Fourier transform as diagonalization of convolution

Question

I've read this in a lot of places but never quite got how this is true or meant. Let's say we have a convolution Operator

$$ A_f(g) = \int f(\tau)g(t-\tau)d\tau $$

and apply it to $g(t)=e^{ikt}$. Then we get

$$ A_f(e^{ikt}) = \mathcal F(f) e^{ikt} $$ or $$ A_f(g) = \mathcal F(f) g(t) $$

I recognize that this can be interpreted as: The convolution operator $A_f$ has the eigenfunction $e^{ikt}$ and the eigenvalue $\mathcal F(f)$. But I don't get the jump from here to "FT diagonalizes a convolution". Usually a Matrix can be diagonalized with the eigenvectors, but I don't get how this applies here. I'm a physics student, so keep it human

Answer 1

Diagonalization is synonymous with finding a basis of eigenvectors. If you have a diagonal matrix, the elements of the standard basis are all clearly eigenvectors, and conversely, if you write the matrix in a basis of eigenvectors it will be diagonal. The exponentials form a basis of eigenvectors,so in this sense they diagonalize convolution.

Here follows a silly piece of heuristics that might help make that less precise.

In a sense, when we think of a function $g(t)$ as specified by its values at each $t$, we're putting it in a basis of $\delta$-functions: $$``g=\sum_t g(t)\delta_t".$$ Here $\delta_t$ is a delta-function supported at $t$ ($\delta_t(x)=\delta(t-x)$), and the $\delta_t$s are be regarded as basis vectors. The $g(t)$ are to be regarded as components with respect to this basis. (Since you're a physics student, it might help to say that these basis vectors are like the $|x\rangle$ basis in QM). Evaluating this at $x$, you'd more normally see it written as $$ g(x)=\int g(t)\delta(x-t) dt. $$

An alternative basis from the $\delta$-functions are exponentials: let's write them as $e_k$, where $e_k(x)=e^{2\pi ikx}$ (your convention may differ). The components in this basis are given by the Fourier transform of $g$: $$ ``g=\sum_k\hat{g}(k)e_k" $$ ($|p\rangle$ basis in QM!) or, more normally $$ g(x)=\int \hat{g}(k) e^{2\pi ikx}dk. $$

So, the Fourier transform gives the components in the basis of exponentials, that is the basis of convolution eigenvectors.

Answer 2

A basis $\{ e_{1},e_{2},\cdots,e_{N} \}$ of an $N$-dimensional vector space $X$ can be thought of as a way to transform a vector $x \in X$ into a coefficient function $\hat{x} : \{ 1,2,3,\cdots,N \}\rightarrow X$, where the inverse map is $(\hat{x})^{\vee} = \hat{x}(1)e_{1}+\cdots+\hat{x}(N)e_{N}=x$. A linear map $L$ on $X$ is diagonalized by the basis $\{e_{1},\cdots,e_{N}\}$ iff $\widehat{Ax}=a\hat{x}$ where $a : \{1,2,3,\cdots,N\}\rightarrow\mathbb{C}$ is a scalar function and $(a\hat{x})(n)=a(n)\hat{x}(n)$ for $1 \le n \le N$. That is, diagonalization of a linear operator turns that operator into a scalar multiplier on the functions $\hat{x}$. If $\{ e_{1},e_{2},\cdots,e_{N}\}$ is an orthonormal basis, then $\hat{x}$ has an explicit representation as an inner-product $\hat{x}(n)=(x,e_{n})$. In this case, $A$ has an orthonormal basis of eigenvectors $\{ e_{1},\cdots,e_{N}\}$ if $A$ is transformed to a multiplier; this is the case for Hermitian matrices $A$. The transform $x\mapsto \hat{x}$ is unitary in the case of an orthonormal basis; that is, inner-product is preserved as $(x,y)=\sum_{n}\hat{x}(n)\hat{y}(n)^{\star}$ or, in other words, $\|x\|^{2} = \sum_{n}|\hat{x}(n)|^{2}$.

That's what the discrete Fourier transform does to $-i\frac{d}{dt}$ on $L^{2}[0,2\pi]$: it turns differentiation into a scalar multiplier on the fourier coefficient functions $\hat{x}$. In fact, $\widehat{-ix'}=a\hat{x}$, where the multiplier function is $a(n)=n$, which means $\widehat{-ix'}(n)=a(n)\hat{x}(n)$. This is a diagonalization of the differentiation operator. This discrete transform is a unitary change of basis because $(x,y)_{L^{2}}=\sum_{n}\hat{x}(n)\hat{y}(n)^{\star}$; equivalently, $\|x\|_{L^{2}}^{2}=\sum_{n}|\hat{x}(n)|^{2}$. So the integral inner-product on $[0,2\pi]$ is transformed to an inner-product on $l^{2}(\mathbb{Z})$. The Fourier transform is $\hat{x}(n)=(x,e_{n})_{L^{2}}$ where $e_{n}=\frac{1}{\sqrt{2\pi}}e^{int}$ diagonalizes $-i\frac{d}{dt}$ with an orthonormal basis of eigenfunctions.

The Fourier transform is a continuous version of the discrete one. This is the continuous version of diagonalization: $$ \widehat{-ix'}(s) = a\hat{x},\;\;\; a(s)=s. $$ This is analogous to the finite-dimensional case at the beginning, except that the sum is replaced by an integral. But the idea is the same: you've chosen a 'basis' where the linear operator becomes a scalar multiplier; the scalar multiplier function is thought of as an eigenvalue multiplier by analogy with the finite-dimensional case. In some sense, you can think of the Fourier transform as a type of inner-product with respect to a continuous distribution of eigenfunctions: $$ \hat{x}(s) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}x(t)e^{-ist}\,dt = (x,e_{s}). $$ You still have the unitary identity: $$ (x,y)_{L^{2}(\mathbb{R})}= \int_{-\infty}^{\infty}\hat{x}(s)\hat{y}(s)^{\star}\,ds=\int_{-\infty}^{\infty}(x,e_{s})(y,e_{s})^{\star}\,ds, $$ or, equivalently, $$ \|x\|^{2}_{L^{2}} = \int_{-\infty}^{\infty}|\hat{x}(s)|^{2}\,ds. $$ In this context, $-i\frac{d}{dt}$ becomes the multiplier $a$ where $a(s)=s$. The operator of convolution by $F$ is diagonalized in this same basis because $\widehat{(F\star x)}=a\hat{x}$, where $a(s)=\hat{F}(s)$. This is the formula you wrote: $e_{s}$ is an eigenfunction of convolution with eigenvalue multiplier $\hat{F}(s)$.