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Equicontinuity at a point if.f. continuous function constant

  1. Dec 11, 2013 #1
    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 [itex]f: [0, \infty) \rightarrow \mathbb{R} [/itex]. For each [itex]n [/itex] define [itex]f_n(x) = f(x^n) [/itex]. Show that the set of continuous function [itex]\{f_1, f_2, \ldots \}[/itex] is equicontinuous at [itex] x=1[/itex] if and only if [itex]f [/itex] is a constant function.

    1. Assume [itex]f = k[/itex] is constant, then [itex] f_n(x) = f(x^n) = k [/itex] for all [itex] x[/itex].

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


    2. Assume [itex] f_n [/itex] is equicontinuous at [itex] x=1 [/itex]. By definition this means

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

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

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

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

    Thanks in advance!
     
  2. jcsd
  3. Dec 11, 2013 #2
    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

    [tex]f(1/2), f(1/4),..., f(1/2^n)[/tex]

    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

    [tex]f(\sqrt{1/2}), f(1/2), ...[/tex]

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

    You see where I'm going??
     
  4. Dec 11, 2013 #3
    Thanks for the reply!

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

    I think you are going for the fact that [itex] f [/itex] 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

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

    I must ask this. If we for example have that

    for every [itex] p_n \rightarrow p [/itex] have [itex] \lim_{n\rightarrow \infty} f(p_n) = constant [/itex] implies that [itex] f [/itex] is constant?
     
  5. Dec 11, 2013 #4

    pasmith

    User Avatar
    Homework Helper

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

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