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Lemma 4 If \( {p}_{n} \geq {11} \) (i. e., \( n \geq 5 \) ), \( {p}_{n}^{2} \) is contained in the first block of \( {M}_{{p}_{n - 1}} \) . (Here the block means the subset of \( {M}_{{p}_{n - 1}} \) corresponding to the \( {D}_{{p}_{n - 1}} \) . ) | Proof First,\n\n\[ \n{p}_{n}^{2} < {p}_{n} + L\left( {P}_{{p}_{n - 1}}\right) \Leftrightarrow {p}_{n}\left( {{p}_{n} - 1}\right) < {p}_{n - 1}{p}_{n - 2}\cdots 5 \cdot 3 \cdot 2.\n\]\n\nIt is easy to check that \( {p}_{k} = {11} \) satisfies the inequality:\n\n\[ \n{p}_{k}\left( {{p}_{k} - 1}\right) < {p}_{k - 1}{p}_{k... | Yes |
Theorem 5 The gap sequence except the last element in the pattern is mirror symmetric. | Proof For example, when constructing \( {D}_{11} \) from \( {D}_{7} \), we delete the numbers \( \left\{ {{11k} \mid k \in {M}_{7}}\right\} \) . We list the divisors \( k \) of the first period. They are\n\n<table><tr><td>1</td><td>11</td><td>13</td><td>17</td><td>19</td><td>23</td><td>29</td><td>31</td><td>37</td><td>... | Yes |
Theorem 7 (i) The multiplicities of skips 2 and 4 are always the same and odd. | Proof (i) In fact, these multiplicities for \( k \) -prime basis, \( k > 1 \), are both precisely equal to\n\n\[ \left( {3 - 2}\right) \cdot \left( {5 - 2}\right) \cdots \left( {{p}_{k} - 2}\right) ,\]\n\n(2)\n\nwhich is always odd as well. For example, there are 15 skips 2 in the pattern \( {P}_{7} \) (see (1)).\n\nFo... | Yes |
Theorem 12 Let \( c \) be a positive integer. Define a characteristic function for each interval \( {I}_{k} = \lbrack \left( {k - 1}\right) c,{kc}), k = 1,2,\cdots \) as follows\n\n\[ \n{\chi }_{{I}_{k}} = \left\{ \begin{matrix} 1, & \text{ if }{I}_{k}\text{ contains one or more primes,} \\ 0, & \text{ otherwise. } \en... | Proof Let \( {J}_{ki}, i = 1,2,\cdots, c \) be the subintervals in \( {I}_{k} \), each \( {J}_{ki} \) has the form \( \lbrack d, d + 1) \) . Then, \( {\chi }_{{I}_{k}} = \mathop{\max }\limits_{{1 \leq i \leq c}}{\chi }_{{J}_{ki}} \) . Thus,\n\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{m}{\chi }_{{I}_{k}} = \mathop{\sum }\l... | Yes |
Let \( \mathbf{G} \) be a generator matrix of a \( q \) -ary self-dual code in the form of (a) and \( \underline{\mathbf{H}} \) arise from \( \bar{Z}\left( \mathbf{G}\right) \) by deleting its all-zero columns. Then \( \underline{\mathbf{H}} \) is a parity check matrices of a LDPC code. | Obviously, \( {\mathbf{I}}_{2n} \) will be remained, when \( \bar{Z}\left( {\mathbf{I}}_{2n}\right) \) delete its all-zero columns. It implies that rows of \( \underline{\mathbf{H}} \) are linear independence and the number of columns of \( \underline{\mathbf{H}} \) is at most\n\n\[ \n{4n}\left( {q - 1}\right) - {2n}\l... | Yes |
Lemma 2.2 The number of all non-zero columns of \( {\left( \mathbf{Z}{\left( {a}_{1}\right) }^{\top },\mathbf{Z}{\left( {a}_{2}\right) }^{\top },\cdots ,\mathbf{Z}{\left( {a}_{m}\right) }^{\top }\right) }^{\top } \) is \( 1\left( {{a}_{1},{a}_{2},\cdots ,{a}_{m}}\right) \) . | Proof Let \( \alpha \) be a primitive element in \( {GF}\left( q\right) \) and \( {a}_{i} \) be a power of \( \alpha \) . If \( {a}_{i} \neq {a}_{j} \), then location vectors of \( {a}_{i} \) and \( {a}_{j} \) are distinct. So 1 stays in different positions of their location vectors. If \( b \in {GF}\left( q\right) \) ... | Yes |
Theorem 2.3 The length \( L \) of the LDPC code corresponding to \( \underline{\mathbf{H}} \) is\n\n\[ L = {2n} + n\mathrm{l}\left( {{\mathbf{\alpha }}_{1},\cdots ,{\mathbf{\alpha }}_{n},{\mathbf{\beta }}_{1},\cdots ,{\mathbf{\beta }}_{n}}\right) + n\mathrm{l}\left( {{\mathbf{\alpha }}_{1},\cdots ,{\mathbf{\alpha }}_{n... | Proof Because \( \mathbf{A} \) and \( \mathbf{B} \) are all circulant, \( {\mathbf{A}}^{\mathrm{T}} \) and \( - {\mathbf{B}}^{\mathrm{T}} \) are circulant. Then all columns of \( \mathbf{A} \) and \( {\mathbf{A}}^{\mathrm{T}} \) are composed of \( \left( {{\mathbf{\alpha }}_{1},\cdots ,{\mathbf{\alpha }}_{n}}\right) \)... | Yes |
Corollary 2.4 \( \frac{1}{2} \leq R \leq \frac{q - 1}{q} \) . | Proof We consider two especial cases. Firstly, if \( \mathbf{A} = {\mathbf{I}}_{n} \) and \( \mathbf{B} = \mathbf{O} \), then the rate reach low bound \( \frac{1}{2} \). Secondly, if first rows of \( \left( {\mathbf{A},\mathbf{B}}\right) \) and \( \left( {\mathbf{A}, - \mathbf{B}}\right) \) contain all \( q - 1 \) nonz... | No |
Theorem 3.1 Assumed that \( u\left( t\right), v\left( t\right) \in C\left\lbrack {a, b}\right\rbrack, u\left( t\right) \leq v\left( t\right) \) and\n\n\[ \n{D}^{\alpha }u\left( t\right) \leq f\left( {t, u\left( t\right) }\right) ,\;{D}^{\alpha }v\left( t\right) \geq f\left( {t, v\left( t\right) }\right) ,\;0 < \alpha <... | Proof Let \( \omega \left( t\right) \in C\left( \left\lbrack {a, b}\right\rbrack \right) \), defined\n\n\[ \nx\left( t\right) = \left\{ \begin{array}{ll} u\left( t\right) , & \omega \left( t\right) < u\left( t\right) , \\ \omega \left( t\right) , & u\left( t\right) \leq \omega \left( t\right) \leq v\left( t\right) , \\... | Yes |
