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Metric-spaces, equicontinuity, proof

  1. Mar 15, 2014 #1
    Hi, can you please check if my proof is correct?

    Proposition
    Assume that a sequence of continuous functions [itex]\{f_n\}[/itex] converges uniformly to [itex]f[/itex].
    [itex]f_n: A\rightarrow B[/itex]. Also assume that [itex]A[/itex] is compact.
    Then [itex]\{f_n\}[/itex] is equicontinuous.

    Proof:
    Given an [itex]\epsilon[/itex] we must find a [itex]\delta[/itex] so that if [itex]x,y \in A,d_A(x,y)< \delta[/itex], then [itex]d_B(f_n(x),f_n(y))<\epsilon[/itex], for all [itex]n[/itex], and for all [itex]x,y[/itex].

    Given [itex]\epsilon[/itex]. There exists an [itex]N[/itex] such that [itex]d_B(f_n(x),f(x))<\epsilon/3[/itex] for all [itex]n\ge N[/itex] and all [itex]x \in A[/itex].
    Since f must be continuous and A is compact, f is uniformly continuous. So there exists a [itex]\delta^*[/itex] so that if [itex]d_A(x,y)<\delta^*\rightarrow d_B(f(x),f(y))<\epsilon/3[/itex]
    For each [itex]i, i <N[/itex], there is a [itex]\delta_i[/itex], so that if [itex]d_A(x,y)<\delta_i \rightarrow d_B(f_i(x),f_i(y))<\epsilon[/itex]. This must be, because each [itex]f_n[/itex] must be uniformly continuous on the compact space A.
    Choose [itex]\delta = min\{\delta^*,\delta_1,\delta_2,\delta_{N-1}\}[/itex].

    Now, for any n, if [itex]d_A(x,y)<\delta \rightarrow d_B(f_n(x),f_n(y))< \epsilon[/itex]. This is so, because if [itex]n < N[/itex] it obviously holds, if [itex]n \ge N[/itex] we have:
    [itex]d_B(f_n(x),f_n(y)) \le d_B(f_n(x),f(x))+d_B(f(x),f(y))+d_B(f(y),f_n(y))< \epsilon/3+\epsilon/3+\epsilon/3 = \epsilon[/itex].
     
    Last edited: Mar 15, 2014
  2. jcsd
  3. Mar 15, 2014 #2

    micromass

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    There are some typo's in your proof, but aside from that, the idea seems to be entirely correct.
     
  4. Mar 15, 2014 #3
    Thanks, do you know if the statement is correct or not?(it could actually be false) It is the "converse"(not entirely converse, some assumptions are omitted) of a statement I read in a book, so I am not entirely sure if it is correct. The original proof states that if the funcitons converges pointwise on a dense subset of A, and A is compact, B is complete, and the functions are equicontinuous, then it converges uniformly on A. So this is kind of the converse of that.
     
    Last edited: Mar 15, 2014
  5. Mar 15, 2014 #4

    micromass

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    The statement is absolutely correct. In fact, if you're familiar with function spaces like ##\mathcal{C}([a,b])## with the uniform norm, then you can prove that compactness is equivalent to being equicontinuous and being closed and being pointswise bounded. This is the Arzela-Ascoli theorem. In particular, if ##f_n\rightarrow f## uniformly, then

    [tex]\{f_n~\vert~n\}\cup \{f\}[/tex]

    is equicontinuous. So it remains equicontinuous if we leave out ##f##.
     
  6. Mar 15, 2014 #5
    Thanks, yes it is the "converse" of a lemma in the book, leading up to Arzela-Ascoli theorem. However I wasn't able to use that theorem on it directly, because the theorem in the book was stated only for [itex]C(X,R^m)[/itex], where X is compact, but in this lemma, Y can be any complete space. But beside from that if you want to show that [itex]\{f_n\}[/itex] is equicontinuous using the arzela-ascoli theorem, then you also have to show that that [itex]\{f_n\}[/itex] is compact, and I presume that is hard?
     
  7. Mar 15, 2014 #6

    micromass

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    No, that is actually very easy in this case if you use the right definition (= open cover definition). You have in any metric space ##X## that if ##x_n\rightarrow x##, then ##\{x_n~\vert~n\}\cup \{x\}## is compact.
     
  8. Mar 15, 2014 #7
    Ah, cool thanks. But the proposition in the first post would then follow from Arzela-Ascoli if B was [itex]R^n[/itex]? Because then [itex]\{f_n\}[/itex] is compact, and hence equicontinuous. But you agree that I can't use the theorem directly when B is any space, or even if B is complete?(A is compact no matter what.)
     
  9. Mar 15, 2014 #8

    micromass

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    Do they state Arzela-Axcoli only for ##B=\mathbb{R}^n##? It's true in a far greater generality. For example, there is a version of the theorem for ##B## an arbitrary metric space.
     
  10. Mar 15, 2014 #9
    Yeah, here is the theorem:
    theorem.png

    Could you change [itex]R^m[/itex] with something more general here?
     
  11. Mar 15, 2014 #10

    micromass

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  12. Mar 15, 2014 #11
    Thanks, I am taking real analysis and have not taken topology, so I must admit I do not understand much of that.
    But if I want to generalize the theorem in my book. The link from Wikipedia says:
    Now this link says that all metric spaces are Hausdorff spaces:
    http://math.stackexchange.com/quest...ll-metric-spaces-are-hausdorff-spaces-correct So that is fine. The problem is the "f is pointwise relatively compact", I guess that if B= [itex]R^m[/itex] "bounded" implies "pointwise relatively compact". But that maybe if B is only a complete metric space, or arbirary metric space, "bounded" may not imply "pointwise relatively compact".


    EDIT: I see that for my original question this may not affect it. Because it is on the other side of the equivalence. So that if we start by saying that [itex]\{f_n\}[/itex] converge uniformly to f, and that A is compact, we must have that the limit is continuous, so that limit is in C(A,B). The open covering property says that [itex]f_n\cup f[/itex] is compact. And since we are on the left side of the Arzela-Ascoli(Metric->Hausdorff) we get that they all must be equicontinious, and we remove f?, så all [itex]f_n[/itex] are equicontinuous?

    Thanks for your help.
     
    Last edited: Mar 15, 2014
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