Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Lemma 2. If an algebra \( A \) of real (resp. complex) functions on \( X \) separates the points of \( X \) and does not vanish on \( X \), then for any two distinct points \( {x}_{1},{x}_{2} \in X \) and any real (resp. complex) numbers \( {c}_{1},{c}_{2} \) there is a function \( f \) in \( A \) such that \( f\left( ... | Proof. It obviously suffices to prove the lemma when \( {c}_{1} = 0,{c}_{2} = 1 \) and when \( {c}_{1} = 1,{c}_{2} = 0 \) . By the symmetry of the hypotheses on \( {x}_{1} \) and \( {x}_{2} \), we consider only the case \( {c}_{1} = 1,{c}_{2} = 0 \) . We begin by remarking that \( A \) contains a special function \( s ... | Yes |
Theorem 3. (Stone \( {}^{8} \) ). Let \( A \) be an algebra of continuous real-valued functions defined on a compact set \( K \) . If \( A \) separates the points of \( K \) and does not vanish on \( K \), then \( A \) is an everywhere-dense subspace of \( C\left( {K,\mathbb{R}}\right) \) . | Proof. Let \( \bar{A} \) be the closure of the set \( A \subset C\left( {K,\mathbb{R}}\right) \) in \( C\left( {K,\mathbb{R}}\right) \), that is, \( \bar{A} \) consists of the continuous functions \( f \in C\left( {K,\mathbb{R}}\right) \) that can be approximated uniformly with arbitrary precision by functions of \( A ... | Yes |
Let \( P = \{ \left( {x, y}\right) \in {\mathbb{R}}^{2}|a \leq x \leq b \land c \leq y \leq d\} \) be a rectangle in the plane \( {\mathbb{R}}^{2} \) . If the function \( f : P \rightarrow \mathbb{R} \) is continuous, that is, if \( f \in \) \( C\left( {P,\mathbb{R}}\right) \), then the function\n\n\[ F\left( y\right) ... | It follows from the uniform continuity of the function \( f \) on the compact set \( P \) that \( {\varphi }_{y}\left( x\right) \mathrel{\text{:=}} f\left( {x, y}\right) \rightrightarrows f\left( {x,{y}_{0}}\right) = : {\varphi }_{{y}_{0}}\left( x\right) \) on \( \left\lbrack {a, b}\right\rbrack \) as \( y \rightarrow ... | Yes |
If a function \( f \) belongs to the class \( {C}^{\left( 1\right) }\left( {U,\mathbb{R}}\right) \) in a neighborhood \( U \) of the point \( {x}_{0} \), then in some neighborhood of \( {x}_{0} \) it can be represented in the form\n\n\[ f\left( x\right) = f\left( {x}_{0}\right) + \varphi \left( x\right) \left( {x - {x}... | Equality (17.3) follows easily from the Newton-Leibniz formula\n\n\[ f\left( {{x}_{0} + h}\right) - f\left( {x}_{0}\right) = {\int }_{0}^{1}{f}^{\prime }\left( {{x}_{0} + {th}}\right) \mathrm{d}t \cdot h \]\n\nand Proposition 1 applied to the function \( F\left( h\right) = {\int }_{0}^{1}{f}^{\prime }\left( {{x}_{0} + ... | Yes |
Proposition 2. If the function \( f : P \rightarrow \mathbb{R} \) is continuous and has a continuous partial derivative with respect to \( y \) on the rectangle \( P = \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{2} \mid }\right. \) \( a \leq x \leq b \land c \leq y \leq d\} \), then the integral (17.2) belongs to \... | Proof. We shall verify directly that if \( {y}_{0} \in \left\lbrack {c, d}\right\rbrack \), then \( {F}^{\prime }\left( {y}_{0}\right) \) can be computed by formula (17.5):\n\n\[ \n\left| {F\left( {{y}_{0} + h}\right) - F\left( {y}_{0}\right) - \left( {{\int }_{a}^{b}\frac{\partial f}{\partial y}\left( {x,{y}_{0}}\righ... | Yes |
