Confusion on the proofs of convolution theorem of Fourier Transform.

Question

The convolution theorem of Fourier transform is stated as follows:

Define $h(x):=f(x)*g(x)$, then we have $$\hat{h}(k)=\hat{f}(k)\hat{g}(k).$$

I have a confusion of the proofs of this theorem. Most proofs go in the following way:

Define $h(x):=f(x)*g(x)$. By definition, we have $$\hat{h}(k)=\int_{-\infty}^{\infty}h(x)e^{-ikx}dx.$$ Then, plugging in the integral expression of $f*g$ into the above integral yields us \begin{align*} \hat{h}(k)&=\int_{-\infty}^{\infty}h(x)e^{-ikx}dx\\ &=\int_{-\infty}^{\infty}\Big(\int_{-\infty}^{\infty}f(x-s)g(s)ds\Big)e^{-ikx}dx\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-s)g(s)e^{-ikx}dsdx.\ \ \ \ \ \ (*) \end{align*}

Then, most proofs directly change integral order in the question $(*)$, so that

\begin{align*} \hat{h}(k)&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-s)g(s)e^{-ikx}dsdx\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-s)g(s)e^{-ikx}dxds\\ &=\int_{-\infty}^{\infty}\Big(\int_{-\infty}^{\infty}f(x-s)e^{-ikx}dx\Big)g(s)ds. \end{align*}

What remains is standard: note that by the shifting property, the inner integral is $$\int_{-\infty}^{\infty}f(x-s)e^{-ikx}dx=\hat{f}(k)e^{-iks},$$ and thus \begin{align*} \hat{h}(k)&=\int_{-\infty}^{\infty}\Big(\int_{-\infty}^{\infty}f(x-s)e^{-ikx}dx\Big)g(s)ds\\ &=\hat{f}(k)\int_{-\infty}^{\infty}g(s)e^{-iks}ds\\ &=\hat{f}(k)\hat{g}(k), \end{align*} where the last equality was obtained by definition. The proof is concluded.

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However, why could we change the integral order of (*) in the first place? Don't we need Fubini? I tried to use Fubini as follows: $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|f(x-s)g(s)e^{-ikx}|dsdx=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|f(x-s)g(s)|dsdx$$ but I do not know how to proceed further. I had expected that I can split the integral so that I would have two $L^{1}-$norm, but since the first integral is $ds$, I could not split the integral..

Do we directly assume the function is nice enough so that we can interchange the integral order? If so, what type of functions do we assume?

Thank you!

Answer 1

Thanks to demonakos' help, it turned out that as long as $f,g\in L^{1}$, the integral order can be interchanged.

The proof is as follows: note that by Tonelli's theorem (in the third equality), \begin{align*} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|f(x-s)g(s)e^{-ikx}|dsdx=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|f(x-s)g(s)|dsdx&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|f(x-s)g(s)|dxds\\ &=\int_{-\infty}^{\infty}\Big(|g(s)|\int_{-\infty}^{\infty}|f(x-s)|dx\Big)ds\\ &=\int_{-\infty}^{\infty}\Big(|f(x-s)\|g\|_{L^{1}(\mathbb{R})})dx\\ &=\|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})}<\infty, \end{align*} the $L^{1}-$norm is finite since $f$ and $g$ were hypothesized to be in $L^{1}$.

Hence, it follows from Fubini's Theorem that we can interchange integration order in $(*)$. That is \begin{align*} \hat{h}(k)&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-s)g(s)e^{-ikx}dsdx\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-s)g(s)e^{-ikx}dxds\\ &=\int_{-\infty}^{\infty}\Big(\int_{-\infty}^{\infty}f(x-s)e^{-ikx}dx\Big)g(s)ds. \end{align*}

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In fact, one could see the Fubini-Tonelli Theorem here: https://en.wikipedia.org/wiki/Fubini%27s_theorem#Tonelli's_theorem_for_non-negative_measurable_function.

Instead of using Tonelli and Fubini seperately, one could directly check $\int\int|f(x-s)g(s)|dxds<\infty$ as above, and using Fubini-Tonelli to directly conclude that the integral order can be interchanged.

Although equivalent, there is still a little bit difference.

Answer 2

Without changing the order of integration:
$ikx=ik(x-s)+iks$ and $u=x-s$ so $h(k)=\int_{-\infty} ^\infty \int_{-\infty} ^\infty f(u)e^{-iku}g(s)e^{-iks}dsdu$$=\hat{f}(k)\hat{g}(k)$.