# 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!!!

#### girdav

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