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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.

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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
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2 Answers 2

up vote 3 down vote accepted
  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.

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:I had not seen your answer until I posted mine. –  Mhenni Benghorbal Jul 19 '12 at 14:16
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