I am stuck on proving the convolution theorem for the product of three functions using the Dirac delta function. Please excuse any nonstandard notation--I am a physics major who has not been formally trained in the convolution theorem.

So, the question:

Let's call them f(x), g(x) and h(x), and let the transform be from x-space to k-space. Assume all the functions are continuous, or discontinuous at only finite points.

Per the formalism of the convolution theorem, we should get the following:

\begin{split} \mathcal{F}\{f(x)\cdot g(x)\cdot h(x)\} & = \mathcal{F}\{f(x)\}*\mathcal{F}\{g(x)\cdot h(x)\}\\ & = \mathcal{F}\{f(x)\}*\mathcal{F}\{g(x)\cdot h(x)\}\\ & = \mathcal{F}\{f(x)\}*(\mathcal{F}\{g(x)\}*\mathcal{F}\{h(x)\})\\ %& =\int^\infty_{-\infty}\tilde{f}(k')\cdot \mathcal{F}\{(g\cdot h)(x)\}[k'-k]dk'\\ %& =\int^\infty_{-\infty}\tilde{f}(k')\cdot \mathcal{F}\{(g\cdot h)(x)\}[k'-k]dk' \end{split}

But if we actually carry out the transform to prove this, using dirac delta, we actually get this:

\begin{split} \mathcal{F}\{f(x)\cdot g(x)\cdot h(x)\} & = \int\limits^\infty_{-\infty}f(x)\cdot g(x)\cdot h(x) \cdot e^{-i k' x}dx\\ & = \int\limits^\infty_{-\infty}\{ \int\limits^\infty_{-\infty}\tilde{f}(k_1) e^{i k_1 x}\frac{dk_1}{2\pi} \int\limits^\infty_{-\infty}\tilde{g}(k_2) e^{i k_2 x}\frac{dk_2}{2\pi} \int\limits^\infty_{-\infty}\tilde{h}(k_3) e^{i k_3 x}\frac{dk_3}{2\pi}\} \cdot e^{-i k' x}dx\\ & = \int\limits^\infty_{-\infty} \int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty} \tilde{f}(k_1)\tilde{g}(k_2)\tilde{h}(k_3) e^{i x (k_1+k_2+k_3-k')} \frac{dk_1dk_2dk_3dx}{(2\pi)^3}\ \end{split}

A definition of Dirac delta is this, from Wikipedia:

\begin{equation} \delta (x-\alpha )={\frac {1}{2\pi }}\int \limits_{-\infty }^{\infty }e^{ip(x-\alpha )} \ dp\end{equation}


\begin{equation} f(t-T)=\int \limits _{-\infty }^{\infty }f(\tau )\delta (\tau -(t-T))d\tau = \int \limits _{-\infty }^{\infty }f(\tau ) \int \limits _{-\infty }^{\infty }\frac{1}{2\pi}e^{i x (\tau+T-t)}dx d\tau \end{equation}

which means

\begin{split} & \int\limits^\infty_{-\infty} \int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty} \tilde{f}(k_1)\tilde{g}(k_2)\tilde{h}(k_3) e^{i x (k_1+k_2+k_3-k')} \frac{dk_1dk_2dk_3dx}{(2\pi)^3}\\\ = & \int\limits^\infty_{-\infty} \int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty} \tilde{f}(k_1)\tilde{g}(k_2)\tilde{h}(k_3) e^{i x (k_2+k_1-k')}e^{i x k_3} \frac{dk_1dk_2dk_3dx}{(2\pi)^3}\\ = & \int\limits^\infty_{-\infty} \int\limits^\infty_{-\infty} \tilde{f}(k_1)\tilde{g}(k'-k_1)\tilde{h}(k_3) e^{i x k_3} \frac{dk_1dk_3}{(2\pi)^2}. \end{split}

Since there is now no $dx$ term, it is impossible to use the Dirac delta function further. But then we'd not be able to possibly obtain what we had above. So this surely is incorrect somewhere?

Especially, if we had treated $g(x)\cdot h(x)$ as one function $j(x)$, then we'd get

\begin{split} & \mathcal{F}\{f(x)\cdot g(x)\cdot h(x)\} = \mathcal{F}\{f(x)\cdot j(x)\}\\ & = \int\limits^\infty_{-\infty}f(x)\cdot j(x) \cdot e^{-i k' x}dx\\ & = \int\limits^\infty_{-\infty} \int\limits^\infty_{-\infty}\tilde{f}(k_1) e^{i k_1 x}\frac{dk_1}{2\pi} \int\limits^\infty_{-\infty}\tilde{j}(k_2) e^{i k_2 x}\frac{dk_2}{2\pi} \cdot e^{-i k' x}dx\\ & = \int\limits^\infty_{-\infty} \int\limits^\infty_{-\infty} \int\limits^\infty_{-\infty}\tilde{f}(k_1)\tilde{j}(k_2) \cdot e^{i x (k_2-(k'-k_1))}\frac{dk_1dk_2dx}{(2\pi)^2}\\ & = \int\limits^\infty_{-\infty} \tilde{f}(k_1)\tilde{j}(k'-k_1) \frac{dk_1}{2\pi}\\ & = \int\limits^\infty_{-\infty} \tilde{f}(k'-k_1)\tilde{j}(k_1) \frac{dk_1}{2\pi}\\ & = \int\limits^\infty_{-\infty} \tilde{f}(k'-k_1)\{ \int\limits^\infty_{-\infty} \int\limits^\infty_{-\infty}\tilde{g}(k_a)e^{i k_a x''} \int\limits^\infty_{-\infty}\tilde{h}(k_b)e^{i k_b x''} e^{-i k'' x''} dx'' \}\frac{dk_1}{2\pi}\\ & = \int\limits^\infty_{-\infty} \tilde{f}(k'-k_1)\{ \int\limits^\infty_{-\infty} \tilde{g}(k_a)\tilde{h}(k''-k_a) \frac{dk_a}{2\pi} \}\frac{dk_1}{2\pi}\\ & = \int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty} \tilde{f}(k'-k_1)\tilde{g}(k_a)\tilde{h}(k''-k_a) \frac{dk_a}{2\pi}\frac{dk_1}{2\pi}\\ \end{split}

