5

If E Is equicontinuous in C(X,R), I need to show that $\bar{E}$ is equicontinuous as well.

Now $\forall f \in \bar{E}, \exists f_n\in E$ s.t. $f_n\rightarrow f$, thus $\forall \epsilon > 0, \exists n_o \text{ s.t. if } n\geq n_0, |f_n(x)-f(x)|< \epsilon $. $$ |f(x)-f(y)|\leq|f(x)-f_{n0}(x)|+|f_{n0}(x)-f_{n0}(y)|+|f_{n0}(y)-f(y)|$$.. whats next?

Cameron Buie
  • 102,994
Klara
  • 1,233
  • For RHS, the first term and the last term are bounded, the middle one can be bounded by equicont condition. And let $n_0$ go to infinity. And I assume the family of equicont functions are uniformly equicont. – Yimin Mar 07 '13 at 16:18

2 Answers2

2

Let $x \in \mathbb{R}$ be given. Fix $\epsilon > 0$. Choose $\delta > 0$ such that $|g(x)-g(y)|<\epsilon$ for all $g \in E$ and $y \in \mathbb{R}$ with $|y-x|<\delta$. Let $f \in \overline{E}$ be given. Choose a sequence $(f_n) \to f$ in $E$. Then for any $y \in \mathbb{R}$ with $|y-x|<\delta$ we have $$ |f(x)-f(y)| = \lim_{n \to \infty} |f_n(x)-f_n(y)| = \limsup_{n \to \infty} |f_n(x)-f_n(y)| \leq \limsup_{n \to \infty} \epsilon= \epsilon. $$ It follows that $\overline{E}$ is equicontinuous.

nullUser
  • 27,877
  • Does the same result hold for uniform equicontinuity? I mean, if $E$ is uniformly equicontinuous, is $\overline{E}$ still uniformly equicontinuous? – Logan Tatham Dec 11 '13 at 20:18
0

For $\delta>0$ and $f\colon X\to\Bbb R$, define the modulus of continuity $$\omega(f,\delta):=\sup\{|f(x)-f(y)|,d(x,y)<\delta,x,y\in X\}.$$ A family $\mathcal F$ of functions from $X$ to $\Bbb R$ is equi-continuous if and only if $$\lim_{\delta\to 0}\sup_{f\in\mathcal F}\omega(f,\delta)=0.$$

In this context, one can show that $\sup_{f\in E}\omega(f,\delta)=\sup_{f\in\overline E}\omega(f,\delta)$ where the closure is taken with respect to the uniform norm.

Davide Giraudo
  • 172,925