"Completeness" for weak convergence of unbounded closed operators on a separable Hilbert space $H$

Question

Let $H$ be a separable Hilbert space with the inner product $\langle, \rangle$ and $\{ T_n \}$ be a sequence of unbounded closed linear operators with a common dense domain $D \subset H$ such that $T_n (D) \subset D$ for all $n$.

Assume that for all $v,w \in D$, \begin{equation} \lim\limits_{n \to \infty} \langle v, T_n w \rangle = \alpha_{v,w} \in \mathbb{C} \end{equation}

Then, I wonder if there exists a unique unbounded closed linear operator $T$ on $H$ with the domain $D$ such that $T(D) \subset D$ and \begin{equation} \alpha_{v,w} = \langle v, T w \rangle \end{equation} for all $v,w \in D$.

If $T_n$ are bounded linear maps with $D=H$, then the answer is obviously yes. However, I wonder what will happen if we merely assume unbounded closed linear maps with a common dense domain. I tried to apply some notions like "strong resolvent convergence" but it only applies to self-adjoint operators.

Could anyone please help me?

Answer 1

Not always. What you are talking about is also called "convergence in the sense of sesquilinear forms", because you are taking a pointwise limit in $D\times D$ of a sequence of sesquilinear forms $\alpha_n(v,w)=\langle v,T_n w\rangle$ with common (dense) domain $D\subset H$, $v,w\in D$. The corresponding limit certainly defines a sesquilinear form $\alpha$ with domain $D$. However, there may not be an unbounded linear operator $T$ on $H$ with domain $D$ such that $\langle v,Tw\rangle=\alpha(v,w)$ for all $v,w\in D$.

A typical counterexample (for non-closed operators) is the following: let $H=L^2(\mathbb{R})$, $D=\mathscr{D}(\mathbb{R})$ and $(T_n f)(x)=\phi_n(x)f(x)$, where $\phi_n(x)=n\phi(nx)$ and $0\leq\phi\in\mathscr{D}(\mathbb{R})$ is such that $\phi(0)>0$ and$\int_{\mathbb{R}}\phi(x)\,dx=1$. Such a $T_n$ is actually bounded (hence closable) and the domain of $\overline{T}_n$ is $H$ for all $n\in\mathbb{N}$ by the closed graph theorem. In this case we have that $\alpha(f,g)=\bar{f}(0)g(0)$ for all $f,g\in D$ and such a sesquilinear form cannot be of the form $\alpha(f,g)=\langle f,Tg\rangle$ for any linear map $T:D\rightarrow H$ because clearly $Tg=g(0)\delta$, where $\delta$ is the Dirac delta distribution.

Answer 2

Even if you can define $T$ by this limit, this operator need not be closed. Take for example $H=\ell^2$ and $T_n$ as multiplication by $(0,\ldots, 0, n, n+1,\ldots)$ on its natural domain $D$. Note that $D$ is independent of $n$. We have $\langle x, T_n y\rangle\to 0$, but the zero operator is not closed on $D$.

These operators do not satisfy your extra assumption that the range is also contained in $D$. This assumption is quite restrictive and perhaps a bit unnatural. For example, it implies that $\sigma(T_n)=\mathbb C$. We can, however, modify the above example and obtain this property also. To do this, start out with multiplication by $(0,2,0,4,0,6,\ldots)$, followed up by a shift. Then take $T_n$ again as a cut off (at $n$) version of this. So $T_n x = (0, \ldots , 0, 2x_n, 0, 4x_{n+2}, 0,\ldots)$, with the non-zero entries sitting in the odd slots now.