Prove that there exists at least on fixed point of T. Hint consider the map $T_k = (1-\frac{1}{k})T$

Question

Let $\Omega$ = closed ball $B_1(0)$ in $\mathbb{R}^n$ with metric d induced by the Euclidean norm. Suppose the mapping $T: \Omega \to \Omega$ satisfies

$d(Tx,Ty) \leq d(x,y)$ for all $x,y \in \Omega$

Prove that there exists at least on fixed point of T. Hint consider the map $T_k = (1-\frac{1}{k})T$

So I first start off with proving T is a contraction, nonetheless, consider

$|T_k(x) - T_k(y)| = |(1-\frac{1}{k})T(x) - (1-\frac{1}{k})T(y)|$

$=|(1-\frac{1}{k})(T(x)-T(y))|$

$\leq |1-\frac{1}{k}||T(x)-T(y)$

However, the answer says that

$|T_k(x) - T_k(y)| \leq (1-\frac{1}{k})^2|x-y|$

Where did the square come from is it because we're dealing with the Euclidean norm or did I do something wrong? Any help would be greatly appreciated

Answer 1

So the inequality condition gives the continuity of $T $. Now use Brouwer fixed-point theorem to get your result.