# Norm of inverse, Banach algebra, Proof of Gelfand-Mazur theorem

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?

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