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Example 54.3.3 If \( A \in \mathcal{L}\left( {H, H}\right) \), then \( A \) is sectorial. | The spectrum \( \sigma \left( A\right) \) is bounded by \( \parallel A\parallel \) and so there is clearly a sector of the above form contained in the resolvent set of \( A \) . As to the estimate 54.3.4, let \( a \) be larger than \( 2\parallel A\parallel \) and let \( {S}_{a\phi } \) be contained in the resolvent set... | Yes |
Theorem 54.3.8 Let \( A \) be a sectorial operator as defined in Definition 54.3.1 for the sector \( {S}_{a,\phi } \) . Then there exists a semigroup \( S\left( t\right) \) for \( t \in \left| {\arg z}\right| \leq r < \left( {\frac{\pi }{2} - \phi }\right) \) which satisfies the following conditions.\n\n1. Then \( S\le... | Proof: Consider the first claim. This follows right away from the formula: \( S\left( t\right) \equiv \frac{1}{2\pi i}{\int }_{{\gamma }_{\varepsilon ,\phi }}{e}^{\lambda t}{\left( \lambda I - A\right) }^{-1}{d\lambda } \) . One can differentiate under the integral sign using the dominated convergence theorem to obtain... | Yes |
Corollary 54.3.11 If for some \( a \in \mathbb{R} \), the numerical values of \( - {aI} + A \) are in the set \( \{ \lambda : \left| \lambda \right| \geq \pi - \phi \} \) where \( 0 < \phi < \pi /2 \), and \( a \in r\left( A\right) \) then \( A \) is sectorial. | Proof: By assumption, \( 0 \in r\left( {-{aI} + A}\right) \) and also from Proposition 54.3.10, for \( \mu \in {S}_{0,{\phi }^{\prime }} \) where \( \pi /2 > {\phi }^{\prime } > \phi \) ,\n\n\[ \n{\left( \left( -aI + A\right) - \mu I\right) }^{-1} \in \mathcal{L}\left( {H, H}\right) ,\begin{Vmatrix}{\left( \left( -aI +... | Yes |
Lemma 54.3.14 The Riesz map is one to one and onto and linear. | Proof: It is obvious it is one to one and linear. The only challenge is to show it is onto. Let \( {z}^{ * } \in {V}^{\prime } \) . If \( {z}^{ * }\left( V\right) = \{ 0\} \), then letting \( z = 0 \), it follows \( {Rz} = {z}^{ * } \) . If \( {z}^{ * }\left( V\right) \neq 0 \), then\n\n\[ \ker \left( {z}^{ * }\right) ... | Yes |
Theorem 54.3.17 For \( {\left( -A\right) }^{-\alpha } \) as defined in Definition 54.3.16\n\n\[{\left( -A\right) }^{-\alpha }{\left( -A\right) }^{-\beta } = {\left( -A\right) }^{-\left( {\alpha + \beta }\right) }\] | Proof: Consider 54.3.17.\n\n\[{\left( -A\right) }^{-\alpha }{\left( -A\right) }^{-\beta } \equiv \frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) }{\int }_{0}^{\infty }{\int }_{0}^{\infty }{t}^{\alpha - 1}{s}^{\beta - 1}S\left( {t + s}\right) {dsdt}\]\n\nChanging variables and using Fubini's theorem wh... | Yes |
Theorem 54.3.19 The definition of \( {\left( -A\right) }^{\alpha } \) is well defined and \( {\left( -A\right) }^{\alpha } \) is densely defined and closed. Also for any \( \alpha > 0 \) , \[ \begin{Vmatrix}{{\left( -A\right) }^{\alpha }S\left( t\right) }\end{Vmatrix} \leq \frac{{C}_{\alpha }}{\delta }\frac{1}{{t}^{\al... | Proof: It is obvious \( {\left( -A\right) }^{\alpha } \) is densely defined because its domain is at least as large as \( D\left( A\right) \) which was assumed to be dense. It is a closed operator because if \( {x}_{n} \in D\left( {\left( -A\right) }^{\alpha }\right) \) and \[ {x}_{n} \rightarrow x,{\left( -A\right) }^... | Yes |
Corollary 54.3.20 Let \( \alpha \in \left( {0,1}\right) \) . Then for all \( \varepsilon > 0 \), there exists a constant \( C\left( {\alpha ,\varepsilon }\right) \) such that\n\n\[ \begin{Vmatrix}{{\left( -A\right) }^{-\alpha }x}\end{Vmatrix} \leq \varepsilon \parallel x\parallel + C\left( {\varepsilon ,\alpha }\right)... | Proof: The first part is done in the above theorem. Let \( S \) be a bounded set and let \( \eta > 0 \) . Then let \( \varepsilon > 0 \) be small enough that for all \( x \in S,\varepsilon \parallel x\parallel < \eta /4 \) . Let \( \left\{ {{\left( -A\right) }^{-1}{x}_{n}}\right\} \) be a \( \eta /\left( {2 + {2C}\left... | Yes |
Proposition 54.3.27 The \( {H}_{\alpha } \) above are Banach spaces and they decrease in \( \alpha \) . Furthermore, if \( {b}_{i} > a \) for \( i = 1,2 \) then the two norms associated with the \( {b}_{i} \) are equivalent. | Proof: That the \( {H}_{\alpha } \) are decreasing was shown above in Theorem 54.3.17. They are Banach spaces because \( {\left( bI - A\right) }^{\alpha } \) is a closed mapping which is also one to one.\n\nIt only remains to verify the claim about the equivalence of the norms. Let \( {b}_{2} > {b}_{1} > a \) . Then if... | Yes |
Lemma 55.2.2 The fractional linear transformation, 55.2.1 can be written as a finite composition of dilations, inversions, and translations. | Proof: Let\n\n\[ \n{S}_{1}\left( z\right) = z + \frac{d}{c},{S}_{2}\left( z\right) = \frac{1}{z},{S}_{3}\left( z\right) = \frac{\left( bc - ad\right) }{{c}^{2}}z \n\] \n\nand \n\n\[ \n{S}_{4}\left( z\right) = z + \frac{a}{c} \n\] \n\nin the case where \( c \neq 0 \) . Then \( f\left( z\right) \) given in 55.2.1 is of t... | Yes |
Corollary 55.2.3 Fractional linear transformations map circles and lines to circles or lines. | Proof: It is obvious that dilations and translations map circles to circles and lines to lines. What of inversions? If inversions have this property, the above lemma implies a general fractional linear transformation has this property as well.\n\nNote that all circles and lines may be put in the form\n\n\[ \alpha \left... | Yes |
Consider the fractional linear transformation, \( w = \frac{z - i}{z + i} \). | First consider what this mapping does to the points of the form \( z = x + {i0} \). Substituting into the expression for \( w \), \[ w = \frac{x - i}{x + i} = \frac{{x}^{2} - 1 - {2xi}}{{x}^{2} + 1}, \] a point on the unit circle. Thus this transformation maps the real axis to the unit circle. The upper half plane is c... | Yes |
Example 55.2.5 Let \( \operatorname{Im}\xi > 0 \) and consider the fractional linear transformation which takes \( \xi \) to \( 0,\bar{\xi } \) to \( \infty \) and 0 to \( \xi /\bar{\xi }, \) . | The equation for \( w \) is\n\n\[ \frac{w - 0}{w - \left( {\xi /\bar{\xi }}\right) } = \frac{z - \xi }{z - 0} \cdot \frac{\bar{\xi } - 0}{\bar{\xi } - \xi } \]\n\nAfter some computations,\n\n\[ w = \frac{z - \xi }{z - \bar{\xi }} \]\n\nNote that this has the property that \( \frac{x - \xi }{x - \bar{\xi }} \) is always... | Yes |
Example 55.2.6 Let \( {z}_{1} = 0,{z}_{2} = 1 \), and \( {z}_{3} = 2 \) and let \( {w}_{1} = 0,{w}_{2} = i \), and \( {w}_{3} = {2i} \) . Then the equation to solve is \[ \frac{w}{w - {2i}} \cdot \frac{-i}{i} = \frac{z}{z - 2} \cdot \frac{-1}{1} \] | Solving this yields \( w = {iz} \) which clearly works. | Yes |
Lemma 55.3.4 Suppose \( F : B\\left( {0,1}\\right) \\rightarrow B\\left( {0,1}\\right), F \) is analytic, and \( F\\left( 0\\right) = 0 \) . Then for all \( z \\in B\\left( {0,1}\\right) \) ,\n\n\[ \n\\left| {F\\left( z\\right) }\\right| \\leq \\left| z\\right| \n\]\n\nand\n\n\[ \n\\left| {{F}^{\\prime }\\left( 0\\righ... | Proof: First note that by assumption, \( F\\left( z\\right) /z \) has a removable singularity at 0 if its value at 0 is defined to be \( {F}^{\\prime }\\left( 0\\right) \) . By the maximum modulus theorem, if \( \\left| z\\right| < r < 1 \)\n\n\[ \n\\left| \\frac{F\\left( z\\right) }{z}\\right| \\leq \\mathop{\\max }\\... | Yes |
