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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 cas... | 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 \... | 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 cas... | 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-Riem... | 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 }... | 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\[ \parall... | 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... | 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 }\rig... | 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{... | 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 }\l... | 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-... | 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}\ri... | 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... | 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|... | 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 }\\... | 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(... | 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| ... | 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)... | 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 m... | 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 m... | Yes |
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 \( {\ps... | 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 ... | 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 ... | 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 - ... | 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(... | 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\;\t... | 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}\ri... | 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}{d... | 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},\ldot... | 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) {... | 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... | 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\d... | 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(ii... | 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... | 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 ... | 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| }^... | 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 depend... | 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, th... | 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... | 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} ... | 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}}^... | 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}^{ *... | 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... | 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... | 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 \( \mathca... | 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... | 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| \lamb... | 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.... | 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 }... | 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 } ... | 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/... | 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 \( {\int }_{0}^{\infty }{e}^{-{4\pi i\rho }{\left| \xi \eta \right| }^{1/2}}{f}_{0}\left( {\rho }^{2}\right) {\rho d\rho } \)... | 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 \r... | 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 inclu... | 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 \( \{ i\} \notin \mathcal{F} \), wh... | 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} ... | Yes |
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 \( A \) who... | 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\n... | 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... | 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 ... | 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}... | 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 t... | 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 \ri... | 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\[ \le... | 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 al... | 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\[ \left| {{s}_{n} - L}\right| < \varepsilon \simeq 0. \]\... | 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 ... | 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... | 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{... | 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 ... | 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) }\... | 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 \( \left( {c - \delta, c + \delta }\right) \) ... | Yes |
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}\righ... | 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 ... | 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,... | 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}\le... | 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 \( \var... | 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)... | 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)... | 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\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... | Yes |
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}... | 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... | 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... | 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... | 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 \( {\De... | 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 \( {f}^{\prime }\left( x\right) \simeq {f}^{\prime }\left( c\right) \)... | 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\l... | 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\righ... | 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 ... | 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 \( \l... | 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 prov... | 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... | 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... | 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 poi... | 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 \... | 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... | 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}_{... | 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 hypernat... | No |
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 - \fr... | 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 s... | 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} ... | 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 + \fr... | 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 \( \ma... | 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( {\l... | 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}\r... | 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... | 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\{ {... | 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{... | 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 * \( T \in \) * \( \math... | Yes |
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