Uniform limit of Lipschitz functions is a Lipschitz function Let $f_n:[0,1] \rightarrow \mathbb{R}$ be a sequence of Lipschitz functions. Each $f_n$ has a Lipschitz constant equal to $M_n>0$. 
Suppose that $f_n$ converges 
uniformly to a function $f$. Then $f$ is Lipschitz.  
My attemp:
For all $x,y \in [0,1], x \neq y$:
$|f(x)-f(y)| \leq |f(x)-f_n(x)| + |f_n(x)-f_n(y)| + |f_n(y)-f(y)|$
For $n\geq n_0$, we have $|f(x)-f_n(x)|<|x-y|$ < and $|f(y)-f_n(y)|<|x-y|$ 
since $f_n$ converges uniformly to $f$.
Thus:
$|f(x)-f(y)| \leq |f(x)-f_{n_{0}}(x)| + |f_{n_{0}}(x)-f_{n_{0}}(y)| + |f_{n_{0}}(y)-f(y)| \leq (2+M_{n_0})|x-y|$
Then $f$ is Lipschitz with constant equal to $2+M_{n_0}$
Am I right? Is there an easier way to solve this problem?
Thank you.
 A: Thiis is not true.  Look at $$f_n(x)=\sum_{k=1}^n{\sin(2^kx)\over k^2}.$$
These converge uniformly to a continous function.  They are all Lipschitz, but the limit is a nondifferentialble "monster."
A: Elementary example: Let $f_n(x) = \sqrt {x+1/n}$ for $x\in [0,1].$ Then each $f_n$ is continuously differentiable on $[0,1],$ hence is Lipschitz there. But $f_n(x)\to f(x)=\sqrt x$ uniformly on $[0,1],$ and $f$ is not Lipschitz there.
A: This is clearly false. For example, you know the theorem of Weierstrass that if $f$ is continuous on $[a,b]$ then there exists a sequence of polynomials $P_n$ so that $P_n\to f$ uniformly on $[a.b]$, right? Those polynomials are Lipschitz on $[a,b]$.
Edit: As mentioned in comments, it is true if the $M_n$ are bounded. In that case there exists $M$ so that $$|f_n(x)-f_n(y)|\le M|x-y|$$for all $x,y$ and $n$. So $$|f(x)-f(y)|=\lim_{n\to\infty}|f_n(x)-f_n(y)|\le M|x-y|.$$(So if the $M_n$ are bounded you don't even need to assume uniform convergence, just pointwise convergence. The point is that if the $M_n$ are bounded then the $f_n$ are equicontinuous; see the Arzela-Ascoli theorem.)
