Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Corollary 5.3 If \( f, g \in \mathcal{S}\left( {\mathbb{R}}^{3}\right) \) and \( \mathcal{R}\left( f\right) = \mathcal{R}\left( g\right) \), then \( f = g \) . | The proof of the corollary follows from an application of the lemma to the difference \( f - g \) and use of the Fourier inversion theorem. | No |
Theorem 5.4 If \( f \in \mathcal{S}\left( {\mathbb{R}}^{3}\right) \), then\n\n\[ \bigtriangleup \left( {{\mathcal{R}}^{ * }\mathcal{R}\left( f\right) }\right) = - 8{\pi }^{2}f \] | We recall that \( \bigtriangleup = \frac{{\partial }^{2}}{\partial {x}_{1}^{2}} + \frac{{\partial }^{2}}{\partial {x}_{2}^{2}} + \frac{{\partial }^{2}}{\partial {x}_{3}^{2}} \) is the Laplacian.\n\nProof. By our previous lemma, we have\n\n\[ \mathcal{R}\left( f\right) \left( {t,\gamma }\right) = {\int }_{-\infty }^{\in... | Yes |
Lemma 1.1 The family \( \left\{ {{e}_{0},\ldots ,{e}_{N - 1}}\right\} \) is orthogonal. In fact,\n\n\[ \left( {{e}_{m},{e}_{\ell }}\right) = \left\{ \begin{array}{ll} N & \text{ if }m = \ell \\ 0 & \text{ if }m \neq \ell \end{array}\right. \] | Proof. We have\n\n\[ \left( {{e}_{m},{e}_{\ell }}\right) = \mathop{\sum }\limits_{{k = 0}}^{{N - 1}}{\zeta }^{mk}{\zeta }^{-\ell k} = \mathop{\sum }\limits_{{k = 0}}^{{N - 1}}{\zeta }^{\left( {m - \ell }\right) k}. \]\n\nIf \( m = \ell \), each term in the sum is equal to 1, and the sum equals \( N \) . If \( m \neq \e... | Yes |
Theorem 1.2 If \( F \) is a function on \( \mathbb{Z}\left( N\right) \), then\n\n\[ F\left( k\right) = \mathop{\sum }\limits_{{n = 0}}^{{N - 1}}{a}_{n}{e}^{{2\pi ink}/N}. \]\n\nMoreover,\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{{N - 1}}{\left| {a}_{n}\right| }^{2} = \frac{1}{N}\mathop{\sum }\limits_{{k = 0}}^{{N - 1}}{\l... | The proof follows directly from (1) once we observe that\n\n\[ {a}_{n} = \frac{1}{N}\left( {F,{e}_{n}}\right) = \frac{1}{\sqrt{N}}\left( {F,{e}_{n}^{ * }}\right) . \] | Yes |
Theorem 1.3 Given \( {\omega }_{N} = {e}^{-{2\pi i}/N} \) with \( N = {2}^{n} \), it is possible to calculate the Fourier coefficients of a function on \( \mathbb{Z}\left( N\right) \) with at most\n\n\[ 4 \cdot {2}^{n}n = {4N}{\log }_{2}\left( N\right) = O\left( {N\log N}\right) \]\n\noperations. | The proof of the theorem consists of using the calculations for \( M \) division points, to obtain the Fourier coefficients for \( {2M} \) division points. Since we choose \( N = {2}^{n} \), we obtain the desired formula as a consequence of a recurrence which involves \( n = O\left( {\log N}\right) \) steps. | No |
Lemma 1.4 If we are given \( {\omega }_{2M} = {e}^{-{2\pi i}/\left( {2M}\right) } \), then\n\n\[ \n\# \left( {2M}\right) \leq 2\# \left( M\right) + {8M}.\n\] | Proof. The calculation of \( {\omega }_{2M},\ldots ,{\omega }_{2M}^{2M} \) requires no more than \( {2M} \) operations. Note that in particular we get \( {\omega }_{M} = {e}^{-{2\pi i}/M} = {\omega }_{2M}^{2} \). The main idea is that for any given function \( F \) on \( \mathbb{Z}\left( {2M}\right) \), we consider two... | Yes |
Lemma 2.1 The set \( \widehat{G} \) is an abelian group under multiplication defined \( {by} \)\n\n\[ \left( {{e}_{1} \cdot {e}_{2}}\right) \left( a\right) = {e}_{1}\left( a\right) {e}_{2}\left( a\right) \;\text{ for all }a \in G. \] | The proof of this assertion is straightforward if one observes that the trivial character plays the role of the unit. We call \( \widehat{G} \) the dual group of \( G \) . | No |
