# Thread: Uniform convergent and Lipschitz functions

1. ## 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!!!

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