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Corollary 2. If \( f \) is continuous at \( x \), its Fourier series either diverges at \( x \) or converges to \( f\left( x\right) \) . | Proof. Only the case of convergence requires formal verification. If the sequence \( {S}_{n}\left( x\right) \) has a limit as \( n \rightarrow \infty \), then the sequence \( {\sigma }_{n}\left( x\right) = \frac{{S}_{0}\left( x\right) + \cdots + {S}_{n}\left( x\right) }{n + 1} \) has that same limit. But by Fejér’s the... | Yes |
Lemma 3. (Differentiation of a Fourier series). If a continuous function \( f \in C\left( {\left\lbrack {-\pi ,\pi }\right\rbrack ,\mathbb{C}}\right) \) assuming equal values at the endpoints of the closed interval \( \left\lbrack {-\pi ,\pi }\right\rbrack \) is piecewise continuously differentiable on \( \left\lbrack ... | Proof. Starting from the definition of the Fourier coefficients (18.44), we find through integration by parts that\n\n\[ \n{c}_{k}\left( {f}^{\prime }\right) = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{f}^{\prime }\left( x\right) {\mathrm{e}}^{-\mathrm{i}{kx}}\mathrm{\;d}x = {\left. \frac{1}{2\pi }f\left( x\right) {\mathrm... | Yes |
Proposition 1. (Connection between smoothness of a function and the rate of decrease of its Fourier coefficients). Let \( f \in {C}^{\left( m - 1\right) }\left( {\left\lbrack {-\pi ,\pi }\right\rbrack ,\mathbb{C}}\right) \) and \( {f}^{\left( j\right) }\left( {-\pi }\right) = {f}^{\left( j\right) }\left( \pi \right), j... | Proof. Relation (18.64) follows from an \( m \) -fold application of Eq. (18.63): \n\n\[ \n{c}_{k}\left( {f}^{\left( m\right) }\right) = \left( {\mathrm{i}k}\right) {c}_{k}\left( {f}^{\left( m - 1\right) }\right) = \cdots = {\left( \mathrm{i}k\right) }^{m}{c}_{k}\left( f\right) . \n\] \n\nSetting \( {\gamma }_{k} = \le... | Yes |
Theorem 5. If the function \( f : \left\lbrack {-\pi ,\pi }\right\rbrack \rightarrow \mathbb{C} \) is such that\n\na) \( f \in {C}^{\left( m - 1\right) }\left\lbrack {-\pi ,\pi }\right\rbrack, m \in \mathbb{N} \) ,\n\nb) \( {f}^{\left( j\right) }\left( {-\pi }\right) = {f}^{\left( j\right) }\left( \pi \right), j = 0,1,... | Proof. We write the partial sum (18.40) of the Fourier series in the compact notation \( \left( {18.40}^{\prime }\right) \) :\n\n\[ {S}_{n}\left( x\right) = \mathop{\sum }\limits_{{-n}}^{n}{c}_{k}\left( f\right) {\mathrm{e}}^{\mathrm{i}{kx}}. \]\n\nAccording to the assumptions on the function \( f \) and Proposition 1 ... | Yes |
Proposition 2. If the function \( f : \left\lbrack {-\pi ,\pi }\right\rbrack \rightarrow \mathbb{C} \) is piecewise continuous, then after integration the correspondence \( f\left( x\right) \sim \mathop{\sum }\limits_{{-\infty }}^{\infty }{c}_{k}\left( f\right) {\mathrm{e}}^{\mathrm{i}{kx}} \) becomes the equality | Proof. Consider the auxiliary function\n\n\[ F\left( x\right) = {\int }_{0}^{x}f\left( t\right) \mathrm{d}t - {c}_{0}\left( f\right) x \]\n\non the interval \( \left\lbrack {-\pi ,\pi }\right\rbrack \) . Obviously \( F \in C\left\lbrack {-\pi ,\pi }\right\rbrack \) . Also \( F\left( {-\pi }\right) = F\left( \pi \right)... | Yes |
Theorem 6. (Completeness of the trigonometric system). Every function \( f \in {\mathcal{R}}_{2}\left\lbrack {-\pi ,\pi }\right\rbrack \) can be approximated arbitrarily closely in mean\na) by functions of compact support in \( \rbrack - \pi ,\pi \lbrack \) that are Riemann integrable over the closed interval \( \left\... | Proof. Since it obviously suffices to prove the theorem for real-valued functions, we confine ourselves to this case.\na) It follows from the definition of the improper integral that\n\n\[{\int }_{-\pi }^{\pi }{f}^{2}\left( x\right) \mathrm{d}x = \mathop{\lim }\limits_{{\delta \rightarrow + 0}}{\int }_{-\pi + \delta }^... | Yes |
