# Hellinger–Toeplitz Theorem

Conventions:

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

Recall that an operator $T:D(T)→H$ in a Hilbert space $H$ is called *symmetric*, if it has the property

Hellinger–Toeplitz Theorem

An everywhere-defined symmetric operator on a Hilbert space is 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 $H$ such that $∥Tx_{n}∥→∞$. Consider the sequence $(f_{n})$ of linear functionals on $H$, defined by

$f_{n}(y)=⟨Tx_{n},y⟩=⟨x_{n},Ty⟩y∈H.$The second identity is due to the symmetry of $T$. Apply Cauchy-Schwarz to both expressions to obtain the inequalities

$∣f_{n}(y)∣≤∥Tx_{n}∥∥y∥and∣f_{n}(y)∣≤∥x_{n}∥∥Ty∥$for each $n∈N$ and $y∈H$. The first inequality shows that the functionals $f_{n}$ are bounded. The second one shows that, for fixed $y$, the sequence $(∣f_{n}(y)∣)$ is bounded by $∥Ty∥$, since $∥x_{n}∥=1$ for all $n$. By the Uniform Boundedness Theorem, $(∥f_{n}∥)$ is a bounded sequence. One has

$∥Tx_{n}∥_{2}=∣f_{n}(Tx_{n})∣≤∥f_{n}∥∥Tx_{n}∥n∈N.$Divide by $∥Tx_{n}∥$ (if nonzero) to obtain $∥Tx_{n}∥≤∥f_{n}∥$ for all but finitely many $n$. Thus $(∥Tx_{n}∥)$ is a bounded sequence, contradicting $∥Tx_{n}∥→∞$.

## 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 $H$ such that the image sequence $(Tx_{n})$ converges as well. Naming the limits $x$ and $z$, respectively, we have

$x_{n}→xandTx_{n}→z.$Continuity of the inner product implies

$⟨x_{n},Ty⟩→⟨x,Ty⟩and⟨Tx_{n},y⟩→⟨z,y⟩$for all $y∈H$. Since $T$ is symmetric, the first assertion can be rewritten as

$⟨Tx_{n},y⟩→⟨Tx,y⟩.$A sequence of complex numbers has at most one limit, hence $⟨Tx,y⟩=⟨z,y⟩$ for all $y$. By the Riesz representation theorem, $Tx=z$.