Let $H=(H,(\cdot,\cdot))$ be a Hilbert space and $A:D(A)\subset H \longrightarrow H$ a linear operator (not necessarily bounded) such that $\overline{D(A)}=H$ and $A \geq 0$, that is,
$$(A(x),x)\geq 0,\: \forall \: x \in D(A).$$
Then $A$ is self-adjoint?
I know that if $A:H \longrightarrow H$ is linear and bounded and $A\geq 0$ then $A$ is self-adjoint. I would like to know if this result is more general?
Best Answer
First, you need $H$ to be a complex Hilbert space for this to be true even for bounded operators; it is false for the real Hilbert space $\mathbb{R}^2$, as witnessed by $A = \left( \begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\right)$.
But even for complex Hilbert spaces, your claim is in general not true. Most glaringly, you have no hypotheses that force $A$ to be closed. But it's not even true for closed operators. Take $H = L^2([0,1])$ and consider the operator $Af=-f''$ defined on the domain $D(A) = H^2_0([0,1])$ (i.e. the completion of $C^\infty_c((0,1))$ in the $H^2$ Sobolev norm). This is a positive closed operator, but it is not self-adjoint, since for instance the constant function 1 is in the domain of $A^*$.
(Indeed, you can get two different self-adjoint extensions of $A$ by allowing either Dirichlet or Neumann boundary conditions. The problem with the domain $H^2_0$ is that it's imposing both boundary conditions, which is too restrictive.)
For a positive result, however (no pun intended), it is true that every densely defined positive unbounded operator on a complex Hilbert space has at least one self-adjoint extension. This is the famous Friedrichs extension theorem.