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 $T:D(T)\to \mathcal{H}$ in a Hilbert space $\mathcal{H}$ is called symmetric, if it has the property

$$⟨Tx,y⟩=⟨x,Ty⟩\forall x,y\in D(T).$$

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.

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Proof using the Uniform Boundedness Theorem

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

$$f_n(y)=⟨T{x}_n,y⟩=⟨x_n,Ty⟩y\in \mathcal{H}.$$

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

$$|f_n(y)|\le \Vert T{x}_n\Vert \Vert y\Vert \text{and}|f_n(y)|\le \Vert x_n\Vert \Vert Ty\Vert$$

for each $n\in \mathbb{N}$ and $y\in \mathcal{H}$. 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 $\Vert Ty\Vert$, since $\Vert x_n\Vert =1$ for all $n$. By the Uniform Boundedness Theorem, $(\Vert f_n\Vert )$ is a bounded sequence. One has

$$\Vert T{x}_n{\Vert }^2=|f_n(T{x}_n)|\le \Vert f_n\Vert \Vert T{x}_n\Vert n\in \mathbb{N}.$$

Divide by $\Vert T{x}_n\Vert$ (if nonzero) to obtain $\Vert T{x}_n\Vert \le \Vert f_n\Vert$ for all but finitely many $n$. Thus $(\Vert T{x}_n\Vert )$ is a bounded sequence, contradicting $\Vert T{x}_n\Vert \to \infty$.

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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 $\mathcal{H}$ such that the image sequence $(T{x}_n)$ converges as well. Naming the limits $x$ and $z$, respectively, we have

$$x_n\to x\text{and}T{x}_n\to z.$$

Continuity of the inner product implies

$$⟨x_n,Ty⟩\to ⟨x,Ty⟩\text{and}⟨T{x}_n,y⟩\to ⟨z,y⟩$$

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

$$⟨T{x}_n,y⟩\to ⟨Tx,y⟩.$$

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