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Theorem 3.1 Suppose \( \widehat{f} \) satisfies the decay condition \( \left| {\widehat{f}\left( \xi \right) }\right| \leq A{e}^{-{2\pi a}\left| \xi \right| } \) for some constants \( a, A > 0 \) . Then \( f\left( x\right) \) is the restriction to \( \mathbb{R} \) of \( a \) function \( f\left( z\right) \) holomorphic ... | Proof. Define\n\n\[ \n{f}_{n}\left( z\right) = {\int }_{-n}^{n}\widehat{f}\left( \xi \right) {e}^{2\pi i\xi z}{d\xi }\n\]\n\nand note that \( {f}_{n} \) is entire by Theorem 5.4 in Chapter 2. Observe also that \( f\left( z\right) \) may be defined for all \( z \) in the strip \( {S}_{b} \) by\n\n\[ \nf\left( z\right) =... | Yes |
Corollary 3.2 If \( \widehat{f}\left( \xi \right) = O\left( {e}^{-{2\pi a}\left| \xi \right| }\right) \) for some \( a > 0 \), and \( f \) vanishes in a non-empty open interval, then \( f = 0 \) . | Since by the theorem \( f \) is analytic in a region containing the real line, the corollary is a consequence of Theorem 4.8 in Chapter 2. In particular, we recover the fact proved in Exercise 21, Chapter 5 in Book I, namely that \( f \) and \( \widehat{f} \) cannot both have compact support unless \( f = 0 \) . | No |
Theorem 3.3 Suppose \( f \) is continuous and of moderate decrease on \( \mathbb{R} \) . Then, \( f \) has an extension to the complex plane that is entire with \( \left| {f\left( z\right) }\right| \leq A{e}^{{2\pi M}\left| z\right| } \) for some \( A > 0 \), if and only if \( \widehat{f} \) is supported in the interva... | One direction is simple. Suppose \( \widehat{f} \) is supported in \( \left\lbrack {-M, M}\right\rbrack \) . Then both \( f \) and \( \widehat{f} \) have moderate decrease, and the Fourier inversion formula applies\n\n\[ f\left( x\right) = {\int }_{-M}^{M}\widehat{f}\left( \xi \right) {e}^{2\pi i\xi x}{d\xi } \]\n\nSin... | Yes |
Theorem 3.4 Suppose \( F \) is a holomorphic function in the sector\n\n\[ S = \{ z : - \pi /4 < \arg z < \pi /4\} \]\n\nthat is continuous on the closure of \( S \) . Assume \( \left| {F\left( z\right) }\right| \leq 1 \) on the boundary of the sector, and that there are constants \( C, c > 0 \) such that \( \left| {F\l... | Proof. The idea is to subdue the \ | No |
Theorem 3.5 Suppose \( f \) and \( \widehat{f} \) have moderate decrease. Then \( \widehat{f}\left( \xi \right) = 0 \) for all \( \xi < 0 \) if and only if \( f \) can be extended to a continuous and bounded function in the closed upper half-plane \( \{ z = x + {iy} : y \geq 0\} \) with \( f \) holomorphic in the inter... | Proof. First assume \( \widehat{f}\left( \xi \right) = 0 \) for \( \xi < 0 \) . By the Fourier inversion formula\n\n\[ f\left( x\right) = {\int }_{0}^{\infty }\widehat{f}\left( \xi \right) {e}^{2\pi ix\xi }{d\xi } \]\n\nand we can extend \( f \) when \( z = x + {iy} \) with \( y \geq 0 \) by\n\n\[ f\left( z\right) = {\... | Yes |
Lemma 1.2 If \( {z}_{1},\ldots ,{z}_{N} \) are the zeros of \( f \) inside the disc \( {D}_{R} \), then\n\n\[{\int }_{0}^{R}\mathfrak{n}\left( r\right) \frac{dr}{r} = \mathop{\sum }\limits_{{k = 1}}^{N}\log \left| \frac{R}{{z}_{k}}\right| .\] | Proof. First we have\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{N}\log \left| \frac{R}{{z}_{k}}\right| = \mathop{\sum }\limits_{{k = 1}}^{N}{\int }_{\left| {z}_{k}\right| }^{R}\frac{dr}{r} \]\n\nIf we define the characteristic function\n\n\[ {\eta }_{k}\left( r\right) = \left\{ \begin{array}{ll} 1 & \text{ if }r > \left| {... | Yes |
Theorem 2.1 If \( f \) is an entire function that has an order of growth \( \leq \rho \) , then:\n\n(i) \( \mathfrak{n}\left( r\right) \leq C{r}^{\rho } \) for some \( C > 0 \) and all sufficiently large \( r \) .\n\n(ii) If \( {z}_{1},{z}_{2},\ldots \) denote the zeros of \( f \), with \( {z}_{k} \neq 0 \), then for a... | Proof. It suffices to prove the estimate for \( \mathfrak{n}\left( r\right) \) when \( f\left( 0\right) \neq 0 \) . Indeed, consider the function \( F\left( z\right) = f\left( z\right) /{z}^{\ell } \) where \( \ell \) is the order of the zero of \( f \) at the origin. Then \( {\mathfrak{n}}_{f}\left( r\right) \) and \(... | Yes |
Proposition 3.1 If \( \sum \left| {a}_{n}\right| < \infty \), then the product \( \mathop{\prod }\limits_{{n = 1}}^{\infty }\left( {1 + {a}_{n}}\right) \) converges. Moreover, the product converges to 0 if and only if one of its factors is 0. | Proof. If \( \sum \left| {a}_{n}\right| \) converges, then for all large \( n \) we must have \( \left| {a}_{n}\right| < 1/2 \) . Disregarding if necessary finitely many terms, we may assume that this inequality holds for all \( n \) . In particular, we can define \( \log \left( {1 + {a}_{n}}\right) \) by the usual pow... | Yes |
Proposition 3.2 Suppose \( \left\{ {F}_{n}\right\} \) is a sequence of holomorphic functions on the open set \( \Omega \) . If there exist constants \( {c}_{n} > 0 \) such that\n\n\[ \sum {c}_{n} < \infty \;\text{ and }\;\left| {{F}_{n}\left( z\right) - 1}\right| \leq {c}_{n}\;\text{ for all }z \in \Omega ,\]\n\nthen:\... | Proof. To prove the first statement, note that for each \( z \) we may argue as in the previous proposition if we write \( {F}_{n}\left( z\right) = 1 + {a}_{n}\left( z\right) \), with \( \left| {{a}_{n}\left( z\right) }\right| \leq {c}_{n} \) . Then, we observe that the estimates are actually uniform in \( z \) because... | Yes |
Theorem 4.1 Given any sequence \( \left\{ {a}_{n}\right\} \) of complex numbers with \( \left| {a}_{n}\right| \rightarrow \infty \) as \( n \rightarrow \infty \), there exists an entire function \( f \) that vanishes at all \( z = {a}_{n} \) and nowhere else. Any other such entire function is of the form \( f\left( z\r... | To begin the proof, note first that if \( {f}_{1} \) and \( {f}_{2} \) are two entire functions that vanish at all \( z = {a}_{n} \) and nowhere else, then \( {f}_{1}/{f}_{2} \) has removable singularities at all the points \( {a}_{n} \) . Hence \( {f}_{1}/{f}_{2} \) is entire and vanishes nowhere, so that there exists... | Yes |
Lemma 4.2 If \( \left| z\right| \leq 1/2 \), then \( \left| {1 - {E}_{k}\left( z\right) }\right| \leq c{\left| z\right| }^{k + 1} \) for some \( c > 0 \) . | Proof. If \( \left| z\right| \leq 1/2 \), then with the logarithm defined in terms of the power series, we have \( 1 - z = {e}^{\log \left( {1 - z}\right) } \), and therefore\n\n\[ \n{E}_{k}\left( z\right) = {e}^{\log \left( {1 - z}\right) + z + {z}^{2}/2 + \cdots + {z}^{k}/k} = {e}^{w}, \n\] \n\nwhere \( w = - \mathop... | Yes |
Lemma 5.2 The canonical products satisfy\n\n\\[ \n\\left| {{E}_{k}\\left( z\\right) }\\right| \\geq {e}^{-c{\\left| z\\right| }^{k + 1}}\\;\\text{ if }\\left| z\\right| \\leq 1/2 \n\\]\n\nand\n\n\\[ \n\\left| {{E}_{k}\\left( z\\right) }\\right| \\geq \\left| {1 - z}\\right| {e}^{-{c}^{\\prime }{\\left| z\\right| }^{k}}... | Proof. If \\( \\left| z\\right| \\leq 1/2 \\) we can use the power series to define the logarithm of \\( 1 - z \\), so that\n\n\\[ \n{E}_{k}\\left( z\\right) = {e}^{\\log \\left( {1 - z}\\right) + \\mathop{\\sum }\\limits_{{n = 1}}^{k}{z}^{n}/n} = {e}^{-\\mathop{\\sum }\\limits_{{n = k + 1}}^{\\infty }{z}^{n}/n} = {e}^... | Yes |
