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Since you didn't specify, I'll assume this is in a Banach space $X$, with $T: X \to Y$ (for another Banach space $Y$) a closed densely-defined operator, and all $x_n \in \mathscr D(T)$. By the unifor …

Perhaps not what you're looking for, but you may be interested in the following result. Let $\cal H$ be a separable infinite-dimensional Hilbert space, and $H$ any self-adjoint unbounded linear opera …

Here's a counterexample (subject perhaps to what you consider "natural"). Take a separable Hilbert space with orthonormal basis $\{u_n : n = 1, 2, \ldots\}$ and the operator $A$ defined by $$ A u_n = …

You mean an infinite-dimensional separable Hilbert space. The answer is no. Suppose $p(z)$ has distinct roots $\alpha_1, \alpha_2$. Define a sequence $x_1, x_2, \ldots$ in the unit sphere of $H$ suc …

Yes, of course. By definition of the adjoint operator, $\{[-A^* y, y]: y \in \mathscr D(A^*)\}$ is the orthogonal complement in $H \oplus H$ of the graph $\{[x, Ax]: x \in \mathscr D(A)\}$ of $A$. …