Let $L$ be an elliptic linear operator on $\mathbb R^n, n\geq3$. For simplicity, let's stick to the following Schrodinger operator
$$
Lu:=-\Delta u+V(x)u
$$
where $V\geq0$ is the electric potential, and $V\in\mathscr{B}$ where $\mathscr{B}$ is some function space (say, for example, $L^{\infty}\_{loc}(\mathbb R^n)$). Now let $f\in C\_{c}^{\infty}(\mathbb R^n)$ be arbitrary. Suppose the space $\mathscr{B}$ is such that the equation
$$
Lu=f
$$
has a unique solution in a weak sense. In this case we can invert the operator $L$ and write
$$
u=L^{-1}f
$$
for $f\in C\_c^{\infty}(\mathbb R^n)$. A locally integrable function $\Gamma(x,y)$ is called the fundamental solution for operator $L$ if it satisfies
$$
L\Gamma(\cdot,y)=\delta\_y
$$
in the sense of distributions. If the operator $L$ is invertible and the fundamental solution exists, then we can write
\begin{equation}
(L^{-1}f)(x)=\int\limits\_{\mathbb R^n}\Gamma(y,x)f(y)\,dy.
\end{equation}


I am interested in the following specific questions:


**a)** Does every invertible operator given by a PDE as in this case necessarily have a kernel? If so, is this kernel necessarily the fundamental solution?


**b)** Does existence of the fundamental solution imply invertibility of $L$?


These questions are motivated by an observation that in some research papers, the fundamental solution is simply said to be the "kernel of the operator $L^{-1}$", without any justification that such a kernel exists, or that such a kernel is necessarily the fundamental solution.


Thanks for your attention!



When does an inverse PDE operator have a kernel (i.e. a fundamental solution?)
Following your assumptions, it seems that the mapping $L^{-1}$ sends linearly and continuously the smooth compactly supported functions into distributions and thus, from the Schwartz (Laurent) kernel theorem, has a distribution kernel $\Gamma(x,y)$. It means that your integral formula holds weakly: for $\phi,\psi\in \mathscr D(\mathbb R^n)$
$$
\langle L^{-1}\phi,\psi\rangle\_{\mathscr D'(\mathbb R^n),\mathscr D(\mathbb R^n)}=
\langle \Gamma(x,y),\psi(x)\otimes\phi(y)\rangle\_{\mathscr D'(\mathbb R^{2n}),\mathscr D(\mathbb R^{2n})}.
$$
N.B. You wrote $\Gamma(y,x)$ but I prefer the more traditional $\Gamma(x,y)$.


On the other hand every constant coefficient differential operator $P(D)$ (non-zero) has a fundamental solution (Malgrange-Ehrenpreis theorem), and that does not imply invertibility of all these operators: take just $\partial/\partial x\_1$ with the huge kernel of distributions of $x\_2,\dots, x\_n$ and fundamental solution
$$
H(x\_1)\otimes \delta\_0(x\_2)\otimes \dots \delta\_0(x\_n),\quad (H=\mathbf 1\_{\mathbb R\_+}).
$$


