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!!!