Let $A \leftarrow C \rightarrow B$ be affinoid $K$-algebras, where $K$ is a non-archimedean field with non-trivial absolute value. Equipping $A$, $B$, $C$ with the supremum seminorms, there is a canonical seminorm $\nu_1$ on $A \otimes_C B$: $$\nu_1(f) = \inf \max_i |a_i|_{\sup} |b_i|_{\sup},$$ the infimum taken over all representations $f = \sum a_i \otimes b_i$. Then $\nu_1$ extends by continuity to a seminorm on the completed tensor product $A \widehat{\otimes}_C B$ (where the latter is endowed with any residue norm). On the other hand, $A \widehat{\otimes}_C B$ is known to be an affinoid algebra in its own right, so has its own supremum seminorm $\nu_2$.

Does $\nu_1 = \nu_2$?

I'm hopeful this is true, but doubt it (although I have no specific counter-example). It's clear that $\nu_2 \le \nu_1$, and the equality holds if and only if $\nu_1$ is power-multiplicative (i.e. $\nu_1(f^n) = \nu_1(f)^n$).

If it helps, we may assume $C$ is a Tate algebra and $\text{Sp} \, A$, $\text{Sp} \, B$ are affinoid subdomains of $\text{Sp} \, C$.

Orthonormalbasen in der nichtarchimedischen Funktionentheoriein Manuscripta Math 1. Maybe that paper yields $\nu_1=\nu_2$ for $K$ algebraically closed? $\endgroup$ – nfdc23 Sep 3 '17 at 21:08