Corollary 5.4 There exists a sequence of radii, \( {r}_{1},{r}_{2},\ldots \), with \( {r}_{m} \rightarrow \infty \), such that\n\n\[ \left| {\mathop{\prod }\limits_{{n = 1}}^{\infty }{E}_{k}\left( {z/{a}_{n}}\right) }\right| \geq {e}^{-c{\left| z\right| }^{s}}\;\text{ for }\left| z\right| = {r}_{m} \]\n | Proof. Since \( \sum {\left| {a}_{n}\right| }^{-k - 1} < \infty \), there exists an integer \( N \) so that\n\n\[ \mathop{\sum }\limits_{{n = N}}^{\infty }{\left| {a}_{n}\right| }^{-k - 1} < 1/{10} \]\n\nTherefore, given any two consecutive large integers \( L \) and \( L + 1 \), we can find a positive number \( r \) w... | Yes |
Lemma 5.5 Suppose \( g \) is entire and \( u = \operatorname{Re}\left( g\right) \) satisfies\n\n\[ u\left( z\right) \leq C{r}^{s}\;\text{ whenever }\left| z\right| = r \]\n\nfor a sequence of positive real numbers \( r \) that tends to infinity. Then \( g \) is a polynomial of degree \( \leq s \) . | Proof. We can expand \( g \) in a power series centered at the origin\n\n\[ g\left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{z}^{n} \]\n\nWe have already proved in the last section of Chapter 3 (as a simple application of Cauchy's integral formulas) that\n\n(6)\n\n\[ \frac{1}{2\pi }{\int }_{0}^{2\pi ... | Yes |
Proposition 1.1 The gamma function extends to an analytic function in the half-plane \( \operatorname{Re}\left( s\right) > 0 \), and is still given there by the integral formula (1). | Proof. It suffices to show that the integral defines a holomorphic function in every strip\n\n\[ \n{S}_{\delta, M} = \{ \delta < \operatorname{Re}\left( s\right) < M\} \n\]\n\nwhere \( 0 < \delta < M < \infty \) . Note that if \( \sigma \) denotes the real part of \( s \), then \( \left| {{e}^{-t}{t}^{s - 1}}\right| = ... | Yes |
Lemma 1.2 If \( \operatorname{Re}\left( s\right) > 0 \), then\n\n\[ \Gamma \left( {s + 1}\right) = {s\Gamma }\left( s\right) \] | Proof. Integrating by parts in the finite integrals gives\n\n\[ {\int }_{\epsilon }^{1/\epsilon }\frac{d}{dt}\left( {{e}^{-t}{t}^{s}}\right) {dt} = - {\int }_{\epsilon }^{1/\epsilon }{e}^{-t}{t}^{s}{dt} + s{\int }_{\epsilon }^{1/\epsilon }{e}^{-t}{t}^{s - 1}{dt} \]\n\nand the desired formula (2) follows by letting \( \... | Yes |
Theorem 1.3 The function \( \Gamma \left( s\right) \) initially defined for \( \operatorname{Re}\left( s\right) > 0 \) has an analytic continuation to a meromorphic function on \( \mathbb{C} \) whose only singularities are simple poles at the negative integers \( s = 0, - 1,\ldots \) . The residue of \( \Gamma \) at \(... | Proof. It suffices to extend \( \Gamma \) to each half-plane \( \operatorname{Re}\left( s\right) > - m \), where \( m \geq 1 \) is an integer. For \( \operatorname{Re}\left( s\right) > - 1 \), we define\n\n\[ \n{F}_{1}\left( s\right) = \frac{\Gamma \left( {s + 1}\right) }{s}.\n\]\n\nSince \( \Gamma \left( {s + 1}\right... | Yes |
Theorem 1.4 For all \( s \in \mathbb{C} \) ,\n\n\[ \Gamma \left( s\right) \Gamma \left( {1 - s}\right) = \frac{\pi }{\sin {\pi s}} \] | Observe that \( \Gamma \left( {1 - s}\right) \) has simple poles at the positive integers \( s = \) \( 1,2,3,\ldots \), so that \( \Gamma \left( s\right) \Gamma \left( {1 - s}\right) \) is a meromorphic function on \( \mathbb{C} \) with simple poles at all the integers, a property also shared by \( \pi /\sin {\pi s} \)... | Yes |
Lemma 1.5 For \( 0 < a < 1,{\int }_{0}^{\infty }\frac{{v}^{a - 1}}{1 + v}{dv} = \frac{\pi }{\sin {\pi a}} \) . | Proof. We observe first that\n\n\[ \n{\int }_{0}^{\infty }\frac{{v}^{a - 1}}{1 + v}{dv} = {\int }_{-\infty }^{\infty }\frac{{e}^{ax}}{1 + {e}^{x}}{dx} \n\]\n\nwhich follows by making the change of variables \( v = {e}^{x} \) . However, using contour integration, we saw in Example 2 of Section 2.1 in Chapter 3, that the... | No |
Theorem 1.7 For all \( s \in \mathbb{C} \) , \n\n\[ \n\frac{1}{\Gamma \left( s\right) } = {e}^{\gamma s}s\mathop{\prod }\limits_{{n = 1}}^{\infty }\left( {1 + \frac{s}{n}}\right) {e}^{-s/n} \n\] | Proof. By the Hadamard factorization theorem and the fact that \( 1/\Gamma \) is entire, of growth order 1, and has simple zeros at \( s = 0, - 1, - 2,\ldots \), we can expand \( 1/\Gamma \) in a Weierstrass product of the form \n\n\[ \n\frac{1}{\Gamma \left( s\right) } = {e}^{{As} + B}s\mathop{\prod }\limits_{{n = 1}}... | Yes |
Proposition 2.1 The series defining \( \zeta \left( s\right) \) converges for \( \operatorname{Re}\left( s\right) > 1 \), and the function \( \zeta \) is holomorphic in this half-plane. | Proof. If \( s = \sigma + {it} \) where \( \sigma \) and \( t \) are real, then\n\n\[\n\left| {n}^{-s}\right| = \left| {e}^{-s\log n}\right| = {e}^{-\sigma \log n} = {n}^{-\sigma }.\n\]\n\nAs a consequence, if \( \sigma > 1 + \delta > 1 \) the series defining \( \zeta \) is uniformly bounded by \( \mathop{\sum }\limits... | Yes |
Theorem 2.2 If \( \operatorname{Re}\left( s\right) > 1 \), then\n\n\[{\pi }^{-s/2}\Gamma \left( {s/2}\right) \zeta \left( s\right) = \frac{1}{2}{\int }_{0}^{\infty }{u}^{\left( {s/2}\right) - 1}\left\lbrack {\vartheta \left( u\right) - 1}\right\rbrack {du}.\n\] | Proof. This and further arguments are based on the observation that\n\n(6)\n\n\[{\int }_{0}^{\infty }{e}^{-\pi {n}^{2}u}{u}^{\left( {s/2}\right) - 1}{du} = {\pi }^{-s/2}\Gamma \left( {s/2}\right) {n}^{-s},\;\text{ if }n \geq 1.\]\n\nIndeed, if we make the change of variables \( u = t/\pi {n}^{2} \) in the integral, the... | Yes |
Theorem 2.3 The function \( \xi \) is holomorphic for \( \operatorname{Re}\left( s\right) > 1 \) and has an analytic continuation to all of \( \mathbb{C} \) as a meromorphic function with simple poles at \( s = 0 \) and \( s = 1 \) . Moreover,\n\n\[ \xi \left( s\right) = \xi \left( {1 - s}\right) \;\text{ for all }s \i... | Proof. The idea of the proof is to use the functional equation for \( \vartheta \) , namely\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{e}^{-\pi {n}^{2}u} = {u}^{-1/2}\mathop{\sum }\limits_{{n = - \infty }}^{\infty }{e}^{-\pi {n}^{2}/u},\;u > 0. \]\n\nWe then could multiply both sides by \( {u}^{\left( {s/2... | Yes |
Theorem 2.4 The zeta function has a meromorphic continuation into the entire complex plane, whose only singularity is a simple pole at \( s = 1 \) . | Proof. A look at (7) provides the meromorphic continuation of \( \zeta \) , namely\n\n\[ \zeta \left( s\right) = {\pi }^{s/2}\frac{\xi \left( s\right) }{\Gamma \left( {s/2}\right) }.\]\n\nRecall that \( 1/\Gamma \left( {s/2}\right) \) is entire with simple zeros at \( 0, - 2, - 4,\ldots \), so the simple pole of \( \xi... | Yes |
