# Equicontinuity at a point if.f. continuous function constant

1. Dec 11, 2013

Hello,

I have a problem I cannot solve. I have been working with problems with convergence of sequences of functions for some time now. But I can't seem to solve most of the problems. Anyway here is my problem:

Consider a continuous function $f: [0, \infty) \rightarrow \mathbb{R}$. For each $n$ define $f_n(x) = f(x^n)$. Show that the set of continuous function $\{f_1, f_2, \ldots \}$ is equicontinuous at $x=1$ if and only if $f$ is a constant function.

1. Assume $f = k$ is constant, then $f_n(x) = f(x^n) = k$ for all $x$.

$|f_n(x) - f_n(y) | = |f(x^n) - f(y^n)| = |k-k| = 0$, so it is indeed equicontinuous everywhere, in particular at 1.

2. Assume $f_n$ is equicontinuous at $x=1$. By definition this means

$\forall \epsilon > 0, \, \exists \delta > 0$ : $|1 - y| < \delta \Rightarrow |f_n(1) - f_n(y)| < \epsilon$ holds for all $n$.

Now $f_n(1) = f(1^n) = f(1)$ for all $n$, let's say $f(1^n) = c$, and $f_n(y) = f(y^n)$. So we can write the above expression:

$\forall \epsilon > 0, \, \exists \delta > 0$ : $|1 - y| < \delta \Rightarrow |f(y^n) - c| < \epsilon$.

Now from this I want to arrive at $f$ is constant. I think I need to use the fact that $f$ is continuous but I can't see how. It feels intuitively right that if $|f(y^n) - c| < \epsilon$ for every $n$, it is arbitrarily close to a constant function along a sequence which for $y < 1$ converges to 0, and for $y > 1$ diverges to $\infty$.

2. Dec 11, 2013

### R136a1

Let's give an intuitive argument. I leave it up to you to get this thing rigorous.

Take $\varepsilon$ small. Find a $\delta$ such that equicontinuity holds at $1$. We know that if $|y-1|<\delta$, then $|f(y^n) - c|<\varepsilon$. And you know this to be true for all $n$. Assume that $y = 1/2$ is of $\delta$-distance of $1$. Then you know that all of

$$f(1/2), f(1/4),..., f(1/2^n)$$

are $\varepsilon$-close to $c$.

Let's take a smaller $\varepsilon^\prime$ now and take a $\delta^\prime$ satisfying equicontinuity. We might not have anymore that $1/2$ is within $\delta^\prime$-distance of $1$. But maybe $\sqrt{1/2}$ or something similar is, and then you know that all of

$$f(\sqrt{1/2}), f(1/2), ...$$

is within $\varepsilon^\prime$-distance of $c$, and this contains the previous sequence.

You see where I'm going??

3. Dec 11, 2013

But I am afraid I still don't see how this can be used to show that $f$ is constant.

I think you are going for the fact that $f$ must be constant since it is arbitrarily close to a constant function along every sequence?
There must be something I'm missing. It's the same with every single problem. I get halfway but then never manage to finish.

$\lim_{x\rightarrow p} f(x) = q$ if and only if $\lim_{n\rightarrow \infty} f(p_n) = q$ whenever $p_n \neq p$ and $\lim_{n\rightarrow \infty} p_n = p$.

I must ask this. If we for example have that

for every $p_n \rightarrow p$ have $\lim_{n\rightarrow \infty} f(p_n) = constant$ implies that $f$ is constant?

4. Dec 11, 2013

### pasmith

It is easier to prove the contrapositive: show that if $f$ is not constant then $(f_n)$ are not equicontinuous at 1.

You will want to recall that if $x > 0$ then
$$\lim_{k \to \infty} x^{1/k} = 1.$$