convergence sequence and continuous functions I have the following question: assume $C$ is a subset (we can assume it is convex and compact) of a Banach space $(X,\|\cdot\|)$, $f:C\longrightarrow C$ continuous and $x_{0}\in C$ a fixed point of $f$. Also, let $(x_{n})_{n\geq 1}\subset C$ be a sequence having a subsequence converging to $x_{0}$, and such that
$$\Big| \|x_{n+1}-x_{0}\|-\|f(x_{n})-f(x_{0})\| \Big|\leq c_{n},$$
for each $n\geq 1$, where $c_{n}$ is a positive sequence of numbers with $c_{n}\rightarrow 0$.
From the above conditions, can we derive that $x_{n}\rightarrow x_{0}$? 
I know that if $f$ is nonexpansive (i.e. $\|f(x)-f(y)\|\leq \|x-y\|$) the assett is true under some additional conditions on the sequence $(c_{n})_{n\geq 1}$,but assuming only the continuity of $f$, it is not clear.
What do you think? 
Many thanks in advance for your comments!

Without additional conditions on the sequence $(c_{n})_{n\geq 1}$ (maybe, that the series $\sum c_n$ converges) the asset may be false even for the segment  $[0,1]$ endowed with the standard metric and the identity map $f$. As a counterexample it suffices to put $x_0=0$ and $\{x_n\}$ any sequence of points of $[0,1]$ such that $\lim_{n\to\infty} |x_{n}-x_{n+1}|=0$,  $\{x_n\}$ contains a subsequence convergent to $x_0$, but  $\{x_n\}$ does not converges to $0$. 
