Analytic elements in non-archimedean geometry Let $(k,|.|)$ be a complete non-archimedean valued field. Let $D$ be the open unit disc over $k$. (Anything I write could be adapted to the case of an open annulus.) The ring $\mathcal{O}(D)$ of analytic functions on $D$ admits an explicit description: it consists of the series $\sum_{n\ge 0} a_n T^n$ such that for any $r<1$, the sequence $(|a_n|r^n)_{n\ge 0}$ tends to 0. If such a function is bounded, its uniform norm is precisely $\sup_{n\ge 0}(|a_n|)$.
For some purposes, it is convenient to consider smaller rings: the ring of bounded functions or the ring of analytic elements, for instance. Let me recall that this latter ring $\mathcal{H}(D)$ is defined as the completion of $k(T) \cap \mathcal{O}(D)$ (i.e. rational functions with no poles in $D$) for the uniform norm on the disc. Remark that this definition depends on the choice of a coordinate $T$ on $D$. Those functions have nice properties: for example, they only have a finite number of zeros on $D$. We refer to Gilles Christol's book (chapter I) for a more detailed account. 
Consider the disc $D$ inside the affine line $\mathbf{A}^1$. Let $X$ be an affinoid domain of $\mathbf{A}^1$ that contains $D$. One may check that any function on $X$ restricts to an analytic element on $D$, whatever the choice of the coordinate (because the approximation property actually holds for any function on $X$). 
The question is the following: if $D$ is embedded inside an arbitrary affinoid space $X$, do functions on $X$ still restrict to analytic elements? 
The issue is that the choice of a coordinate in the definition of $\mathcal{H}(D)$ makes it difficult to use outside the affine line. I would appreciate any idea that helps recognizing analytic elements on a disc inside an arbitrary curve. 
 A: Vladimir Berkovich has indicated to me that the answer is no: on a general curve, functions need not restrict to analytic elements. Let me copy here his argument.
The algebra of analytic elements $\mathcal{H}(D)$ can be defined as the completion of the inductive limit of the $k$-affinoid algebras $\mathcal{A}_X$ with respect to the supremum norm on $D$ taken over all affinoid domains $X$ in the affine line that contain $D$. It follows that the spectrum of $\mathcal{H}(D)$ is the closure of $D$, i.e. the union of $D$ with the Gauss point $x$, and the non-Archimedean field $\mathscr{H}'(x)$ associated to $x$ with respect to the Banach algebra $\mathcal{H}(D)$ coincides with the similar field $\mathscr{H}(x)$ of the Gauss point in the affine line. In particular, the residue field of $\mathscr{H}'(x)$ is the field of rational functions on the projective line over the residue field of $k$.
Let now $Y$ be the analytification of the generic fiber of a smooth projective curve $Z$ over the ring of integers of $k$. The space $Y$ contains a unique "generic point" $y$, whose image under the reduction map is the generic point  of the closed fiber of $Z$, and the complement of $y$ in $Y$ is a disjoint union of open unit discs (assume for simplicity that $k$ is algebraically closed). Pick up one of them $D$, and consider the similar Banach algebra $\mathcal{H}'(D)$ which is the completion of the intersection of $k(Z)$ and $\mathcal{O}(D)$ with respect to the supremum norm on $D$. It coincides with the completion of the inductive limit of the $k$-affinoid algebras $A_X$ with respect to the supremum norm on $D$ taken over all affinoid domains $X$ in $Y$ that contain $D$. The spectrum of $\mathcal{H}'(D)$ coincides with the closure of $D$ in $Y$, i.e. the union of $D$ with the point $y$, and the non-Archimedean field $\mathscr{H}'(y)$ associated to $y$ with respect to the Banach algebra $\mathcal{H}'(D)$ coincides with the similar field $\mathscr{H}(y)$ of $y$ with respect to $Y$. It follows that the residue field of $\mathscr{H}'(y)$ is the field of rational functions on the closed fiber of $Z$. If the genus of $Z$ is positive, then so is the genus of its closed fiber. This implies that the functions on $X$ (in $Y$) do not necessarily restrict to analytic elements on $D$.