Let us verify that the function \( u\left( x\right) = {\int }_{0}^{\pi }\cos \left( {{n\varphi } - x\sin \varphi }\right) \mathrm{d}\varphi \) satisfies Bessel’s equation \( {x}^{2}{u}^{\prime \prime } + x{u}^{\prime } + \left( {{x}^{2} - {n}^{2}}\right) u = 0 \) . | Indeed, after carrying out the differentiation with formula (17.5) and making simple transformations we find\n\n\[ - {x}^{2}{\int }_{0}^{\pi }{\sin }^{2}\varphi \cos \left( {{n\varphi } - x\sin \varphi }\right) \mathrm{d}\varphi + x{\int }_{0}^{\pi }\sin \varphi \sin \left( {{n\varphi } - x\sin \varphi }\right) \mathrm... | Yes |
The complete elliptic integrals\n\n\[ E\left( k\right) = {\int }_{0}^{\pi /2}\sqrt{1 - {k}^{2}{\sin }^{2}\varphi }\mathrm{d}\varphi ,\;K\left( k\right) = {\int }_{0}^{\pi /2}\frac{\mathrm{d}\varphi }{\sqrt{1 - {k}^{2}{\sin }^{2}\varphi }} \]\n\nas functions of the parameter \( k,0 < k < 1 \), called the modulus of the ... | Let us verify, for example, the first of these. By formula (17.5)\n\n\[ \frac{\mathrm{d}E}{\mathrm{\;d}k} = - {\int }_{0}^{\pi /2}k{\sin }^{2}\varphi \cdot {\left( 1 - {k}^{2}{\sin }^{2}\varphi \right) }^{-1/2}\mathrm{\;d}\varphi = \]\n\n\[ = \frac{1}{k}{\int }_{0}^{\pi /2}{\left( 1 - {k}^{2}{\sin }^{2}\varphi \right) ... | Yes |
Let\n\n\\[ F\\left( \\alpha \\right) = {\\int }_{0}^{\\pi /2}\\ln \\left( {{\\alpha }^{2} - {\\sin }^{2}\\varphi }\\right) \\mathrm{d}\\varphi \\;\\left( {\\alpha > 1}\\right) . | According to formula (17.5)\n\n\\[ {F}^{\\prime }\\left( \\alpha \\right) = {\\int }_{0}^{\\pi /2}\\frac{{2\\alpha }\\mathrm{d}\\varphi }{{\\alpha }^{2} - {\\sin }^{2}\\varphi } = \\frac{\\pi }{\\sqrt{{\\alpha }^{2} - 1}}, \n\nfrom which we find \\( F\\left( \\alpha \\right) = \\pi \\ln \\left( {\\alpha + \\sqrt{{\\alp... | Yes |
Let\n\n\[ \n{F}_{n}\left( x\right) = \frac{1}{\left( {n - 1}\right) !}{\int }_{0}^{x}{\left( x - t\right) }^{n - 1}f\left( t\right) \mathrm{d}t \]\n\nwhere \( n \in \mathbb{N} \) and \( f \) is a function that is continuous on the interval of integration. Let us verify that \( {F}_{n}^{\left( n\right) }\left( x\right) ... | For \( n = 1 \) we have \( {F}_{1}\left( x\right) = {\int }_{0}^{x}f\left( t\right) \mathrm{d}t \) and \( {F}_{1}^{\prime }\left( x\right) = f\left( x\right) \) .\n\nBy formula (17.8) we find for \( n > 1 \) that\n\n\[ \n{F}_{n}^{\prime }\left( x\right) = \frac{1}{\left( {n - 1}\right) !}{\left( x - x\right) }^{n - 1}f... | Yes |
Proposition 3. If the function \( f : P \rightarrow \mathbb{R} \) is continuous in the rectangle \( P = \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{2} \mid a \leq x \leq b \land c \leq y \leq d}\right\} \), then the integral (17.2) is integrable over the closed interval \( \left\lbrack {c, d}\right\rbrack \) and th... | Proof. From the point of view of multiple integrals, Eq. (17.9) is an elementary version of Fubini's theorem. However, we shall give a proof of (17.9) that justifies it independently of Fubini's theorem.\n\nConsider the functions\n\n\[ \n\varphi \left( u\right) = {\int }_{c}^{u}\left( {{\int }_{a}^{b}f\left( {x, y}\rig... | Yes |
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 un... | Yes |