Lemma 2.3 Let\n\n\[ L\left( x\right) = \left( {{x}^{2} + \frac{53}{210} - \frac{2117}{{423360}{x}^{4}}}\right) \ln \left( {1 + \frac{1}{{12}{x}^{3} + \frac{24}{7}x - \frac{1}{2}}}\right) \]\n\nand\n\n\[ U\left( x\right) = \left( {{x}^{2} + \frac{53}{210} - \frac{2117}{{423360}{x}^{4}} + \frac{46075}{{5588352}{x}^{6}}}\... | Proof Define the function \( M\left( x\right) \) by\n\n\[ M\left( x\right) = L\left( x\right) - \left( {\frac{1}{12x} - \frac{1}{{360}{x}^{3}} + \frac{1}{{1260}{x}^{5}} - \frac{1}{{1680}{x}^{7}} + \frac{943}{{6096384}{x}^{9}}}\right) . \]\n\nDifferentiation yields\n\n\[ \frac{{M}^{\prime }\left( x\right) }{{2x} + \frac... | Yes |
Theorem 3.2 For \( x \geq 2 \)\n\n\[ \sqrt{2\pi x}{\left( \frac{x}{\mathrm{e}}\right) }^{x}{\left( 1 + \frac{1}{{12}{x}^{3} + \frac{24}{7}x - \frac{1}{2}}\right) }^{{x}^{2} + \frac{53}{210}\frac{2117}{{423360}{x}^{4}}} < \Gamma \left( {x + 1}\right) \]\n\n\[ < \sqrt{2\pi x}{\left( \frac{x}{\mathrm{e}}\right) }^{x}{\lef... | Proof The lower bound is obtained by considering the function \( f\left( x\right) \) defined for \( x \geq 2 \) by\n\n\[ f\left( x\right) = \ln \Gamma \left( x\right) - \left( {x - \frac{1}{2}}\right) \ln x + x - \frac{1}{2}\ln \left( {2\pi }\right) - L\left( x\right) ,\]\n\nwhere \( L\left( x\right) \) is define in (2... | Yes |
Lemma 2.4 Let \( \tau > 0 \) and \( 0 < \beta < 1 \) . Then \( h\left( t\right) = {\left( t - \tau \right) }^{\beta } - {t}^{\beta } \) is strictly increasing on \( \lbrack \tau , + \infty ) \) and \( h\left( t\right) \in \left\lbrack {-{\tau }^{\beta },0}\right) \) . | Proof For \( t > \tau \), we have \( h\left( t\right) = {\left( t - \tau \right) }^{\beta } - {t}^{\beta } > - {\tau }^{\beta } \) and\n\n\[ \n{h}^{\prime }\left( t\right) = \frac{\beta }{{\left( t - \tau \right) }^{1 - \beta }} - \frac{\beta }{{t}^{1 - \beta }} > 0.\n\]\n\nTherefore, for \( t \geq \tau, h\left( t\righ... | Yes |
Theorem 3.9 Let \( {b}_{2} \) be a real constant and satisfy\n\n(i) \( {b}_{2} > \frac{\Gamma \left( {\alpha + 1}\right) \ln \left( {1 + 2\parallel \mathbf{C}{\parallel }_{0}}\right) }{{\tau }^{\alpha }} + \parallel \mathbf{A}{\parallel }_{0} + \parallel \mathbf{B}{\parallel }_{0} \) ,\n\n(ii) \( \left( {1 + 2\parallel... | Proof Applying the Property 2.2, we can obtain\n\n\[ \mathbf{x}\left( t\right) = \mathbf{\varphi }\left( 0\right) + \frac{1}{\Gamma \left( \alpha \right) }{\int }_{0}^{t}{\left( t - s\right) }^{\alpha - 1}\left\lbrack {\mathbf{A}\mathbf{x}\left( s\right) + \mathbf{B}\mathbf{x}\left( {s - \tau }\right) + \mathbf{G}\math... | Yes |
Theorem 2.1 Suppose that \( f \) is an integrable function on the circle with \( \widehat{f}\left( n\right) = 0 \) for all \( n \in \mathbb{Z} \) . Then \( f\left( {\theta }_{0}\right) = 0 \) whenever \( f \) is continuous at the point \( {\theta }_{0} \) . | Proof. We suppose first that \( f \) is real-valued, and argue by contradiction. Assume, without loss of generality, that \( f \) is defined on \( \left\lbrack {-\pi ,\pi }\right\rbrack \), that \( {\theta }_{0} = 0 \), and \( f\left( 0\right) > 0 \) . The idea now is to construct a family of trigonometric polynomials ... | No |
Corollary 2.3 Suppose that \( f \) is a continuous function on the circle and that the Fourier series of \( f \) is absolutely convergent, \( \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\left| {\widehat{f}\left( n\right) }\right| < \infty \) . Then, the Fourier series converges uniformly to \( f \), that is,\n\n\[... | Proof. Recall that if a sequence of continuous functions converges uniformly, then the limit is also continuous. Now observe that the assumption \( \sum \left| {\widehat{f}\left( n\right) }\right| < \infty \) implies that the partial sums of the Fourier\n\nseries of \( f \) converge absolutely and uniformly, and theref... | Yes |
Corollary 2.4 Suppose that \( f \) is a twice continuously differentiable function on the circle. Then\n\n\[ \widehat{f}\left( n\right) = O\left( {1/{\left| n\right| }^{2}}\right) \;\text{ as }\left| n\right| \rightarrow \infty , \]\n\nso that the Fourier series of \( f \) converges absolutely and uniformly to \( f \) ... | Proof. The estimate on the Fourier coefficients is proved by integrating by parts twice for \( n \neq 0 \) . We obtain\n\n\[ {2\pi }\widehat{f}\left( n\right) = {\int }_{0}^{2\pi }f\left( \theta \right) {e}^{-{in\theta }}{d\theta } \]\n\n\[ = {\left\lbrack f\left( \theta \right) \cdot \frac{-{e}^{-{in\theta }}}{in}\rig... | Yes |
