# Prove the Fourier transform is a unitary linear operator

I am trying to prove that the inverse of the fourier transform is equal to its adjoint (i.e. it is a unitary linear operator). I am working with the inner product $\langle s_1,s_2 \rangle=\int_{-\infty}^{\infty}s_1^*(t)s_2(t)dt$.

The Fourier transform (and inverse (I do not require the proof of the inverse)) is as follows:

$\mathcal{F}(s(t))=\tilde{s}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}s(t)e^{-i\omega t} dt$

$\mathcal{F}^{-1}(\tilde{s}(\omega))=s(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\tilde{s}(\omega)e^{i\omega t} d\omega$

I know the adjoint is defined such that $\langle \mathcal{F}(s_1),s_2 \rangle=\langle s_1,\mathcal{F}^*(s_2) \rangle$. It is from here, however, that I am struggling to progress. I am trying to find out the adjoint of the Fourier transform so that I can then show that it equals the inverse Fourier transform above.

• A map $U$ between Hilbert spaces is a unitary if and only if it is a surjective isometry. Proving that should be easier. – Prahlad Vaidyanathan Sep 10 '15 at 6:38

Suppose that both $f$ and $g$ are in $L^{1}$. Then $\hat{f}$ and $\hat{g}$ are bounded and, therefore, $\hat{f}g$, $f\hat{g}$ are in $L^{1}$. And, \begin{align} \int_{-\infty}^{\infty}\hat{f}(s)g(s)ds & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(t)e^{-ist}dt g(s)ds \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)\int_{-\infty}^{\infty}e^{-ist}g(s)ds \\ & = \int_{-\infty}^{\infty}f(t)\hat{g}(t)dt. \end{align} Suppose $f,g \in \mathcal{C}_{c}^{\infty}(\mathbb{R})$ (i.e., are compactly supported $C^{\infty}$ functions.) Then $f,g,\mathcal{F}f,\mathcal{F}g$ are in $L^{1}$, and the above gives $$\int_{-\infty}^{\infty}\mathcal{F}f\overline{\mathcal{F}g}ds= \int_{-\infty}^{\infty}\mathcal{F}f\mathcal{F^{-1}}\overline{g}ds = \int_{-\infty}^{\infty}f\mathcal{F}\mathcal{F^{-1}}\overline{g}dt = \int_{-\infty}^{\infty}f\overline{g}dt$$ Therefore, $\|\mathcal{F}f\|=\|f\|$. Replace $g$ by $\mathcal{F}^{-1}g$ in the above: $$(\mathcal{F}f,g)=(f,\mathcal{F}^{-1}g).$$ Therefore $\mathcal{F}$, $\mathcal{F}^{-1}$ extend by continuity to isometries on $L^{2}(\mathbb{R})$, and the last equality similarly extends, which gives $\mathcal{F}^{\star}=\mathcal{F}^{-1}$ and $(\mathcal{F}^{-1})^{\star}=\mathcal{F}$. So $\mathcal{F}$ is unitary.
I suppose what you mean is that it is unitary when considered as an operator on $L^2(\mathbb R ^n)$. The usual procedure is to define it on dense subspace, shows that on this subspace it is isometry w.r.t. $L^2$ scalar product and extend it to isometry on whole space by continuity. This dense subspace is usually space of Schwartz class functions $\mathcal S$ (smooth with decay at infinity faster than $|\mathbf x|^{-k}$ for all integer k) on which Fourier transform is homeomorphism (this can be proved by closed graph theorem). That it is isometry w.r.t. $L^2$ scalar product is shown by the following calculation. Suppose $f,g$ are Schwartz and calculate their scalar product:
$(f,g)_{L^2} = \int _{\mathbb R^n} d^n x f(x)^* g(x) = \int _{\mathbb R^n} d^n x f(x)^* (2 \pi) ^{-\frac{n}{2}} \int _{\mathbb R ^n} d^n k \hat g (k) e ^{ik\cdot x}=$
$=(2\pi)^{-\frac{n}{2}} \int _{\mathbb R^n} d^n k \hat g (k) \left (\int _{\mathbb R^n} d^n x f (x) e^{-ikx} \right)^* = (\hat f , \hat g)_{L^2}$