Corollary 2.6 For \( \operatorname{Re}\left( s\right) > 0 \) we have\n\n\[ \zeta \left( s\right) - \frac{1}{s - 1} = H\left( s\right) \]\n\nwhere \( H\left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{\delta }_{n}\left( s\right) \) is holomorphic in the half-plane \( \operatorname{Re}\left( s\right) > 0 \) . | Turning to the corollary, we assume first that \( \operatorname{Re}\left( s\right) > 1 \) . We let \( N \) tend to infinity in formula (8) of the proposition, and observe that by the estimate \( \left| {{\delta }_{n}\left( s\right) }\right| \leq \left| s\right| /{n}^{\sigma + 1} \) we have the uniform convergence of th... | Yes |
Proposition 2.7 Suppose \( s = \sigma + {it} \) with \( \sigma, t \in \mathbb{R} \). Then for each \( {\sigma }_{0} \), \( 0 \leq {\sigma }_{0} \leq 1 \), and every \( \epsilon > 0 \), there exists a constant \( {c}_{\epsilon } \) so that\n\n(i) \( \left| {\zeta \left( s\right) }\right| \leq {c}_{\epsilon }{\left| t\ri... | In particular, the proposition implies that \( \zeta \left( {1 + {it}}\right) = O\left( {\left| t\right| }^{\epsilon }\right) \) as \( \left| t\right| \) tends to infinity, \( {}^{3} \) and the same estimate also holds for \( {\zeta }^{\prime } \). For the proof, we use Corollary 2.6. Recall the estimate \( \left| {{\d... | Yes |
Lemma 1.3 If \( \operatorname{Re}\left( s\right) > 1 \), then\n\n\[ \log \zeta \left( s\right) = \mathop{\sum }\limits_{{p, m}}\frac{{p}^{-{ms}}}{m} = \mathop{\sum }\limits_{{n = 1}}^{\infty }{c}_{n}{n}^{-s} \]\n\nfor some \( {c}_{n} \geq 0 \) . | Proof. Suppose first that \( s > 1 \) . Taking the logarithm of the Euler product formula, and using the power series expansion for the logarithm\n\n\[ \log \left( \frac{1}{1 - x}\right) = \mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{{x}^{m}}{m} \]\n\nwhich holds for \( 0 \leq x < 1 \), we find that\n\n\[ \log \zeta ... | Yes |
Lemma 1.4 If \( \theta \in \mathbb{R} \), then \( 3 + 4\cos \theta + \cos {2\theta } \geq 0 \) . | This follows at once from the simple observation\n\n\[ 3 + 4\cos \theta + \cos {2\theta } = 2{\left( 1 + \cos \theta \right) }^{2}. \] | Yes |
Corollary 1.5 If \( \sigma > 1 \) and \( t \) is real, then\n\n\[ \log \left| {{\zeta }^{3}\left( \sigma \right) {\zeta }^{4}\left( {\sigma + {it}}\right) \zeta \left( {\sigma + {2it}}\right) }\right| \geq 0. \] | Proof. Let \( s = \sigma + {it} \) and note that\n\n\[ \operatorname{Re}\left( {n}^{-s}\right) = \operatorname{Re}\left( {e}^{-\left( {\sigma + {it}}\right) \log n}\right) = {e}^{-\sigma \log n}\cos \left( {t\log n}\right) = {n}^{-\sigma }\cos \left( {t\log n}\right) .\n\]\n\nTherefore,\n\n\[ \log \left| {{\zeta }^{3}\... | Yes |
If \( \psi \left( x\right) \sim x \) as \( x \rightarrow \infty \), then \( \pi \left( x\right) \sim x/\log x \) as \( x \rightarrow \infty \). | The argument here is elementary. By definition, it suffices to prove the following two inequalities:\n\n(4)\n\n\[1 \leq \mathop{\liminf }\limits_{{x \rightarrow \infty }}\pi \left( x\right) \frac{\log x}{x}\;\text{ and }\;\mathop{\limsup }\limits_{{x \rightarrow \infty }}\pi \left( x\right) \frac{\log x}{x} \leq 1.\]\n... | Yes |
Proposition 2.2 If \( {\psi }_{1}\left( x\right) \sim {x}^{2}/2 \) as \( x \rightarrow \infty \), then \( \psi \left( x\right) \sim x \) as \( x \rightarrow \infty \) , and therefore \( \pi \left( x\right) \sim x/\log x \) as \( x \rightarrow \infty \) . | Proof. By Proposition 2.1, it suffices to prove that \( \psi \left( x\right) \sim x \) as \( x \rightarrow \infty \) . This will follow quite easily from the fact that if \( \alpha < 1 < \beta \) , then\n\n\[ \frac{1}{\left( {1 - \alpha }\right) x}{\int }_{\alpha x}^{x}\psi \left( u\right) {du} \leq \psi \left( x\right... | Yes |
Proposition 2.3 For all \( c > 1 \n\n(6)\n\n\[ \n{\psi }_{1}\left( x\right) = \frac{1}{2\pi i}{\int }_{c - i\infty }^{c + i\infty }\frac{{x}^{s + 1}}{s\left( {s + 1}\right) }\left( {-\frac{{\zeta }^{\prime }\left( s\right) }{\zeta \left( s\right) }}\right) {ds}. \n\] | To make the proof of this formula clear, we isolate the necessary contour integrals in a lemma. | No |
Lemma 2.4 If \( c > 0 \), then\n\n\[ \n\frac{1}{2\pi i}{\int }_{c - i\infty }^{c + i\infty }\frac{{a}^{s}}{s\left( {s + 1}\right) }{ds} = \left\{ \begin{array}{ll} 0 & \text{ if }0 < a \leq 1 \\ 1 - 1/a & \text{ if }1 \leq a \end{array}\right.\n\]\n\nHere, the integral is over the vertical line \( \operatorname{Re}\lef... | Proof. First note that since \( \left| {a}^{s}\right| = {a}^{c} \), the integral converges. We suppose first that \( 1 \leq a \), and write \( a = {e}^{\beta } \) with \( \beta = \log a \geq 0 \) . Let\n\n\[ f\left( s\right) = \frac{{a}^{s}}{s\left( {s + 1}\right) } = \frac{{e}^{s\beta }}{s\left( {s + 1}\right) }.\]\n\... | Yes |
Proposition 1.1 If \( f : U \rightarrow V \) is holomorphic and injective, then \( {f}^{\prime }\left( z\right) \neq 0 \) for all \( z \in U \) . In particular, the inverse of \( f \) defined on its range is holomorphic, and thus the inverse of a conformal map is also holomorphic. | Proof. We argue by contradiction, and suppose that \( {f}^{\prime }\left( {z}_{0}\right) = 0 \) for some \( {z}_{0} \in U \) . Then\n\n\[ f\left( z\right) - f\left( {z}_{0}\right) = a{\left( z - {z}_{0}\right) }^{k} + G\left( z\right) \;\text{ for all }z\text{ near }{z}_{0}, \]\n\nwith \( a \neq 0, k \geq 2 \) and \( G... | Yes |
Theorem 1.2 The map \( F : \mathbb{H} \rightarrow \mathbb{D} \) is a conformal map with inverse \( G : \mathbb{D} \rightarrow \mathbb{H} \) . | Proof. First we observe that both maps are holomorphic in their respective domains. Then we note that any point in the upper half-plane is closer to \( i \) than to \( - i \), so \( \left| {F\left( z\right) }\right| < 1 \) and \( F \) maps \( \mathbb{H} \) into \( \mathbb{D} \) . To prove that \( G \) maps into the upp... | Yes |
Lemma 1.3 Let \( V \) and \( U \) be open sets in \( \mathbb{C} \) and \( F : V \rightarrow U \) a holomorphic function. If \( u : U \rightarrow \mathbb{C} \) is a harmonic function, then \( u \circ F \) is harmonic on \( V \) . | Proof. The thrust of the lemma is purely local, so we may assume that \( U \) is an open disc. We let \( G \) be a holomorphic function in \( U \) whose real part is \( u \) (such a \( G \) exists by Exercise 12 in Chapter 2, and is determined up to an additive constant). Let \( H = G \circ F \) and note that \( u \cir... | No |
Lemma 2.1 Let \( f : \mathbb{D} \rightarrow \mathbb{D} \) be holomorphic with \( f\left( 0\right) = 0 \) . Then\n\n(i) \( \left| {f\left( z\right) }\right| \leq \left| z\right| \) for all \( z \in \mathbb{D} \) .\n\n(ii) If for some \( {z}_{0} \neq 0 \) we have \( \left| {f\left( {z}_{0}\right) }\right| = \left| {z}_{0... | Proof. We first expand \( f \) in a power series centered at 0 and convergent in all of \( \mathbb{D} \)\n\n\[ f\left( z\right) = {a}_{0} + {a}_{1}z + {a}_{2}{z}^{2} + \cdots . \]\n\nSince \( f\left( 0\right) = 0 \) we have \( {a}_{0} = 0 \), and therefore \( f\left( z\right) /z \) is holomorphic in \( \mathbb{D} \) (s... | Yes |
