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Lemma 3.8 If \( s > 1 \), then\n\n\[ \mathop{\prod }\limits_{\chi }L\left( {s,\chi }\right) \geq 1 \]\n\nwhere the product is taken over all Dirichlet characters. In particular the product is real-valued. | Proof. We have shown earlier that for \( s > 1 \)\n\n\[ L\left( {s,\chi }\right) = \exp \left( {\mathop{\sum }\limits_{p}{\log }_{1}\left( \frac{1}{1 - \chi \left( p\right) {p}^{-s}}\right) }\right) .\n\nHence,\n\n\[ \mathop{\prod }\limits_{\chi }L\left( {s,\chi }\right) = \exp \left( {\mathop{\sum }\limits_{\chi }\mat... | Yes |
Lemma 3.9 The following three properties hold:\n\n(i) If \( L\\left( {1,\\chi }\\right) = 0 \), then \( L\\left( {1,\\bar{\\chi }}\\right) = 0 \) .\n\n(ii) If \( \\chi \) is non-trivial and \( L\\left( {1,\\chi }\\right) = 0 \), then\n\n\[ \n\\left| {L\\left( {s,\\chi }\\right) }\\right| \\leq C\\left| {s - 1}\\right| ... | Proof. The first statement is immediate because \( L\\left( {1,\\bar{\\chi }}\\right) = \\overline{L\\left( {1,\\chi }\\right) } \) . The second statement follows from the mean-value theorem since \( L\\left( {s,\\chi }\\right) \) is continuously differentiable for \( s > 0 \) when \( \\chi \) is non-trivial. Finally, ... | Yes |
Proposition 3.10 If \( N \) is a positive integer, then:\n\n(i) \( \mathop{\sum }\limits_{{1 \leq n \leq N}}\frac{1}{n} = {\int }_{1}^{N}\frac{dx}{x} + O\left( 1\right) = \log N + O\left( 1\right) \).\n\n(ii) More precisely, there exists a real number \( \gamma \), called Euler’s constant, so that\n\n\[ \mathop{\sum }\... | Proof. It suffices to establish the more refined estimate given in part (ii). Let\n\n\[ {\gamma }_{n} = \frac{1}{n} - {\int }_{n}^{n + 1}\frac{dx}{x} \]\n\nSince \( 1/x \) is decreasing, we clearly have\n\n\[ 0 \leq {\gamma }_{n} \leq \frac{1}{n} - \frac{1}{n + 1} \leq \frac{1}{{n}^{2}} \]\n\nso the series \( \mathop{\... | Yes |
Proposition 3.11 If \( N \) is a positive integer, then\n\n\[ \mathop{\sum }\limits_{{1 \leq n \leq N}}\frac{1}{{n}^{1/2}} = {\int }_{1}^{N}\frac{dx}{{x}^{1/2}} + {c}^{\prime } + O\left( {1/{N}^{1/2}}\right) \]\n\n\[ = 2{N}^{1/2} + c + O\left( {1/{N}^{1/2}}\right) \text{.} \] | The proof is essentially a repetition of the proof of the previous proposition, this time using the fact that\n\n\[ \left| {\frac{1}{{n}^{1/2}} - \frac{1}{{\left( n + 1\right) }^{1/2}}}\right| \leq \frac{C}{{n}^{3/2}} \]\n\nThis last inequality follows from the mean-value theorem applied to \( f\left( x\right) = {x}^{-... | Yes |
Theorem 3.12 If \( k \) is a positive integer, then\n\n\[ \n\frac{1}{N}\mathop{\sum }\limits_{{k = 1}}^{N}d\left( k\right) = \log N + O\left( 1\right) \n\]\n\nMore precisely,\n\n\[ \n\frac{1}{N}\mathop{\sum }\limits_{{k = 1}}^{N}d\left( k\right) = \log N + \left( {{2\gamma } - 1}\right) + O\left( {1/{N}^{1/2}}\right) ,... | Proof. Let \( {S}_{N} = \mathop{\sum }\limits_{{k = 1}}^{N}d\left( k\right) \) . We observed that summing \( F = 1 \) along hyperbolas gives \( {S}_{N} \) . Summing vertically, we find\n\n\[ \n{S}_{N} = \mathop{\sum }\limits_{{1 \leq m \leq N}}\mathop{\sum }\limits_{{1 \leq n \leq N/m}}1 \n\]\n\nBut \( \mathop{\sum }\l... | Yes |
Lemma 3.14 \( \mathop{\sum }\limits_{{n \mid k}}\chi \left( n\right) \geq \left\{ \begin{array}{ll} 0 & \text{ for all }k \\ 1 & \text{ if }k = {\ell }^{2}\text{ for some }\ell \in \mathbb{Z}. \end{array}\right. \) | The proof of the lemma is simple. If \( k \) is a power of a prime, say \( k = {p}^{a} \), then the divisors of \( k \) are \( 1, p,{p}^{2},\ldots ,{p}^{a} \) and\n\n\[ \mathop{\sum }\limits_{{n \mid k}}\chi \left( n\right) = \chi \left( 1\right) + \chi \left( p\right) + \chi \left( {p}^{2}\right) + \cdots + \chi \left... | Yes |
Lemma 3.15 For all integers \( 0 < a < b \) we have\n\n(i) \( \mathop{\sum }\limits_{{n = a}}^{b}\frac{\chi \left( n\right) }{{n}^{1/2}} = O\left( {a}^{-1/2}\right) \), \n\n(ii) \( \mathop{\sum }\limits_{{n = a}}^{b}\frac{\chi \left( n\right) }{n} = O\left( {a}^{-1}\right) \). | Proof. This argument is similar to the proof of Proposition 3.4; we use summation by parts. Let \( {s}_{n} = \mathop{\sum }\limits_{{1 \leq k \leq n}}\chi \left( k\right) \), and remember that \( \left| {s}_{n}\right| \leq q \) for all \( n \) . Then\n\n\[ \mathop{\sum }\limits_{{n = a}}^{b}\frac{\chi \left( n\right) }... | Yes |
Proposition 1.1 If \( f \) and \( g \) are integrable on \( \left\lbrack {a, b}\right\rbrack \), then:\n\n(i) \( f + g \) is integrable, and \( {\int }_{a}^{b}f\left( x\right) + g\left( x\right) {dx} = {\int }_{a}^{b}f\left( x\right) {dx} + {\int }_{a}^{b}g\left( x\right) {dx} \) . | Proof. For property (i) we may assume that \( f \) and \( g \) are real-valued. If \( P \) is a partition of \( \left\lbrack {a, b}\right\rbrack \), then\n\n\[ \mathcal{U}\left( {P, f + g}\right) \leq \mathcal{U}\left( {P, f}\right) + \mathcal{U}\left( {P, g}\right) \;\text{ and }\;\mathcal{L}\left( {P, f + g}\right) \... | Yes |