Corollary 1. If the function \( f \) in the integral (17.10) is continuous on the set \( \left\lbrack {a,\omega \left\lbrack {\times \left\lbrack {c, d}\right\rbrack \text{ and the integral (17.10) converges for every }y \in }\right\rbrack c, d}\right\rbrack \) but diverges for \( y = c \) or \( y = d \), then it conve... | 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\[ \lef... | 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 \( y \in Y \). If the inequality \( \left| {f... | Proof. This follows from the estimates \[ \left| {{\int }_{{b}_{1}}^{{b}_{2}}f\left( {x, y}\right) \mathrm{d}x}\right| \leq {\int }_{{b}_{1}}^{{b}_{2}}\left| {f\left( {x, y}\right) }\right| \mathrm{d}x \leq {\int }_{{b}_{1}}^{{b}_{2}}g\left( {x, y}\right) \mathrm{d}x \] and Cauchy’s criterion for uniform convergence of... | Yes |
The integral\n\n\[ \n{\int }_{0}^{\infty }\frac{\cos {\alpha x}}{1 + {x}^{2}}\mathrm{\;d}x \n\]\n\nconverges uniformly on the the whole set \( \mathbb{R} \) of values of the parameter \( \alpha \) | since \( \left| \frac{\cos {\alpha x}}{1 + {x}^{2}}\right| \leq \frac{1}{1 + {x}^{2}} \), and the integral \( {\int }_{0}^{\infty }\frac{\mathrm{d}x}{1 + {x}^{2}} \) converges. | 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\n\n\[ \n{\int }_{0}^{\infty }\sin x{\mathrm{e}}^{-t{x}^{2}}\mathrm{\;d}x \n\]\n\nas follows from Proposition 2 and the results of Example 3, converges uniformly on every set of the form ... | 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 zero as a limit point. | 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 uniform c... | 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}\r... | Yes |
The integral\n\n\[ \n{\int }_{1}^{+\infty }\frac{\sin x}{{x}^{\alpha }}\mathrm{d}x \n\] | as follows from the Cauchy criterion and the Abel-Dirichlet test for convergence of improper integrals, converges only for \( \alpha > 0 \) . Setting \( f\left( {x,\alpha }\right) = \sin x \) , \( \left. {g\left( {x,\alpha }\right) = {x}^{-\alpha }\text{, we see that the pair}\left. {\alpha }_{1}\right) ,{\beta }_{1}}\... | Yes |
The integral\n\n\[ \n{\int }_{0}^{+\infty }\frac{\sin x}{x}{\mathrm{e}}^{-{xy}}\mathrm{\;d}x \n\]\n\nconverges 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 se... | No |
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\le... | 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. ... | Yes |
Let \( Y = \{ y \in \mathbb{R} \mid y > 0\} \) and \[ f\left( {x, y}\right) = \left\{ \begin{matrix} 1/y, & \text{ if }0 \leq x \leq y, \\ 0, & \text{ if }y < x. \end{matrix}\right. \] Obviously, \( f\left( {x, y}\right) \rightrightarrows 0 \) on the interval \( 0 \leq x < + \infty \) as \( y \rightarrow + \infty \) . ... | Using Dini's theorem (Proposition 2 of Sect. 16.3), we can obtain the following sometimes useful corollary of Proposition 4. | No |
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 \( a \leq x < \omega . \)\n\nIf\n\na) the function \( f\left( {x, y}\right) \) is monotonically increasing as \( y \) increase... | Proof. It follows from Dini’s theorem that \( f\left( {x, y}\right) \rightrightarrows \varphi \left( x\right) \) on each closed interval \( \left\lbrack {a, b}\right\rbrack \subset \lbrack a,\omega \lbrack \) .\n\nIt follows from the inequalities \( 0 \leq f\left( {x, y}\right) \leq \varphi \left( x\right) \) and the W... | 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 |
Proposition 5. If\n\na) the function \( f\left( {x, y}\right) \) is continuous on the set \( \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{2} \mid a \leq x < \omega }\right. \) \( \land c \leq y \leq d\} \), and\n\nb) the integral \( F\left( y\right) = {\int }_{a}^{\omega }f\left( {x, y}\right) \mathrm{d}x \) converg... | Proof. It follows from hypothesis a) that for any \( b \in \lbrack a,\omega \lbrack \) the proper integral\n\n\[