Proposition 3.1 Suppose that \( f, g \), and \( h \) are \( {2\pi } \) -periodic integrable functions. Then:\n\n(i) \( f * \left( {g + h}\right) = \left( {f * g}\right) + \left( {f * h}\right) \) . | Proof. Properties (i) and (ii) follow at once from the linearity of the integral. | No |
Lemma 3.2 Suppose \( f \) is integrable on the circle and bounded by \( B \) . Then there exists a sequence \( {\left\{ {f}_{k}\right\} }_{k = 1}^{\infty } \) of continuous functions on the circle so that\n\n\[ \mathop{\sup }\limits_{{x \in \left\lbrack {-\pi ,\pi }\right\rbrack }}\left| {{f}_{k}\left( x\right) }\right... | Using this result, we may complete the proof of the proposition as follows. Apply Lemma 3.2 to \( f \) and \( g \) to obtain sequences \( \left\{ {f}_{k}\right\} \) and \( \left\{ {g}_{k}\right\} \) of approximating continuous functions. Then\n\n\[ f * g - {f}_{k} * {g}_{k} = \left( {f - {f}_{k}}\right) * g + {f}_{k} *... | No |
Theorem 4.1 Let \( {\left\{ {K}_{n}\right\} }_{n = 1}^{\infty } \) be a family of good kernels, and \( f \) an integrable function on the circle. Then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {f * {K}_{n}}\right) \left( x\right) = f\left( x\right) \]\n\nwhenever \( f \) is continuous at \( x \) . If ... | Proof of Theorem 4.1. If \( \epsilon > 0 \) and \( f \) is continuous at \( x \), choose \( \delta \) so that \( \left| y\right| < \delta \) implies \( \left| {f\left( {x - y}\right) - f\left( x\right) }\right| < \epsilon \) . Then, by the first property of good kernels, we can write\n\n\[ \left( {f * {K}_{n}}\right) \... | Yes |
Lemma 5.1 We have\n\n\[ \n{F}_{N}\left( x\right) = \frac{1}{N}\frac{{\sin }^{2}\left( {{Nx}/2}\right) }{{\sin }^{2}\left( {x/2}\right) }, \]\n\nand the Fejér kernel is a good kernel. | The proof of the formula for \( {F}_{N} \) (a simple application of trigonometric identities) is outlined in Exercise 15. To prove the rest of the lemma, note that \( {F}_{N} \) is positive and \( \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{F}_{N}\left( x\right) {dx} = 1 \), in view of the fact that a similar identity holds ... | No |
Corollary 5.3 If \( f \) is integrable on the circle and \( \widehat{f}\left( n\right) = 0 \) for all \( n \) , then \( f = 0 \) at all points of continuity of \( f \) . | The proof is immediate since all the partial sums are 0 , hence all the Cesàro means are 0 . | No |
Corollary 5.4 Continuous functions on the circle can be uniformly approximated by trigonometric polynomials. | This means that if \( f \) is continuous on \( \left\lbrack {-\pi ,\pi }\right\rbrack \) with \( f\left( {-\pi }\right) = f\left( \pi \right) \) and \( \epsilon > 0 \), then there exists a trigonometric polynomial \( P \) such that\n\n\[ \left| {f\left( x\right) - P\left( x\right) }\right| < \epsilon \;\text{ for all }... | Yes |
Lemma 5.5 If \( 0 \leq r < 1 \), then\n\n\[ \n{P}_{r}\left( \theta \right) = \frac{1 - {r}^{2}}{1 - {2r}\cos \theta + {r}^{2}}. \n\] | Proof. The identity \( {P}_{r}\left( \theta \right) = \frac{1 - {r}^{2}}{1 - {2r}\cos \theta + {r}^{2}} \) has already been derived in Section 1.1. Note that\n\n\[ \n1 - {2r}\cos \theta + {r}^{2} = {\left( 1 - r\right) }^{2} + {2r}\left( {1 - \cos \theta }\right) . \n\]\n\nHence if \( 1/2 \leq r \leq 1 \) and \( \delta... | Yes |
Lemma 1.2 (Best approximation) If \( f \) is integrable on the circle with Fourier coefficients \( {a}_{n} \), then\n\n\[ \begin{Vmatrix}{f - {S}_{N}\left( f\right) }\end{Vmatrix} \leq \begin{Vmatrix}{f - \mathop{\sum }\limits_{{\left| n\right| \leq N}}{c}_{n}{e}_{n}}\end{Vmatrix} \]\n\nfor any complex numbers \( {c}_{... | Proof. This follows immediately by applying the Pythagorean theorem to\n\n\[ f - \mathop{\sum }\limits_{{\left| n\right| \leq N}}{c}_{n}{e}_{n} = f - {S}_{N}\left( f\right) + \mathop{\sum }\limits_{{\left| n\right| \leq N}}{b}_{n}{e}_{n} \]\n\nwhere \( {b}_{n} = {a}_{n} - {c}_{n} \) . | Yes |
Theorem 1.4 (Riemann-Lebesgue lemma) If \( f \) is integrable on the circle, then \( \widehat{f}\left( n\right) \rightarrow 0 \) as \( \left| n\right| \rightarrow \infty \) . | An equivalent reformulation of this proposition is that if \( f \) is integrable on \( \left\lbrack {0,{2\pi }}\right\rbrack \), then\n\n\[ \n{\int }_{0}^{2\pi }f\left( \theta \right) \sin \left( {N\theta }\right) {d\theta } \rightarrow 0\;\text{ as }N \rightarrow \infty \n\]\n\nand\n\n\[ \n{\int }_{0}^{2\pi }f\left( \... | Yes |
Lemma 1.5 Suppose \( F \) and \( G \) are integrable on the circle with\n\n\[ F \sim \sum {a}_{n}{e}^{in\theta }\;\text{ and }\;G \sim \sum {b}_{n}{e}^{in\theta }.\]\n\nThen\n\n\[ \frac{1}{2\pi }{\int }_{0}^{2\pi }F\left( \theta \right) \overline{G\left( \theta \right) }{d\theta } = \mathop{\sum }\limits_{{n = - \infty... | Proof. The proof follows from Parseval's identity and the fact that\n\n\[ \left( {F, G}\right) = \frac{1}{4}\left\lbrack {\parallel F + G{\parallel }^{2} - \parallel F - G{\parallel }^{2} + i\left( {\parallel F + {iG}{\parallel }^{2} - \parallel F - {iG}{\parallel }^{2}}\right) }\right\rbrack \]\n\nwhich holds in every... | No |