Lemma 1.2 If \( f \) is real-valued integrable on \( \left\lbrack {a, b}\right\rbrack \) and \( \varphi \) is a real-valued continuous function on \( \mathbb{R} \), then \( \varphi \circ f \) is also integrable on \( \left\lbrack {a, b}\right\rbrack \) . | Proof. Let \( \epsilon > 0 \) and remember that \( f \) is bounded, say \( \left| f\right| \leq M \) . Since \( \varphi \) is uniformly continuous on \( \left\lbrack {-M, M}\right\rbrack \) we may choose \( \delta > 0 \) so that if \( s, t \in \left\lbrack {-M, M}\right\rbrack \) and \( \left| {s - t}\right| < \delta \... | Yes |
Proposition 1.3 A bounded monotonic function \( f \) on an interval \( \left\lbrack {a, b}\right\rbrack \) is integrable. | Proof. We may assume without loss of generality that \( a = 0, b = 1 \) , and \( f \) is monotonically increasing. Then, for each \( N \), we choose the uniform partition \( {P}_{N} \) given by \( {x}_{j} = j/N \) for all \( j = 0,\ldots, N \) . If \( {\alpha }_{j} = \) \( f\left( {x}_{j}\right) \), then we have\n\n\[ ... | Yes |
Proposition 1.4 Let \( f \) be a bounded function on the compact interval \( \left\lbrack {a, b}\right\rbrack \) . If \( c \in \left( {a, b}\right) \), and if for all small \( \delta > 0 \) the function \( f \) is integrable on the intervals \( \left\lbrack {a, c - \delta }\right\rbrack \) and \( \left\lbrack {c + \del... | Proof. Suppose \( \left| f\right| \leq M \) and let \( \epsilon > 0 \) . Choose \( \delta > 0 \) (small) so that \( {4\delta M} \leq \epsilon /3 \) . Now let \( {P}_{1} \) and \( {P}_{2} \) be partitions of \( \left\lbrack {a, c - \delta }\right\rbrack \) and \( \lbrack c + \) \( \delta, b\rbrack \) so that for each \(... | Yes |
Lemma 1.6 The union of countably many sets of measure 0 has measure 0. | Proof. Say \( {E}_{1},{E}_{2},\ldots \) are sets of measure 0, and let \( E = { \cup }_{i = 1}^{\infty }{E}_{i} \) . Let \( \epsilon > 0 \), and for each \( i \) choose open interval \( {I}_{i,1},{I}_{i,2},\ldots \) so that\n\n\[ \n{E}_{i} \subset \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{I}_{i, k}\;\text{ and }\;\m... | Yes |
Lemma 1.8 If \( \epsilon > 0 \), then the set \( {A}_{\epsilon } \) is closed and therefore compact. | Proof. The argument is simple. Suppose \( {c}_{n} \in {A}_{\epsilon } \) converges to \( c \) and assume that \( c \notin {A}_{\epsilon } \) . Write \( \operatorname{osc}\left( {f, c}\right) = \epsilon - \delta \) where \( \delta > 0 \) . Select \( r \) so that \( \operatorname{osc}\left( {f, c, r}\right) < \epsilon - ... | No |
Theorem 2.1 Let \( f \) be a continuous function defined on a closed rectangle \( R \subset {\mathbb{R}}^{d} \) . Suppose \( R = {R}_{1} \times {R}_{2} \) where \( {R}_{1} \subset {\mathbb{R}}^{{d}_{1}} \) and \( {R}_{2} \subset {\mathbb{R}}^{{d}_{2}} \) with \( d = {d}_{1} + {d}_{2} \) . If we write \( x = \left( {{x}... | Proof. The continuity of \( F \) follows from the uniform continuity of \( f \) on \( R \) and the fact that\n\n\[ \n\left| {F\left( {x}_{1}\right) - F\left( {x}_{1}^{\prime }\right) }\right| \leq {\int }_{{R}_{2}}\left| {f\left( {{x}_{1},{x}_{2}}\right) - f\left( {{x}_{1}^{\prime },{x}_{2}}\right) }\right| d{x}_{2}.\n... | Yes |
Theorem 2.2 Suppose \( A \) and \( B \) are compact subsets of \( {\mathbb{R}}^{d} \) and \( g : A \rightarrow B \) is a diffeomorphism of class \( {C}^{1} \) . If \( f \) is continuous on \( B \) , then\n\n\[ \n{\int }_{g\left( A\right) }f\left( x\right) {dx} = {\int }_{A}f\left( {g\left( y\right) }\right) \left| {\de... | The proof of this theorem consists first of an analysis of the special situation when \( g \) is a linear transformation \( L \) . In this case, if \( R \) is a rectangle, then\n\n\[ \n\left| {g\left( R\right) }\right| = \left| {\det \left( L\right) }\right| \left| R\right|\n\]\nwhich explains the term \( \left| {\det ... | Yes |
Lemma 1 Suppose that there exists some \( \omega \in \Omega \) such that \( \rho \left( \widetilde{\omega }\right) > 0 \) and \( M\left( \widetilde{\omega }\right) \) is a \( K - {R}_{0} \) matrix. Then the function \( f \) given in (3) is coercive, that is, \( f\left( \mathbf{x}\right) \) tends to \( + \infty \) as \(... | Proof It is obvious that\n\n\[ f\left( \mathbf{x}\right) = \mathrm{E}\left\lbrack {\parallel \Phi \left( {\mathbf{x},\omega }\right) {\parallel }^{2} + \parallel F\left( {\mathbf{x},\omega }\right) {\parallel }^{2}}\right\rbrack \geq \mathrm{E}\left\lbrack {\parallel \Phi \left( {\mathbf{x},\omega }\right) {\parallel }... | No |
Theorem 1 Under Assumption 1, for any fixed \( \mathbf{x} \), we have\n\n\[ f\left( \mathbf{x}\right) = \mathop{\lim }\limits_{{k \rightarrow \infty }}{f}_{k}\left( \mathbf{x}\right) ,\;\text{ w. p. }1. \] | Proof By (4), we only need to show the integrability of the function \( \parallel \Phi \left( {\mathbf{x},\omega }\right) {\parallel }^{2} + \parallel F\left( {\mathbf{x},\omega }\right) {\parallel }^{2} \) over \( \Omega \) . Since \( \Omega \) is a nonempty and compact set in a finite dimensional Euclidean space, and... | Yes |
