Continuous function $f:X\to X$ such that $d(f(x), f(y)) < d(x, y)$ whenever $x\neq y$, has a fixed point in complete metric space.

Question

Problem Prove or disprove the following:

Let $f : X \to X$ be a continuous function from a complete metric space $(X, d)$ into itself such that $d(f(x), f(y)) < d(x, y)$ whenever $x\neq y$. Then $f$ has a fixed point.

A fixed point of a function is an element of the function's domain that is mapped to itself by the function.

Attempt If we define $f(x)=x+\frac{1}{x} $ with domain in $ [1,\infty )$ then $d(f(x),f(y))<d(x,y)$. Then we can easily see that $f$ hasn't fixed point.

Is it correct enough?

Don't confuse with @Darman's answer. Actually that was my previous attempt.

@Darman's answer is perfectly correct.

Answer 1

No. With your assumption your state isn't true.

If you define $f(x)=x+\frac{1}{x} $ with domain in $ [1,\infty )$ then $d(f(x),f(y))<d(x,y)$, because

$|x+\frac{1}{x}-y-\frac{1}{y}|<|x-y| \Leftrightarrow|\frac{yx^2+y-xy^2-x}{xy}|<|x-y| \Leftrightarrow \frac{|xy(x-y)-(x-y)|}{|xy|}<|x-y| \Leftrightarrow |(xy-1)||x-y|<|xy||x-y| \Leftrightarrow |xy-1|<|xy| \Leftrightarrow -1<0$

So this function have the assumption but if $f(x_0)=x_0$,

$x_0+\frac{1}{x_0}=x_0 $ and this is a contradiction

Answer 2

To add on to @Darman's answer, here's another counterexample to the claim.

Let $X = \{ \text{partial sums of the harmonic series}\} = \left\{ 1, 1 + \frac12, 1+\frac12 + \frac 13, \dots \right\} \subseteq \mathbb R$ with the usual metric.

This space is complete: We know that any subset of $\mathbb R$ is complete iff it is closed. So suppose $(x_n)$ is a sequence in $X$ with $x_n \to x \in \mathbb R$. Then $x_n$ must be bounded, so it can sample only finitely many elements of $X$ (since the harmonic series diverges). Take $\varepsilon$ less than the minimal distance between any two of the finite set of elements, and we get that $x_n$ must in fact be eventually constant. So $x \in X$ and $X$ is closed.

Next let $f: X \to X$ simply map one element to the next, i.e. $1 \mapsto 1 + \frac 12, 1 + \frac 12 \mapsto 1 + \frac 12 + \frac 13$, etc. (check $f$ continuous!). We have $$ |x - y| = \sum_{n = M+1}^N \frac 1n \implies |f(x) - f(y)| = \sum_{n = M+1}^N \frac 1{n+1} < |x - y|, $$ so indeed $d(f(x), f(y)) < d(x, y)$ is satisfied for any $x \neq y$. But it's clear that $f$ has no fixed point.

When you're constructing the set $X = \{S_1, S_2, \dots \}$ with such a sequence $(S_n)$, one really important thing to take note of is that you want $$ \sup_{n\in \mathbb N} \frac{d(f(S_n), f(S_{n+1}))}{d(S_n, S_{n+1})} = 1, $$ for if you could find a $\lambda \in [0, 1)$ s.t. $d(f(x), f(y)) \leq \lambda d(x, y)$, then the contraction mapping theorem says that $f$ does have a fixed point. Indeed, in this example, $$ \frac{d(f(S_n), f(S_{n+1}))}{d(S_n, S_{n+1})} = \frac{n+1}{n+2} \uparrow 1 \quad \text{as } n \to \infty, $$ which is really nice, so it is not as arbitrary as it may initially seem.

It is useful to note that if you replace "complete" with "compact" (or, equivalently in $\mathbb R$, "sequentially compact") in the statement, then the claim would be true. See this post.