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_{-inf}^{inf} e^{itx} e^{-x^2/2} = sqrt{2 pi} e^{-t^2/2}

Thus if we let h_0 = e^{-x^2}/2 then we have h_0^ = h_0 (where ^ denotes the fourier transform)

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

This satisfies:

h_n’ – xh_n = -h_{n+1}

So taking the Fourier transform we get

ix h_n^ – i (h_n^)’ = -h_{n+1}^

So, h_n and (-i)^n h_n^ satisfy the same recurrence relation. Further h_0^ = h_0.

Hence we have that 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 span.

On this span we have the map h -> h^ is a linear map with each h_n an eigenvector. Further 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 h_n^^(x) = (-1)^n h_n(-x)

And hence 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^{-x^2 / 2}

It will thus suffice to prove the following: Suppose f is in L^2 and Int x^n e^{-x^2 / 2} f(x) dx = 0 for every x. Then f = 0.

But this is just an easy 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). 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 the more important one.