for all $v$ implies

Theorem.

Let $H$ be a unitary space, $T$ be a self-adjoint linear 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 assume they 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. Ungar Publishing, 1955, chap VI.

Title	for all $v$ implies
Canonical name	vertlangleTvvranglevertleqmuVertVVert2ForAllVImpliesVertTVertleqmu
Date of creation	2013-03-22 15:25:33
Last modified on	2013-03-22 15:25:33
Owner	Gorkem (3644)
Last modified by	Gorkem (3644)
Numerical id	16
Author	Gorkem (3644)
Entry type	Theorem
Classification	msc 46C05