Is the Gauss norm for $p$-adic power series a supremum of values taken over the unit disk?

Let $$A$$ be the ring of convergent power series over $$\mathbf{Q}_p$$ in one variable. That is, elements of $$A$$ are formal series $$f(t) = \sum_{m=0}^\infty a_m t^m$$ such that the coefficients $$a_m$$ tend to zero in $$\mathbf{Q}_p$$ as $$m \rightarrow \infty$$.

We define the Gauss norm on $$A$$ by setting $$\Vert f(t) \Vert = \sup_m |a_m|$$ It is obvious that for any point $$z \in \mathbf{Z}_p$$, we have $$|f(z)| \leq \Vert f(t) \Vert$$.

Failed attempt (see edit below): In fact, this upper bound should be attained i.e. there should be some $$z \in \mathbf{Z}_p$$ with $$|f(z)| = \Vert f(t) \Vert$$.

How do we show the existence of such a $$z$$?

Edit: So it turns out the above is not true. The supremum is not always necessarily attained. Here is a revised question that is hopefully more true.

Let's replace $$\mathbf{Q}_p$$ with its completed algebraic closure $$\mathbf{C}_p$$. The definition of convergent power series and Gauss norms for $$\mathbf{C}_p$$ are unchanged. Now we have

$$\sup_{z \in \mathcal{O}_{\mathbf{C}_p}} |f(z)| \leq \Vert f(t) \Vert$$

Can we upgrade this inequality to an equality?

I have dropped the requirement that the upper bound on the right side be attained by some $$z \in \mathcal{O}_{\mathbf{C}_p}$$.

• The max is not attained for $z \in \mathbb{Z}_p$. Counterexample, $f(t)= t^p-t$. Dec 3, 2020 at 14:53
• @Merosity What if we replace $\mathbf{Q}_p$ with its algebraic closure (or the completion of its algebraic closure) $K$? Will it be attained in $\mathcal{O}_K$ then?
– Alex
Dec 3, 2020 at 15:08

$$\|f\| = \sup_n |a_n|_p = \sup_{b\in O_K} |f(b)|_p, \qquad O_K=\bigcup_{n\ge 1}\Bbb{Z}_p[\zeta_{p^n-1}]$$ Proof: $$\sup_{b\in O_K} |f(b)|_p\le \|f\|$$, and $$\|f\| = |p^m|_p$$ means that $$f=p^m g+p^{m+1}h$$ where $$g\in \Bbb{Z}_p[x],h\in A\cap \Bbb{Z}_p[[x]]$$ and $$g\bmod p$$ is a non-zero, this polynomial has finitely many roots in the field $$O_K/(p)$$ thus for some $$b\in O_K$$, $$g(b) \in O_K^\times$$ and $$|f(b)|_p = |p^m g(b)|_p = \|f\|$$.
We can replace $$K$$ by any extension of $$\Bbb{Q}_p$$ with an infinite residue field.