Theorem 2.2 If \( f \) is an automorphism of the disc, then there exist \( \theta \in \) \( \mathbb{R} \) and \( \alpha \in \mathbb{D} \) such that\n\n\[ f\left( z\right) = {e}^{i\theta }\frac{\alpha - z}{1 - \bar{\alpha }z}. \] | Proof. Since \( f \) is an automorphism of the disc, there exists a unique complex number \( \alpha \in \mathbb{D} \) such that \( f\left( \alpha \right) = 0 \) . Now we consider the automorphism \( g \) defined by \( g = f \circ {\psi }_{\alpha } \) . Then \( g\left( 0\right) = 0 \), and the Schwarz lemma gives\n\n\[ ... | Yes |
Corollary 3.2 Any two proper simply connected open subsets in \( \mathbb{C} \) are conformally equivalent. | Clearly, the corollary follows from the theorem, since we can use as an intermediate step the unit disc. Also, the uniqueness statement in the theorem is straightforward, since if \( F \) and \( G \) are conformal maps from \( \Omega \) to \( \mathbb{D} \) that satisfy these two conditions, then \( H = F \circ {G}^{-1}... | Yes |
Theorem 3.3 Suppose \( \mathcal{F} \) is a family of holomorphic functions on \( \Omega \) that is uniformly bounded on compact subsets of \( \Omega \) . Then:\n\n(i) \( \mathcal{F} \) is equicontinuous on every compact subset of \( \Omega \) . | The theorem really consists of two separate parts. The first part says that \( \mathcal{F} \) is equicontinuous under the assumption that \( \mathcal{F} \) is a family of holomorphic functions that is uniformly bounded on compact subsets of \( \Omega \) . The proof follows from an application of the Cauchy integral for... | Yes |
Lemma 3.4 Any open set \( \Omega \) in the complex plane has an exhaustion. | Proof. If \( \Omega \) is bounded, we let \( {K}_{\ell } \) denote the set of all points in \( \Omega \) at distance \( \geq 1/\ell \) from the boundary of \( \Omega \) . If \( \Omega \) is not bounded, let \( {K}_{\ell } \) denote the same set as above except that we also require \( \left| z\right| \leq \ell \) for al... | Yes |
Proposition 3.5 If \( \Omega \) is a connected open subset of \( \mathbb{C} \) and \( \left\{ {f}_{n}\right\} \) a sequence of injective holomorphic functions on \( \Omega \) that converges uniformly on every compact subset of \( \Omega \) to a holomorphic function \( f \), then \( f \) is either injective or constant. | Proof. We argue by contradiction and suppose that \( f \) is not injective, so there exist distinct complex numbers \( {z}_{1} \) and \( {z}_{2} \) in \( \Omega \) such that \( f\left( {z}_{1}\right) = \) \( f\left( {z}_{2}\right) \) . Define a new sequence by \( {g}_{n}\left( z\right) = {f}_{n}\left( z\right) - {f}_{n... | Yes |
Proposition 4.1 Suppose \( S\left( z\right) \) is given by (5).\n\n(i) If \( \mathop{\sum }\limits_{{k = 1}}^{n}{\beta }_{k} = 2 \), and \( \mathfrak{p} \) denotes the polygon whose vertices are given (in order) by \( {a}_{1},\ldots ,{a}_{n} \), then \( S \) maps the real axis onto \( \mathfrak{p} - \left\{ {a}_{\infty... | Proof. We assume that \( \mathop{\sum }\limits_{{k = 1}}^{n}{\beta }_{k} = 2 \) . If \( {A}_{k} < x < {A}_{k + 1} \) when \( 1 \leq k \leq n - 1 \), then\n\n\[ \n{S}^{\prime }\left( x\right) = \mathop{\prod }\limits_{{j \leq k}}{\left( x - {A}_{j}\right) }^{-{\beta }_{j}}\mathop{\prod }\limits_{{j > k}}{\left( x - {A}_... | Yes |
Theorem 4.2 If \( F : \mathbb{D} \rightarrow P \) is a conformal map, then \( F \) extends to a continuous bijection from the closure \( \overline{\mathbb{D}} \) of the disc to the closure \( \bar{P} \) of the polygonal region. In particular, \( F \) gives rise to a bijection from the boundary of the disc to the bounda... | The main point consists in showing that if \( {z}_{0} \) belongs to the unit circle, then \( \mathop{\lim }\limits_{{z \rightarrow {z}_{0}}}F\left( z\right) \) exists. To prove this, we need a preliminary result, which uses the fact that if \( f : U \rightarrow f\left( U\right) \) is conformal, then\n\n\[ \operatorname... | No |
For each \( 0 < r < 1/2 \), let \( {C}_{r} \) denote the circle centered at \( {z}_{0} \) of radius \( r \) . Suppose that for all sufficiently small \( r \) we are given two points \( {z}_{r} \) and \( {z}_{r}^{\prime } \) in the unit disc that also lie on \( {C}_{r} \) . If we let \( \rho \left( r\right) = \left| {f\... | Proof. If not, there exist \( 0 < c \) and \( 0 < R < 1/2 \) such that \( c \leq \rho \left( r\right) \) for all \( 0 < r \leq R \) . Observe that\n\n\[ f\left( {z}_{r}\right) - f\left( {z}_{r}^{\prime }\right) = {\int }_{\alpha }{f}^{\prime }\left( \zeta \right) {d\zeta } \]\n\nwhere the integral is taken over the arc... | Yes |
Lemma 4.4 Let \( {z}_{0} \) be a point on the unit circle. Then \( F\left( z\right) \) tends to a limit as \( z \) approaches \( {z}_{0} \) within the unit disc. | Proof. If not, there are two sequences \( \left\{ {{z}_{1},{z}_{2},\ldots }\right\} \) and \( \left\{ {{z}_{1}^{\prime },{z}_{2}^{\prime },\ldots }\right\} \) in the unit disc that converge to \( {z}_{0} \) and are so that \( F\left( {z}_{n}\right) \) and \( F\left( {z}_{n}^{\prime }\right) \) converge to two distinct ... | Yes |
Lemma 4.5 The conformal map \( F \) extends to a continuous function from the closure of the disc to the closure of the polygon. | Proof. By the previous lemma, the limit\n\n\[ \mathop{\lim }\limits_{{z \rightarrow {z}_{0}}}F\left( z\right) \]\n\nexists, and we define \( F\left( {z}_{0}\right) \) to be the value of this limit. There remains to prove that \( F \) is continuous on the closure of the unit disc. Given \( \epsilon \), there exists \( \... | Yes |
Theorem 4.7 If \( F \) is a conformal map from the upper half-plane to the polygonal region \( P \) and maps the points \( {A}_{1},\ldots ,{A}_{n - 1},\infty \) to the vertices of \( \mathfrak{p} \), then there exist constants \( {C}_{1} \) and \( {C}_{2} \) such that\n\n\[ F\left( z\right) = {C}_{1}{\int }_{0}^{z}\fra... | Proof. After a preliminary translation, we may assume that \( {A}_{j} \neq 0 \) for \( j = 1,\ldots, n - 1 \) . Choose a point \( {A}_{n}^{ * } > 0 \) on the real line, and consider the fractional linear map defined by\n\n\[ \Phi \left( z\right) = {A}_{n}^{ * } - \frac{1}{z} \]\n\nThen \( \Phi \) is an automorphism of ... | Yes |
Theorem 1.2 An entire doubly periodic function is constant. | Proof. The function is completely determined by its values on \( {P}_{0} \) and since the closure of \( {P}_{0} \) is compact, we conclude that the function is bounded on \( \mathbb{C} \), hence constant by Liouville’s theorem in Chapter 2. | Yes |
Theorem 1.3 The total number of poles of an elliptic function in \( {P}_{0} \) is always \( \geq 2 \) . | Proof. Suppose first that \( f \) has no poles on the boundary \( \partial {P}_{0} \) of the fundamental parallelogram. By the residue theorem we have\n\n\[ {\int }_{\partial {P}_{0}}f\left( z\right) {dz} = {2\pi i}\sum \operatorname{res}f \]\n\nand we contend that the integral is 0 . To see this, we simply use the per... | Yes |
