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The integral\n\n\[ \n{\int }_{1}^{+\infty }\frac{\mathrm{d}x}{{x}^{2} + {y}^{2}} \n\]\n\nconverges uniformly on the entire set \( \mathbb{R} \) of values of the parameter \( y \in \mathbb{R} \) | since for every \( y \in \mathbb{R} \)\n\n\[ \n{\int }_{b}^{+\infty }\frac{\mathrm{d}x}{{x}^{2} + {y}^{2}} \leq {\int }_{b}^{+\infty }\frac{\mathrm{d}x}{{x}^{2}} = \frac{1}{b} < \varepsilon \n\]\n\nprovided \( b > 1/\varepsilon \) . | Yes |
The integral \[ {\int }_{0}^{+\infty }{\mathrm{e}}^{-{xy}}\mathrm{\;d}x \] obviously converges only when \( y > 0 \) . Moreover it converges uniformly on every set \( \left\{ {y \in \mathbb{R} \mid y \geq {y}_{0} > 0}\right\} \) . | Indeed, if \( y \geq {y}_{0} > 0 \), then \[ 0 \leq {\int }_{b}^{+\infty }{\mathrm{e}}^{-{xy}}\mathrm{\;d}x = \frac{1}{y}{\mathrm{e}}^{-{by}} \leq \frac{1}{{y}_{0}}{\mathrm{e}}^{-b{y}_{0}} \rightarrow 0\;\text{ as }b \rightarrow + \infty . \] | Yes |
Let us show that each of the integrals\n\n\[ \Phi \left( x\right) = {\int }_{0}^{+\infty }{x}^{\alpha }{y}^{\alpha + \beta + 1}{\mathrm{e}}^{-\left( {1 + x}\right) y}\mathrm{\;d}y, \]\n\n\[ F\left( y\right) = {\int }_{0}^{+\infty }{x}^{\alpha }{y}^{\alpha + \beta + 1}{\mathrm{e}}^{-\left( {1 + x}\right) y}\mathrm{\;d}x... | For the remainder of the integral \( \Phi \left( x\right) \) we find immediately that\n\n\[ 0 \leq {\int }_{b}^{+\infty }{x}^{\alpha }{y}^{\alpha + \beta + 1}{\mathrm{e}}^{-\left( {1 + x}\right) y}\mathrm{\;d}y = \]\n\n\[ = {\int }_{b}^{+\infty }{\left( xy\right) }^{\alpha }{\mathrm{e}}^{-{xy}}{y}^{\beta + 1}{\mathrm{e... | Yes |
Proposition 1 (Cauchy criterion) A necessary and sufficient condition for the improper integral (17.10) depending on the parameter \( y \in Y \) to converge uniformly on a set \( E \subset Y \) is that for every \( \varepsilon > 0 \) there exist a neighborhood \( {U}_{\lbrack a,\omega \lbrack }\left( \omega \right) \) ... | Proof Inequality (17.15) is equivalent to the relation \( \left| {{F}_{{b}_{2}}\left( y\right) - {F}_{{b}_{2}}\left( y\right) }\right| < \varepsilon \), so that Proposition 1 is an immediate corollary of the form (17.13) for the definition of uniform convergence of the integral (17.10) and the Cauchy criterion for unif... | Yes |
Corollary 1 If the function fin the integral (17.10) is continuous on the set \( \lbrack a,\omega \lbrack \times \) \( \left\lbrack {c, d}\right\rbrack \) and the integral (17.10) converges for every \( y \in \rbrack c, d\lbrack \) but diverges for \( y = c \) or \( y = d \), then it converges nonuniformly on the inter... | Proof If the integral (17.10) diverges at \( y = c \), then by the Cauchy criterion for convergence of an improper integral there exists \( {\varepsilon }_{0} > 0 \) such that in every neighborhood \( {U}_{\lbrack a,\omega \lbrack }\left( \omega \right) \) there exist numbers \( {b}_{1},{b}_{2} \) for which\n\n\[ \left... | Yes |
The integral\n\n\[ \n{\int }_{0}^{+\infty }{\mathrm{e}}^{-t{x}^{2}}\mathrm{\;d}x \n\]\n\nconverges for \( t > 0 \) and diverges at \( t = 0 \), hence it demonstrably converges nonuniformly on every set of positive numbers having 0 as a limit point. In particular, it converges nonuniformly on the whole set \( \{ t \in \... | In this case, one can easily verify these statements directly:\n\n\[ \n{\int }_{b}^{+\infty }{\mathrm{e}}^{-t{x}^{2}}\mathrm{\;d}x = \frac{1}{\sqrt{t}}{\int }_{b\sqrt{t}}^{+\infty }{\mathrm{e}}^{-{u}^{2}}\mathrm{\;d}u \rightarrow + \infty \;\text{ as }t \rightarrow + 0. \n\] | Yes |
Proposition 2 (The Weierstrass test) Suppose the functions \( f\left( {x, y}\right) \) and \( g\left( {x, y}\right) \) are integrable with respect to \( x \) on every closed interval \( \left\lbrack {a, b}\right\rbrack \subset \lbrack a,\omega \lbrack \) for each value of \( y \in Y \) .\n\nIf the inequality \( \left| ... | Proof This follows from the estimates\n\n\[ \left| {{\int }_{{b}_{1}}^{{b}_{2}}f\left( {x, y}\right) \mathrm{d}x}\right| \leq {\int }_{{b}_{1}}^{{b}_{2}}\left| {f\left( {x, y}\right) }\right| \mathrm{d}x \leq {\int }_{{b}_{1}}^{{b}_{2}}g\left( {x, y}\right) \mathrm{d}x \]\n\nand Cauchy's criterion for uniform convergen... | Yes |
Example 6 In view of the inequality \( \left| {\sin x{\mathrm{e}}^{-t{x}^{2}}}\right| \leq {\mathrm{e}}^{-t{x}^{2}} \), the integral \[ {\int }_{0}^{\infty }\sin x{\mathrm{e}}^{-t{x}^{2}}\mathrm{\;d}x \] | as follows from Proposition 2 and the results of Example 3, converges uniformly on every set of the form \( \left\{ {t \in \mathbb{R} \mid t \geq {t}_{0} > 0}\right\} \) . Since the integral diverges for \( t = 0 \), on the basis of the Cauchy criterion we conclude that it cannot converge uniformly on any set having ze... | Yes |
Proposition 3 (Abel-Dirichlet test) Assume that the functions \( f\left( {x, y}\right) \) and \( g\left( {x, y}\right) \) are integrable with respect to \( x \) at each \( y \in Y \) on every closed interval \( \left\lbrack {a, b}\right\rbrack \subset \) \( \lbrack a,\omega \lbrack \) . A sufficient condition for unifo... | Proof Applying the second mean-value theorem for the integral, we write \[ {\int }_{{b}_{1}}^{{b}_{2}}\left( {f \cdot g}\right) \left( {x, y}\right) \mathrm{d}x = g\left( {{b}_{1}, y}\right) {\int }_{{b}_{1}}^{\xi }f\left( {x, y}\right) \mathrm{d}x + g\left( {{b}_{2}, y}\right) {\int }_{\xi }^{{b}_{2}}f\left( {x, y}\ri... | Yes |
The integral \[ {\int }_{1}^{+\infty }\frac{\sin x}{{x}^{\alpha }}\mathrm{d}x \] | as follows from the Cauchy criterion and the Abel-Dirichlet test for convergence of improper integrals, converges only for \( \alpha > 0 \) . Setting \( f\left( {x,\alpha }\right) = \sin x, g\left( {x,\alpha }\right) = \) \( {x}^{-\alpha } \), we see that the pair \( \left. {\alpha }_{1}\right) ,{\beta }_{1} \) ) of hy... | Yes |
The integral \[ {\int }_{0}^{\infty }\frac{\sin x}{x}{\mathrm{e}}^{-{xy}}\mathrm{\;d}x \] converges uniformly on the set \( \{ y \in \mathbb{R} \mid y \geq 0\} \) . | Proof First of all, on the basis of the Cauchy criterion for convergence of the improper integral one can easily conclude that for \( y < 0 \) this integral diverges. Now assuming \( y \geq 0 \) and setting \( f\left( {x, y}\right) = \frac{\sin x}{x}, g\left( {x, y}\right) = {\mathrm{e}}^{-{xy}} \), we see that the sec... | Yes |
Proposition 4 Let \( f\left( {x, y}\right) \) be a family of functions depending on a parameter \( y \in Y \) that are integrable, possibly in the improper sense, on the interval \( a \leq x < \omega \), and let \( {\mathcal{B}}_{Y} \) be a base in \( Y \) . If a) for every \( b \in \lbrack a,\omega \lbrack \) \( f\lef... | Proof The proof reduces to checking the following diagram:  The left vertical limiting passage follows from hypothesis a) and the theorem on passage to the limit under a proper integral sign (see Theorem 3 of Sect. 1... | Yes |
