y\right) . \]\n\nBut if \( x, y \in {}^{ * }B \), then by compactness \( x \simeq r \in B \) and \( y \simeq s \in B \) for some \( r, s \) . Thus if \( x \simeq y \), then \( r \simeq s \), and so \( r = s \), as both are real. Hence by continuity of \( f \) at \( r \in B, f\left( x\right) \simeq f\left( r\right) \) and \( f\left( y\right) \simeq f\left( r\right) \), whence \( f\left( x\right) \simeq f\left( y\right) \) as desired. | Yes |
Theorem 11.3.1 Any nonempty internal subset of \( {}^{ * }\mathbb{N} \) has a least member. | Proof. Let \( \left\lbrack {A}_{n}\right\rbrack \) be a nonempty internal subset of \( {}^{ * }\mathbb{N} \) . Then by the observations above we can assume that for each \( n \in \mathbb{N} \), \[ \varnothing \neq {A}_{n} \subseteq \mathbb{N} \] and so \( {A}_{n} \) has a least member \( {r}_{n} \) . This defines a point \( \left\lbrack {r}_{n}\right\rbrack \in {}^{ * }\mathbb{R} \) with \[ \left\{ {n \in \mathbb{N} : {r}_{n} \in {A}_{n}}\right\} = \mathbb{N} \in \mathcal{F} \] so \( \left\lbrack {r}_{n}\right\rbrack \in \left\lbrack {A}_{n}\right\rbrack \) . Moreover, if \( \left\lbrack {s}_{n}\right\rbrack \in \left\lbrack {A}_{n}\right\rbrack \), then \[ \left\{ {n \in \mathbb{N} : {s}_{n} \in {A}_{n}}\right\} \in \mathcal{F}\;\text{and}\;\left\{ {n \in \mathbb{N} : {s}_{n} \in {A}_{n}}\right\} \subseteq \left\{ {n \in \mathbb{N} : {r}_{n} \leq {s}_{n}}\right\} , \] leading to the conclusion \( \left\lbrack {r}_{n}\right\rbrack \leq \left\lbrack {s}_{n}\right\rbrack \) in * \( \mathbb{R} \) . Hence \( \left\lbrack {A}_{n}\right\rbrack \) indeed has a least member, namely the hyperreal number \( \left\lbrack {r}_{n}\right\rbrack \) determined by the sequence of least members of the sets \( {A}_{n} \) . Writing \ | Yes |
Theorem 11.3.2 (Internal Induction) If \( X \) is an internal subset of \( {}^{ * }\mathbb{N} \) that contains 1 and is closed under the successor function \( n \mapsto n + 1 \) , then \( X = {}^{ * }\mathbb{N} \) . | Proof. Let \( Y = {}^{ * }\mathbb{N} - X \) . Then \( Y \) is internal \( \left( {{11.2}\left( 1\right) }\right) \), so if it is nonempty, it has a least element \( n \) . Then \( n \neq 1 \), as \( 1 \in X \), so \( n - 1 \in * \mathbb{N} \) . But now \( n - 1 \notin Y \), as \( n \) is least in \( Y \), so \( n - 1 \in X \), and therefore \( n = \left( {n - 1}\right) + 1 \) is in \( X \) by closure under successor. This contradiction forces us to conclude that \( Y = \varnothing \), and so \( X = {}^{ * }\mathbb{N} \) . | Yes |
Theorem 11.4.1 Let \( X \) be an internal subset of \( {}^{ * }\mathbb{N} \) and \( k \in \mathbb{N} \) . If \( n \in X \) for all \( n \in \mathbb{N} \) with \( k \leq n \), then there is an unlimited \( K \in * \mathbb{N} \) with \( n \in X \) for all \( n \in {}^{ * }\mathbb{N} \) with \( k \leq n \leq K \) . | Proof. If all unlimited hypernaturals are in \( X \), then any unlimited \( K \in {}^{ * }\mathbb{N} \) will do. Otherwise there are unlimited hypernaturals not in \( X \) . If we can show that there is a least such unlimited number \( H \), then all unlimited numbers smaller than \( H \) will be in \( X \), giving the desired result.