# 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.

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