Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.