Uniform convergent and Lipschitz functions

**tttcomrader** · November 2nd 2012, 08:15 AM

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!!!

**girdav** · November 2nd 2012, 09:50 AM

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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Uniform convergent and Lipschitz functions

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