TITLE: Operator norm of positive operator.
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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