Example 9 Let \( Y = \{ y \in \mathbb{R} \mid y > 0\} \) and\n\n\[ f\left( {x, y}\right) = \left\{ \begin{array}{ll} 1/y, & \text{ if }0 \leq x \leq y \\ 0, & \text{ if }y < x \end{array}\right.\]\n\nObviously, \( f\left( {x, y}\right) \rightrightarrows 0 \) on the interval \( 0 \leq x < + \infty \) as \( y \rightarrow... | \[ {\int }_{0}^{+\infty }f\left( {x, y}\right) \mathrm{d}x = {\int }_{0}^{y}f\left( {x, y}\right) \mathrm{d}x = {\int }_{0}^{y}\frac{1}{y}\mathrm{\;d}x = 1,\]\n\nand therefore Eq. (17.17) does not hold in this case. | Yes |
Corollary 2 Suppose that the real-valued function \( f\left( {x, y}\right) \) is nonnegative at each value of the real parameter \( y \in Y \subset \mathbb{R} \) and continuous on the interval \( a \leq x < \omega \) . If a) the function \( f\left( {x, y}\right) \) is monotonically increasing as \( y \) increases and t... | Proof It follows from Dini’s theorem that \( f\left( {x, y}\right) \rightrightarrows \varphi \left( x\right) \) on each closed interval \( \left\lbrack {a, b}\right\rbrack \subset \lbrack a,\omega \lbrack \) . It follows from the inequalities \( 0 \leq f\left( {x, y}\right) \leq \varphi \left( x\right) \) and the Weier... | Yes |
In Example 3 of Sect. 16.3 we verified that the sequence of functions \( {f}_{n}\left( x\right) = n\left( {1 - {x}^{1/n}}\right) \) is monotonically increasing on the interval \( 0 < x \leq 1 \), and \( {f}_{n}\left( x\right) \nearrow \ln \frac{1}{x} \) as \( n \rightarrow + \infty \) . | Hence, by Corollary 2\n\n\[\n\mathop{\lim }\limits_{{n \rightarrow \infty }}{\int }_{0}^{1}n\left( {1 - {x}^{1/n}}\right) \mathrm{d}x = {\int }_{0}^{1}\ln \frac{1}{x}\mathrm{\;d}x.\n\] | No |
For a fixed value \( \alpha > 0 \) the integral\n\n\[ \n{\int }_{0}^{+\infty }{x}^{\alpha }{\mathrm{e}}^{-{xy}}\mathrm{\;d}x \n\]\n\nconverges uniformly with respect to the parameter \( y \) on every interval of the form \( \left\{ {y \in \mathbb{R} \mid y \geq {y}_{0} > 0}\right\} \) . | This follows from the estimate \( 0 \leq {x}^{\alpha }{\mathrm{e}}^{-{xy}} < {x}^{\alpha }{\mathrm{e}}^{-x{y}_{0}} < \) \( {\mathrm{e}}^{-x\frac{{y}_{0}}{2}} \), which holds for all sufficiently large \( x \in \mathbb{R} \) . | Yes |
Example 13 Let us compute the Dirichlet integral\n\n\\[ \n{\\int }_{0}^{+\\infty }\\frac{\\sin x}{x}\\mathrm{\\;d}x \n\\] | To do this we return to the integral (17.18), and we remark that for \\( y > 0 \\)\n\n\\[ \n{F}^{\\prime }\\left( y\\right) = - {\\int }_{0}^{+\\infty }\\sin x{\\mathrm{e}}^{-{xy}}\\mathrm{\\;d}x \n\\]\n\n(17.20)\n\nsince the integral (17.20) converges uniformly on every set of the form \\( \\{ y \\in \\mathbb{R} \\mid... | Yes |
Example 14 Consider the function \( f\left( {x, y}\right) = \left( {2 - {xy}}\right) {xy}{\mathrm{e}}^{-{xy}} \) on the set \( \{ \left( {x, y}\right) \in \) \( \left. {{\mathbb{R}}^{2} \mid 0 \leq x < + \infty \land 0 \leq y \leq 1}\right\} \) . Using the primitive \( {u}^{2}{\mathrm{e}}^{-u} \) of the function \( (2 ... | \[ 0 = {\int }_{0}^{1}\mathrm{\;d}y{\int }_{0}^{+\infty }\left( {2 - {xy}}\right) {xy}{\mathrm{e}}^{-{xy}}\mathrm{\;d}x \neq {\int }_{0}^{+\infty }\mathrm{d}x{\int }_{0}^{1}\left( {2 - {xy}}\right) {xy}{\mathrm{e}}^{-{xy}}\mathrm{\;d}y = 1. \] | Yes |
Example 15 Computing the integral \[ {\int }_{A}^{+\infty }\frac{{x}^{2} - {y}^{2}}{{\left( {x}^{2} + {y}^{2}\right) }^{2}}\mathrm{\;d}x \] | \[ {\int }_{A}^{+\infty }\frac{{x}^{2} - {y}^{2}}{{\left( {x}^{2} + {y}^{2}\right) }^{2}}\mathrm{\;d}x = - {\left. \frac{x}{{x}^{2} + {y}^{2}}\right| }_{A}^{+\infty } = \frac{A}{{A}^{2} + {y}^{2}} < \frac{1}{A} \] for \( A > 0 \) shows at the same time that for every fixed value of \( A > 0 \) it converges uniformly wi... | Yes |
For \( \alpha > 0 \) and \( \beta > 0 \) the iterated integral\n\n\[ \n{\int }_{0}^{+\infty }\mathrm{d}y{\int }_{0}^{+\infty }{x}^{\alpha }{y}^{\alpha + \beta - 1}{\mathrm{e}}^{-\left( {1 + x}\right) y}\mathrm{\;d}x = {\int }_{0}^{+\infty }{y}^{\beta }{\mathrm{e}}^{-y}\mathrm{\;d}y{\int }_{0}^{+\infty }{\left( xy\right... | Just as Corollary 3 followed from Proposition 7, we can deduce the following corollary from Proposition 8. | No |
Corollary 4 If\na) the function \( f\left( {x, y}\right) \) is continuous on the set\n\n\[ P = \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{2} \mid a \leq x < \omega \land c \leq y \leq \widetilde{\omega }}\right\} ,\;\text{ and }\n\]\n\nb) is nonnegative on \( P \), and\n\nc) the two integrals\n\n\[ F\left( y\right... | Proof Reasoning as in the proof of Corollary 3, we conclude from hypotheses a), b), and c) and Dini's theorem that hypothesis b) of Proposition 8 holds in this case. Since \( f \geq 0 \), hypothesis d) here is the same as hypothesis c) of Proposition 8 . Thus all the hypotheses of Proposition 8 are satisfied, and so Eq... | Yes |
Example 17 By changing the order of integration in two improper integrals, let us show that\n\n\[ \n{\int }_{0}^{+\infty }{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x = \frac{1}{2}\sqrt{\pi }\n\]\n\n(17.26)\n\nThis is the famous Euler-Poisson integral. | Proof We first observe that for \( y > 0 \)\n\n\[ \n\mathcal{J} \mathrel{\text{:=}} {\int }_{0}^{+\infty }{\mathrm{e}}^{-{u}^{2}}\mathrm{\;d}u = y{\int }_{0}^{+\infty }{\mathrm{e}}^{-{\left( xy\right) }^{2}}\mathrm{\;d}x\n\]\n\nand that the value of the integral in (17.26) is the same whether it is taken over the half-... | Yes |
\[ {\int }_{0}^{\pi /2}{\sin }^{\alpha - 1}\varphi {\cos }^{\beta - 1}\varphi \mathrm{d}\varphi = \frac{1}{2}B\left( {\frac{\alpha }{2},\frac{\beta }{2}}\right) . \] | Proof To prove this, it suffices to make the change of variable \( {\sin }^{2}\varphi = x \) in the integral.\n\nUsing formula (17.44), we can express the integral (17.45) in terms of the gamma function. In particular, taking account of (17.43), we obtain\n\n\[ {\int }_{0}^{\pi /2}{\sin }^{\alpha - 1}\varphi \mathrm{d}... | Yes |
If we assume that the \( \left( {\left( {n - 1}\right) \text{-dimensional}}\right) \) volume of the \( \left( {n - 1}\right) \) -dimensional ball of radius \( r \) is expressed by the formula \( {V}_{n - 1}\left( r\right) = {c}_{n - 1}{r}^{n - 1} \), then, integrating over sections (see Example 3 of Sect. 11.4), we obt... | \[ {V}_{n}\left( r\right) = {\int }_{-r}^{r}{c}_{n - 1}{\left( {r}^{2} - {x}^{2}\right) }^{\frac{n - 1}{2}}\mathrm{\;d}x = \left( {{c}_{n - 1}{\int }_{-\pi /2}^{\pi /2}{\cos }^{n}\varphi \mathrm{d}\varphi }\right) \cdot {r}^{n}, \] that is, \( {V}_{n}\left( r\right) = {c}_{n}{r}^{n} \), where \[ {c}_{\eta } = 2{c}_{n -... | Yes |
