Let $f , g$ be two non-negative functions in $L^1(\mathbb{R})$ such that $$ m(\{x \in \mathbb{R} : f(x) > 0\}) , m(\{x \in \mathbb{R} : g(x) > 0\}) > 0\mbox{.} $$ Show that exists a measurable set $A \subset \mathbb{R}$, with $m(A) > 0$, such that $f*g > 0$ on $A$. Begining, I though $A$ could have the form of $$ A_1 = \{x \in \mathbb{R} : f(x) > 0\} \cap \{x \in \mathbb{R} : g(x) > 0\} $$ or $$ A_2 = \{x \in \mathbb{R} : f(x) g(x) > 0\} $$ but $A_1$ or $A_2$ can have measure zero; in fact, take $f = {\chi}_{(0 , 1)}$ and $g = {\chi}_{(- 1 , 0)}$. Can you help me to find $A$? Thank you very much.
The convolution is positive in a set
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real-analysis
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0Yes I forgot write this in the question thank you I change the enunciate – 2017-01-27
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0Look at translations of one of the functions and define a parameter dependent family of $A$s – 2017-01-27
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0But if $f(x) > 0$, can happen $f(x - y) = 0$. – 2017-01-27
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0That's not a problem. Look at the counter example to your approach that you gave and see what happens when you translate one of your functions. – 2017-01-27
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2Look at $$\int_{\mathbb{R}} (f\ast g)(x)\,dx.$$ – 2017-01-27
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0@Daniel that's clever. I'm not sure if OP has Tonelli yet though. – 2017-01-27
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0What is OP? and what can I do with this integral? – 2017-01-27
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0Do you have a theorem for interchanging integrals? – 2017-01-27
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0Tonelli and Fubini of course. – 2017-01-27
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0Oh goodie. Use Tonelli on the integral Daniel gave. Note that convolution is given by an integral so you'll end up changing the order. – 2017-01-27
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0In this case i would use Tonelli. But I don't obtain anything later using Tonelli theorem on double integral. – 2017-01-27