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Thread: Uniform convergent and Lipschitz functions

  1. #1
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    Uniform convergent and Lipschitz functions

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

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

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

    My solution:

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

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

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

    Now, fix $\displaystyle n \geq N$,

    Consider $\displaystyle |f(x)-f(y)|$

    $\displaystyle = |f(x)-f_n(x)+f_n(x)-f_n(y)+f_n(y)-f(y)|$

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

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

    Therefore $\displaystyle f$ is Lipschitz.

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

    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 $\displaystyle L_n$ simply by $\displaystyle L$.

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