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I'm trying to determine whether or not the standard proof that the spectrum of a point in a unital Banach algebra is non-empty can be adapted to prove the same thing over certain non-Archimedean fields, in particular the field $\hat{\mathbb{C}}$ of complex Hahn series.

The standard proof involves showing that if $\sigma(a)$ is empty, then for each $\psi\in A^*$, the map $\lambda\mapsto\psi((\lambda1_A-a)^{-1}):\mathbb{C}\to\mathbb{C}$ is bounded, entire and vanishes at infinity, so vanishes on the whole of $\mathbb{C}$ by Liouville's Theorem, which leads to a contradiction.

There are some analogues of Liouville's Theorem that work over non-Archimedean fields e.g. Theorem 1.4.5 here and one in Schikhof's Ultrametric Calculus, although these may not be strong enough to apply to $f$.

For example, even if $f$ vanishes at infinity, we still need to show that it is bounded on a closed ball centred at the origin. But in this non-Archimedean case, closed balls are non-compact, so it might be the case that $f$ is unbounded and a stronger form of Liouville's Theorem would be required.

Alternatively, I am looking for counterexamples to disprove this. Any advice as to how I might resolve this issue would be much appreciated. :)

A related question of mine: Literature on non-Archimedean analogues of basic complex analysis results

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    $\begingroup$ This paper, in Example 2.4, gives an example of a bounded operator on a $\mathbb C_p$-Banach space with empty spectrum. So it certainly doesn't hold in general. I suspect you can adapt it to work over other fields. $\endgroup$
    – Wojowu
    Dec 27, 2021 at 22:18
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    $\begingroup$ For a complete non-archimedean valued field $K$, a power series $\sum a_nx^n$ on $K$ with infinite radius of convergence is bounded on each disc in $K$. You don't need compactness. Indeed, if the radius of convergence is infinite then $|a_n|^{1/n} \to 0$ as $n \to \infty$. So for each $r > 0$, we have $|a_n|^{1/n} \leq 1/r$ for $n \geq N_r$ (bound depends on $r$). Thus $|x| \leq r \Rightarrow |a_nx^n| = (|a_n|^{1/n}|x|)^n \leq 1$ for $n \geq N_r$. Hence $|a_nx^n| \leq \max\{|a_n|r^n, 1 : n \leq N_r\}$, which is a uniform upper bound on $|\sum a_nx^n|$ for all $x$ such that $|x| \leq r$. $\endgroup$
    – KConrad
    Dec 28, 2021 at 1:40
  • $\begingroup$ @KConrad I understand your argument, but how do we know that (in the non-Archimedean case) an entire function can be expanded as a power series on the whole of $K$? Virtually all of the literature I have encountered on 'non-Archimedean complex analysis' starts with power series, yet here I have a function that is holomorphic in the sense of complex differentiable. It can certainly be expanded as a power series for $|\lambda|>\|a\|$ by Carl Neumann Criterion, but surely a 'holomorphic iff analytic' type result is needed? $\endgroup$
    – Very Forgetful Functor
    Jan 3, 2022 at 22:33
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    $\begingroup$ How exactly do you define "entire function"? In nonarchimedean context, one usually defines them as functions represented by a globally convergent power series. The reason one does it is that in nonarchimedean world, locally analytic functions are very weirdly behaved because underlying spaces are totally disconnected. It is not true that for a function analytic at every point, the Taylor expansions have infinite radius of convergence, and if it does, it need not converge to the original function everywhere. $\endgroup$
    – Wojowu
    Jan 3, 2022 at 22:44
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    $\begingroup$ I agree with Wojowu: nonarchimedean entire functions are defined as being expressible by a single power series with infinite radius of convergence. There is no result like "differentiable implies power series representation" for functions on a disc in non-archimedean analysis. After all, the function on a nonarchimedean field that's 1 when $|x| < 1$ and 0 when $|x| \geq 1$ is locally expressible as a power series (it's locally constant!) but it sure isn't expressible as a single power series. If you want to prove something has a global power series representation, try to prove it directly. $\endgroup$
    – KConrad
    Jan 4, 2022 at 0:37

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