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Prove/disprove that g(x) = sup f(x), f in F, continuous

  1. Aug 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Let X be a compact metric space, F a family of uniformly bounded, equicontinuous real-valued functions on X.

    Is the function
    g(x) := supf ∈ F f(x)
    necessarily continuous?

    3. The attempt at a solution
    The hypotheses about the function family seem to point to some use of Ascoli's theorem. If we have that g(x) = sup f(x) for f in the family then we have an increasing sequence fn(x) -> g(x) for fixed x.

    I tried starting with proving that g is continuous: by Ascoli's theorem this corresponding sequence fn has a (uniformly) convergent subsequence; in particular, g(x) is attained at x by some function f in the closure of F. Perhaps there's a continuity argument to be made there? Any function in the closure of F would have to be continuous because X is compact and the set of bounded continuous functions with compact domain is closed. But the connection to sup f is not clear since sup f need not be the uniform limit of any sequence in F.

    I also tried a triangulation argument like this:
    Let xn → x. There's a sequence of functions in F such that fk(x) approaches g(x) pointwise. Choose ε > 0, then [fk(x) - g(x)| < ε/3 for sufficiently large k. Since the fk are equicontinuous there exists δ > 0 such that |xn - x| < δ implies [fk(xn) - fk(x)| < ε / 3 for any k.

    If I could only guarantee [fk(xn) - g(xn)| < ε/3 for sufficiently large k and n, I will have arrived. But I just can't bridge the gap. Again, it looks like I need some sort of continuity-based argument here, but I don't know how I can make one without assuming that g is continuous.

    Then I try to think of counterexamples, but the rather dense hypotheses make counterexamples a little hard to visualize for me.
     
  2. jcsd
  3. Aug 8, 2015 #2

    Zondrina

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    I'm sorry, but could you explain exactly what this line means:

    $$g(x) := \text{sup}_{f \in F} f(x)$$

    You're choosing one of the functions from ##F##, and treating ##f(x)## like a set?

    Did you mean:

    $$g(x) := \text{sup}\{f \in F \}$$

    As in ##g(x)## is chosen to be the biggest function in the set ##F##?

    I really think this should read ##g(x) := \text{max}\{f \in F \}## anyway.
     
  4. Aug 8, 2015 #3
    No -- I meant that, at each individual x, we choose the supremum of the set of values at x.
    For example, if F = {the functions x → ax: a ∈ (0, 1)}, then g(0.5) = 0.5 and g(-0.5) = 0.
     
  5. Aug 9, 2015 #4

    Zondrina

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    So ##g(x)## takes on the value of one of the equicontinuous functions ##f_i \in F, \space 1 \leq i \leq n## for each ##x##. The function ##f_i## chosen for the value of ##x## must maximize ##g(x) = f_i(x)##.

    We could write this as ##g(x) := \displaystyle \sup_{1 \leq i \leq n} f_i(x) = \sup \{ f_1(x), f_2(x), \cdots, f_n(x) \}##.

    So we choose a value of ##x##, then we choose a value of ##i## such that ##f_i(x)## is maximized.

    How do you usually go about showing a function is continuous? How does this apply to ##g(x)##? I think you want to show:

    $$\displaystyle \lim_{x \to a} g(x) = \displaystyle \lim_{x \to a} \displaystyle \sup_{1 \leq i \leq n} f_i(x) = \sup \{ f_1(a), f_2(a), \cdots, f_n(a) \}$$

    More formally:

    $$\forall \epsilon > 0, \exists \delta > 0 \space | \space 0 < | x - a | < \delta \Rightarrow |g(x) - \sup \{ f_1(a), f_2(a), \cdots, f_n(a) \}| < \varepsilon$$

    You know each ##f_i## is equicontinuous, and you also know ##|\sup \{ f_1(a), f_2(a), \cdots, f_n(a) \}| \leq M## because the functions in ##F## are uniformly bounded.
     
    Last edited: Aug 9, 2015
  6. Aug 9, 2015 #5

    pasmith

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    There is nothing in the question to suggest that [itex]F[/itex] is finite or even countable.
     
  7. Aug 9, 2015 #6

    Zondrina

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    Yes, I suppose this would change the notation in my last post a bit. I was intending to mean there are infinitely many of these functions, but I did not express it as such. It's hard to think when you first wake up with no coffee, my apologies.

    The previous post I wrote should read something like:

     
  8. Aug 9, 2015 #7

    WWGD

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    Never mind, sorry, I don't see the delete option.
     
  9. Aug 10, 2015 #8

    Zondrina

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    After sleeping on this a bit, I think the notation used in the original post is okay. I can't understand why, but I looked at it differently yesterday than I do today. The limit in my last post should have read:

    $$\displaystyle \lim_{x \to a} g(x) = \displaystyle \lim_{x \to a} \displaystyle \sup_{f \in F} f(x) = \sup \{ f_1(a), f_2(a), f_3(a), ... \}$$

    Still, the point is to show ##|g(x) - \sup \{f_1(a), f_2(a), f_3(a), ... \}| < \varepsilon##.

    A graphical argument might help to see what is really happening.
     
    Last edited: Aug 10, 2015
  10. Aug 11, 2015 #9

    haruspex

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    Throughout this thread I see references to equicontinuous functions, but as far as I am aware there's no such beast. It's the family that's equicontinuous, and from that fact, it seems to me the desired result follows fairly swiftly.
     
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