Lemma 2.2 Let \( G \) be a finite abelian group, and \( e : G \rightarrow \mathbb{C} - \{ 0\} \) a multiplicative function, namely \( e\left( {a \cdot b}\right) = e\left( a\right) e\left( b\right) \) for all \( a, b \in G \) . Then \( e \) is a character. | Proof. The group \( G \) being finite, the absolute value of \( e\left( a\right) \) is bounded above and below as \( a \) ranges over \( G \) . Since \( \left| {e\left( {b}^{n}\right) }\right| = {\left| e\left( b\right) \right| }^{n} \), we conclude that \( \left| {e\left( b\right) }\right| = 1 \) for all \( b \in G \)... | Yes |
Theorem 2.3 The characters of \( G \) form an orthonormal family with respect to the inner product defined above. | Since \( \left| {e\left( a\right) }\right| = 1 \) for any character, we find that\n\n\[\n\left( {e, e}\right) = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{a \in G}}e\left( a\right) \overline{e\left( a\right) } = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{a \in G}}{\left| e\left( a\right) \right| }^{2} = 1... | No |
Lemma 2.4 If \( e \) is a non-trivial character of the group \( G \), then \( \mathop{\sum }\limits_{{a \in G}}e\left( a\right) = 0. \) | Proof. Choose \( b \in G \) such that \( e\left( b\right) \neq 1 \) . Then we have\n\n\[ e\left( b\right) \mathop{\sum }\limits_{{a \in G}}e\left( a\right) = \mathop{\sum }\limits_{{a \in G}}e\left( b\right) e\left( a\right) = \mathop{\sum }\limits_{{a \in G}}e\left( {ab}\right) = \mathop{\sum }\limits_{{a \in G}}e\lef... | Yes |
Theorem 2.5 The characters of a finite abelian group \( G \) form a basis for the vector space of functions on \( G \) . | There are several proofs of this theorem. One consists of using the structure theorem for finite abelian groups we have mentioned earlier, which states that any such group is the direct product of cyclic groups, that is, groups of the type \( \mathbb{Z}\left( N\right) \) . Since cyclic groups are self-dual, using this ... | No |
Lemma 2.6 Suppose \( \\left\\{ {{T}_{1},\\ldots ,{T}_{k}}\\right\\} \) is a commuting family of unitary transformations on the finite-dimensional inner product space \( V \) ; that is,\n\n\[ \n{T}_{i}{T}_{j} = {T}_{j}{T}_{i}\\;\\text{ for all }i, j.\n\]\n\nThen \( {T}_{1},\\ldots ,{T}_{k} \) are simultaneously diagonal... | Proof. We use induction on \( k \) . The case \( k = 1 \) is simply the spectral theorem. Suppose that the lemma is true for any family of \( k - 1 \) commuting unitary transformations. The spectral theorem applied to \( {T}_{k} \) says that \( V \) is the direct sum of its eigenspaces\n\n\[ \nV = {V}_{{\\lambda }_{1}}... | Yes |
Theorem 2.7 Let \( G \) be a finite abelian group. The characters of \( G \) form an orthonormal basis for the vector space \( V \) of functions on \( G \) equipped with the inner product\n\n\[ \left( {f, g}\right) = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{a \in G}}f\left( a\right) \overline{g\left( a\right) ... | In particular, any function \( f \) on \( G \) is equal to its Fourier series\n\n\[ f = \mathop{\sum }\limits_{{e \in \widehat{G}}}\widehat{f}\left( e\right) e \] | Yes |
Theorem 2.8 If \( f \) is a function on \( G \), then \( \parallel f{\parallel }^{2} = \mathop{\sum }\limits_{{e \in \widehat{G}}}{\left| \widehat{f}\left( e\right) \right| }^{2} \) . | Proof. Since the characters of \( G \) form an orthonormal basis for the vector space \( V \), and \( \left( {f, e}\right) = \widehat{f}\left( e\right) \), we have that\n\n\[ \parallel f{\parallel }^{2} = \left( {f, f}\right) = \mathop{\sum }\limits_{{e \in \widehat{G}}}\left( {f, e}\right) \overline{\widehat{f}\left( ... | Yes |
