Contraction Map on Compact Normed Space has a Fixed Point Let $K$  be a compact normed space  and $f:K\rightarrow K$ such that $$\|f(x)-f(y)\|<\|x-y\|\quad\quad\forall\,\, x, y\in K, x\neq y.$$ Prove that $f$ has a fixed point.

Here is a proof by contradiction:
First we note that if $f(x)=x$ and $f(x')=x'$, we must have $x=x'$ (otherwise $\|f(x')-f(x)\| =\|x'-x\| < \|x'-x\|$, which is a contradiction). Hence any fixed point is unique.
Choose $x_1 \in K$ and let $x_{n+1} = f(x_n)$. If $x_{n+1} = x_n$ for some $n$, we have a fixed point, so assume that $x_{n+1} \neq x_n$ for all $n$. Note that $\|x_{n+2}-x_{n+1}\| < \|x_{n+1} - x_{n}\|$ for all $n$.
Since $K$ is compact, we can find a subsequence such that $x_{n_k} \to x$ and $x_{n_k+1} \to x'$. If $x' = x$ we have found a fixed point, so suppose $x \neq x'$. Then we have $\|f(x')-f(x) \| < \|x'-x\|$. Let $\lambda = \frac{\|f(x')-f(x) \|}{\|x'-x\|}$, and note that $\lambda <1$.
By continuity, for some $\beta \in (\lambda,1)$ we have $\|f(y')-f(y) \| \le \beta \|y'-y\|$ for $y'$ close to $x'$ and $y$ close to $x$.
Let $d_n = x_{n+1}-x_n$. Then we have $\|d_{n+1} \| < \|d_n\|$ for all $n$, and $\|d_{n_{k+1}}\| \le \beta \|d_{n_k} \|$ infinitely often. Hence $d_n \to 0$, which is a contradiction.