{F}_{b}\left( y\right) = {\int }_{a}^{b}f\left( {x, y}\right) \mathrm{d}x
\]\n\nis a continuous function on \( \left\lbrack {c, d}\right\rbrack \) (see Proposition 1 of Sect. 17.1).\n\nBy hypothesis b) we ha... | Yes |
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 \( x \in \mathbb{R} \) . | No |
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\\( \\left( {17.20}\\right) \\)\n\nsince the integral (17.20) converges uniformly on every set of the form \\( \\{ ... | Yes |
Consider the function \( f\left( {x, y}\right) = \left( {2 - {xy}}\right) {xy}{\mathrm{e}}^{-{xy}} \) on the set \( \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{2} \mid 0 \leq x < + \infty \land 0 \leq y \leq 1}\right\} \) . | Using the primitive \( {u}^{2}{\mathrm{e}}^{-u} \) of the function \( \left( {2 - u}\right) u{\mathrm{e}}^{-u} \), it is easy to compute directly that\n\n\[ 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 }_... | No |
Corollary 3. If\na) the function \( f\\left( {x, y}\\right) \) is continuous on the set \( P = \\left\\{ {\\left( {x, y}\\right) \\in {\\mathbb{R}}^{2} \\mid }\\right. \) \( a \\leq x < \\omega \\land c \\leq y \\leq d\\} \) and\nb) nonnegative on \( P \), and\nc) the integral \( F\\left( y\\right) = {\\int }_{a}^{\\om... | Proof. It follows from hypothesis a) that for every \( b \\in \\lbrack a,\\omega \\lbrack \) the integral\n\n\[ {F}_{b}\\left( y\\right) = {\\int }_{a}^{b}f\\left( {x, y}\\right) \\mathrm{d}x \]\n\nis continuous with respect to \( y \) on the closed interval \( \\left\\lbrack {c, d}\\right\\rbrack \) .\n\nIt follows fr... | Yes |
Computing the integral \[ {\int }_{A}^{+\infty }\frac{{x}^{2} - {y}^{2}}{{\left( {x}^{2} + {y}^{2}\right) }^{2}}\mathrm{\;d}x \] for \( A > 0 \) | \[ {\int }_{A}^{+\infty }\frac{{x}^{2} - {y}^{2}}{{\left( {x}^{2} + {y}^{2}\right) }^{2}}\mathrm{\;d}x = - {\left. \frac{x}{{x}^{2} + {y}^{2}}\right| }_{A}^{+\infty } = \frac{A}{{A}^{2} + {y}^{2}} < \frac{1}{A} \] | 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... | Thus, in this case hypotheses a) and c) of Proposition 8 hold. The fact that both conditions of b) hold for this integral was verified in Example 3. Hence by Proposition 8 we have the equality\n\n\[ \n{\int }_{0}^{+\infty }\mathrm{d}y{\int }_{0}^{+\infty }{x}^{\alpha }{y}^{\alpha + \beta + 1}{\mathrm{e}}^{-\left( {1 + ... | No |
Corollary 4. If\na) the function \( f\left( {x, y}\right) \) is continuous on the set\n\n\[ \nP = \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{2} \mid a \leq x < \omega \land c \leq y \leq \widetilde{\omega }}\right\} ,\;\text{ and }\n\]\n\nb) is nonnegative on \( P \), and\n\nc) the two integrals\n\n\[ \nF\left( y\... | 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 ... | 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 ob... | \[ {V}_{n}\left( r\right) = {\int }_{-r}^{r}{c}_{n - 1}{\left( {r}^{2} - {x}^{2}\right) }^{\frac{n - 1}{2}}\mathrm{\;d}x = \left( {{c}_{n - 1}{\int }_{-\pi /2}^{\pi /2}{\cos }^{n}\varphi \mathrm{d}\varphi }\right) \cdot {r}^{n}, \] that is, \( {V}_{n}\left( r\right) = {c}_{n}{r}^{n} \), where \[ {c}_{n} = 2{c}_{n - 1}{... | 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{... | 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 \( v : \mathbb{R} \rightarrow \mathbb{C} \) .\n\n1) The functions \( {\left| u\right| }^{2} \) and \( {\left| v\right| }^{2} \... | Proof. 