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Lemma 14.5.\n\n\\[ \n\\Gamma \\left( {x, y, z}\\right) = \\frac{{\\Delta }_{x + y + z}!{\\Delta }_{x - 1}!{\\Delta }_{y - 1}!{\\Delta }_{z - 1}!}{{\\Delta }_{y + z - 1}!{\\Delta }_{z + x - 1}!{\\Delta }_{x + y - 1}!}. \n\\] | Proof. Consider the equations depicted in Figure 14.4; as usual a symbol beside a line is a count of the number of parallel arcs that it represents. The first equality follows from the defining relation of Figure 13.6 for \\( {f}^{\\left( y + z - 1\\right) } \\) (together with \\( {f}^{\\left( z\\right) }{e}_{z - 1} = ... | Yes |
Lemma 14.7. Let \( \left( {a, b, c}\right) \) be admissible and let \( A \) be a primitive \( 4{r}^{\text{th }} \) root of unity. Then \( {\tau }_{a, b, c}^{ * } \) is non-zero if and only if \( a + b + c \leq 2\left( {r - 2}\right) \) . | Proof. \( \mathcal{S}{D}^{\prime } \) has a base consisting of all diagrams in \( {D}^{\prime } \) with no crossing. For all but one of these diagrams there is an arc from a point of one of the three specified subsets (for example, that with \( a \) points) to another point of the same subset. As usual (using \( {f}^{\... | Yes |
Lemma 14.9. Suppose \( A \) is not a root of unity. \( A \) base for \( {Q}_{a, b, c, d} \) is the set of elements as in Figure 14.10 (the boundary of the disc is not shown), where \( j \) takes all values such that \( \left( {a, b, j}\right) \) and \( \left( {c, d, j}\right) \) are both admissible. | Proof. Note that the proposed base elements each consist of two triads glued together; there is an \( {f}^{\left( j\right) } \) on the central line. Certainly \( {Q}_{a, b, c, d} \) is spanned by all elements of the form shown in Figure 14.11, where the lines all represent multiple parallel arcs, for, as usual, any oth... | Yes |
Lemma 14.10. Suppose \( A \) is a primitive \( 4{r}^{\text{th }} \) root of unity. A base for \( {Q}_{a, b, c, d}^{ * } \) (this being \( {Q}_{a, b, c, d} \) regarded as maps of diagrams outside the disc) is the set of elements as in Figure 14.10 where \( j \) takes all values such that \( \left( {a, b, j}\right) \) an... | Proof. The proof that the given elements span is the same as in Lemma 14.9 with a small modification. Now, \( {f}^{\left( n\right) } \) does not exist for \( n \geq r \) . However, \( {f}^{\left( r - 1\right) } \) is the zero map of outsides. Thus working in this dual context, any diagram as in the above proof, with at... | Yes |
Lemma 14.11. In \( \mathcal{S}\left( {{S}^{1} \times I}\right) ,{S}_{a}\left( \alpha \right) {S}_{b}\left( \alpha \right) = \mathop{\sum }\limits_{c}{S}_{c}\left( \alpha \right) \) where the summation is over all \( c \) such that \( \left( {a, b, c}\right) \) is admissible. If \( \bar{A} \) is a primitive \( 4{r}^{\te... | Proof. In fact the first part of this lemma is almost immediate. This is because it is a result on Chebyshev polynomials that \( {S}_{a}\left( x\right) {S}_{b}\left( x\right) = \mathop{\sum }\limits_{c}{S}_{c}\left( x\right) \), the sum being over all \( c \) such that \( \left( {a, b, c}\right) \) is admissible. This ... | Yes |
Theorem 14.12. Let \( {F}_{g} \) be the closed orientable surface of genus \( g \) and let \( A \) be a primitive \( 4{r}^{\text{th }} \) root of unity, \( r \geq 3 \) . Then \( {\mathcal{I}}_{A}\left( {{S}^{1} \times {F}_{g}}\right) \) is an integer. It is \( r - 1 \) when \( g = 1 \) . Otherwise it is the number of w... | Proof. The 3-manifold \( {S}^{1} \times {F}_{g} \) is obtained by surgery on a link that consists of \( g \) copies of the Borromean rings summed together on one component, each component having the zero framing. (Proving this is an interesting exercise.) A diagram \( D \) for such a link is obtained by taking \( g \) ... | No |