\n\nTo spell this out: if \( {}^{ * }\mathbb{N} - X \) has unlimited members, then these must be greater than \( k \), and so the set\n\n\[ Y = \left\{ {n \in {}^{ * }\mathbb{N} : k < n \in {}^{ * }\mathbb{N} - X}\right\} \]\n\nis nonempty. But \( Y \) is internal, by the algebra of internal sets, since it is equal to\n\n\[ \left( {*\mathbb{N}-\{ 1,\ldots, k\} }\right) \cap \left( {*\mathbb{N} - X}\right) \]\n\nHence \( Y \) has a least element \( H \) by the internal least number principle. Then \( H \) is a hypernatural that is greater than \( k \) but not in \( X \), so it must be the case that \( H \notin \mathbb{N} \), because of our hypothesis that all limited \( n \geq k \) are in \( X \) . Thus \( H \) is unlimited. Then \( K = H - 1 \) is unlimited and meets the requirements of the theorem: \( H \) is the least hypernatural greater than \( k \) that is not in \( X \), so every \( n \in {}^{ * }\mathbb{N} \) with \( k \leq n \leq H - 1 \) does belong to \( X \) . | Yes |
Theorem 11.5.1 If a nonempty internal subset of \( {}^{ * }\mathbb{R} \) is bounded above/ below, then it has a least upper/greatest lower bound in \( {}^{ * }\mathbb{R} \) . | Proof. We treat the case of upper bounds. In effect, the point of the proof is to show that the least upper bound of a bounded internal set \( \left\lbrack {A}_{n}\right\rbrack \) is the hyperreal number determined by the sequence of least upper bounds of the \( {A}_{n} \) ’s:\n\n\[ \operatorname{lub}\left\lbrack {A}_{n}\right\rbrack = \left\lbrack {\operatorname{lub}{A}_{n}}\right\rbrack \]\n\nMore precisely, it is enough to require that \( \mathcal{F} \) -almost all \( {A}_{n} \) ’s have least upper bounds to make this work.\n\nSuppose then that a nonempty internal set \( \left\lbrack {A}_{n}\right\rbrack \) has an upper bound \( \left\lbrack {r}_{n}\right\rbrack . \) Write \( {A}_{n} \leq x \) to mean that \( x \) is an upper bound of \( {A}_{n} \) in \( \mathbb{R} \), and put\n\n\[ J = \left\{ {n \in \mathbb{N} : {A}_{n} \leq {r}_{n}}\right\} \]\n\nWe want \( J \in \mathcal{F} \) . If not, then \( {J}^{c} \in \mathcal{F} \) . But if \( n \in {J}^{c} \), there exists some \( {a}_{n} \) with \( {r}_{n} < {a}_{n} \in {A}_{n} \) . This leads to the conclusion \( \left\lbrack {r}_{n}\right\rbrack < \left\lbrack {a}_{n}\right\rbrack \in \left\lbrack {A}_{n}\right\rbrack \) , contradicting the fact that \( \left\lbrack {r}_{n}\right\rbrack \) is an upper bound of \( \left\lbrack {A}_{n}\right\rbrack \) .\n\nIt follows that \( J \in \mathcal{F} \) . Since \( \left\lbrack {A}_{n}\right\rbrack \neq \varnothing \), this then implies\n\n\[ {J}^{\prime } = \left\{ {n \in \mathbb{N} : \varnothing \neq {A}_{n} \leq {r}_{n}}\right\} \in \mathcal{F}. \]\n\nNow, if \( n \in {J}^{\prime } \), then \( {A}_{n} \) is a nonempty subset of \( \mathbb{R} \) bounded above (by \( {r}_{n} \) ), and so by the order-completeness of \( \mathbb{R},{A}_{n} \) has a least upper bound \( {s}_{n} \in \mathbb{R} \) . Then if \( \left\lbrack {b}_{n}\right\rbrack \in \left\lbrack {A}_{n}\right\rbrack \) ,\n\n\[ \left\{ {n \in \mathbb{N} : {b}_{n} \in {A}_{n}}\right\} \cap {J}^{\prime } \subseteq \left\{ {n \in \mathbb{N} : {b}_{n} \leq {s}_{n}}\right\} \]\n\nleading to \( \left\lbrack {b}_{n}\right\rbrack \leq \left\lbrack {s}_{n}\right\rbrack \), and showing that \( \left\lbrack {s}_{n}\right\rbrack \) is an upper bound of \( \left\lbrack {A}_{n}\right\rbrack \) . Finally, if \( \left\lbrack {t}_{n}\right\rbrack \) is any other upper bound of \( \left\lbrack {A}_{n}\right\rbrack \), then \( \left\{ {n : {A}_{n} \leq {t}_{n}}\right\} \in \mathcal{F} \) by the same argument as for \( \left\lbrack {r}_{n}\right\rbrack \), and\n\n\[ \left\{ {n \in \mathbb{N} : {A}_{n} \leq {t}_{n}}\right\} \cap {J}^{\prime } \subseteq \left\{ {n \in \mathbb{N} : {s}_{n} \leq {t}_{n}}\right\} \]\n\nso we get \( \left\lbrack {s}_{n}\right\rbrack \leq \left\lbrack {t}_{n}\right\rbrack \) . This shows that \( \left\lbrack {s}_{n}\right\rbrack \) is indeed the least upper bound of \( \left\lbrack {A}_{n}\right\rbrack \) in \( {}^{ * }\mathbb{R} \) . | Yes |
Theorem 11.8.1 Let \( X \) be an internal subset of \( {}^{ * }\mathbb{N} \), and let \( K \in {}^{ * }\mathbb{N} \) be unlimited. If every unlimited hypernatural \( H \leq K \) belongs to \( X \), then there is some \( k \in \mathbb{N} \) such that every limited \( n \) with \( k \leq n \) belongs to \( X \) . | Proof. For \( M, N \in * \mathbb{N} \) with \( M \leq N \), let\n\n\[ \lfloor M, N\rfloor = \left\{ {z \in {}^{ * }\mathbb{N} : M \leq z \leq N}\right\} \]\n\nbe the interval in \( {}^{ * }\mathbb{N} \) between \( M \) and \( N \) . Our hypothesis is that \( \lfloor H, K\rfloor \subseteq X \) for all unlimited hypernatural \( H \leq K \) . What we want to show is that \( \lfloor k, K\rfloor \subseteq X \) for some \( k \in \mathbb{N} \) . To put this more symbolically, we want to show that the set\n\n\[ Y = \left\{ {k \in {}^{ * }\mathbb{N} : \lfloor k, K\rfloor \subseteq X}\right\} \]\n\nhas a limited member.\n\nNow, if \( Y \) is internal, then by the internal least number principle it has a least element \( k \), and such a \( k \) must belong to \( \mathbb{N} \), because if it were unlimited, then \( k - 1 \) would be unlimited, so by our hypothesis \( k - 1 \) would also be in \( Y \) but less than \( k \) .\n\nIt thus suffices to show that \( Y \) is internal. But if \( \varphi \left( {x, y, A}\right) \) is the formula\n\n\[ x \in \mathbb{N} \land x \leq y \land \forall z \in \mathbb{N}\left( {x \leq z \leq y \rightarrow z \in A}\right) ,\]\nexpressing \ | Yes |
Theorem 11.9.1 If \( X \) is an internal subset of \( {}^{ * }\mathbb{R} \) that contains all points that are infinitely close to \( b \in {}^{ * }\mathbb{R} \), then there is a positive real \( \varepsilon \) such that \( X \) contains all points that are within \( \varepsilon \) of \( b \) . | Proof. Our hypothesis is that \( \operatorname{hal}\left( b\right) \subseteq X \) . For \( k \in * \mathbb{N} \), let \( \left( {b - \frac{1}{k}, b + \frac{1}{k}}\right) \) be the hyperreal interval\n\n\[ \left\{ {z \in {}^{ * }\mathbb{R} : \left| {z - b}\right| < \frac{1}{k}}\right\} . \]\n\nNow,\n\n\[ \left( {b - \frac{1}{k}, b + \frac{1}{k}}\right) \subseteq X \]\n\nwhenever \( k \) is unlimited, because in this case \( \frac{1}{k} \) is infinitesimal, and so\n\n\[ \left( {b - \frac{1}{k}, b + \frac{1}{k}}\right) \subseteq \operatorname{hal}\left( b\right) \subseteq X \]\n\nby our hypothesis. Thus the set\n\n\[ Y = \left\{ {k \in {}^{ * }\mathbb{N} : \left( {b - \frac{1}{k}, b + \frac{1}{k}}\right) \subseteq