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Is C^l closed in C^0?

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Is [itex]C^{k}[a,b][/itex] closed in [itex]C^{0}[a,b]?[/itex]

    3. The attempt at a solution

    [itex]C^{k}[a,b][/itex] is obviously a subset of [itex]C^{0}[a,b][/itex].

    My gut feeling says no. I thought the best way would be to construct a function [itex]f_{n}(x)[/itex] which converges to [itex]f(x)[/itex] and where [itex]f_{n}(x)[/itex] is in [itex]C^{k}[a,b][/itex] but [itex]f(x)[/itex] is not.

    I thought maybe [itex]f_{n}(x)=x^{k+1}sin(\frac{1}{nx})[/itex] would do it since it's not k+1 differentiable at 0. But then [itex]f(x)=0[/itex] which can be differentiated infinitely (since each derivative is 0).
  2. jcsd
  3. Apr 12, 2012 #2
    You are definitely on right track. Try writing your sequence as a series
    [itex] f_n = \sum_{i=0}^n a_n \sin(b_n x) [/itex] and then choose an and bn so that limit of fn exists but f'n diverges.

    ... Or just google "Weierstrass function", if you're lazy :)
  4. Apr 12, 2012 #3


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    If you're asking such questions, then you should always say which metric you're working with.
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