Have a look at the fixed-point theorem ( the result holds for any f: M-> M where M is a complete metric space and f is such that |f(x)-f(y)| <= K*|x-y| where K<1 ).

It is quite classical am I wrong? Unicity : If $x \neq y$ are such that $f(x)=x$ and $f(y)=y$ then $|x-y| = |f(x)-f(y)| < c|x-y| < |x-y|$ which is absurd. Existence : Let $a \in A$. Let $x_0 = a$ and $x_{n+1} = f(x_n).$ (1) Thus $|x_{n+1}-x_n | < c |x_n - x_{n-1}|.$ Using that $c<1$, it is now easy to deduce that the sequence has the Cauchy property. Since $\mathbb {R}^m$ is a complete space, we deduce that the sequence has a limit, say $L$. Since $A$ is closed and each of the $x_i$ is in $A$, it follows that $L$ is in $A$. That $L = f(L)$ is obvious from (1). Pierre.