It is clear from geometric considerations that \( \mathrm{d}{V}_{n}\left( r\right) = {S}_{n - 1}\left( r\right) \mathrm{d}r \) , where \( {S}_{n - 1}\left( r\right) \) is the \( \left( {n - 1}\right) \) -dimensional surface area of the sphere bounding the \( n \) -dimensional ball of radius \( r \) in \( {\mathbb{R}}^{... | Thus \( {S}_{n - 1}\left( r\right) = \frac{\mathrm{d}{V}_{n}}{\mathrm{\;d}r}\left( r\right) \), and, taking account of (17.47), we obtain\n\n\[ \n{S}_{n - 1}\left( r\right) = \frac{2{\pi }^{\frac{n}{2}}}{\Gamma \left( \frac{n}{2}\right) }{r}^{n - 1}.\n\] | Yes |
Proposition 1 Each of the conditions listed below is sufficient for the existence of the convolution \( u * v \) of locally integrable functions \( u : \mathbb{R} \rightarrow \mathbb{C} \) and \( v : \mathbb{R} \rightarrow C \) . | 1) By the Cauchy-Bunyakovskii inequality\n\n\[ \n{\left( {\int }_{\mathbb{R}}\left| u\left( y\right) v\left( x - y\right) \right| \mathrm{d}y\right) }^{2} \leq {\int }_{\mathbb{R}}{\left| u\right| }^{2}\left( y\right) \mathrm{d}y{\int }_{\mathbb{R}}{\left| v\right| }^{2}\left( {x - y}\right) \mathrm{d}y \n\] \n\nfrom w... | Yes |
Proposition 2 If the convolution \( u * v \) exists, then the convolution \( v * u \) also exists, and the following equality holds:\n\n\[ u * v = v * u. \] | Proof Making the change of variable \( x - y = z \) in (17.49), we obtain\n\n\[ u * v\left( x\right) \mathrel{\text{:=}} {\int }_{-\infty }^{+\infty }u\left( y\right) v\left( {x - y}\right) \mathrm{d}y = {\int }_{-\infty }^{+\infty }v\left( z\right) u\left( {x - z}\right) \mathrm{d}z = : v * u\left( x\right) . \] | Yes |
Proposition 3 If the convolution \( u * v \) of the functions \( u \) and \( v \) exists, then the following equalities hold:\n\n\[ \n{T}_{{x}_{0}}\left( {u * v}\right) = {T}_{{x}_{0}}u * v = u * {T}_{{x}_{0}}v. \n\] | Proof If we recall the physical meaning of formula (17.48), the first of these equalities becomes obvious, and the second can then be obtained from the symmetry of convolution. Nevertheless, let us give a formal verification of the first equality:\n\n\[ \n\left( {T}_{{x}_{0}}\right) \left( {u * v}\right) \left( x\right... | Yes |
Proposition 4 If \( u \) is a locally integrable function and \( v \) is a \( {C}_{0}^{\left( m\right) } \) function of compact support \( \left( {0 \leq m \leq + \infty }\right) \), then \( \left( {u * v}\right) \in {C}^{\left( m\right) } \), and | Proof When \( u \) is a continuous function, the proposition follows immediately from what was just proved above. In its general form it can be obtained if we also keep in mind the observation made in Problem 6 of Sect. 17.1. | No |
Consider the sequence of functions \[ {\Delta }_{n}\left( x\right) = \left\{ \begin{array}{ll} \frac{{\left( 1 - {x}^{2}\right) }^{n}}{{\int }_{\left| x\right| < 1}{\left( 1 - {x}^{2}\right) }^{n}\mathrm{\;d}x} & \text{ for }\left| x\right| \leq 1, \\ 0 & \text{ for }\left| x\right| > 1. \end{array}\right. } \] | To establish that this family is an approximate identity we need only verify that condition c) of Definition 4 holds in addition to a) and b). But for every \( \varepsilon \in \rbrack 0,1\rbrack \) we have \[ 0 \leq {\int }_{\varepsilon }^{1}{\left( 1 - {x}^{2}\right) }^{n}\mathrm{\;d}x \leq {\int }_{\varepsilon }^{1}{... | Yes |
Example 4 Let\n\n\\[ \n{\\Delta }_{n}\\left( x\\right) = \\left\\{ \\begin{array}{ll} {\\cos }^{2n}\\left( x\\right) /{\\int }_{-\\pi /2}^{\\pi /2}{\\cos }^{2n}\\left( x\\right) \\mathrm{d}x & \\text{ for }\\left| x\\right| \\leq \\pi /2, \\ \\ 0 & \\text{ for }\\left| x\\right| > \\pi /2. \\end{array}\\right.\n\\]\n\n... | We remark first of all that\n\n\\[ \n{\\int }_{0}^{\\pi /2}{\\cos }^{2n}x\\mathrm{\\;d}x = \\frac{1}{2}B\\left( {n + \\frac{1}{2},\\frac{1}{2}}\\right) = \\frac{1}{2}\\frac{\\Gamma \\left( {n + \\frac{1}{2n}}\\right) }{\\Gamma \\left( n\\right) } \\cdot \\frac{\\Gamma \\left( \\frac{1}{2}\\right) }{n} > \\frac{\\Gamma ... | Yes |
Proposition 5 Let \( f : \mathbb{R} \rightarrow \mathbb{C} \) be a bounded function and \( \left\{ {{\Delta }_{\alpha };\alpha \in A}\right\} \) an approximate identity as \( \alpha \rightarrow \omega \) . If the convolution \( f * {\Delta }_{\alpha } \) exists for every \( \alpha \in A \) and the function \( f \) is u... | Proof Suppose \( \left| {f\left( x\right) }\right| \leq M \) on \( \mathbb{R} \) . Given a number \( \varepsilon > 0 \), we choose \( \rho > 0 \) in accordance with Definition 5 and denote the \( \rho \) -neighborhood of 0 in \( \mathbb{R} \) by \( U\left( 0\right) \) .\n\nTaking account of the symmetry of convolution,... | Yes |
Corollary 1 Every continuous function of compact support on \( \mathbb{R} \) can be uniformly approximated by infinitely differentiable functions. | Proof Let verify that \( {C}_{0}^{\left( \infty \right) } \) is everywhere dense in \( {C}_{0} \) in this sense.\n\nWe let, for example,\n\n\[ \varphi \left( x\right) = \left\{ \begin{array}{ll} k \cdot \exp \left( {-\frac{1}{1 - {x}^{2}}}\right) & \text{ for }\left| x\right| < 1, \\ 0 & \text{ for }\left| x\right| \ge... | Yes |
Every continuous function on a closed interval can be uniformly approximated on that interval by an algebraic polynomial. | Proof Since polynomials map to polynomials under a linear change of variable while the continuity and uniformity of the approximation of functions are preserved, it suffices to verify Corollary 2 on any convenient interval \( \left\lbrack {a, b}\right\rbrack \subset \mathbb{R} \) . For that reason we shall assume \( 0 ... | Yes |
The family of functions \( {\Delta }_{y}\left( x\right) = \frac{1}{\pi } \cdot \frac{y}{{x}^{2} + {y}^{2}} \) is an approximate identity on \( \mathbb{R} \) as \( y \rightarrow + 0 \) | since \( {\Delta }_{y} > 0 \) for \( y > 0 \) ,\n\n\[ \n{\int }_{-\infty }^{\infty }{\Delta }_{y}\left( x\right) \mathrm{d}x = {\left. \frac{1}{\pi }\arctan \left( \frac{x}{y}\right) \right| }_{x = - \infty }^{+\infty } = 1 \n\] \n\nand for every \( \rho > 0 \) we have \n\n\[ \n{\int }_{-\rho }^{\rho }{\Delta }_{y}\lef... | Yes |
The family of functions \( {\Delta }_{t} = \frac{1}{2\sqrt{\pi t}}{\mathrm{e}}^{-\frac{{x}^{2}}{4t}} \) is an approximate identity on \( \mathbb{R} \) as \( t \rightarrow + 0 \). | Indeed, we certainly have \( {\Delta }_{t} > 0 \) and \( {\int }_{-\infty }^{+\infty }{\Delta }_{t}\left( x\right) = 1 \), since \( {\int }_{-\infty }^{+\infty }{e}^{-{v}^{2}}\mathrm{\;d}v = \sqrt{\pi } \) (the Euler-Poisson integral). Finally, for every \( \rho > 0 \) we have\n\n\[{\int }_{-\rho }^{\rho }\frac{1}{2\sq... | Yes |