Proposition 1.4 If \( {\Omega }_{1} \supset {\Omega }_{2} \supset \cdots \supset {\Omega }_{n} \supset \cdots \) is a sequence of non-empty compact sets in \( \mathbb{C} \) with the property that\n\n\[ \operatorname{diam}\left( {\Omega }_{n}\right) \rightarrow 0\;\text{ as }n \rightarrow ∞ ,\]\n\nthen there exists a un... | Proof. Choose a point \( {z}_{n} \) in each \( {\Omega }_{n} \) . The condition \( \operatorname{diam}\left( {\Omega }_{n}\right) \rightarrow 0 \) says precisely that \( \left\{ {z}_{n}\right\} \) is a Cauchy sequence, therefore this sequence converges to a limit that we call \( w \) . Since each set \( {\Omega }_{n} \... | Yes |
Theorem 2.1 A continuous function on a compact set \( \Omega \) is bounded and attains a maximum and minimum on \( \Omega \) . | This is of course analogous to the situation of functions of a real variable, and we shall not repeat the simple proof here. | No |
Proposition 2.3 If \( f \) is holomorphic at \( {z}_{0} \), then\n\n\[ \n\frac{\partial f}{\partial \bar{z}}\left( {z}_{0}\right) = 0\;\text{ and }\;{f}^{\prime }\left( {z}_{0}\right) = \frac{\partial f}{\partial z}\left( {z}_{0}\right) = 2\frac{\partial u}{\partial z}\left( {z}_{0}\right) .\n\]\n\nAlso, if we write \(... | Proof. Taking real and imaginary parts, it is easy to see that the Cauchy-Riemann equations are equivalent to \( \partial f/\partial \bar{z} = 0 \) . Moreover, by our earlier observation\n\n\[ \n{f}^{\prime }\left( {z}_{0}\right) = \frac{1}{2}\left( {\frac{\partial f}{\partial x}\left( {z}_{0}\right) + \frac{1}{i}\frac... | Yes |
Theorem 2.4 Suppose \( f = u + {iv} \) is a complex-valued function defined on an open set \( \Omega \) . If \( u \) and \( v \) are continuously differentiable and satisfy the Cauchy-Riemann equations on \( \Omega \), then \( f \) is holomorphic on \( \Omega \) and \( {f}^{\prime }\left( z\right) = \partial f/\partial... | Proof. Write\n\n\[ u\left( {x + {h}_{1}, y + {h}_{2}}\right) - u\left( {x, y}\right) = \frac{\partial u}{\partial x}{h}_{1} + \frac{\partial u}{\partial y}{h}_{2} + \left| h\right| {\psi }_{1}\left( h\right) \]\n\nand\n\n\[ v\left( {x + {h}_{1}, y + {h}_{2}}\right) - v\left( {x, y}\right) = \frac{\partial v}{\partial x... | Yes |
Theorem 2.5 Given a power series \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{z}^{n} \), there exists \( 0 \leq R \leq \infty \) such that:\n\n(i) If \( \left| z\right| < R \) the series converges absolutely.\n\n(ii) If \( \left| z\right| > R \) the series diverges.\n\nMoreover, if we use the convention that \( ... | Proof. Let \( L = 1/R \) where \( R \) is defined by the formula in the statement of the theorem, and suppose that \( L \neq 0,\infty \) . (These two easy cases are left as an exercise.) If \( \left| z\right| < R \), choose \( \epsilon > 0 \) so small that\n\n\[ \left( {L + \epsilon }\right) \left| z\right| = r < 1 \]\... | No |
Corollary 2.7 A power series is infinitely complex differentiable in its disc of convergence, and the higher derivatives are also power series obtained by termwise differentiation. | We have so far dealt only with power series centered at the origin. More generally, a power series centered at \( {z}_{0} \in \mathbb{C} \) is an expression of the form\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( z - {z}_{0}\right) }^{n}. \]\n\nThe disc of convergence of \( f \) is ... | Yes |
Proposition 3.1 Integration of continuous functions over curves satisfies the following properties:\n\n(i) It is linear, that is, if \( \alpha ,\beta \in \mathbb{C} \), then\n\n\[ \n{\int }_{\gamma }\left( {{\alpha f}\left( z\right) + {\beta g}\left( z\right) }\right) {dz} = \alpha {\int }_{\gamma }f\left( z\right) {dz... | Proof. The first property follows from the definition and the linearity of the Riemann integral. The second property is left as an exercise. For the third, note that\n\n\[ \n\left| {{\int }_{\gamma }f\left( z\right) {dz}}\right| \leq \mathop{\sup }\limits_{{t \in \left\lbrack {a, b}\right\rbrack }}\left| {f\left( {z\le... | No |
Theorem 3.2 If a continuous function \( f \) has a primitive \( F \) in \( \Omega \), and \( \gamma \) is a curve in \( \Omega \) that begins at \( {w}_{1} \) and ends at \( {w}_{2} \), then\n\n\[{\int }_{\gamma }f\left( z\right) {dz} = F\left( {w}_{2}\right) - F\left( {w}_{1}\right)\] | Proof. If \( \gamma \) is smooth, the proof is a simple application of the chain rule and the fundamental theorem of calculus. Indeed, if \( z\left( t\right) : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{C} \) is a parametrization for \( \gamma \), then \( z\left( a\right) = {w}_{1} \) and \( z\left( b\right) ... | Yes |
Corollary 3.3 If \( \gamma \) is a closed curve in an open set \( \Omega \), and \( f \) is continuous and has a primitive in \( \Omega \), then\n\n\[{\int }_{\gamma }f\left( z\right) {dz} = 0\] | This is immediate since the end-points of a closed curve coincide. | Yes |
Corollary 3.4 If \( f \) is holomorphic in a region \( \Omega \) and \( {f}^{\prime } = 0 \), then \( f \) is constant. | Proof. Fix a point \( {w}_{0} \in \Omega \) . It suffices to show that \( f\left( w\right) = f\left( {w}_{0}\right) \) for all \( w \in \Omega \) .\n\nSince \( \Omega \) is connected, for any \( w \in \Omega \), there exists a curve \( \gamma \) which joins \( {w}_{0} \) to \( w \) . Since \( f \) is clearly a primitiv... | Yes |
