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Uniform Convergence & Boundedness

  1. Apr 12, 2010 #1
    "Let fk be functions defined on Rn converging uniformly to a function f. IF each fk is bounded, say by Ak, THEN f is bounded."

    fk converges to f uniformly =>||fk - f|| ->0 as k->∞
    Also, we know|fk(x)|≤ Ak for all k, for all x

    But why does this imply that f is bounded? I don't see why it is necessarily true.

    Any help is appreciated!
     
    Last edited: Apr 12, 2010
  2. jcsd
  3. Apr 12, 2010 #2

    mathman

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    |f|=|f-fk+fk| ≤ |f-fk| + |fk|≤ ε + Ak

    You need uniform for ε to be usable for all x.
     
  4. Apr 13, 2010 #3
    Why does it imply f is bounded? Is ε fixed here?

    Also, why do we need uniform convergence? (why is pointwise convergence not enough...I still don't see why)

    Can someone explain this, please? Thanks!
     
  5. Apr 13, 2010 #4

    mathman

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    Since the norm is defined to be the sup over all values of x
    We can write:
    ||f||=||f-fk+fk|| ≤ ||f-fk|| + ||fk||≤ ε + Ak
    Therefore ||f||≤ ε + Ak, which is a bound. ε will depend on k, but not on x.

    The uniform convergence means that we can use one ε to be a bound for ||f-fk||. If it was simply pointwise convergence, ε would be a function of x and might not be bounded.
     
    Last edited: Apr 13, 2010
  6. Apr 13, 2010 #5
    Sorry, I don't think it makes sense. ε is supposed to be given, always. And we're supposed to find N such that for k>N, ... will be < ε

    Also, Ak depends on fk. What we want is a bound M which is a constant and does not depend on k, right?
     
    Last edited: Apr 13, 2010
  7. Apr 14, 2010 #6

    Landau

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    You need to show that f is bounded, i.e. find some positive real number M such that for all x we have |f(x)|<M. Here M cannot depend on x, but it may ver well depend on k (why not?).

    mathman found such M in his first post.
     
  8. Apr 14, 2010 #7
    put ε =1 and k = N+1
     
  9. Apr 14, 2010 #8

    mathman

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    (Uniform convergence) For every ε > 0 there exists an N so that for all k > N, ||f-fk|| < ε.
    (bounded) ||fk|| < Ak. (Use any k > N)

    Therefore ||f|| < ε + Ak. Thus, in plain English, f is bounded!! This proof does not determine the value of the minimum bound.
     
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