Example 7 Consider a point mass \( m \) that can move along the axis and is attached to one end of an elastic spring whose other end is fixed at the origin; let \( k \) be the elastic constant of the spring. Suppose that a time-dependent force \( f\left( t\right) \) begins to act on the point resting at the origin, mov... | \[ m\ddot{x} + {kx} = f \] (17.57) where \( x\left( t\right) \) is the coordinate of the point (its displacement from its equilibrium position) at time \( t \) . Under these conditions the function \( x\left( t\right) \) is uniquely determined by the function \( f \), and the solution \( x\left( t\right) \) of the diff... | Yes |
Let \( f \in C\left( {\mathbb{R},\mathbb{R}}\right) \) . As our test functions, we choose functions in \( {C}_{0} \) (continuous functions of compact support on \( \mathbb{R} \) ). A function \( f \) generates the following functional, which acts on \( {C}_{0} \) :\n\n\[ \langle f,\varphi \rangle \mathrel{\text{:=}} {\... | Using approximate identities consisting of functions of compact support, one can easily see that \( \langle f,\varphi \rangle \equiv 0 \) on \( {C}_{0} \) if and only if \( f\left( x\right) \equiv 0 \) on \( \mathbb{R} \) . | Yes |
The functional \( \delta \in \mathcal{L}\left( {{C}_{0};\mathbb{R}}\right) \) is defined by the relation\n\n\[ \langle \delta ,\varphi \rangle \mathrel{\text{:=}} \delta \left( \varphi \right) \mathrel{\text{:=}} \varphi \left( 0\right) ,\] | We can verify (see Problem 7) that no locally integrable function \( f \) on \( \mathbb{R} \) can represent the functional \( \delta \) in the form (17.59). | No |
Suppose a unit mass (or unit charge) is distributed on \( \mathbb{R} \). If this distribution is sufficiently regular, in the sense that it has, for example, a continuous or integrable density \( \rho \left( x\right) \) on \( \mathbb{R} \), the interaction of the mass \( M \) with other objects described by functions \... | \[ M\left( \varphi \right) = {\int }_{\mathbb{R}}\rho \left( x\right) \varphi \left( x\right) \mathrm{d}x. \] If the distribution is singular, for example, the whole mass \( M \) is concentrated at a single point, then by \ | No |
Let us see how the distribution \( \delta \cdot g \) acts, where \( g \in {C}^{\left( \infty \right) } \) . | In accordance with the definition (17.61) and the definition of \( \delta \), we obtain\n\n\[ \langle \delta \cdot g,\varphi \rangle \mathrel{\text{:=}} \langle \delta, g \cdot \varphi \rangle \mathrel{\text{:=}} \left( {g \cdot \varphi }\right) \left( 0\right) = g\left( 0\right) \cdot \varphi \left( 0\right) . \] | Yes |
Example 12 If \( f \in {C}^{\left( 1\right) } \), the derivative of \( f \) in the classical sense equals its derivative in the distribution sense (provided, naturally, the classical function is identified with the regular generalized function corresponding to it). | This follows from a comparison of relations (17.62) and (17.63), in which the right-hand sides are equal if the distribution \( F \) is generated by the function \( f \) . | No |
Regarding it as a generalized function, let us find the derivative \( {H}^{\prime } \) of this function, which is discontinuous in the classical sense. | From the definition of the regular generalized function \( H \) corresponding to the Heaviside function and relation (17.63) we find\n\n\[ \left\langle {{H}^{\prime },\varphi }\right\rangle \mathrel{\text{:=}} - \left\langle {H,{\varphi }^{\prime }}\right\rangle \mathrel{\text{:=}} - {\int }_{-\infty }^{+\infty }H\left... | Yes |
Let us compute \( \left\langle {{\delta }^{\prime },\varphi }\right\rangle \) : | \n\[
\left\langle {{\delta }^{\prime },\varphi }\right\rangle \mathrel{\text{:=}} - \left\langle {\delta ,{\varphi }^{\prime }}\right\rangle = - {\varphi }^{\prime }\left( 0\right)
\] | Yes |
Let us show that \( \left\langle {{\delta }^{\left( n\right) },\varphi }\right\rangle = {\left( -1\right) }^{n}{\varphi }^{\left( n\right) }\left( 0\right) \) . | For \( n = 0 \) this is the definition of the \( \delta \) -function.\n\nWe have seen in Example 14 that this equality holds for \( n = 1 \) .\n\nWe now prove it by induction, assuming that it has been established for a fixed value \( n \in \mathbb{N} \) . Using definition (17.63), we find\n\n\[ \left\langle {{\delta }... | Yes |
Suppose the function \( f : \mathbb{R} \rightarrow \mathbb{C} \) is continuously differentiable for \( x < 0 \) and for \( x > 0 \), and suppose the one-sided limits \( f\left( {-0}\right) \) and \( f\left( {+0}\right) \) of the function exist at 0 . We denote the quantity \( f\left( {+0}\right) - f\left( {-0}\right) \... | Proof Indeed,\n\n\[ \n\left\langle {{f}^{\prime },\varphi }\right\rangle = - \left\langle {f,{\varphi }^{\prime }}\right\rangle = - {\int }_{-\infty }^{+\infty }f\left( x\right) {\varphi }^{\prime }\left( x\right) \mathrm{d}x = \n\]\n\n\[ \n= - \left( {{\int }_{-\infty }^{0} + {\int }_{0}^{+\infty }}\right) \left( {f\l... | Yes |
Proposition 2 If \( Y \) is a domain in \( {\mathbb{R}}^{m}, f \in C\left( {X \times Y}\right) \), and \( \frac{\partial f}{\partial {y}^{i}} \in C\left( {X \times Y}\right) \), then the function \( F \) is differentiable with respect to \( {y}^{i} \) in \( Y \), where \( y = \left( {{y}^{1},\ldots ,{y}^{i},\ldots ,{y}... | \[ \frac{\partial F}{\partial {y}^{i}}\left( y\right) = {\int }_{X}\frac{\partial f}{\partial {y}^{i}}\left( {x, y}\right) \mathrm{d}x. \] | No |
Proposition 3 If \( X \) and \( Y \) are measurable compact subsets of \( {\mathbb{R}}^{n} \) and \( {\mathbb{R}}^{m} \) respectively, while \( f \in C\left( {X \times Y}\right) \), then \( F \in C\left( Y\right) \subset \mathcal{R}\left( Y\right) \), and\n\n\[ \n{\int }_{Y}F\left( y\right) \mathrm{d}y \mathrel{\text{:... | We note that the values of the function \( f \) here may lie in any normed vector space \( Z \) . The most important special cases occur when \( Z \) is \( \mathbb{R},\mathbb{C},{\mathbb{R}}^{n} \), or \( {\mathbb{C}}^{n} \) . In these cases the verification of Propositions 1-3 obviously reduce to the case of their pro... | No |
The integral\n\n\[ F\left( \lambda \right) = {\iint }_{{\mathbb{R}}^{2}}{\mathrm{e}}^{-\lambda \left( {{x}^{2} + {y}^{2}}\right) }\mathrm{d}x\mathrm{\;d}y \] | results from the limiting passage\n\n\[ {\iint }_{{\mathbb{R}}^{2}}{\mathrm{e}}^{-\lambda \left( {{x}^{2} + {y}^{2}}\right) }\mathrm{d}x\mathrm{\;d}y \mathrel{\text{:=}} \mathop{\lim }\limits_{{\varepsilon \rightarrow + 0}}{\iint }_{{x}^{2} + {y}^{2} \leq 1/{\varepsilon }^{2}}{\mathrm{e}}^{-\lambda \left( {{x}^{2} + {y... | Yes |
