# Operator norm of positive operator.

I'm studying Reed and Simon's "Methods of Modern Mathematical Physics" Vol. 1 (http://www.math.bme.hu/~balint/oktatas/fun/notes/Reed_Simon_Vol1.pdf). In the proof of the square root lemma (p.196) they use the equation

$\|I-A\|=\sup\limits_{|\varphi|=1}|((I-A)\varphi,\varphi)|,$

where $A$ is a bounded positive operator on a Hilbert space $\mathcal{H}$ and $I$ is the identity operator in $\mathcal{H}$. I understand that for any bounded operator $T$, the Cauchy-Schwarz imequality implies that

$\|T\|\geq\sup\limits_{|\varphi|=1}|(T\varphi,\varphi)|$.

But I am not able to prove the other inequality. Under what hypothesis is it true?

• I think so, as positive implies self-adjoint. – Bruno Suzuki Oct 15 '15 at 13:56
• So for bounded self adjoint opeartors in Hilbert space you have $||A||=\sup_{||x||=1}|\langle Ax,x\rangle|$ – uniquesolution Oct 15 '15 at 14:01
• @uniquesolution is that a fact? Do you happen to know the proof? Or, do you have a reference? – Omnomnomnom Oct 15 '15 at 14:02
• Yes, it is a standard fact. See here: lemma 8.26. math.ucdavis.edu/~hunter/book/ch8.pdf – uniquesolution Oct 15 '15 at 14:03
• @uniquesolution excellent! – Omnomnomnom Oct 15 '15 at 14:07

Note, that $T := (I - A)$ is self-adjoint. Let $M := \sup_{\def\norm#1{\left\|#1\right\|}\norm x = 1} \def\abs#1{\left|#1\right|}\def\<#1>{\left<#1\right>}\abs{\<Ax,x>}$, we will show that $\norm A \le M$. Let $x,y \in H$, then \begin{align*} \<T(x+y),x+y> - \<T(x-y),x-y> &= 2\<Tx,y> + 2\<Ty,x>\\ &= 2\<Tx,y> + 2\<x,Tx>\\ &= 4\Re\<Tx,y> \end{align*} Hence, due to the parallelogram identity, \begin{align*} 4\Re\<Tx,y> &\le M\norm{x+y}^2 + M\norm{x-y}^2\\ &= 2M\bigl(\norm x^2 +\norm y^2 \bigr) \end{align*} Hence, we have $$\Re \<Tx, y> \le M, \qquad \norm x, \norm y \le 1$$ Now, given $y$ choose $\lambda \in \mathbf C$ with $\abs \lambda = 1$ such that $\Re \<Tx,\lambda y> = \abs{\<Tx, y>}$, then $\abs{\<Tx,y>} \le M$, hence $$\norm A = \sup_{\norm x, \norm y \le 1}\abs{\<Tx,y>} \le M.$$

• Can you please explain why $(I - A)$ is self-adjoint? I am getting that it is self-adjoit iff $A$ is self-adjoint. – Vadim Nov 25 '15 at 20:45
• And $A$ is positive, hence self-adjoint. – martini Nov 25 '15 at 21:39
• Ok, I didn't notice that A is positive. Thank you – Vadim Nov 26 '15 at 1:02

Denote numerical range: $$\mathcal{W}(A):=\{\langle A\hat\varphi,\hat\varphi\rangle:\|\hat\varphi\|=1\}$$

By Cauchy-Schwarz: $$\|\mathcal{W}(A)\|\leq\|A\|1\cdot1=\|A\|$$

For bounded operators: $$A\in\mathcal{B}(\mathcal{H}):\quad\sigma(A)\subseteq\overline{\mathcal{W}(A)}$$

For normal operators:* $$N^*N=NN^*:\quad\|N\|=\|\sigma(N)\|$$

All together gives: $$\|N\|=\|\sigma(N)\|\leq\|\mathcal{W}(N)\|\leq\|N\|$$ Concluding equality.

*Hint: Neumann-Series