Limit of recursive sequence with error term converging to zero Consider a function $f(\cdot)$ and the recursively defined sequence $(x_k)$: $$x_{k+1} = f(x_k)$$ Suppose that for any initial condition $x_0$, $x_k \to x$ where the limit $x$ is independent of $x_0$.
Suppose we have the sequences $(y_k)$ and $(e_k)$, which are related by $$y_{k+1} = f(y_k) + e_k$$ and we also have $e_k \to 0$.
Conjecture: $y_k \to x$.
I am fairly certain this conjecture must hold, either as written under a certain set of conditions on $f$ (I think continuity may be sufficient), but I have been unable to prove it either way.
EDIT: Hagen von Eitzen provided some examples that show that as I originally posed it, my conjecture is false. However, for the situation I am interested in, I have a bit more knowledge. The domain and codomain of $f$ is the set of all positive semidefinite matrices. I do not have $f$ continuous in general, but I do assume that $f$ is continuous at $x$. This implies that $x$ is a fixed point of $f$; that is, $f(x) = x$. Moreover, $x$ is the only fixed point of $f$.
 A: Continuity alone is not sufficient:
Consider $f\colon]-1,1[\to[-1,1[$, $x\mapsto x^2$.  Then $x_k\to 0$ for any staring value $x_0$, but with $y_k:=1-\frac1k$, we also have $e_k\to 0$.
How did I find this? $y_k\to 1$ where $1$ is arepelling fixpoint opf $f$, but itself is not part of the domain of $f$ (so that $x_0=1$ is not allowed).
The same can be seen with $f\colon\Bbb R\to\Bbb R$, $x\mapsto x+\frac{x^3}{x^4+1}$ and $y_k=\sqrt k$, say.
A: Here is another counter-example.
Let $f(x) = \max\{0,\sqrt{x^2-1}\}$. It is easy to check that $f$ is continuous and $x_k \to 0$ for all $x_0 \in [0, \infty)$. On the other hand, if $y_k = \sqrt{k}$, then $e_k := y_{k+1} - f(y_k) \to 0$ while $y_k \to \infty$.
Here the idea is clear: $f$ is chosen so that $f(x) \approx x$ for $x$ with large norm. Then as soon as $y_{k+1} \approx y_k$ and $\|y_k\| \gg 1$, we expect that $y_{k+1} \approx f(y_k)$ as well. But we can design $(y_k)$ so that the difference $y_{k+1} - y_k$ accumulates to create a quite wild behavior, such as $\|y_k\| \to \infty$.