Theorem 14.13. If \( p \) and \( q \) are coprime positive integers, then the Jones polynomial of the \( \left( {p, q}\right) \) -torus knot is\n\n\[ \n{t}^{\left( {p - 1}\right) \left( {q - 1}\right) /2}{\left( 1 - {t}^{2}\right) }^{-1}\left( {1 - {t}^{p + 1} - {t}^{q + 1} + {t}^{p + q}}\right) .\n\] | Proof. Consider the diagram of Figure 14.23, which shows \( p \) arcs traversing a rectangle. Suppose \( q \) copies of this are placed side by side and the result is closed up by joining the \( p \) points on the left to those on the right, using \( p \) crossing-free arcs encircling an annulus \( {S}^{1} \times I \) ... | Yes |
Lemma 15.1. Suppose that \( p \) and \( q \) are two arcs in \( {\mathbb{R}}^{2} \) meeting only at their end points \( A \) and \( B \), and let \( R \) be the compact region bounded by \( p \cup q \) . Suppose that \( {t}_{1},{t}_{2},\ldots {t}_{n} \) are arcs in \( R \), each meeting \( p \cup q \) at just its end p... | Proof. Proceed by induction on the number \( n \) of arcs. The result is trivial if \( n = 1 \), so assume \( n > 1 \) . Amongst the end points of the \( {t}_{i} \) that lie on \( p \), let \( X \) be the nearest to \( A \) . If then \( X \) is an end of \( {t}_{j} \), let \( {B}^{\prime } \) be the other end of \( {t}... | Yes |
Theorem 15.5. There exists a function \[ \Lambda : \left\{ {\text{ Unoriented links diagrams in }{S}^{2}}\right\} \rightarrow \mathbb{Z}\left\lbrack {{a}^{\pm 1},{z}^{\pm 1}}\right\rbrack \] that is defined uniquely by the following: (i) \( \Lambda \left( U\right) = 1 \), where \( U \) is the zero-crossing diagram of t... | Proof. Note that, when considering a crossing in an unoriented diagram, it has no claim to be termed \( {D}_{ + } \) rather than \( {D}_{ - } \) in the above notation. However, this never matters, since \( {D}_{ + } \) and \( {D}_{ - } \) feature symmetrically in the formula \( \left( {\star \star }\right) \) ; the tre... | Yes |
Proposition 16.6. For an oriented link \( L \) , \[ V\left( L\right) = F\left( L\right) \text{ when }\left( {a, z}\right) = \left( {-{t}^{-3/4},\;\left( {{t}^{-1/4} + {t}^{1/4}}\right) }\right) , \] \[ {\left( V\left( L\right) \right) }^{2} = {\left( -1\right) }^{\# L - 1}F\left( L\right) \text{ when }t = - {q}^{-2},\;... | Proof. Underlying the Jones polynomial is the Kauffman bracket. Reverting to the notation for that, given in Definition 3.1, \[ \langle > < \rangle = A\langle > < \rangle + {A}^{-1}\langle < \rangle \] \[ \langle > < \rangle = {A}^{-1}\langle > < \rangle + A\langle > < \rangle . \] Adding these equations gives \[ \lang... | Yes |
Proposition 16.9. Suppose \( L \) is an oriented link with \( \# L \) components. Then \( (1 - \) \( \# L) \) is the lowest power both of \( m \) in \( P\left( L\right) \) and of \( z \) in \( F\left( L\right) \), and | \[ {\left\lbrack {z}^{\# L - 1}F\left( L\right) \right\rbrack }_{\left( {a, z}\right) = \left( {l,0}\right) } = {\left\lbrack {\left( -m\right) }^{\# L - 1}P\left( L\right) \right\rbrack }_{m = 0}. \] | No |
Theorem 16.11. If an oriented link \( L \) is the closure of \( \xi \in {B}_{n} \), let \( T\left( L\right) \) be defined to be \( T\left( \xi \right) \) . This is a well-defined link invariant. | Proof. Because \( T \) is essentially a trace function, if \( \xi ,\eta \in {B}_{n} \) then \( T\left( {{\eta }^{-1}{\xi \eta }}\right) = \n\n\( T\left( \xi \right) \) . Using the properties of \( \mu \), it is easy to show that \( T\left( {\xi {\sigma }_{n}}\right) = T\left( {\xi {\sigma }_{n}^{-1}}\right) = \n\n\( T\... | No |
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\[ \n{F}_{N}\left( x\right) = \frac{1}{N}\frac{{\sin }^{2}\left( {{Nx}/2}\right) }{{\sin }^{2}\left( {x/2}\right) }\n\]\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 |
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. | No |
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. | No |
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 |