X}\right\} \]\n\ncontains all unlimited members of \( {}^{ * }\mathbb{N} \) . Hence by underflow we could conclude that \( \left( {b - \frac{1}{k}, b + \frac{1}{k}}\right) \subseteq X \) for some \( k \in \mathbb{N} \), and thereby complete the proof by putting \( \varepsilon = \frac{1}{k} \), provided that \( Y \) is internal. But applying internal set definition with \( \varphi \left( {x, y, A}\right) \) as the formula\n\n\[ x \in \mathbb{N} \land \left( {\forall z \in \mathbb{R}}\right) \left( {\left| {z - y}\right| < \frac{1}{x} \rightarrow z \in A}\right) \]\n\n(expressing \ | Yes |
Theorem 11.10.1 The intersection of a decreasing sequence\n\n\\[ \n{X}^{1} \supseteq {X}^{2} \supseteq \cdots \supseteq {X}^{k} \supseteq \cdots\n\\]\n\nof nonempty internal sets is always nonempty :\n\n\\[ \n\mathop{\\bigcap }\\limits_{{k \\in \\mathbb{N}}}{X}^{k} \\neq \\varnothing .\n\\] | Proof. This is a delicate analysis of the ultrapower construction, involving a kind of diagonalisation argument, that is not easy to motivate intuitively.\n\nFor each \\( k \\in \\mathbb{N} \\), let \\( {X}^{k} = \\left\\lbrack {A}_{n}^{k}\\right\\rbrack \\), so that \\( {X}^{k} \\) is the internal set defined by the sequence \\( \\left\\langle {{A}_{n}^{k} : n \\in \\mathbb{N}}\\right\\rangle \\) of subsets of \\( \\mathbb{R} \\) . Then by Section 11.2 the sets \\( \\left\\{ {n \\in \\mathbb{N} : {A}_{n}^{k} \\neq \\varnothing }\\right\\} \\) and \\( \\left\\{ {n \\in \\mathbb{N} : {A}_{n}^{k} \\supseteq {A}_{n}^{k + 1}}\\right\\} \\) belong to \\( \\mathcal{F} \\) . Hence if\n\n\\[ \n{J}^{k} = \\left\\{ {n \\in \\mathbb{N} : {A}_{n}^{1} \\supseteq \\cdots \\supseteq {A}_{n}^{k} \\neq \\varnothing }\\right\\}\n\\]then by closure of \\( \\mathcal{F} \\) under finite intersections it follows for each \\( k \\in \\mathbb{N} \\) that \\( {J}^{k} \\in \\mathcal{F} \\) . Note that \\( {J}^{1} \\supseteq {J}^{2} \\supseteq \\cdots \\) .\n\nWe want to define a hyperreal \\( \\left\\lbrack {s}_{n}\\right\\rbrack \\) that belongs to every \\( {X}^{k} \\) . This will require that for each \\( k \\) we have \\( {s}_{n} \\in {A}_{n}^{k} \\) for \\( \\mathcal{F} \\) -almost all \\( n \\) . We will arrange this to work for almost all \\( n \\geq k \\), in the sense that\n\n\\[ \n\\{ n \\in \\mathbb{N} : k \\leq n\\} \\cap {J}^{k} \\subseteq \\left\\{ {n \\in \\mathbb{N} : {s}_{n} \\in {A}_{n}^{k}}\\right\\} .\n\\]\n\n(iii)\n\nBut the set \\( \\{ n \\in \\mathbb{N} : k \\leq n\\} \\) is cofinite in \\( \\mathbb{N} \\), and so belongs to the nonprincipal ultrafilter \\( \\mathcal{F} \\) . Since also \\( {J}^{k} \\in \\mathcal{F} \\) ,(iii) then yields \\( \\{ n \\in \\mathbb{N} \\) : \\( \\left. {{s}_{n} \\in {A}_{n}^{k}}\\right\\} \\in \\mathcal{F} \\), and therefore \\( \\left\\lbrack {s}_{n}\\right\\rbrack \\in {X}^{k} \\) as desired.\n\nIt thus remains to define \\( {s}_{n} \\) fulfilling (iii). For \\( n \\in {J}^{1} \\) let\n\n\\[ \n{k}_{n} = \\max \\left\\{ {k : k \\leq n\\text{ and }n \\in {J}^{k}}\\right\\} .\n\\]\n\n(iv)\n\nThen \\( n \\in {J}^{{k}_{n}} \\), so by the definition of \\( {J}^{{k}_{n}} \\) we can choose some \\( {s}_{n} \\in {A}_{n}^{{k}_{n}} \\) , and hence\n\n\\[ \n{s}_{n} \\in {A}_{n}^{1} \\cap \\cdots \\cap {A}_{n}^{{k}_{n}}\n\\]\n\n(v)\n\nFor \\( n \\notin {J}^{1} \\) let \\( {s}_{n} \\) be arbitrary. Now, to prove (iii), observe that if \\( k \\leq n \\) and \\( n \\in {J}^{k} \\), then by (iv), \\( k \\leq {k}_{n} \\), and so by (v), \\( {s}_{n} \\in {A}_{n}^{k} \\) . | Yes |
