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I guess that this is an easy question, but I don't have a very solid math background.

I'm trying to prove that if $f,g \in L^1(\mathbb{R})$, then $h = f \star g \in L^1(\mathbb{R})$.

So, I have:

$ h(t) = \int_{-\infty}^\infty f(u)g(t-u)du$

Then I do:

\begin{eqnarray*} \int_{-\infty}^\infty |h(t)|dt &=& \int_{-\infty}^\infty \left|\int_{-\infty}^\infty f(u)g(t-u)du\right|dt \\ &\leq& \int_{-\infty}^\infty \int_{-\infty}^\infty |f(u)| |g(t-u)|du dt \end{eqnarray*}

And here comes the question. I need to apply Fubini's theorem so I can change the order of integration. But to prove the conditions of applicability of Fubini's, I have to integrate with respect to $t$ first. This sounds circular to me and I can't find a way to do this that convinces me.

Thanks.

Franco Victorio
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1 Answers1

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Hint: Use Fubini-Tonelli Theorem instead of Fubini's Theorem, which only requires that your integrant is non-negative, and has the same conclusion as ordinary Fubini's Theorem.

The statement of Fubini-Tonelli Theorem is:

Let $f:\mathbb{R}^d\to [0,\infty]$ be measurable, and $x\mapsto \int f_x\,d\lambda_2$ and $y\mapsto \int f^y\,d\lambda_1$ are Borel measurable. Then $$\int_{\mathbb{R}^{d_1}}\bigg(\int_{\mathbb{R}^{d_2}}f_x\,d\lambda_2\bigg)\,d\lambda_1=\int_{\mathbb{R}^{d_2}}\bigg(\int_{\mathbb{R}^{d_1}}f^y\,d\lambda_1\bigg)\,d\lambda_2=\int_{\mathbb{R}^{d_1+d_2}}f\,d\lambda.$$

OnoL
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