David R. MacIver's Blog

Silly proofs 2

I swear this was supposed to be Silly proofs three, but obviously my memories of having done two silly proofs are misleading.

This proof isn’t actually that silly. It’s a proof of the $L^2$ version of the Fourier inversion theorem.

We start by noting the following important result:

$\int_{-\infty}^{\infty} e^{itx} e^{-\frac{1}{2}x^2} = \sqrt{2 \pi} e^{-\frac{1}{2}t^2}$

Thus if we let $h_0 = e^{-\frac{1}{2}x^2}$ then we have $\hat{h_0} = h_0 $ (where $\hat{f}$ denotes the fourier transform of $f$)

Let $h_n(x) = (-1)^n e^{x^2/2} \frac{d^n}{dx^n} e^{-x^2}$

This satisfies:

$h_n - xh_n = -h_{n+1}$

So taking the Fourier transform we get

$ix \hat{h_n} - i \frac{d}{dx} \hat{h_n} = -\hat{h_{n+1}}$

So, $h_n$ and $(-i)^n h_n^$ satisfy the same recurrence relation. Further $\hat{h_0} = h_0$

Hence we have that $\hat{h_n} = (-i)^n h_n$.

Now, the functions $h_n$ are orthogonal members of $L^2$, and so form an orthonormal basis for their closed span.

On this span we have the map $h \to \hat{h}$ is an isometric linear map with each $h_n$ an eigenvector. Further $\hat{\hat{h_n}} = (-1)^n h_n$. Thus the fourier transform is a linear isometry from this space to itself.

Now, $h_n$ is odd iff n is odd and even iff n is even. i.e. $h_n(-x) = (-1)^n h_n$

Thus $\hat{\hat{ h_n(x)} } = (-1)^n h_n(-x)$.

And hence $\hat{\hat{h}}(x) = (-1)^n h(-x)$ for any $h$ in the span. As both sides are continuous, it will thus suffice to show that the span of the $h_n$ is dense.

Exercise: The span of the $h_n$ is precisely the set of functions of the form $p(x) e^{- \frac{1}{2} x^2 }$, where $p$ is some polynomial.
It will thus suffice to prove the following: Suppose $f$ is in $L^2$ and $\int x^n e^{-\frac{1}{2}x^2 } f(x) dx = 0$ for every x. Then $f = 0$.

But this is just an application of the density of the polynomial functions in $L^2[a, b] $: pick a big enough interval so that the integral of $|f(x)|^2$ over that interval is within $\epsilon^2$ of $||f||^2$, and this shows that the integral of $|f(x)|^2$ over that interval is $0$. Thus $||f||_2 < \epsilon$, which was arbitrary, hence $||f||_2 = 0$. (Note: When editing this for the new blog site I noticed that this proof is wrong. I haven’t been able to fix it yet, but will update this when I do).
I’ve dodged numerous details here, like how the $L^2$ Fourier transform is actually defined, but this really can be turned into a fully rigorous proof - nothing in this is wrong, just a little fudged. The problem as I see it is that - while the $L^2$ Fourier theory is very pretty and cool - this doesn’t really convert well to a proof of the $L^1$ case, which is in many ways the more important one.