Hellinger–Toeplitz theorem

Remarks

Proof idea: by the closed graph theorem, it suffices to show that $A$ is closed. If $x_n \to x$ and $A x_n \to z$, then for every $y \in H$, $$\ip{z}{y} = \lim \ip{A x_n}{y} = \lim \ip{x_n}{Ay} = \ip{x}{Ay} = \ip{Ax}{y},$$ so $z = Ax$.

Example: if $A$ is given on $\ell^2$ by a real symmetric infinite matrix and is defined on all of $\ell^2$, the theorem says $A$ is automatically bounded.

Completeness of $H$ is needed. On the space of finitely supported sequences with the $\ell^2$ inner product, the diagonal map $(x_n) \mapsto (n x_n)$ is defined everywhere and symmetric. It is unbounded, since it sends the $n$-th unit vector to $n$ times itself.

Going deeper

For a student

The theorem is a short application of the closed graph theorem, and it is one reason why the theory of unbounded self-adjoint operators works with dense, proper domains. It also shows that a bounded operator is self-adjoint exactly when it is symmetric on all of $H$; no extra boundedness hypothesis is needed. A related form: a linear map $A \colon H \to H$ that has a formal adjoint $B \colon H \to H$, with $\ip{Ax}{y} = \ip{x}{By}$ for all $x, y$, is bounded, by the same proof.