Proposition 3. (Uniqueness of Fourier series). Let \( f \) and \( g \) be two functions in \( {\mathcal{R}}_{2}\left\lbrack {-\pi ,\pi }\right\rbrack \) . Then\n\na) if the trigonometric series\n\n\[ \frac{{a}_{0}}{2} + \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}\cos {kx} + {b}_{k}\sin {kx}\;\left( { = \mathop{\su... | Proof. Assertion a) is actually a special case of the general fact that the expansion of a vector in an orthogonal system is unique. The inner product, as we know (see Lemma 1b) shows immediately that the coefficients of such an expansion are the Fourier coefficients and no others.\n\nAssertion b) can be obtained from ... | Yes |
Between the volume \( V \) of a domain in the Euclidean space \( {E}^{n} \) , \( n \geq 2 \), and the \( \left( {n - 1}\right) \) -dimensional surface area \( F \) of the hypersurface that bounds it, the following relation holds:\n\n\[ \n{n}^{n}{v}_{n}{V}^{n - 1} \leq {F}^{n} \n\]\n\ncalled the isoperimetric inequality... | The name \ | No |
Let us find the function having the following spectrum of compact support:\n\n\[ c\\left( \\alpha \\right) = \\left\\{ \\begin{array}{ll} h, & \\text{ if }\\left| \\alpha \\right| \\leq a, \\\\ 0, & \\text{ if }\\left| \\alpha \\right| > a. \\end{array}\\right. \] | By formula (18.82) we find, for \( t \\neq 0 \)\n\n\[ f\\left( t\\right) = {\\int }_{-a}^{a}h{\\mathrm{e}}^{\\mathrm{i}{\\alpha t}}\\mathrm{\\;d}\\alpha = h\\frac{{\\mathrm{e}}^{\\mathrm{i}{\\alpha t}} - {\\mathrm{e}}^{-\mathrm{i}{\\alpha t}}}{\\mathrm{i}t} = {2h}\\frac{\\sin {at}}{t}, \]\n\nand when \( t = 0 \), we ob... | Yes |
Let \( P \) be a device having the following properties: it is a linear signal transform, that is, \( P\left( {\mathop{\sum }\limits_{j}{a}_{j}{f}_{j}}\right) = \mathop{\sum }\limits_{j}{a}_{j}P\left( {f}_{j}\right) \), and it preserves the periodicity of a signal, that is, \( P\left( {e}^{\mathrm{i}{\omega t}}\right) ... | Representing the signal \( f\left( t\right) \) as the Fourier integral (18.82) and using the linearity of the device and the integral, we find\n\n\[ x\left( t\right) = P\left( f\right) \left( t\right) = {\int }_{-\infty }^{\infty }c\left( \omega \right) p\left( \omega \right) {\mathrm{e}}^{\mathrm{i}{\omega t}}\mathrm{... | Yes |
Let us find the Fourier transform of \( f\left( t\right) = \frac{\sin {at}}{t} \) (assuming \( f\left( 0\right) = a \in \mathbb{R}) \). | \[ \mathcal{F}\left\lbrack f\right\rbrack \left( \alpha \right) = \mathop{\lim }\limits_{{A \rightarrow + \infty }}\frac{1}{2\pi }{\int }_{-A}^{A}\frac{\sin {at}}{t}{\mathrm{e}}^{-\mathrm{i}{\alpha t}}\mathrm{\;d}t = \] \[ = \mathop{\lim }\limits_{{A \rightarrow + \infty }}\frac{1}{2\pi }{\int }_{-A}^{A}\frac{\sin {at}... | Yes |
Lemma 1. If the function \( f : \mathbb{R} \rightarrow \mathbb{C} \) is locally integrable and absolutely integrable on \( \mathbb{R} \), then\na) its Fourier transform \( \mathcal{F}\left\lbrack f\right\rbrack \left( \xi \right) \) is defined for every value \( \xi \in \mathbb{R} \) ;\nb) \( \mathcal{F}\left\lbrack f\... | Proof. We have already noted that \( \left| {f\left( x\right) {\mathrm{e}}^{\mathrm{i}{x\xi }}}\right| \leq \left| {f\left( x\right) }\right| \), from which it follows that the integral (18.86) converges absolutely and uniformly with respect to \( \xi \in \mathbb{R} \) . This fact simultaneously proves parts a) and c).... | Yes |
Let us find the Fourier transform of the function \( f\left( t\right) = {\mathrm{e}}^{-{t}^{2}/2} \) | \[ \mathcal{F}\left\lbrack f\right\rbrack \left( \alpha \right) = {\int }_{-\infty }^{+\infty }{\mathrm{e}}^{-{t}^{2}/2}{\mathrm{e}}^{-\mathrm{i}{\alpha t}}\mathrm{\;d}t = {\int }_{-\infty }^{+\infty }{\mathrm{e}}^{-{t}^{2}/2}\cos {\alpha t}\mathrm{\;d}t. \]\n\nDifferentiating this last integral with respect to the par... | Yes |
