# Thread: Uniform convergent and Lipschitz functions

1. ## 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)|

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)|

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!!!

2. ## 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 $f(x)=\sqrt x$.