Theorem 2.1 Let \( f \) be an integrable function on the circle which is differentiable at a point \( {\theta }_{0} \) . Then \( {S}_{N}\left( f\right) \left( {\theta }_{0}\right) \rightarrow f\left( {\theta }_{0}\right) \) as \( N \) tends to infinity. | Proof. Define\n\n\[ F\left( t\right) = \left\{ \begin{array}{ll} \frac{f\left( {{\theta }_{0} - t}\right) - f\left( {\theta }_{0}\right) }{t} & \text{ if }t \neq 0\text{ and }\left| t\right| < \pi \\ - {f}^{\prime }\left( {\theta }_{0}\right) & \text{ if }t = 0. \end{array}\right. \]\n\nFirst, \( F \) is bounded near 0... | Yes |
Theorem 2.2 Suppose \( f \) and \( g \) are two integrable functions defined on the circle, and for some \( {\theta }_{0} \) there exists an open interval \( I \) containing \( {\theta }_{0} \) such that\n\n\[ f\left( \theta \right) = g\left( \theta \right) \;\text{ for all }\theta \in I. \]\n\nThen \( {S}_{N}\left( f\... | Proof. The function \( f - g \) is 0 in \( I \), so it is differentiable at \( {\theta }_{0} \), and we may apply the previous theorem to conclude the proof. | No |
Lemma 2.3 Suppose that the Abel means \( {A}_{r} = \mathop{\sum }\limits_{{n = 1}}^{\infty }{r}^{n}{c}_{n} \) of the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{c}_{n} \) are bounded as \( r \) tends to 1 (with \( r < 1 \) ). If \( {c}_{n} = O\left( {1/n}\right) \), then the partial sums \( {S}_{N} = \mathop{\s... | Proof. Let \( r = 1 - 1/N \) and choose \( M \) so that \( n\left| {c}_{n}\right| \leq M \) . We estimate the difference\n\n\[ \n{S}_{N} - {A}_{r} = \mathop{\sum }\limits_{{n = 1}}^{N}\left( {{c}_{n} - {r}^{n}{c}_{n}}\right) - \mathop{\sum }\limits_{{n = N + 1}}^{\infty }{r}^{n}{c}_{n} \n\]\n\nas follows:\n\n\[ \n\left... | Yes |
Lemma 2.4\n\n\[ \n{S}_{M}\left( {P}_{N}\right) = \left\{ \begin{array}{ll} {P}_{N} & \text{ if }M \geq {3N} \\ {\widetilde{P}}_{N} & \text{ if }M = {2N} \\ 0 & \text{ if }M < N \end{array}\right. \n\] | This is clear from what has been said above and from Figure 3. \n\nFigure 3. Breaking symmetry in Lemma 2.4\n\nThe effect is that when \( M = {2N} \), the operator \( {S}_{M} \) breaks the symmetry of \( {P}_{N} \), bu... | Yes |
Theorem 2.1 If \( \gamma \) is irrational, then the sequence of fractional parts \( \langle \gamma \rangle ,\langle {2\gamma }\rangle ,\langle {3\gamma }\rangle ,\ldots \) is equidistributed in \( \lbrack 0,1) \) . | In particular, \( \langle {n\gamma }\rangle \) is dense in \( \lbrack 0,1) \), and we get Kronecker’s theorem as a corollary. In Figure 2 we illustrate the set of points \( \langle \gamma \rangle ,\langle {2\gamma }\rangle \) , \( \langle {3\gamma }\rangle ,\ldots ,\langle {N\gamma }\rangle \) for three different value... | Yes |
Lemma 2.2 If \( f \) is continuous and periodic of period 1, and \( \gamma \) is irrational, then\n\n\[ \n\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}f\left( {n\gamma }\right) \rightarrow {\int }_{0}^{1}f\left( x\right) {dx}\;\text{ as }N \rightarrow \infty .\n\] | The proof of the lemma is divided into three steps.\n\nStep 1. We first check the validity of the limit in the case when \( f \) is one of the exponentials \( 1,{e}^{2\pi ix},\ldots ,{e}^{2\pi ikx},\ldots \) . If \( f = 1 \), the limit\nsurely holds. If \( f = {e}^{2\pi ikx} \) with \( k \neq 0 \), then the integral is... | Yes |
Corollary 2.3 The conclusion of Lemma 2.2 holds for every function \( f \) which is Riemann integrable in \( \left\lbrack {0,1}\right\rbrack \), and periodic of period 1 . | Proof. Assume \( f \) is real-valued, and consider a partition of the interval \( \left\lbrack {0,1}\right\rbrack \), say \( 0 = {x}_{0} < {x}_{1} < \cdots < {x}_{N} = 1 \) . Next, define \( {f}_{U}\left( x\right) = \) \( \mathop{\sup }\limits_{{{x}_{j - 1} \leq y \leq {x}_{j}}}f\left( y\right) \) if \( x \in \left\lbr... | Yes |
Lemma 3.3 If \( {2N} = {2}^{n} \), then\n\n\[{\bigtriangleup }_{2N}\left( f\right) - {\bigtriangleup }_{N}\left( f\right) = {2}^{-{n\alpha }}{e}^{i{2}^{n}x}.\] | This follows from our previous observation (6) because \( {\bigtriangleup }_{2N}\left( f\right) = \) \( {S}_{2N}\left( f\right) \) and \( {\bigtriangleup }_{N}\left( f\right) = {S}_{N}\left( f\right) \) . | No |
