# $|f'_n(x)| \leq 1, \forall x, n$ then $\lim_n f_n$ is continuous

Let $f_n : \mathbb{R} \rightarrow \mathbb{R}$ be differentiable with $|f'_n(x)| \leq 1, \forall x, n$ and $f_n \rightarrow g$. Prove that $g$ is continuous.

$|g(x) - g(y)| = \lim_n |f_n (x) - f_n(y)|$. By mean value theorem, there is a $\xi_n \in (x,y)$ s.t. $|f_n (x) - f_n(y)| = |f'(\xi_n)| |x-y| \leq |x-y|$, so $|g(x) - g(y)| \leq |x-y|$ and $g$ continuous.

I've seen a much longer proof of this fact, this looks right to me but I'm not completely sure...

• Do we assume the limit exist or do you have to prove that it does? Because IMO you need to assume it. – Patrick Da Silva Aug 12 '13 at 13:15
• You need some assumption here : Take $f_n(x) = n$ for all $x$. – Prahlad Vaidyanathan Aug 12 '13 at 13:17

If you assume that $g$ is well defined (i.e. that the limit defining $g$ exists everywhere), then your proof shows way stronger than just the continuity of $g$ ; it is a 1-Lipschitz function. Which seems natural to expect because the $f_n$ functions are all $1$-Lipschitz too.