Corollary 1. Let \( f : \mathbb{R} \rightarrow \mathbb{C} \) be a continuous absolutely integrable function. If the function \( f \) is differentiable at each point \( x \in \mathbb{R} \) or has finite one-sided derivatives or satisfies a Hölder condition, then it is represented by its Fourier integral. | Hence for functions of these classes both equalities (18.80) and (18.82) or (18.98) and (18.99) hold, and we have thus proved the inversion formula for the Fourier transform for such functions. | Yes |
Assume that the signal \( v\left( t\right) = P\left( f\right) \left( t\right) \) emerging from the device \( P \) considered in Example 2 is known, and we wish to find the input signal \( f\left( t\right) \) entering the device \( P \). | In Example 2 we have shown that \( f \) and \( v \) are connected by the relation\n\n\[ v\left( t\right) = {\int }_{-\infty }^{\infty }c\left( \omega \right) p\left( \omega \right) {\mathrm{e}}^{\mathrm{i}{\omega t}}\mathrm{\;d}\omega \]\n\nwhere \( c\left( \omega \right) = \mathcal{F}\left\lbrack f\right\rbrack \left(... | Yes |
Let \( a > 0 \) and\n\n\[ f\left( x\right) = \left\{ \begin{matrix} {\mathrm{e}}^{-{ax}} & \text{ for }x > 0, \\ 0 & \text{ for }x \leq 0. \end{matrix}\right. \]\n\nThen\n\n\[ \mathcal{F}\left\lbrack f\right\rbrack \left( \xi \right) = \frac{1}{2\pi }{\int }_{0}^{+\infty }{\mathrm{e}}^{-{ax}}{\mathrm{e}}^{-\mathrm{i}{\... | \[ \mathcal{F}\left\lbrack f\right\rbrack \left( \xi \right) = \frac{1}{2\pi }{\int }_{0}^{+\infty }{\mathrm{e}}^{-{ax}}{\mathrm{e}}^{-\mathrm{i}{\xi x}}\mathrm{\;d}x = \frac{1}{2\pi }\frac{1}{a + \mathrm{i}\xi }. \] | Yes |
Lemma 2. Let \( f : \mathbb{R} \rightarrow \mathbb{C} \) be a continuous function having a locally piecewise continuous derivative \( {f}^{\prime } \) on \( \mathbb{R} \) . Given this,\n\na) if the function \( {f}^{\prime } \) is integrable on \( \mathbb{R} \), then \( f\left( x\right) \) has a limit both as \( x \righ... | Proof. Under these restrictions on the functions \( f \) and \( {f}^{\prime } \) the Newton-Leibniz formula holds\n\n\[ f\left( x\right) = f\left( 0\right) + {\int }_{0}^{x}{f}^{\prime }\left( t\right) \mathrm{d}t \]\n\nIn conditions a) the right-hand side of this equality has a limit both as \( x \rightarrow + \infty ... | Yes |
Proposition 1. (Connection between the smoothness of a function and the rate of decrease of its Fourier transform.) If \( f \in {C}^{\left( k\right) }\left( {\mathbb{R},\mathbb{C}}\right) \left( {k = 0,1,\ldots }\right) \) and all the functions \( f,{f}^{\prime },\ldots ,{f}^{\left( k\right) } \) are absolutely integra... | Proof. If \( k = 0 \), then a) holds trivially and b) follows from the Riemann-Lebesgue lemma.\n\nLet \( k > 0 \) . By Lemma 2 the functions \( f,{f}^{\prime },\ldots ,{f}^{\left( k - 1\right) } \) tend to zero as \( x \rightarrow \infty \) . Taking this into account, we integrate by parts,\n\n\[ \widehat{{f}^{\left( k... | Yes |
Proposition 2. (The connection between the rate of decrease of a function and the smoothness of its Fourier transform). If a locally integrable function \( f : \mathbb{R} \rightarrow \mathbb{C} \) is such that the function \( {x}^{k}f\left( x\right) \) is absolutely integrable on \( \mathbb{R} \), then\na) the Fourier ... | Proof. For \( k = 0 \) relation (18.103) holds trivially, and the continuity of \( \widehat{f}\left( \xi \right) \) has already been proved in Lemma 1. If \( k > 0 \), then for \( n < k \) we have the estimate \( \left| {{x}^{n}f\left( x\right) }\right| \leq \left| {{x}^{k}f\left( x\right) }\right| \) at infinity, from... | Yes |
Lemma 3. The restriction of the Fourier transform to \( S \) is a vector-space automorphism of \( S \) . | Proof. We first show that \( \left( {f \in S}\right) \Rightarrow \left( {\widehat{f} \in S}\right) \) .\n\nTo do this we first remark that by Proposition 2a we have \( \widehat{f} \in {C}^{\left( \infty \right) }\left( {\mathbb{R},\mathbb{C}}\right) . \)\n\nWe then remark that the operation of multiplication by \( {x}^... | Yes |
