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Let $(X, \rho)$ be a compact metric space.

Suppose that $T: X \rightarrow X$ and for all $u \neq v \in X$,

$$ \rho(T(u),T(v)) < \rho(u,v) $$

Then show that $T$ has a unique fixed point.

I am thinking to first show existence, and then show that it has to be unique. To show existence, there is a hint to use the fact that $x$ is fixed if and only if $\rho(T(x),x) = 0$. It is similar to the contraction mapping principle, but not enough that you can use it to prove this.

Also, if the above strict inequality is replaced by a weak inequality ($\leq$) then must $T$ have a fixed point?

I am thinking no, given that the requirement for a fixed point in the contraction mapping principle is that there be a $c \in (0,1)$ such that $ \rho(T(u),T(v)) \leq c\rho(u,v)$. I am thinking that a counterexample would be enough to show that this is not true.

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    For the inequality $\le$ take say an irrational rotation on the circle. – 2017-02-03

2 Answers 2

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Hint: Fix $x_0\in X$. Recursively define a sequence by $x_{n+1}=T(x_n)$. By compactness, it has a convergent subsequence $\{x_{n_k}\}$.

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    Do you mean $x_{n+1} = T(x_{n})$? – 2017-02-03
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    Omitting parentheses is the usual convention. Yes, this is so. But, telling the truth, I do prefer $T(x_n)$ instead of $Tx_n$. – 2017-02-03
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    how do you deduce that the limit is a fixed point? – 2017-02-04
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Use compactness to show that there exists some $C < 1$ such that $$\rho(T(u),T(v)) \leq C \rho(u,v)$$

Indeed, if such constant doesn't exist, for each $n$ there would exists some $u_n, v_n$ such that $$\rho(T(u_n),T(v_n)) \geq (1-\frac{1}{n})\rho(u_n,v_n)$$

Now pick a subsequence $u_{k_n}\to u$ of $u_n$ and a subsequence $v_{l_n}\to v$ of $v_{k_n}$ and show that $$\rho(T(u),T(v)) \geq C \rho(u,v)$$