Example 2 Suppose, as always, that \( B\left( {a, r}\right) = \left\{ {x \in {\mathbb{R}}^{n}\left| \right| x - a \mid < r}\right\} \) is the ball of radius \( r \) with center at \( a \in {\mathbb{R}}^{n} \), and let \( y \in {\mathbb{R}}^{n} \) . Consider the integral\n\n\[ F\left( y\right) = {\int }_{B\left( {0,1}\r... | Passing to polar coordinates in \( {\mathbb{R}}^{n} \), we verify that this integral converges only for \( \alpha < 1 \) . If the value \( \alpha < 1 \) is fixed, the integral converges uniformly with respect to the parameter \( y \) on every compact set \( Y \subset {\mathbb{R}}^{n} \), since \( \left| {x - y}\right| ... | Yes |
Example 3 As is known, the potential of a unit charge located at the point \( x \in {\mathbb{R}}^{3} \) is expressed by the formula \( U\left( {x, y}\right) = \frac{1}{\left| x - y\right| } \), where \( y \) is a variable point of \( {\mathbb{R}}^{3} \). If the charge is now distributed in a bounded region \( X \subset... | The role of the parameter in this last integral is played by the variable point \( y \in {\mathbb{R}}^{3} \). If the point \( y \) lies in the exterior of the set \( X \), the integral (17.75) is a proper integral; but if \( y \in \bar{X} \), then \( \left| {x - y}\right| \rightarrow 0 \) as \( X \ni x \rightarrow y \)... | Yes |
Let us now verify that the function \( U\left( y\right) \) - the potential (17.75) - really does have a partial derivative \( \frac{\partial U}{\partial {y}^{i}} \) and that \( \frac{\partial U}{\partial {y}^{i}}\left( y\right) = {V}_{i}\left( y\right) \) . | To do this it obviously suffices to verify that\n\n\[ \n{\int }_{a}^{b}{V}_{i}\left( {{y}^{1},{y}^{2},{y}^{3}}\right) \mathrm{d}{y}^{i} = {\left. U\left( {y}^{1},{y}^{2},{y}^{3}\right) \right| }_{{y}^{i} = a}^{b}.\n\] \n\nBut in fact,\n\n\[ \n{\int }_{a}^{b}{V}_{i}\left( y\right) \mathrm{d}{y}^{i} = {\int }_{a}^{b}\mat... | Yes |
Suppose a charge is distributed on a smooth compact surface \( S \subset {\mathbb{R}}^{3} \) with surface density \( v\left( x\right) \). The potential of such a charge distribution is called a single-layer potential and is obviously represented by the surface integral\n\n\[ U\left( y\right) = {\int }_{S}\frac{v\left( ... | Suppose \( v \) is a bounded function. Then for \( y \notin S \) this integral is proper, and the function \( U\left( y\right) \) is infinitely differentiable outside \( S \). But if \( y \in S \), the integral has an integrable singularity at the point \( y \). The singularity is integrable because the surface \( S \)... | Yes |
Example 8 A generalization of the \( \delta \) -function (corresponding, for example, to a unit charge located at the origin in \( {\mathbb{R}}^{n} \) ) is the following generalized function \( {\delta }_{S} \) (corresponding to a distribution of charge over a piecewise-smooth surface \( S \) with a distribution of uni... | \[ \left\langle {{\delta }_{S},\varphi }\right\rangle \mathrel{\text{:=}} {\int }_{S}\varphi \left( x\right) \mathrm{d}\sigma \] | Yes |
If \( \mu \in D \), then \( \mu {\delta }_{S} \) is a generalized function acting according to the rule\n\n\[ \left\langle {\mu {\delta }_{S},\varphi }\right\rangle = {\int }_{S}\varphi \left( x\right) \mu \left( x\right) \mathrm{d}\sigma \] | If the function \( \mu \left( x\right) \) were defined only on the surface \( S \), Eq. (17.81) could be regarded as the definition of the generalized function \( \mu {\delta }_{S} \) . By natural analogy, the generalized function introduced in this way is called a single layer on the surface \( S \) with density \( \m... | No |
Now consider an operator \( D = \mathop{\sum }\limits_{m}{a}_{m}{D}^{m} \), where \( m = \left( {{m}_{1},\ldots ,{m}_{n}}\right) \) is a multi-index, \( {D}^{m} = {\left( \frac{\partial }{\partial {x}^{1}}\right) }^{{m}_{1}} \cdot \ldots \cdot {\left( \frac{\partial }{\partial {x}^{n}}\right) }^{{m}_{n}},{a}_{m} \) are... | The transpose or adjoint of \( D \) is the operator usually denoted \( {}^{t}D \) or \( {D}^{ * } \) and defined by the relation\n\n\[ \langle {DF},\varphi \rangle = : \left\langle {F,{}^{t}{D\varphi }}\right\rangle \]\n\nwhich must hold for all \( \varphi \in \mathcal{D} \) and \( F \in {\mathcal{D}}^{\prime } \) . St... | Yes |
We shall now show that if \( f \) is regarded as a generalized function, then the following important formula holds in the sense of differentiation of generalized functions:\n\n\[ \frac{\partial f}{\partial {x}^{i}} = \left\{ \frac{\partial f}{\partial {x}^{i}}\right\} + {\left( \int f\right) }_{S}\cos {\alpha }_{i}{\d... | Proof Formula (17.83) generalizes Eq. (17.64), which we use to derive it.\n\nFor definiteness we consider the case \( i = 1 \) . Then\n\n\[ \left\langle {\frac{\partial f}{\partial {x}^{1}},\varphi }\right\rangle \mathrel{\text{:=}} - \left\langle {f,\frac{\partial \varphi }{\partial {x}^{1}}}\right\rangle = - {\int }_... | Yes |
Let \( G \) be a finite domain in \( {\mathbb{R}}^{n} \) bounded by a piecewise-smooth surface \( S \). Let \( \mathbf{A} = \left( {{A}^{1},\ldots ,{A}^{n}}\right) \) be a vector field that is continuous in \( \bar{G} \) and such that the function \( \operatorname{div}\mathbf{A} = \mathop{\sum }\limits_{{i = 1}}^{n}\fr... | Relation (17.84) is equality of generalized functions. Let us apply it to the function \( \psi \in {C}_{0}^{\left( \infty \right) } \) equal to 1 on \( G \) (the existence and construction of such a function has been discussed more than once previously). Since for every function \( \varphi \in \mathcal{D} \)\n\n\[ \lan... | Yes |
We consider the vector field \( \mathbf{A} = \frac{x}{{\left| x\right| }^{3}} \) defined in \( {\mathbb{R}}^{3} \smallsetminus 0 \) and show that in the space \( {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{3}\right) \) of generalized functions we have the equality\n\n\[ \operatorname{div}\frac{x}{{\left| x\right| }^{3}... | We remark first that for \( x \neq 0 \) we have \( \operatorname{div}\frac{x}{{\left| x\right| }^{3}} = 0 \) in the classical sense.\n\nNow, using successively the definition of div \( \mathbf{A} \) in the form (17.85), the definition of an improper integral, the equality div \( \frac{x}{{\left| x\right| }^{3}} = 0 \) ... | Yes |
Example 14 We verify that the regular generalized function \( E\left( x\right) = - \frac{1}{{4\pi }\left| x\right| } \) in \( {D}^{\prime }\left( {\mathbb{R}}^{3}\right) \) is a fundamental solution of the Laplacian \( \Delta = {\left( \frac{\partial }{\partial {x}^{1}}\right) }^{2} + {\left( \frac{\partial }{\partial ... | Indeed, \( \Delta = \operatorname{div}\operatorname{grad} \), and \( \operatorname{grad}E\left( x\right) = \frac{x}{{4\pi }{\left| x\right| }^{3}} \) for \( x \neq 0 \), and therefore the equality div grad \( E = \delta \) follows from relation (17.87). | No |
