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Let $(X, d)$ a metric space, $F: X \rightarrow \mathbb{R}$ a continuous and bounded function, and for all $n \in \mathbb{N}$ $\alpha_n: X \rightarrow X$ a function such that $(\sup_{x \in X} d(\alpha_n(x), x): n \in \mathbb{N})$ converges to zero. Let $F_n = F \circ \alpha_n$. Show that $F_n$ is equicontinuous and converges uniformly.

Any help for this problem, thanks!

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    Well, as stated it is obviously false. Take $X=\mathbb{R}$ and $a_n(x)=x+\frac{1}{n}.$ Obviously, $F(x+\frac{1}{n})$ need not to converge uniformly. Just think of smth. oscillating at $\infty.$2012-09-29
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    I changed \mbox{sup} to \sup. That is standard. Among the effects of that notation are that when you write $a\sup b$, proper spacing precedes and follows "sup", and when it's in a "displayed" rather than an "inline" setting, the subscript appears directly below "sup", thus: $\displaystyle a\sup_{x\in S} f(x)$.2012-09-30

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