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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...

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

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