1) By the Cauchy-Bunyakovskii inequality\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\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\righ... | 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 \( {}^{3} \n\n\[ \n{D}^{k}\left( {u * v}\right) = u * \left( {{D}^{k}v}... | Proof. When \( u \) is a continuous funcction, 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\n\n\[ \n{\Delta }_{n}\left( x\right) = \left\{ \begin{matrix} \frac{{\left( 1 - {x}^{2}\right) }^{n}}{\mathop{\int }\limits_{{\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{matrix}\... | But for every \( \varepsilon \in \rbrack 0,1\rbrack \) we have\n\n\[ \n0 \leq {\int }_{\varepsilon }^{1}{\left( 1 - {x}^{2}\right) }^{n}\mathrm{\;d}x \leq {\int }_{\varepsilon }^{1}{\left( 1 - {\varepsilon }^{2}\right) }^{n}\mathrm{\;d}x = \n\]\n\n\[ \n= {\left( 1 - {\varepsilon }^{2}\right) }^{n}\left( {1 - \varepsilo... | Yes |
Example 4. Let\n\n\\[ \n{\\Delta }_{n}\\left( x\\right) = \\left\\{ \\begin{matrix} {\\cos }^{2n}\\left( x\\right) /\\mathop{\\int }\\limits_{{-\\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{matrix... | 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 ... | 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{matrix} k \cdot \exp \left( {-\frac{1}{1 - {x}^{2}}}\right) & \text{ for }\left| x\right| < 1, \\ 0 & \text{ for }\left| x\right| \geq ... | Yes |
Corollary 2. (The Weierstrass approximation theorem). 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 \( \mathop{\int }\limits_{{-\infty }}^{{+\infty }}{e}^{-{v}^{2}}\mathrm{\;d}v = \sqrt{\pi } \) (the Euler-Poisson integral). Finally, for every \( \rho > 0 \) we have\n\n\[ \n{\int }_{-\rho ... | Yes |
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, moving it alo... | \[ m\ddot{x} + {kx} = f \] (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\[ \langle \delta ,\varphi \rangle \mathrel{\text{:=}} \delta \left( \varphi \right) \mathrel{\text{:=}} \varphi \left( 0\right) \]\n\nwhich must hold for every function \( \varphi \in {C}_{0} \) . | 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 |
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\[\n\langle \delta \cdot g,\varphi \rangle \mathrel{\text{:=}} \langle \delta, g \cdot \varphi \rangle \mathrel{\text{:=}} \left( {g \cdot \varphi }\right) \left( 0\right) = g\left( 0\right) \cdot \varphi \left( 0\right) .\n\] | Yes |
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 righthand sides are equal if the distribution \( F \) is generated by the function \( f \) . | No |
Take the Heaviside \( {}^{6} \) function\n\n\[ H\left( x\right) = \left\{ \begin{array}{ll} 0 & \text{ for }x < 0 \\ 1 & \text{ for }x \geq 0 \end{array}\right. \]\n\nsometimes called the unit step. Regarding it as a generalized function, let us find the derivative \( {H}^{\prime } \) of this function, which is discont... | 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 \( \left\langle {{\delta }^{\left( n\right) },\varphi }\right\rangle = {\left( -1\right) }^{n}{\varphi }^{\left( n\right) }\left( 0\right) \) . | Proof. 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\[ \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\[ \left\langle {{f}^{\prime },\varphi }\right\rangle = - \left\langle {f,{\varphi }^{\prime }}\right\rangle = - {\int }_{-\infty }^{+\infty }f\left( x\right) {\varphi }^{\prime }\left( x\right) \mathrm{d}x = \]\n\n\[ = - \left( {{\int }_{-\infty }^{0} + {\int }_{0}^{+\infty }}\right) \left( {f\left( ... | Yes |
