Self adjointness of an elliptic differential operator

Question

Let $A:D(a)\to L^2(\mathbb{R}^n)$ be an elliptic partial differential operator $$ A(f)=\sum_{i,j=1}^{\infty}\partial_{x_j}(a_{ij}(x)\partial_{x_i}f) $$ where $a_{ij}\in C^{\infty}_b(\mathbb{R}^n)$, this means they are bounded continuously differentiable functions with bounded derivative of all orders. Assume that there is a $c>0$ such that for every $y=(y_1,\ldots,y_n)$ $$ \sum_{i,j=1}^{n}a_{ij}(x)y_iy_j>c|y|^2,\ \text{for every $x$.} $$ Assume also that $a_{ij}=a_{ji}$. The question is how to prove that, on the domain $D(A)=H^2(\mathbb{R}^n)$, the operator $A$ is selfadjoint.

Answer 1

We recall definitions first. Let $H$ be a Hilbert space. A linear operator $A\colon D(A)\subset H\to H$ is said to be symmetric if \begin{equation} (Af, g)=(f, Ag),\qquad \forall f, g\in D(A), \end{equation} and it is said to be self-adjoint if \begin{equation} D(A)=D(A^\star)\quad \text{and}\quad (Af, g)=(f, Ag),\quad \forall f,g\in D(A). \end{equation} Here $D(A^\star)$ is the biggest domain on which the adjoint operator $A^\star$ can be defined, that is \begin{equation} D(A^\star)=\left\{g\in H\ |\ \text{the map }f\in D(A)\mapsto (Af, g)\ \text{is a continuous linear functional}\right\}\end{equation} A self-adjoint operator is necessarily closed, meaning that its graph $\{(f, Af)\in D(A)\times H\}$ is a closed subset of $H\times H$. This is equivalent to the fact that $D(A)$ is a Banach space when equipped with the graph norm $$\lVert f\rVert_{G(A)}=\lVert f\rVert+\lVert Af\rVert,\qquad f\in D(A).$$

To prove that a symmetric operator is self-adjoint one can use the following criterion, named basic criterion of self-adjointess on Reed & Simon's Methods of Modern Mathematical Physics, volume I, Theorem VIII.3.

Theorem. Let $A$ be symmetric. Then the following three statements are equivalent:

1. $A$ is self-adjoint;
2. $A$ is closed and $\ker(A^\star\pm i)=\{0\}$;
3. $A\pm i$ are both surjective.

We will show that the operator $A$ given in the question is self-adjoint by proving that it satisfies property 2. First of all we note that $A$ is closed, because from elliptic regularity theory (see for example Evans's book on PDE, 2nd edition, §6.3.5) we have for the graph norm $$\lVert f\rVert_{G(A)}=\lVert f\rVert_{L^2}+\lVert Af\rVert_{L^2}\ge C\lVert f\rVert_{H^2},\qquad \forall f\in D(A),$$ so a $\lVert\cdot \rVert_{G(A)}$-Cauchy sequence is a $\lVert \cdot \rVert_{H^2}$-Cauchy sequence also and $D(A)$ is a Banach space with the graph norm. We now turn to the equation \begin{equation}\tag{1} (A^\star + i)f=0. \end{equation} We claim that any $f\in D(A^\star)$ satisfying \eqref{1} is necessarily null. Indeed, again by regularity theory we can observe that $f\in C^\infty$. So \begin{equation} 0=(A^\star+i)f=(A+i)f, \end{equation} and because of symmetry we have $\lVert (A+i)f\rVert_{L^2}^2=\lVert Af\rVert_{L^2}^2+\lVert f\rVert_{L^2}^2$, from which we conclude that $f\equiv 0$. We have thus proved that $A$ satisfies all properties of the basic criterion of self-adjointness.