Conventions:

• Hilbert spaces are complex.
• The inner product is anti-linear in its first argument.
• Operators are linear and possibly unbounded.

## Statement

Hellinger–Toeplitz theorem

An everywhere-defined symmetric operator on a Hilbert space is bounded.

A more verbose version of this statement:

Hellinger–Toeplitz theorem

If an operator $T$ in a Hilbert space $\hilb{H}$ is defined everywhere, that is on all of $\hilb{H}$, and symmetric, i.e. satisfies

$\innerp{Tx}{y} = \innerp{x}{Ty} \quad \forall x,y \in \hilb{H},$

then $T$ is necessarily bounded.

Consequently, a symmetric Hilbert space operator that is (truly) unbounded cannot be defined everywhere.

## Proof using the Uniform Boundedness Theorem

Assume that $T$ is not bounded. Then there exists a sequence $(x_n)$ of unit vectors in $\hilb{H}$ such that $\norm{Tx_n} \to \infty$. Consider the sequence $(f_n)$ of linear functionals on $\hilb{H}$, defined by

$f_n(y) = \innerp{Tx_n}{y} = \innerp{x_n}{Ty} \quad y \in \hilb{H}.$

The second identity is due to the symmetry of $T$. Apply Cauchy-Schwarz to both expressions to obtain the inequalities

$\abs{f_n(y)} \le \norm{Tx_n} \norm{y} \quad \text{and} \quad \abs{f_n(y)} \le \norm{x_n} \norm{Ty}$

for each $n \in \NN$ and $y \in \hilb{H}$. The first inequality shows that the functionals $f_n$ are bounded. The second one shows that, for fixed $y$, the sequence $(\abs{f_n(y)})$ is bounded by $\norm{Ty}$, since $\norm{x_n} = 1$ for all $n$. By the Uniform Boundedness Theorem, $(\norm{f_n})$ is a bounded sequence. One has

$\norm{Tx_n}^2 = \abs{f_n(Tx_n)} \le \norm{f_n} \norm{Tx_n} \quad n \in \NN.$

Divide by $\norm{Tx_n}$ (if nonzero) to obtain $\norm{Tx_n} \le \norm{f_n}$ for all but finitely many $n$. Thus $(\norm{Tx_n})$ is a bounded sequence, contradicting $\norm{Tx_n} \to \infty$. $\square\enspace$

## Proof using the Closed Graph Theorem

By the Closed Graph Theorem, it is sufficient to show that the graph of $T$ is closed. Let $(x_n)$ be a convergent sequence of vectors in $\hilb{H}$ such that the image sequence $(Tx_n)$ converges as well. Naming the limits $x$ and $z$, respectively, we have

$x_n \to x \quad \text{and} \quad Tx_n \to z.$

Continuity of the inner product implies

$\innerp{x_n}{Ty} \to \innerp{x}{Ty} \quad \text{and} \quad \innerp{Tx_n}{y} \to \innerp{z}{y}$

for all $y \in \hilb{H}$. Since $T$ is symmetric, the first assertion can be rewritten as

$\innerp{Tx_n}{y} \to \innerp{Tx}{y}.$

A sequence of complex numbers has at most one limit, hence $\innerp{Tx}{y} = \innerp{z}{y}$ for all $y$. By the Riesz representation theorem, $Tx=z$. $\square\enspace$