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.

Solution 1:

The sequence $K \supset f(K) \supset f(f(K)) \supset \cdots$ is a nested sequence of compact sets. Then their intersection is non-empty.

Denote the intersection by $A$. Then since

$$A =\cap f^{(n)}(K)$$

it is easy to show that $f(A)=A$.

Note there is a typo in the problem, the inequality can only hold for $x \neq y$.

Since $A$ is non-empty and compact, by This Question you can find $a_1,a_2 \in A$ so that

$$\|a_1-a_2\| = \operatorname{diam}(A) \,.$$

I claim that $a_1=a_2$ which shows that $A$ consists of a single point.

Indeed, as $f(A)=A$, pick $x_1,x_2 \in A$ so that $f(x_i)=a_i$. If we assume by contradiction that $a_1 \neq a_2$ then

$$\operatorname{diam}(A)=\|a_1-a_2\| = \| f(x_1)-f(x_2) \| < \|x_1-x_2 \|$$

but this is a contradiction, as $x_1,x_2 \in A$.

As $A$ is a single point set and $f(A)=A$ we are done.

Solution 2:

Edit: Note that $f$ is continuous. (Why?) Define $$g(x)=\lVert f(x)-x\rVert$$ for all $x\in K$. Use the compactness of $K$ to show that $g$ obtains a non-negative minimum. If $f$ has no fixed point, we can then derive a contradiction. (Apologies for my earlier, bogus approach.) You can further show that $f$ has a unique fixed point, if you like.

Solution 3:

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.