Corollary 1.2 If \( f \) is holomorphic in an open set \( \Omega \) that contains a rectangle \( R \) and its interior, then\n\n\[{\int }_{R}f\left( z\right) {dz} = 0\] | This is immediate since we first choose an orientation as in Figure 2 and note that\n\n\[{\int }_{R}f\left( z\right) {dz} = {\int }_{{T}_{1}}f\left( z\right) {dz} + {\int }_{{T}_{2}}f\left( z\right) {dz}.\] | Yes |
Theorem 2.2 (Cauchy's theorem for a disc) If \( f \) is holomorphic in a disc, then\n\n\[{\int }_{\gamma }f\left( z\right) {dz} = 0\]\n\nfor any closed curve \( \gamma \) in that disc. | Proof. Since \( f \) has a primitive, we can apply Corollary 3.3 of Chapter 1. | No |
Corollary 2.3 Suppose \( f \) is holomorphic in an open set containing the circle \( C \) and its interior. Then\n\n\[{\int }_{C}f\left( z\right) {dz} = 0\] | Proof. Let \( D \) be the disc with boundary circle \( C \) . Then there exists a slightly larger disc \( {D}^{\prime } \) which contains \( D \) and so that \( f \) is holomorphic on \( {D}^{\prime } \) . We may now apply Cauchy’s theorem in \( {D}^{\prime } \) to conclude that \( {\int }_{C}f\left( z\right) {dz} = 0.... | Yes |
Theorem 4.1 Suppose \( f \) is holomorphic in an open set that contains the closure of a disc D. If \( C \) denotes the boundary circle of this disc with the positive orientation, then\n\n\[ f\left( z\right) = \frac{1}{2\pi i}{\int }_{C}\frac{f\left( \zeta \right) }{\zeta - z}{d\zeta }\;\text{ for any point }z \in D. \... | Proof. Fix \( z \in D \) and consider the \ | No |
If \( f \) is holomorphic in an open set \( \Omega \), then \( f \) has infinitely many complex derivatives in \( \Omega \) . Moreover, if \( C \subset \Omega \) is a circle whose interior is also contained in \( \Omega \), then\n\n\[ \n{f}^{\left( n\right) }\left( z\right) = \frac{n!}{2\pi i}{\int }_{C}\frac{f\left( \... | The proof is by induction on \( n \), the case \( n = 0 \) being simply the Cauchy integral formula. Suppose that \( f \) has up to \( n - 1 \) complex derivatives and that\n\n\[ \n{f}^{\left( n - 1\right) }\left( z\right) = \frac{\left( {n - 1}\right) !}{2\pi i}{\int }_{C}\frac{f\left( \zeta \right) }{{\left( \zeta - ... | Yes |
Corollary 4.3 (Cauchy inequalities) If \( f \) is holomorphic in an open set that contains the closure of a disc \( D \) centered at \( {z}_{0} \) and of radius \( R \) , then\n\n\[ \left| {{f}^{\left( n\right) }\left( {z}_{0}\right) }\right| \leq \frac{n!\parallel f{\parallel }_{C}}{{R}^{n}} \]\n\nwhere \( \parallel f... | Proof. Applying the Cauchy integral formula for \( {f}^{\left( n\right) }\left( {z}_{0}\right) \), we obtain\n\n\[ \left| {{f}^{\left( n\right) }\left( {z}_{0}\right) }\right| = \left| {\frac{n!}{2\pi i}{\int }_{C}\frac{f\left( \zeta \right) }{{\left( \zeta - {z}_{0}\right) }^{n + 1}}{d\zeta }}\right| \]\n\n\[ = \frac{... | Yes |
Theorem 4.4 Suppose \( f \) is holomorphic in an open set \( \Omega \) . If \( D \) is a disc centered at \( {z}_{0} \) and whose closure is contained in \( \Omega \), then \( f \) has a power series expansion at \( {z}_{0} \) \n\n\[ \nf\left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( z - {z}_{... | Proof. Fix \( z \in D \) . By the Cauchy integral formula, we have \n\n(10) \n\n\[ \nf\left( z\right) = \frac{1}{2\pi i}{\int }_{C}\frac{f\left( \zeta \right) }{\zeta - z}{d\zeta } \n\] \n\nwhere \( C \) denotes the boundary of the disc and \( z \in D \) . The idea is to write \n\n(11) \n\n\[ \n\frac{1}{\zeta - z} = \f... | Yes |
Corollary 4.5 (Liouville's theorem) If \( f \) is entire and bounded, then \( f \) is constant. | Proof. It suffices to prove that \( {f}^{\prime } = 0 \), since \( \mathbb{C} \) is connected, and we may then apply Corollary 3.4 in Chapter 1.\n\nFor each \( {z}_{0} \in \mathbb{C} \) and all \( R > 0 \), the Cauchy inequalities yield\n\n\[ \left| {{f}^{\prime }\left( {z}_{0}\right) }\right| \leq \frac{B}{R} \]\n\nwh... | Yes |
Corollary 4.6 Every non-constant polynomial \( P\left( z\right) = {a}_{n}{z}^{n} + \cdots + {a}_{0} \) with complex coefficients has a root in \( \mathbb{C} \) . | Proof. If \( P \) has no roots, then \( 1/P\left( z\right) \) is a bounded holomorphic function. To see this, we can of course assume that \( {a}_{n} \neq 0 \), and write\n\n\[ \frac{P\left( z\right) }{{z}^{n}} = {a}_{n} + \left( {\frac{{a}_{n - 1}}{z} + \cdots + \frac{{a}_{0}}{{z}^{n}}}\right) \]\n\nwhenever \( z \neq... | Yes |
Every polynomial \( P\left( z\right) = {a}_{n}{z}^{n} + \cdots + {a}_{0} \) of degree \( n \geq \) 1 has precisely \( n \) roots in \( \mathbb{C} \) . If these roots are denoted by \( {w}_{1},\ldots ,{w}_{n} \) , then \( P \) can be factored as \[ P\left( z\right) = {a}_{n}\left( {z - {w}_{1}}\right) \left( {z - {w}_{2... | Proof. By the previous result \( P \) has a root, say \( {w}_{1} \) . Then, writing \( z = \left( {z - {w}_{1}}\right) + {w}_{1} \), inserting this expression for \( z \) in \( P \), and using the binomial formula we get \[ P\left( z\right) = {b}_{n}{\left( z - {w}_{1}\right) }^{n} + \cdots + {b}_{1}\left( {z - {w}_{1}... | Yes |