Proposition 6. a) Every generalized function \( F \in {\mathcal{D}}^{\prime } \) is infinitely differentiable. | Proof. a) \( \left\langle {{F}^{\left( m\right) },\varphi }\right\rangle \mathrel{\text{:=}} - \left\langle {{F}^{\left( m - 1\right) },{\varphi }^{\prime }}\right\rangle \mathrel{\text{:=}} {\left( -1\right) }^{m}\left\langle {F,{\varphi }^{\left( m\right) }}\right\rangle \) . | Yes |
Proposition 2. If \( Y \) is a domain in \( {\mathbb{R}}^{m}, f \in C\left( {X \times Y}\right) \), and \( \frac{\partial f}{\partial {y}^{i}} \in C(X \times \) \( Y) \), 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\[{\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 pr... | Yes |
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 |
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}\right) }\fr... | 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 \subse... | 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}\ma... | Yes |
Suppose a charge is distributed on a smooth compact surface \( S \subset {\mathbb{R}}^{3} \) with surface density \( \nu \left( x\right) \). 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{\nu \... | The singularity is integrable because the surface \( S \) is smooth and differs by little from a piece of the plane \( {\mathbb{R}}^{2} \) near the point \( y \in S \) ; and we know that a singularity of type \( 1/{r}^{\alpha } \) is integrable in the plane if \( \alpha < 2 \). Using Proposition 5, we can turn this gen... | Yes |
We shall show that these functions, regarded as regular distributions in \( {\mathbb{R}}^{n} \), converge to the \( \delta \) -function on \( {\mathbb{R}}^{n} \) as \( t \rightarrow + 0 \) . | For the proof it suffices to verify that the family of functions \( {\Delta }_{t} \) is an approximate identity in \( {\mathbb{R}}^{n} \) as \( t \rightarrow + 0 \) .\n\nUsing a change of variable, reduction of the multiple integral to an iterated integral, and the value of the Euler-Poisson integral, we find\n\n\[{\in... | Yes |
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 unit surface ... | \[\n\left\langle {{\delta }_{S},\varphi }\right\rangle \mathrel{\text{:=}} {\int }_{S}\varphi \left( x\right) \mathrm{d}\sigma\n\] | Yes |
If \( \mu \in \mathcal{D} \), then \( \mu {\delta }_{S} \) is a generalized function acting according to the rule \[ \left\langle {\mu {\delta }_{S},\varphi }\right\rangle = {\int }_{S}\varphi \left( x\right) \mu \left( x\right) \mathrm{d}\sigma \] | 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, \( {D}^{m} = {\left( \frac{\partial }{\partial {x}^{1}}\right) }^{{m}_{1}}\cdots \cdot {\left( \frac{\partial }{\partial {x}^{n}}\right) }^{{m}_{n}},{a}_{m} \) 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\[ \frac{\partial f}{\partial {x}^{i}} = \left\{ \frac{\partial f}{\partial {x}^{i}}\right\} + {\left( \int f\right) }_{S}\cos {\alpha }_{i}{\del... | 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}\f... | 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 \( \operatorname{div}\mathbf{A} \) in the form (17.85), the definition of an improper integral, the equality \( \operatorname{div}\frac{x}{{\l... | Yes |