Theorem 1.4 Every elliptic function of order \( m \) has \( m \) zeros in \( {P}_{0} \) . | Proof. Assuming first that \( f \) has no zeros or poles on the boundary of \( {P}_{0} \), we know by the argument principle in Chapter 3 that\n\n\[ \n{\int }_{\partial {P}_{0}}\frac{{f}^{\prime }\left( z\right) }{f\left( z\right) }{dz} = {2\pi i}\left( {{\mathcal{N}}_{\mathfrak{z}} - {\mathcal{N}}_{\mathfrak{p}}}\righ... | Yes |
Lemma 1.5 The two series\n\n\[ \mathop{\sum }\limits_{{\left( {n, m}\right) \neq \left( {0,0}\right) }}\frac{1}{{\left( \left| n\right| + \left| m\right| \right) }^{r}}\;\text{ and }\;\mathop{\sum }\limits_{{n + {m\tau } \in {\Lambda }^{ * }}}\frac{1}{{\left| n + m\tau \right| }^{r}} \]\n\nconverge if \( r > 2 \) . | Recall that according to the Note at the end of Chapter 7, the question whether a double series converges absolutely is independent of the order of summation. In the present case, we shall first sum in \( m \) and then in \( n \) .\n\nFor the first series, the usual integral comparison can be applied. \( {}^{2} \) For ... | Yes |
Theorem 1.6 The function \( \wp \) is an elliptic function that has periods 1 and \( \tau \), and double poles at the lattice points. | Proof. It remains only to prove that \( \wp \) is periodic with the correct periods. To do so, note that the derivative is given by differentiating the series for \( \wp \) termwise so\n\n\[{\wp }^{\prime }\left( z\right) = - 2\mathop{\sum }\limits_{{n, m \in \mathbb{Z}}}\frac{1}{{\left( z + n + m\tau \right) }^{3}}.\]... | Yes |
Theorem 1.7 The function \( {\left( {\wp }^{\prime }\right) }^{2} \) is the cubic polynomial in \( \wp \)\n\n\[{\left( {\wp }^{\prime }\right) }^{2} = 4\left( {\wp - {e}_{1}}\right) \left( {\wp - {e}_{2}}\right) \left( {\wp - {e}_{3}}\right) .\] | Proof. The only roots of \( F\left( z\right) = \left( {\wp \left( z\right) - {e}_{1}}\right) \left( {\wp \left( z\right) - {e}_{2}}\right) \left( {\wp \left( z\right) - {e}_{3}}\right) \) in the fundamental parallelogram have multiplicity 2 and are at the points \( 1/2,\tau /2 \), and \( \left( {1 + \tau }\right) /2 \)... | Yes |
Lemma 1.9 Every even elliptic function \( F \) with periods 1 and \( \tau \) is a rational funcion of \( \wp \) . | Proof. If \( F \) has a zero or pole at the origin it must be of even order, since \( F \) is an even function. As a consequence, there exists an integer \( m \) so that \( F{\wp }^{m} \) has no zero or pole at the lattice points. We may therefore assume that \( F \) itself has no zero or pole on \( \Lambda \) .\n\nOur... | Yes |
Theorem 2.1 Eisenstein series have the following properties:\n\n(i) The series \( {E}_{k}\left( \tau \right) \) converges if \( k \geq 3 \), and is holomorphic in the upper half-plane.\n\n(ii) \( {E}_{k}\left( \tau \right) = 0 \) if \( k \) is odd.\n\n(iii) \( {E}_{k}\left( \tau \right) \) satisfies the following trans... | Proof. By Lemma 1.5 and the remark after it, the series \( {E}_{k}\left( \tau \right) \) converges absolutely and uniformly in every half-plane \( \operatorname{Im}\left( \tau \right) \geq \delta > 0 \) , whenever \( k \geq 3 \) ; hence \( {E}_{k}\left( \tau \right) \) is holomorphic in the upper half-plane \( \operato... | Yes |
Theorem 2.2 For \( z \) near 0, we have\n\n\[ \wp \left( z\right) = \frac{1}{{z}^{2}} + 3{E}_{4}{z}^{2} + 5{E}_{6}{z}^{4} + \cdots \]\n\n\[ = \frac{1}{{z}^{2}} + \mathop{\sum }\limits_{{k = 1}}^{\infty }\left( {{2k} + 1}\right) {E}_{{2k} + 2}{z}^{2k}. \] | Proof. From the definition of \( \wp \), if we note that we may replace \( \omega \) by \( - \omega \) without changing the sum, we have\n\n\[ \wp \left( z\right) = \frac{1}{{z}^{2}} + \mathop{\sum }\limits_{{\omega \in {\Lambda }^{ * }}}\left\lbrack {\frac{1}{{\left( z + \omega \right) }^{2}} - \frac{1}{{\omega }^{2}}... | Yes |
Corollary 2.3 If \( {g}_{2} = {60}{E}_{4} \) and \( {g}_{3} = {140}{E}_{6} \), then\n\n\[{\left( {\wp }^{\prime }\right) }^{2} = 4{\wp }^{3} - {g}_{2}\wp - {g}_{3}\] | Note that this identity is another version of Theorem 1.7, and it allows one to express the symmetric functions of the \( {e}_{j} \) ’s in terms of the Eisenstein series. | No |
Lemma 2.4 If \( k \geq 2 \) and \( \operatorname{Im}\left( \tau \right) > 0 \), then\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\frac{1}{{\left( n + \tau \right) }^{k}} = \frac{{\left( -2\pi i\right) }^{k}}{\left( {k - 1}\right) !}\mathop{\sum }\limits_{{\ell = 1}}^{\infty }{\ell }^{k - 1}{e}^{{2\pi i\tau }... | Proof. This identity follows from applying the Poisson summation formula to \( f\left( z\right) = 1/{\left( z + \tau \right) }^{k} \) ; see Exercise 7 in Chapter 4. | No |
Theorem 2.5 If \( k \geq 4 \) is even, and \( \operatorname{Im}\left( \tau \right) > 0 \), then\n\n\[ \n{E}_{k}\left( \tau \right) = {2\zeta }\left( k\right) + \frac{2{\left( -1\right) }^{k/2}{\left( 2\pi \right) }^{k}}{\left( {k - 1}\right) !}\mathop{\sum }\limits_{{r = 1}}^{\infty }{\sigma }_{k - 1}\left( r\right) {e... | Proof. First observe that \( {\sigma }_{k - 1}\left( r\right) \leq r{r}^{k - 1} = {r}^{k} \) . If \( \operatorname{Im}\left( \tau \right) = t \), then whenever \( t \geq {t}_{0} \) we have \( \left| {e}^{2\pi ir\tau }\right| \leq {e}^{-{2\pi r}{t}_{0}} \), and we see that the series in the theorem is absolutely converg... | Yes |
Proposition 1.1 The function \( \Theta \) satisfies the following properties:\n\n(i) \( \Theta \) is entire in \( z \in \mathbb{C} \) and holomorphic in \( \tau \in \mathbb{H} \) .\n\n(ii) \( \Theta \left( {z + 1 \mid \tau }\right) = \Theta \left( {z \mid \tau }\right) \) .\n\n(iii) \( \Theta \left( {z + \tau \mid \tau... | Proof. Suppose that \( \operatorname{Im}\left( \tau \right) = t \geq {t}_{0} > 0 \) and \( z = x + {iy} \) belongs to a bounded set in \( \mathbb{C} \), say \( \left| z\right| \leq M \) . Then, the series defining \( \Theta \) is absolutely and uniformly convergent, since\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{... | Yes |
For each fixed \( \tau \in \mathbb{H} \), the quotient\n\n\[ \n{\left( \log \Theta \left( z \mid \tau \right) \right) }^{\prime \prime } = \frac{\Theta \left( {z \mid \tau }\right) {\Theta }^{\prime \prime }\left( {z \mid \tau }\right) - {\left( {\Theta }^{\prime }\left( z \mid \tau \right) \right) }^{2}}{\Theta {\left... | Let \( F\left( z\right) = {\left( \log \Theta \left( z \mid \tau \right) \right) }^{\prime } = \Theta {\left( z \mid \tau \right) }^{\prime }/\Theta \left( {z \mid \tau }\right) \) . Differentiating the identities (ii) and (iii) of Proposition 1.1 gives \( F\left( {z + 1}\right) = F\left( z\right) \) , \( F\left( {z + ... | Yes |
Theorem 1.6 If \( \tau \in \mathbb{H} \), then\n\n\[ \Theta \left( {z \mid - 1/\tau }\right) = \sqrt{\frac{\tau }{i}}{e}^{{\pi i\tau }{z}^{2}}\Theta \left( {{z\tau } \mid \tau }\right) \;\text{ for all }z \in \mathbb{C}. \] | Proof. It suffices to prove this formula for \( z = x \) real and \( \tau = {it} \) with \( t > 0 \), since for each fixed \( x \in \mathbb{R} \), the two sides of equation (5) are holomorphic functions in the upper half-plane which then agree on the positive imaginary axis, and hence must be equal everywhere. Also, fo... | Yes |
