$\|P(A)^* P(A)\| = \|(\overline{P}P)(A)\|$ where $A$ is a bounded self adjoint operator and $P$ is a polynomial in $A$

Question

I am following the proof of a lemma in Reed & Simon that states:

Let $A$ be a bounded self adjoint operator. Then $$\|P(A)\| = \sup_{\lambda \in \sigma(A)} |P(\lambda)|.$$

In their proof the give the following equality

$$\|P(A)\|^2 = \|P(A)^*P(A)\| = \|(\overline{P}P)(A)\|.$$

How did they obtain the equality $\|P(A)^* P(A)\| = \|(\overline{P}P)(A)\|$? Where does this come from?

Moreover, how does $A$ being self-adjoint imply $P(A)$ is self adjoint?

Answer 1

For a fixed element of an algebra $A$, the map $P \mapsto P(A)$ is a morphism of algebras, which means that for all $P,Q$, $(P+Q)(A) = P(A) + Q(A)$ and $(PQ)(A) = P(A)Q(A)$. Moreover, we have that $P(A)^* = \overline{P}(A^*)$.

The proof of these facts is straightforward (using the coefficients of the polynomials) and can be found in any book on linear algebra.

From the above,

• assuming $A$ is self-adjoint, $P(A)^*P(A) = \overline{P}(A^*)P(A) = \overline{P}(A)P(A) = (\overline{P}P)(A)$;

• if $P$ is a real-valued polynomial, if $A = A^*$, then $P(A) ^* = \overline{P}(A^*) = P(A)$, so $P(A)$ is also self-adjoint.

Answer 2

For every bounded operator $B:X\to X,$ where $X$ is a Banach space, and every polynomial $P(z)$ there holds $$\sigma(P(B))=P(\sigma (B))\quad (1)$$ If $B$ is a normal bounded operator on a Hilbert space $\mathcal{H}$ we have $$\|B\|=\max_{\lambda\in \sigma(B)}|\lambda|\quad(2)$$ Let $A\in B(\mathcal{H})$ be a normal operator. Then for any polynomial $P(z)$ the operator $P(A)$ is normal. Hence applying $(2)$ and $(1)$ gives $$\|P(A)\|=\max_{\mu\in\sigma(P(A))}|\mu|=\max_{\lambda\in\sigma(A)}|P(\lambda)|$$