Abstract

Bohr's inequality says that if $f(z) = \sum^{\infty}_{n = 0} a_n z^n$ is a bounded analytic function on the closed unit disc, then $\sum^{\infty}_{n = 0} \lvert a_n\rvert r^n \leq \Vert f\Vert_{\infty}$ for $0 \leq r \leq 1/3$ and that $1/3$ is sharp. In this paper we give an operator-theoretic proof of Bohr's inequality that is based on von Neumann's inequality. Since our proof is operator-theoretic, our methods extend to several complex variables and to non-commutative situations.We obtain Bohr type inequalities for the algebras of bounded analytic functions and the multiplier algebras of reproducing kernel Hilbert spaces on various higher-dimensional domains, for the non-commutative disc algebra ${\mathcal A}_n$, and for the reduced (respectively full) group C*-algebra of the free group on $n$ generators.We also include an application to Banach algebras. We prove that every Banach algebra has an equivalent norm in which it satisfies a non-unital version of von Neumann's inequality.2000 Mathematical Subject Classification: 47A20, 47A56.

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