en>Heron: definition was entirely wrong; rewrote

2013-09-26T11:01:12Z

definition was entirely wrong; rewrote

← Older revision		Revision as of 13:01, 26 September 2013
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	~~Hello~~ and ~~welcome~~. ~~My title is Numbers Wunder~~. ~~To do aerobics~~ is a ~~thing that~~ I'~~m completely addicted~~ to. ~~For years he~~'s ~~been residing~~ in ~~North Dakota and his family enjoys it~~. ~~Managing people is his profession~~.<br><br>~~Feel free~~ to ~~surf~~ to ~~my web-site - std testing at home~~ ([~~http~~://~~blogzaa~~.~~com~~/~~blogs~~/~~post~~/~~9199 mouse click the up coming webpage~~])		In [[mathematical analysis]], the '''Schur test''', named after German mathematician [[Issai Schur]], is a bound on the <math>L^2\to L^2</math> [[operator norm]] of an [[integral operator]] in terms of its [[Schwartz kernel]] (see [[Schwartz kernel theorem]]).

			Here is one version.<ref>[[Paul Richard Halmos]] and Viakalathur Shankar Sunder, ''Bounded integral operators on <math>L^{2}</math> spaces'', Ergebnisse der Mathematik und ihrer Grenzgebiete (Results in Mathematics and Related Areas), vol. 96., Springer-Verlag, Berlin, 1978. Theorem 5.2.</ref> Let <math>X,\,Y</math> be two [[measurable space]]s (such as <math>\mathbb{R}^n</math>). Let <math>\,T</math> be an [[integral operator]] with the non-negative Schwartz kernel <math>\,K(x,y)</math>, <math>x\in X</math>, <math>y\in Y</math>:

			:<math>T f(x)=\int_Y K(x,y)f(y)\,dy.</math>

			If there exist functions <math>\,p(x)>0</math> and <math>\,q(x)>0</math> and numbers <math>\,\alpha,\beta>0</math> such that

			:<math> (1)\qquad \int_Y K(x,y)q(y)\,dy\le\alpha p(x) </math>

			for [[almost everywhere\|almost all]] <math>\,x</math> and

			:<math> (2)\qquad \int_X p(x)K(x,y)\,dx\le\beta q(y)</math>

			for almost all <math>\,y</math>, then <math>\,T</math> extends to a [[continuous operator]] <math>T:L^2\to L^2</math> with the [[operator norm]]

			:<math> \Vert T\Vert_{L^2\to L^2} \le\sqrt{\alpha\beta}.</math>

			Such functions <math>\,p(x)</math>, <math>\,q(x)</math> are called the Schur test functions.

			In the original version, <math>\,T</math> is a matrix and <math>\,\alpha=\beta=1</math>.<ref>[[I. Schur]], ''Bemerkungen zur Theorie der Beschränkten Bilinearformen mit unendlich vielen Veränderlichen'', J. reine angew. Math. 140 (1911), 1–28.</ref>

			==Common usage and Young's inequality==

			A common usage of the Schur test is to take <math>\,p(x)=q(x)=1.</math> Then we get:

			:<math>
			\Vert T\Vert^2_{L^2\to L^2}\le
			\sup_{x\in X}\int_Y\|K(x,y)\| \, dy
			\cdot
			\sup_{y\in Y}\int_X\|K(x,y)\| \, dx.
			</math>

			This inequality is valid no matter whether the Schwartz kernel <math>\,K(x,y)</math> is non-negative or not.

			A similar statement about <math>L^p\to L^q</math> operator norms is known as [[Young's inequality]]:<ref>Theorem 0.3.1 in: C. D. Sogge, ''Fourier integral operators in classical analysis'', Cambridge University Press, 1993. ISBN 0-521-43464-5</ref>

			if

			:<math>\sup_x\Big(\int_Y\|K(x,y)\|^r\,dy\Big)^{1/r} + \sup_y\Big(\int_X\|K(x,y)\|^r\,dx\Big)^{1/r}\le C,</math>

			where <math>r\,</math> satisfies <math>\frac 1 r=1-\Big(\frac 1 p-\frac 1 q\Big)</math>, for some <math>1\le p\le q\le\infty</math>, then the operator <math>Tf(x)=\int_Y K(x,y)f(y)\,dy</math> extends to a continuous operator <math>T:L^p(Y)\to L^q(X)</math>, with <math>\Vert T\Vert_{L^p\to L^q}\le C.</math>

			==Proof==

			Using the [[Cauchy–Schwarz inequality]] and the inequality (1), we get:

			:<math>
			\begin{align} \|Tf(x)\|^2=\left\|\int_Y K(x,y)f(y)\,dy\right\|^2
			&\le \left(\int_Y K(x,y)q(y)\,dy\right)
			\left(\int_Y \frac{K(x,y)f(y)^2}{q(y)} dy\right)\\
			&\le\alpha p(x)\int_Y \frac{K(x,y)f(y)^2}{q(y)} \, dy.
			\end{align}
			</math>

			Integrating the above relation in <math>x</math>, using [[Fubini's Theorem]], and applying the inequality (2), we get:

			:<math> \Vert T f\Vert_{L^2}^2
			\le \alpha \int_Y \left(\int_X p(x)K(x,y)\,dx\right) \frac{f(y)^2}{q(y)} \, dy
			\le\alpha\beta \int_Y f(y)^2 dy =\alpha\beta\Vert f\Vert_{L^2}^2. </math>

			It follows that <math>\Vert T f\Vert_{L^2}\le\sqrt{\alpha\beta}\Vert f\Vert_{L^2}</math> for any <math>f\in L^2(Y)</math>.

			==See also==

			* [[Hardy–Littlewood–Sobolev inequality]]

			==References==
			<references />

			[[Category:Inequalities]]

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2012-07-13T07:54:45Z

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Hello and welcome. My title is Numbers Wunder. To do aerobics is a thing that I'm completely addicted to. For years he's been residing in North Dakota and his family enjoys it. Managing people is his profession.<br><br>Feel free to surf to my web-site - std testing at home ([http://blogzaa.com/blogs/post/9199 mouse click the up coming webpage])

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en>Heron: definition was entirely wrong; rewrote

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