Theorem 4.8 Suppose \( f \) is a holomorphic function in a region \( \Omega \) that vanishes on a sequence of distinct points with a limit point in \( \Omega \). Then \( f \) is identically 0. | Proof. Suppose that \( {z}_{0} \in \Omega \) is a limit point for the sequence \( {\left\{ {w}_{k}\right\} }_{k = 1}^{\infty } \) and that \( f\left( {w}_{k}\right) = 0 \). First, we show that \( f \) is identically zero in a small disc containing \( {z}_{0} \). For that, we choose a disc \( D \) centered at \( {z}_{0}... | Yes |
Theorem 5.1 Suppose \( f \) is a continuous function in the open disc \( D \) such that for any triangle \( T \) contained in \( D \)\n\n\[{\int }_{T}f\left( z\right) {dz} = 0\]\n\nthen \( f \) is holomorphic. | Proof. By the proof of Theorem 2.1 the function \( f \) has a primitive \( F \) in \( D \) that satisfies \( {F}^{\prime } = f \) . By the regularity theorem, we know that \( F \) is indefinitely (and hence twice) complex differentiable, and therefore \( f \) is holomorphic. | Yes |
Theorem 5.2 If \( {\left\{ {f}_{n}\right\} }_{n = 1}^{\infty } \) is a sequence of holomorphic functions that converges uniformly to a function \( f \) in every compact subset of \( \Omega \), then \( f \) is holomorphic in \( \Omega \) . | Proof. Let \( D \) be any disc whose closure is contained in \( \Omega \) and \( T \) any triangle in that disc. Then, since each \( {f}_{n} \) is holomorphic, Goursat’s theorem implies\n\n\[{\int }_{T}{f}_{n}\left( z\right) {dz} = 0\;\text{ for all }n.\]\n\nBy assumption \( {f}_{n} \rightarrow f \) uniformly in the cl... | Yes |
Theorem 5.3 Under the hypotheses of the previous theorem, the sequence of derivatives \( {\left\{ {f}_{n}^{\prime }\right\} }_{n = 1}^{\infty } \) converges uniformly to \( {f}^{\prime } \) on every compact subset of \( \Omega \) . | Proof. We may assume without loss of generality that the sequence of functions in the theorem converges uniformly on all of \( \Omega \) . Given \( \delta > 0 \) , let \( {\Omega }_{\delta } \) denote the subset of \( \Omega \) defined by\n\n\[ \n{\Omega }_{\delta } = \left\{ {z \in \Omega : \overline{{D}_{\delta }}\le... | Yes |
Theorem 5.4 Let \( F\left( {z, s}\right) \) be defined for \( \left( {z, s}\right) \in \Omega \times \left\lbrack {0,1}\right\rbrack \) where \( \Omega \) is an open set in \( \mathbb{C} \) . Suppose \( F \) satisfies the following properties:\n\n(i) \( F\left( {z, s}\right) \) is holomorphic in \( z \) for each \( s \... | Proof. For each \( n \geq 1 \), we consider the Riemann sum\n\n\[ {f}_{n}\left( z\right) = \left( {1/n}\right) \mathop{\sum }\limits_{{k = 1}}^{n}F\left( {z, k/n}\right) .\n\nThen \( {f}_{n} \) is holomorphic in all of \( \Omega \) by property (i), and we claim that on any disc \( D \) whose closure is contained in \( ... | Yes |
Theorem 5.5 (Symmetry principle) If \( {f}^{ + } \) and \( {f}^{ - } \) are holomorphic functions in \( {\Omega }^{ + } \) and \( {\Omega }^{ - } \) respectively, that extend continuously to \( I \) and\n\n\[{f}^{ + }\left( x\right) = {f}^{ - }\left( x\right) \;\text{ for all }x \in I,\]\nthen the function \( f \) defi... | Proof. One notes first that \( f \) is continuous throughout \( \Omega \) . The only difficulty is to prove that \( f \) is holomorphic at points of \( I \) . Suppose \( D \) is a disc centered at a point on \( I \) and entirely contained in \( \Omega \) . We prove that \( f \) is holomorphic in \( D \) by Morera’s the... | Yes |
Theorem 5.6 (Schwarz reflection principle) Suppose that \( f \) is a holomorphic function in \( {\Omega }^{ + } \) that extends continuously to \( I \) and such that \( f \) is real-valued on \( I \) . Then there exists a function \( F \) holomorphic in all of \( \Omega \) such that \( F = f \) on \( {\Omega }^{ + } \)... | Proof. The idea is simply to define \( F\left( z\right) \) for \( z \in {\Omega }^{ - } \) by\n\n\[ F\left( z\right) = \overline{f\left( \bar{z}\right) }.\]\n\nTo prove that \( F \) is holomorphic in \( {\Omega }^{ - } \) we note that if \( z,{z}_{0} \in {\Omega }^{ - } \), then \( \bar{z},\overline{{z}_{0}} \in {\Omeg... | Yes |
Lemma 5.8 Suppose \( f \) is holomorphic in an open set \( \Omega \), and \( K \subset \Omega \) is compact. Then, there exists finitely many segments \( {\gamma }_{1},\ldots ,{\gamma }_{N} \) in \( \Omega - K \) such that\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{n = 1}}^{N}\frac{1}{2\pi i}{\int }_{{\gamma }_{n... | Proof. Let \( d = c \cdot d\left( {K,{\Omega }^{c}}\right) \), where \( c \) is any constant \( < 1/\sqrt{2} \), and consider a grid formed by (solid) squares with sides parallel to the axis and of length \( d \) .\n\nWe let \( \mathcal{Q} = \left\{ {{Q}_{1},\ldots ,{Q}_{M}}\right\} \) denote the finite collection of s... | Yes |
Lemma 5.9 For any line segment \( \gamma \) entirely contained in \( \Omega - K \), there exists a sequence of rational functions with singularities on \( \gamma \) that approximate the integral \( {\int }_{\gamma }f\left( \zeta \right) /\left( {\zeta - z}\right) {d\zeta } \) uniformly on \( K \) . | Proof. If \( \gamma \left( t\right) : \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{C} \) is a parametrization for \( \gamma \), then\n\n\[ \n{\int }_{\gamma }\frac{f\left( \zeta \right) }{\zeta - z}{d\zeta } = {\int }_{0}^{1}\frac{f\left( {\gamma \left( t\right) }\right) }{\gamma \left( t\right) - z}{\gamma }^{\... | Yes |