Proposition 1.1 The integral of a function of moderate decrease defined by (5) satisfies the following properties:\n\n(i) Linearity: if \( f, g \in \mathcal{M}\left( \mathbb{R}\right) \) and \( a, b \in \mathbb{C} \), then\n\n\[ \n{\int }_{-\infty }^{\infty }\left( {{af}\left( x\right) + {bg}\left( x\right) }\right) {d... | We say a few words about the proof. Property (i) is immediate. To verify property (ii), it suffices to see that\n\n\[ \n{\int }_{-N}^{N}f\left( {x - h}\right) {dx} - {\int }_{-N}^{N}f\left( x\right) {dx} \rightarrow 0\;\text{ as }N \rightarrow \infty . \n\]\n\nSince \( {\int }_{-N}^{N}f\left( {x - h}\right) {dx} = {\in... | Yes |
Proposition 1.2 If \( f \in \mathcal{S}\left( \mathbb{R}\right) \) then:\n\n(i) \( f\left( {x + h}\right) \rightarrow \widehat{f}\left( \xi \right) {e}^{2\pi ih\xi } \) whenever \( h \in \mathbb{R} \) .\n\n(ii) \( f\left( x\right) {e}^{-{2\pi ixh}} \rightarrow \widehat{f}\left( {\xi + h}\right) \) whenever \( h \in \ma... | Proof. Property (i) is an immediate consequence of the translation invariance of the integral, and property (ii) follows from the definition. Also, the third property of Proposition 1.1 establishes (iii).\n\nIntegrating by parts gives\n\n\[{\int }_{-N}^{N}{f}^{\prime }\left( x\right) {e}^{-{2\pi ix\xi }}{dx} = {\left\l... | Yes |
Theorem 1.3 If \( f \in \mathcal{S}\left( \mathbb{R}\right) \), then \( \widehat{f} \in \mathcal{S}\left( \mathbb{R}\right) \) . | The proof is an easy application of the fact that the Fourier transform interchanges differentiation and multiplication. In fact, note that if \( f \in \) \( \mathcal{S}\left( \mathbb{R}\right) \), its Fourier transform \( \widehat{f} \) is bounded; then also, for each pair of non-negative integers \( \ell \) and \( k ... | Yes |
Theorem 1.4 If \( f\left( x\right) = {e}^{-\pi {x}^{2}} \), then \( \widehat{f}\left( \xi \right) = f\left( \xi \right) \) . | Proof. Define\n\n\[ F\left( \xi \right) = \widehat{f}\left( \xi \right) = {\int }_{-\infty }^{\infty }{e}^{-\pi {x}^{2}}{e}^{-{2\pi ix\xi }}{dx} \]\n\nand observe that \( F\left( 0\right) = 1 \), by our previous calculation. By property (v) in Proposition 1.2, and the fact that \( {f}^{\prime }\left( x\right) = - {2\pi... | Yes |
Corollary 1.7 If \( f \in \mathcal{S}\left( \mathbb{R}\right) \), then\n\n\[ \left( {f * {K}_{\delta }}\right) \left( x\right) \rightarrow f\left( x\right) \;\text{ uniformly in }x\text{ as }\delta \rightarrow 0. \] | Proof. First, we claim that \( f \) is uniformly continuous on \( \mathbb{R} \) . Indeed, given \( \epsilon > 0 \) there exists \( R > 0 \) so that \( \left| {f\left( x\right) }\right| < \epsilon /4 \) whenever \( \left| x\right| \geq R \) . Moreover, \( f \) is continuous, hence uniformly continuous on the compact int... | Yes |
Proposition 1.8 If \( f, g \in \mathcal{S}\left( \mathbb{R}\right) \), then\n\n\[{\int }_{-\infty }^{\infty }f\left( x\right) \widehat{g}\left( x\right) {dx} = {\int }_{-\infty }^{\infty }\widehat{f}\left( y\right) g\left( y\right) {dy}.\] | To prove the proposition, we need to digress briefly to discuss the interchange of the order of integration for double integrals. Suppose \( F\left( {x, y}\right) \) is a continuous function in the plane \( \left( {x, y}\right) \in {\mathbb{R}}^{2} \) . We will assume the following decay condition on \( F \) :\n\n\[ \l... | Yes |
Theorem 1.9 (Fourier inversion) If \( f \in \mathcal{S}\left( \mathbb{R}\right) \), then\n\n\[ f\left( x\right) = {\int }_{-\infty }^{\infty }\widehat{f}\left( \xi \right) {e}^{2\pi ix\xi }{d\xi }.\] | Proof. We first claim that\n\n\[ f\left( 0\right) = {\int }_{-\infty }^{\infty }\widehat{f}\left( \xi \right) {d\xi }\n\nLet \( {G}_{\delta }\left( x\right) = {e}^{-{\pi \delta }{x}^{2}} \) so that \( \widehat{{G}_{\delta }}\left( \xi \right) = {K}_{\delta }\left( \xi \right) \) . By the multiplication formula we get\n... | Yes |
Proposition 1.11 If \( f, g \in \mathcal{S}\left( \mathbb{R}\right) \) then:\n\n(i) \( f * g \in \mathcal{S}\left( \mathbb{R}\right) \).\n\n(ii) \( f * g = g * f \).\n\n(iii) \( \widehat{\left( f * g\right) }\left( \xi \right) = \widehat{f}\left( \xi \right) \widehat{g}\left( \xi \right) \). | Proof. To prove that \( f * g \) is rapidly decreasing, observe first that for any \( \ell \geq 0 \) we have \( \mathop{\sup }\limits_{x}{\left| x\right| }^{\ell }\left| {g\left( {x - y}\right) }\right| \leq {A}_{\ell }{\left( 1 + \left| y\right| \right) }^{\ell } \), because \( g \) is rapidly decreasing (to check thi... | Yes |
Theorem 1.12 (Plancherel) If \( f \in \mathcal{S}\left( \mathbb{R}\right) \) then \( \parallel \widehat{f}\parallel = \parallel f\parallel \) . | Proof. If \( f \in \mathcal{S}\left( \mathbb{R}\right) \) define \( {f}^{b}\left( x\right) = \overline{f\left( {-x}\right) } \) . Then \( \widehat{{f}^{b}}\left( \xi \right) = \overline{\widehat{f}\left( \xi \right) } \) . Now let \( h = f * {f}^{b} \) . Clearly, we have\n\n\[ \n\widehat{h}\left( \xi \right) = {\left| ... | Yes |
