Banach fixed point theorem and $\sum d(T^n(x),T^n(y))<\infty$ Let $T:X \to X$ be a map on a complete non-empty metric space.
Assume that for all $x$ and $y$ in $X$, $\sum_n d(T^n(x),T^n(y))<\infty$. Then $T$ has a unique fixed point.
guess: I assume that the existence and unicity of a fixed point can be shown directly with the standard Banach fixed point theorem, by a suitable choice of the metric that makes the map $T$ a contraction.
 A: Why isn't the following a counterexample?
Let $X=\{0\}\cup\{1/2^n:n\in\mathbb{N}\}$ and endow $X$ with the absolute-value-metric. Then $X$ is complete. Define $T:X\to X$ by $T(0)=1/2$ and $T(1/2^n)=1/2^{n+1}$. Then $T$ has no fixed point. Now $|T^n(1/2^l)-T^n(1/2^m)|\leq1/2^n$ and every point has this form after at least one application of $T$. So the summability condition should hold too.
A: I am assuming map means continuous function. Otherwise Michael Greinecker's answer gives a counterexample.
Existence:
Take any point $x_0 \in X$ and define inductively $x_{n+1} := T(x_n)$. Given $\epsilon>0$ and assuming $m>n>N$, we have $$d(x_n,x_m) \leq \sum_{k=0}^{m-n-1} d(x_{n+k}, x_{n+k+1}) \leq \sum_{k=0}^\infty d(x_{N+k}, x_{N+k+1}) = \sum_{k=N}^\infty d(T^k(x_0),T^k(x_1)) \leq \epsilon$$ for sufficiently large $N$. By completeness there exists $\bar{x}\in X$ with $\lim_{n\to\infty}x_n=\bar{x}$ and by continuity of $T$ we get 
$$T(\bar{x}) = T(\lim_{n\to \infty} x_n) = \lim_{n\to \infty}T(x_n) = \lim_{n\to \infty} x_{n+1} = \bar{x}.$$
Uniqueness:
Let $\bar{x}$ and $\tilde{x}$ be fixed points of $T$, then $\sum_{k=0}^\infty d(T^k(\bar{x}),T^k(\tilde{x})) = \sum_{k=0}^\infty d(\bar{x},\tilde{x}) < \infty$, so $d(\bar{x},\tilde{x})=0$, hence $\bar{x} = \tilde{x}$.