Let us verify that the function \[ E\left( {x, t}\right) = \frac{H\left( t\right) }{{\left( 2a\sqrt{\pi t}\right) }^{n}}{\mathrm{e}}^{-\frac{{\left| x\right| }^{2}}{4{a}^{2}t}} \] where \( x \in {\mathbb{R}}^{n}, t \in \mathbb{R} \), and \( H \) is the Heaviside function (that is, we set \( E\left( {x, t}\right) = 0 \)... | When \( t > 0 \), we have \( E \in {C}^{\left( \infty \right) }\left( {\mathbb{R}}^{n + 1}\right) \) and by direct differentiation we verify that \[ \left( {\frac{\partial }{\partial t} - {a}^{2}\Delta }\right) E = 0\;\text{ when }t > 0. \] Taking this fact into account along with the result of Example 7, we obtain for... | Yes |
Let us show that the function\n\n\[ E\left( {x, t}\right) = \frac{1}{2a}H\left( {{at} - \left| x\right| }\right) \]\n\nwhere \( a > 0, x \in {\mathbb{R}}_{x}^{1}, t \in {\mathbb{R}}_{t}^{1} \), and \( H \) is the Heaviside function, satisfies the equation\n\n\[ \left( {\frac{{\partial }^{2}}{\partial {t}^{2}} - {a}^{2}... | Let \( \varphi \in \mathcal{D}\left( {\mathbb{R}}^{2}\right) \) . Using the abbreviation \( {▱}_{a} \mathrel{\text{:=}} \frac{{\partial }^{2}}{\partial {t}^{2}} - {a}^{2}\frac{{\partial }^{2}}{\partial {x}^{2}} \), we find\n\n\[ \left\langle {{▱}_{a}E,\varphi }\right\rangle = \left\langle {E,{▱}_{a}\varphi }\right\rang... | Yes |
Using the function \( E\left( {x, t}\right) \) of Example 16, one can thus present the solution\n\n\[ u\left( {x, t}\right) = \frac{1}{2a}{\int }_{0}^{t}\mathrm{\;d}\tau {\int }_{x - a\left( {t - \tau }\right) }^{x + a\left( {t - \tau }\right) }f\left( {\xi ,\tau }\right) \mathrm{d}\xi \]\nof the equation\n\n\[ \frac{{... | By direct differentiation of the resulting integral with respect to the parameters, one can easily verify that \( u\left( {x, t}\right) \) is indeed a solution of the equation \( {▱}_{a}u = f \) . | No |
Thus, from the point of view of generalized functions one could pose the question of the solution of the equation \( \frac{\partial u}{\partial t} - {\Delta u} = f \) taking as \( f\left( {x, t}\right) \) the generalized function \( \varphi \left( x\right) \cdot \delta \left( t\right) \), where \( \varphi \in \mathcal{... | The formal substitution of such a function \( f \) under the integral sign leads to the relation\n\n\[ u\left( {x, t}\right) = {\int }_{{\mathbb{R}}^{n}}\frac{\varphi \left( \xi \right) }{{\left\lbrack 2a\sqrt{\pi t}\right\rbrack }^{n}}{\mathrm{e}}^{-\frac{{\left| x - \xi \right| }^{2}}{4{a}^{2}t}}\mathrm{\;d}\xi . \]\... | Yes |
Example 19 Finally, recalling the fundamental solution of the Laplace operator obtained in Example 14, we find the solution\n\n\[ u\\left( x\\right) = {\\int }_{{\\mathbb{R}}^{n}}\\frac{f\\left( \\xi \\right) \\mathrm{d}\\xi }{\\left| x - \\xi \\right| }\n\]\n\nof the Poisson equation \( {\\Delta u} = - {4\\pi f} \), w... | If the function \( f \) is taken as \( v\\left( x\\right) {\\delta }_{S} \), where \( S \) is a piecewise smooth surface in \( {\\mathbb{R}}^{3} \), formal substitution into the integral leads to the function\n\n\[ u\\left( x\\right) = {\\int }_{S}\\frac{v\\left( \\xi \\right) \\mathrm{d}\\sigma \\left( \\xi \\right) }... | Yes |
Example 3 We remark that for \( \alpha \neq \beta \)\n\n\[{\int }_{0}^{l}\sin {\alpha x}\sin {\beta x}\mathrm{\;d}x = \frac{1}{2}\left( {\frac{\sin \left( {\alpha - \beta }\right) l}{\alpha - \beta } - \frac{\sin \left( {\alpha + \beta }\right) l}{\alpha + \beta }}\right) = \] | \[= \cos {\alpha l}\cos {\beta l} \cdot \frac{\beta \tan {\alpha l} - \alpha \tan {\beta l}}{{\alpha }^{2} - {\beta }^{2}}.\] | Yes |
Consider the equation\n\n\[ \left( {\frac{{\mathrm{d}}^{2}}{\mathrm{\;d}{x}^{2}} + q\left( x\right) }\right) u\left( x\right) = {\lambda u}\left( x\right) \]\n\nwhere \( q \in {C}^{\left( \infty \right) }\left( {\left\lbrack {a, b}\right\rbrack ,\mathbb{R}}\right) \) and \( \lambda \) is a numerical coefficient. Let us... | Indeed, integrating by parts, we find that\n\n\[ {\int }_{a}^{b}\left\lbrack {\left( {\frac{{\mathrm{d}}^{2}}{\mathrm{\;d}{x}^{2}} + q\left( x\right) }\right) {u}_{i}\left( x\right) }\right\rbrack {u}_{j}\left( x\right) \mathrm{d}x = {\int }_{a}^{b}{u}_{i}\left( x\right) \left\lbrack {\left( {\frac{{\mathrm{d}}^{2}}{\m... | Yes |
The process of orthogonalizing the linearly independent system \( \{ 1, x \) , \( \left. {{x}^{2},\ldots }\right\} \) in \( {\mathcal{R}}_{2}\left( {\left\lbrack {-1,1}\right\rbrack ,\mathbb{R}}\right) \) leads to the system of orthogonal polynomials known as the Legendre polynomials. | One can verify by direct computation that these polynomials are orthogonal on the closed interval \( \left\lbrack {-1,1}\right\rbrack \) . Taking Rodrigues’ formula as the definition of the polynomial \( {P}_{n}\left( x\right) \), let us verify that the system of Legendre polynomials \( \left\{ {{P}_{n}\left( x\right) ... | Yes |
Lemma 1 (Continuity of the inner product) Let \( \langle \) , \( \rangle : X \rightarrow \mathbb{C} \) be an inner product in the complex vector space \( X \) . Then a) the function \( \left( {x, y}\right) \mapsto \langle x, y\rangle \) is continuous jointly in the two variables; | Proof Assertion a) follows from the Cauchy-Bunyakovskii inequality (see Sect. 10.1): \[ {\left| \left\langle x - {x}_{0}, y - {y}_{0}\right\rangle \right| }^{2} \leq {\begin{Vmatrix}x - {x}_{0}\end{Vmatrix}}^{2} \cdot {\begin{Vmatrix}y - {y}_{0}\end{Vmatrix}}^{2}. \] | Yes |
Example 6 Let \( X = {\mathcal{R}}_{2}\left( {\left\lbrack {-\pi ,\pi }\right\rbrack ,\mathbb{R}}\right) \) . Consider the orthogonal system\n\n\[ \n\{ 1,\cos {kx},\sin {kx};k \in \mathbb{N}\} \n\]\n\nof Example 1. To the function \( f \in {\mathcal{R}}_{2}\left( {\left\lbrack {-\pi ,\pi }\right\rbrack ,\mathbb{R}}\rig... | Let us set \( f\left( x\right) = x \) . Then \( {a}_{k} = 0, k = 0,1,2,\ldots \), and \( {b}_{k} = {\left( -1\right) }^{k + 1}\frac{2}{k}, k = \) \( 1,2,\ldots \) . Hence in this case we obtain\n\n\[ \nf\left( x\right) = x \sim \mathop{\sum }\limits_{{k = 1}}^{\infty }{\left( -1\right) }^{k + 1}\frac{2}{k}\sin {kx}. \n... | Yes |
Example 7 Let us consider the orthogonal system \( \left\{ {{\mathrm{e}}^{ikx};k \in \mathbb{Z}}\right\} \) of Example 1 in the space \( {\mathcal{R}}_{2}\left( {\left\lbrack {-\pi ,\pi }\right\rbrack ,\mathbb{C}}\right) \) . Let \( f \in {\mathcal{R}}_{2}\left( {\left\lbrack {-\pi ,\pi }\right\rbrack ,\mathbb{C}}\righ... | \[ {c}_{k}\left( f\right) = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left( x\right) {\mathrm{e}}^{-{ikx}}\mathrm{\;d}x\left( { = \frac{\left\langle f\left( x\right) ,{\mathrm{e}}^{ikx}\right\rangle }{\left\langle {\mathrm{e}}^{ikx},{\mathrm{e}}^{ikx}\right\rangle }}\right) . \] | Yes |
