von Neumann's inequality

Remarks

Proof idea: by Sz.-Nagy's dilation theorem there is a unitary $U$ on a larger space $K \supseteq H$ with $T^n = P_H U^n|_H$ for $n \ge 0$, so $p(T) = P_H \, p(U)|_H$. Hence $\norm{p(T)} \le \norm{p(U)}$. Since $U$ is unitary, its spectrum lies on the unit circle. So $\norm{p(U)}$ is the maximum of $\abs{p}$ on the spectrum, which is at most the maximum of $\abs{p}$ on the closed unit disc.

Example: for the nilpotent contraction $T = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $p(z) = 1 + z$, one has $p(T) = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ with norm $\tfrac{1 + \sqrt 5}{2} \approx 1.618$, which is at most $\max_{\abs{z} \le 1} \abs{1 + z} = 2$.

The Hilbert space structure matters. On a Banach space a contraction need not satisfy the inequality. The inequality extends to two commuting contractions (Andô's theorem) but fails in general for three commuting contractions.

Going deeper

For a student

The inequality lets one define $f(T)$ for every $f$ in the disc algebra, the uniform closure of the polynomials on the closed disc, with $\norm{f(T)} \le \norm{f}_\infty$. This is the start of the holomorphic functional calculus of contractions developed by Sz.-Nagy and Foias. Its failure for three commuting contractions shows that multivariable operator theory differs from the one-variable theory.