Product of two locally integrable function is locally integrable?

Question

I have to solve this question : Let f and g be two locally summable functions (f and g $\in L^1_{loc}$), show that $\int_0^x | f(t)g(x-t) |dt < +\infty$

I was thinking of showing that the function $fg \in L^1_{loc}$, but according to this answer, the product of 2 Lebesgue integrable functions is Lebesgue integrable only if at least one of the functions is bounded.

But how can I know that it is the case here ?

How can I prove above inequality otherwise ?

Thanks for any help you can provide.

Peter

Answer 1

I suppose you are taking $x \geq 0$. You can only prove the result for almost all $x$, not for every $x$.

Let $f_1(x)=f(x)$ for $x \geq 0%$ and $0$ for $x <0$. Simailarly, $g_1(x)=g(x)$ for $x \geq 0$ and $0$ for $x <0$. Then $f_1$ and $g_1$ are interable functions and you want to prove that convolution of two integrable functions is finite almost everywhere. This is a standard application of Fubini's Theorem. Ref: Rudin's RCA.

$f(x)=g(x)=\frac 1 {\sqrt x}$ for $0<x<1$, $0$ elsewhere gives two locally integrable functions whose product is not locally integrable.