# $f$ has a fixed point if \begin{align} \|f(x)-f(y)\|<\|x-y\|\text{ for all }x,y \in E, x\ne y \ . \end{align}

Let $$E=[a,b]\times[c,d]$$ be a subset in $$\Bbb R^2$$, and $$f:E\to E$$ be a function satisfying \begin{align} \|f(x)-f(y)\|<\|x-y\|\text{ for all }x,y \in E, x\ne y \ . \end{align} Prove that there is a point $$x_0\in E$$ such that $$f(x_0)=x_0$$.

I know the case when the contraction constant is $$0\le c<1$$, but this is a case when $$c=1$$. Thanks for any suggestion.

After reading the comments below, I think I have some idea. Choose $$x_0 \in E$$, and define $$x_{n+1}=f(x_n)$$ for $$n\in \Bbb N$$. Then $$\{x_n\}$$ is a sequence in $$E$$, which is compact, $$\{x_n\}$$ has a convergent subsequence, say $$x_{n_k}\to x$$. By assumption, $$f$$ is continuous on $$E$$, so $$\lim_\limits{k\rightarrow\infty}f(x_{n_k})=f(\lim_\limits{k\rightarrow\infty}x_{n_k})=f(x)$$. On the other hand, $$\lim_\limits{k\rightarrow\infty}f(x_{n_k})=\lim_\limits{k\rightarrow\infty}x_{n_{k+1}}=x$$. Hence, we get $$x=f(x)$$.

• Notice that your metric space is compact. Dec 7, 2020 at 6:01
• Moreover, it is better not to use the notation $||\cdot||$ because $E$ is not a vector space. Dec 7, 2020 at 6:18
• @DannyPak-KeungChan $E$ is a subset of a normed space, so there is no issue. Dec 7, 2020 at 6:20
• I have edited my post. Dec 7, 2020 at 6:33
• @Kavi Rama Murthy: Such a $c<1$ may not exist apriori, e.g. the map $x\mapsto x^2$ from $[0,\frac{1}{2}]$ to itself. Dec 7, 2020 at 6:36

Let $$g:E\rightarrow[0,\infty)$$ be defined by $$g(x)=d(x,f(x))$$. Clearly $$g$$ is continuous. Since $$E$$ is compact, $$g$$ attains it minimum. Choose $$x_0\in E$$ such that $$g(x_0)=\min g(E):=\xi$$. If $$x_{0}\neq f(x_{0})$$, then $$g(f(x_{0}))=d(f(x_{0}),f(f(x_{0}))), a contradiction.
Consider $$x_0$$ such that $$d(x_0, f(x_0))$$ is smallest.