spectral mapping type norm identity for self adjoint operator I am currently trying to understand the spectral theorem as given in "Functional Analysis" (Vol.1) by Reed and Simon. Leading to its proof is a preliminary Lemma where I got stuck. It says
Let $P(x) = \sum_{n = 1}^N a_nx^n$ be a polynomial, let $P(A) := \sum_{n = 1}^N a_nA^n$. Then $\|P(A)\| = \sup_{\lambda \sigma(A)} |P(\lambda)|$.
The proof starts by stating that
$$
\|P(A)\|^2 = \|P(A)^*P(A)\| = \|\bar PP(A)\|
$$
I am fine with the last inequality (since $P(A)^* = \sum_{n = 1}^N \bar a_nA^n$) , however for the first one I can only verify that
$$
\|P(A)\|^2 = \langle P(A),P(A)\rangle = \sqrt{\langle P(A),P(A)\rangle^2} = \sqrt{\langle P(A),P(A)\rangle\langle \bar P(A),\bar P(A)\rangle}
$$
whereas
$$
\|P(A)^*P(A)\| = \|\bar P(A) P(A)\| = \sqrt{\langle \bar P(A) P(A),\bar P(A) P(A)\rangle}
$$
and I am somehow missing the last step that says these are equal.
Many thanks for your help!
 A: For your question
The statement $$\| P(A)\|^2 = \|P(A)^* P(A)\|$$ holds since, for any bounded operator $X$ on a Hilbert space $H$ the following holds $\|X\|^2 = \|X^*X\|$ (C*-identity). 
Proof
Let $h \in H$ have norm less or equal $1$, 
$$\|Xh\|^2 = \left<Xh, Xh\right> = \left<X^*Xh, h\right> \leq \|X^*Xh\|\|h\| \leq \|X^*X\|\leq \|X^* \|\|X\|.$$
Hence $$\|X\|^2 \leq \|X^*X\| \leq \|X^*\|\|X\|.$$
so $\|X\| \leq \|X^*\|$.
 Now, use th fact that $X=X^{**}$ so use $X^*$ in place for $X$. You will get that
$\|X^*\| \leq \|X\|$ thus
$$ \|X\|^2\leq \|X^*X\|\leq \|X\|^2.$$
Here is a different proof of lemma from Reed and Simon:
It is easy to check that if $A$ is a normal operator then $\| A \| = r(A) := \sup_{\lambda \in \sigma(A)}| \lambda|$ (for selfadjoint one is even easier to verify that), $r$ is called a spectral radious of $A$ .
Using this and spectral mapping theorem ( http://planetmath.org/spectralmappingtheorem> ) we obtain 
$$\| P(A) \| = r(p(A)) = \sup_{\lambda \in \sigma(P(A))}|\lambda| = \sup_{\lambda \in P(\sigma(A))}|\lambda| = \sup_{\lambda \in \sigma(A)}|p(\lambda)|.$$