Corollary 11.10.2 If \( \left\{ {{X}_{n} : n \in \mathbb{N}}\right\} \) is a collection of internal sets and \( X \) is internal, then:\n\n(1) \( \mathop{\bigcap }\limits_{{n \in \mathbb{N}}}{X}_{n} \neq \varnothing \) if \( \left\{ {{X}_{n} : n \in \mathbb{N}}\right\} \) has the finite intersection property. | (1) Let \( {Y}^{k} = {X}_{1} \cap \cdots \cap {X}_{k} \) . Then \( {Y}^{1} \supseteq {Y}^{2} \supseteq \cdots \), and each \( {Y}^{k} \) is internal by 11.2(1). The finite intersection property implies that \( {Y}^{k} \neq \varnothing \), so by the above theorem there is some hyperreal that belongs to every \( {Y}^{k} \), and hence to every \( {X}_{k} \) . | Yes |
Theorem 11.13.1 If \( X \) is internal, then \( \operatorname{sh}\left( X\right) \) is closed. | Proof. Let \( r \in \mathbb{R} \) be a closure point of \( \operatorname{sh}\left( X\right) \) . We need to show that \( r \in \operatorname{sh}\left( X\right) \), i.e., \( r \) is the shadow of some \( y \in X \) .\n\nNow, for each \( n \in \mathbb{N} \), the hyperreal open interval \( \left( {r - \frac{1}{n}, r + \frac{1}{n}}\right) \) meets \( \operatorname{sh}\left( X\right) \) in some real point \( {s}_{n} \) that must be the shadow of some \( {x}_{n} \in X \) . Hence \( {x}_{n} \simeq {s}_{n} \in \left( {r - \frac{1}{n}, r + \frac{1}{n}}\right) \), so the internal set\n\n\[ \begin{array}{l} {X}_{n} = X \cap \left( {r - \frac{1}{n}, r + \frac{1}{n}}\right) \end{array} \]\n\ncontains \( {x}_{n} \) and is thereby nonempty . The \( {X}_{n} \) ’s form a decreasing sequence, so by countable saturation there is a point \( y \) in their intersection. Then \( y \in X \) and \( \left| {y - r}\right| < \frac{1}{n} \) for all \( n \in \mathbb{N} \), so \( y \simeq r \) . Hence \( r = \operatorname{sh}\left( y\right) \in \operatorname{sh}\left( X\right) . \)\n\nThis shows that \( \operatorname{sh}\left( X\right) \) contains all its closure points and so is closed. | Yes |
Lemma 11.14.1 Every hyper-open set is a union of hyperreal open intervals. | Proof. Let \( A = \left\lbrack {A}_{n}\right\rbrack \) be hyper-open. Take a point \( r = \left\lbrack {r}_{n}\right\rbrack \) in \( A \) . Then we find that the set\n\n\[ J = \left\{ {n \in \mathbb{N} : {r}_{n} \in {A}_{n}\text{ and }{A}_{n}\text{ is open in }\mathbb{R}}\right\} \]\n\nbelongs to the ultrafilter \( \mathcal{F} \) . Our task is to show that \( r \) belongs to some hyperreal interval \( \left( {a, b}\right) \) that is included in \( A \) .\n\nNow, if \( n \in J \), then there is some real interval \( \left( {{a}_{n},{b}_{n}}\right) \subseteq \mathbb{R} \) with\n\n\[ {r}_{n} \in \left( {{a}_{n},{b}_{n}}\right) \subseteq {A}_{n} \]\n\nSince \( J \in \mathcal{F} \), this is enough to specify \( a \) as the hyperreal number \( \left\lbrack {a}_{n}\right\rbrack \) and \( b \) as \( \left\lbrack {b}_{n}\right\rbrack \) . Working with the properties of \( \mathcal{F} \), in a now familiar way, we can then show that \( \left\lbrack {a}_{n}\right\rbrack < \left\lbrack {r}_{n}\right\rbrack < \left\lbrack {b}_{n}\right\rbrack \), and also that \( \left\lbrack {s}_{n}\right\rbrack \in \left\lbrack {A}_{n}\right\rbrack \) whenever \( \left\lbrack {a}_{n}\right\rbrack < \left\lbrack {s}_{n}\right\rbrack < \left\lbrack {b}_{n}\right\rbrack \), so that\n\n\[ r \in \left( {a, b}\right) \subseteq A \]\n\nas desired. | Yes |
