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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 $\lim\limits_{n\rightarrow \infty}\sup\limits_{x\in K}d_Y(f(x),f_n(x))=0$)

  • $Lip(f_n)\rightarrow 1$

Question: does it follow $Lip(f)=1$? Can you motivate your answer?

I am sorry if I do not show my reasoning, but I can not even understand if the statement is true or not

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Let $X = Y = \mathbb R$ and $f_n = 1$ on $[-n,n]$, $f_n = 0$ on $[-n-1, n+1]^c$ and affine linear with slope $\pm 1$ on $[-n-1,-n] \cup [n,n+1]$. Then $Lip(f_n) = 1$, the uniform convergence on compact sets holds with $f = 1$, but $Lip(f) = 0$.

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    $\begingroup$ No problem! Much more minor: should it be $f=1$ in the last sentence? $\endgroup$ – preferred_anon Jun 11 '18 at 20:45
  • $\begingroup$ You are again correct, I should go to bed... I updated it again. $\endgroup$ – pcp Jun 11 '18 at 20:49
  • $\begingroup$ @pcp thank you! your example was really enlightening! Do you think it could happen $Lip(f)>1$? I will post another question about that $\endgroup$ – User29983 Jun 12 '18 at 7:16

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