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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?

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  • $\begingroup$ I think so, as positive implies self-adjoint. $\endgroup$
    – Bruno Suzuki
    Oct 15, 2015 at 13:56
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    $\begingroup$ So for bounded self adjoint opeartors in Hilbert space you have $||A||=\sup_{||x||=1}|\langle Ax,x\rangle|$ $\endgroup$
    – uniquesolution
    Oct 15, 2015 at 14:01
  • $\begingroup$ @uniquesolution is that a fact? Do you happen to know the proof? Or, do you have a reference? $\endgroup$
    – Ben Grossmann
    Oct 15, 2015 at 14:02
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    $\begingroup$ Yes, it is a standard fact. See here: lemma 8.26. math.ucdavis.edu/~hunter/book/ch8.pdf $\endgroup$
    – uniquesolution
    Oct 15, 2015 at 14:03
  • $\begingroup$ @uniquesolution excellent! $\endgroup$
    – Ben Grossmann
    Oct 15, 2015 at 14:07

2 Answers 2

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Note, that $T := (I - A)$ is self-adjoint. Let $M := \sup_{\def\norm#1{\left\|#1\right\|}\norm x = 1} \def\abs#1{\left|#1\right|}\def\<#1>{\left<#1\right>}\abs{\<Tx,x>}$, we will show that $\norm T \le M$. 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 $$ \norm T = \sup_{\norm x, \norm y \le 1}\abs{\<Tx,y>} \le M. $$

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  • $\begingroup$ Can you please explain why $(I - A)$ is self-adjoint? I am getting that it is self-adjoit iff $A$ is self-adjoint. $\endgroup$
    – Vadim
    Nov 25, 2015 at 20:45
  • $\begingroup$ And $A $ is positive, hence self-adjoint. $\endgroup$
    – martini
    Nov 25, 2015 at 21:39
  • $\begingroup$ Ok, I didn't notice that A is positive. Thank you $\endgroup$
    – Vadim
    Nov 26, 2015 at 1:02
  • $\begingroup$ Could you please explain to me why \begin{aligned} 4 \Re\langle T x, y\rangle & \leq M\|x+y\|^2+M\|x-y\|^2 \\ & =2 M\left(\|x\|^2+\|y\|^2\right) \end{aligned} $\endgroup$
    – Tung Nguyen
    Feb 15 at 17:30
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    $\begingroup$ @AnneBauval Thank you a lot. I have understood everything perfectly. $\endgroup$
    – Tung Nguyen
    Feb 17 at 7:33
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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

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