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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.

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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.

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