Theorem 1.6 The collection \( {\left\{ {K}_{\delta }\right\} }_{\delta > 0} \) is a family of good kernels 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 |
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... | 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 |
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 |
Theorem 2.7 Suppose \( u \) is continuous on the closure of the upper half-plane \( \overline{{\mathbb{R}}_{ + }^{2}} \), satisfies \( \bigtriangleup u = 0 \) for \( \left( {x, y}\right) \in {\mathbb{R}}_{ + }^{2}, u\left( {x,0}\right) = 0 \), and \( u\left( {x, y}\right) \) vanishes at infinity. Then \( u = 0 \) . | To prove Theorem 2.7 we argue by contradiction. Considering separately the real and imaginary parts of \( u \), we may suppose that \( u \) itself is real-valued, and is somewhere strictly positive, say \( u\left( {{x}_{0},{y}_{0}}\right) > 0 \) for some \( {x}_{0} \in \mathbb{R} \) and \( {y}_{0} > 0 \) . We shall see... | Yes |
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 |
Theorem 3.3 The heat kernel on the circle is the periodization of the heat kernel on the real line: | \[ {H}_{t}\left( x\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{\mathcal{H}}_{t}\left( {x + n}\right) \] | 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 \( f \in \mathcal{S}\left( {\mathbb{R}}^{d}\right) \). (i) \( f\left( {x + h}\right) \rightarrow \widehat{f}\left( \xi \right) {e}^{{2\pi i\xi } \cdot h} \) whenever \( h \in {\mathbb{R}}^{d} \). (ii) \( f\left( x\right) {e}^{-{2\pi ixh}} \rightarrow \widehat{f}\left( {\xi + h}\right) \) whenever \(... | 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 \( b \) are complex numbers and \( \alpha \) is real. Then\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\nThis follows directly becaus... | Yes |
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\[ \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) = {... | 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 |
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\[ \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.... | Yes |
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 |
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 2.8 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\[ \n{ax} + {by} = d\text{.} \n\] | 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\[ \n{ax} + {by} = s.\n\]\n\nClearly, any ... | 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\\[ \n\\mathop{\\sum }\\limits_{p}1/p \n\\]\n\n diverges, 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\\[ \n- \\mathop{\\sum }\\limits_{p}\\log \\left( {1 - 1/{p}^{s}}\\right) = \\log \\zeta \\left( s... | 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 \... | Yes |
Theorem 2.3 If \( \chi \) is a nontrivial Dirichlet character, then the sum\n\n\[ \mathop{\sum }\limits_{p}\frac{\chi \left( p\right) }{{p}^{s}} \]\n\nremains bounded as \( s \rightarrow {1}^{ + } \) . | The proof of Theorem 2.3 requires the introduction of the \( L \) -functions, to which we now turn. | 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 |
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 |
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 |
Proposition 3.13 The following statements are true:\n\n(i) \( {S}_{N} \geq c\log N \) for some constant \( c > 0 \) .\n\n(ii) \( {S}_{N} = 2{N}^{1/2}L\left( {1,\chi }\right) + O\left( 1\right) \) . | It suffices to prove the proposition, since the assumption \( L\left( {1,\chi }\right) = 0 \) would give an immediate contradiction.\n\nWe first sum along hyperbolas. Observe that\n\n\[ \mathop{\sum }\limits_{{{nm} = k}}\frac{\chi \left( n\right) }{{\left( nm\right) }^{1/2}} = \frac{1}{{k}^{1/2}}\mathop{\sum }\limits_{... | 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 |
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