let us assume that we have an arbitrary system of linearly independent vectors \( {x}_{1},\ldots ,{x}_{n} \) in \( X \) and are seeking the best approximation of a given vector \( x \in X \) by linear combinations \( \mathop{\sum }\limits_{{k = 1}}^{n}{\alpha }_{k}{x}_{k} \) of vectors of the system. | Since we can use the orthogonalization process to construct an orthonormal system \( {e}_{1},\ldots ,{e}_{n} \) that generates the same space \( L \) that is generated by the vectors \( {x}_{1},\ldots ,{x}_{n} \), we can conclude from the extremal property of the Fourier coefficients that there exists a unique vector \... | Yes |
If \( X = {E}^{3} \) and \( {e}_{1},{e}_{2},{e}_{3} \) is a basis in \( {E}^{3} \), then the system \( \left\{ {{e}_{1},{e}_{2},{e}_{2}}\right\} \) is complete in \( X \). | The system \( \left\{ {{e}_{1},{e}_{2}}\right\} \) is not complete in \( X \), but it is complete relative to the set \( L\left\{ {{e}_{1},{e}_{2}}\right\} \) or any subset \( E \) of it. | No |
Let us regard the sequence of functions \( 1, x,{x}^{2},\ldots \) as a system of vectors \( \left\{ {{x}^{k};k = 0,1,2,\ldots }\right\} \) in the space \( {\mathcal{R}}_{2}\left( {\left\lbrack {a, b}\right\rbrack ,\mathbb{R}}\right) \) or \( {\mathcal{R}}_{2}\left( {\left\lbrack {a, b}\right\rbrack ,\mathbb{C}}\right) ... | Indeed, for any function \( f \in C\left\lbrack {a, b}\right\rbrack \) and for every number \( \varepsilon > 0 \), the Weierstrass approximation theorem implies that there exists an algebraic polynomial \( P\left( x\right) \) such that \( \mathop{\max }\limits_{{x \in \left\lbrack {a, b}\right\rbrack }}\left| {f\left( ... | Yes |
If we remove one function, for example the function 1, from the system \( \{ 1,\cos {kx},\sin {kx};k \in \mathbb{N}\} \), the remaining system \( \{ \cos {kx},\sin {kx};k \in \mathbb{N}\} \) is no longer complete in \( {\mathcal{R}}_{2}\left( {\left\lbrack {-\pi ,\pi }\right\rbrack ,\mathbb{C}}\right) \) or \( {\mathca... | Indeed, by the extremal property of the Fourier coefficients the best approximation of the function \( f\left( x\right) \equiv 1 \) among all the finite linear combinations\n\n\[ \n{T}_{n}\left( x\right) = \mathop{\sum }\limits_{{k = 1}}^{n}\left( {{a}_{k}\cos {kx} + {b}_{k}\sin {kx}}\right) \n\]\n\nof any length \( n ... | Yes |
We now observe that in \( X = L\left\{ {e,{e}_{1},{e}_{2},\ldots }\right\} \) there is no nonzero vector orthogonal to all the vectors \( {e}_{1},{e}_{2},\ldots \) . | Indeed, let \( x \in X \), that is, \( x = {\alpha e} + \mathop{\sum }\limits_{{k = 1}}^{n}{\alpha }_{k}{e}_{k} \), and let \( \left\langle {x,{e}_{k}}\right\rangle = 0, k = 1,2,\ldots \) . Then \( \left\langle {x,{e}_{n + 1}}\right\rangle = \frac{\alpha }{{2}^{n + 1}} = 0 \), that is, \( \alpha = 0 \) . But then \( {\... | Yes |
Let us regard the closed interval \( \left\lbrack {0, l}\right\rbrack \) as the equilibrium position of a homogeneous elastic string fastened at the endpoints of this interval, but otherwise free and capable of making small transverse oscillations about this equilibrium position. Let \( u\left( {x, t}\right) \) be a fu... | To solve such problems there exists a very natural procedure called the method of separation of variables or the Fourier method in mathematics. It consists of the following. The solution \( u\left( {x, t}\right) \) is sought in the form of a series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{X}_{n}\left( x\right) {T}_... | Yes |
If \( f \) is a continuous function in \( U\left( x\right) \) satisfying the Hölder condition\n\n\[ \left| {f\left( {x + t}\right) - f\left( x\right) }\right| \leq M{\left| t\right| }^{\alpha },\;0 < \alpha \leq 1, \] | then, since the estimate\n\n\[ \left| \frac{f\left( {x + t}\right) - f\left( x\right) }{t}\right| \leq \frac{M}{{\left| t\right| }^{1 - \alpha }} \]\n\nnow holds, the function \( f \) satisfies the Dini conditions at \( x \) . | Yes |
If a function is piecewise continuously differentiable on a closed interval, then it satisfies the Hölder conditions with exponent \( \alpha = 1 \) at every point of the interval. | as follows from Lagrange's finite-increment (mean-value) theorem. Hence, by Example 1, such a function satisfies Dini's conditions at every point of the interval. At the endpoints of the interval, of course only the corresponding one-sided pair of Dini's conditions needs to be verified. | No |
Theorem 3 (Sufficient conditions for convergence of a Fourier series at a point) Let \( f : \mathbb{R} \rightarrow \mathbb{C} \) be a function of period \( {2\pi } \) that is absolutely integrable on the closed interval \( \left\lbrack {-\pi ,\pi }\right\rbrack \) . If \( f \) satisfies the Dini conditions at a point \... | Proof By relations (18.52) and (18.50)\n\n\[ {S}_{n}\left( x\right) - \frac{f\left( {x}_{ - }\right) + f\left( {x}_{ + }\right) }{2} = \]\n\n\[ = \frac{1}{\pi }{\int }_{0}^{\pi }\frac{\left( {f\left( {x - t}\right) - f\left( {x}_{ - }\right) }\right) + \left( {f\left( {x + t}\right) - f\left( {x}_{ + }\right) }\right) ... | Yes |
Example 5 In Example 6 of Sect. 18.1 we found the Fourier series\n\n\\[ \nx \sim \mathop{\sum }\limits_{{k = 1}}^{\infty }2\frac{{\\left( -1\\right) }^{k + 1}}{k}\\sin {kx} \n\\]\n\n(18.60)\n\nfor the function \\( f\\left( x\\right) = x \\) on the closed interval \\( \\left\\lbrack {-\\pi ,\\pi }\\right\\rbrack \\) . E... | \\[ \n\\mathop{\\sum }\\limits_{{k = 1}}^{\\infty }2\\frac{{\\left( -1\\right) }^{k + 1}}{k}\\sin {kx} = \\left\\{ \\begin{array}{ll} x, & \\text{ if }\\left| x\\right| < \\pi \\\\ 0, & \\text{ if }\\left| x\\right| = \\pi \\end{array}\\right. \n\\] | Yes |
Example 6 Let \( \alpha \in \mathbb{R} \) and \( \left| \alpha \right| < 1 \) . Consider the \( {2\pi } \) -periodic function \( f\left( x\right) \) defined on the closed interval \( \left\lbrack {-\pi ,\pi }\right\rbrack \) by the formula \( f\left( x\right) = \cos {\alpha x} \). | By formulas (18.35) and (18.36) we find its Fourier coefficients\n\n\[ \n{a}_{n}\left( f\right) = \frac{1}{\pi }{\int }_{-\pi }^{\pi }\cos {\alpha x}\cos {nx}\mathrm{\;d}x = \frac{{\left( -1\right) }^{n}\sin {\pi \alpha }}{\pi } \cdot \frac{2\alpha }{{\alpha }^{2} - {n}^{2}}, \n\]\n\n\[ \n{b}_{n}\left( f\right) = \frac... | Yes |