Lemma 5.10 If \( {K}^{c} \) is connected and \( {z}_{0} \notin K \), then the function \( 1/\left( {z - {z}_{0}}\right) \) can be approximated uniformly on \( K \) by polynomials. | Proof. First, we choose a point \( {z}_{1} \) that is outside a large open disc \( D \) centered at the origin and which contains \( K \) . Then\n\n\[ \frac{1}{z - {z}_{1}} = - \frac{1}{{z}_{1}}\frac{1}{1 - z/{z}_{1}} = \mathop{\sum }\limits_{{n = 1}}^{\infty } - \frac{{z}^{n}}{{z}_{1}^{n + 1}} \]\n\nwhere the series c... | Yes |
Theorem 1.1 Suppose that \( f \) is holomorphic in a connected open set \( \Omega \) , has a zero at a point \( {z}_{0} \in \Omega \), and does not vanish identically in \( \Omega \) . Then there exists a neighborhood \( U \subset \Omega \) of \( {z}_{0} \), a non-vanishing holomorphic function \( g \) on \( U \), and ... | Proof. Since \( \Omega \) is connected and \( f \) is not identically zero, we conclude that \( f \) is not identically zero in a neighborhood of \( {z}_{0} \) . In a small disc centered at \( {z}_{0} \) the function \( f \) has a power series expansion\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }... | Yes |
Theorem 1.2 If \( f \) has a pole at \( {z}_{0} \in \Omega \), then in a neighborhood of that point there exist a non-vanishing holomorphic function \( h \) and a unique positive integer \( n \) such that\n\n\[ f\left( z\right) = {\left( z - {z}_{0}\right) }^{-n}h\left( z\right) . \] | Proof. By the previous theorem we have \( 1/f\left( z\right) = {\left( z - {z}_{0}\right) }^{n}g\left( z\right) \) , where \( g \) is holomorphic and non-vanishing in a neighborhood of \( {z}_{0} \), so the result follows with \( h\left( z\right) = 1/g\left( z\right) \) . | Yes |
Theorem 1.3 If \( f \) has a pole of order \( n \) at \( {z}_{0} \), then\n\n\[ f\left( z\right) = \frac{{a}_{-n}}{{\left( z - {z}_{0}\right) }^{n}} + \frac{{a}_{-n + 1}}{{\left( z - {z}_{0}\right) }^{n - 1}} + \cdots + \frac{{a}_{-1}}{\left( z - {z}_{0}\right) } + G\left( z\right) ,\] \n\nwhere \( G \) is a holomorphi... | Proof. The proof follows from the multiplicative statement in the previous theorem. Indeed, the function \( h \) has a power series expansion\n\n\[ h\left( z\right) = {A}_{0} + {A}_{1}\left( {z - {z}_{0}}\right) + \cdots \]\n\nso that\n\n\[ f\left( z\right) = {\left( z - {z}_{0}\right) }^{-n}\left( {{A}_{0} + {A}_{1}\l... | Yes |
Theorem 1.4 If \( f \) has a pole of order \( n \) at \( {z}_{0} \), then\n\n\[ \n{\operatorname{res}}_{{z}_{0}}f = \mathop{\lim }\limits_{{z \rightarrow {z}_{0}}}\frac{1}{\left( {n - 1}\right) !}{\left( \frac{d}{dz}\right) }^{n - 1}{\left( z - {z}_{0}\right) }^{n}f\left( z\right) .\n\] | The theorem is an immediate consequence of formula (1), which implies\n\n\[ \n{\left( z - {z}_{0}\right) }^{n}f\left( z\right) = {a}_{-n} + {a}_{-n + 1}\left( {z - {z}_{0}\right) + \cdots + {a}_{-1}{\left( z - {z}_{0}\right) }^{n - 1} + \n\]\n\n\[ \n+ G\left( z\right) {\left( z - {z}_{0}\right) }^{n}\text{.} \n\] | Yes |
Theorem 2.1 Suppose that \( f \) is holomorphic in an open set containing a circle \( C \) and its interior, except for a pole at \( {z}_{0} \) inside \( C \) . Then\n\n\[ \n{\int }_{C}f\left( z\right) {dz} = {2\pi i}{\operatorname{res}}_{{z}_{0}}f \n\] | Proof. Once again, we may choose a keyhole contour that avoids the pole, and let the width of the corridor go to zero to see that\n\n\[ \n{\int }_{C}f\left( z\right) {dz} = {\int }_{{C}_{\epsilon }}f\left( z\right) {dz} \n\]\n\nwhere \( {C}_{\epsilon } \) is the small circle centered at the pole \( {z}_{0} \) and of ra... | Yes |
Corollary 2.2 Suppose that \( f \) is holomorphic in an open set containing a circle \( C \) and its interior, except for poles at the points \( {z}_{1},\ldots ,{z}_{N} \) inside C. Then\n\n\[{\int }_{C}f\left( z\right) {dz} = {2\pi i}\mathop{\sum }\limits_{{k = 1}}^{N}{\operatorname{res}}_{{z}_{k}}f.\] | For the proof, consider a multiple keyhole which has a loop avoiding each one of the poles. Let the width of the corridors go to zero. In the limit, the integral over the large circle equals a sum of integrals over small circles to which Theorem 2.1 applies. | No |
Corollary 2.3 Suppose that \( f \) is holomorphic in an open set containing a toy contour \( \gamma \) and its interior, except for poles at the points \( {z}_{1},\ldots ,{z}_{N} \) inside \( \gamma \) . Then\n\n\[ \n{\int }_{\gamma }f\left( z\right) {dz} = {2\pi i}\mathop{\sum }\limits_{{k = 1}}^{N}{\operatorname{res}... | The proof consists of choosing a keyhole appropriate for the given toy contour, so that, as we have seen previously, we can reduce the situation to integrating over small circles around the poles where Theorem 2.1 applies. | No |