Which would be correct, and thus different from what would be obtained from just directly inverse-transforming all three functions individually. Essentially we have gotten $\mathcal{F}\{f(x)\cdot g(x)\cdot h(x)\} \neq \mathcal{F}\{f(x)\cdot g(x)\cdot h(x)\},$ which is of course not possible, right?

  • $\begingroup$ Go back and check your differentials. I'm fairly certain you change order of integration without justification, and that you're using $dx$ multiple times within the same integral. $\endgroup$ – adfriedman Mar 8 '17 at 1:12
  • $\begingroup$ @user153126 Could you maybe point out where you think there is a mistake? The only place where I change the order of integration at all is when applying the Dirac delta, and there I think that is fine...? And I would only use $dx$ once ever, which is why I encounter a problem when directly inverse-expanding each function--because I run out of $dx$ to use. In the second case I simply do the Fourier expansion again for $j(x)$ ($j(k_1)$ is really the Fourier expansion of $j(x)$), which is how it generates a new dx for me to use, and how I manage to escape that problem. $\endgroup$ – Rethliopuks Mar 8 '17 at 22:33
  • $\begingroup$ Which brings my problem: why would two mathematically equally legitimate and sensible routes (it seems) to apply the same method yield two different results? That would violate the law of identity so there has to be a mistake somewhere. Specifically, by the result obtained via the formalism of the convolution theorem, the mistake must be in the first one--but I can't figure it out. $\endgroup$ – Rethliopuks Mar 8 '17 at 22:41

The problem in the proof is where you claim that \begin{align} \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \tilde{f}(k_1)\tilde{g}(k_2)\tilde{h}(k_3) e^{ix(k_1+k_2-k')} e^{ixk_3}\frac{dk_1dk_2dk_3dx}{(2\pi)^3}\\ = \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} \tilde{f}(k_1)\tilde{g}(k'-k_1)\tilde{h}(k_3)e^{ixk_3} \frac{dk_1 dk_3}{(2\pi)^2} \end{align}

You have somehow pulled $e^{ixk_3}$ out of the integral over $x$. This would be like claiming $\int x^2 \;dx = \int x\cdot x\;dx = x\int x dx$.

In fact, you don't need the Dirac delta here at all. Given that you know the definitions of the Fourier and inverse Fourier

\begin{align} \mathcal{F}\{f(x)g(x)h(x)\}(k) &= \int\limits_{-\infty}^{\infty} f(x)g(x)h(x) e^{-ikx} dx\\ &= \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} \mathcal{F}\{g\cdot h\}(k_1)e^{i k_1x} \frac{d k_1}{2\pi}f(x) e^{-ikx} dx\\ &= \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} \mathcal{F}\{g\cdot h\}(k_1) \,f(x)\,e^{ik_1x - ik x} \frac{d k_1 dx}{2\pi} \\ &\overset{(*)}{=} \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} \mathcal{F}\{g\cdot h\}(k_1) \,f(x)\,e^{-ix (k- k_1)} \frac{d x \,dk_1}{2\pi} \\ &= \int\limits_{-\infty}^{\infty} \mathcal{F}\{g\cdot h\}(k_1) \int\limits_{-\infty}^{\infty} \,f(x)\,e^{-ix (k- k_1)} \frac{d x }{2\pi} dk_1 \\ &= \int\limits_{-\infty}^{\infty} \mathcal{F}\{g\cdot h\}(k_1) \mathcal{F}\{f\}(k-k_1) dk_1 \\ &= \left( \mathcal{F}\{f\} * \mathcal{F}\{g\cdot h\} \right)(k) \end{align}

and we may then finish by applying the same process again to $\mathcal{F}\{g\cdot h\}$.

Note that the bounds of integration being swapped at $(*)$ is not always possible. Fubini's Theorem gives a sufficient condition. For instance, it holds if $f,g,h$ satisfy $$ \int\limits_{-\infty}^{\infty} |f(x)|dx < \infty,\quad \int\limits_{-\infty}^{\infty} |g(x)|dx < \infty,\quad\text{and} \int\limits_{-\infty}^{\infty} |h(x)|dx < \infty$$


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