Proposition 2.9.54B698DACE987

Proof. Let $g\in L^2$ be the inverse $L^2$ transform of $\widehat\mu $, so that $\widehat g=\widehat\mu $. For $\phi \in \mathcal S$ we have

The first equality follows from the inversion formula for $\phi $ and Fubini's theorem. For the second, we apply Parseval's identity to $\phi $ and $\overline g$; it extends from $\mathcal S$ to $L^2$ because the $L^2$ transform is unitary. We also use that the transform of $\overline g$ at $\xi $ is $\overline{\widehat g(-\xi )}$; this holds for Schwartz functions by definition, and it extends to $L^2$ because the maps $g\mapsto \widehat{\overline g}$ and $g\mapsto \overline{\widehat g(-\, \cdot \, )}$ are continuous from $L^2$ to $L^2$.

Consequently, for every test function $\phi $ the number $\int \phi g\, \mathrm d x=\int \phi \, \mathrm d\mu $ is real when $\phi $ is real and nonnegative when $\phi \geq 0$. By the Cauchy–Schwarz inequality on compact sets, $g$ is locally integrable, so it defines a positive distribution. By proposition 2.5.3(3) its mollifications $g*\rho _\varepsilon $ are nonnegative, and they converge to $g$ in $L^1$ on compact sets by lemma 2.5.1(2). Hence $g$ is real and nonnegative almost everywhere. The measures $\mu $ and $g\, \mathrm d x$ are Radon by proposition 2.2.2, and they define the same distribution; by proposition 2.5.3(4), $\mu =g\, \mathrm d x$. In particular $\int g\, \mathrm d x=\mu (\mathbb R^ d)<\infty $, and $\| g\| _2=\| \widehat\mu \| _2$ because the $L^2$ transform is unitary. $\square$