Norm of Hermitian Operator/Corollary

Corollary to Norm of Hermitian Operator

Let $\mathbb F \in \set {\R, \C}$.

Let $\HH$ be a Hilbert space over $\mathbb F$.

Let $A : \HH \to \HH$ be a bounded Hermitian operator.

Let $\innerprod \cdot \cdot_\HH$ denote the inner product on $\HH$.

Suppose that:

$\forall h \in \HH: \innerprod {A h} h_\HH = 0$

Then $A$ is the zero operator $\mathbf 0$.

Proof

Let $\norm \cdot_\HH$ denote the inner product norm on $\HH$.

Let $\norm A$ denote the norm of $A$.

$\norm A = \sup \set {\size {\innerprod {A h} h_\HH}: h \in \HH, \norm h_\HH = 1}$

By definition of inner product norm:

$\forall h \in \HH: \innerprod {A h} h_\HH = 0$

Hence, in particular:

$\innerprod {A h} h_\HH = 0$

for all $h \in \HH$ such that $\norm h_\HH = 1$.

So:

$\set {\size {\innerprod {A h} h_\HH}: h \in \HH, \norm h_\HH = 1} = \set 0$

giving:

$\norm A = \sup \set 0$

Hence from the definition of supremum:

$\norm A = 0$
$A$ is the zero operator.

$\blacksquare$