Equicontinuity at a point if.f. continuous function constant

[adam512]

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.

Thanks in advance!

 

[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??

 

[adam512]

Thanks for the reply!

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.

I have a question about this: In my textbook it says that

\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?

 

[pasmith]

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??

Thanks for the reply!

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

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 &gt; 0 then
<br /> \lim_{k \to \infty} x^{1/k} = 1.<br />