Hellinger-Toeplitz Theorem

Theorem

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.

Let $T : \HH \to \HH$ be a Hermitian operator.

That is:

$\innerprod {T x} y = \innerprod x {T y}$ for each $x, y \in \HH$.

Then:

$T$ is bounded.

Proof 1

Let $\struct {\HH \times \HH, \norm \cdot_{\HH \times \HH} }$ be the direct product of $\HH$ with itself, with the direct product norm.

From the Closed Graph Theorem, it suffices to show that:

$G_T = \set {\tuple {x, T x} \in \HH \times \HH : x \in \HH}$

is closed in $\struct {\HH \times \HH, \norm \cdot_{\HH \times \HH} }$.

Let $\sequence {\tuple {x_n, T x_n} }_{n \mathop \in \N}$ be a sequence in $G_T$ converging to $\tuple {x, y}$ in $\struct {\HH \times \HH, \norm \cdot_{\HH \times \HH} }$.

From Convergence in Direct Product Norm, we have:

$\sequence {x_n}_{n \mathop \in \N}$ converges to $x$ in $X$

and:

$\sequence {y_n}_{n \mathop \in \N}$ converges to $y$ in $Y$.

We show that $y = T x$.

Let $z \in \HH$.

Then we have:

$\innerprod {T x_n} z = \innerprod {x_n} {T z}$

Taking $n \to \infty$, we have:

$\innerprod y z = \innerprod x {T z}$

from Inner Product is Continuous.

Since $T$ is Hermitian, we have:

$\innerprod x {T z} = \innerprod {T x} z$

Then we have:

$\innerprod y z = \innerprod {T x} z$ for all $z \in \HH$.

So from Inner Product is Sesquilinear, we have:

$\innerprod {y - T x} z = 0$ for all $z \in \HH$.

In particular:

$\innerprod {y - T x} {y - T x} = 0$

so:

$y = T x$

So $G_T$ is closed in $\struct {\HH \times \HH, \norm \cdot_{\HH \times \HH} }$.

So, by the Closed Graph Theorem, $T$ is bounded.

$\blacksquare$

Proof 2

Aiming for a contradiction, suppose suppose that $T$ is not bounded.

Then there does not exist $C > 0$ such that:

$\norm {T x}_\HH \le C$ for each $x \in \HH$ with $\norm x_\HH = 1$.

That is, for each $n \in \N$ there exists $y_n \in \HH$ such that:

$\norm {T y_n}_\HH \ge n$

with $\norm {y_n}_\HH = 1$.

For each $n \in \N$, define the linear functional $f_n : \HH \to \HH$ by:

$\map {f_n} x = \innerprod x {T y_n}$

Then, from the Riesz Representation Theorem (Hilbert Spaces), we have that $f_n$ is a bounded linear functional with:

$\norm {f_n}_{\HH^\ast} = \norm {T y_n}_\HH \ge n$

So:

$\ds \sup_n \norm {f_n}_{\HH^\ast} = \infty$

Then, from the Principle of Condensation of Singularities, there exists $x \in X$ such that:

$\ds \sup_n \cmod {\map {f_n} x} = \infty$

Since $T$ is Hermitian, we have:

$\map {f_n} x = \innerprod {T x} {y_n}$ for each $n \in \N$.

Then, we have:

\(\ds \cmod {\map {f_n} x}\)

\(=\)

\(\ds \cmod {\innerprod {T x} {y_n} }\)

\(\ds \)

\(\le\)

\(\ds \norm {T x}_\HH \norm {y_n}_\HH\)

Cauchy-Bunyakovsky-Schwarz Inequality

\(\ds \)

\(=\)

\(\ds \norm {T x}_\HH\)

But then we have:

$\ds \sup_n \cmod {\map {f_n} x} \le \norm {T x}_\HH < \infty$

a contradiction.

So $T$ is bounded.

$\blacksquare$

Source of Name

This entry was named for Ernst David Hellinger and Otto Toeplitz.