Hellinger-Toeplitz Theorem/Proof 1

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

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.

$\blacksquare$