# How to derive Fourier inversion formula from the idea of Schwartz's kernel's theorem?

I am following Melrose's notes (page 16) on microlocal analysis. He used the idea of Schwartz kernel theorem to prove the Fourier inversion formula for Schwartz class functions. I cannot really follow why from $$K(x,y)=c\delta(x-y)$$ we would conclude $$G\circ F(\phi)=c\phi$$.

Melrose proposes using $$K(\phi)=(2\pi)^{-n}\int\int\int e^{iy\cdot \epsilon-ix\cdot \epsilon}\phi(y,x)dxd\epsilon dy$$ as the kernel of the Fourier inversion formula. Then we would have $$(G\circ F(\phi))(\zeta)=\int (G\circ F(\phi))(y)\zeta(y)dy=\int \zeta(y)(2\pi)^{-n}(\int e^{iy\cdot \epsilon}\int e^{-ix\cdot \epsilon}\psi(x)dxd\epsilon)dy$$Melrose's original formula is $$(G\circ F(\phi))(\zeta)=\int \zeta(y)(2\pi)^{-n}(\int e^{iy\cdot \epsilon}\int e^{-ix\cdot \epsilon}\psi(x)dxd\epsilon)dyd\epsilon dy$$ To me this should be a typo since $$G\circ F(\phi)$$ is a function in $$S(\bf R^{n})$$ and cannot be evaluated at $$\zeta$$. Further Melrose claimed this one to be equal to $$K(\zeta\boxtimes \psi)=K(\zeta(y)\psi(x))=(2\pi)^{-n}\int\int\int e^{iy\cdot \epsilon-ix\cdot \epsilon}\zeta(y)\psi(x)dxd\epsilon dy$$ It seems the only way to properly define $$K$$ as a distribution is $$\langle K(x,y),\zeta(y)\phi(x)\rangle=K(\zeta\boxtimes\phi)$$ But I still do not get why $$K(x,y)=c\delta(x-y)$$ gives us the desired result. Assume this we would have the left hand side to be $$c\zeta(x)\psi(x)$$, but it is not clear to me how this implies anything meaningful in the right hand side. Namely the term

$$(2\pi)^{-n}\int\int\int \zeta(y)\psi(x)dxd\epsilon dy$$ does not make sense as in the triple integral $$\int_{\bf R^{n}}d\epsilon=\infty$$. And even if this does make sense, how do we use it to prove Fourier inversion formula? I feel I am at lost.

As to your first query, $G\circ F(\phi)$ is a distribution, so you can evaluate it at $\zeta$. It will be $\int G\circ F(\phi)(x)\zeta(x)dx$.
Now, when you have shown $K(x, y) = c\delta(x - y)$, you can see that $$K(\zeta\boxtimes\phi) = \int\int c\delta(x - y)\zeta(x)\phi(y)dxdy = \int c\zeta(x)\phi(x)dx$$
Now you can evaluate that $c = 1$. If the above expression is equal to $$\int G\circ F(\phi(x))\zeta(x)dx$$ that means $$\int \phi(x)\zeta(x)dx = \int G\circ F(\phi(x))\zeta(x)dx$$ for all $\phi$ and $\zeta$, which immediately gives you what you want.