Prove that if $f:[0,1]\to [0,1]$ is a continuous function, then the sequence of iterates $x_{n+1} = f(x_n), x_0$ is arbitrary, converges iff $\lim\limits_{n\to\infty} (x_{n+1}-x_n) = 0.$
The only if part is straightforward since $\lim\limits_{n\to\infty} x_n$ is a fixed point of $f$ by continuity. Now suppose $\lim\limits_{n\to\infty} (x_{n+1}-x_n) = 0.$ Suppose for a contradiction that the sequence $\{x_n\}$ does not converge. Then there exists $\epsilon > 0$ so that for all real values L, $\forall N, \exists n\ge N$ so that $|x_n-L| \ge \epsilon.$ Also, for all $\epsilon > 0,\exists N, \forall n\ge N, |f(x_n)-x_n|<\epsilon$. We have from the continuity of $f$ that for all $\epsilon > 0,\exists \delta > 0$ so that $|x-y|<\delta\Rightarrow |f(x)-f(y)| < \epsilon$. Also, if there is some $y\in (0,1)$ with $|f(y) - y| > k$ for some $k$, then by the continuity of $f$, there is some $\delta > 0$ so that $|f(x)-x| > k$ for all $x$ with $|x-y|<\delta$. But I'm not sure what to do from here.
