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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.

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

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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}= $

$=(2\pi)^{-\frac{n}{2}} \int _{\mathbb R^n} d^n k \hat g (k) \left (\int _{\mathbb R^n} d^n x f (x) e^{-ikx} \right)^* = (\hat f , \hat g)_{L^2}$

With hat over function denotes its Fourier transform. All manipulations here are justified as functions under integral are very regular, by assumption. This shows that Fourier transform preserves scalar products.

That was a very brief overview which might be difficult or not depending on your background; I'd rather answer specific questions if something was unclear rather than explain everything in most basic terms from the scratch.

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