Closure of positive operator is positive

Question

Let $\mathcal{H}$ be a Hilbert space and $A:\mathcal{D}(A)\to\mathcal{H}$ be a densely-defined, closable linear operator. It is called positive (or better non-negative), if $\langle Ax,x\rangle\in\mathbb{R}$ for all $x\in\mathcal{D}(A)$ and

$$\langle Ax,x\rangle\geq 0\qquad\qquad\forall x\in\mathcal{D}(A)$$

Now, consider the closure $\overline{A}:\mathcal{D}(\overline{A})\to\mathcal{H}$. My question:

If $A$ is positive, does this imply that also $\overline{A}$ is positive?

By definition, $\omega\in\mathcal{D}(\overline{A})$, if there exists a sequence $(\omega_{n})_{n}$ in $\mathcal{D}(A)$ converging to $\omega$ such that $(A\omega_{n})_{n}$ is convergent. In this case, we set $\overline{A}\omega:=\lim_{n\to\infty} A\omega_{n}$. My first idea was trying to prove that

$$\langle A\omega_{n},\omega_{n}\rangle\to\langle\overline{A}\omega,\omega\rangle\qquad \mathrm{as}\quad n\to\infty$$

because this would prove the claim (a non-negative series in $\mathbb{R}$ always converges to a non-negative element). However, I am not sure whether the limit above is correct, since in general, if $a_{n}$ and $b_{n}$ are two convergent sequences with limits $a$ and $b$, respectively, then in general it is not true that $\langle a_{n},b_{n}\rangle$ converges to $\langle a,b\rangle$.

Answer 1

To see that the convergence

$$\langle A\omega_{n},\omega_{n}\rangle\to\langle\overline{A}\omega,\omega\rangle$$

we make the following estimates:

$$\vert \langle A\omega_{n},\omega_{n}\rangle-\langle\overline{A}\omega,\omega\rangle\vert = \vert \langle A\omega_{n},\omega_{n}\rangle-\langle\overline{A}\omega,\omega+\omega_{n}-\omega_{n}\rangle\vert =\\=\vert\langle A\omega_{n}-\overline{A}\omega,\omega_{n}\rangle-\langle\overline{A}\omega,\omega-\omega_{n}\rangle\vert\leq \underbrace{\Vert A\omega_{n}-\overline{A}\omega\Vert}_{\to 0}\cdot\underbrace{\Vert\omega_{n}\Vert}_{\to \Vert\omega\Vert}+\Vert\overline{A}\omega\Vert\cdot\underbrace{\Vert\omega-\omega_{n}\Vert}_{\to 0}\xrightarrow{n\to\infty}0$$

where we used the triangle inequality, the Cauchy-Schwarz inequality as well as the fact that convergence implies convergence of norms, i.e. $\omega_{n}\to\omega$ implies $\Vert\omega_{n}\Vert\to\Vert\omega\Vert$ (by the inverse triangle inequality).

Answer 2

By definition $(x,y)$ belongs to the graph of $\overline{A}$ if there exists a sequence $x_n\in D(A)$ such that $x_n\to x$ and $Ax_n\to y.$ Then $$\langle \overline{A}x,x\rangle=\langle y,x\rangle =\lim_n\langle Ax_n,x_n\rangle\ge 0$$ The second equality follows from the continuity of the inner product as the function of two variables. The latter follows from $$\langle a_n,b_n\rangle -\langle a,b\rangle \\ =\langle a_n-a,b_n-b\rangle +\langle a,b_n-b\rangle +\langle a_n-a,b\rangle$$ and the Cauchy-Schwarz inequality applied to each of the three terms.