# Different proofs for the $L^2$ isometry of the Fourier transform on $\Bbb R$

Over the years I have come across several different proofs for the $L^2$ isometry of the Fourier transform (as it exists as an integral operator on $\Bbb R$). Often the traditional proofs hinge on the use of the Schwartz space or $L^1(\Bbb R)\cap L^2(\Bbb R)$ in conjunction with the convolution theorem. I recently came up with a proof that doesn't use the convolution theorem at all and is instead somewhat of a "pure" $L^2$ approach while still working on $L^1(\Bbb R)\cap L^2(\Bbb R)$. I am sure it is not a new proof, but I will post it as an answer below.

This led me to wonder just how many different fairly well-known proofs there are for the $L^2$ isometry of the Fourier transform on $\Bbb R$. I was hoping that we could collect some of them in a big list. Of course replicating a full proof is not advisable since they can be quite lengthy, but a brief, rough outline would be quite appreciated along with a text or PDF reference for the full proof.

One of the issues with integral transform theory is that proof techniques vary wildly from one integral transform to another. What I am hoping to accomplish with this list is that it will provide a wealth of approaches for showing $L^2$ properties of integral transforms.

• The Hermite function proof of the Plancherel Theorem requires knowing that the hermite functions form a complete orthonormal basis of $L^{2}(\mathbb{R})$, which is non-trivial. However, Norbert Wiener takes this approach (and others) to the Plancherel Theorem in his 1932 book "The Fourier Integral and Certain of Its Applications." – DisintegratingByParts Nov 12 '15 at 21:01
• It's not hard to prove completeness using some complex analysis and Fourier transforms (particularly that it's an $L^2$ isometry), but if you want to avoid that, the proofs are not very fun at all. They are not difficult, but they are very laborious. I found a proof in Orthogonal Functions. – Cameron Williams Nov 12 '15 at 21:03

The proof I came up with for the $L^2$ isometry of the Fourier transform on $L^1(\Bbb R)\cap L^2(\Bbb R)$ hinges on the $L^2$ completeness of the Hermite-Gauss functions. It is well-known that these are eigenfunctions of the Fourier transform however the proofs using the Hermite-Gauss functions immediately jump to an $L^2$ theory and neglect the integral operator aspect of the Fourier transform, i.e. the $L^1(\Bbb R)$ aspect. The convention for the Fourier transform I am using is

$$\mathcal{F}f(k) = \frac{1}{\sqrt{2\pi}}\int_{\Bbb R} e^{-ikx}f(x)\,dx.$$

The Hermite polynomials $H_n$ are given by the recursion relation

$$H_n(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}.$$

The Hermite-Gauss functions are given by $H_n(x)e^{-\frac{x^2}{2}}$. It is not hard to prove that these are eigenfunctions of the Fourier transform with the following recursion relation:

$$\left(x-\frac{d}{dx}\right)\left(H_n(x)e^{-\frac{x^2}{2}}\right) = H_{n+1}(x)e^{-\frac{x^2}{2}}.$$

We proceed by induction on $n$. It is well-known that the Gaussian is an eigenfunction of the Fourier transform with eigenvalue $1$ and this takes care of the base case.

Suppose then that $H_n(x)e^{-\frac{x^2}{2}}$ is an eigenfunction with eigenvalue $(-i)^n$, then we wish to show that $H_{n+1}(x)e^{-\frac{x^2}{2}}$ is also an eigenfunction with eigenvalue $(-i)^{n+1}$. Taking the Fourier transform of this, we get

\begin{align} \int_{\Bbb R} e^{-ikx} H_{n+1}(x)e^{-\frac{x^2}{2}}\,dx &= \int_{\Bbb R} e^{-ikx} \left(x-\frac{d}{dx}\right)\left(H_n(x)e^{-\frac{x^2}{2}}\right)\,dx \\ &= \left(i\frac{d}{dk}-ik\right)\int_{\Bbb R} e^{-ikx}H_n(x)e^{-\frac{x^2}{2}}\,dx \\ &= -i(-i)^n\left(k-\frac{d}{dk}\right)\left(H_n(k)e^{-\frac{k^2}{2}}\right) \\ &= (-i)^{n+1} H_{n+1}(k)e^{-\frac{k^2}{2}}. \end{align}

Thus the Hermite-Gauss functions are eigenfunctions of the Fourier transform with eigenvalues $\pm 1,\pm i$. It remains to see that they are mutually orthogonal. Suppose then that $m < n$.

$$\int_{\Bbb R}H_m(x)e^{-\frac{x^2}{2}} H_n(x)e^{-\frac{x^2}{2}}\,dx = \int_{\Bbb R} H_m(x) \frac{d^n}{dx^n} e^{-x^2}\,dx.$$

Since $H_m$ is an $m$th degree polynomial, an integration by parts $m$ times will leave a constant multiple of an $n-m$th derivative of $e^{-\frac{x^2}{2}}$ and so the integral is zero, i.e. if $m\neq n$, $H_m(x)e^{-\frac{x^2}{2}}$ and $H_n(x)e^{-\frac{x^2}{2}}$ are orthogonal.