The function \( {\mathrm{e}}^{-{\left| x\right| }^{2}} \), where \( {\left| x\right| }^{2} = {x}_{1}^{2} + \cdots + {x}_{n}^{2} \), and all the functions in \( {C}_{0}^{\left( \infty \right) }\left( {{\mathbb{R}}^{n},\mathbb{C}}\right) \) of compact support belong to \( S \) . | If \( f \in S \), then integral in relation (18.104) obviously converges absolutely and uniformly with respect to \( \xi \) on the entire space \( {\mathbb{R}}^{n} \) . Moreover, if \( f \in S \) , then by standard rules this integral can be differentiated as many times as desired with respect to any of the variables \... | No |
Let us find the Fourier transform of the function \( \exp \left( {-{\left| x\right| }^{2}/2}\right) \) . | In the present case, using Fubini's theorem and Example 4, we find\n\n\[ \n\frac{1}{{\left( 2\pi \right) }^{n/2}}{\int }_{{\mathbb{R}}^{n}}{\mathrm{e}}^{-{\left| x\right| }^{2}/2} \cdot {\mathrm{e}}^{-\mathrm{i}\left( {\xi, x}\right) }\mathrm{d}x = \n\]\n\n\[ \n= \mathop{\prod }\limits_{{j = 1}}^{n}\frac{1}{\sqrt{2\pi ... | Yes |
The period\n\n\[ T = 4\sqrt{\frac{l}{g}}{\int }_{0}^{\pi /2}\frac{\mathrm{d}\theta }{\sqrt{1 - {k}^{2}{\sin }^{2}\theta }} \]\n\nof oscillations of a pendulum is connected with the maximal angle of deviation \( {\varphi }_{0} \) from its equilibrium position via the parameter \( {k}^{2} = {\sin }^{2}\frac{{\varphi }_{0... | If the oscillations are small, that is, \( {\varphi }_{0} \approx 0 \), we obtain the simple formula\n\n\[ T \approx {2\pi }\sqrt{\frac{l}{g}} \]\n\nfor the period of such oscillations. | Yes |
Suppose a restoring force acting on a particle \( m \) is returning it to its equilibrium position and that the force is proportional to the displacement (a spring with spring constant \( k \), for example). Suppose also that the resisting force of the medium is proportional to the square of the velocity (with coeffici... | If the medium \ | No |
If \( \pi \left( x\right) \) is the number of primes not larger than \( x \in \mathbb{R} \), then, as is known (see Sect. 3.2), for large \( x \) the quantity \( \pi \left( x\right) \) can be found with small relative error by the formula | \[ \pi \left( x\right) \approx \frac{x}{\ln x}. \] | No |
The labor involved in computing the values of \( n \) ! or \( \ln n \) ! increase as \( n \in \mathbb{N} \) increases. We shall use the fact that \( n \) is large, however, and obtain under that assumption a convenient asymptotic formula for computing \( \ln n \) ! approximately. | It follows from the obvious relations\n\n\[ \n{\int }_{1}^{n}\ln x\mathrm{\;d}x = \mathop{\sum }\limits_{{k = 2}}^{n}{\int }_{k - 1}^{k}\ln x\mathrm{\;d}x < \mathop{\sum }\limits_{{k = 1}}^{n}\ln k < \mathop{\sum }\limits_{{k = 2}}^{n}{\int }_{k}^{k + 1}\ln x\mathrm{\;d}x = {\int }_{2}^{n + 1}\ln x\mathrm{\;d}x \]\n\nt... | Yes |
We shall show that as \( x \rightarrow + \infty \) the function\n\n\[ \n{f}_{n}\left( x\right) = {\int }_{1}^{x}\frac{{\mathrm{e}}^{t}}{{t}^{n}}\mathrm{\;d}t\;\left( {n \in \mathbb{R}}\right) \n\]\n\nis asymptotically equivalent to the function \( {g}_{n}\left( x\right) = {x}^{-n}{\mathrm{e}}^{x} \) . | Since \( {g}_{n}\left( x\right) \rightarrow \) \( + \infty \) as \( x \rightarrow + \infty \), applying L’Hôpital’s rule we find\n\n\[ \n\mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{f}_{n}\left( x\right) }{{g}_{n}\left( x\right) } = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{f}_{n}^{\prime }\lef... | Yes |
Let us find the asymptotic behavior of the function\n\n\\[ f\\left( x\\right) = {\\int }_{1}^{x}\\frac{{\\mathrm{e}}^{t}}{t}\\mathrm{\\;d}t \\]\n\nmore precisely. It differs from the exponential integral\n\n\\[ \\operatorname{Ei}\\left( x\\right) = {\\int }_{-\\infty }^{x}\\frac{{\\mathrm{e}}^{t}}{t}\\mathrm{\\;d}t \\]... | Integrating by parts, we obtain\n\n\\[ f\\left( x\\right) = {\\left. \\frac{{\\mathrm{e}}^{t}}{t}\\right| }_{1}^{x} + {\\int }_{1}^{x}\\frac{{\\mathrm{e}}^{t}}{{t}^{2}}\\mathrm{\\;d}t = {\\left. \\left( \\frac{{\\mathrm{e}}^{t}}{t} + \\frac{{\\mathrm{e}}^{t}}{{t}^{2}}\\right) \\right| }_{1}^{x} + {\\int }_{1}^{x}\\frac... | Yes |
