Proof, [Corollary 1.2, Po02]. By Sz.-Nagy’s Dilation Theorem, there exists a Hilbert space $K \supset H$ and a unitary operator $U \in B(K)$ such that $T^{n} = P_{H}U^{n}|_{H}$ for all $n \in \natp$, where $P_{H}$ is the orthogonal projection onto $H$.

Thus for any $p \in \complex[x]$, $p(T) = P_{H}p(U)|_{H}$. As $U$ is unitary, the continuous functional calculus shows that $\norm{p(U)}_{B(K)}\le \sup_{z \in B_\complex(0, 1)}|p(z)|$. Therefore $\norm{p(T)}_{B(H)}\le \sup_{z \in B_\complex(0, 1)}|p(z)|$.$\square$