# Convolution theorem for other transforms

The Fourier transform is an integral transform with turns any function into a superposition of sinusoidal waves. The convolution theorem states the astonishing property that if you convolve two functions, this is equivalent to simply multiplying the corresponding transformed functions. Since the inverse Fourier transform is nearly identical to the Fourier transform, it is perhaps unsurprising that the same holds in the other direction.

Do all integral transforms have this fascinating property? Or is it unique to the Fourier transform? (And if the Fourier transform is the only integral transform in all existence that has this property, why is that?)

-
Correction:"The convolution theorem states the astonishing property that if you convolve two functions, this is equivalent to" should be "The convolution theorem states the Fourier transform of the convolution of two functions is equal to" – AD. Nov 22 '13 at 19:20
You should read about commutative Banach algebras and the Gelfand transform. – AD. Nov 22 '13 at 19:21

The answer depends on how you define convolution. For example, consider (real-valued) vectors $$x = \bigr (x[{\bf 0}], x[{\bf 1}], \ldots x[{\bf 2^n-1}]\bigr)$$ of length $2^n$ with $x[{\bf i}]$ being thought of not as the $i$-th element, $0 \leq i \leq 2^n-1$, but rather as the element in the $\bf i$-th position where $\bf i$ is the $n$-bit standard binary representation of $i$. Then, the term-by-term product of the Walsh-Hadamard transforms $xH$ and $yH$ of $x$ and $y$ respectively is the Walsh-Hadamard transform of their convolution which is defined as $$(x \star y)[\bf k] = \sum_{\bf \ell} x[\bf \ell]y[\bf k \oplus \bf \ell]$$ where the sum is over all binary $n$-tuples $\bf \ell$. Here $H$ is the $2^n\times 2^n$ Hadamard matrix in the Sylvester form.
On the other hand, the Walsh-Hadamard transform does not support the "usual" cyclic convolution $\displaystyle (x\star y)[k] = \sum_\ell x[\ell]y[k-\ell]$ of vectors where $k-l$ would be interpreted modulo $2^n$.