show that a function $A$ such that $ \rho (Ax,Ay)< \rho (x,y) $ $ \forall x\neq y $ not necessary has a fix point $ (Ax\neq x \space \forall x )$ I don´t know an  example wich   $ \rho (Ax,Ay)< \rho (x,y) $    $  \forall x\neq y $   is not sufficient for the existence of a fixed point .
can anybody help me? please
 A: $f:(0,1)\to (0,1)$ with $f(x)=x/2$.
A: Even simpler example, although not on whole $\mathbb R$:
$$f\!: \mathbb R\setminus\{0\}\to\mathbb R,\quad f(x) = \frac 12 x.$$
(This is what ThePortakal said in commment.)
A: HINT: Take your space to be $\Bbb R$, and try $A(x)=x-f(x)$, where $f$ is positive and increasing and satisfies 
$$\frac{f(y)-f(x)}{y-x}<1$$
whenever $x<y$. You can get such an $f$ by tinkering with the arctangent function.
A: I was about to answer this question here, but it got marked as a duplicate of this one. Thus, I will answer it on this thread.

Let $T$ be a mapping of a complete metric space $X$ into itself. How do I show that the condition $d(Tx, Ty) < d(x, y)$, where $x \neq y$, does not suffice for the existence of a fixed point for $T$?

Let $$F(y) = \int_{-\infty}^y e^{-\pi x^2} dx.$$The relevant properties of $F$ are that $F$ is strictly positive, strictly increasing in $y$, and$$0 < F(y_1) - f(y_2) \le y_1 - y_2$$whenever $y_1 > y_2$. (Convergence of this integral is evident by comparing to $e^{-|x|}$ on $|x| > 1$, and monotonicity because the integrand is positive. The last part follows from the Mean Value Theorem because $F'(y) = e^{-\pi y^2}$ and $0 < e^{-\pi y^2} \le 1$.) We then claim that $T(y) = y- F(y)$ is strictly decreasing: if $y_1 > y_2$, then the inequalities$$0 < F(y_1) - F(y_2) \le y_1 - y_2$$are equivalent to $$y_1 - y_2 > y_1 - y_2 - (F(y_1) - F(y_2)) \ge 0.$$The middle quantity above is $T(y_1) - T(y_2)$ so $T$ decreases distances on $\mathbb{R}$, but clearly $T$ has no fixed points because $T(y) < y$ for all $y$.