Proposition 3. (Integration of asymptotic equalities). Let \( f \) be a continuous function on the interval \( I = \lbrack a,\omega \left\lbrack \left( {\text{or}I = \rbrack \omega, a}\right) \right\rbrack \) . a) If the function \( g\left( x\right) \) is continuous and nonnegative on \( I \) and the integral \( {\int ... | Proof. a) If \( f\left( x\right) = O\left( {g\left( x\right) }\right) \) as \( I \ni x \rightarrow \omega \), there exists \( {x}_{0} \in I \) and a constant \( M \) such that \( \left| {f\left( x\right) }\right| \leq {Mg}\left( x\right) \) for \( x \in \left\lbrack {{x}_{0},\omega \lbrack }\right. \) . It follows that... | Yes |
The function \( f\left( x\right) = {\mathrm{e}}^{-x}\sin \left( {\mathrm{e}}^{x}\right) \) is continuously differentiable on \( \mathbb{R} \) and is an asymptotic zero with respect to the asymptotic sequence \( \left\{ \frac{1}{{x}^{n}}\right\} \) as \( x \rightarrow + \infty \) . | The derivatives of the functions \( \frac{1}{{x}^{n}} \), up to a constant factor, again have the form \( \frac{1}{{x}^{k}} \) . However the function \( \widetilde{{f}^{\prime }}\left( x\right) = - {\mathrm{e}}^{x}\sin \left( {\mathrm{e}}^{x}\right) + \cos \left( {\mathrm{e}}^{x}\right) \) not only fails to be an asymp... | Yes |
Proposition 4. Let 0 be a limit point of \( E \) and let\n\n\[ f\left( x\right) \simeq {a}_{0} + {a}_{1}x + {a}_{2}{x}^{2} + \cdots ,\]\n\n\[ \text{as}E \ni x \rightarrow 0\text{.}\]\n\n\[ g\left( x\right) \simeq {b}_{0} + {b}_{1}x + {b}_{2}{x}^{2} + \cdots \]\n\nThen as \( E \ni x \rightarrow 0 \) ,\n\na) \( \left( {{... | Proof. a) This is a special case of Proposition 2.\n\nb) Using the properties of \( o\left( \text{ ) (see Proposition 4 of Sect. 3.2), we find }\right) \) that\n\n\[ \left( {f \cdot g}\right) \left( x\right) = f\left( x\right) \cdot g\left( x\right) =\n\n\[ = \left( {{a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n} + o\lef... | Yes |
If \( U \) is a neighborhood (or one-sided neighborhood) of infinity in \( \mathbb{R} \) and the function \( f \) is continuous in \( U \) and has the asymptotic expansion\n\n\[ f\left( x\right) \simeq {a}_{0} + \frac{{a}_{1}}{x} + \frac{{a}_{2}}{{x}^{2}} + \cdots + \frac{{a}_{n}}{{x}^{n}} + \cdots \text{ as }U \ni x \... | The convergence of the integral is obvious, since\n\n\[ f\left( t\right) - {a}_{0} - \frac{{a}_{1}}{t} \sim \frac{{a}_{2}}{{t}^{2}}\text{ as }U \ni t \rightarrow \infty . \]\n\nIt remains only to integrate the asymptotic expansion\n\n\[ f\left( t\right) - {a}_{0} - \frac{{a}_{1}}{t} \simeq \frac{{a}_{2}}{{t}^{2}} + \fr... | Yes |
Corollary 2. If in addition to the hypotheses of Corollary 1 it is known that \( f \in {C}^{\left( 1\right) }\left( U\right) \) and \( {f}^{\prime } \) admits the asymptotic expansion\n\n\[ \n{f}^{\prime }\left( x\right) \simeq {a}_{0}^{\prime } + \frac{{a}_{1}^{\prime }}{x} + \frac{{a}_{2}^{\prime }}{{x}^{2}} + \cdots... | Proof. Since \( {f}^{\prime }\left( x\right) = {a}_{0}^{\prime } + \frac{{a}_{1}^{\prime }}{x} + O\left( {1/{x}^{2}}\right) \) as \( U \ni x \rightarrow \infty \), we have\n\n\[ \nf\left( x\right) = f\left( {x}_{0}\right) + {\int }_{{x}_{0}}^{x}{f}^{\prime }\left( t\right) \mathrm{d}t = {a}_{0}^{\prime }x + {a}_{1}^{\p... | Yes |
Example 1. The Laplace transform\n\n\[ L\\left( f\\right) \\left( \\xi \\right) = {\\int }_{0}^{+\\infty }f\\left( x\\right) {\\mathrm{e}}^{-{\\xi x}}\\mathrm{\\;d}x \] | is a special case of a Laplace integral. | No |