Theorem 11.14.4 If \( B \) is an internal set, then \( B \) is \( S \) -open if and only if it contains the halo of each of its points. | Proof. We have already observed that an S-open set is a union of halos.\n\nConversely, assume that \( \operatorname{hal}\left( r\right) \subseteq B \) whenever \( r \in B \) . For such an \( r \) , consider the set\n\n\[ X = \left\{ {n \in {}^{ * }\mathbb{N} : \left( {\forall x \in {}^{ * }\mathbb{R}}\right) \left( {\left| {r - x}\right| < \frac{1}{n} \rightarrow x \in B}\right) }\right\} .\n\]\n\nSince \( B \) is internal, it follows by the internal set definition principle that \( X \) is internal. Moreover, since \( \operatorname{hal}\left( r\right) \subseteq B \), it follows that \( \mathrm{X} \) contains every unlimited hypernatural \( n \), because for such an \( n,\left| {r - x}\right| < \frac{1}{n} \) implies \( x \in \) hal \( \left( r\right) \) . Hence by underflow, \( X \) must contain some standard \( n \in \mathbb{N} \), so \( B \) includes the real-radius interval \( \left( {r - \frac{1}{n}, r + \frac{1}{n}}\right) \) . But then since \( \frac{1}{n} \) is real,\n\n\[ r \in \left( \left( {r - \frac{1}{n}, r + \frac{1}{n}}\right) \right) \subseteq \left( {r - \frac{1}{n}, r + \frac{1}{n}}\right) \subseteq B.\n\]\n\nThis shows that \( B \) is the union of S-neighbourhoods, and is thereby S-open. | Yes |
Theorem 12.1.1 Let \( \left\langle {{f}_{n} : n \in \mathbb{N}}\right\rangle \) and \( \left\langle {{g}_{n} : n \in \mathbb{N}}\right\rangle \) be sequences of partial functions from \( \mathbb{R} \) to \( \mathbb{R} \) . Then the internal functions \( \left\lbrack {f}_{n}\right\rbrack \) and \( \left\lbrack {g}_{n}\right\rbrack \) are equal if and only if\n\n\[ \left\{ {n \in \mathbb{N} : {f}_{n} = {g}_{n}}\right\} \in \mathcal{F}. \]\n | Proof. Let \( {J}_{fg} = \left\{ {n \in \mathbb{N} : {f}_{n} = {g}_{n}}\right\} \), and suppose \( {J}_{fg} \in \mathcal{F} \) . Now in general, two functions are equal precisely when they have the same domain and assign the same values to all members of that domain. Thus\n\n\[ {J}_{fg} \subseteq \left\{ {n \in \mathbb{N} : \operatorname{dom}{f}_{n} = \operatorname{dom}{g}_{n}}\right\} \]\n\nleading by \( {11.2}\left( 3\right) \) to the conclusion that the internal sets \( \left\lbrack {\operatorname{dom}{f}_{n}}\right\rbrack \) and \( \left\lbrack {\operatorname{dom}{g}_{n}}\right\rbrack \) are equal, i.e., \( \operatorname{dom}\left\lbrack {f}_{n}\right\rbrack = \operatorname{dom}\left\lbrack {g}_{n}\right\rbrack \) . But for \( \left\lbrack {r}_{n}\right\rbrack \in \operatorname{dom}\left\lbrack {f}_{n}\right\rbrack \) ,\n\n\[ {J}_{fg} \cap \left\{ {n \in \mathbb{N} : {r}_{n} \in \operatorname{dom}{f}_{n}}\right\} \subseteq \left\{ {n \in \mathbb{N} : {f}_{n}\left( {r}_{n}\right) = {g}_{n}\left( {r}_{n}\right) }\right\} \]\n\nwhich leads to \( \left\lbrack {f}_{n}\right\rbrack \left( \left\lbrack {r}_{n}\right\rbrack \right) = \left\lbrack {g}_{n}\right\rbrack \left( \left\lbrack {r}_{n}\right\rbrack \right) \) . Hence \( \left\lbrack {f}_{n}\right\rbrack = \left\lbrack {g}_{n}\right\rbrack \) .