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!

  • 1
    $\begingroup$ Consider the bilinear form $\Gamma(u,v) = \langle u, L^{-1}v \rangle$. It is also a linear form on the tensor product $\Gamma(u,v) = \Gamma(u\otimes v)$. When $u$ and $v$ belong to a reasonable function space, so does their tensor product. As long as $\Gamma$ is bounded as a linear form, you can recover the integral kernel from a corresponding Riesz representation theorem. $\endgroup$ Oct 15, 2016 at 20:05

1 Answer 1


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 $$ H(x_1)\otimes \delta_0(x_2)\otimes \dots \delta_0(x_n),\quad (H=\mathbf 1_{\mathbb R_+}). $$

  • $\begingroup$ Thanks for your answer; I don't have much experience with the tensor product, so could you please expand exactly what is meant by the RHS of your first equation? $\endgroup$
    – Lentes
    Oct 16, 2016 at 18:12
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    $\begingroup$ The rhs of the first equation is simply the bracket of duality between the distribution $\Gamma$ and the smooth compactly supported function $(x,y)\mapsto \psi(x)\phi(y)$. $\endgroup$
    – Bazin
    Oct 19, 2016 at 13:24

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