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title: Adjoint Operators
parent: Unbounded Operators
nav_order: 1
description: >
The Hellinger–Toeplitz Theorem states that an everywheredefined symmetric
operator on a Hilbert space is bounded. We give a proof using the Uniform
Boundedness Theorem. We give another proof using the Closed Graph Theorem.
# spellchecker:dictionaries latex
# spellchecker:words Hellinger Toeplitz innerp hilb Schwarz functionals enspace Riesz

# {{ page.title }}
Conventions:
 Hilbert spaces are complex.
 The inner product is antilinear in its first argument.
 Operators are linear and possibly unbounded.
## Statement
{: .theoremtitle }
> Hellinger–Toeplitz theorem
>
> An everywheredefined symmetric operator on a Hilbert space is bounded.
A more verbose version of this statement:
{: .theoremtitle }
> 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 CauchySchwarz 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 $\norm{Tx_n} \to \infty$.
{{ site.qed }}

## 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, $Tx=z$.
{{ site.qed }}
