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Totally Bounded Subset

  1. May 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Let G = { f [itex]\in[/itex] C[0,1] : [itex]^{0}_{1}[/itex][itex]\int[/itex]|f(x)|dx [itex]\leq[/itex] 1 }
    Endowed with the metric d(f,h) = [itex]^{0}_{1}[/itex][itex]\int[/itex]|f(x)-h(x)|dx. Is G totally bounded? Prove or provide counterexample

    2. Relevant Theorems

    Arzela-Ascoli Theorem, Theorems relating to compactness, equicontinuity etc and

    Let M be a compact metric space. A subset of
    C(M) is totally bounded iff it is bounded and equicontinuous. (in terms of the sup metric)

    3. The attempt at a solution
    I'm not to sure whether G is totally bounded or not to begin with. If I was working in the sup metric the subset is not totally bounded as I can find a family of functions which are not equicontinuous in G. However I'm not sure how or if this translates to the integral metric defined above. Any help would be much appreciated!
  2. jcsd
  3. May 14, 2012 #2


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    Try to find a sequence of functions [itex]f_i(x)[/itex] such that [itex]d(f_i,f_j)=1[/itex] for all [itex]i \ne j[/itex]. Step functions will do. What would that tell you about total boundedness?
  4. May 14, 2012 #3
    Such a sequence could not be totally bounded since if I took a finite cover of the space with balls radius say half then each function in the sequence must be in a separate ball and as the sequence is infinite this leads to a contradiction. But I can't think of a sequence functions that are continuous on [0,1] with this property?
  5. May 14, 2012 #4


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    Can you think of a way to define f1 so that it's integral is 1/2, but it is only nonzero on [1/2,1]. Same thing for f2 but it is only nonzero on [1/4,1/2], put f3 on [1/8,1/4] etc.
    Last edited: May 14, 2012
  6. May 14, 2012 #5
    Yeah just figured it out thanks for your help!
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