Lemma 2 The sequence of functions \[ {\Delta }_{n}\left( x\right) = \left\{ \begin{array}{ll} \frac{1}{2\pi }{\mathcal{F}}_{n}\left( x\right) , & \text{ if }\left| x\right| \leq \pi , \\ 0, & \text{ if }\left| x\right| > \pi \end{array}\right. \] is an approximate identity on \( \mathbb{R} \) . | Proof The nonnnegativity of \( {\Delta }_{n}\left( x\right) \) is clear. Equality (18.50) enables us to conclude that \[ {\int }_{-\infty }^{\infty }{\Delta }_{n}\left( x\right) \mathrm{d}x = {\int }_{-\pi }^{\pi }{\Delta }_{n}\left( x\right) \mathrm{d}x = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{\mathcal{F}}_{n}\left( x\... | Yes |
Theorem 4 (Fejér) Let \( f : \mathbb{R} \rightarrow \mathbb{C} \) be a function of period \( {2\pi } \) that is absolutely integrable on the closed interval \( \left\lbrack {-\pi ,\pi }\right\rbrack \) . If\n\na) \( f \) is uniformly continuous on the set \( E \subset \mathbb{R} \), then\n\n\[ \n{\sigma }_{n}\left( x\r... | Proof Statements b) and c) are special cases of a).\n\nStatement a) itself is a special case of the general Proposition 5 of Sect. 17.4 on the convergence of a convolution, since\n\n\[ \n{\sigma }_{n}\left( x\right) = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left( {x - t}\right) {\mathcal{F}}_{n}\left( t\right) \mathrm{d... | Yes |
Corollary 1 (Weierstrass' theorem on approximation by trigonometric polynomials) If a function \( f : \left\lbrack {-\pi ,\pi }\right\rbrack \rightarrow \mathbb{C} \) is continuous on the closed interval \( \left\lbrack {-\pi ,\pi }\right\rbrack \) and \( f\left( {-\pi }\right) = f\left( \pi \right) \), then this funct... | Proof Extending \( f \) as a function of period \( {2\pi } \), we obtain a continuous \( {2\pi } \) -periodic function on \( \mathbb{R} \), to which the trigonometric polynomials \( {\sigma }_{n}\left( x\right) \) converge uniformly by Fejér's theorem. | Yes |
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 theo... | 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\lbr... | 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}}^{-{ikx}}\mathrm{\;d}x = {\left. \frac{1}{2\pi }f\left( x\right) {\mathrm{e}}^{-{ik... | 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... | Proof Relation (18.64) follows from an \( m \) -fold application of Eq. (18.63):\n\n\[ \n{c}_{k}\left( {f}^{\left( m\right) }\right) = \left( {ik}\right) {c}_{k}\left( {f}^{\left( m - 1\right) }\right) = \cdots = {\left( ik\right) }^{m}{c}_{k}\left( f\right) .\n\] \n\nSetting \( {\gamma }_{k} = \left| {{c}_{k}\left( {f... | 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\[ {S}_{n}\left( x\right) = \mathop{\sum }\limits_{{-n}}^{n}{c}_{k}\left( f\right) {\mathrm{e}}^{ikx}. \]\n\nAccording to the assumptions on the function \( f \) and Proposition 1 we have \( \... | Yes |
Proposition 2 If the function \( f : \left\lbrack {-\pi ,\pi }\right\rbrack \rightarrow \mathbb{C} \) is piecewise continuous, then after integration the correspondence \( f\left( x\right) \sim \mathop{\sum }\limits_{{-\infty }}^{\infty }{c}_{k}\left( f\right) {\mathrm{e}}^{ikx} \) becomes the equality | \[ {\int }_{0}^{x}f\left( t\right) \mathrm{d}t = {c}_{0}\left( f\right) x + \mathop{\sum }\limits_{{-\infty }}^{\infty }\frac{{c}_{k}\left( f\right) }{ik}\left( {{\mathrm{e}}^{ikx} - 1}\right) ,\] where the prime indicates that the term with index \( k = 0 \) is omitted from the sum; the summation is the limit of the s... | 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{\sum ... | 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 P... | 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\n(18.72)\n\ncalled the isoperimetric inequ... | The name \ | No |
Let us find the function having the following spectrum of compact support:\n\n\[ c\\left( \\alpha \\right) = \\left\\{ \\begin{array}{ll} h, & \\text{ if }\\left| \\alpha \\right| \\leq a \\\\ 0, & \\text{ if }\\left| \\alpha \\right| > a \\end{array}\\right. \] | Proof By formula (18.82) we find, for \( t \\neq 0 \)\n\n\[ f\\left( t\\right) = {\\int }_{-a}^{a}h{\\mathrm{e}}^{i\\alpha t}\\mathrm{\\;d}\\alpha = h\\frac{{\\mathrm{e}}^{i\\alpha t} - {\\mathrm{e}}^{-{i\\alpha t}}}{it} = {2h}\\frac{\\sin {at}}{t}, \]\n\nand when \( t = 0 \), we obtain \( f\\left( 0\\right) = {2ha} \)... | Yes |
Let us assume that we know the spectral characteristic \( p\left( \omega \right) \) of the device \( P \) and the signal \( f\left( t\right) \) that enters the device; we ask how to find the signal \( x\left( t\right) = P\left( f\right) \left( t\right) \) that emerges from the device. | 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}}^{i\omega t}\mathrm{\;d}\omega ... | Yes |
Let us find the Fourier transform of \( f\left( t\right) = \frac{\sin {at}}{t} \) (assuming \( f\left( 0\right) = \) \( a \in \mathbb{R}) \) . | \[ \mathcal{F}\left\lbrack f\right\rbrack \left( \alpha \right) = \mathop{\lim }\limits_{{A \rightarrow + \infty }}\frac{1}{2\pi }{\int }_{-A}^{A}\frac{\sin {at}}{t}{\mathrm{e}}^{-{i\alpha t}}\mathrm{\;d}t = \] \[ = \mathop{\lim }\limits_{{A \rightarrow + \infty }}\frac{1}{2\pi }{\int }_{-A}^{A}\frac{\sin {at}\cos {\al... | 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\r... | Proof We have already noted that \( \left| {f\left( x\right) {\mathrm{e}}^{ix\xi }}\right| \leq \left| {f\left( x\right) }\right| \), 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).\n\nPart d) ... | Yes |
Let us find the Fourier transform of the function \( f\left( t\right) = {\mathrm{e}}^{-{t}^{2}/2} \) : | \[ \mathcal{F}\left\lbrack f\right\rbrack \left( \alpha \right) = {\int }_{-\infty }^{+\infty }{\mathrm{e}}^{-{t}^{2}/2}{\mathrm{e}}^{-{i\alpha t}}\mathrm{\;d}t = {\int }_{-\infty }^{+\infty }{\mathrm{e}}^{-{t}^{2}/2}\cos {\alpha t}\mathrm{\;d}t. \]\n\nDifferentiating this last integral with respect to the parameter \(... | 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\[ v\left( t\right) = {\int }_{-\infty }^{\infty }c\left( \omega \right) p\left( \omega \right) {\mathrm{e}}^{i\omega t}\mathrm{\;d}\omega \]\n\nwhere \( c\left( \omega \right) = \mathcal{F}\left\lbrack f\right\rbrack \left( \omega \ri... | Yes |
Example 6 Let \( a > 0 \) and\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} {\mathrm{e}}^{-{ax}} & \text{ for }x > 0 \\ 0 & \text{ for }x \leq 0 \end{array}\right. \]\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}}^{-{i... | In discussing the definition of the Fourier transform, we have already noted a number of its obvious properties in Part b of the present subsection. We note further that if \( {f}_{ - }\left( x\right) \mathrel{\text{:=}} f\left( {-x}\right) \), then \( \mathcal{F}\left\lbrack {f}_{ - }\right\rbrack \left( \xi \right) =... | Yes |
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