for all $v$ implies

Theorem.

Let $H$ be a unitary space, $T$ be a self-adjointlinear operator and $\mu \mathrm{\ge }\mathrm{0}$. If $\mathrm{|}\mathrm{⟨}T\mathit{}v\mathrm{,}v\mathrm{⟩}\mathrm{|}\mathrm{\le }\mu \mathit{}{\mathrm{\parallel }v\mathrm{\parallel }}^{\mathrm{2}}$ for all $v\mathrm{\in }H$ then $T$ is a bounded operator and$\mathrm{\parallel }T\mathrm{\parallel }\mathrm{\le }\mu$.

Proof.

We will show that $\parallel Tv\parallel \le \mu \parallel v\parallel$ for all $v\in H$.This is trivial if $\parallel Tv\parallel$ or $\parallel v\parallel$ is zero, so assumethey are not. Let $\lambda$ be any positive number.

${\parallel Tv\parallel }^2$$=⟨Tv,Tv⟩$

$={\displaystyle \frac{1}{4}}\left[⟨T\left(\lambda v+{\displaystyle \frac{1}{\lambda }}Tv\right),\left(\lambda v+{\displaystyle \frac{1}{\lambda }}Tv\right)⟩-⟨T\left(\lambda v-{\displaystyle \frac{1}{\lambda }}Tv\right),\left(\lambda v-{\displaystyle \frac{1}{\lambda }}Tv\right)⟩\right]$

$\le {\displaystyle \frac{\mu }{4}}\left[{\parallel \lambda v+{\displaystyle \frac{1}{\lambda }}Tv\parallel }^2+{\parallel \lambda v-{\displaystyle \frac{1}{\lambda }}Tv\parallel }^2\right]$

$\le {\displaystyle \frac{\mu }{2}}\left[{\lambda }^2{\parallel v\parallel }^2+{\displaystyle \frac{1}{{\lambda }^2}}{\parallel Tv\parallel }^2\right]$

Now if we put ${\lambda }^2={\displaystyle \frac{\parallel Tv\parallel }{\parallel v\parallel }}$ we get ${\parallel Tv\parallel }^2\le \mu \parallel Tv\parallel \parallel v\parallel$ hence $\parallel Tv\parallel \le \mu \parallel v\parallel$.∎

Reference:

F. Riesz and B. Sz-Nagy, Functional Analysis, F. UngarPublishing, 1955, chap VI.