Hellinger-Toeplitz Theorem/Proof 2

Proof
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:

But then we have:


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

a contradiction.

So $T$ is bounded.