Corollary 1.7 If \( \operatorname{Im}\left( \tau \right) > 0 \), then \( \theta \left( {-1/\tau }\right) = \sqrt{\tau /i}\theta \left( \tau \right) \) . | Note that if \( \tau = {it} \), then \( \theta \left( \tau \right) = \vartheta \left( t\right) \), and the above relation is precisely the functional equation for \( \vartheta \) which appeared in Chapter 4. | No |
Corollary 1.8 If \( \tau \in \mathbb{H} \), then\n\n\[ \theta \left( {1 - 1/\tau }\right) = \sqrt{\frac{\tau }{i}}\mathop{\sum }\limits_{{n = - \infty }}^{\infty }{e}^{{\pi i}{\left( n + 1/2\right) }^{2}\tau } \]\n\n\[ = \sqrt{\frac{\tau }{i}}\left( {2{e}^{{\pi i\tau }/4} + \cdots }\right) . \]\n\nThe second identity m... | Proof. First, we note that \( n \) and \( {n}^{2} \) have the same parity, so\n\n\[ \theta \left( {1 + \tau }\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{\left( -1\right) }^{n}{e}^{{i\pi }{n}^{2}\tau } = \Theta \left( {1/2 \mid \tau }\right) \]\n\nhence \( \theta \left( {1 - 1/\tau }\right) = \Theta \lef... | Yes |
Proposition 1.9 If \( \operatorname{Im}\left( \tau \right) > 0 \), then \( \eta \left( {-1/\tau }\right) = \sqrt{\tau /i}\eta \left( \tau \right) \) . | Proof. From the product formula for the theta function, we may write with \( q = {e}^{\pi i\tau } \), \[ \Theta \left( {z \mid \tau }\right) = \left( {1 + q{e}^{-{2\pi iz}}}\right) \mathop{\prod }\limits_{{n = 1}}^{\infty }\left( {1 - {q}^{2n}}\right) \left( {1 + {q}^{{2n} - 1}{e}^{2\pi iz}}\right) \left( {1 + {q}^{{2n... | Yes |
Theorem 2.1 If \( \left| x\right| < 1 \), then \( \mathop{\sum }\limits_{{n = 0}}^{\infty }p\left( n\right) {x}^{n} = \mathop{\prod }\limits_{{k = 1}}^{\infty }\frac{1}{1 - {x}^{k}} \) . | Formally, we can write each fraction as\n\n\[ \frac{1}{1 - {x}^{k}} = \mathop{\sum }\limits_{{m = 0}}^{\infty }{x}^{km} \]\n\nand multiply these out together to obtain \( p\left( n\right) \) as the coefficient of \( {x}^{n} \) . Indeed, when we group together equal integers in a partition of \( n \), this partition can... | Yes |
Proposition 2.2 \( \mathop{\prod }\limits_{{n = 1}}^{\infty }\left( {1 - {x}^{n}}\right) = \mathop{\sum }\limits_{{k = - \infty }}^{\infty }{\left( -1\right) }^{k}{x}^{\frac{k\left( {{3k} + 1}\right) }{2}} \) . | Proof. If we set \( x = {e}^{2\pi iu} \), then we can write\n\n\[ \mathop{\prod }\limits_{{n = 1}}^{\infty }\left( {1 - {x}^{n}}\right) = \mathop{\prod }\limits_{{n = 1}}^{\infty }\left( {1 - {e}^{2\pi inu}}\right) \]\n\nin terms of the triple product\n\n\[ \mathop{\prod }\limits_{{n = 1}}^{\infty }\left( {1 - {q}^{2n}... | Yes |
Theorem 3.1 If \( n \geq 1 \), then \( {r}_{2}\left( n\right) = 4\left( {{d}_{1}\left( n\right) - {d}_{3}\left( n\right) }\right) \) . | To prove the theorem, we first establish a crucial relationship that identifies the generating function of the sequence \( {\left\{ {r}_{2}\left( n\right) \right\} }_{n = 1}^{\infty } \) with the square of the \( \theta \) function, namely\n\n(6)\n\n\[ \theta {\left( \tau \right) }^{2} = \mathop{\sum }\limits_{{n = 0}}... | Yes |
Proposition 3.2 The identity \( {r}_{2}\left( n\right) = 4\left( {{d}_{1}\left( n\right) - {d}_{3}\left( n\right) }\right), n \geq 1 \), is equivalent to the identities (7) \[ \theta {\left( \tau \right) }^{2} = 2\mathop{\sum }\limits_{{n = - \infty }}^{\infty }\frac{1}{{q}^{n} + {q}^{-n}} = 1 + 4\mathop{\sum }\limits_... | Proof. We note first that both series converge absolutely since \( \left| q\right| < 1 \) , and the first equals the second, because \( 1/\left( {{q}^{n} + {q}^{-n}}\right) = {q}^{\left| n\right| }/\left( {1 + {q}^{2\left| n\right| }}\right) \) . Since \( {\left( 1 + {q}^{2n}\right) }^{-1} = \left( {1 - {q}^{2n}}\right... | Yes |
Theorem 3.4 Suppose \( f \) is a holomorphic function in the upper half-plane that satisfies:\n\n(i) \( f\left( {\tau + 2}\right) = f\left( \tau \right) \),\n\n(ii) \( f\left( {-1/\tau }\right) = f\left( \tau \right) \),\n\n(iii) \( f\left( \tau \right) \) is bounded,\n\nthen \( f \) is constant. | For the proof of this theorem, we introduce the following subset of the closed upper half-plane, which is defined by\n\n\[ \mathcal{F} = \{ \tau \in \overline{\mathbb{H}} : \left| {\operatorname{Re}\left( \tau \right) }\right| \leq 1\text{ and }\left| \tau \right| \geq 1\} \]\n\nand illustrated in Figure 1.\n\n![eba412... | No |
Every point in the upper half-plane can be mapped into \( \mathcal{F} \) using repeatedly one or another of the following fractional linear transformations or their inverses:\n\n\[ \n{T}_{2} : \tau \mapsto \tau + 2,\;S : \tau \mapsto - 1/\tau .\n\] | Proof of Lemma 3.5. Let \( \tau \in \mathbb{H} \) . If \( g \in G \) with \( g\left( \tau \right) = \left( {{a\tau } + b}\right) /\left( {{c\tau } + d}\right) \) , then \( c \) and \( d \) are integers, and by (9) we may choose a \( {g}_{0} \in G \) such that \( \operatorname{Im}\left( {{g}_{0}\left( \tau \right) }\rig... | Yes |
Theorem 3.6 Every positive integer is the sum of four squares, and moreover\n\n\[ \n{r}_{4}\left( n\right) = 8{\sigma }_{1}^{ * }\left( n\right) \;\text{ for all }n \geq 1.\n\] | As before, we relate the sequence \( \left\{ {{r}_{4}\left( n\right) }\right\} \) via its generating function to an appropriate power of the function \( \theta \), which in this case is its fourth power. The result is that\n\n\[ \n\theta {\left( \tau \right) }^{4} = \mathop{\sum }\limits_{{n = 0}}^{\infty }{r}_{4}\left... | No |
Proposition 3.7 The assertion \( {r}_{4}\left( n\right) = 8{\sigma }_{1}^{ * }\left( n\right) \) is equivalent to the identity\n\n\[ \theta {\left( \tau \right) }^{4} = \frac{-1}{{\pi }^{2}}{E}_{2}^{ * }\left( \tau \right) ,\;\text{ where }\tau \in \mathbb{H}. \] | Proof. It suffices to prove that if \( q = {e}^{\pi i\tau } \), then\n\n\[ \frac{-1}{{\pi }^{2}}{E}_{2}^{ * }\left( \tau \right) = 1 + \mathop{\sum }\limits_{{k = 1}}^{\infty }8{\sigma }_{1}^{ * }\left( k\right) {q}^{k}. \]\n\nFirst, recall the forbidden Eisenstein series that we considered in the last section of the p... | Yes |
Proposition 3.8 The function \( {E}_{2}^{ * }\left( \tau \right) \) defined in the upper half-plane has the following properties:\n\n(i) \( {E}_{2}^{ * }\left( {\tau + 2}\right) = {E}_{2}^{ * }\left( \tau \right) \).\n\n(ii) \( {E}_{2}^{ * }\left( \tau \right) = - {\tau }^{-2}{E}_{2}^{ * }\left( {-1/\tau }\right) \).\n... | The periodicity (i) of \( {E}_{2}^{ * } \) is immediate from the definition. The proofs of the other properties of \( {E}_{2}^{ * } \) are a little more involved. | No |
