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Let $f: [a,b] \to[a,b]$ such that: $$ \lvert f(x)-f(y)\rvert \le K\lvert x-y\rvert$$ when $0

Now I need to prove that the series defined: $a_{n+1}=f(a_n) , a_0 = x$ for $x \in [a,b]$ converges to this fixed point. I know that given it does converge, then the fixed point is the limit. Any help is proving it actually converges? Thanks!

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    Hint : Is it a Cauchy sequence ? Notice also that $\sum_{n \geq 0} K^n$ converges. – 2017-01-13
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    Do you mean the sequence (not the series) $(a_n)_n,$ or the series $\sum_n a_n$? (I assume the former) – 2017-01-13
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    How did you prove that there is a fixed point without looking at a sequence of such sort? – 2017-01-13
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    This is, of course, a very special case of Banach Fixed Point Principle: if the metric space $(X,d)$ is complete and $f:X\to X$ fulfills for some $K\in(0,1)$ the condition $d\bigl(f(x),f(y)\bigr)\le Kd(x,y)$ for all $x,y\in X$, then $f$ admits the unique fixed point. The proof goes exactly in the same way as in the case of a closed interval. – 2017-01-13
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    Intermediate value theorem – 2017-01-13
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    @user401516: of course, this argument gives us the existence of a fixed point, not the uniqueness. This is gueranteed by the limit construction, which is not always possible. It is, whenever $f$ is contractive, as discussed here. – 2017-01-13
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    @szw1710 no need for a sequence. Assume two distinct fixed points, apply the function to both, and then use that $K <1$ to conclude that this is impossible. – 2017-01-13
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    @Clement C. - yes, of course :) Thanks for this valuable remark. – 2017-01-13

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Idea: show it is Cauchy. For this, you can use that for any $n,m\geq 0$ $$\rvert a_{n+m}-a_n\rvert \leq K \lvert a_{n+m-1}-a_{n-1} \rvert \leq \dots \leq K^n \lvert a_{m}-a_{0} \rvert \leq K^n \lvert b-a \rvert. $$ Now, $K\in(0,1)$...

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    great! but I still dont know how to end it, I mean we got to a point where $K^n\lvert b-a\rvert$ converges, but how do we show that its smaller than any $\epsilon$ ? – 2017-01-13
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    Because $K^n\xrightarrow[n\to\infty]{}0$, there exists $N\geq 0$ such that $0\leq K^n \leq \frac{\varepsilon}{\lvert b-a\rvert }$ for all $n\geq N$. – 2017-01-13