There is this proof for the integral of convolution between two functions:

$$\begin{align}\int_{-\infty}^{\infty} (f*g)(x)dx&=\int_{-\infty}^{\infty}\left [ \int_{-\infty}^{\infty}f(x-\xi)g(\xi)d\xi \right ] dx \\&=\int_{-\infty}^{\infty}g(\xi)\left [ \int_{-\infty}^{\infty}f(x-\xi)dx \right ] d\xi \\ &=\int_{-\infty}^{\infty}g(\xi)\left [ \int_{-\infty}^{\infty}f(\eta)d\eta \right ] d\xi\\&=\int_{-\infty}^{\infty}g(\xi) d\xi \int_{-\infty}^{\infty}f(\eta)d\eta\end{align}$$

What confuses me is the way author has easily changed the order of terms under integral sign. I'll appreciate any explanation.

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    $\begingroup$ Please what are the hypotheses on $f,g$? $\endgroup$ Dec 22, 2014 at 20:17
  • $\begingroup$ They are both continuous, piecewise smooth, and absolutely integrable functions. $\endgroup$
    – keyan3d
    Dec 22, 2014 at 20:21
  • $\begingroup$ Ok, so they are in $L^1(\mathbb{R})$. You may use Fubini-Tonelli Theorem. $\endgroup$ Dec 22, 2014 at 20:22
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    $\begingroup$ Thanks a lot, I'm still trying to figure out how you did a similar argument for integration of F with respect to xi. $\endgroup$
    – keyan3d
    Dec 22, 2014 at 23:42
  • $\begingroup$ You may notice that, by an easy change of variable $x-\xi \to u$, you get that $f*g=g*f$, so you can conclude concerning $F$ with respect to $\xi$. $\endgroup$ Dec 22, 2014 at 23:47

1 Answer 1


Hint. You may use Fubini-Tonelli theorem.

Let $f,g \in L^1(\mathbb{R})$. Set $F(x,\xi):=f(x-\xi)g(\xi)$. Then $$\int_{\mathbb{R}}|F(x,\xi)|\,dx=|g(\xi)|\int_{\mathbb{R}}|f(x-\xi)|\,dx=|g(\xi)|\int_{\mathbb{R}}|f(x)|\,dx=|g(\xi)|\: \Vert f\Vert _1$$ giving $$ \int_{\mathbb{R}}\left(\int_{\mathbb{R}}|F(x,\xi)|\,dx\right)d\xi=\int_{\mathbb{R}}\Vert f\Vert _1|g(\xi)|\,d\xi=\Vert f\Vert _1\Vert g\Vert _1 <\infty. $$ In the same manner, $$ \int_{\mathbb{R}}\left(\int_{\mathbb{R}}|F(x,\xi)|\,d\xi\right)dx=\Vert f\Vert _1\Vert g\Vert _1 <\infty $$ Thus using Fubini-Tonelli theorem, we have $\displaystyle F \in L^1(\mathbb{R} \times \mathbb{R})$ and $$ \int_{\mathbb{R} \times \mathbb{R}}|F(x,\xi)|\,d(x,\xi)=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}|F(x,\xi)|\,dx\right)d\xi=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}|F(x,\xi)|\,d\xi\right)dx $$ Moreover, $\displaystyle \Vert f*g \Vert _1 \leq \Vert f\Vert _1\Vert g\Vert _1$ , since $$ \Vert f*g \Vert _1=\!\int_{\mathbb{R}}\!|(f*g)(x)|\,dx=\!\int_{\mathbb{R}}\left|\int_{\mathbb{R}}F(x,\xi)d\xi\right|dx\leq \!\!\int_{\mathbb{R}}\!\left(\int_{\mathbb{R}}|F(x,\xi)|d\xi\right)\!dx=\Vert f\Vert _1\Vert g\Vert _1. $$


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