sequences and series - uniform converges Let $f_n:[a,b] \rightarrow \mathbb{R}$ a sequence of Lipschitz uniform functions, that is, there is $K>0$ such that $|f_n(x) - f_n(y)| \leq K|x-y|, \forall x,y \in \mathbb{R}, \forall n \in \mathbb{N}.$ Show that if $f_n$ converges pointwise to $f$ then $f_n \longrightarrow f$ uniformly.
This is an exercise of Real Analysis II, I think I need to show that this is a cauchy sequences but I am not sure.
Can you give me a way to solve it?
 A: Hints:


*

*Since $f_{n}(y_{i}) \to f(y_{i})$ there is $n_{y_{i}}$ such that $m,n > n_{y_{i}}$ implies $|f_{n}(y_{i}) - f_{m}(y_{i})| < \epsilon.$ 

*Since each $f_{n}$ is uniformly continuous, $|f_{n}(x) - f_{n}(y_{i})| < \epsilon$ whenever $|x-y_{i}| < \delta$.

*Since $[a,b]$ is compact, $[a,b] \subset B_{\delta}(y_{1})\cup \cdots \cup B_{\delta}(y_{k})$ for some $y_{1},...,y_{k}$.

*Note that
$$|f_{n}(x) - f_{m}(x)| \leq |f_{n}(x) - f_{n}(y_{i})| + |f_{n}(y_{i}) - f_{m}(y_{i})| + |f_{m}(y_{i}) - f_{m}(x)| < 2\epsilon + |f_{n}(y_{i})-f_{m}(y_{i})|.$$

*Now, choose some $N$ that works for all $x$.
A: Ok i will prove first that $f$ is Lipschtiz let $x,y\in [a,b]$ then there exists $N_1,N_2 \in \mathbb{N}$ such that $|f_n(x)-f(x)| < \frac{\epsilon}{2}, |f_n(y)-f(y)|<\frac{\epsilon}{2}$ for all $n> \max\{N_1,N_2\}$ then $|f(x) - f(y) | = |f(x) - f_n(x) + f_n(x) - f_n(y) + f_n(y) - f(y) | < \epsilon + k|x-y|$ since $\epsilon$ is arbitrary we can let it be zero. 
Given $\epsilon>0$ let $m\in \mathbb{N}$ such that $$\frac{b-a}{m} < \frac{\epsilon}{3k}$$
call $\frac{b-a}{m} = \delta$  where $k$ is the Lipschtiz constant. 
Now let $x_1,x_2,\cdots , x_m$ be middle points in the subintervals $[a,a+\delta) , [a+\delta, a+2\delta),\cdots ,[a+(m-1)\delta, b]$ now for each of these $x_i$ let $N_i$ be the natural number where $$|f_n(x_i) - f(x_i) | < \frac{\epsilon}{3}, \text{ for } n > N_i$$ 
Let $N=\max\{N_1, N_2, \cdots , N_m\}$. Finally for any $y\in [a,b]$ let $x_j$ be the middle element of the subinterval where $y$ belongs then for $n>N$
$$|f_n(y) - f(y) | = |f_n(y) - f_n(x_j) + f_n(x_j) - f(x_j) + f(x_j) - f(y)|$$
$$|f_n(y) - f(y) | \leq k |y-x_j| + \frac{\epsilon}{3} + k |x_j - y|< \epsilon$$
Now if $y$ is one of the $x_i$ how can you solve it? 