Theorem 55.3.6 Let \( \Omega \neq \mathbb{C} \) for \( \Omega \) a region and suppose \( \Omega \) has the square root property. Then for \( {z}_{0} \in \Omega \) there exists \( h : \Omega \rightarrow B\left( {0,1}\right) \) such that \( h \) is one to one, onto, analytic, and \( h\left( {z}_{0}\right) = 0 \) . | Proof: Define \( \mathcal{F} \) to be the set of functions, \( f \) such that \( f : \Omega \rightarrow B\left( {0,1}\right) \) is one to one and analytic. The first task is to show \( \mathcal{F} \) is nonempty. Then, using Montel’s theorem it will be shown there is a function in \( \mathcal{F}, h \), such that \( \le... | Yes |
Lemma 55.3.7 Let \( \Omega \) be a simply connected region with \( \Omega \neq \mathbb{C} \) . Then \( \Omega \) has the square root property. | Proof: Let \( f \) and \( \frac{1}{f} \) both be analytic on \( \Omega \) . Then \( \frac{{f}^{\prime }}{f} \) is analytic on \( \Omega \) so by Corollary 51.7.23, there exists \( \widetilde{F} \), analytic on \( \Omega \) such that \( {\widetilde{F}}^{\prime } = \frac{{f}^{\prime }}{f} \) on \( \Omega \) . Then \( {\l... | Yes |
Corollary 55.3.8 (Riemann mapping theorem) Let \( \Omega \) be a simply connected region with \( \Omega \neq \mathbb{C} \) and let \( {z}_{0} \in \Omega \) . Then there exists a function, \( f : \Omega \rightarrow B\left( {0,1}\right) \) such that \( f \) is one to one, analytic, and onto with \( f\left( {z}_{0}\right)... | Proof: From Theorem 55.3.6 and Lemma 55.3.7 there exists a function, \( f : \Omega \rightarrow \) \( B\left( {0,1}\right) \) which is one to one, onto, and analytic such that \( f\left( {z}_{0}\right) = 0 \) . The assertion that \( {f}^{-1} \) is analytic follows from the open mapping theorem. | Yes |
Theorem 55.4.2 Suppose \( f \) is analytic on \( B\left( {a, r}\right) \) and the power series\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }{a}_{k}{\left( z - a\right) }^{k} \]\n\nhas radius of convergence \( r \) . Then there exists a singular point on \( \partial B\left( {a, r}\right) \) . | Proof: If not, then for every \( z \in \partial B\left( {a, r}\right) \) there exists \( {\delta }_{z} > 0 \) and \( {g}_{z} \) analytic on \( B\left( {z,{\delta }_{z}}\right) \) such that \( {g}_{z} = f \) on \( B\left( {z,{\delta }_{z}}\right) \cap B\left( {a, r}\right) \) . Since \( \partial B\left( {a, r}\right) \)... | Yes |
Lemma 55.4.4 Suppose \( \left( {f, B\left( {0, r}\right) }\right) \) for \( r < 1 \) is a function element and \( \left( {f, B\left( {0, r}\right) }\right) \) can be analytically continued along every curve in \( B\left( {0,1}\right) \) that starts at 0 . Then there exists an analytic function, \( g \) defined on \( B\... | Proof: Let\n\n\[ R = \sup \left\{ {{r}_{1} \geq r}\right. \text{such that there exists}{g}_{{r}_{1}}\]\n\n\[ \text{analytic on}B\left( {0,{r}_{1}}\right) \text{which agrees with}f\text{on}B\left( {0, r}\right) \text{.}\} \]\n\nDefine \( {g}_{R}\left( z\right) \equiv {g}_{{r}_{1}}\left( z\right) \) where \( \left| z\rig... | Yes |
Theorem 55.4.5 Let \( \Omega \) be a simply connected proper subset of \( \mathbb{C} \) and suppose \( \left( {f, B\left( {a, r}\right) }\right) \) is a function element with \( B\left( {a, r}\right) \subseteq \Omega \) . Suppose also that this function element can be analytically continued along every curve through a.... | Proof: By the Riemann mapping theorem, there exists \( h : \Omega \rightarrow B\left( {0,1}\right) \) which is analytic, one to one and onto such that \( f\left( a\right) = 0 \) . Since \( h \) is an open map, there exists \( \delta > 0 \) such that\n\n\[ B\left( {0,\delta }\right) \subseteq h\left( {B\left( {a, r}\rig... | Yes |
Corollary 55.4.6 Suppose \( \left( {f, B\left( {a, r}\right) }\right) \) is a function element with \( B\left( {a, r}\right) \subseteq \mathbb{C} \) . Suppose also that this function element can be analytically continued along every curve through a. Then there exists \( G \) analytic on \( \mathbb{C} \) such that \( G ... | Proof: Let \( {\Omega }_{1} \equiv \{ z \in \mathbb{C} : a + {it} : t > a\} \) and \( {\Omega }_{2} \equiv \{ z \in \mathbb{C} : a - {it} : t > a\} \) . Here is a picture of \( {\Omega }_{1} \) .\n\n\n\nA picture o... | Yes |
Lemma 55.5.1 Let \( f \) be analytic on a region containing \( \overline{B\left( {0, r}\right) } \) and suppose\n\n\[ \left| {{f}^{\prime }\left( 0\right) }\right| = b > 0, f\left( 0\right) = 0, \]\n\nand \( \left| {f\left( z\right) }\right| \leq M \) for all \( z \in \overline{B\left( {0, r}\right) } \) . Then \( f\le... | Proof: By assumption,\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }{a}_{k}{z}^{k},\left| z\right| \leq r. \]\n\n\( \left( {55.5.16}\right) \)\n\nThen by the Cauchy integral formula for the derivative,\n\n\[ {a}_{k} = \frac{1}{2\pi i}{\int }_{\partial B\left( {0, r}\right) }\frac{f\left( w\right) }{... | Yes |
Lemma 55.5.3 Let \( f \) be analytic on an open set containing \( \overline{B\left( {{z}_{0}, R}\right) } \) and suppose \( \left| {{f}^{\prime }\left( {z}_{0}\right) }\right| > 0 \) . Then there exists \( a \in B\left( {{z}_{0}, R}\right) \) such that\n\n\[ f\left( {B\left( {{z}_{0}, R}\right) }\right) \supseteq B\lef... | Proof: You look at \( g\left( z\right) \equiv f\left( {{z}_{0} + z}\right) - f\left( {z}_{0}\right) \) for \( z \in B\left( {0, R}\right) \) . Then \( {g}^{\prime }\left( 0\right) = \) \( {f}^{\prime }\left( {z}_{0}\right) \) and so by Lemma 55.5.2 there exists \( {a}_{1} \in B\left( {0, R}\right) \) such that\n\n\[ g\... | Yes |
Theorem 55.5.5 If \( h \) is an entire function which omits two values then \( h \) is a constant. | Proof: Suppose the two values omitted are \( a \) and \( b \) and that \( h \) is not constant. Let \( f\left( z\right) = \left( {h\left( z\right) - a}\right) /\left( {b - a}\right) \) . Then \( f \) omits the two values 0 and 1 . Let \( H \) be defined in Lemma 55.5.4. Then \( H\left( z\right) \) is clearly not of the... | Yes |
Corollary 55.5.6 If \( f \) is a meromophic function defined on \( \mathbb{C} \) which omits three distinct values, \( a, b, c \), then \( f \) is a constant. | Proof: Let \( \phi \left( z\right) \equiv \frac{z - a}{z - c}\frac{b - c}{b - a} \) . Then \( \phi \left( c\right) = \infty ,\phi \left( a\right) = 0 \), and \( \phi \left( b\right) = 1 \) . Now consider the function, \( h = \phi \circ f \) . Then \( h \) misses the three points \( \infty ,0 \), and 1 . Since \( h \) i... | Yes |
Theorem 55.5.8 Let \( f \) be analytic on \( B\left( {{z}_{0}, R}\right) \) and suppose that \( f \) does not take on either of the two distinct values a or \( b \) . Also suppose \( \left| {f\left( {z}_{0}\right) }\right| \leq \beta \) . Then letting \( \theta \in \left( {0,1}\right) \), it follows\n\n\[ \left| {f\lef... | Proof: First you can reduce to the case where the two values are 0 and 1 by considering\n\n\[ h\left( z\right) \equiv \frac{f\left( z\right) - a}{b - a}. \]\n\nIf there exists an estimate of the desired sort for \( h \), then there exists such an estimate for \( f \) . Of course here the function, \( M \) would depend ... | Yes |
