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Sequences of Lipschitz Functions

  1. May 24, 2009 #1
    Let [tex])<C<\infty[/tex] and [tex]a,b \in \mathbb{R}[/tex]. Also let
    [tex]Lip_{C}\left(\left[a,b\right]\right) := \left\{f:\left[a,b\right]\rightarrow \mathbb{R} | \left|f(x) - f(y)\right| \leq C \left|x-y\right| \forall x,y \in \left[a,b\right]\right\}
    [/tex].

    Let [tex]\left(f_{n}\right) _{n \in \mathbb(N)} [/tex] be a sequence of functions with [tex]f_{n} \in Lip_{C}\left(\left[a,b\right]\right)[/tex] for all n.

    i) Show that if [tex]\left(f_{n}\right) _{n \in \mathbb(N)} [/tex] converges uniformly to a function [tex] f : \left[a,b\right]\rightarrow \mathbb{R}[/tex], then [tex] f \in Lip_{C}\left(\left[a,b\right]\right)[/tex].

    ii) Is [tex]Lip_{C}\left(\left[a,b\right]\right)[/tex] a sub vector space of [tex]B\left(\left[a,b\right]\right) := \left\{f:\left[a,b\right]\rightarrow \mathbb{R} | f [/tex] bounded }?


    ____________________________________
    Useful formula:

    The sequence converges uniformly, so:
    For all [tex]\epsilon > 0[/tex], there exists [tex]N \in \mathbb{N}[/tex] so that for all [tex]n,m > N[/tex], [tex]|f_{n}(x)-f_{m}(x)|<\epsilon[/tex] for all [tex] x \in \left[a,b\right]\[/tex].

    For (i), I honestly have no idea.

    It seems that I can just say:

    [tex]|f(x)-f(y)|[/tex]
    [tex]= |\lim_{n \to \infty} f_{n}(x) - \lim_{n \to \infty} f_{n}(y)|[/tex]
    [tex]= |\lim_{n \to \infty} (f_{n}(x) - f_{n}(y))|[/tex]
    [tex]= \lim_{n \to \infty} |(f_{n}(x) - f_{n}(y))| \leq C |x-y|[/tex]

    But I know that has to be wrong. It's not even using the uniformity.


    For (ii) I know the answer is of course not.

    Let [tex] f(x):= Cx[/tex]. Then [tex] 2 f \notin Lip_{C}\left(\left[a,b\right]\right)[/tex], since

    [tex]|2f(x)-2f(y)| = 2C|x-y| > C|x-y|[/tex].




    Help on part (i) please?
     
  2. jcsd
  3. May 24, 2009 #2
    Hmm.. after a bit more thinking....
    I think that the sequence only needs to converge pointwise (not uniformly) to a function in order for the limit function to also be in LipC([a,b]).

    So, my proof for (i) pretty much works. (I think) Please tell me if I am wrong?!

    Also, interestingly, I think that a pointwise convergent series of functions in LipC([a,b]) is also uniformly convergent.
     
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