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In a proof of the Gelfand-Mazur theorem, I read that $$ \lim_{\lambda\to\infty}\|(a-\lambda 1)^{-1}\|=0 $$ (where it is assumed for the sake of contradiction that the inverses in the norm exist for all $\lambda\in\mathbb{C}$)

How can this be proven?

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For $\lvert\lambda\rvert > \lVert a\rVert$, we can write

$$(a-\lambda 1)^{-1} = -\lambda^{-1} (1 - \lambda^{-1}a)^{-1} = -\lambda^{-1}\sum_{\nu = 0}^\infty \lambda^{-\nu} a^\nu,$$

and the triangle inequality together with the submultiplicativity of the norm yields

$$\lVert (a-\lambda 1)^{-1}\rVert \leqslant \frac{1}{\lvert\lambda\rvert} \sum_{\nu=0}^\infty \biggl(\frac{\lVert a\rVert}{\lvert\lambda\rvert}\biggr)^\nu = \frac{1}{\lvert \lambda\rvert - \lVert a\rVert}.$$

This part is true in any Banach algebra, for $\lvert\lambda\rvert > \lVert a\rVert$, the inverse of $a-\lambda 1$ exists, and its norm is bounded by $\dfrac{1}{\lvert\lambda\rvert - \lVert a\rVert}$.

And thus, if an $a$ had empty spectrum, that would give rise to a non-constant bounded entire holomorphic function.

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