Example 2. Laplace himself applied his method to integrals of the form \( \mathop{\int }\limits_{a}^{b}f\left( x\right) {\varphi }^{n}\left( x\right) \;\mathrm{d}x \), where \( n \in \mathbb{N} \) and \( \varphi \left( x\right) > 0 \) on \( \rbrack a, b\lbrack \) . | Such an integral is also a special case of a general Laplace integral (19.1), since \( {\varphi }^{n}\left( x\right) = \exp \left( {n\ln \varphi \left( x\right) }\right) \). We shall be interested in the asymptotics of the integral (19.1) for large values of the parameter \( \lambda \), more precisely as \( \lambda \ri... | "No" |
Example 3. Let \( {x}_{0} = a,{S}^{\prime }\left( a\right) \neq 0 \), and \( f\left( a\right) \neq 0 \), which happens, for example, when the function \( S\left( x\right) \) is monotonically decreasing on \( \left\lbrack {a, b}\right\rbrack \) . Under these conditions \( f\left( x\right) = f\left( a\right) + o\left( 1\... | Carrying out the idea of Laplace’s method, for a small \( \varepsilon > 0 \) and \( \lambda \rightarrow + \infty \), we find that\n\n\[ F\left( \lambda \right) \sim {\int }_{a}^{a + \varepsilon }f\left( x\right) {\mathrm{e}}^{{\lambda S}\left( x\right) }\mathrm{d}x \sim f\left( a\right) {\mathrm{e}}^{{\lambda S}\left( ... | Yes |
Let \( a < {x}_{0} < b \) . Then \( {S}^{\prime }\left( {x}_{0}\right) = 0 \), and we assume that \( {S}^{\prime \prime }\left( {x}_{0}\right) \neq \) 0, that is, \( {S}^{\prime \prime }\left( {x}_{0}\right) < 0 \), since \( {x}_{0} \) is a maximum. | Using the expansions \( f\left( x\right) = f\left( {x}_{0}\right) + o\left( {x - {x}_{0}}\right) \) and \( S\left( x\right) = S\left( {x}_{0}\right) + \) \( \frac{1}{2}{S}^{\prime \prime }\left( {x}_{0}\right) {\left( x - {x}_{0}\right) }^{2} + o\left( {\left( x - {x}_{0}\right) }^{2}\right) \), which hold as \( x \rig... | Yes |
If \( {x}_{0} = a \), but \( {S}^{\prime }\left( {x}_{0}\right) = 0 \) and \( {S}^{\prime \prime }\left( {x}_{0}\right) < 0 \), then, reasoning as in Example 4, we find this time that | \[ F\left( \lambda \right) \sim {\int }_{a}^{a + \varepsilon }f\left( x\right) {\mathrm{e}}^{{\lambda S}\left( x\right) }\mathrm{d}x \sim f\left( {x}_{0}\right) {\mathrm{e}}^{{\lambda S}\left( {x}_{0}\right) }{\int }_{0}^{\varepsilon }{\mathrm{e}}^{\frac{1}{2}\lambda {S}^{\prime \prime }\left( {x}_{0}\right) {t}^{2}}\m... | Yes |
Lemma 1. (Exponential estimate). Let \( M = \mathop{\sup }\limits_{{a < x < b}}S\left( x\right) < \infty \), and suppose that for some value \( {\lambda }_{0} > 0 \) the integral (19.1) converges absolutely. Then it converges absolutely for every \( \lambda \geq {\lambda }_{0} \) and the following estimate holds for su... | Proof. Indeed, for \( \lambda \geq {\lambda }_{0} \) , \n\n\[ \left| {F\left( \lambda \right) }\right| = \left| {{\int }_{a}^{b}f\left( x\right) {\mathrm{e}}^{{\lambda S}\left( x\right) }\mathrm{d}x}\right| = \left| {{\int }_{a}^{b}f\left( x\right) {\mathrm{e}}^{{\lambda }_{0}S\left( x\right) }{\mathrm{e}}^{\left( {\la... | Yes |
Lemma 2. (Estimate of the contribution of a maximum point). Suppose the integral (19.1) converges absolutely for some value \( \lambda = {\lambda }_{0} \), and suppose that in the interior or on the boundary of the interval \( I \) there is a point \( {x}_{0} \) at which \( S\left( {x}_{0}\right) = \mathop{\sup }\limit... | Proof. For a fixed \( \varepsilon > 0 \) let us take any neighborhood \( {U}_{I}\left( {x}_{0}\right) \) inside which \( \left| {f\left( x\right) }\right| \geq \frac{1}{2}\left| {f\left( {x}_{0}\right) }\right| \) and \( S\left( {x}_{0}\right) - \varepsilon \leq S\left( x\right) \leq \bar{S}\left( {x}_{0}\right) \) . A... | Yes |
Proposition 1. (Localization principle). Suppose the integral (19.1) converges absolutely for a value \( \lambda = {\lambda }_{0} \), and suppose that inside or on the boundary of the interval \( I \) of integration the function \( S\left( x\right) \) has a unique point \( {x}_{0} \) of absolute maximum, that is, outsi... | Proof. It follows from Lemma 2 that if the neighborhood \( {U}_{I}\left( {x}_{0}\right) \) is sufficiently small, then the following inequality holds ultimately as \( \lambda \rightarrow + \infty \) for every \( \varepsilon > 0 \)\n\n\[ \left| {{F}_{{U}_{I}\left( {x}_{0}\right) }\left( \lambda \right) }\right| > {\math... | Yes |
