Changing order of integration in the Fourier inversion formula for the Schwartz functions

Question

In the proof of the Fourier inversion formula for $\mathcal S(\mathbb R^n)$ functions, it is observed that in the integral $$\int_{\mathbb R^n} \widehat f(\xi)e^{i\xi\cdot x}\,\mathrm d\xi = \int_{\mathbb R^n}\int_{\mathbb R^n} f(y) e^{-iy\cdot\xi} e^{i\xi\cdot x}\,\mathrm dy\mathrm d\xi$$ one cannot change the order of integration, because then the inner integral might diverge. I'm probably missing something simple, but how is that so?

The inner integral is bounded by $$\int_{\mathbb R^n} |f(y)|\,\mathrm dy$$ which should be convergent, since Schwartz functions decay faster than any negative power of $x$. Then my doubt is whether it's the outer integral that wouldn't converge after the change. Can someone shed some light here?

Answer 1

I think I found out where my reasoning was wrong. Consider the simple counter-example on $\mathbb R$: $$f(x) = e^{-x^2}$$ which is a function belonging to $\mathcal S(\mathbb R)$. If we change the order of integration, we obtain $$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-y^2}e^{-iy\xi}e^{ix\xi}\ \mathrm d\xi\mathrm dy = \int_{-\infty}^{+\infty} e^{-y^2}\int_{-\infty}^{+\infty} e^{i(x-y)\xi}\,\mathrm d\xi\mathrm dy$$ and the inner integral does not converge because $e^{i(x-y)\xi}$ is a combination of sines and cosines.