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I am trying to prove that for two integrable functions $f,g: \mathbb{R}^n \rightarrow \mathbb{R}$ the function $y \mapsto f(x-y)g(y)$ is integrable for almost every $x$. By using the holder inequality I reduced this to showing that if a function is integrable then also its square is integrable but after browsing a bit I found this so I guess this leads nowhere. Any hints are welcomed.

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  • $\begingroup$ This is called the convolution. It is very important. Try to derive some of the other properties found at en.wikipedia.org/wiki/Convolution $\endgroup$ Dec 28, 2015 at 3:00
  • $\begingroup$ The function $x\mapsto\sqrt x$ on the interval $(0,1)$ is integrable but its square is not. ${}\qquad{}$ $\endgroup$ Dec 28, 2015 at 3:44

1 Answer 1

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Hint: consider $\int\int |f(x-y)g(y)| \, dy \, dx$, and use Tonelli's theorem to reverse the order of integration. If you can show this integral is finite, then $\int |f(x-y)g(y)| \, dy$ is finite for almost every $x$.

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  • $\begingroup$ Tonelli says $$\int\int |f(x-y)g(y)| \, dy \, dx = \int\int |f(x-y)g(y)| \, dx \, dy.$$ The inside integral is plainly finite if $f$ is integrable and $g(y)$ is finite. But how does that imply the outside integral is finite? ${}\qquad{}$ $\endgroup$ Dec 28, 2015 at 6:42
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    $\begingroup$ Once you've switched the order of integration, $|g(y)|$ is just a constant in the inner integral, so you may bring it outside the (inner) integral (but still inside the outer integral)... $\endgroup$
    – kccu
    Dec 28, 2015 at 13:43
  • $\begingroup$ But you still have the factor $y\mapsto \int |f(x-y)|\,dx$, so the integrability of $g$ is not enough at that point. ${}\qquad{}$ $\endgroup$ Dec 28, 2015 at 17:00
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    $\begingroup$ You also need to use the translation-invariance of the Lebesgue integral and the integrability of $f$. $\endgroup$
    – kccu
    Dec 28, 2015 at 20:47
  • $\begingroup$ aha! I guess I was distracted by other aspects of the problem. ${}\qquad{}$ $\endgroup$ Dec 28, 2015 at 22:27

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