composition of local Lipschitz function and uniform convegence

Question

let $G$ be a local Lipschitz function and $f_{n}$ a sequence of function uniformly convergent to $f:\Re\rightarrow \Re$

if $|f|\le M\in \Re$.

How can i prove that the composition $G(f_{n})$ is uniformly convergent to $G(f)$. I used the definition of local Lipschitz but I don't know how to use the fact that f is limited. Can someone show me the proof?

Answer 1

Since $f_n \to f$ uniformly, and $|f|\leq M$, then there exists an index $N\in\mathbb{N}$ such that $|f_n|\leq M+1$ for every $n\geq N$.

Since $G$ is locally Lipschitz, there exists a constant $L> 0$ (depending on $M$) such that $$ |G(x) - G(y)| \leq L |x-y| \qquad \forall x,y\in\mathbb{R},\ |x|, |y| \leq M+1. $$ Hence, for every $n\geq N$, $$ |G(f_n) - G(f)| \leq L |f_n - f|, $$ so that the uniform convergence of $G(f_n)$ to $G(f)$ follows.