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 $x=f(x)$.

  • 2
    $\begingroup$ Notice that your metric space is compact. $\endgroup$ Dec 7, 2020 at 6:01
  • $\begingroup$ Moreover, it is better not to use the notation $||\cdot||$ because $E$ is not a vector space. $\endgroup$ Dec 7, 2020 at 6:18
  • 1
    $\begingroup$ @DannyPak-KeungChan $E$ is a subset of a normed space, so there is no issue. $\endgroup$
    – copper.hat
    Dec 7, 2020 at 6:20
  • $\begingroup$ I have edited my post. $\endgroup$
    – Steven Lu
    Dec 7, 2020 at 6:33
  • $\begingroup$ @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. $\endgroup$
    – orangeskid
    Dec 7, 2020 at 6:36

2 Answers 2


Following orangeskid's hints, the proof is straight forward:

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})))<d(x_{0},f(x_{0}))=\xi$, a contradiction.

  • $\begingroup$ Thanks, I got it. $\endgroup$
    – Steven Lu
    Dec 7, 2020 at 7:12


Consider $x_0$ such that $d(x_0, f(x_0))$ is smallest.


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