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Let $Z$ be the set of all antiderivatives of continuous functions $[0, 1] \rightarrow [0, 1]$, i.e.

$$Z = {f : [0, 1] \rightarrow \mathbb R| f': [0, 1] \rightarrow [0, 1]}$$

is continuous

If I Let $(f_n)_{ n \in \mathbb N} \in Z^{\mathbb N}$ be a sequence of functions $f_n \in Z$ with $f_n(0) = 0$ for all $n ∈ \mathbb N$.

Is there a subsequence of $(f_n)_{n\in \mathbb N}$ which converges uniformly to some continuous function??

Student146
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  • Note that such functions are all monotonous and $f_n(1)- f_n(0)\leq1$ due to the mean value theorem. And since $f_n(0)=0$, this makes $f_n:[0,1]\to[0,1]$ as well. So your sequence is bounded as an element of the Banach space $C^1([0,1])$. You might use some of these thoughts to go forward. (But no guarantee) – Vercassivelaunos Aug 17 '20 at 12:45

1 Answers1

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The family is equi-contnuous by MVT and uniformly bounded. By Arzela-Ascoli Theorem every sequence in this family has a subsequence which converges uniformly. of course the lookout function is continuous.

Kavi Rama Murthy
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