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Sequences of Lipschitz Functions

  • #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?
 

Answers and Replies

  • #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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