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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.