Theorem 55.5.10 \( \left( {\widehat{\mathbb{C}}, d}\right) \) is a compact, hence complete metric space. | Proof: Suppose \( \left\{ {z}_{n}\right\} \) is a sequence in \( \widehat{\mathbb{C}} \) . This means \( \left\{ {\theta \left( {z}_{n}\right) }\right\} \) is a sequence in \( {S}^{2} \) which is compact. Therefore, there exists a subsequence, \( \left\{ {\theta {z}_{{n}_{k}}}\right\} \) and a point, \( z \in {S}^{2} \... | Yes |
Theorem 55.5.11 Let \( \Omega \) be an open subset of \( \mathbb{C} \) and let \( f : \Omega \rightarrow \widehat{\mathbb{C}} \) be meromorphic. Then \( f \) is continuous with respect to the metric, \( d \) on \( \widehat{\mathbb{C}} \) . | Proof: Let \( {z}_{n} \rightarrow z \) where \( z \in \Omega \) . Then if \( z \) is a pole, it follows from Theorem 51.7.11 that\n\n\[ d\left( {f\left( {z}_{n}\right) ,\infty }\right) \equiv d\left( {f\left( {z}_{n}\right), f\left( z\right) }\right) \rightarrow 0. \]\n\nIf \( z \) is not a pole, then \( f\left( {z}_{n... | Yes |
Theorem 55.5.15 Suppose \( K \) is a nonempty compact subset of \( {\mathbb{R}}^{n} \) and \( A \subseteq \) \( C\left( {K, X}\right) \), is uniformly bounded and uniformly equicontinuous where \( X \) is a locally compact complete metric space. Then if \( \left\{ {f}_{k}\right\} \subseteq A \), there exists a function... | \[ \mathop{\lim }\limits_{{l \rightarrow \infty }}{\rho }_{K}\left( {{f}_{{k}_{l}}, f}\right) = 0 \] | No |
Corollary 55.5.18 Suppose \( f \) is entire and nonconstant and not a polynomial. Then \( f \) assumes every complex value infinitely many times with the possible exception of one. | Proof: Since \( f \) is entire, \( f\left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{z}^{n} \) . Define for \( z \neq 0 \) , \[ g\left( z\right) \equiv f\left( \frac{1}{z}\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( \frac{1}{z}\right) }^{n}. \] Thus 0 is an isolated essential sing... | Yes |
Theorem 56.1.3 Let \( K \subseteq \Omega \) where \( K \) is compact and \( \Omega \) is open. Then there exist oriented closed curves, \( {\gamma }_{k} \) such that \( {\gamma }_{k}^{ * } \cap K = \varnothing \) but \( {\gamma }_{k}^{ * } \subseteq \Omega \), such that for all \( z \in K \) ,\n\n\[ f\left( z\right) = ... | Proof: This follows from Theorem 51.7.25 and the Cauchy integral formula. As shown in the proof, you can assume the \( {\gamma }_{k} \) are linear mappings but this is not important. | No |
Lemma 56.1.4 Let \( K \) be a compact subset of an open set, \( \Omega \) and let \( f \) be analytic on \( \Omega \) . Then there exists a rational function, \( Q \) whose poles are not in \( K \) such that\n\n\[ \parallel Q - f{\parallel }_{K,\infty } < \varepsilon \] | Proof: By Theorem 56.1.3 there are oriented curves, \( {\gamma }_{k} \) described there such that for all \( z \in K \) ,\n\n\[ f\left( z\right) = \frac{1}{2\pi i}\mathop{\sum }\limits_{{k = 1}}^{p}{\int }_{{\gamma }_{k}}\frac{f\left( w\right) }{w - z}{dw}. \]\n\n\( \left( {56.1.2}\right) \)\n\nDefining \( g\left( {w, ... | Yes |
Theorem 56.1.5 Suppose \( \mathop{\sum }\limits_{{i = r}}^{\infty }{a}_{i} \) and \( \mathop{\sum }\limits_{{j = r}}^{\infty }{b}_{j} \) both converge absolutely \( {}^{1} \) . Then\n\n\[ \left( {\mathop{\sum }\limits_{{i = r}}^{\infty }{a}_{i}}\right) \left( {\mathop{\sum }\limits_{{j = r}}^{\infty }{b}_{j}}\right) = ... | Proof: Let \( {p}_{nk} = 1 \) if \( r \leq k \leq n \) and \( {p}_{nk} = 0 \) if \( k > n \) . Then\n\n\[ {c}_{n} = \mathop{\sum }\limits_{{k = r}}^{\infty }{p}_{nk}{a}_{k}{b}_{n - k + r} \]\n\n--- \n\nAlso,\n\n\[ \mathop{\sum }\limits_{{k = r}}^{\infty }\mathop{\sum }\limits_{{n = r}}^{\infty }{p}_{nk}\left| {a}_{k}\r... | Yes |
Lemma 56.1.6 Let \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}\left( z\right) \) and \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{b}_{n}\left( z\right) \) be two convergent series for \( z \in \) \( K \) which satisfy the conditions of the Weierstrass \( M \) test. Thus there exist positive constants, \( {A}_{n} ... | Proof:\n\n\[ \n\left| {{c}_{n}\left( z\right) }\right| \leq \mathop{\sum }\limits_{{k = 0}}^{n}\left| {{a}_{n - k}\left( z\right) }\right| \left| {{b}_{k}\left( z\right) }\right| \leq \mathop{\sum }\limits_{{k = 0}}^{n}{A}_{n - k}{B}_{k}.\n\]\n\nAlso,\n\n\[ \n\mathop{\sum }\limits_{{n = 0}}^{\infty }\mathop{\sum }\limi... | Yes |
Theorem 56.1.9 Let \( K \) be a compact subset of an open set, \( \Omega \) and let \( \left\{ {b}_{j}\right\} \) be a set which consists of one point from each component of \( \widehat{\mathbb{C}} \smallsetminus K \) . Let \( f \) be analytic on \( \Omega \) . Then for each \( \varepsilon > 0 \), there exists a ration... | Proof: By Lemma 56.1.4 there exists a rational function of the form \[ R\left( z\right) = \mathop{\sum }\limits_{{k = 1}}^{M}\frac{{A}_{k}}{{w}_{k} - z} \] where the \( {w}_{k} \) are elements of components of \( \mathbb{C} \smallsetminus K \) and \( {A}_{k} \) are complex numbers such that \[ \parallel R - f{\parallel... | Yes |
Lemma 56.1.10 Let \( \Omega \) be an open set in \( \mathbb{C} \) . Then there exists a sequence of compact sets, \( \left\{ {K}_{n}\right\} \) such that\n\n\[ \Omega = { \cup }_{k = 1}^{\infty }{K}_{n},\cdots ,{K}_{n} \subseteq \operatorname{int}{K}_{n + 1}\cdots ,\]\n\n(56.1.10)\n\nand for any \( K \subseteq \Omega \... | Proof: Let\n\[ {V}_{n} \equiv \{ z : \left| z\right| > n\} \cup \mathop{\bigcup }\limits_{{z \notin \Omega }}B\left( {z,\frac{1}{n}}\right) .\n\]\n\nThus \( \{ z : \left| z\right| > n\} \) contains the point, \( \infty \) . Now let\n\n\[ {K}_{n} \equiv \widehat{\mathbb{C}} \smallsetminus {V}_{n} = \mathbb{C} \smallsetm... | No |
Theorem 56.1.11 (Runge) Let \( \Omega \) be an open set, and let \( A \) be a set which has one point in each component of \( \widehat{\mathbb{C}} \smallsetminus \Omega \) and let \( f \) be analytic on \( \Omega \) . Then there exists a sequence of rational functions, \( \left\{ {R}_{n}\right\} \) having poles only in... | Proof: Let \( {K}_{n} \) be the compact sets of Lemma 56.1.10 where each component of \( \widehat{\mathbb{C}} \smallsetminus {K}_{n} \) contains a component of \( \widehat{\mathbb{C}} \smallsetminus \Omega \) . It follows each component of \( \widehat{\mathbb{C}} \smallsetminus {K}_{n} \) contains a point of \( A \) . ... | Yes |
Corollary 56.1.12 Let \( \Omega \) be simply connected and \( f \) analytic on \( \Omega \) . Then there exists a sequence of polynomials, \( \left\{ {p}_{n}\right\} \) such that \( {p}_{n} \rightarrow f \) uniformly on compact sets of \( \Omega \) . | Proof: By definition of what is meant by simply connected, \( \widehat{\mathbb{C}} \smallsetminus \Omega \) is connected and so there are no bounded components of \( \widehat{\mathbb{C}} \smallsetminus \Omega \) . Therefore, in the proof of Theorem 56.1.11 when you use Theorem 56.1.9, you can always have \( {R}_{n} \) ... | No |
Lemma 56.2.6 Suppose \( \gamma \) is a closed continuous bounded variation curve in an open set, \( \Omega \) which is homotopic to a point. Then if \( a \notin \Omega \), it follows \( n\left( {a,\gamma }\right) = 0 \) . | Proof: Let \( H \) be the homotopy described above. The problem with this is that it is not known that \( H\left( {\alpha , \cdot }\right) \) is of bounded variation. There is no reason it should be. Therefore, it might not make sense to take the integral which defines the winding number. There are various ways around ... | Yes |
