Simple proof that $\|p(A)\|\le \sup_{|z|\le 1}|p(z)|$ for polynomials $p$ and $\|A\| \le 1$.

Question

Let $\mathcal{H}$ be a complex Hilbert space, and let $A$ be a bounded operator linear operator on $\mathcal{H}$ with $\|A\| \le 1$. It is known that $\|p(A)\|\le \sup_{|z|=1}|p(z)|$ for all complex polynomials $p(z)$. Does anyone know a short proof of this fact?

Answer 1

More generally, you can work with functions $p$ which are analytic in some open neighbourhood of the unit disc. The proof is based on the fact that you can find a norm-analytic map $$A(\cdot)\colon \mathbb{C}\to \mathscr{B}(\mathcal{H}\oplus \mathcal{H})$$ such that $A(0)=A\oplus 0$ and $A(z)$ is unitary for $|z|=1$. Indeed, simply put

$$A(z) = U(z)BU(z)$$

where

$$U(z) = \left( \begin{array}{cc}I & 0 \\ 0 & zI \end{array}\right)\text{ and }B(z) = \left( \begin{array}{cc}A & (I-|A^*|^2)^{1/2} \\ (I-|A|^2)^{1/2} & -A^* \end{array}\right).$$

Let $F(z) = p(A(z))$. In particular, $$F(0) = \left( \begin{array}{cc}p(A) & 0 \\ 0 & p(0) \end{array}\right).$$

Now use the maximum modulus principle for operator-valued analytic functions to get:

$$ \begin{array}{lcl} \|p(A)\| & \leqslant & \|F(0)\| \\ &\leqslant & \max_{|z|=1} \|F(z)\| \\ &\leqslant & \max_{|z|=1} |p(z)|.\end{array}$$