We verify that the regular generalized function \( E\left( x\right) = - \frac{1}{{4\pi }\left| x\right| } \) in \( {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{3}\right) \) is a fundamental solution of the Laplacian \( \mathit{Δ} = {\left( \frac{\partial }{\partial {x}^{1}}\right) }^{2} + {\left( \frac{\partial }{\parti... | 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 a... | 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\[ u\left( {x, t}\right) = \frac{1}{2a}{\int }_{0}^{t}\mathrm{\;d}\tau {\int }_{x - a\left( {t - \tau }\right) }^{x + a\left( {t - \tau }\right) }f\left( {\xi ,\tau }\right) \mathrm{d}\xi \]\n\nof the equation\n\n\[ \frac... | which is the convolution \( f * E \) of the functions \( f \) and \( E \) and necessarily exists under the assumption, for example, that the function \( f \) is continuous. By direct differentiation of the resulting integral with respect to the parameters, one can easily verify that \( u\left( {x, t}\right) \) is indee... | Yes |
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 \[ u\left( {x, t}\right) = {\int }_{{\mathbb{R}}^{n}}\frac{\varphi \left( \xi \right) }{{\left\lbrack 2a\sqrt{\pi t}\right\rbrack }^{n}}{\mathrm{e}}^{-\frac{{\left| x - \xi \right| }^{2}}{4{a}^{2}t}}\mathrm{\;d}\xi . \] App... | Yes |
Let \( {I}_{x} \) be an interval in \( {\mathbb{R}}^{m} \) and \( {I}_{y} \) an interval in \( {\mathbb{R}}^{n} \), and let \( \left\{ {{f}_{i}\left( x\right) }\right\} \) be an orthogonal system of functions in \( {\mathcal{R}}_{2}\left( {{I}_{x},\mathbb{R}}\right) \) and \( \left\{ {{g}_{j}\left( y\right) }\right\} \... | as follows from Fubini’s theorem | No |
We remark that for \( \alpha \neq \beta \)\n\n\[ \n{\int }_{0}^{l}\sin {\alpha x}\sin {\beta x}\mathrm{\;d}x = \frac{1}{2}\left( {\frac{\sin \left( {\alpha - \beta }\right) l}{\alpha - \beta } - \frac{\sin \left( {\alpha + \beta }\right) l}{\alpha + \beta }}\right) =\n\]\n\n\[ \n= \cos {\alpha l}\cos {\beta l} \cdot \f... | Hence, if \( \alpha \) and \( \beta \) are such that \( \frac{\tan {\alpha l}}{\alpha } = \frac{\tan {\beta l}}{\beta } \), the original integral equals zero. Consequently, if \( {\xi }_{1} < {\xi }_{2} < \cdots < {\xi }_{n} < \cdots \) is a sequence of roots of the equation \( \tan {\xi l} = {c\xi } \), where \( c \) ... | 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 \( \left\{ {1, x,{x}^{2},\ldots }\right\} \) 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\n\na) the function \( \left( {x, y}\right) \mapsto \langle x, y\rangle \) is continuous jointly in the two variables;\n\nb) if \( x = \mathop{\sum }\lim... | Proof. Assertion a) follows from the Cauchy-Bunyakovskii inequality (see Sect. 10.1):\n\n\[ \n{\left| \left\langle x - {x}_{0}, y - {y}_{0}\right\rangle \right| }^{2} \leq {\begin{Vmatrix}x - {x}_{0}\end{Vmatrix}}^{2} \cdot {\begin{Vmatrix}y - {y}_{0}\end{Vmatrix}}^{2}.\n\]\n\nAssertion \( b \) ) follows from \( a \) )... | Yes |
To the function \( f \in {\mathcal{R}}_{2}\left( {\left\lbrack {-\pi ,\pi }\right\rbrack ,\mathbb{R}}\right) \) there corresponds a Fourier series\n\n\[ f \sim \frac{{a}_{0}\left( f\right) }{2} + \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}\left( f\right) \cos {kx} + {b}_{k}\left( f\right) \sin {kx} \]\nin this sys... | 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\[ f\left( x\right) = x \sim \mathop{\sum }\limits_{{k = 1}}^{\infty }{\left( -1\right) }^{k + 1}\frac{2}{k}\sin {kx}. \] | Yes |