\n\nFor the converse, suppose that \( {J}_{fg} \notin \mathcal{F} \) . Now, \( {J}_{fg}^{c} \) is a subset of the union\n\n\[ \left\{ {n \in \mathbb{N} : \operatorname{dom}{f}_{n} \neq \operatorname{dom}{g}_{n}}\right\} \cup \left\{ {n \in \mathbb{N} : \operatorname{dom}{f}_{n} = \operatorname{dom}{g}_{n}\text{ but }{f}_{n} \neq {g}_{n}}\right\} ,\]\n\nso either \( \left\{ {n : \operatorname{dom}{f}_{n} \neq \operatorname{dom}{g}_{n}}\right\} \in \mathcal{F} \), whence \( \operatorname{dom}\left\lbrack {f}_{n}\right\rbrack \neq \operatorname{dom}\left\lbrack {g}_{n}\right\rbrack \) and so \( \left\lbrack {f}_{n}\right\rbrack \neq \left\lbrack {g}_{n}\right\rbrack \), or else\n\n\[ J = \left\{ {n \in \mathbb{N} : \operatorname{dom}{f}_{n} = \operatorname{dom}{g}_{n}\text{ but }{f}_{n} \neq {g}_{n}}\right\} \in \mathcal{F}. \]\n\nBut for \( n \in J \) there exists some \( {r}_{n} \) with \( {f}_{n}\left( {r}_{n}\right) \neq {g}_{n}\left( {r}_{n}\right) \) . This leads to \( \left\lbrack {f}_{n}\right\rbrack \left( \left\lbrack {r}_{n}\right\rbrack \right) \neq \left\lbrack {g}_{n}\right\rbrack \left( \left\lbrack {r}_{n}\right\rbrack \right) \), and so \( \left\lbrack {f}_{n}\right\rbrack \neq \left\lbrack {g}_{n}\right\rbrack \) . | Yes |
Theorem 12.5.1 An internal set \( A \) is hyperfinite with internal cardinality \( N \) if and only if there is an internal bijection \( f : \{ 1,\ldots, N\} \rightarrow A \) . | Proof. Let \( A = \left\lbrack {A}_{n}\right\rbrack \) . If \( A \) is hyperfinite with internal cardinality \( N = \) \( \left\lbrack {N}_{n}\right\rbrack \), then we may suppose that for each \( n \in \mathbb{N},{A}_{n} \) is a finite set of cardinality \( {N}_{n} \) . Thus there is a bijection \( {f}_{n} : \left\{ {1,\ldots ,{N}_{n}}\right\} \rightarrow {A}_{n} \) . Let \( f = \) \( \left\lbrack {f}_{n}\right\rbrack \) . Then \( f \) is an internal function with domain \( \{ 1,\ldots, N\} \) that is injective \( \left( {{12.2}\left( 4\right) }\right) \) and has range \( A\left( {{12.2}\left( 1\right) }\right) \) .\n\nConversely, suppose that \( f = \left\lbrack {f}_{n}\right\rbrack \) is an internal bijection from \( \{ 1,\ldots, N\} \) onto \( A \) . Then\n\n\[ \left\lbrack {\operatorname{dom}{f}_{n}}\right\rbrack = \operatorname{dom}\left\lbrack {f}_{n}\right\rbrack = \{ 1,\ldots, N\} = \left\lbrack \left\{ {1,\ldots ,{N}_{n}}\right\} \right\rbrack ,\]\n\nso for \( \mathcal{F} \) -almost all \( n \) ,\n\n\[ \operatorname{dom}{f}_{n} = \left\{ {1,\ldots ,{N}_{n}}\right\} \]\n\n(i)\n\nAlso, as \( A \) is the image of \( \{ 1,\ldots, N\} \) under \( \left\lbrack {f}_{n}\right\rbrack \), Exercise 12.2(1) implies that \( A = \left\lbrack {{f}_{n}\left( \left\{ {1,\ldots ,{N}_{n}}\right\} \right) }\right\rbrack \), so\n\n\[ {f}_{n}\left( \left\{ {1,\ldots ,{N}_{n}}\right\} \right) = {A}_{n} \]\n\n(ii)\n\nfor \( \mathcal{F} \) -almost all \( n \) . Finally, by \( {12.2}\left( 4\right) \),\n\n\[ {f}_{n}\text{is injective} \]\n\n(iii)\n\nfor \( \mathcal{F} \) -almost all \( n \) . Then the set \( J \) of those \( n \in \mathbb{N} \) satisfying (i)-(iii) must belong to \( \mathcal{F} \) . But for \( n \in J,{A}_{n} \) is finite of cardinality \( {N}_{n} \) . Hence \( A \) is hyperfinite of internal cardinality \( N \) . | Yes |
