$\delimiter"026A30C\delimiter"426830ATv,v\delimiter"526930B\delimiter"026A30C% \leq{\mu}\delimiter"026B30Dv\delimiter"026B30D^{2}$ for all $v$ implies $\delimiter"026B30DT\delimiter"026B30D\leq{\mu}$

Theorem.

Let $H$ be a unitary space, $T$ be a self-adjoint linear operator and $\mu\geq 0$. If $|\left\langle Tv,v\right\rangle|\leq\mu\|v\|^{2}$ for all $v\in H$ then $T$ is a bounded operator and $\|T\|\leq\mu$.

Proof.

We will show that $\left\|Tv\right\|\leq\mu\left\|v\right\|$ for all $v\in H$. This is trivial if $\left\|Tv\right\|$ or $\left\|v\right\|$ is zero, so assume they are not. Let $\lambda$ be any positive number.

 $\displaystyle\left\|Tv\right\|^{2}$ $\displaystyle=\left\langle Tv,Tv\right\rangle$ $\displaystyle=\frac{1}{4}\left[\left\langle T\left(\lambda v+\frac{1}{\lambda}% Tv\right),\left(\lambda v+\frac{1}{\lambda}Tv\right)\right\rangle-\left\langle T% \left(\lambda v-\frac{1}{\lambda}Tv\right),\left(\lambda v-\frac{1}{\lambda}Tv% \right)\right\rangle\right]$ $\displaystyle\leq\frac{\mu}{4}\left[\left\|\lambda v+\frac{1}{\lambda}Tv\right% \|^{2}+\left\|\lambda v-\frac{1}{\lambda}Tv\right\|^{2}\right]$ $\displaystyle\leq\frac{\mu}{2}\left[\lambda^{2}\left\|v\right\|^{2}+\frac{1}{% \lambda^{2}}\left\|Tv\right\|^{2}\right]$

Now if we put $\displaystyle\lambda^{2}=\frac{\left\|Tv\right\|}{\left\|v\right\|}$ we get $\left\|Tv\right\|^{2}\leq\mu\left\|Tv\right\|\left\|v\right\|$ hence $\left\|Tv\right\|\leq\mu\left\|v\right\|$. ∎

Reference:

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

Title $\delimiter"026A30C\delimiter"426830ATv,v\delimiter"526930B\delimiter"026A30C% \leq{\mu}\delimiter"026B30Dv\delimiter"026B30D^{2}$ for all $v$ implies $\delimiter"026B30DT\delimiter"026B30D\leq{\mu}$ vertlangleTvvranglevertleqmuVertVVert2ForAllVImpliesVertTVertleqmu 2013-03-22 15:25:33 2013-03-22 15:25:33 Gorkem (3644) Gorkem (3644) 16 Gorkem (3644) Theorem msc 46C05