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 [tex]L^2[/tex] version of the Fourier inversion theorem.

We start by noting the following important result:

[tex]\int_{-\infty}^{\infty} e^{itx} e^{-\frac{1}{2}x^2} = \sqrt{2 \pi} e^{-\frac{1}{2}t^2}[/tex]

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

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

This satisfies:

[tex]h_n – xh_n = -h_{n+1}[/tex]

So taking the Fourier transform we get

[tex]ix \hat{h_n} – i \frac{d}{dx} \hat{h_n} = -\hat{h_{n+1}}[/tex]

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

Hence we have that [tex]\hat{h_n} = (-i)^n h_n[/tex].

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

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

Now, [tex]h_n[/tex] is odd iff n is odd and even iff n is even. i.e. [tex]h_n(-x) = (-1)^n h_n[/tex]

Thus [tex]\hat{\hat{ h_n(x)} } = (-1)^n h_n(-x)[/tex].

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

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

But this is just an application of the density of the polynomial functions in [tex]L^2[a, b] [/tex]: pick a big enough interval so that the integral of [tex]|f(x)|^2[/tex] over that interval is within [tex]\epsilon^2[/tex] of [tex]||f||^2[/tex], and this shows that the integral of [tex]|f(x)|^2[/tex] over that interval is [tex]0[/tex]. Thus [tex]||f||_2 < \epsilon[/tex], which was arbitrary, hence [tex]||f||_2 = 0[/tex]. (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 [tex]L^2[/tex] 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 [tex]L^2[/tex] Fourier theory is very pretty and cool – this doesn’t really convert well to a proof of the [tex]L^1[/tex] case, which is in many ways the more important one.

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