Theorem 2.1 Given \( f \in \mathcal{S}\left( \mathbb{R}\right) \), let\n\n\[ u\left( {x, t}\right) = \left( {f * {\mathcal{H}}_{t}}\right) \left( x\right) \;\text{ for }t > 0 \]\n\nwhere \( {\mathcal{H}}_{t} \) is the heat kernel. Then:\n\n(i) The function \( u \) is \( {C}^{2} \) when \( x \in \mathbb{R} \) and \( t >... | Proof. Because \( u = f * {\mathcal{H}}_{t} \), taking the Fourier transform in the \( x \) - variable gives \( \widehat{u} = \widehat{f}{\widehat{\mathcal{H}}}_{t} \), and so \( \widehat{u}\left( {\xi, t}\right) = \widehat{f}\left( \xi \right) {e}^{-4{\pi }^{2}{\xi }^{2}t} \) . The Fourier inversion formula gives\n\n\... | Yes |
Corollary 2.2 \( u\left( {\cdot, t}\right) \) belongs to \( \mathcal{S}\left( \mathbb{R}\right) \) uniformly in \( t \), in the sense that for any \( T > 0 \)\n\n(9)\n\n\[ \mathop{\sup }\limits_{\substack{{x \in \mathbb{R}} \\ {0 < t < T} }}{\left| x\right| }^{k}\left| {\frac{{\partial }^{\ell }}{\partial {x}^{\ell }}u... | Proof. This result is a consequence of the following estimate:\n\n\[ \left| {u\left( {x, t}\right) }\right| \leq {\int }_{\left| y\right| \leq \left| x\right| /2}\left| {f\left( {x - y}\right) }\right| {\mathcal{H}}_{t}\left( y\right) {dy} + {\int }_{\left| y\right| \geq \left| x\right| /2}\left| {f\left( {x - y}\right... | Yes |
Theorem 2.3 Suppose \( u\left( {x, t}\right) \) satisfies the following conditions:\n\n(i) \( u \) is continuous on the closure of the upper half-plane.\n\n(ii) \( u \) satisfies the heat equation for \( t > 0 \) .\n\n(iii) \( u \) satisfies the boundary condition \( u\left( {x,0}\right) = 0 \) .\n\n(iv) \( u\left( {\c... | Proof. We define the energy at time \( t \) of the solution \( u\left( {x, t}\right) \) by\n\n\[ E\left( t\right) = {\int }_{\mathbb{R}}{\left| u\left( x, t\right) \right| }^{2}{dx} \]\n\nClearly \( E\left( t\right) \geq 0 \) . Since \( E\left( 0\right) = 0 \) it suffices to show that \( E \) is a decreasing function, ... | Yes |
Lemma 2.4 The following two identities hold:\n\n\[ \n{\int }_{-\infty }^{\infty }{e}^{-{2\pi }\left| \xi \right| y}{e}^{2\pi i\xi x}{d\xi } = {\mathcal{P}}_{y}\left( x\right) \n\]\n\n\[ \n{\int }_{-\infty }^{\infty }{\mathcal{P}}_{y}\left( x\right) {e}^{-{2\pi ix\xi }}{dx} = {e}^{-{2\pi }\left| \xi \right| y}. \n\] | Proof. The first formula is fairly straightforward since we can split the integral from \( - \infty \) to 0 and 0 to \( \infty \) . Then, since \( y > 0 \) we have\n\n\[ \n{\int }_{0}^{\infty }{e}^{-{2\pi \xi y}}{e}^{2\pi i\xi x}{d\xi } = {\int }_{0}^{\infty }{e}^{{2\pi i}\left( {x + {iy}}\right) \xi }{d\xi } = {\left\... | Yes |
Lemma 2.5 The Poisson kernel is a good kernel on \( \mathbb{R} \) as \( y \rightarrow 0 \) . | Proof. Setting \( \xi = 0 \) in the second formula of the lemma shows that \( {\int }_{-\infty }^{\infty }{\mathcal{P}}_{y}\left( x\right) {dx} = 1 \), and clearly \( {\mathcal{P}}_{y}\left( x\right) \geq 0 \), so it remains to check the last property of good kernels. Given a fixed \( \delta > 0 \), we may change varia... | Yes |
Theorem 2.6 Given \( f \in \mathcal{S}\left( \mathbb{R}\right) \), let \( u\left( {x, y}\right) = \left( {f * {\mathcal{P}}_{y}}\right) \left( x\right) \) . Then:\n\n(i) \( u\left( {x, y}\right) \) is \( {C}^{2} \) in \( {\mathbb{R}}_{ + }^{2} \) and \( \bigtriangleup u = 0 \) .\n\n(ii) \( u\left( {x, y}\right) \righta... | Proof. The proofs of parts (i), (ii), and (iii) are similar to the case of the heat equation, and so are left to the reader. Part (iv) is a consequence of two easy estimates whenever \( f \) is of moderate decrease. First, we have\n\n\[ \left| {\left( {f * {\mathcal{P}}_{y}}\right) \left( x\right) }\right| \leq C\left(... | No |
Lemma 2.8 (Mean-value property) Suppose \( \Omega \) is an open set in \( {\mathbb{R}}^{2} \) and let \( u \) be a function of class \( {C}^{2} \) with \( \bigtriangleup u = 0 \) in \( \Omega \) . If the closure of the disc centered at \( \left( {x, y}\right) \) and of radius \( R \) is contained in \( \Omega \), then\... | Proof. Let \( U\left( {r,\theta }\right) = u\left( {x + r\cos \theta, y + r\sin \theta }\right) \) . Expressing the Laplacian in polar coordinates, the equation \( \bigtriangleup u = 0 \) then implies\n\n\[ 0 = \frac{{\partial }^{2}U}{\partial {\theta }^{2}} + r\frac{\partial }{\partial r}\left( {r\frac{\partial U}{\pa... | Yes |
Theorem 3.1 (Poisson summation formula) If \( f \in \mathcal{S}\left( \mathbb{R}\right) \), then\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }f\left( {x + n}\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\widehat{f}\left( n\right) {e}^{2\pi inx}. \]\n\nIn particular, setting \( x = 0 \) we have\n\... | Proof. To check the first formula it suffices, by Theorem 2.1 in Chapter 2, to show that both sides (which are continuous) have the same Fourier coefficients (viewed as functions on the circle). Clearly, the \( {m}^{\text{th }} \) Fourier coefficient of the right-hand side is \( \widehat{f}\left( m\right) \) . For the ... | Yes |
