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 | | From: | daniel.wolff at csfb.com | | Subject: | Integral Inequality | | Date: | 23 Jan 2005 19:47:24 -0800 |
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 | I have seen in the literature the following inequality but have no Idea
where the 4 comes from.
for f,g\inL^2 and \episilon>0 we have
\int_Bigomega |f||g|dx\leq\episilon
||f||^2_{L^2(\Bigomega)}+1/(4\episilon)||g||^2_{L^2(\Bigomega)}.
Now, using Young's inequality, following the Cauchy-Schwarz with weight
\episilon, I have
\int_Bigomega |f||g|dx\leq\episilon
||f||^2_{L^2(\Bigomega)}+1/(2\episilon)||g||^2_{L^2(\Bigomega)}.
But, I cannot demonstrate that an additional factor of 1/2 to the right
term still preserves the inequality.
Any hints?
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| | From: | Robert Israel | | Subject: | Re: Integral Inequality | | Date: | 24 Jan 2005 05:48:07 GMT |
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| In article <1106538444.379612.102000@f14g2000cwb.googlegroups.com>,
wrote:
>I have seen in the literature the following inequality but have no Idea
>where the 4 comes from.
>for f,g\inL^2 and \episilon>0 we have
>
>\int_Bigomega |f||g|dx\leq\episilon
>||f||^2_{L^2(\Bigomega)}+1/(4\episilon)||g||^2_{L^2(\Bigomega)}.
By Cauchy-Schwarz,
int |f| |g| dx <= ||f|| ||g||, and this is of course best possible.
So the only way your inequality can be true is
||f|| ||g|| <= epsilon ||f||^2 + 1/(4 epsilon) ||g||^2
which comes from factoring:
epsilon a^2 - a b + 1/(4 epsilon) b^2 = epsilon (a - b/(2 epsilon))^2 >= 0
where a = ||f||, b = ||g||.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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