Example 7. Let us consider the orthogonal system \( \left\{ {{\mathrm{e}}^{\mathrm{i}{kx}};k \in \mathbb{Z}}\right\} \) 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 ,\mat... | \[ {c}_{k}\left( f\right) = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left( x\right) {\mathrm{e}}^{-\mathrm{i}{kx}}\mathrm{\;d}x\left( { = \frac{\left\langle f\left( x\right) ,{\mathrm{e}}^{\mathrm{i}{kx}}\right\rangle }{\left\langle {\mathrm{e}}^{\mathrm{i}{kx}},{\mathrm{e}}^{\mathrm{i}{kx}}\right\rangle }}\right) . \] | 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 |
Example 12. 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... | Proof. Indeed, by Lemma 3 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 \( n \) is given by the trigonometric po... | Yes |
Consider the space \( {l}_{2} \) (see Sect. 10.1) of real sequences \( a = \) \( \left( {{a}^{1},{a}^{2},\ldots }\right) \) for which \( \mathop{\sum }\limits_{{j = 1}}^{\infty }{\left( {a}^{j}\right) }^{2} < \infty \) . We define the inner product of the vectors \( a = \left( {{a}^{1},{a}^{2},\ldots }\right) \) and \(... | We note that the vector \( {e}_{0} = \left( {1,0,0,\ldots }\right) \) obviously cannot be obtained as a finite linear combination of vectors in the system \( e,{e}_{1},{e}_{2},\ldots \) , and therefore it does not belong to \( X \) . At the same time, it can be approximated as closely as desired in \( {l}_{2} \) by suc... | Yes |
Example 15. The Fourier method.\n\nLet 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 ... | 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 |
Theorem 2. (Localization principle). Let \( f \) and \( g \) be real- or complex-valued locally integrable functions on \( \rbrack - \pi ,\pi \lbrack \) and absolutely integrable on the whole interval (possibly in the improper sense). | If the functions \( f \) and \( g \) are equal in any (arbitrarily small) neighborhood of the point \( \left. {{x}_{0} \in }\right\rbrack - \pi ,\pi \lbrack \), then their Fourier series\n\n\[ f\left( x\right) \sim \mathop{\sum }\limits_{{-\infty }}^{{+\infty }}{c}_{k}\left( f\right) {\mathrm{e}}^{\mathrm{i}{kx}},\;g\l... | 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 |
In Example 6 of Sect. 18.1 we found the Fourier series\n\n\[ x \sim \mathop{\sum }\limits_{{k = 1}}^{\infty }2\frac{{\left( -1\right) }^{k + 1}}{k}\sin {kx} \]\n\n(18.60)\n\nfor the function \( f\left( x\right) = x \) on the closed interval \( \left\lbrack {-\pi ,\pi }\right\rbrack \) . Extending the function \( f\left... | \[ \mathop{\sum }\limits_{{k = 1}}^{\infty }2\frac{{\left( -1\right) }^{k + 1}}{k}\sin {kx} = \left\{ \begin{array}{ll} x, & \text{ if }\left| x\right| < \pi , \\ 0, & \text{ if }\left| x\right| = \pi . \end{array}\right. \] | 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\n\n\\[ \n{\\Delta }_{n}\\left( x\\right) = \\left\\{ \\begin{matrix} \\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{matrix}\\right.\n\\]\n\nis an approximate identity on \\( \\mat... | Proof. The nonnnegativity of \\( {\\Delta }_{n}\\left( x\\right) \\) is clear.\n\nEquality (18.50) enables us to conclude that\n\n\\[ \n{\\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 }... | 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... | 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{... | 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 fun... | 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 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.