I was reading Stein's book on Fourier analysis, and in particular the proof that for Schwarz functions, the Fourier operator $\mathcal{F}:\mathcal{S}(\mathbb{R})\to \mathcal{S}(\mathbb{R})$ and the operator $\mathcal{F}^* :\mathcal{S}(\mathbb{R})\to \mathcal{S}(\mathbb{R})$ are inverse operators, where :
$\mathcal{F}(f)(\xi)=\int_{-\infty}^{\infty}f(x)e^{-2\pi i x\xi}dx$
$\mathcal{F}^*(g)(\xi)=\int_{-\infty}^{\infty}g(\xi)e^{2\pi i x\xi}d\xi$
He shows that $\mathcal{F}^*\circ \mathcal{F}=I$ and then he says that since $\mathcal{F}(f)(y)=\mathcal{F}(f)(-y)$, (call this trick $(*)$) we also have $\mathcal{F}\circ \mathcal{F}^*=I$. But I don't really understand this. Indeed, if I start from the first equality, I have $\mathcal{F}^*\circ \mathcal{F}(f)=f$ so $\mathcal{F}^*\circ \mathcal{F}(f)(-y)=f(-y)$, in other words $\mathcal{F}^*( \mathcal{F}(f))(-y)=f(-y)$ so by the trick $(*)$, we get $\mathcal{F}(\mathcal{F}(f))(y)=f(-y)$ which doesn't really help us no?
Because $\mathcal{F}^*\circ\mathcal{F}=I$, then $\mathcal{F}^*$ is surjective, which also means that $\mathcal{F}$ is surjective because $(\mathcal{F}f)(-y)=(\mathcal{F}^*f)(y)$. Similarly, $\mathcal{F}$ is injective because $\mathcal{F}^*\circ\mathcal{F}=I$. Therefore $\mathcal{F}$ is invertible with inverse $\mathcal{F}^*$, which gives $\mathcal{F}\circ\mathcal{F}^*=I$.