Theorem 1.1 (Euclid’s algorithm) For any integers \( a \) and \( b \) with \( b > 0 \), there exist unique integers \( q \) and \( r \) with \( 0 \leq r < b \) such that\n\n\[ a = {qb} + r. \] | Proof. First we prove the existence of \( q \) and \( r \) . Let \( S \) denote the set of all non-negative integers of the form \( a - {qb} \) with \( q \in \mathbb{Z} \) . This set is non-empty and in fact \( S \) contains arbitrarily large positive integers since \( b \neq 0 \) . Let \( r \) denote the smallest elem... | Yes |
Theorem 1.2 If \( \gcd \left( {a, b}\right) = d \), then there exist integers \( x \) and \( y \) such that\n\n\[{ax} + {by} = d\text{.}\] | Proof. Consider the set \( S \) of all positive integers of the form \( {ax} + {by} \) where \( x, y \in \mathbb{Z} \), and let \( s \) be the smallest element in \( S \) . We claim that \( s = \) \( d \) . By construction, there exist integers \( x \) and \( y \) such that\n\n\[{ax} + {by} = s.\]\n\nClearly, any divis... | Yes |
Corollary 1.3 Two positive integers \( a \) and \( b \) are relatively prime if and only if there exist integers \( x \) and \( y \) such that \( {ax} + {by} = 1 \) . | Proof. If \( a \) and \( b \) are relatively prime, two integers \( x \) and \( y \) with the desired property exist by Theorem 1.2. Conversely, if \( {ax} + {by} = 1 \) holds and \( d \) is positive and divides both \( a \) and \( b \), then \( d \) divides 1, hence \( d = 1 \) . | Yes |
Corollary 1.4 If a and \( c \) are relatively prime and \( c \) divides \( {ab} \), then \( c \) divides \( b \) . In particular, if \( p \) is a prime that does not divide \( a \) and \( p \) divides \( {ab} \), then \( p \) divides \( b \) . | Proof. We can write \( 1 = {ax} + {cy} \), so multiplying by \( b \) we find \( b = \) \( {abx} + {cby} \) . Hence \( c \mid b \) . | Yes |
Corollary 1.5 If \( p \) is prime and \( p \) divides the product \( {a}_{1}\cdots {a}_{r} \), then \( p \) divides \( {a}_{i} \) for some \( i \) . | Proof. By the previous corollary, if \( p \) does not divide \( {a}_{1} \), then \( p \) divides \( {a}_{2}\cdots {a}_{r} \), so eventually \( p \mid {a}_{i} \) . | No |
Theorem 1.6 Every positive integer greater than 1 can be factored uniquely into a product of primes. | Proof. First, we show that such a factorization is possible. We do so by proving that the set \( S \) of positive integers \( > 1 \) which do not have a factorization into primes is empty. Arguing by contradiction, we assume that \( S \neq \varnothing \) . Let \( n \) be the smallest element of \( S \) . Since \( n \) ... | Yes |
Theorem 1.7 There are infinitely many primes. | Proof. Suppose not, and denote by \( {p}_{1},\ldots ,{p}_{n} \) the complete set of primes. Define\n\n\[ N = {p}_{1}{p}_{2}\cdots {p}_{n} + 1 \]\n\nSince \( N \) is larger than any \( {p}_{i} \), the integer \( N \) cannot be prime. Therefore, \( N \) is divisible by a prime that belongs to our list. But this is also a... | Yes |
