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$\text{Let $(X,d) $ be a compact metric space and $f:X\longrightarrow X$ a continuous operator}$
$\text{such that } d\big(f(x),f(y)\big)<d(x,y)\text{ } \forall x,y\in X, x\neq y.$
$\text{I know that there exists a unique fixed point $x_0\in X: f(x_0)=x_0$},$
$\text{however, how can we show that: }$ $$\lim_{n\to\infty}f^n(x)=x_0 \text{ }\forall x\in X$$ $\text{Where }f^n(x):=f\color{black}{\underbrace{ ofofo\dots o f(x) }_{ \text{$n$ times } }}.$

$$\bullet \text{ My approach: $X$ is compact, hence $\exists$ subsequence $f^{k_n}(x) :k_n\ge n$ s.t}$$ $$f^{k_n}(x) \longrightarrow a\in X,\text{ $f$ is continuous, thus $f^{k_n +1}(x)\longrightarrow f(a)\implies f(a)=a\implies a=x_0$}$$

$\text{How can i continue from this point on, any ideas?}$

Rick Sanchez C-666
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    An alternative approach is to directly consider the sequence $(d(f^n(x), x_0))$. Note that $d(f^{n + 1}(x), x_0) = d(f(f^n(x)), f(x_0)) \le d(f^n(x), x_0)$, so that sequence is decreasing, and therefore it converges. But your subsequence argument shows that it must have a subsequence converging to $0$, so it converges to $0$. But this means $f^n(x) \to x_0$ :D – Izaak van Dongen Feb 24 '21 at 20:42
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    Actually I'm not sure I follow your proof that $f(a) = a$. You show that the subsequence $f^{k_n}(x) \to a$ and that the different subsequence $f^{k_n + 1}(x) \to f(a)$. But you haven't shown that the whole sequence converges, so you can't use uniqueness of limits to conclude that $f(a) = a$. Etienne's answer at https://math.stackexchange.com/questions/1684121/contractive-operators-on-compact-spaces?rq=1 fills this part in, I think. – Izaak van Dongen Feb 25 '21 at 01:11
  • @IzaakvanDongen You are totaly right, my approach is incorrect, thank you for the comment. – Rick Sanchez C-666 Feb 25 '21 at 13:59
  • @IzaakvanDongen, looks good. I tried to come up with an argument to save the original approach, but could not find one, so my answer is obsoleted as well. Hopefully at least showing Uryhson's subsequence principle will help you sometime. Again, thanks for the comment Izaak. – DoubleDip Feb 28 '21 at 14:13

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You're almost there: To finish you can use the Uryshon subsequence principle which can be found here: https://www.math.ucla.edu/~hmkhang24/131BHW19/Notes.pdf

Indeed for any $x \in X$ take the sequence $(f^n(x))_k$. Take a subsequence $(f^n_k(x))_k$. Then with this subsequence you use your argument (compactness $\implies$ convergent subsequence which converges to $x_0$). Then use Uryshon's subsequence principle (also called subsubsequence principle) since you took an arbitrary subsequence of the original sequence.

Alternatively you can argue by contradiction (which amounts to using the subsequence principle). Suppose your sequence does not converge to $x_0$, then there is a subsequence which does not converge to $x_0$, for some $x \in X$. But this subsequence has a convergent subsequence due to compactness and then you use the argument that you gave.

DoubleDip
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