Denoting the (normalized) Hermite-Gauss functions by $\psi_n$, from the above property, $\|\mathcal{F}\psi_n\| = \|(-i)^n \psi_n\| = \|\psi_n\|$. As noted above, the Hermite-Gauss functions are complete in $L^2(\Bbb R)$ which can be proved without the Fourier transform (though most modern proofs do use the Fourier transform).

Consider then $f\in L^1(\Bbb R)\cap L^2(\Bbb R)$. On this space, the Fourier transform still acts as an integral operator. Moreover,

$$\int_{\Bbb R^2}|e^{-ikx}f(x)\psi_n(k)|\,dx\,dk < \infty$$

since $f,\psi_n\in L^1(\Bbb R)$. Making use of Fubini's theorem, we have that

$$\int_{\Bbb R} \mathcal{F}f(k)\psi_n(k)\,dx = \int_{\Bbb R} f(x)\mathcal{F}\psi_n(x) = (-i)^n \int_{\Bbb R} f(x)\psi_n(x)\,dx.$$

Thus

$$\sum_{n=0}^{\infty} |\langle \mathcal{F}f,\psi_n\rangle|^2 = \sum_{n=0}^{\infty} |(-i)^n\langle f,\psi_n\rangle|^2 = \|f\|^2 < \infty$$

and so $\mathcal{F}f\in L^2(\Bbb R)$ and moreover $\|\mathcal{F}f\| = \|f\|.$ Thus $\mathcal{F}(L^1(\Bbb R)\cap L^2(\Bbb R))\subseteq L^2(\Bbb R)$ as claimed.

Spectral theory gives a nice proof of the Plancherel Theorem. The operator studied is $L=\frac{1}{i}\frac{d}{dx}$ on $L^{2}(\mathbb{R})$. The resolvent of $L$ is an integral operator $$R(\lambda)f=(L-\lambda I)^{-1}f= \left\{\begin{array}{cc} i\int_{-\infty}^{x}e^{i\lambda(x-u)}f(u)du, & \Im\lambda > 0 \\ -i\int_{x}^{\infty}e^{i\lambda(x-u)}f(u)dt, & \Im\lambda < 0. \end{array} \right.$$ For any $f \in L^{2}$, the function $g=R(\lambda)f$ is the unique absolutely continuous solution of $$\left(\frac{1}{i}\frac{d}{dx}-\lambda I\right)g=f, \;\;\; g \in L^{2}.$$ This can be verified directly.

For any resolvent of a selfadjoint operator, $$\lim_{R\rightarrow\infty}\lim_{s\downarrow 0}\frac{1}{2\pi i}\int_{-R}^{R}\langle R(r+is)f-R(r-is)f,f\rangle dr=\|f\|^{2}$$ In this case, \begin{align} \|f\|^{2} & =\lim_{R\rightarrow\infty}\lim_{s\downarrow 0}\frac{1}{2\pi}\int_{-R}^{R}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}e^{-iru}e^{-s|x-u|}f(u)du\right) \overline{e^{-irx}f(x)}dx dr\\ & =\lim_{R\rightarrow\infty}\int_{-R}^{R}|\hat{f}(r)|^{2}dr = \int_{-\infty}^{\infty}|\hat{f}(r)|^{2}dr = \|\hat{f}\|^{2}. \end{align} This technique works for the classical integral transforms.

• Very nice and very clean! I always suspected there was a way to do it this way but I couldn't quite see how to put it together. Would you mind putting in a reference for the resolvent result? – Cameron Williams Nov 12 '15 at 23:16
• It's called Stone's Formula. You can boil it down to the study of finite positive Borel measures $\mu$ on $\mathbb{R}$, and the holomorphic function $F(\lambda)=\int_{-\infty}^{\infty}\frac{d\mu(t)}{t-\lambda}$. You can integrate the following directly using $\tan^{-1}$: $\frac{1}{2\pi i}\int_{-R}^{R}\{F(r+is)-F(r-is)\}dr$ and show that as $s\downarrow 0$, and then as $R\rightarrow\infty$, you $\mu(\mathbb{R})$. Graph the $\tan^{-1}$ terms and you'll see that you get convergence to $\frac{1}{2}\{\chi_{[-R,R]}+\chi_{(-R,R)}\}$ in the integrand. It's something worth working through once. – DisintegratingByParts Nov 12 '15 at 23:37
• I'd never heard of Stone's formula though I think I've seen something rather similar to it in scattering theory arguments. It's pretty cool. Thanks for the info. – Cameron Williams Nov 12 '15 at 23:59