Proposition 2. (Reduction). Suppose the interval of integration \( I = \left\lbrack {a, b}\right\rbrack \) in the integral (19.1) is finite and the following conditions hold:\n\na) \( f, S \in C\left( {I,\mathbb{R}}\right) \) ;\n\nb) \( \mathop{\max }\limits_{{x \in I}}S\left( x\right) \) is attained only at the one po... | Proof. Using the localization principle, we replace the integral (19.1) with the integral over a neighborhood \( {I}_{x} = {U}_{I}\left( {x}_{0}\right) \) of \( {x}_{0} \) in which the hypotheses of Lemma 3 hold. Making the change of variable \( x = \varphi \left( y\right) \), we obtain\n\n\[ {\int }_{{I}_{x}}f\left( x... | Yes |
Lemma 4. \( \left( {\text{Watson}}^{3}\right) \) . Let \( \alpha > 0,\beta > 0,0 < a \leq \infty \), and \( f \in C\left( {\left\lbrack {0, a}\right\rbrack ,\mathbb{R}}\right) \) . Then with respect to the asymptotics of the integral\n\n\[ W\left( \lambda \right) = {\int }_{0}^{a}{x}^{\beta - 1}f\left( x\right) {\mathr... | Proof. We represent the integral (19.12) as a sum of integrals over the interval \( \rbrack 0,\varepsilon \rbrack \) and \( \lbrack \varepsilon, a\lbrack \), where \( \varepsilon \) is an arbitrarily small positive number.\n\nBy Lemma 1\n\n\[ \left| {{\int }_{\varepsilon }^{a}{x}^{\beta - 1}f\left( x\right) {\mathrm{e}... | Yes |
Consider the Laplace transform\n\n\[ F\\left( \\lambda \\right) = {\\int }_{0}^{+\\infty }f\\left( x\\right) {\\mathrm{e}}^{-{\\lambda x}}\\mathrm{\\;d}x \]\n\nwhich we have already encountered in Example 1. If this integral converges absolutely for some value \( \\lambda = {\\lambda }_{0} \) and the function \( f \) i... | \[ F\\left( \\lambda \\right) \\simeq \\mathop{\\sum }\\limits_{{k = 0}}^{\\infty }{f}^{\\left( k\\right) }\\left( 0\\right) {\\lambda }^{-\\left( {k + 1}\\right) }\\text{ as }\\lambda \\rightarrow + \\infty . \] | Yes |
Theorem 1. (A typical principal term of the asymptotics). Suppose the interval of integration \( I = \left\lbrack {a, b}\right\rbrack \) in the integral (19.1) is finite, \( f, S \in C\left( {I,\mathbb{R}}\right) \) , and \( \mathop{\max }\limits_{{x \in I}}S\left( x\right) \) is attained only at one point \( {x}_{0} \... | Proof. Using the localization principle and making the change of variable \( x = \varphi \left( y\right) \) shown in Lemma 3, according to the reduction in Proposition 2, we arrive at the following relations:\n\n\[ \text{a)}F\left( \lambda \right) = {\mathrm{e}}^{{\lambda S}\left( {x}_{0}\right) }\left( {{\int }_{0}^{\... | Yes |
The asymptotics of the gamma function. The function\n\n\[ \n\Gamma \left( {\lambda + 1}\right) = {\int }_{0}^{+\infty }{t}^{\lambda }{\mathrm{e}}^{-t}\mathrm{\;d}t\;\left( {\lambda > - 1}\right) \n\] | can be represented as a Laplace integral\n\n\[ \n\Gamma \left( {\lambda + 1}\right) = {\int }_{0}^{+\infty }{\mathrm{e}}^{-t}{\mathrm{e}}^{\lambda \ln t}\mathrm{\;d}t \n\]\n\nand if for \( \lambda > 0 \) we make the change of variable \( t = {\lambda x} \), we arrive at the integral\n\n\[ \n\Gamma \left( {\lambda + 1}\... | Yes |
The asymptotics of the Bessel function\n\n\[ \n{I}_{n}\left( x\right) = \frac{1}{\pi }{\int }_{0}^{\pi }{\mathrm{e}}^{x\cos \theta }\cos {n\theta }\mathrm{d}\theta \n\]\n\nwhere \( n \in \mathbb{N} \) . | Here \( f\left( \theta \right) = \cos {n\theta }, S\left( \theta \right) = \cos \theta ,\mathop{\max }\limits_{{0 \leq x \leq \pi }}S\left( \theta \right) = S\left( 0\right) = 1 \) , \( {S}^{\prime }\left( 0\right) = 0 \), and \( {S}^{\prime \prime }\left( 0\right) = - 1 \), so that by assertion c) of Theorem 1\n\n\[ \... | Yes |
