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.
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 $b \ne f(b)$.