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Math Help - Uniform convergent and Lipschitz functions

  1. #1
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    Uniform convergent and Lipschitz functions

    Let (f_n) be a uniform convergence sequence in C^0([a,b], \mathbb {R} ) . Suppose that each f_n is Lipschitz with Lipschitz constant L_n.

    a) If all of L_n are the same, prove that  \lim _{n \rightarrow \infty } f_n is also Lipschitz.

    b) If not all of the L_n are the same, must the limit be Lipschitz?

    My solution:

    a) Let  \lim _{n \rightarrow \infty } f_n = f , for all x,y \in [a,b] , we wish to show that there exist some constant, say L, we will have |f(x)-f(y)|<L|x-y|

    Since f_n uniformly convergence to f, given  \epsilon > 0, there exist N \in \mathbb {N} such that |f_n(x)-f(x)|< \epsilon \ \ \ \ \ \forall n \geq N, \forall x \in [a,b]

    Now, since f_n are Lipschitz, we then have |f_n(x)-f(x)| \leq L_n|x-y| \ \ \ \ \forall x,y \in [a,b]

    Now, fix n \geq N,

    Consider  |f(x)-f(y)|

     = |f(x)-f_n(x)+f_n(x)-f_n(y)+f_n(y)-f(y)|

     \leq |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|

     \leq \epsilon + L_n|x-y|+ \epsilon \leq L_n|x-y|

    Therefore f is Lipschitz.

    b) Now, suppose that L_i \neq L_j for some  i \neq j , we can pick L = \min L_n , then we will still have |f(x)-f(y)|<L|x-y|

    I'm pretty sure I'm wrong somewhere here, because there is no way this problem can be this easy. Any help please? Thank you!!!
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  2. #2
    Super Member girdav's Avatar
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    Re: Uniform convergent and Lipschitz functions

    What you did for a) is correct (except the last inequality), and you can denote L_n simply by L.

    For b), if it was true, then for all f, approximate it uniformly by polynomials on [a,b]. These functions are Lipschitz. This would mean that each continuous function on [a,b] is Lipschitz, which is not true, for example take f(x)=\sqrt x.
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