Theorem 56.2.7 The following are equivalent for an open set, \( \Omega \) .\n\n1. \( \Omega \) is homeomorphic to the unit disk, \( B\left( {0,1}\right) \) .\n\n2. Every closed curve contained in \( \Omega \) is homotopic to a point in \( \Omega \) .\n\n3. If \( z \notin \Omega \), and if \( \gamma \) is a closed bound... | Proof: \( 1 \Rightarrow 2 \) . Assume 1 and let \( \gamma \) be a closed curve in \( \Omega \) . Let \( h \) be the homeomorphism, \( h : B\left( {0,1}\right) \rightarrow \Omega \) . Let \( H\left( {\alpha, t}\right) = h\left( {\alpha \left( {{h}^{-1}\gamma \left( t\right) }\right) }\right) \) . This works.\n\n\( 2 \Ri... | Yes |
Lemma 57.1.1 For \( z \in \mathbb{C} \) , \[ \left| {{e}^{z} - 1}\right| \leq \left| z\right| {e}^{\left| z\right| } \] | Proof: Consider 57.1.3. \[ \left| {{e}^{z} - 1}\right| = \left| {\mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{z}^{k}}{k!}}\right| \leq \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{\left| z\right| }^{k}}{k!} = {e}^{\left| z\right| } - 1 \leq \left| z\right| {e}^{\left| z\right| } \] the last inequality holding by ... | Yes |
Corollary 57.1.3 For \( {E}_{p} \) defined above and \( \left| z\right| \leq 1/2 \) , \[ \left| {{E}_{p}\left( z\right) - 1}\right| \leq 3{\left| z\right| }^{p + 1}. \] | Proof: From elementary calculus, \( \ln \left( {1 - x}\right) = - \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{x}^{n}}{n} \) for all \( \left| x\right| < 1 \) . Therefore, for \( \left| z\right| < 1 \) , \[ \log \left( {1 - z}\right) = - \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{z}^{n}}{n},\log \left( {\left( ... | Yes |
Theorem 57.1.4 Let \( \\left\\{ {z}_{n}\\right\\} \) be a sequence of nonzero complex numbers which have no limit point in \( \\mathbb{C} \) and let \( \\left\\{ {p}_{n}\\right\\} \) be a sequence of nonnegative integers such that\n\n\[ \n\\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }{\\left( \\frac{R}{\\left| {z}_{n}\\... | Proof: Since \( \\left\\{ {z}_{n}\\right\\} \) has no limit point, it follows \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\left| {z}_{n}\\right| = \\infty \) . Therefore, if \( {p}_{n} = n - 1 \) the condition,57.1.5 holds for this choice of \( {p}_{n} \) . Now by Theorem 57.0.2, the infinite product in thi... | Yes |
Lemma 57.1.6 Let \( \Omega \) be an open set. Also let \( \left\{ {z}_{n}\right\} \) be a sequence of points in \( \Omega \) which is bounded and which has no point repeated more than finitely many times such that \( \left\{ {z}_{n}\right\} \) has no limit point in \( \Omega \) . Then there exist \( \left\{ {w}_{n}\rig... | Proof: Since \( \partial \Omega \) is closed, there exists \( {w}_{n} \in \partial \Omega \) such that \( \operatorname{dist}\left( {{z}_{n},\partial \Omega }\right) = \) \( \left| {{z}_{n} - {w}_{n}}\right| \) . Now if there is a subsequence, \( \left\{ {z}_{{n}_{k}}\right\} \) such that \( \left| {{z}_{{n}_{k}} - {w}... | Yes |
Corollary 57.1.7 Let \( \\left\\{ {z}_{n}\\right\\} \) be a sequence of complex numbers contained in \( \\Omega \), an open subset of \( \\mathbb{C} \) which has no limit point in \( \\Omega \) . Suppose each \( {z}_{n} \) is repeated no more than finitely many times. Then there exists a function \( f \) which is analy... | Proof: There is nothing to prove if \( \\left\\{ {z}_{n}\\right\\} \) is finite. You just let \( f\\left( z\\right) = \) \( \\mathop{\\prod }\\limits_{{j = 1}}^{m}\\left( {z - {z}_{j}}\\right) \) where \( \\left\\{ {z}_{n}\\right\\} = \\left\\{ {{z}_{1},\\cdots ,{z}_{m}}\\right\\} \) . Pick \( w \\in \\Omega \\smallset... | Yes |
Theorem 57.1.8 Suppose \( Q \) is a meromorphic function on an open set, \( \Omega \) . Then there exist analytic functions on \( \Omega, f\left( z\right) \) and \( g\left( z\right) \) such that \( Q\left( z\right) = f\left( z\right) /g\left( z\right) \) for all \( z \) not in the set of poles of \( Q \) . | Proof: Let \( Q \) have a pole of order \( m\left( z\right) \) at \( z \) . Then by Corollary 57.1.7 there exists an analytic function, \( g \) which has a zero of order \( m\left( z\right) \) at every \( z \in \Omega \) . It follows \( {gQ} \) has a removable singularity at the poles of \( Q \) . Therefore, there is a... | Yes |
Corollary 57.1.9 Suppose \( \Omega \) is a region and \( Q \) is a meromorphic function defined on \( \Omega \) such that the set, \( \{ z \in \Omega : Q\left( z\right) = c\} \) has a limit point in \( \Omega \) . Then \( Q\left( z\right) = c \) for all \( z \in \Omega \) . | Proof: From Theorem 57.1.8 there are analytic functions, \( f, g \) such that \( Q = \frac{f}{g} \) . Therefore, the zero set of the function, \( f\left( z\right) - {cg}\left( z\right) \) has a limit point in \( \Omega \) and so \( f\left( z\right) - {cg}\left( z\right) = 0 \) for all \( z \in \Omega \) . This proves t... | Yes |
Theorem 57.2.1 Let \( f \) be analytic on \( \mathbb{C}, f\left( 0\right) \neq 0 \), and let the zeros of \( f \), be \( \left\{ {z}_{k}\right\} \), listed according to order. (Thus if \( z \) is a zero of order \( m \), it will be listed \( m \) times in the list, \( \left\{ {z}_{k}\right\} \) .) Choosing nonnegative ... | Proof: \( \left\{ {z}_{n}\right\} \) cannot have a limit point because if there were a limit point of this sequence, it would follow from Theorem 51.5.3 that \( f\left( z\right) = 0 \) for all \( z \), contradicting the hypothesis that \( f\left( 0\right) \neq 0 \) . Hence \( \mathop{\lim }\limits_{{n \rightarrow \inft... | Yes |
Corollary 57.2.2 Let \( f \) be analytic on \( \mathbb{C}, f \) has a zero of order \( m \) at 0, and let the other zeros of \( f \) be \( \left\{ {z}_{k}\right\} \), listed according to order. (Thus if \( z \) is a zero of order \( l \) , it will be listed \( l \) times in the list, \( \left\{ {z}_{k}\right\} \) .) Al... | Proof: Since \( f \) has a zero of order \( m \) at 0, it follows from Theorem 51.5.3 that \( \left\{ {z}_{k}\right\} \) cannot have a limit point in \( \mathbb{C} \) and so you can apply Theorem 57.2.1 to the function, \( f\left( z\right) /{z}^{m} \) which has a removable singularity at 0 . This proves the corollary. | No |
Let \( {\gamma }_{N} \) be the contour which goes from \( - N - \frac{1}{2} - {Ni} \) horizontally to \( N + \frac{1}{2} - {Ni} \) and from there, vertically to \( N + \frac{1}{2} + {Ni} \) and then horizontally to \( - N - \frac{1}{2} + {Ni} \) and finally vertically to \( - N - \frac{1}{2} - {Ni} \) . Thus the contou... | First you show that \( \cot {\pi z} \) is bounded on this contour. This is easy using the formula for \( \cot \left( z\right) = \frac{{e}^{iz} + {e}^{-{iz}}}{{e}^{iz} - {e}^{-{iz}}} \) . Therefore, \( {I}_{N} \rightarrow 0 \) as \( N \rightarrow \infty \) because the integrand is of order \( 1/{N}^{2} \) while the diam... | Yes |
Find an interesting formula for \( \tan \left( {\pi z}\right) \) . | This is easy to obtain from the formula for \( \cot \left( {\pi z}\right) \) . \n\n\[ \cot \left( {\pi \left( {z + \frac{1}{2}}\right) }\right) = - \tan {\pi z} \] \n\nfor \( z \) real and therefore, this formula holds for \( z \) complex also. Therefore, for \( z + \frac{1}{2} \) not an integer \n\n\[ \pi \cot \left( ... | No |
