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I am trying to prove the following result:

Let $(X,d)$ be a compact metric space, and suppose that $f:X\to X$ satisfies:

$$ d(f(x),f(y))<d(x,y)\,,\forall x,y\in X\,... (1)$$

Then, there is a unique $x^* \in X$ such that $f(x^*)=x^*$.

I am aware there are a couple proofs in here, but I tried a different approach, namely: a direct proof. And I would like some feedback about it.

Proof: Let $x_{0}\in X$ arbitrary but fixed. Let us define the sequence $(x_{k})_{k=0}^{\infty}$ by

$$ x_{k}:=\begin{cases}x_{0}&\text{ if } k=0\\ f(x_{k-1})&\text{ if } k\geq1 \end{cases} $$

That is: we are defining recursively the compostion of $f\circ f$.

Now, since $X$ is compact, there is a subsequence $(y_{l})_{l=0}^{\infty}:=(x_{k_{l}})_{l=0}^{\infty}$ of $(x_{k})_{k=0}^{\infty}$ such that:

$$ \lim_{l\to\infty}y_{l}\to y^* $$

We claim that $y^*$ is a fixed point for the sequence $y_{l}$

This is because:

$$ f(y^*)=f(\lim_{l\to\infty}{y_{l}})=\lim_{l\to\infty}f(y_{l})=\lim_{l\to\infty}y_{l+1}=y^* $$

Where the last two equalities are because:

  1. f is continuous (because $(1)$)
  2. The tail of the sequence has the same limit.

Now, we claim that $x^*:=y^*$ is a fixed point for $(x_{k})_{k=0}^{\infty}$.

Let $(z_{k})_{k=0}^{\infty}$ be defined as follows:

$$ z_{k}:=d(x_{k},x^*) $$

Observe that, $d(x_{k},x^*)\geq d(x_{k+1},x^*)\geq0$ applying the function $f$ and $(1)$.

Then, $(z_{k})_{k=1}^{\infty}$ is a decreasing and bounded by $0$ sequence. Thus it converges to $0$. (I think this step requires a deeper treatment).

Then $x_{k}\to x^*$ and similarly to the sequence $y_{l}$ we show that $x^*$ is a fixed point for $f$.

To show that $x^*$ is unique, suppose that $w^*$ is a different fixed point for $f$. Then, since $w^*\neq x^*$:

$$ 0\leq d(x^*,w^*)<d(x^*,w^*) $$

But that is a contradiction. Then $x^*=w^*$.

$\square$

René Quijada
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  • Your proof is correct! However, "$y^$ is a fixed point for $(x_k)_{k\ge 0}$" does not make any sense. But it is also not required. You have already exhibited the existence of some fixed point $y^$. In the last part, you have argued that $f$ can have at most one fixed point. This completes the proof. – Raghav Jan 18 '22 at 04:29
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    The proof above has a flaw, it is not true in general that $f(y_l) = y_{l+1}$ because $y_l$ is a subsequence of $x_k$. – copper.hat Jan 18 '22 at 05:19
  • @copper.hat you are right. I'll change that line. – René Quijada Jan 18 '22 at 05:22
  • Another flaw, just because something decreases and is bounded below by 0,doesn't mean it converges to 0. Take for example the sequence $\frac{1}{n}+1$. It is decreasing and bounded below by 0 however it converges to 1. – Manatee Pink Jan 18 '22 at 10:25
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    There does not exist a function with $d(f(x),f(y))<d(x,y)$ for all $x$ and $y$. (consider the case $x=y$.) – David C. Ullrich Jan 18 '22 at 11:24
  • In particular, this answer in that duplicate uses the same line of proof and have all the missing pieces. – Arctic Char Jan 18 '22 at 16:45

1 Answers1

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If $X\ne\emptyset$ there obviously cannot exist a function satisfying the hypothesis (considering $x=y$ leads to $0<0$). Hence strictly speaking the result is true, but only vacuously; it's also true that if $f$ is as above then I am the king of France.

A less silly version:

Exercise. If $(X,d)$ is a compact metric space and $f:X\to X$ is such that $d(f(x),f(y))<d(x,y)$ whenever $x\ne y$ then $f$ has a unique fixed point.

Proof. Define $D(x)=d(x, f(x))$. Since $D$ is continuous there exists $x_0$ with $D(x_0)\le D(x)$ for all $x$. Hence $f(x_0)=x_0$, because if not then $$D(f(x_0))=d(f(x_0),f(f(x_0))<d(x_0,f(x_0))=D(x_0).$$Uniqueness is still trivial.

David C. Ullrich
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