Sequence of continuous functions with bounded derivative

Let $f_n$ be a sequence of continuous functions on $[0,1]$, and continuously differentiable on $(0,1)$. Assume $|f_n|\le 1$ and $f_n'\le 1$ $\forall x\in [0,1]$ and $n$. Then

1. $f_n$ is a convergent sequence in $C[0,1]$

2. $f_n$ has a convergent subsequence in $C[0,1]$

well, by Bolzano-Weirstrass theorem (every bounded sequence has a convergent subsequence) we can say $2$ is correct, I am not able to say true or false against $1$, please help.

• How about just choosing constant functions alternating between two values for number 1? Also, for 2, Bolzano-Weierstrass applies to subsets of the real numbers, not function spaces. – Seth Jul 19 '12 at 13:41
• Bolzano-Weierstrass only gives a subsequence which works for a point. By diagonal method, you can show that there is a subsequence which works for countably many point, for example the rational of $[0,1]$, but in general, without equi-continuity you wont be able to show that it works for each point of $[0,1]$. – Davide Giraudo Jul 19 '12 at 14:00

1. Consider the case $f_n(x)=(-1)^n$ for all integer $n$ and all $x\in [0,1]$.
2. Take $f_n(x):=-x^n$. Then $|f_n(x)|\leq 1$ and $f'_n(x)=-nx^{n-1}\leq 0\leq 1$. We have that $f_n$ converges pointwise to the function which is $0$ in $[0,1)$ and $-1$ at $1$. A uniformly converging subsequence would converge to this map, which is not possible.
However, if we replace the condition "$\forall x\in[0,1], f'_n(x)\leq 1$" by "$\forall x\in[0,1], |f'_n(x)|\leq 1$", Arzelà-Ascoli theorem applies.