Hellinger-Toeplitz Theorem/Proof 2

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

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.

$\blacksquare$