Operator norm of a convolution

Question

Consider the operator on $L^2(\Bbb R)$, $f\rightarrow f*g$, where $g\geq 0$ is some $L^1$ function. Show the operator is a bounded linear operator with operator norm equal to $||g||_1$.

Showing this actually is a linear operator was not hard, and the operator norm is less than $||g||_1$ by Young's inequality, but I have had no luck trying to show the operator norm is actually $||g||_1$.

@julien How do I guarantee convergence in $L^p$ with some approximation? I know how I can guarantee convergence point wise almost everywhere, but that's it. — PVAL-inactive, Nov 03 '13 at 22:00
@mike I don't see any approach related to the fourier transform... — PVAL-inactive, Nov 04 '13 at 01:52

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

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This question Limit of convolution answers the question for general convolution operator on $L^p$.

edited Apr 13 '17 at 12:20

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answered Apr 13 '15 at 03:11

digiboy1

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Operator norm of a convolution

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