Theorem 3.2 \( {s}^{-1/2}\vartheta \left( {1/s}\right) = \vartheta \left( s\right) \) whenever \( s > 0 \) . | The proof of this identity consists of a simple application of the Poisson summation formula to the pair\n\n\[ f\left( x\right) = {e}^{-{\pi s}{x}^{2}}\;\text{ and }\;\widehat{f}\left( \xi \right) = {s}^{-1/2}{e}^{-\pi {\xi }^{2}/s}. \] | Yes |
Corollary 3.4 The kernel \( {H}_{t}\left( x\right) \) is a good kernel for \( t \rightarrow 0 \) . | Proof. We already observed that \( {\int }_{\left| x\right| \leq 1/2}{H}_{t}\left( x\right) {dx} = 1 \) . Now note that \( {H}_{t} \geq 0 \), which is immediate from the above formula since \( {\mathcal{H}}_{t} \geq 0 \) . Finally, we claim that when \( \left| x\right| \leq 1/2 \) ,\n\n\[ \n{H}_{t}\left( x\right) = {\m... | Yes |
Theorem 3.5 \( {P}_{r}\left( {2\pi x}\right) = \mathop{\sum }\limits_{{n \in \mathbb{Z}}}{\mathcal{P}}_{y}\left( {x + n}\right) \) where \( r = {e}^{-{2\pi y}} \) . | This is again an immediate corollary of the Poisson summation formula applied to \( f\left( x\right) = {\mathcal{P}}_{y}\left( x\right) \) and \( \widehat{f}\left( \xi \right) = {e}^{-{2\pi }\left| \xi \right| y} \) . Of course, here we use the Poisson summation formula under the assumptions that \( f \) and \( \wideha... | Yes |
Theorem 4.1 Suppose \( \psi \) is a function in \( \mathcal{S}\left( \mathbb{R}\right) \) which satisfies the normalizing condition \( {\int }_{-\infty }^{\infty }{\left| \psi \left( x\right) \right| }^{2}{dx} = 1 \) . Then\n\n\[ \left( {{\int }_{-\infty }^{\infty }{x}^{2}{\left| \psi \left( x\right) \right| }^{2}{dx}}... | Proof. The second inequality actually follows from the first by replacing \( \psi \left( x\right) \) by \( {e}^{-{2\pi ix}{\xi }_{0}}\psi \left( {x + {x}_{0}}\right) \) and changing variables. To prove the first inequality, we argue as follows. Beginning with our normalizing assumption \( \int {\left| \psi \right| }^{2... | Yes |
Proposition 2.1 Let \( f \in \mathcal{S}\left( {\mathbb{R}}^{d}\right) \) .\n\n(i) \( f\left( {x + h}\right) \rightarrow \widehat{f}\left( \xi \right) {e}^{{2\pi i\xi } \cdot h} \) whenever \( h \in {\mathbb{R}}^{d} \) .\n\n(ii) \( f\left( x\right) {e}^{-{2\pi ixh}} \rightarrow \widehat{f}\left( {\xi + h}\right) \) whe... | The first five properties are proved in the same way as in the one-dimensional case. To verify the last property, simply change variables \( y = {Rx} \) in the integral. Then, recall that \( \left| {\det \left( R\right) }\right| = 1 \), and \( {R}^{-1}y \cdot \xi = y \cdot {R\xi } \), because \( R \) is a rotation. | Yes |
Corollary 2.3 The Fourier transform of a radial function is radial. | This follows at once from property (vi) in the last proposition. Indeed, the condition \( f\left( {Rx}\right) = f\left( x\right) \) for all \( R \) implies that \( \widehat{f}\left( {R\xi }\right) = \widehat{f}\left( \xi \right) \) for all \( R \), thus \( \widehat{f} \) is radial whenever \( f \) is. | Yes |
A solution of the Cauchy problem for the wave equation is\n\n\[ u\left( {x, t}\right) = {\int }_{{\mathbb{R}}^{d}}\left\lbrack {\widehat{f}\left( \xi \right) \cos \left( {{2\pi }\left| \xi \right| t}\right) + \widehat{g}\left( \xi \right) \frac{\sin \left( {{2\pi }\left| \xi \right| t}\right) }{{2\pi }\left| \xi \right... | Proof. We first verify that \( u \) solves the wave equation. This is straightforward once we note that we can differentiate in \( x \) and \( t \) under the integral sign (because \( f \) and \( g \) are both Schwartz functions) and therefore \( u \) is at least \( {C}^{2} \). On the one hand we differentiate the expo... | Yes |
Theorem 3.2 If \( u \) is the solution of the wave equation given by formula (3), then \( E\left( t\right) \) is conserved, that is,\n\n\[ E\left( t\right) = E\left( 0\right) ,\;\text{ for all }t \in \mathbb{R}. \] | The proof requires the following lemma.\n\nLemma 3.3 Suppose a and | No |
Lemma 3.3 Suppose a and \( b \) are complex numbers and \( \alpha \) is real. Then\n\n\[ \n{\left| a\cos \alpha + b\sin \alpha \right| }^{2} + {\left| -a\sin \alpha + b\cos \alpha \right| }^{2} = {\left| a\right| }^{2} + {\left| b\right| }^{2}.\n\] | This follows directly because \( {e}_{1} = \left( {\cos \alpha ,\sin \alpha }\right) \) and \( {e}_{2} = \left( {-\sin \alpha ,\cos \alpha }\right) \) are a pair of orthonormal vectors, hence with \( Z = \left( {a, b}\right) \in {\mathbb{C}}^{2} \), we have\n\n\[ \n{\left| Z\right| }^{2} = {\left| Z \cdot {e}_{1}\right... | Yes |