Theorem 57.3.1 Let \( P \equiv {\left\{ {z}_{k}\right\} }_{k = 1}^{\infty } \) be a set of points in \( \mathbb{C} \), which has no limit point. For each \( {z}_{k} \), consider\n\n\[ \mathop{\sum }\limits_{{j = 0}}^{{m}_{k}}{a}_{j}^{k}{\left( z - {z}_{k}\right) }^{j} \]\n\n(57.3.18)\n\nThen there exists an analytic fu... | Proof: By the Weierstrass product theorem, Theorem 57.1.4, there exists an analytic function, \( f \) defined on all of \( \Omega \) such that \( f \) has a zero of order \( {m}_{k} + 1 \) at \( {z}_{k} \) . Consider this \( {z}_{k} \) Thus for \( z \) near \( {z}_{k} \), \n\n\[ f\left( z\right) = \mathop{\sum }\limits... | Yes |
Corollary 57.3.2 Let \( P \equiv {\left\{ {z}_{k}\right\} }_{k = 1}^{\infty } \) be a set of points in \( \Omega \), an open set such that \( P \) has no limit points in \( \Omega \) . For each \( {z}_{k} \), consider\n\n\[ \mathop{\sum }\limits_{{j = 0}}^{{m}_{k}}{a}_{j}^{k}{\left( z - {z}_{k}\right) }^{j} \]\n\n(57.3... | Proof: The proof is identical to the above except you use the versions of the Mittag-Leffler theorem and Weierstrass product which pertain to open sets. | No |
Corollary 57.3.5 Every finitely generated ideal in \( H\left( \Omega \right) \) for \( \Omega \) an open set is a principal ideal. | Proof: Let \( \left\lbrack {{g}_{1},\cdots ,{g}_{n}}\right\rbrack \) be a finitely generated ideal in \( H\left( \Omega \right) \) . Let \( \left\{ {U}_{k}\right\} \) be the components of \( \Omega \) . Then applying the above to each component, there exists \( {h}_{k} \in H\left( {U}_{k}\right) \) such that restrictin... | Yes |
Lemma 57.4.1\n\[ \n{\int }_{-\pi }^{\pi }\ln \left| {1 - {e}^{i\theta }}\right| {d\theta } = 0 \n\] | Proof: First note that the only problem with the integrand occurs when \( \theta = 0 \) . However, this is an integrable singularity so the integral will end up making sense. Letting \( z = {e}^{i\theta } \), you could get the above integral as a limit as \( \varepsilon \rightarrow 0 \) of the following contour integra... | Yes |
Lemma 57.4.2 Let \( u \) be harmonic on \( B\left( {0, r + \varepsilon }\right) \) . Then\n\n\[ u\left( 0\right) = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }u\left( {r{e}^{i\theta }}\right) {d\theta } \] | Proof: For a harmonic function, \( u \) defined on \( B\left( {0, r + \varepsilon }\right) \), there exists an analytic function, \( h = u + {iv} \) where\n\n\[ v\left( {x, y}\right) \equiv {\int }_{0}^{y}{u}_{x}\left( {x, t}\right) {dt} - {\int }_{0}^{x}{u}_{y}\left( {t,0}\right) {dt}. \]\n\nBy the Cauchy integral the... | Yes |
Corollary 57.4.3 Suppose \( f \) is analytic on \( B\left( {0, r + \varepsilon }\right) \) and has no zeros on \( \overline{B\left( {0, r}\right) } \) . Then \[ \ln \left| {f\left( 0\right) }\right| = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }\ln \left| {f\left( {r{e}^{i\theta }}\right) }\right| \] | What if \( f \) has some zeros on \( \left| z\right| = r \) but none on \( B\left( {0, r}\right) \) ? It turns out 57.4.21 is still valid. Suppose the zeros are at \( {\left\{ r{e}^{i{\theta }_{k}}\right\} }_{k = 1}^{m} \), listed according to multiplicity. Then let \[ g\left( z\right) = \frac{f\left( z\right) }{\matho... | Yes |
Lemma 57.4.4 Suppose \( f \) is analytic on \( B\left( {0, r + \varepsilon }\right) \) and has no zeros on \( B\left( {0, r}\right) \) . Then\n\n\[ \ln \left| {f\left( 0\right) }\right| = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }\ln \left| {f\left( {r{e}^{i\theta }}\right) }\right| \] | With this preparation, it is now not too hard to prove Jensen's formula. Suppose there are \( n \) zeros of \( f \) in \( B\left( {0, r}\right) ,{\left\{ {a}_{k}\right\} }_{k = 1}^{n} \), listed according to multiplicity, none equal to zero. Let\n\n\[ F\left( z\right) \equiv f\left( z\right) \mathop{\prod }\limits_{{i ... | Yes |
Theorem 57.4.5 Let \( f \) be analytic on \( B\left( {0, r + \varepsilon }\right) \) and suppose \( f\left( 0\right) \neq 0 \) . If the zeros of \( f \) in \( B\left( {0, r}\right) \) are \( {\left\{ {a}_{k}\right\} }_{k = 1}^{n} \), listed according to multiplicity, then\n\n\[ \ln \left| {f\left( 0\right) }\right| = -... | Proof: From the above discussion and Lemma 57.4.4,\n\n\[ \ln \left| {F\left( 0\right) }\right| = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }\ln \left| {f\left( {r{e}^{i\theta }}\right) }\right| {d\theta } \]\n\nBut \( F\left( 0\right) = f\left( 0\right) \mathop{\prod }\limits_{{i = 1}}^{n}\frac{r}{{a}_{i}} \) and so \( \ln \... | Yes |
Theorem 57.5.2 Suppose \( f \) is an analytic function on \( B\left( {0,1}\right), f\left( 0\right) \neq 0 \), and \( \left| {f\left( z\right) }\right| \leq M \) for all \( z \in B\left( {0,1}\right) \) . Suppose also that the zeros of \( f \) are \( {\left\{ {\alpha }_{k}\right\} }_{k = 1}^{\infty } \), listed accordi... | Proof: If there are only finitely many zeros, there is nothing to prove so assume there are infinitely many. Also let the zeros be listed such that \( \left| {\alpha }_{n}\right| \leq \left| {\alpha }_{n + 1}\right| \cdots \) Let \( n\left( r\right) \) denote the number of zeros in \( B\left( {0, r}\right) \) . By Jens... | Yes |
Corollary 57.5.3 Suppose \( f \) is an analytic function on \( B\left( {0,1}\right) \) and \( \left| {f\left( z\right) }\right| \leq M \) for all \( z \in B\left( {0,1}\right) \) . Suppose also that the nonzero zeros \( {}^{4} \) of \( f \) are \( {\left\{ {\alpha }_{k}\right\} }_{k = 1}^{\infty } \), listed according ... | Proof: Suppose \( f \) has a zero of order \( m \) at 0 . Then consider the analytic function, \( g\left( z\right) \equiv f\left( z\right) /{z}^{m} \) which has the same zeros except for 0 . The argument goes the same way except here you use \( g \) instead of \( f \) and only consider \( r > {r}_{0} > 0 \) .\n\n\( {}^... | Yes |
Theorem 58.1.2 Let \( f \) be a meromorphic function and let \( M \) be the module of periods. Then if \( M \) has a limit point, then \( f \) equals a constant. If this does not happen then either there exists \( {w}_{1} \in M \) such that \( \mathbb{Z}{w}_{1} = M \) or there exist \( {w}_{1},{w}_{2} \in \) \( M \) su... | Proof: Suppose \( f \) is meromorphic and \( M \) has a limit point, \( {w}_{0} \) . By Theorem 57.1.8 on Page 1892 there exist analytic functions, \( p, q \) such that \( f\left( z\right) = \frac{p\left( z\right) }{q\left( z\right) } \) . Now pick \( {z}_{0} \) such that \( {z}_{0} \) is not a pole of \( f \) . Then l... | Yes |
Theorem 58.1.4 Suppose \( f \) is an elliptic function which has no poles. Then \( f \) is constant. | Proof: Since \( f \) has no poles it is analytic. Now consider the parallelograms determined by the vertices, \( m{w}_{1} + n{w}_{2} \) for \( m, n \in \mathbb{Z} \) . By periodicity of \( f \) it must be bounded because its values are identical on each of these parallelograms. Therefore, it equals a constant by Liouvi... | Yes |
Theorem 58.1.6 The sum of the residues of any elliptic function, \( f \) equals zero on every \( {P}_{a} \) if \( a \) is chosen so that there are no poles on \( \partial {P}_{a} \) . | Proof: Choose \( a \) such that there are no poles of \( f \) on the boundary of \( {P}_{a} \) . By periodicity,\n\n\[ \n{\int }_{\partial {P}_{a}}f\left( z\right) {dz} = 0 \n\]\n\nbecause the integrals over opposite sides of the parallelogram cancel out because the values of \( f \) are the same on these sides and the... | Yes |
