Lipschitz constant of limit of functions part 2

Question

This question follows from my other question Lipschitz constant of limit of functions.

Consider two metric spaces $(X,d_X)$ and $(Y,d_Y)$ and define the lipschitz constant of every continuous function $f:X\rightarrow Y$ as

$$Lip(f):=\sup\limits_{x\neq y}\frac{d_Y(f(x),f(y))}{d_X(x,y)}$$

Consider a sequence of continuous functions $f_n:X\rightarrow Y$ such that

• there is a $k>1$ such that for every $n\in \mathbb{N}$ it is $Lip(f_n)\le k$

• $\{f_n\}$ has limit $f:X\rightarrow Y$ for the uniform convergence on compact sets (this means that for every $K\subset X$ compact it results $\lim\limits_{n\rightarrow \infty}\sup\limits_{x\in K}d_Y(f(x),f_n(x))=0$)

• $Lip(f_n)\rightarrow 1$

User pcp showed that it is not true $Lip(f)=1$, but showed an example when it happens $Lip(f)=0$. It seems to me that his counter-example only works for proving $Lip(f)<1$, so my other question is the following.

Question: Can it happen $Lip(f)>1$? Can you motivate your answer?

Answer 1

let $\epsilon >0$. Then $Lip(f_n) <1+\epsilon$ for $n$ sufficiently large. Hence $d_Y(f_n(x),f_n(y)) \leq (1+\epsilon ) d_X(x,y)$ for all $x,y$ for $n$ sufficiently large.. Letting $n \to \infty$ we get $d_Y(f(x),f(y)) \leq (1+\epsilon ) d_X(x,y)$ for all $x,y$. Letting $\epsilon \to 0$ we conclude that $Lip(f) \leq 1$.