Theorem 2.9.4 (Plancherel)4C82DCD17551

Proof. Let $f\in \mathcal S$ and $\varepsilon >0$. Insert the factor $\mathrm e^{-\pi \varepsilon ^2|\xi |^2}$ into the inversion integral. By Fubini's theorem and lemma 2.9.2 with $t=\varepsilon ^2$,

The functions $g_\varepsilon $ are positive with integral one, and the substitution $y=\varepsilon y'$ shows that their mass outside a fixed ball $B(0,\delta )$ tends to zero as $\varepsilon \to 0$. Since $f$ is bounded, and uniformly continuous because its gradient is bounded, splitting the convolution integral into the regions $|y|<\delta $ and $|y|\geq \delta $ shows that $f*g_\varepsilon \to f$ uniformly. On the left-hand side, $\widehat f\in \mathcal S\subset L^1$ by lemma 2.9.2, so dominated convergence lets $\varepsilon \to 0$ under the frequency integral. This proves the inversion formula.

Write $\mathcal F$ for the Fourier transform on $\mathcal S$, $\mathcal F^*$ for the inverse integral transform $\mathcal F^*g(x)=\int g(\xi )\mathrm e^{2\pi \mathrm i x\cdot \xi }\, \mathrm d\xi $, and $Rf(x)=f(-x)$. The substitution $x\mapsto -x$ shows that $\mathcal F^*=R\mathcal F=\mathcal FR$ on $\mathcal S$. In particular $\mathcal F^*$ maps $\mathcal S$ into itself, and both compositions $\mathcal F^*\mathcal F$ and $\mathcal F\mathcal F^*$ equal $R\mathcal F\mathcal F$, which is the identity on $\mathcal S$ by the inversion formula. For $f,g\in \mathcal S$, applying inversion to $g$ and Fubini's theorem, we obtain Parseval's identity

the interchange of integrals is justified because the double integral of the absolute value is at most $\| f\| _1\| \widehat g\| _1$.

By lemma 2.5.1(5), applied to Lebesgue measure, $C_ c^\infty (\mathbb R^ d)$, and hence $\mathcal S$, is dense in $L^2$. By Parseval's identity, $\mathcal F$ maps sequences in $\mathcal S$ that are Cauchy in $L^2$ to Cauchy sequences, and $L^2$ is complete; therefore $\mathcal F$ extends uniquely from $\mathcal S$ to a bounded operator on $L^2$, and this extension is an isometry. Since $\mathcal F^*=R\mathcal F$, the inverse integral transform extends in the same way. Both compositions of the two extensions are continuous and equal to the identity on the dense subspace $\mathcal S$, hence everywhere. Thus the extension is unitary, with the stated inverse.

Finally let $f\in L^1\cap L^2$. We choose $f_ j\in C_ c^\infty $ converging to $f$ both in $L^1$ and in $L^2$: first replace $f$ by $f\mathbf1_{B(0,R)}$ with $R$ large, and then mollify; by lemma 2.5.1(2) the mollifications converge in both norms, since their supports lie in a fixed compact set. The integral transforms of the $f_ j$ converge uniformly to that of $f$, because the difference is bounded by $\| f_ j-f\| _1$, while by isometry they converge in $L^2$ to the $L^2$ transform of $f$. A subsequence of an $L^2$ convergent sequence converges almost everywhere: choose it so that the squared $L^2$ errors are summable, and apply Tonelli's theorem to the sum of the squared errors. Hence the two transforms of $f$ agree almost everywhere. $\square$