# Fourier Inversion formula on $L^2$

The Fourier transform is defined by

$\mathcal{F}f(\xi)=\int_{\mathbb{R}^n} e^{-ix\cdot \xi} f(x) dx$

If we restrict to Schwartz functions on $\mathbb{R}^n$, then the Fourier transform has an inverse:

$\mathcal{G}g(x)=\int_{\mathbb{R}^n} e^{ix\cdot \xi} g(\xi) d\xi$

Can one define the inverse Fourier transform on functions in $L^2(\mathbb{R}^n)$? The Fourier transform of functions in $L^2$ is defined to be the limit of the Fourier transform of Schwartz functions. Does defining the inverse Fourier transform by such a limit work?

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Yes, by density. Another, but related way to see this is to define the Fourier transform on tempered distributions by duality, and see that it indeed agrees with the definition by density. –  timur Sep 4 '12 at 1:19
Thanks, I'll have to look into the definition by tempered distributions. –  gabbering Sep 4 '12 at 1:32
See here. $L_2$ functions are not Fourier transformable in general. –  Mhenni Benghorbal Sep 4 '12 at 1:37
On the other hand, if $f \in L_1 \cap L_2$, then yes. –  Mhenni Benghorbal Sep 4 '12 at 1:47
@Sebastian that's because you can. This is the content of Plancherel's theorem: There is a unique linear isometry on $L^{2}$ which agrees with the definition of the usual Fourier transform on $L^{1}\cap L^{2}$ –  jmracek Sep 4 '12 at 2:04

Let us denote by $F$ the extension of the Fourier transform to $L^2(\mathbb{R}^n)$ as described in the OP, and denote by $G$ the analogous extension of the inverse Fourier transform to $L^2(\mathbb{R}^n)$. We know that $FG=I$ and $GF=I$ on the Schwarz functions; the question is if it is true on $L^2(\mathbb{R}^n)$. The answer is yes, and the answer below is in a more general topological space setting, because this abstraction illustrates the underlying phenomenon more clearly. Apart from continuity, the main reason why it is true is the density of the Schwarz functions in $L^2$ and the Hausdorff property of $L^2$ (because its topology is induced by a norm).
Claim. Let $X$ and $Y$ be topological spaces, where $Y$ is also Hausdorff. Let $A,B:X\to Y$ be continuous maps, satisfying $Ax=Bx$ for all $x\in S$ for some dense subset $S\subset X$. Then $A=B$.
Proof. Pick $x\in X\setminus S$, and let $a=A(x)$ and $b=B(x)$. Consider arbitrary neighbourhoods $V$ and $W$ of $a$ and $b$, respectively. Then by continuity, there is an open neighbourhood $U$ of $x$ such that $A(U)\subset V$ and $B(U)\subset W$. By density, there is $s\in U\cap S$, hence $V\ni A(s)=B(s)\in W$, meaning that $V\cap W\neq\varnothing$. Since $V$ and $W$ were arbitrary, and $Y$ is Hausdorff, we conclude that $a=b$.