9
$\begingroup$

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.

$\endgroup$
6
  • 1
    $\begingroup$ Please what are the hypotheses on $f,g$? $\endgroup$ Dec 22 '14 at 20:17
  • $\begingroup$ They are both continuous, piecewise smooth, and absolutely integrable functions. $\endgroup$
    – keyan3d
    Dec 22 '14 at 20:21
  • $\begingroup$ Ok, so they are in $L^1(\mathbb{R})$. You may use Fubini-Tonelli Theorem. $\endgroup$ Dec 22 '14 at 20:22
  • 1
    $\begingroup$ 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 '14 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 '14 at 23:47
10
$\begingroup$

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 $$ \Vert f*g \Vert _1=\!\int_{\mathbb{R}}\!|(f*g)(x)|\,dx=\!\int_{\mathbb{R}}\left|\int_{\mathbb{R}}F(x,\xi)d\xi\right|dx\leq \!\!\int_{\mathbb{R}}\!\left(\int_{\mathbb{R}}|F(x,\xi)|d\xi\right)\!dx=\Vert f\Vert _1\Vert g\Vert _1. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.