Lemma 3.9 The functions \( F \) and \( \widetilde{F} \) satisfy:\n\n(a) \( F\left( {-1/\tau }\right) = {\tau }^{2}\widetilde{F}\left( \tau \right) \),\n\n(b) \( F\left( \tau \right) - \widetilde{F}\left( \tau \right) = {2\pi i}/\tau \),\n\n(c) \( F\left( {-1/\tau }\right) = {\tau }^{2}F\left( \tau \right) - {2\pi i\tau... | Proof. Property (a) follows directly from the definitions of \( F \) and \( \widetilde{F} \), and the identity\n\n\[{\left( n + m\left( -1/\tau \right) \right) }^{2} = {\tau }^{-2}{\left( -m + n\tau \right) }^{2}.\n\nTo prove property (b), we invoke the functional equation for the Dedekind eta function which was establ... | Yes |
Theorem 1.1 \( {J}_{\nu }\left( s\right) = \sqrt{\frac{2}{\pi s}}\cos \left( {s - \frac{\pi \nu }{2} - \frac{\pi }{4}}\right) + O\left( {s}^{-3/2}\right) \) as \( s \rightarrow \infty \) . | In view of the formula for \( {J}_{\nu }\left( s\right) \) it suffices to investigate\n\n(3)\n\n\[ I\left( s\right) = {\int }_{-1}^{1}{e}^{isx}{\left( 1 - {x}^{2}\right) }^{\nu - 1/2}{dx} \]\n\nand to this end we consider the analytic function \( f\left( z\right) = {e}^{isz}{\left( 1 - {z}^{2}\right) }^{\nu - 1/2} \) i... | Yes |
Proposition 1.2 Suppose a and \( m \) are fixed, with \( a > 0 \) and \( m > - 1 \) . Then as \( s \rightarrow \infty \n\n\[{\int }_{0}^{a}{e}^{-{sx}}{x}^{m}{dx} = {s}^{-m - 1}\Gamma \left( {m + 1}\right) + O\left( {e}^{-{cs}}\right) ,\]\n\nfor some positive \( c \) . | Proof. The fact that \( m > - 1 \) guarantees that \( {\int }_{0}^{a}{e}^{-{sx}}{x}^{m}{dx} = \) \( \mathop{\lim }\limits_{{\epsilon \rightarrow 0}}{\int }_{\epsilon }^{a}{e}^{-{sx}}{x}^{m}{dx} \) exists. Then, we write\n\n\[{\int }_{0}^{a}{e}^{-{sx}}{x}^{m}{dx} = {\int }_{0}^{\infty }{e}^{-{sx}}{x}^{m}{dx} - {\int }_{... | Yes |
Proposition 1.3 Suppose a and \( m \) are fixed, with \( a > 0 \) and \( - 1 < m < 0 \) . Then as \( \left| s\right| \rightarrow \infty \) with \( \operatorname{Re}\left( s\right) \geq 0 \) , \[ {\int }_{0}^{a}{e}^{-{sx}}{x}^{m}{dx} = {s}^{-m - 1}\Gamma \left( {m + 1}\right) + O\left( {1/\left| s\right| }\right) . \] (... | Proof. We begin by showing that when \( \operatorname{Re}\left( s\right) \geq 0, s \neq 0 \) , \[ {\int }_{0}^{\infty }{e}^{-{sx}}{x}^{m}{dx} = \mathop{\lim }\limits_{{N \rightarrow \infty }}{\int }_{0}^{N}{e}^{-{sx}}{x}^{m}{dx} \] exists and equals \( {s}^{-m - 1}\Gamma \left( {m + 1}\right) \) . If \( N \) is large, ... | Yes |
Proposition 2.1 Under the above assumptions, with \( s > 0 \) and \( s \rightarrow \infty \) ,\n\n\[{\int }_{a}^{b}{e}^{-{s\Phi }\left( x\right) }\psi \left( x\right) {dx} = {e}^{-{s\Phi }\left( {x}_{0}\right) }\left\lbrack {\frac{A}{{s}^{1/2}} + O\left( \frac{1}{s}\right) }\right\rbrack ,\]\n\nwhere\n\n\[A = \sqrt{2\p... | Proof. By replacing \( \Phi \left( x\right) \) by \( \Phi \left( x\right) - \Phi \left( {x}_{0}\right) \) we may assume that \( \Phi \left( {x}_{0}\right) = 0 \) . Since \( {\Phi }^{\prime }\left( {x}_{0}\right) = 0 \), we note that\n\n\[ \frac{\Phi \left( x\right) }{{\left( x - {x}_{0}\right) }^{2}} = \frac{{\Phi }^{\... | Yes |
Proposition 2.2 With the same assumptions on \( \Phi \) and \( \psi \), the relation (8) continues to hold if \( \left| s\right| \rightarrow \infty \) with \( \operatorname{Re}\left( s\right) \geq 0 \) . | Proof. We proceed as before to the equation (9), and obtain the appropriate asymptotic for the first term, by virtue of Proposition 1.3, with \( m = - 1/2 \) . To deal with the rest we start with an observation. If \( \Psi \) and \( \psi \) are given on an interval \( \left\lbrack {\bar{a},\bar{b}}\right\rbrack \), are... | Yes |
Theorem 3.1 Suppose \( u > 0 \) . Then as \( u \rightarrow \infty \) , (i) \( \operatorname{Ai}\left( {-u}\right) = {\pi }^{-1/2}{u}^{-1/4}\cos \left( {\frac{2}{3}{u}^{3/2} - \frac{\pi }{4}}\right) \left( {1 + O\left( {1/{u}^{3/4}}\right) }\right) \) . | To consider the first case, we make the change of variables \( x \mapsto {u}^{1/2}x \) in the defining integral with \( s = - u \) . This gives \[ \operatorname{Ai}\left( {-u}\right) = {u}^{1/2}{I}_{ - }\left( {u}^{3/2}\right) \] where \[ {I}_{ - }\left( t\right) = \frac{1}{2\pi }{\int }_{-\infty }^{\infty }{e}^{{it}\l... | Yes |
Theorem 1.1 A region \( \Omega \) is holomorphically simply connected if and only if \( \Omega \) is simply connected. | Proof. One direction is simply the version of Cauchy's theorem in Corollary 5.3, Chapter 3. Conversely, suppose that \( \Omega \) is holomorphically simply connected. If \( \Omega = \mathbb{C} \), then it is clearly simply connected. If \( \Omega \) is not all of \( \mathbb{C} \), recall that the Riemann mapping theore... | No |
Lemma 1.3 Let \( \gamma \) be a closed curve in \( \mathbb{C} \). (i) If \( z \notin \gamma \), then \( {W}_{\gamma }\left( z\right) \in \mathbb{Z} \). (ii) If \( z \) and \( w \) belong to the same open connected component in the complement of \( \gamma \), then \( {W}_{\gamma }\left( z\right) = {W}_{\gamma }\left( w\... | Proof. To see why (i) is true, suppose that \( \gamma : \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{C} \) is a parametrization for the curve, and let \[ G\left( t\right) = {\int }_{0}^{t}\frac{{\gamma }^{\prime }\left( s\right) }{\gamma \left( s\right) - z}{ds}. \] Then \( G \) is continuous and, except possibl... | Yes |
Theorem 1.4 A bounded region \( \Omega \) is simply connected if and only if \( {W}_{\gamma }\left( z\right) = 0 \) for any closed curve \( \gamma \) in \( \Omega \) and any point \( z \) not in \( \Omega \) . | Proof. If \( \Omega \) is simply connected and \( z \notin \Omega \), then \( f\left( \zeta \right) = 1/\left( {\zeta - z}\right) \) is holomorphic in \( \Omega \), and Cauchy’s theorem gives \( {W}_{\gamma }\left( z\right) = 0 \) . For the converse, it suffices to prove that the complement of \( \Omega \) is connected... | Yes |
Lemma 1.5 Let \( w \) be any point in \( {F}_{1} \) . Under the above assumptions, there exists a finite collection of closed squares \( \mathcal{Q} = \left\{ {{Q}_{1},\ldots ,{Q}_{n}}\right\} \) that belong to a uniform grid \( \mathcal{G} \) of the plane, and are such that:\n\n(i) \( w \) belongs to the interior of \... | Proof of the lemma. Since \( {F}_{2} \) is closed, the sets \( {F}_{1} \) and \( {F}_{2} \) are at a finite non-zero distance \( d \) from one another. Now consider a uniform grid \( {\mathcal{G}}_{0} \) of the plane consisting of closed squares of side length which is much smaller than \( d \), say \( < d/{100} \), an... | Yes |
Theorem 2.2 Let \( \Gamma \) be a curve in the plane which is simple, closed, and piecewise-smooth. Then, the complement of \( \Gamma \) consists of two disjoint connected open sets. Precisely one of these regions is bounded and simply connected; it is called the interior of \( \Gamma \) and denoted by \( \Omega \) . T... | Moreover, with the appropriate orientation for \( \Gamma \), we have\n\n\[ \n{W}_{\Gamma }\left( z\right) = \left\{ \begin{array}{ll} 1 & \text{ if }z \in \Omega \\ 0 & \text{ if }z \in \mathcal{U} \end{array}\right. \n\] | No |