Theorem 3.1 (Riemann's theorem on removable singularities) Suppose that \( f \) is holomorphic in an open set \( \Omega \) except possibly at a point \( {z}_{0} \) in \( \Omega \) . If \( f \) is bounded on \( \Omega - \left\{ {z}_{0}\right\} \), then \( {z}_{0} \) is a removable singularity. | Proof. Since the problem is local we may consider a small disc \( D \) centered at \( {z}_{0} \) and whose closure is contained in \( \Omega \) . Let \( C \) denote the boundary circle of that disc with the usual positive orientation. We shall prove that if \( z \in D \) and \( z \neq {z}_{0} \), then under the assumpt... | Yes |
Corollary 3.2 Suppose that \( f \) has an isolated singularity at the point \( {z}_{0} \) . Then \( {z}_{0} \) is a pole of \( f \) if and only if \( \left| {f\left( z\right) }\right| \rightarrow \infty \) as \( z \rightarrow {z}_{0} \) . | Proof. If \( {z}_{0} \) is a pole, then we know that \( 1/f \) has a zero at \( {z}_{0} \), and therefore \( \left| {f\left( z\right) }\right| \rightarrow \infty \) as \( z \rightarrow {z}_{0} \) . Conversely, suppose that this condition holds. Then \( 1/f \) is bounded near \( {z}_{0} \), and in fact \( 1/\left| {f\le... | Yes |
Theorem 3.3 (Casorati-Weierstrass) Suppose \( f \) is holomorphic in the punctured disc \( {D}_{r}\left( {z}_{0}\right) - \left\{ {z}_{0}\right\} \) and has an essential singularity at \( {z}_{0} \) . Then, the image of \( {D}_{r}\left( {z}_{0}\right) - \left\{ {z}_{0}\right\} \) under \( f \) is dense in the complex p... | Proof. We argue by contradiction. Assume that the range of \( f \) is not dense, so that there exists \( w \in \mathbb{C} \) and \( \delta > 0 \) such that\n\n\[ \left| {f\left( z\right) - w}\right| > \delta \;\text{ for all }z \in {D}_{r}\left( {z}_{0}\right) - \left\{ {z}_{0}\right\} . \]\n\nWe may therefore define a... | Yes |
Theorem 3.4 The meromorphic functions in the extended complex plane are the rational functions. | Proof. Suppose that \( f \) is meromorphic in the extended plane. Then \( f\left( {1/z}\right) \) has either a pole or a removable singularity at 0, and in either case it must be holomorphic in a deleted neighborhood of the origin, so that the function \( f \) can have only finitely many poles in the plane, say at \( {... | Yes |
Theorem 4.3 (Rouché’s theorem) Suppose that \( f \) and \( g \) are holomorphic in an open set containing a circle \( C \) and its interior. If\n\n\[ \left| {f\left( z\right) }\right| > \left| {g\left( z\right) }\right| \;\text{ for all }z \in C, \]\nthen \( f \) and \( f + g \) have the same number of zeros inside the... | Proof. For \( t \in \left\lbrack {0,1}\right\rbrack \) define\n\n\[ {f}_{t}\left( z\right) = f\left( z\right) + \operatorname{tg}\left( z\right) \]\n\nso that \( {f}_{0} = f \) and \( {f}_{1} = f + g \) . Let \( {n}_{t} \) denote the number of zeros of \( {f}_{t} \) inside the circle counted with multiplicities, so tha... | Yes |
Theorem 4.4 (Open mapping theorem) If \( f \) is holomorphic and nonconstant in a region \( \Omega \), then \( f \) is open. | Proof. Let \( {w}_{0} \) belong to the image of \( f \), say \( {w}_{0} = f\left( {z}_{0}\right) \) . We must prove that all points \( w \) near \( {w}_{0} \) also belong to the image of \( f \) .
Define \( g\left( z\right) = f\left( z\right) - w \) and write
\[
g\left( z\right) = \left( {f\left( z\right) - {w}_{0}}\... | Yes |
Theorem 4.5 (Maximum modulus principle) If \( f \) is a non-constant holomorphic function in a region \( \Omega \), then \( f \) cannot attain a maximum in \( \Omega \) . | Proof. Suppose that \( f \) did attain a maximum at \( {z}_{0} \) . Since \( f \) is holomorphic it is an open mapping, and therefore, if \( D \subset \Omega \) is a small disc centered at \( {z}_{0} \), its image \( f\left( D\right) \) is open and contains \( f\left( {z}_{0}\right) \) . This proves that there are poin... | Yes |
Corollary 4.6 Suppose that \( \Omega \) is a region with compact closure \( \bar{\Omega } \) . If \( f \) is holomorphic on \( \Omega \) and continuous on \( \bar{\Omega } \) then\n\n\[ \mathop{\sup }\limits_{{z \in \Omega }}\left| {f\left( z\right) }\right| \leq \mathop{\sup }\limits_{{z \in \bar{\Omega } - \Omega }}\... | In fact, since \( f\left( z\right) \) is continuous on the compact set \( \bar{\Omega } \), then \( \left| {f\left( z\right) }\right| \) attains its maximum in \( \bar{\Omega } \) ; but this cannot be in \( \Omega \) if \( f \) is non-constant. If \( f \) is constant, the conclusion is trivial. | Yes |
Theorem 5.2 Any holomorphic function in a simply connected domain has a primitive. | Proof. Fix a point \( {z}_{0} \) in \( \Omega \) and define\n\n\[ F\left( z\right) = {\int }_{\gamma }f\left( w\right) {dw} \]\n\nwhere the integral is taken over any curve in \( \Omega \) joining \( {z}_{0} \) to \( z \) . This definition is independent of the curve chosen, since \( \Omega \) is simply connected, and ... | Yes |
Corollary 5.3 If \( f \) is holomorphic in the simply connected region \( \Omega \) , then\n\n\[{\int }_{\gamma }f\left( z\right) {dz} = 0\]\n\nfor any closed curve \( \gamma \) in \( \Omega \) . | This is immediate from the existence of a primitive. | No |