Theorem 58.1.7 Let \( {P}_{a} \) be a period parallelogram for a nonconstant elliptic function, \( f \) which has order equal to \( m \) . Then \( f \) assumes every value in \( f\left( {P}_{a}\right) \) exactly \( m \) times. | Proof: Let \( c \in f\left( {P}_{a}\right) \) and consider \( {P}_{{a}^{\prime }} \) such that \( {f}^{-1}\left( c\right) \cap {P}_{{a}^{\prime }} = {f}^{-1}\left( c\right) \cap {P}_{a} \) and \( {P}_{{a}^{\prime }} \) contains the same poles and zeros of \( f - c \) as \( {P}_{a} \) but \( {P}_{{a}^{\prime }} \) has n... | Yes |
Lemma 58.1.11 Define\n\n\[ \phi \left( \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \right) \equiv \frac{{az} + b}{{cz} + d}. \]\n\nThen\n\n\[ \phi \left( {AB}\right) = \phi \left( A\right) \circ \phi \left( B\right) ,\]\n\n\( \left( {58.1.3}\right) \)\n\n\( \phi \left( A\right) \left( z\right) = z \) if ... | Proof: The equation in 58.1.3 is just a simple computation. Now suppose \( \phi \left( A\right) \left( z\right) = z \) . Then letting \( A = \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \), this requires\n\n\[ {az} + b = z\left( {{cz} + d}\right) \]\n\nand so \( {az} + b = c{z}^{2} + {dz} \) . Since this ... | Yes |
Theorem 58.1.12 If \( f \) is a nonconstant elliptic function with a basis \( \left\{ {{w}_{1},{w}_{2}}\right\} \) for the module of periods, then \( \left\{ {{w}_{1}^{\prime },{w}_{2}^{\prime }}\right\} \) is another basis, if and only if there exists a unimodular transformation, \( \left( \begin{array}{ll} a & b \\ c... | Proof: Since \( \left\{ {{w}_{1},{w}_{2}}\right\} \) is a basis, there exist integers, \( a, b, c, d \) such that 58.1.4 holds. It remains to show the transformation determined by the matrix is unimodular. Taking conjugates,\n\n\[ \left( \frac{\overline{{w}_{1}^{\prime }}}{{w}_{2}^{\prime }}\right) = \left( \begin{arra... | Yes |
Lemma 58.1.14 The numbers, \( \wp \left( {{w}_{1}/2}\right) ,\wp \left( {{w}_{2}/2}\right) \), and \( \wp \left( {\left( {{w}_{1} + {w}_{2}}\right) /2}\right) \) are distinct. | Proof: Choose \( {P}_{a} \), a period parallelogram which contains the pole 0, and the points \( {w}_{1}/2,{w}_{2}/2 \), and \( \left( {{w}_{1} + {w}_{2}}\right) /2 \) but no other pole of \( \wp \left( z\right) \) . Also \( \partial {P}_{a}^{ * } \) does not contain any zeros of the elliptic function, \( z \rightarrow... | Yes |
Lemma 58.1.16 \( \lambda \left( \tau \right) = \lambda \left( {\tau }^{\prime }\right) \) if and only if\n\n\[{\tau }^{\prime } = \frac{{a\tau } + b}{{c\tau } + d}\]\n\nwhere 58.1.12 holds. | Proof: It only remains to verify that if \( \wp \left( {{w}_{1}^{\prime }/2}\right) = \wp \left( {{w}_{1}/2}\right) \) then it is necessary that\n\n\[ \frac{{w}_{1}^{\prime }}{2} - \frac{{w}_{1}}{2} \in M \]\n\nwith a similar requirement for \( {w}_{2} \) and \( {w}_{2}^{\prime } \) . If \( \frac{{w}_{1}^{\prime }}{2} ... | Yes |
Lemma 58.1.18 The following functional equations hold for \( \lambda \) . | \[ \lambda \left( {1 + \tau }\right) = \frac{\lambda \left( \tau \right) }{\lambda \left( \tau \right) - 1},1 = \lambda \left( \tau \right) + \lambda \left( \frac{-1}{\tau }\right) \] (58.1.25) \[ \lambda \left( {\tau + 2}\right) = \lambda \left( \tau \right) \] (58.1.26) \( \lambda \left( z\right) = \lambda \left( w\r... | Yes |
Lemma 58.1.20 \( \mathop{\lim }\limits_{{b \rightarrow \infty }}\lambda \left( {a + {ib}}\right) {e}^{-{i\pi }\left( {a + {ib}}\right) } = {16} \) uniformly in \( a \in \mathbb{R} \) . | Proof: From 58.1.30 and Lemma 58.1.19, this lemma will be proved if it is shown\n\n\[ \mathop{\lim }\limits_{{b \rightarrow \infty }}\left( {\frac{2}{{\cos }^{2}\left( {\pi \left( \frac{1}{2}\right) \left( {a + {ib}}\right) }\right) } - \frac{2}{{\sin }^{2}\left( {\pi \left( \frac{1}{2}\right) \left( {a + {ib}}\right) ... | Yes |
Corollary 58.1.21 \( \mathop{\lim }\limits_{{b \rightarrow \infty }}\lambda \left( {a + {ib}}\right) = 0 \) uniformly in \( a \in \mathbb{R} \) . Also \( \lambda \left( {ib}\right) \) for \( b > 0 \) is real and is between 0 and 1, \( \lambda \) is real on the line, \( {l}_{2} \) and on the curve, \( C \) and \( \matho... | Proof: From Lemma 58.1.20,\n\n\[ \left| {\lambda \left( {a + {ib}}\right) {e}^{-{i\pi }\left( {a + {ib}}\right) } - {16}}\right| < 1 \]\n\nfor all \( a \) provided \( b \) is large enough. Therefore, for such \( b \) ,\n\n\[ \left| {\lambda \left( {a + {ib}}\right) }\right| \leq {17}{e}^{-{\pi b}}. \]\n\n58.1.28 proves... | No |
Theorem 58.1.22 Let \( \Omega \) be the domain described above. Then \( \lambda \) maps \( \Omega \) one to one and onto the upper half plane of \( \mathbb{C},\{ z \in \mathbb{C} \) such that \( \operatorname{Im}\left( z\right) > 0\} \) . Also, the line \( \lambda \left( {l}_{1}\right) = \left( {0,1}\right) ,\lambda \l... | Proof: Let \( \operatorname{Im}\left( w\right) > 0 \) and denote by \( \gamma \) the oriented contour described above and illustrated in the above picture. Then the winding number of \( \lambda \circ \gamma \) about \( w \) equals 1. Thus\n\n\[ \n\frac{1}{2\pi i}{\int }_{\lambda \circ \gamma }\frac{1}{z - w}{dz} = 1 \n... | Yes |
Lemma 58.1.25 If \( \operatorname{Im}\left( \tau \right) > 0 \) then there exists a unimodular \( \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \) such that\n\n\[ \frac{c + {d\tau }}{a + {b\tau }} \]\n\nis contained in the interior of \( Q \) . In fact, \( \left| \frac{c + {d\tau }}{a + {b\tau }}\right| \g... | Proof: Letting a basis for the module of periods of \( \wp \) be \( \{ 1,\tau \} \), it follows from Theorem 58.1.2 on Page 1918 that there exists a basis for the same module of periods, \( \left\{ {{w}_{1}^{\prime },{w}_{2}^{\prime }}\right\} \) with the property that for \( {\tau }^{\prime } = {w}_{2}^{\prime }/{w}_{... | Yes |
Corollary 58.1.26 \( {\lambda }^{\prime }\left( \tau \right) \neq 0 \) for all \( \tau \) in the upper half plane, denoted by \( {P}_{ + } \) . | Proof: Let \( \tau \in {P}_{ + } \) . By Lemma 58.1.25 there exists \( \phi \) a unimodular transformation and \( {\tau }^{\prime } \) in the interior of \( Q \) such that \( {\tau }^{\prime } = \phi \left( \tau \right) \) . Now from the definition of \( \lambda \) in terms of the \( {e}_{i} \), there is at worst a per... | Yes |
Theorem 58.2.1 Let \( f \) be meromorphic on \( \mathbb{C} \) and suppose \( f \) misses three distinct points, \( a, b, c \) . Then \( f \) is a constant function. | Proof: Let \( \phi \left( z\right) \equiv \frac{z - a}{z - c}\frac{b - c}{b - a} \) . Then \( \phi \left( c\right) = \infty ,\phi \left( a\right) = 0 \), and \( \phi \left( b\right) = 1 \) . Now consider the function, \( h = \phi \circ f \) . Then \( h \) misses the three points \( \infty ,0 \), and 1 . Since \( h \) i... | Yes |
Lemma 59.1.2 Let \( \left( {\Omega ,\mathcal{F},\lambda }\right) \) be a measure space and let \( \left\{ {A}_{i}\right\} \) be a sequence of measurable sets satisfying\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{\infty }\lambda \left( {A}_{i}\right) < \infty \]\n\nThen letting \( S \) denote the set of \( \omega \in \Omega... | Proof: \( S = { \cap }_{k = 1}^{\infty }{ \cup }_{m = k}^{\infty }{A}_{m} \) . Therefore, \( S \) is measurable and also\n\n\[ \lambda \left( S\right) \leq \lambda \left( {{ \cup }_{m = k}^{\infty }{A}_{m}}\right) \leq \mathop{\sum }\limits_{{m = k}}^{\infty }\lambda \left( {A}_{k}\right) \]\n\nand this converges to 0 ... | Yes |