Lemma 1.8 The exponential and logarithm functions satisfy the following properties:\n\n(i) \( {e}^{\log x} = x \) .\n\n(ii) \( \log \left( {1 + x}\right) = x + E\left( x\right) \) where \( \left| {E\left( x\right) }\right| \leq {x}^{2} \) if \( \left| x\right| < 1/2 \) .\n\n(iii) If \( \log \left( {1 + x}\right) = y \)... | Proof. Property (i) is standard. To prove property (ii) we use the power series expansion of \( \log \left( {1 + x}\right) \) for \( \left| x\right| < 1 \), that is,\n\n(2)\n\n\[ \log \left( {1 + x}\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n + 1}}{n}{x}^{n}. \]\n\nThen we have\n\n\[ ... | Yes |
Proposition 1.9 If \( {A}_{n} = 1 + {a}_{n} \) and \( \sum \left| {a}_{n}\right| \) converges, then the product \( \mathop{\prod }\limits_{n}{A}_{n} \) converges, and this product vanishes if and only if one of its factors \( {A}_{n} \) vanishes. Also, if \( {a}_{n} \neq 1 \) for all \( n \), then \( \mathop{\prod }\li... | Proof. If \( \sum \left| {a}_{n}\right| \) converges, then for all large \( n \) we must have \( \left| {a}_{n}\right| < \) \( 1/2 \) . Disregarding finitely many terms if necessary, we may assume that this inequality holds for all \( n \) . Then we may write the partial products as follows:\n\n\[ \mathop{\prod }\limit... | Yes |
Theorem 1.10 For every \( s > 1 \), we have\n\n\[ \zeta \left( s\right) = \mathop{\prod }\limits_{p}\frac{1}{1 - 1/{p}^{s}} \]\n\nwhere the product is taken over all primes. | Proof. Suppose \( M \) and \( N \) are positive integers with \( M > N \). Observe now that any positive integer \( n \leq N \) can be written uniquely as a product of primes, and that each prime must be less than or equal to \( N \) and repeated less than \( M \) times. Therefore\n\n\[ \mathop{\sum }\limits_{{n = 1}}^... | Yes |
Proposition 1.11 The series\n\n\[ \mathop{\sum }\limits_{p}1/p \]\n\n\ndiverges, when the sum is taken over all primes p. | Proof. We take logarithms of both sides of the Euler formula. Since \( \log x \) is continuous, we may write the logarithm of the infinite product as the sum of the logarithms. Therefore, we obtain for \( s > 1 \)\n\n\[ - \mathop{\sum }\limits_{p}\log \left( {1 - 1/{p}^{s}}\right) = \log \zeta \left( s\right) \]\n\nSin... | Yes |
Lemma 2.2 The Dirichlet characters are multiplicative. Moreover, \[ {\delta }_{\ell }\left( m\right) = \frac{1}{\varphi \left( q\right) }\mathop{\sum }\limits_{\chi }\overline{\chi \left( \ell \right) }\chi \left( m\right) \] where the sum is over all Dirichlet characters. | With the above lemma we have taken our first step towards a proof of the theorem, since this lemma shows that \[ \mathop{\sum }\limits_{{p \equiv \ell }}\frac{1}{{p}^{s}} = \mathop{\sum }\limits_{p}\frac{{\delta }_{\ell }\left( p\right) }{{p}^{s}} \] \[ = \frac{1}{\varphi \left( q\right) }\mathop{\sum }\limits_{\chi }\... | Yes |
Theorem 2.4 If \( s > 1 \), then\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\chi \left( n\right) }{{n}^{s}} = \mathop{\prod }\limits_{p}\frac{1}{\left( 1 - \chi \left( p\right) {p}^{-s}\right) } \]\n\nwhere the product is over all primes. | Assuming this theorem for now, we can follow Euler's argument formally: taking the logarithm of the product and using the fact that \( \log \left( {1 + x}\right) = x + O\left( {x}^{2}\right) \) whenever \( x \) is small, we would get\n\n\[ \log L\left( {s,\chi }\right) = - \mathop{\sum }\limits_{p}\log \left( {1 - \chi... | No |
