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If $X$ is a compact metric space and $f$ is a function from $X$ to $X$ such that $$d(f(x),f(y))<d(x,y)$$ for all distinct points $x$ and $y$ in $X$. How do I show that $f(x)=x$ for some $x\in X$?

If it were a question about real numbers, I would have defined $g(x)=f(x)-x$ and proceeded, but subtraction may not be defined for a general metric space.

I also tried to prove by contradiction, but didn't work.

Martund
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    Hint: minimise $x\mapsto d(x,f(x))$. – Gary Mar 30 '22 at 02:03
  • BThis is the banach fixed point theorem... – Clemens Bartholdy Mar 30 '22 at 02:04
  • @Buraian In some sense yes. Banach requires a multiplier $0<q<1$ on the RHS. If $X$ was not compact, the statement in the question would not hold. On the other hand Banach (with $q$) does not require compactness. – Gary Mar 30 '22 at 02:05
  • You can see this on an intuitive level by constructing the sequence $ \left( f(x),f(y) \right), \left( f^2( x) , f^2(y) \right).....$ You can see the valuation of the distance function of the pair of point on this sequence keeps decreasing as you take higher and higher composition. This means that $x$ and $y$ are coalesce to one point with higher and higher compositions – Clemens Bartholdy Mar 30 '22 at 02:08
  • @Gary I realized that, after proving the function $x\mapsto d(x,f(x))$ is a continuous function. So it should attain its minimum on a compact set. But I don't know what to do further. – Martund Mar 30 '22 at 02:11
  • @Martund Assume the minimum is attained at $x$. What if $f(x) \neq x$? Hint: use the inequality in the problem to show that in that case $x$ was not a minimiser. – Gary Mar 30 '22 at 02:12
  • @Gary Oh I thought the interval were closed on Banach. My bad. – Clemens Bartholdy Mar 30 '22 at 02:12
  • Another one: https://math.stackexchange.com/q/2034383/42969 – Martin R Mar 30 '22 at 02:37

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Set $M = \inf\{ d(x,f(x)): x\in X\}$.

Note that there exists a point $p\in X$ so that $d(p,f(p)) = M$. This fact is true because for each $n\in \mathbb{N}$ there exists $p_{n} \in X$ so that $d(p_{n},f(p_{n})) \leq M+\frac{1}{n}$. A convergent subsequence of $p_{n_{1}},p_{n_{2}},...$ will converge to some $p$ due to compactness and note that for each $j \in \mathbb{N}$ we have

$$d(p,f(p)) \leq d(p,p_{n_{j}}) +d(p_{n_{j}},f(p_{n_{j}}))+d(f(p_{n_{j}}),f(p))$$

$$\leq d(p,p_{n_{j}}) +d(p_{n_{j}},f(p_{n_{j}}))+d(p_{n_{j}},p)$$

$$\leq M \text{ when taking }j \rightarrow \infty.$$

Now if $M > 0$ then there exists a point $p$ so that $d(p,f(p)) = M$ and $d(f(p),f(f(p))) < d(p,f(p)) < M$. This is a contradiction. Hence $M = 0$ and there is a point $p$ with $d(p,f(p)) = 0 \Rightarrow p = f(p)$.

acreativename
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