Basic inequality in a Banach Algebra

Question

Suppose $x$ is an element of a Banach Algebra. Let $U$ be an open set in $\mathbb{C}$ containing the spectrum of $x$. Is it true, that inf $\{\|(x-\lambda1_A)^{-1}\|^{-1}:\lambda \in \mathbb{C}-U \}>0?$

Answer 1

Yes, it's true. In particular, note that $V = \Bbb C - U$ is a closed subset of $\Bbb C$. As such, the continuous function $$ f:V \to [0,\infty)\\ \lambda \mapsto \|(x - \lambda 1)^{-1}\|^{-1} $$ Satisfies $\lim_{|\lambda| \to \infty} \|(x - \lambda 1)^{-1}\|^{-1} = \infty$. As such, there exists an $R > 0$ such that $|\lambda| > R \implies f(\lambda) > 1$. By continuity, $f$ must attain a minimum over the compact set $\{\lambda \in V : |\lambda| \leq R\}$. However, since we cannot have $f(\lambda) = 0$, this minimum cannot be zero.

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For that limit: note that $$ \lim_{|\lambda| \to \infty} \|(x - \lambda 1)^{-1}\|^{-1} = \lim_{|\lambda| \to \infty} |\lambda| \cdot \|(\frac 1{\lambda}x - 1)^{-1}\|^{-1} $$ but $\|(\frac 1{\lambda}x - 1)^{-1}\| \to \|1\| = 1$.