Proof of Convolution Theorem for three functions, using Dirac delta 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}
So
\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?
 A: 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
$$ \int\limits_{-\infty}^{\infty} |f(x)|dx < \infty,\quad
\int\limits_{-\infty}^{\infty} |g(x)|dx < \infty,\quad\text{and}
\int\limits_{-\infty}^{\infty} |h(x)|dx < \infty$$