Proposition 3.1 The logarithm function \( {\log }_{1} \) satisfies the following properties:\n\n(i) If \( \left| z\right| < 1 \), then\n\n\[ \n{e}^{{\log }_{1}\left( \frac{1}{1 - z}\right) } = \frac{1}{1 - z}.\n\]\n\n(ii) If \( \left| z\right| < 1 \), then\n\n\[ \n{\log }_{1}\left( \frac{1}{1 - z}\right) = z + {E}_{1}\... | Proof. To establish the first property, let \( z = r{e}^{i\theta } \) with \( 0 \leq r < 1 \) , and observe that it suffices to show that\n\n(5)\n\n\[ \n\left( {1 - r{e}^{i\theta }}\right) {e}^{\mathop{\sum }\limits_{{k = 1}}^{\infty }{\left( r{e}^{i\theta }\right) }^{k}/k} = 1.\n\]\n\nTo do so, we differentiate the le... | Yes |
Proposition 3.2 If \( \sum \left| {a}_{n}\right| \) converges, and \( {a}_{n} \neq 1 \) for all \( n \), then\n\n\[ \mathop{\prod }\limits_{{n = 1}}^{\infty }\left( \frac{1}{1 - {a}_{n}}\right) \]\n\nconverges. Moreover, this product is non-zero. | Proof. For \( n \) large enough, \( \left| {a}_{n}\right| < 1/2 \), so we may assume without loss of generality that this inequality holds for all \( n \geq 1 \) . Then\n\n\[ \mathop{\prod }\limits_{{n = 1}}^{N}\left( \frac{1}{1 - {a}_{n}}\right) = \mathop{\prod }\limits_{{n = 1}}^{N}{e}^{{\log }_{1}\left( \frac{1}{1 -... | Yes |
Proposition 3.3 Suppose \( {\chi }_{0} \) is the trivial Dirichlet character,\n\n\[ \n{\chi }_{0}\left( n\right) = \left\{ \begin{array}{ll} 1 & \text{ if }n\text{ and }q\text{ are relatively prime,} \\ 0 & \text{ otherwise,} \end{array}\right.\n\]\n\nand \( q = {p}_{1}^{{a}_{1}}\cdots {p}_{N}^{{a}_{N}} \) is the prime... | Proof. The identity follows at once on comparing the Dirichlet and Euler product formulas. The final statement holds because \( \zeta \left( s\right) \rightarrow \infty \) as \( s \rightarrow {1}^{ + } \) | Yes |
Lemma 3.5 If \( \chi \) is a non-trivial Dirichlet character, then\n\n\[ \left| {\mathop{\sum }\limits_{{n = 1}}^{k}\chi \left( n\right) }\right| \leq q,\;\text{ for any }k. \] | Proof. First, we recall that\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{q}\chi \left( n\right) = 0 \]\n\nIn fact, if \( S \) denotes the sum and \( a \in {\mathbb{Z}}^{ * }\left( q\right) \), then the multiplicative property of the Dirichlet character \( \chi \) gives\n\n\[ \chi \left( a\right) S = \sum \chi \left( a\right... | Yes |
Proposition 3.6 If \( s > 1 \), then\n\n\[ \n{e}^{{\log }_{2}L\left( {s,\chi }\right) } = L\left( {s,\chi }\right) \n\]\n\nMoreover\n\n\[ \n{\log }_{2}L\left( {s,\chi }\right) = \mathop{\sum }\limits_{p}{\log }_{1}\left( \frac{1}{1 - \chi \left( p\right) /{p}^{s}}\right) .\n\] | Proof. Differentiating \( {e}^{-{\log }_{2}L\left( {s,\chi }\right) }L\left( {s,\chi }\right) \) with respect to \( s \) gives\n\n\[ \n- \frac{{L}^{\prime }\left( {s,\chi }\right) }{L\left( {s,\chi }\right) }{e}^{-{\log }_{2}L\left( {s,\chi }\right) }L\left( {s,\chi }\right) + {e}^{-{\log }_{2}L\left( {s,\chi }\right) ... | Yes |
Theorem 3.7 If \( \chi \neq {\chi }_{0} \), then \( L\left( {1,\chi }\right) \neq 0 \) . | There are several proofs of this fact, some involving algebraic number theory (among them Dirichlet's original argument), and others involving complex analysis. Here we opt for a more elementary argument that requires no special knowledge of either of these areas. The proof splits in two cases, depending on whether \( ... | No |