Proposition 59.1.3 Suppose \( {E}_{i} \) is a separable Banach space. Then if \( {B}_{i} \) is a Borel set of \( {E}_{i} \), it follows \( \mathop{\prod }\limits_{{i = 1}}^{n}{B}_{i} \) is a Borel set in \( \mathop{\prod }\limits_{{i = 1}}^{n}{E}_{i} \) . | Proof: An easy way to do this is to consider the projection maps.\n\n\[{\pi }_{i}\mathbf{x} \equiv {x}_{i}\]\n\nThen these projection maps are continuous. Hence for \( U \) an open set,\n\n\[{\pi }_{i}^{-1}\left( U\right) \equiv \mathop{\prod }\limits_{{j = 1}}^{n}{A}_{j},{A}_{j} = {E}_{j}\text{ if }j \neq i\text{ and ... | Yes |
Proposition 59.1.5 Let \( \left( {\Omega ,\mathcal{S},\mu }\right) \) be a measure space and let \( \mathbf{X} : \Omega \rightarrow Z \) where \( Z \) is a separable Banach space. Then \( \mathbf{X} \) is strongly measurable if and only if \( {\mathbf{X}}^{-1}\left( U\right) \in \mathcal{S} \) for all \( U \) open in \... | Proof: To begin with, let \( D\left( {a, r}\right) \) be the closure of the open ball \( B\left( {a, r}\right) \) . By Lemma 21.1.6, there exists \( \left\{ {f}_{i}\right\} \subseteq {B}^{\prime } \), the unit ball in \( {Z}^{\prime } \) such that\n\n\[ \parallel z{\parallel }_{Z} = \mathop{\sup }\limits_{i}\left\{ \le... | Yes |
Proposition 59.1.6 If \( \mathbf{X} : \Omega \rightarrow Z \) is measurable, then \( \sigma \left( \mathbf{X}\right) \) equals the smallest \( \sigma \) algebra such that \( \mathbf{X} \) is measurable with respect to it. Also if \( {X}_{i} \) are random variables having values in separable Banach spaces \( {Z}_{i} \),... | Proof: Let \( \mathcal{G} \) denote the smallest \( \sigma \) algebra such that \( \mathbf{X} \) is measurable with respect to this \( \sigma \) algebra. By definition \( {\mathbf{X}}^{-1} \) (open) \( \in \mathcal{G} \) . Furthermore, the set of all \( E \) such that \( {\mathbf{X}}^{-1}\left( E\right) \in \mathcal{G}... | Yes |
Lemma 59.1.9 Let \( \mu \) be a finite measure on a \( \sigma \) algebra containing \( \mathcal{B}\left( X\right) \), the Borel sets of \( X \), a separable complete metric space. Then if \( C \) is a closed set,\n\n\[ \mu \left( C\right) = \sup \{ \mu \left( K\right) : K \subseteq C\text{and}K\text{is compact.}\} \] | Proof: Let \( \left\{ {a}_{k}\right\} \) be a countable dense subset of \( C \) . Thus \( { \cup }_{k = 1}^{\infty }B\left( {{a}_{k},\frac{1}{n}}\right) \supseteq C \) . Therefore, there exists \( {m}_{n} \) such that\n\n\[ \mu \left( {C \smallsetminus { \cup }_{k = 1}^{{m}_{n}}\overline{B\left( {{a}_{k},\frac{1}{n}}\r... | Yes |
Proposition 59.1.12 For \( \mathbf{X} \) a random vector defined above, \( \mathbf{X} \) having values in a complete separable metric space \( Z \), then \( {\lambda }_{\mathbf{X}} \) is inner and outer regular and Borel. | Proof: The regularity claims are established above. It remains to verify 59.1.1. Since \( h \in {L}^{1}\left( {Z, E}\right) \), it follows there exists a sequence of simple functions \( \left\{ {h}_{n}\right\} \) such that \[ {h}_{n}\left( \mathbf{x}\right) \rightarrow h\left( \mathbf{x}\right) ,{\int }_{Z}\begin{Vmatr... | Yes |
Lemma 59.2.2 The sets, \( \mathcal{E},{\mathcal{E}}_{J} \) defined above form an algebra of sets of \( \mathop{\prod }\limits_{{t \in I}}{M}_{t} \) . | Proof: First consider \( {\mathcal{R}}_{J} \) . If \( \mathbf{A},\mathbf{B} \in {\mathcal{R}}_{J} \), then \( \mathbf{A} \cap \mathbf{B} \in {\mathcal{R}}_{J} \) also. Is \( \mathbf{A} \smallsetminus \mathbf{B} \) a finite disjoint union of sets of \( {\mathcal{R}}_{J} \) ? It suffices to verify that \( {\pi }_{J}\left... | Yes |
Theorem 59.2.3 For each finite set\n\n\[ J = \left( {{t}_{1},\cdots ,{t}_{n}}\right) \subseteq I \]\n\nsuppose there exists a Borel probability measure, \( {\nu }_{J} = {\nu }_{{t}_{1}\cdots {t}_{n}} \) defined on the Borel sets of \( \mathop{\prod }\limits_{{t \in J}}{M}_{t} \) such that the following consistency cond... | Proof: Let \( \mathcal{E} \) be the algebra of sets defined in Definition 14.4.1. I want to define a measure on \( \mathcal{E} \) . For \( \mathbf{F} \in \mathcal{E} \), there exists \( J \) such that \( \mathbf{F} \) is the finite disjoint unions of sets of \( {\mathcal{R}}_{J} \) . Define\n\n\[ {P}_{0}\left( \mathbf{... | Yes |
Lemma 59.3.2 Suppose \( {B}_{i} \in {\mathcal{F}}_{i} \) for \( i \in I \) . Then for any \( m \in \mathbb{N} \)\n\n\[ P\left( {{ \cap }_{k = 1}^{m}{B}_{{i}_{k}}}\right) = \mathop{\prod }\limits_{{k = 1}}^{m}P\left( {B}_{{i}_{k}}\right) . \]\n | Proof: The proof is by induction on the number \( l \) of the \( {B}_{{i}_{k}} \) which are not equal to \( {A}_{{i}_{k}} \) . First suppose \( l = 0 \) . Then the above assertion is true by assumption. Suppose it is so for some \( l \) and there are \( l + 1 \) sets not equal to \( {A}_{{i}_{k}} \) . If any equals \( ... | Yes |
Lemma 59.3.4 Suppose the set of random variables, \( {\left\{ {\mathbf{X}}_{i}\right\} }_{i \in I} \) is independent. Also suppose \( {I}_{1} \subseteq I \) and \( j \notin {I}_{1} \) . Then the \( \sigma \) algebras \( \sigma \left( {{\mathbf{X}}_{i} : i \in {I}_{1}}\right) ,\sigma \left( {\mathbf{X}}_{j}\right) \) ar... | Proof: Let \( B \in \sigma \left( {\mathbf{X}}_{j}\right) \) . I want to show that for any \( A \in \sigma \left( {{\mathbf{X}}_{i} : i \in {I}_{1}}\right) \) , it follows that \( P\left( {A \cap B}\right) = P\left( A\right) P\left( B\right) \) . Let \( \mathcal{K} \) consist of finite intersections of sets of the form... | Yes |
Lemma 59.3.5 If \( {\left\{ {\mathbf{X}}_{k}\right\} }_{k = 1}^{r} \) are independent random variables having values in \( Z \) a separable metric space, and if \( {g}_{k} \) is a Borel measurable function, then \( {\left\{ {g}_{k}\left( {\mathbf{X}}_{k}\right) \right\} }_{k = 1}^{r} \) is also independent. Furthermore... | Proof: First consider the claim about \( {\left\{ {g}_{k}\left( {\mathbf{X}}_{k}\right) \right\} }_{k = 1}^{r} \) . Letting \( O \) be an open set in \( Z \) ,\n\n\[ {\left( {g}_{k} \circ {\mathbf{X}}_{k}\right) }^{-1}\left( O\right) = {\mathbf{X}}_{k}^{-1}\left( {{g}_{k}^{-1}\left( O\right) }\right) = {\mathbf{X}}_{k}... | Yes |
Lemma 59.3.6 Let \( \\left( {\\Omega ,\\mathcal{F}}\\right) \) be a measure space and let \( {X}_{i} : \\Omega \\rightarrow {E}_{i} \) where \( {E}_{i} \) is a separable Banach space. Suppose also that \( X : \\Omega \\rightarrow F \) where \( F \) is a separable Banach space. Then \( X \) is \( \\sigma \\left( {{X}_{1... | Proof: First suppose \( X\\left( \\omega \\right) = f{\\mathcal{X}}_{W}\\left( \\omega \\right) \) where \( f \\in F \) and \( W \\in \\sigma \\left( {{X}_{1},\\cdots ,{X}_{m}}\\right) \) . Then by Proposition 59.1.6, \( W \) is of the form \( {\\left( {X}_{1},\\cdots ,{X}_{m}\\right) }^{-1}\\left( B\\right) \\equiv {\... | Yes |
Lemma 59.4.2 Let \( E \) be a separable real Banach space. Sets of the form\n\n\[ \left\{ {x \in E : {x}_{i}^{ * }\left( x\right) \leq {\alpha }_{i}, i = 1,2,\cdots, m}\right\} \]\n\nwhere \( {x}_{i}^{ * } \in {D}^{\prime } \), a dense subspace of the unit ball of \( {E}^{\prime } \) and \( {\alpha }_{i} \in \lbrack - ... | Proof: The sets described are obviously a \( \pi \) system. I want to show \( \sigma \left( \mathcal{K}\right) \) contains the closed balls because then \( \sigma \left( \mathcal{K}\right) \) contains the open balls and hence the open sets and the result will follow. Let \( {D}^{\prime } \) be described in Lemma 21.1.6... | Yes |
