# Restrictions of affinoid Functions from wide open neighbourhoods

Let $$X=\operatorname{Sp}(A)$$ be an affinoid $$K$$-space, where $$K$$ is a p-adic field. Suppose that $$X$$ lies in the interior of another affinoid $$K$$-space $$X'=\operatorname{Sp}(B)$$. Recall that this means $$X$$ is an affinoid subdomain of $$X'$$ and $$B$$ has a system of affinoid generators $$f_1,..., f_n$$ over $$K$$ such that

$$X \subset \{ y \in X': \vert {f_i(y)} \vert <1 \}.$$

Now, let $$U$$ be an affinoid subdomain of $$X'$$ such that $$X \subset \subset_{X'} U$$, where this notation means that $$U$$ is a wide open neighbourhood of $$X$$ in $$X'$$ in the sense of Exercise 7.1.12 in Fresnel and van der Put's book on rigid analytic geometry. I believe that if $$f,g \in \mathcal{O}_{X'}(U)$$ are such that $$f \vert_X = g \vert_X$$ then there should exist an affinoid subdomain $$W \subset U$$ in $$X'$$ with $$X \subset \subset_{X'} W$$ such that $$f \vert_W = g \vert_W$$.

Irritatingly, I cannot prove this. All my proof attempts have been via contradiction because I cannot see any way to construct $$W$$ explictly. I have attempted to use characterisations of relative compactness in the associated Berkovich space $$\mathcal{M}(X)$$ of $$X$$ in order to reduce the problem to a topological one. I also thought that viewing $$U$$ as a strict neighbourhood of $$X$$ in $$X'$$ might be useful (see Fresnel and van der Put, Exercises 7.1.12, 7.7.1). However, neither of these approaches have allowed me to prove the result, so I am asking for help.

If anybody can point me to a proof of this result, or a counter-example, I would be very grateful. Thank you very much in advance - this has really been bugging me!

I'm not sure if your example can be answered in the affirmative in general, but when $$U$$ is smooth and connected one has the following.
Let $$K$$ be a non-trivially valued, non-Archimedean valuation field of characteristic zero and $$U$$ a $$K$$-affinoid variety. If $$U$$ is smooth and connected, and $$X \subset U$$ is any affinoid subdomain then $$f|_X = g|_X$$ implies that $$f = g$$ on the whole of $$U$$. This is a non-archimedean counterpart to the principle of analytic continuation, proven by Ardakov-Ben Bassat in https://arxiv.org/pdf/1612.01924.pdf , Lemma 4.2.