Let $x\_n$ be a increasing sequence of negative real numbers that converge to $0$. Let $f$ be a function defined on $x\_n$ such that $f(x\_n)$ is a increasing sequence of negative numbers that converge to zero. Can I find a $C^\infty$ or real-analytic function $g$ defined on a ball around zero such that $g(x\_n)=f(x\_n)$ for all $n$ larger than some positive integer? 



smooth/real analytic interpolation of monotonic sequence
As fedja says in the comments, the answer is "No" in general. However, there is a known condition on when it is possible to interpolate a sequence by a smooth function. This can be found in Section 2.8 of Chapter I of Kriegl and Michor's [The Convenient Setting for Global Analysis](http://www.ams.org/online_bks/surv53/) (though it is not said where the concept originates from).


The set up is as follows: we have a locally convex topological vector space $E$, and a sequence $(x\_n)$ in $E$. We say that $(x\_n)$ **converges fast** or **falls fast** to $x \in E$ if, for each $k \in \mathbb{N}$, the sequence $n^k(x\_n - x)$ is bounded. Then the result is:


**Special Curve Lemma:** Let $(x\_n)$ be a sequence which converges fast to $x$ in $E$. Then the infinite polygon through the $(x\_n)$ can be parameterised as a smooth curve $c \colon \mathbb{R} \to E$ such that $c(1/n) = x\_n$ and $c(0) = x$.


Although this is only an "if", it shouldn't be hard to check whether the "only if" holds or not. For a quick proof of the above, see Kriegl and Michor's book (p16 in the printed version; it's free online via the above link).


This uses the notion of smoothness in LCTVS's that Kriegl and Michor describe (in great detail) in their book. For finite dimensional vector spaces, it is the same as the usual one.


