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I am looking for a sequence of functions that satisfies these properties:

$1$. Let $I=[0,1]$. $f_n\in C(I)$ and are continuously differentiable

$2$. $f_n(0)=0$ for every $n$

$3$. $|f_n'(x)|\le \dfrac{1}{1-x}$ for every $x\in (0,1)$

$4$. The sequence $(f_n)$ has no uniformly convergent subsequence on $I$

I would like for $f_n$ to be either not equicontinuous (when considering the sequence as a family of functions), or for them to not be uniformly bounded. I can come up with a few examples that work barring condition $(2)$, but none with all of them. Any ideas?

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  • $\begingroup$ What is an example at your hand that fits (2)? Please also show it in your question. $\endgroup$ – Megadeth Sep 26 '15 at 4:21
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    $\begingroup$ You are not looking for a counterexample to Arzela-Ascoli, for there are none: the theorem is true. You should try to find a more precise title! $\endgroup$ – Mariano Suárez-Álvarez Sep 26 '15 at 4:56
  • $\begingroup$ Well, no: there is no failure of the theorem: that is precisely why the theorem is true!. What you want is something like «Failure of the conclusion of Arela-Ascoli when its hypothesis does not hold» or, much better, «An example showing the necessity of the hypothesis of the Arzela-Ascoli theorem». $\endgroup$ – Mariano Suárez-Álvarez Sep 26 '15 at 5:03
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    $\begingroup$ (Please, do not delete questions for which there are already perfectly good answers!) $\endgroup$ – Mariano Suárez-Álvarez Sep 26 '15 at 5:07
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Let $f_n(x)$ be the polynomial $$f_n(x)=\sum_{k=1}^n \frac{x^k}{k}.$$

It is clear that 1) and 2) are satisfied. 3) holds because $f_n'(x)$ is a partial sum of the infinite series for $1/(1-x)$.

To prove 4), suppose for contradiction that a subsequence of $\{f_n\}$ converged uniformly to some function $g$. Then $g$ is a continuous function which vanishes on $[0,1)$, so $g(1)=0$. On the other hand, $f_n(1)\to\infty$ (because the harmonic series diverges). This contradicts subsequential uniform convergence, so property 4) holds.

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    $\begingroup$ Thanks, I was lazy and included only the highest degree monomial in the Taylor polynomial for $-\ln(1-x)$. I fixed it by including the entire Taylor poylnomial. Good catch! $\endgroup$ – pre-kidney Sep 26 '15 at 8:21

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