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  "body": "\n## Statement\n\nLet $H$ be a complex [[Hilbert space|HM2HHE3J51]] and let $T \\colon H \\to H$ be a bounded linear operator with $\\norm{T} \\le 1$.\nThen for every [[polynomial|1G0B73WPX8]] $p$ with complex coefficients,\n$$\\norm{p(T)} \\le \\sup \\set{\\abs{p(z)} : z \\in \\C, \\ \\abs{z} \\le 1}.$$\n\n## Remarks\n\nProof idea: by [[Sz.-Nagy's dilation theorem|YB7Q3EK0ND]] 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 $p(T) = P_H \\, p(U)|_H$.\nHence $\\norm{p(T)} \\le \\norm{p(U)}$.\nSince $U$ is unitary, its spectrum lies on the unit circle.\nSo $\\norm{p(U)}$ is the maximum of $\\abs{p}$ on the spectrum, which is at most the maximum of $\\abs{p}$ on the closed unit disc.\n\nExample: for the nilpotent contraction $T = \\begin{pmatrix} 0 \u0026 1 \\\\ 0 \u0026 0 \\end{pmatrix}$ and $p(z) = 1 + z$, one has $p(T) = \\begin{pmatrix} 1 \u0026 1 \\\\ 0 \u0026 1 \\end{pmatrix}$ with norm $\\tfrac{1 + \\sqrt 5}{2} \\approx 1.618$, which is at most $\\max_{\\abs{z} \\le 1} \\abs{1 + z} = 2$.\n\nThe Hilbert space structure matters.\nOn a Banach space a contraction need not satisfy the inequality.\nThe inequality extends to two commuting contractions (Andô's theorem) but fails in general for three commuting contractions.\n\n## Explanations\n\n### L1 human\n\nIf an operator never stretches vectors, then a polynomial in the operator is no bigger than the same polynomial is on the unit disc of the complex plane.\nSo a contraction behaves, for polynomial expressions, no worse than a complex number of modulus at most $1$.\n\n### L2 human\n\nThe inequality lets one define $f(T)$ for every $f$ in the disc algebra, the uniform closure of the polynomials on the closed disc, with $\\norm{f(T)} \\le \\norm{f}_\\infty$.\nThis is the start of the holomorphic functional calculus of contractions developed by Sz.-Nagy and Foias.\nIts failure for three commuting contractions shows that multivariable operator theory differs from the one-variable theory.\n",
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