I am reading a proof claiming that every partial differential operator $P(D)$ has a fundamental solution $E$. It says that "if we have a distribution $u$ on $R^n$ with $u(P(D)\phi)=\phi(0)$ for $\phi\in\mathscr{D}$, then the distribution $E=\check{u}$ satisfies the definition of a fundamental solution."
Perhaps I am being dense at the moment, but I really don't see how this makes sense, namely why $\check{u}$ (inverse Fourier transform) is even defined for a distribution. After all the inverse Fourier transform does not map $\mathscr{D}$ into itself. Can someone help me with this? Thanks