A doubt on a problem involving continuous function on a compact metric space

Question

Let $X$ be a compact metric space with metric $d$ and let $f \in C (X, X)$ and such that $d (f(a ),f(b))\ge d (a, b)$ for all $a$ and $b$ in $X.$ Show that $d(f(a), f(b)) = d(a, b)$ for all $a$ and $b$ in $X.$

My attempt:

$X$ is compact implies that $f(X)$ is also compact, since $f$ is continuous. Also, $f$ is uniformly continuous.

On the contrary, let us assume that there exists a pair of points $(x,y) \in X^2$ such that $d (f(x ),f(y)) > d (x, y).$ Since, $X$ is compact, diameter of $X$ is finite, say $M.$ And there exits points $x_0,y_0 \in X$ such that $d(x_0,y_0)=M.$

Suppose that $(x_0,y_0)=(x,y),$ we are done with the contradiction that $d (f(x_0 ),f(y_0)) > d (x_0, y_0)>M,$ since diameter of $f(X) \subseteq X$ is atmost M.

Now we are left with the case where $d(x,y) < M.$

I am stuck here. Please help. Thanks in advance.

Answer 1

I am skipping some of the details that are easy to fill. Also, I use sup instead of max but it does not matter for this proof. The set $f(X)$ is a compact subset of X. Hence, $\sup_{(x,y)\in X^2} d(f(x),f(y))\leq \sup_{(x,y)\in X^2} d(x,y)$. Observe that $X^2$ is also compact and the left hand side is supremum taken over a subset of $X^2$, so the sup cannot dominate.

However, from the property of the metric in the question it also follows that $\sup_{(x,y)\in X^2} d(f(x),f(y))\geq \sup_{(x,y)\in X^2} d(x,y)$.

So, $\sup_{(x,y)\in X^2} d(f(x),f(y))= \sup_{(x,y)\in X^2} d(x,y)$ must hold.

In fact, for any closed subset $A$ of $X^2$ we must have, $\sup_{(x,y)\in A} d(f(x),f(y))= \sup_{(x,y)\in A} d(x,y)$ must hold.

If there is some $a,b \in X$ such that $d(f(a),f(b))> d(a,b)$ we simply define the closed set $A=\{(x,y):d(x,y)\leq d(a,b)\}$ and note that $\sup_{(x,y)\in A} d(x,y)=d(a,b)<d(f(a),f(b))\leq \sup_{(x,y)\in A} d(f(x),f(y))$. This contradiction completes the argument.

Answer 2

Verify that $G(x)=d(f(x),x)$ is a continuous real valued function on the compact set X. Hence, G attains a maximum and a minimum. Let $x^*$ be the point of maximum of $G$. Assume that $G(x^*)>0$. By the property of the metric given in the question and the fact that $x^*$ is the point where $G$ attains the maximum, we must have $G(f(x^*))=G(x^*)$ and $G(x^*)>0$ also implies that the sequence $<f^t(x^*)>$ has distinct points in X, that is, $f^t(x^*)\neq f^{t+1}(x^*)$ . Since X is a compact metric space, it is sequentially compact and hence $<f^t(x^*)>$ has a convergent subsequence. In fact given that the distance between points of the sequence $t,t+1$ is at least as large as that of $t-1, t$ the sequence itself must converge. But the distance between any two consecutive points of this sequence is bounded below by $G(x^*)>0$, which leads to the desired contradiction. Hence, $G(x)=d(f(x),x)\leq G(x^*)=0$ implies $f(x)=x$ for all $x\in X$.