# Sequence of Lipschitz functions

Let $\{f_{n}\}$ be a sequence of positive continuous functions on $\mathbb R$; $f_{n}:\mathbb R\to \mathbb R$, for all $n\geq 1$, with the folloing properties:

(1) $\{f_{n}\}$ is uniformly bounded by some constant $C>0$,

(2) $\{f_{n}\}$ is uniformly Lipschitz on $\mathbb R$ (so it is uniformly continuous)

(3) $\{f\,'_{n}\}$ is uniformly bounded (sequence of the derivatives), which I think follows from (2).

Does this sequence converges on $\mathbb R$ (or at least contains a subsequence which converges), uniformly on compact sets, or pointwise, to some continuous function $f$ ? If not, what extra (possible) condition must the sequence have to converge to some continuous function?

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In your link, at the middle of the page, it says " Euclidean spaces The Arzelà–Ascoli theorem holds, more generally, if the functions ƒn take values in d-dimensional Euclidean space Rd", does this mean that the result apply in $\mathbb R$, and the convergence is uniform in $\mathbb R$? And, what is the R-valued version of the Arzelà–Ascoli theorem ? –  Berry Jun 17 '12 at 16:25
@Berry You can however look at recursively defined subsequences $f^{n}_k:[-n,n]\rightarrow \mathbb{R}$ -- i.e., $f^n_k$ is a subsequence of $f^{n-1}_k$, restricted to $[n,n]$, to arrive a continuous limit function on $\mathbb{R}$. It'd be necessary to check what kind of convergence you'd get, though. Uhm, yes, what's the problem with target $\mathbb{R}^n$? –  user20266 Jun 17 '12 at 16:26
You are looking at $f_k:\mathbb{R} \rightarrow \mathbb{R}$. That is perfectly fine, if you restrict to compact sets (e.g. closed intervals). The comment on the page just says that the target space may be replaced by a more general space. –  user20266 Jun 17 '12 at 16:32