# Existence of fixed point given a mapping

Let $A$ be a closed and bounded subset of $\mathbb{R}^{n}$. Given that $f: A \to A$ satisfies $d(f(x),f(y)) < d(x,y)$ for all $x,y \in A$, $x\neq y$ where $d$ is the usual Euclidean metric. Show that $f$ has a unique fixed point.

My observations:

• Since $\mathbb{R^{n}}$ is complete, then $(A,d)$ is complete since $A$ is closed;

• Since $A$ is bounded, then $A \subset B_{R}$ for some $R > 0$;

• Banach Fixed Point Theorem guarantees a unique fixed point if $f$ is contraction on $A$.

Please correct me if I am wrong. My quandry here is showing that $f$ is a contraction, i.e.

$$d(f(x),f(y)) \leqslant k\cdot d(x,y), \hspace{3mm} k \in (0,1)$$

I'm looking for some hints or tips. Thank you.

• Maybe you are looking at some kind of Brouwer fixed point theorem ? – Svetoslav Feb 18 '16 at 8:43

I do not know if there is a direct way to show the $f$ is a contraction, but the following works to show that $f$ has a fixed point:
Note that $f$ is Lipschitz and hence continuous. Define $g \colon A \to \mathbf R$ by $$g(x) = d\bigl(x, f(x)\bigr)$$ As $A$ is compact, $g$ attains its minimum on $A$, say at $a$. If $a$ is no fixed point, then $a \ne f(a)$, hence \begin{align*} g\bigl(f(a)\bigr) &= d\bigl(f(a), f^2(a)\bigr)\\ &< d\bigl(a, f(a)\bigr)\\ &= g(a) \end{align*} contradicting that $a$ is a minimum. Hence $a = f(a)$.
To see that $a$ is unique, let $a \ne b$, then $$d\bigl(a,f(b)\bigr) = d\bigl(f(a), f(b)\bigr) < d(a,b)$$ Hence $b \ne f(b)$.
• There is no direct way to show that $f$ is a contraction. As a counterexample consider $f : [0,1]\to \Bbb{R},x \mapsto x^2 / 2$. The mean value theorem implies that $f$ satisfies the assumptions, but 1 is the best Lipschitz constant for $f$. – PhoemueX Feb 18 '16 at 18:16
• Could you explain why the statement contradicts that $a$ is a minimum? I do not quite follow. – clocktower Feb 21 '16 at 2:56
• If $g (fa)< g (a)$, then $g (a)$ is not minimal. – martini Mar 8 '16 at 12:42