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What I'm trying to prove is that for $f_1,f_2 \in \mathcal{L}_1$ the function $y \mapsto f_1(y)f_2(x-y)$ is integrable for almost all $x$, or:

$$ F(x) = \int f_1(y)f_2(x-y)\,dy < \infty \text{ almost all }x.$$

I've already shown that the convolution is integrable, but thanks to Fubini thats easy. Here there are less tools to use. I've also looked at Hölder's inequality, but nothing seems to quite work.

Any help/tips would be much appreciated!

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I wonder if you can get by with an assumption that $f_1\in\mathcal{L}_1$ and much weaker assumptions on $f_2$? (I hesitate to suggest that it's enough to assume $f_2$ is measurable.) –  Michael Hardy Dec 16 '11 at 2:02
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1 Answer 1

Hint: look up Young's Inequality. I posted a proof here.

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From what I see Young's Inequality gives a bound on the measure of the convolution, but I want to show the convolution itself is bounded for almost all x. –  BallzofFury Dec 15 '11 at 20:51
    
an integrable function $F(x)$ is finite for almost all $x$. If the integral for $F(x)$ didn't converge for a set of positive measure, then $F(x)$ would not be integrable. –  robjohn Dec 15 '11 at 21:43
    
Awsome! I was just trying to figure out something along the lines of $\iint f_1*f_2\,dy\,dx < \infty \Rightarrow \int f_1*f_2\,dy < \infty$. –  BallzofFury Dec 15 '11 at 21:53
    
(Much appreciated) –  BallzofFury Dec 15 '11 at 21:58
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