Lemma 2.5 Suppose \( z \) is a point which does not belong to the smooth curve \( {\Gamma }_{0} \), but that is closer to an interior point of the curve than to either of its end-points. Then \( z \) belongs to \( {\Gamma }_{\epsilon } \) for some \( \epsilon \neq 0 \). More precisely, if \( {z}_{0} \in {\Gamma }_{0} \... | Proof. For \( t \) in a neighborhood of \( {t}_{0} \) the fact that \( \gamma \) is differentiable guarantees that \[ z - \gamma \left( t\right) = z - \gamma \left( {t}_{0}\right) - {\gamma }^{\prime }\left( {t}_{0}\right) \left( {t - {t}_{0}}\right) + o\left( \left| {t - {t}_{0}}\right| \right) . \] Since \( {z}_{0} =... | Yes |
Proposition 2.6 Let \( A \) and \( B \) denote the two end-points of a simple smooth curve \( {\Gamma }_{0} \), and suppose that \( K \) is a compact set that satisfies either \[ {\Gamma }_{0} \cap K = \varnothing \;\text{ or }\;{\Gamma }_{0} \cap K = A \cup B. \] If \( z \notin {\Gamma }_{0} \) and \( w \notin {\Gamma... | Proof. By the previous lemma, consider \( {z}_{0} = \gamma \left( {t}_{0}\right) \) and \( {w}_{0} = \gamma \left( {s}_{0}\right) \) that are interior points of \( {\Gamma }_{0} \) closest to \( z \) and \( w \), respectively. Then \[ z = \gamma \left( {t}_{0}\right) + i{\epsilon }_{0}{\gamma }^{\prime }\left( {t}_{0}\... | Yes |
Lemma 2.7 Let \( {\Gamma }_{0} \) be a simple smooth curve. There exists \( {\kappa }_{2} > 0 \) so that the set \( N \), which consists of points of the form \( z = \gamma \left( L\right) + \epsilon {e}^{i\theta }{\gamma }^{\prime }\left( L\right) \) with \( - \pi /2 \leq \theta \leq \pi /2 \) and \( 0 < \epsilon < {\... | Proof. The argument is similar to the one given in the proof of Lemma 2.4. First, we note that\n\n\[ \gamma \left( L\right) + \epsilon {e}^{i\theta }{\gamma }^{\prime }\left( L\right) - \gamma \left( t\right) = {\int }_{t}^{L}\left\lbrack {{\gamma }^{\prime }\left( u\right) - {\gamma }^{\prime }\left( L\right) }\right\... | Yes |
Proposition 2.8 Let \( A \) denote an end-point of the simple smooth curve \( {\Gamma }_{0} \), and suppose that \( K \) is a compact set that satisfies either\n\n\[ \n{\Gamma }_{0} \cap K = \varnothing \;\text{ or }\;{\Gamma }_{0} \cap K = A.\n\]\n\nIf \( z \notin {\Gamma }_{0} \) and \( w \notin {\Gamma }_{0} \) are ... | We only provide an outline of the argument, which is similar to the proof of Proposition 2.6. It suffices to consider the case when \( z \) and \( w \) lie on opposite sides of \( {\Gamma }_{0} \) and \( A = \gamma \left( 0\right) \) . First, we may join\n\n\[ \n{z}_{\epsilon } = \gamma \left( {t}_{0}\right) + {i\epsil... | No |
Theorem 2.9 If a function \( f \) is holomorphic in an open set that contains a simple closed piecewise-smooth curve \( \Gamma \) and its interior, then\n\n\[{\int }_{\Gamma }f = 0\] | Let \( \mathcal{O} \) denote an open set on which \( f \) is holomorphic, and which contains \( \Gamma \) and its interior \( \Omega \) . The idea is to construct a closed curve \( \Lambda \)\nin \( \Omega \) that is so close to \( \Gamma \) that \( {\int }_{\Gamma }f = {\int }_{\Lambda }f \) . Then, the integral on th... | Yes |
Lemma 2.10 Let \( \gamma : \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{C} \) be a simple smooth curve. Then, for all sufficiently small \( \delta > 0 \) the circle \( {C}_{\delta } \) centered at \( \gamma \left( 0\right) \) and of radius \( \delta \) intersects \( \gamma \) in precisely one point. | Proof. We may assume that \( \gamma \left( 0\right) = 0 \) . Since \( \gamma \left( 0\right) \neq \gamma \left( 1\right) \) it is clear that for each small \( \delta > 0 \), the circle \( {C}_{\delta } \) intersects \( \gamma \) in at least one point. If the conclusion in the lemma is false, we can find a sequence of p... | Yes |
Lemma 1.1 If a rectangle is the almost disjoint union of finitely many other rectangles, say \( R = \mathop{\bigcup }\limits_{{k = 1}}^{N}{R}_{k} \), then\n\n\[ \left| R\right| = \mathop{\sum }\limits_{{k = 1}}^{N}\left| {R}_{k}\right| \] | Proof. We consider the grid formed by extending indefinitely the sides of all rectangles \( {R}_{1},\ldots ,{R}_{N} \) . This construction yields finitely many rectangles \( {\widetilde{R}}_{1},\ldots ,{\widetilde{R}}_{M} \), and a partition \( {J}_{1},\ldots ,{J}_{N} \) of the integers between 1 and \( M \), such that... | Yes |
Lemma 1.2 If \( R,{R}_{1},\ldots ,{R}_{N} \) are rectangles, and \( R \subset \mathop{\bigcup }\limits_{{k = 1}}^{N}{R}_{k} \), then\n\n\[ \left| R\right| \leq \mathop{\sum }\limits_{{k = 1}}^{N}\left| {R}_{k}\right| \] | The main idea consists of taking the grid formed by extending all sides of the rectangles \( R,{R}_{1},\ldots ,{R}_{N} \), and noting that the sets corresponding to the \( {J}_{k} \) (in the above proof) need not be disjoint any more. | No |
Theorem 1.3 Every open subset \( \mathcal{O} \) of \( \mathbb{R} \) can be writen uniquely as a countable union of disjoint open intervals. | Proof. For each \( x \in \mathcal{O} \), let \( {I}_{x} \) denote the largest open interval containing \( x \) and contained in \( \mathcal{O} \) . More precisely, since \( \mathcal{O} \) is open, \( x \) is contained in some small (non-trivial) interval, and therefore if\n\n\[ \n{a}_{x} = \inf \{ a < x : \left( {a, x}... | Yes |
Theorem 1.4 Every open subset \( \mathcal{O} \) of \( {\mathbb{R}}^{d}, d \geq 1 \), can be written as a countable union of almost disjoint closed cubes. | Proof. We must construct a countable collection \( \mathcal{Q} \) of closed cubes whose interiors are disjoint, and so that \( \mathcal{O} = \mathop{\bigcup }\limits_{{Q \in \mathcal{Q}}}Q \) .\n\nAs a first step, consider the grid in \( {\mathbb{R}}^{d} \) formed by taking all closed cubes of side length 1 whose verti... | Yes |
Property 2 If \( {m}_{ * }\left( E\right) = 0 \), then \( E \) is measurable. In particular, if \( F \) is a subset of a set of exterior measure 0, then \( F \) is measurable. | By Observation 3 of the exterior measure, for every \( \epsilon > 0 \) there exists an open set \( \mathcal{O} \) with \( E \subset \mathcal{O} \) and \( {m}_{ * }\left( \mathcal{O}\right) \leq \epsilon \) . Since \( \left( {\mathcal{O} - E}\right) \subset \mathcal{O} \) , monotonicity implies \( {m}_{ * }\left( {\math... | No |
Property 3 A countable union of measurable sets is measurable. | Suppose \( E = \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{E}_{j} \), where each \( {E}_{j} \) is measurable. Given \( \epsilon > 0 \), we may choose for each \( j \) an open set \( {\mathcal{O}}_{j} \) with \( {E}_{j} \subset {\mathcal{O}}_{j} \) and \( {m}_{ * }\left( {{\mathcal{O}}_{j} - {E}_{j}}\right) \leq \epsil... | Yes |
Property 4 Closed sets are measurable. | First, we observe that it suffices to prove that compact sets are measurable. Indeed, any closed set \( F \) can be written as the union of compact sets, say \( F = \mathop{\bigcup }\limits_{{k = 1}}^{\infty }F \cap {B}_{k} \), where \( {B}_{k} \) denotes the closed ball of radius \( k \) centered at the origin; then P... | Yes |
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