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Recall that the convolution of two functions is given by $$f*g(y)=\int f(x)g(y-x)dx.$$ The well known inequality known as Young's inequality, say that $$\|f*g\|_r\leq\|f\|_p\cdot\|g\|_q $$ provided $\frac 1p + \frac 1q = 1 + \frac 1r$ and $1\le p,q,r\le\infty$. Obvious implications is that

  • $L^1$ is a Banach algebra
  • $L^\infty$ is an $L^1$-module with respect to convolution

Do you have any deeper applications/examples?

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    Could I convince you to write $(f*g)(y)$ rather than just $f*g$ on the definition (but obviously not in the inequality)?2011-11-17
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    @MichaelHardy Thanks!2011-11-17
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    You could make that $L^\infty$ an $L^r$, of course.2011-11-17
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    @t.b. Sure, thanks and yes - I just took the first examples that popped up in my head. ;)2011-11-17
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    Could I convince you to write $(f\star g)(y)$ instead of $f\star g(y)$? I have always disliked this notation, just as $f\circ g(x)$ instead of $(f\circ g)(x)$.2011-11-17
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    @wildildildlife Well, it is a matter of taste I agree, I am used to $f*g(x)$. (Could I convince you to write $*$ instead of $\star$?)2011-11-18

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