Equicontinuity Question

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April 1st 2011, 12:41 PM
Haven

Equicontinuity Question

Let $f$ be a continuous function on $\mathbb{R}$ . Let $f_n = f(nt)$ for $n \in \mathbb{N}$ be equicontinuous on [0,1]. I.e., $\forall \epsilon >0, \exists \delta >0$ such that if $|x-y| < \delta$ , then $|f_n(x) - f_n(y)| < \epsilon$ for all n.

What can we conclude about $f$ ?

All I am able to get is that since each $f_n$ is defined on a compact set, then each $f_n$ is pointwise bounded. So, $\{f_n \}$ is uniformly bounded. So, $f$ is uniformly bounded.

Is there something else that I can conclude?
April 1st 2011, 02:03 PM
Drexel28

Quote:

Originally Posted by Haven

Let $f$ be a continuous function on $\mathbb{R}$ . Let $f_n = f(nt)$ for $n \in \mathbb{N}$ be equicontinuous on [0,1]. I.e., $\forall \epsilon >0, \exists \delta >0$ such that if $|x-y| < \delta$ , then $|f_n(x) - f_n(y)| < \epsilon$ for all n.

What can we conclude about $f$ ?

All I am able to get is that since each $f_n$ is defined on a compact set, then each $f_n$ is pointwise bounded. So, $\{f_n \}$ is uniformly bounded. So, $f$ is uniformly bounded.

Is there something else that I can conclude?

You can show that $f$ must be constant. Can you prove that?
April 1st 2011, 11:53 PM
Haven

Quote:

Originally Posted by Drexel28

You can show that $f$ must be constant. Can you prove that?

Actually, I am having a bit of difficult getting showing it is constant. I consider $x \in \mathbb{R}$ and let $f(x) = c$ . Now I suppose $\exists y\in \real$ such that $f(y) \neq c$ . Now let $n$ be a natural number larger than y. It follows that $f_n(y/n)= f(y) \neq c$ .

However, I am having troubles with the $\delta$ 's in this case. for a given epsilon, Does the intermediate value theorem guarantee that I can find $z$ such that $x<z<y$ and $z-x < \delta$ ?
April 2nd 2011, 07:04 AM
girdav

You can write that $|f(x)-f(y)| =|f_n\left(\dfrac xn\right)-f_n\left(\dfrac yn\right)|$ . Take a $\varepsilon >0$ . We can find a $n$ such that $\left|\dfrac xn-\dfrac yn\right|\leq \delta$ hence $|f(x)-f(y)|\leq\varepsilon$ . Now you can conclude.