Theorem 6.1 Suppose that \( \Omega \) is simply connected with \( 1 \in \Omega \), and \( 0 \notin \) \( \Omega \) . Then in \( \Omega \) there is a branch of the logarithm \( F\left( z\right) = {\log }_{\Omega }\left( z\right) \) so that\n\n(i) \( F \) is holomorphic in \( \Omega \) ,\n\n(ii) \( {e}^{F\left( z\right) ... | Proof. We shall construct \( F \) as a primitive of the function \( 1/z \) . Since \( 0 \notin \Omega \), the function \( f\left( z\right) = 1/z \) is holomorphic in \( \Omega \) . We define\n\n\[ \n{\log }_{\Omega }\left( z\right) = F\left( z\right) = {\int }_{\gamma }f\left( w\right) {dw} \n\]\n\nwhere \( \gamma \) i... | Yes |
Theorem 7.1 The coefficients of the power series expansion of \( f \) are given by\n\n\[ \n{a}_{n} = \frac{1}{{2\pi }{r}^{n}}{\int }_{0}^{2\pi }f\left( {{z}_{0} + r{e}^{i\theta }}\right) {e}^{-{in\theta }}{d\theta }\n\]\n\nfor all \( n \geq 0 \) and \( 0 < r < R \) . Moreover,\n\n\[ \n0 = \frac{1}{{2\pi }{r}^{n}}{\int ... | Proof. Since \( {f}^{\left( n\right) }\left( {z}_{0}\right) = {a}_{n}n \) !, the Cauchy integral formula gives\n\n\[ \n{a}_{n} = \frac{1}{2\pi i}{\int }_{\gamma }\frac{f\left( \zeta \right) }{{\left( \zeta - {z}_{0}\right) }^{n + 1}}{d\zeta }\n\]\n\nwhere \( \gamma \) is a circle of radius \( 0 < r < R \) centered at \... | Yes |
Corollary 7.3 If \( f \) is holomorphic in a disc \( {D}_{R}\left( {z}_{0}\right) \), and \( u = \operatorname{Re}\left( f\right) \) , then\n\n\[ u\left( {z}_{0}\right) = \frac{1}{2\pi }{\int }_{0}^{2\pi }u\left( {{z}_{0} + r{e}^{i\theta }}\right) {d\theta },\;\text{ for any }0 < r < R. \] | Recall that \( u \) is harmonic whenever \( f \) is holomorphic, and in fact, the above corollary is a property enjoyed by every harmonic function in the disc \( {D}_{R}\left( {z}_{0}\right) \) . This follows from Exercise 12 in Chapter 2, which shows that every harmonic function in a disc is the real part of a holomor... | No |
Theorem 2.1 If \( f \) belongs to the class \( {\mathfrak{F}}_{a} \) for some \( a > 0 \), then \( \left| {\widehat{f}\left( \xi \right) }\right| \leq B{e}^{-{2\pi b}\left| \xi \right| } \) for any \( 0 \leq b < a \) . | Proof. Recall that \( \widehat{f}\left( \xi \right) = {\int }_{-\infty }^{\infty }f\left( x\right) {e}^{-{2\pi ix\xi }}{dx} \). The case \( b = 0 \) simply says that \( \widehat{f} \) is bounded, which follows at once from the integral defining \( \widehat{f} \), the assumption that \( f \) is of moderate decrease, and... | Yes |
Lemma 2.3 If \( A \) is positive and \( B \) is real, then \( {\int }_{0}^{\infty }{e}^{-\left( {A + {iB}}\right) \xi }{d\xi } = \) \( \frac{1}{A + {iB}} \) . | Proof. Since \( A > 0 \) and \( B \in \mathbb{R} \), we have \( \left| {e}^{-\left( {A + {iB}}\right) \xi }\right| = {e}^{-{A\xi }} \), and the integral converges. By definition\n\n\[ \n{\int }_{0}^{\infty }{e}^{-\left( {A + {iB}}\right) \xi }{d\xi } = \mathop{\lim }\limits_{{R \rightarrow \infty }}{\int }_{0}^{R}{e}^{... | Yes |
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\[ \n{\int }_{0}^{R}\mathfrak{n}\left( r\right) \frac{dr}{r} = \mathop{\sum }\limits_{{k = 1}}^{N}\log \left| \frac{R}{{z}_{k}}\right| .\n\] | Proof. First we have\n\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{N}\log \left| \frac{R}{{z}_{k}}\right| = \mathop{\sum }\limits_{{k = 1}}^{N}{\int }_{\left| {z}_{k}\right| }^{R}\frac{dr}{r}\n\]\n\nIf we define the characteristic function\n\n\[ \n{\eta }_{k}\left( r\right) = \left\{ \begin{array}{ll} 1 & \text{ if }r > \le... | 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\[ \n\sum {c}_{n} < \infty \;\text{ and }\;\left| {{F}_{n}\left( z\right) - 1}\right| \leq {c}_{n}\;\text{ for all }z \in \Omega ,\n\]\n\nth... | 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 an... | 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}^{w}. ... | Yes |
Corollary 5.4 There exists a sequence of radii, \( {r}_{1},{r}_{2},\ldots \), with \( {r}_{m} \rightarrow \infty \), such that\n\n\[ \left| {\mathop{\prod }\limits_{{n = 1}}^{\infty }{E}_{k}\left( {z/{a}_{n}}\right) }\right| \geq {e}^{-c{\left| z\right| }^{s}}\;\text{ for }\left| z\right| = {r}_{m} \] | 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 \( s... | 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\[{\pi }^{-s/2}\Gamma \left( {s/2}\right) \zeta \left( s\right) = \frac{1}{2}{\int }_{0}^{\infty }{u}^{\left( {s/2}\right) - 1}\left\lbrack {\vartheta \left( u\right) - 1}\right\rbrack {du}.\] | 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, \[ \xi \left( s\right) = \xi \left( {1 - s}\right) \;\text{ for all }s \in \... | Proof. The idea of the proof is to use the functional equation for \( \vartheta \) , namely \[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{e}^{-\pi {n}^{2}u} = {u}^{-1/2}\mathop{\sum }\limits_{{n = - \infty }}^{\infty }{e}^{-\pi {n}^{2}/u},\;u > 0. \] We then could multiply both sides by \( {u}^{\left( {s/2}\righ... | 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 |
Proposition 2.1 If \( \psi \left( x\right) \sim x \) as \( x \rightarrow \infty \), then \( \pi \left( x\right) \sim x/\log x \) as \( x \rightarrow \infty \) . | Proof. 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... | 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 |
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