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!

  • $\begingroup$ 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 $\infty.$ $\endgroup$ – leshik Sep 29 '12 at 23:59
  • $\begingroup$ 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: $\displaystyle a\sup_{x\in S} f(x)$. $\endgroup$ – Michael Hardy Sep 30 '12 at 0:33

For equicontinuity: I assume you want the $\alpha_n$ to be continuous. Fix $x \in X$ and $\epsilon > 0$. By continuity of $F$, there is a ball $B_\delta(x)$ so that $F(B_\delta(x)) \subset B_\epsilon(F(x))$. The idea is that for large n, the $\alpha_n$ won't move a point in $B_\frac{\delta}{2}(x)$ out of $B_\delta(x)$, and so $F_n(B_\frac{\delta}{2}(x)) \subset B_\epsilon(F(x))$ should hold for large n, and we have $B_\epsilon(F(x)) \subset B_{2\epsilon} (F_n(x))$ for large n, as well. So we've found $\delta$ so that $F_n(B_\frac{\delta}{2}(x)) \subset B_{2\epsilon} (F_n(x))$ for large n. We can deal with the finitely many other $F_n$ by taking intersections. You might also want to clean up the $\epsilon$'s and $\delta$'s some.

I believe you need something like uniform continuity on $F$ to show that the sequence converges uniformly, however. Take $X = (0,1)$ and $F(x) = \cos(\frac{1}{x})$ and $$\alpha_n(x) = \begin{cases} \frac{1}{\frac{1}{x} + \pi n} \mbox{ for } x \leq \frac{1}{n} \\ x + \frac{1}{n}\bigg(\frac{1}{1+\pi} - 1\bigg) \mbox{ for } x \geq \frac{1}{n} \end{cases}.$$ Then for any $n$, we have $F(\frac{1}{2\pi n}) = 1$ while $F_n(\frac{1}{2\pi n}) = F(\frac{1}{(2n+1)\pi}) = -1$, and so we do not have uniform convergence.


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