Lemma 59.4.4 Let \( \\mathcal{K} \) be a \( \\pi \) system of sets of \( E \), a separable real Banach space and let \( \\left( {\\Omega ,\\mathcal{F}, P}\\right) \) be a probability space and \( X : \\Omega \\rightarrow E \) be a random variable. Then\n\n\[ \n{X}^{-1}\\left( {\\sigma \\left( \\mathcal{K}\\right) }\\ri... | Proof: First note that \( {X}^{-1}\\left( {\\sigma \\left( \\mathcal{K}\\right) }\\right) \) is a \( \\sigma \) algebra which contains \( {X}^{-1}\\left( \\mathcal{K}\\right) \) and so it contains \( \\sigma \\left( {{X}^{-1}\\left( \\mathcal{K}\\right) }\\right) \) . Thus\n\n\[ \n{X}^{-1}\\left( {\\sigma \\left( \\mat... | Yes |
Theorem 59.5.1 Let \( {X}_{i} \) be a random variable having values in \( E \) a real separable Banach space. The random variables \( {\left\{ {X}_{i}\right\} }_{i \in I} \) are independent if whenever\n\n\[ \left\{ {{i}_{1},\cdots ,{i}_{n}}\right\} \subseteq I \]\n\n\( {m}_{{i}_{1}},\cdots ,{m}_{{i}_{n}} \) are positi... | Proof: It is necessary to show that the events \( {X}_{{i}_{j}}^{-1}\left( {B}_{{i}_{j}}\right) \) are independent events whenever \( {B}_{ij} \) are Borel sets. By Lemma 59.4.1 and the above Lemma 59.4.2, it suffices to verify that the events\n\n\[ {X}_{{i}_{j}}^{-1}\left( {{\mathbf{g}}_{{m}_{{i}_{j}}}^{-1}\left( {C}_... | Yes |
Theorem 59.6.1 Suppose \( {A}_{n} \in {\mathcal{F}}_{n} \) where the \( \sigma \) algebras \( {\left\{ {\mathcal{F}}_{n}\right\} }_{n = 1}^{\infty } \) are independent. Suppose also that\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }P\left( {A}_{k}\right) = \infty \]\n\nThen\n\n\[ P\left( {{ \cap }_{n = 1}^{\infty }{ ... | Proof: It suffices to verify that\n\n\[ P\left( {{ \cup }_{n = 1}^{\infty }{ \cap }_{m = n}^{\infty }{A}_{m}^{C}}\right) = 0 \]\n\nwhich can be accomplished by showing\n\n\[ P\left( {{ \cap }_{m = n}^{\infty }{A}_{m}^{C}}\right) = 0 \]\nfor each \( n \) . The sets \( \left\{ {A}_{k}^{C}\right\} \) satisfy \( {A}_{k}^{C... | Yes |
Lemma 59.6.3 Suppose \( {\left\{ {\mathcal{F}}_{n}\right\} }_{n = 1}^{\infty } \) are independent \( \sigma \) algebras and suppose \( A \) is a tail event and \( {A}_{{k}_{i}} \in {\mathcal{F}}_{{k}_{i}}, i = 1,\cdots, m \) are given sets. Then\n\n\[ P\left( {{A}_{{k}_{1}} \cap \cdots \cap {A}_{{k}_{m}} \cap A}\right)... | Proof: Let \( \mathcal{K} \) be the \( \pi \) system consisting of finite intersections of the form\n\n\[ {B}_{{m}_{1}} \cap {B}_{{m}_{2}} \cap \cdots \cap {B}_{{m}_{j}} \]\n\nwhere \( {m}_{i} \in {\mathcal{F}}_{{k}_{i}} \) for \( {k}_{i} > \max \left\{ {{k}_{1},\cdots ,{k}_{m}}\right\} \equiv N \) . Thus \( \sigma \le... | Yes |
Theorem 59.6.4 Suppose the \( \sigma \) algebras, \( {\left\{ {\mathcal{F}}_{n}\right\} }_{n = 1}^{\infty } \) are independent and suppose \( A \) is a tail event. Then \( P\left( A\right) \) either equals 0 or 1. | Proof: Let \( A \in \mathcal{T} \) . I want to show that \( P\left( A\right) = P{\left( A\right) }^{2} \) . Let \( \mathcal{K} \) denote sets of the form \( {A}_{{k}_{1}} \cap \cdots \cap {A}_{{k}_{m}} \) for some \( m,{A}_{{k}_{j}} \in {\mathcal{F}}_{{k}_{j}} \) where each \( {k}_{j} > n \) . Thus \( \mathcal{K} \) is... | Yes |
Theorem 59.6.5 Let \( \left\{ {\mathbf{X}}_{k}\right\} \) be a sequence of independent random variables having values in \( Z \) a Banach space. Then\n\n\[ A \equiv \left\{ {\omega : \left\{ {{\mathbf{X}}_{k}\left( \omega \right) }\right\} \text{ converges }}\right\} \]\n\n is a tail event of the independent \( \sigma ... | Proof: Since \( Z \) is complete, \( A \) is the same as the set where \( \left\{ {{\mathbf{X}}_{k}\left( \omega \right) }\right\} \) is a Cauchy sequence. This set is\n\n\[ { \cap }_{n = 1}^{\infty }{ \cap }_{p = 1}^{\infty }{ \cup }_{m = p}^{\infty }{ \cap }_{l, k \geq m}\left\{ {\omega : \left| \right| {\mathbf{X}}_... | Yes |
Lemma 59.7.1 If \( \mathbf{Y},\mathbf{X} \) are independent random variables having values in a real separable Hilbert space, \( H \) with \( E\left( {\left| \mathbf{X}\right| }^{2}\right), E\left( {\left| \mathbf{Y}\right| }^{2}\right) < \infty \), then\n\n\[{\int }_{\Omega }\left( {\mathbf{X},\mathbf{Y}}\right) {dP} ... | Proof: Let \( \left\{ {\mathbf{e}}_{k}\right\} \) be a complete orthonormal basis. Thus\n\n\[{\int }_{\Omega }\left( {\mathbf{X},\mathbf{Y}}\right) {dP} = {\int }_{\Omega }\mathop{\sum }\limits_{{k = 1}}^{\infty }\left( {\mathbf{X},{\mathbf{e}}_{k}}\right) \left( {\mathbf{Y},{\mathbf{e}}_{k}}\right) {dP}\]\n\nNow\n\n\[... | Yes |
Theorem 59.7.3 Let \( {\left\{ {\mathbf{X}}_{k}\right\} }_{k = 1}^{\infty } \) be independent random vectors having values in a separable real Hilbert space and suppose \( E\left( \left| {\mathbf{X}}_{k}\right| \right) < \infty \) for each \( k \) and \( E\left( {\mathbf{X}}_{k}\right) = \mathbf{0} \) . Suppose also th... | Proof: Let \( \varepsilon > 0 \) be given. By Kolmogorov’s inequality, Theorem 59.7.2, it follows that for \( p \leq m < n \)\n\n\[ P\left( \left\lbrack {\mathop{\max }\limits_{{m \leq k \leq n}}\left| {\mathop{\sum }\limits_{{j = m}}^{k}{\mathbf{X}}_{j}}\right| \geq \varepsilon }\right\rbrack \right) \leq \frac{1}{{\v... | Yes |
Lemma 59.7.4 Suppose \( {s}_{k} \rightarrow s \) . Then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{n}\mathop{\sum }\limits_{{k = 1}}^{n}{s}_{k} = s \] | Proof: Consider the first part. Since \( {s}_{k} \rightarrow s \), it follows there is some constant, \( C \) such that \( \left| {s}_{k}\right| < C \) for all \( k \) and \( \left| s\right| < C \) also. Choose \( K \) so large that if \( k \geq K \) , then for \( n > K \) ,\n\n\[ \left| {s - {s}_{k}}\right| < \varepsi... | Yes |
Theorem 59.7.5 Suppose \( \left\{ {\mathbf{X}}_{k}\right\} \) are independent random variables and \( E\left( \left| {\mathbf{X}}_{k}\right| \right) < \) \( \infty \) for each \( k \) and \( E\left( {\mathbf{X}}_{k}\right) = {\mathbf{m}}_{k} \) . Suppose also\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{\infty }\frac{1}{{j}^... | Proof: Consider the sum\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{\infty }\frac{{\mathbf{X}}_{j} - {\mathbf{m}}_{j}}{j} \]\n\nThis sum converges a.e. because of 59.7.12 and Theorem 59.7.3 applied to the random vectors \( \left\{ \frac{{\mathbf{X}}_{j} - {\mathbf{m}}_{j}}{j}\right\} \) . Therefore, from Lemma 59.7.4 it fol... | Yes |
Theorem 59.8.4 Let \( \mathbf{X} \) and \( \mathbf{Y} \) be random vectors with values in \( {\mathbb{R}}^{p} \) and suppose \( E\left( {e}^{i\mathbf{t} \cdot \mathbf{X}}\right) = E\left( {e}^{i\mathbf{t} \cdot \mathbf{Y}}\right) \) for all \( \mathbf{t} \in {\mathbb{R}}^{p} \) . Then \( {\lambda }_{\mathbf{X}} = {\lam... | Proof: For \( \psi \in \mathcal{G} \), let \( {\lambda }_{\mathbf{X}}\left( \psi \right) \equiv {\int }_{{\mathbb{R}}^{p}}{\psi d}{\lambda }_{\mathbf{X}} \) and \( {\lambda }_{\mathbf{Y}}\left( \psi \right) \equiv {\int }_{{\mathbb{R}}^{p}}{\psi d}{\lambda }_{\mathbf{Y}} \) . Thus both \( {\lambda }_{\mathbf{X}} \) and... | Yes |
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