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\[ \left| {L\left( {s,\chi }\right) }\right| \leq C\left| {s - 1}\right| \;\text{ when }1 \leq... | 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, the last sta... | 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\[ \frac{1}{N}\mathop{\sum }\limits_{{k = 1}}^{N}d\left( k\right) = \log N + O\left( 1\right) \]\n\nMore precisely,\n\n\[ \frac{1}{N}\mathop{\sum }\limits_{{k = 1}}^{N}d\left( k\right) = \log N + \left( {{2\gamma } - 1}\right) + O\left( {1/{N}^{1/2}}\right) ,\]\n\n... | 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\[ {S}_{N} = \mathop{\sum }\limits_{{1 \leq m \leq N}}\mathop{\sum }\limits_{{1 \leq n \leq N/m}}1 \]\n\nBut \( \mathop{\sum }\limit... | 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 - ... | Yes |
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 ... | No |
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\[ \n\operatorname{diam}\left( {\Omega }_{n}\right) \rightarrow 0\;\text{ as }n \rightarrow \infty ,\n\]\n\nthen there ex... | 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 n... | 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. | No |
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\[ \n{\int }_{R}f\left( z\right) {dz} = 0 \n\] | This is immediate since we first choose an orientation as in Figure 2 and note that\n\n\[ \n{\int }_{R}f\left( z\right) {dz} = {\int }_{{T}_{1}}f\left( z\right) {dz} + {\int }_{{T}_{2}}f\left( z\right) {dz}. \n\] | 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 |
Corollary 4.2 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}\... | Proof. 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( \... | 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\[ f\left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( z - {z}_{0}\... | Proof. Fix \( z \in D \) . By the Cauchy integral formula, we have\n\n(10)\n\n\[ f\left( z\right) = \frac{1}{2\pi i}{\int }_{C}\frac{f\left( \zeta \right) }{\zeta - z}{d\zeta } \]\n\nwhere \( C \) denotes the boundary of the disc and \( z \in D \) . The idea is to write\n\n(11)\n\n\[ \frac{1}{\zeta - z} = \frac{1}{\zet... | 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 clo... | 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\[ \n{f}^{ + }\left( x\right) = {f}^{ - }\left( x\right) \;\text{ for all }x \in I, \n\]\n\nthen the function \( f... | 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\[ f\left( z\right) = \frac{{a}_{-n}}{{\left( z - {z}_{0}\right) }^{n}} + \frac{{a}_{-n + 1}}{{\left( z - {z}_{0}\right) }^{n - 1}} + \cdots + \frac{{a}_{-1}}{\left( z - {z}_{0}\right) } + G\left( z\right) ,\] | 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\[{\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) . | The theorem is an immediate consequence of formula (1), which implies\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\[+ G\left( z\right) {\left( z - {z}_{0}\right) }^{n}. | No |
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 rad... | 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\[ \left| {f\left( z\right) - w}\right| > \delta \;\text{ for all }z \in {D}_{r}\left( {z}_{0}\right) - \left\{ {z}_{0}\right\} .\n\]\n\nWe may therefore define ... | 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\[ \left| {f\left( z\right) }\right| > \left| {g\left( z\right) }\right| \;\text{ for all }z \in C, \]\n\nthen \( f \) and \( f + g \) have the same number of zeros inside t... | 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 \( f \) .\n\nDefine \( g\left( z\right) = f\left( z\right) - w \) and write\n\n\[ g\left( z\right) = \left( {f\left( z\right) - {w}_{... | 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 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.