Let \( f \in {C}^{\left( 1\right) }\left( {\left\lbrack {a, b}\right\rbrack ,\mathbb{R}}\right), S \in {C}^{\left( 2\right) }\left( {\left\lbrack {a, b}\right\rbrack ,\mathbb{R}}\right) \), with \( S\left( x\right) > 0 \) on \( \left\lbrack {a, b}\right\rbrack \), and \( \mathop{\max }\limits_{{a \leq x \leq b}}S\left(... | \[ \mathcal{F}\left( \lambda \right) = {\varepsilon f}\left( {x}_{0}\right) \sqrt{\frac{2\pi }{-{S}^{\prime \prime }\left( {x}_{0}\right) }}{\left\lbrack S\left( {x}_{0}\right) \right\rbrack }^{\lambda + 1/2}{\lambda }^{-1/2}\left\lbrack {1 + O\left( {\lambda }^{-1/2}\right) }\right\rbrack ,\] where \( \varepsilon = 1 ... | Yes |
The asymptotics of the Legendre polynomials\n\n\[ \n{P}_{n}\left( x\right) = \frac{1}{\pi }{\int }_{0}^{\pi }{\left( x + \sqrt{{x}^{2} - 1}\cos \theta \right) }^{n}\mathrm{\;d}\theta \]\n\nin the domain \( x > 1 \) as \( n \rightarrow \infty, n \in \mathbb{N} \), can be obtained as a special case of the preceding examp... | \n\[ \nS\left( \theta \right) = x + \sqrt{{x}^{2} - 1}\cos \theta ,\;\mathop{\max }\limits_{{0 \leq \theta \leq \pi }}S\left( \theta \right) = S\left( 0\right) = x + \sqrt{{x}^{2} - 1}, \]\n\n\[ \n{S}^{\prime }\left( 0\right) = 0,\;{S}^{\prime \prime }\left( 0\right) = - \sqrt{{x}^{2} - 1}. \]\n\nThus,\n\n\[ \n{P}_{n}\... | Yes |
Theorem 2. (Asymptotic expansion). Let \( I = \left\lbrack {a, b}\right\rbrack \) be a finite interval, \( f, S \in C\left( {I,\mathbb{R}}\right) \), and assume \( \mathop{\max }\limits_{{x \in I}}S\left( x\right) \) is attained only at the point \( {x}_{0} \in I \) and \( f, S \in {C}^{\left( \infty \right) }\left( {{... | Proof. It follows from Lemma 1 that under these hypotheses the integral (19.1) can be replaced by an integral over an arbitrarily small neighborhood of \( {x}_{0} \) up to a quantity of the form \( {\mathrm{e}}^{{\lambda S}\left( {x}_{0}\right) }O\left( {\lambda }^{-\infty }\right) \) as \( \lambda \rightarrow \infty \... | Yes |
The asymptotic behavior of the function\n\n\[ \operatorname{Erf}\left( x\right) = {\int }_{x}^{+\infty }{\mathrm{e}}^{-{u}^{2}}\mathrm{\;d}u \]\n\nas \( x \rightarrow + \infty \) | is easy to obtain through integration by parts:\n\n\[ \operatorname{Erf}\left( x\right) = \frac{{\mathrm{e}}^{-{x}^{2}}}{2x} - \frac{1}{2}{\int }_{x}^{+\infty }{u}^{-2}{\mathrm{e}}^{-{u}^{2}}\mathrm{\;d}u = \frac{{e}^{-{x}^{2}}}{2x} - \frac{3{\mathrm{e}}^{-{x}^{2}}}{{2}^{2}{x}^{3}} + {\int }_{x}^{+\infty }{u}^{-4}{\mat... | 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\[ \n{F}_{N}\left( x\right) = \frac{1}{N}\frac{{\sin }^{2}\left( {{Nx}/2}\right) }{{\sin }^{2}\left( {x/2}\right) } \n\]\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\[\nF\left( t\right) = \left\{ \begin{array}{ll} \frac{f\left( {{\theta }_{0} - t}\right) - f\left( {\theta }_{0}\right) }{t} & \text{ if }t \neq 0\text{ and }\left| t\right| < \pi \\ - {f}^{\prime }\left( {\theta }_{0}\right) & \text{ if }t = 0. \end{array}\right.\n\]\n\nFirst, \( F \) is bounded near... | 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 |
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. | 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 |
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 |
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\[ \n\mathop{\sup }\limits_{\substack{{x \in \mathbb{R}} \\ {0 < t < T} }}{\left| x\right| }^{k}\left| {\frac{{\partial }^{\ell }}{\partial {x}^{\ell }... | Proof. This result is a consequence of the following estimate:\n\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}\rig... | 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 \( f \in \mathcal{S}\left( {\mathbb{R}}^{d}\right) \). (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 |
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 |
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