Theorem 12.6.1 An internal set \( A = \left\lbrack {A}_{n}\right\rbrack \) is hyperfinite if and only if every injective internal function \( f \) whose domain includes \( A \) and has \( f\left( A\right) \subseteq A \) must in fact have \( f\left( A\right) = A \) . | Proof. Suppose \( A \) is hyperfinite. Let \( f = \left\lbrack {f}_{n}\right\rbrack \) be an internal injective function with \( A \subseteq \operatorname{dom}f \) and \( f\left( A\right) \subseteq A \) . Then each of the following is true for \( \mathcal{F} \) -almost all \( n \in \mathbb{N} \) :\n\n\[ \n{A}_{n}\text{is finite,}\n\]\n\n\[ \n{A}_{n} \subseteq \operatorname{dom}{f}_{n}\n\]\n\n\[ \n{f}_{n}\left( {A}_{n}\right) \subseteq {A}_{n},\;\text{ cf. }{12.2}\left( 1\right) ,{11.2}\left( 2\right)\n\]\n\n\( {f}_{n} \) is injective.\n\nThus the set \( J \) of those \( n \in \mathbb{N} \) satisfying all of these conditions must belong to \( \mathcal{F} \) . But\n\n\[ \nJ \subseteq \left\{ {n \in \mathbb{N} : {f}_{n}\left( {A}_{n}\right) = {A}_{n}}\right\}\n\]\n\nby the standard pigeonhole principle, and so \( f\left( A\right) = \left\lbrack {{f}_{n}\left( {A}_{n}\right) }\right\rbrack = \left\lbrack {A}_{n}\right\rbrack = A \) .\n\nFor the converse, suppose \( A \) is not hyperfinite. It follows that\n\n\[ \n{J}^{\prime } = \left\{ {n \in \mathbb{N} : {A}_{n}\text{ is infinite }}\right\} \in \mathcal{F}.\n\]\n\nBut for each \( n \in {J}^{\prime } \) there is an injective function \( {f}_{n} : {A}_{n} \rightarrow {A}_{n} \) and some \( {r}_{n} \in {A}_{n} - {f}_{n}\left( {A}_{n}\right) \) . Let \( f = \left\lbrack {f}_{n}\right\rbrack \) . This makes \( f \) an internal function with domain \( A \) that is injective \( \left( {{12.2}\left( 4\right) }\right) \) and has \( f\left( A\right) = \left\lbrack {{f}_{n}\left( {A}_{n}\right) }\right\rbrack \subseteq A \), while \( \left\lbrack {r}_{n}\right\rbrack \in A - f\left( A\right) \) and so \( f\left( A\right) \neq A \) . | Yes |
Any internal set belongs to a standard set that is transitive. Hence * \( \mathbb{U} \) is strongly transitive, and in particular, every member of an internal set is internal. | Proof. Let \( A \) be internal, with \( A \in {}^{ * }B \) for some \( B \in \mathbb{U} \) . By strong transitivity of \( \mathbb{U} \) there is a transitive \( T \in \mathbb{U} \) with \( B \subseteq T \) . But as we have seen, transitivity is preserved by the transfer map, so the standard set * \( T \in \) * \( \mathbb{U} \) is transitive. Also, \( {}^{ * }B \subseteq {}^{ * }T \), so \( A \in {}^{ * }T \), establishing the first part of the theorem. But then \( A \subseteq {}^{ * }T \), and every member of \( {}^{ * }T \) is internal by definition, so belongs to * \( \mathbb{U} \) . Thus \( A \subseteq {}^{ * }T \subseteq {}^{ * }\mathbb{U} \), completing the proof that * \( \mathbb{U} \) is strongly transitive.\n\nThe assertion that every member of an internal set is internal is now just the statement that \( {}^{ * }\mathbb{U} \) is transitive. | Yes |
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