Lemma 3.4 If \( f \in \mathcal{S}\left( {\mathbb{R}}^{3}\right) \) and \( t \) is fixed, then \( {M}_{t}\left( f\right) \in \mathcal{S}\left( {\mathbb{R}}^{3}\right) \) . Moreover, \( {M}_{t}\left( f\right) \) is indefinitely differentiable in \( t \), and each \( t \) -derivative also belongs to \( \mathcal{S}\left( {... | Proof. Let \( F\left( x\right) = {M}_{t}\left( f\right) \left( x\right) \) . To show that \( F \) is rapidly decreasing, start with the inequality \( \left| {f\left( x\right) }\right| \leq {A}_{N}/\left( {1 + {\left| x\right| }^{N}}\right) \) which holds for every fixed \( N \geq 0 \) . As a simple consequence, wheneve... | Yes |
Lemma 3.5 \( \frac{1}{4\pi }{\int }_{{S}^{2}}{e}^{-{2\pi i\xi } \cdot \gamma }{d\sigma }\left( \gamma \right) = \frac{\sin \left( {{2\pi }\left| \xi \right| }\right) }{{2\pi }\left| \xi \right| } \) . | Proof. Note that the integral on the left is radial in \( \xi \) . Indeed, if \( R \) is a rotation then\n\n\[{\int }_{{S}^{2}}{e}^{-{2\pi iR}\left( \xi \right) \cdot \gamma }{d\sigma }\left( \gamma \right) = {\int }_{{S}^{2}}{e}^{-{2\pi i\xi } \cdot {R}^{-1}\left( \gamma \right) }{d\sigma }\left( \gamma \right) = {\in... | Yes |
Theorem 3.6 The solution when \( d = 3 \) of the Cauchy problem for the wave equation\n\n\[ \bigtriangleup u = \frac{{\partial }^{2}u}{\partial {t}^{2}}\;\text{ subject to }\;u\left( {x,0}\right) = f\left( x\right) \;\text{ and }\;\frac{\partial u}{\partial t}\left( {x,0}\right) = g\left( x\right) \]\n\nis given by\n\n... | Proof. Consider first the problem\n\n\[ \bigtriangleup u = \frac{{\partial }^{2}u}{\partial {t}^{2}}\;\text{ subject to }\;u\left( {x,0}\right) = 0\;\text{ and }\;\frac{\partial u}{\partial t}\left( {x,0}\right) = g\left( x\right) . \]\n\nThen by Theorem 3.1, we know that its solution \( {u}_{1} \) is given by\n\n\[ {u... | Yes |
Theorem 3.7 A solution of the Cauchy problem for the wave equation in two dimensions with initial data \( f, g \in \mathcal{S}\left( {\mathbb{R}}^{2}\right) \) is given by\n\n\[ u\left( {x, t}\right) = \frac{\partial }{\partial t}\left( {t{\widetilde{M}}_{t}\left( f\right) \left( x\right) }\right) + t{\widetilde{M}}_{t... | Formally, the identity in the theorem arises as follows. If we start with an initial pair of functions \( f \) and \( g \) in \( \mathcal{S}\left( {\mathbb{R}}^{2}\right) \), we may consider the corresponding functions \( \widetilde{f} \) and \( \widetilde{g} \) on \( {\mathbb{R}}^{3} \) that are merely extensions of \... | Yes |
Proposition 5.1 If \( f \in \mathcal{S}\left( {\mathbb{R}}^{3}\right) \), then for each \( \gamma \) the definition of \( {\int }_{{\mathcal{P}}_{t,\gamma }}f \) is independent of the choice of \( {e}_{1} \) and \( {e}_{2} \) . Moreover\n\n\[{\int }_{-\infty }^{\infty }\left( {{\int }_{{\mathcal{P}}_{t,\gamma }}f}\righ... | Proof. If \( {e}_{1}^{\prime },{e}_{2}^{\prime } \) is another choice of basis vectors so that \( \gamma ,{e}_{1}^{\prime },{e}_{2}^{\prime } \) is orthonormal, consider the rotation \( R \) in \( {\mathbb{R}}^{2} \) which takes \( {e}_{1} \) to \( {e}_{1}^{\prime } \) and \( {e}_{2} \) to \( {e}_{2}^{\prime } \) . Cha... | Yes |
Lemma 5.2 If \( f \in \mathcal{S}\left( {\mathbb{R}}^{3}\right) \), then \( \mathcal{R}\left( f\right) \left( {t,\gamma }\right) \in \mathcal{S}\left( \mathbb{R}\right) \) for each fixed \( \gamma \) . Moreover, \[ \widehat{\mathcal{R}}\left( f\right) \left( {s,\gamma }\right) = \widehat{f}\left( {s\gamma }\right) \] | Proof. Since \( f \in \mathcal{S}\left( {\mathbb{R}}^{3}\right) \), for every positive integer \( N \) there is a constant \( {A}_{N} < \infty \) so that \[ {\left( 1 + \left| t\right| \right) }^{N}{\left( 1 + \left| u\right| \right) }^{N}\left| {f\left( {{t\gamma } + u}\right) }\right| \leq {A}_{N} \] if we recall tha... | Yes |
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. | It is important to remark that this identity is an analytic expression of the fundamental theorem of arithmetic. In fact, each factor of the product \( 1/\left( {1 - {p}^{-s}}\right) \) can be written as a convergent geometric series\n\n\[ 1 + \frac{1}{{p}^{s}} + \frac{1}{{p}^{2s}} + \cdots + \frac{1}{{p}^{Ms}} + \cdot... | Yes |
Proposition 1.11 The series\n\n\[ \mathop{\sum }\limits_{p}1/p \]\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 |
Proposition 3.4 If \( \chi \) is a non-trivial Dirichlet character, then the series\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }\chi \left( n\right) /{n}^{s} \]\n\nconverges for \( s > 0 \), and we denote its sum by \( L\left( {s,\chi }\right) \) . Moreover:\n\n(i) The function \( L\left( {s,\chi }\right) \) is cont... | We first isolate the key cancellation property that non-trivial Dirichlet characters possess, which accounts for the behavior of the \( L \) -function described in the proposition. | No |
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 |
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