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Show that the functions are not equicontinuous

  1. Feb 14, 2015 #1
    1. The problem statement, all variables and given/known data

    By using the Ascoli-Arzela theorem, show that the functions fn(z) = zn in Δ(1)n = 1, 2,..., are not equicontinuous.

    2. Relevant equations

    A family F of complex-valued functions on A is called equicontinuous if ∀ε > 0, ∃δ > 0 such that |f(z) - f(w)| < ε, ∀z, w ∈ A with |z - w| < δ, ∀ f ∈ F.

    Of course, the unit disk is the set {z : |z| < 1}

    3. The attempt at a solution

    There's actually a bar over the complex unit disk symbol. Is it the conjugate set? I'm not quite sure.
     
  2. jcsd
  3. Feb 15, 2015 #2

    Dick

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    I suspect the bar over the symbol means closure of the unit disk. I.e. {z:|z|<=1}. It's pretty easy to show that family of function is not equicontinuous directly. I don't think you really need Arzela-Ascoli to prove it. But think about what the limit looks like on the real interval [0,1].
     
    Last edited: Feb 15, 2015
  4. Feb 15, 2015 #3
    Really? The closure? In another part, the professor used the bar to indicate the limit supremum of a subsequence in defining the radius of convergence. Okay. Let me take another look at it.
     
  5. Feb 15, 2015 #4
    The closure of the unit disk would include the boundary and of course the unit disk itself is an open set. You mean the limn→∞zn where z ∈ [0,1]? Of course, it would tend to zero everywhere except at 1.
     
  6. Feb 15, 2015 #5

    Dick

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    What does that tell you about equicontinuity? And I'm pretty sure it's really the closure because that family of functions is equicontinuous on the open ball. Tell me why.
     
  7. Feb 15, 2015 #6
    If you choose any of the interior points their limits satisfy the arbitrary closeness requirement, but if you select a boundary point and an interior point, they diverge. I think that's probably correct.
     
  8. Feb 15, 2015 #7

    Dick

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    You should probably state that in a more precise way. And I retract my statement about the functions being equicontinuous on the open ball of radius 1. They are equicontinous on an open ball with radius less than 1.
     
  9. Feb 15, 2015 #8
    Right. Okay. Thanks again for the help. The homework is due before midnight, so I'll tidy it up later.
     
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