Let $f_n$ be a sequence of differentiable functions on $[0,1]$ such that $|f_n'(x)| \le M$ (the absolute value of the derivative of $f_n$ at $x$) for all $n$ and for all $x$ in $[0,1]$. Show that $f_n$ has a uniformly convergent subsequence.

Partial solution: The $f_n$s are equicontinuous (by the mean value theorem). How to prove that the $f_n$s are pointwise bounded, so that we can use Arzela-Ascoli?

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    $\begingroup$ As the problem stands, you cannot; it may be false. For example, let $f_n(x)=n$. Is there anything else you know about $\{f_n\}$? $\endgroup$ – Nate Eldredge Feb 7 '12 at 13:49

As stated, the sequence need not be uniformly (in your words, pointwise, but I have not heard this term used this way before) bounded, and in fact need not have a uniformly convergent subsequence. Take for example the sequence of functions $f_n(x)=n$. I suspect that you are missing an additional assumption, of the form "each $f_n$ satisfies $f_n(0)=0$". If this is the case, then we can apply the Fundamental Theorem of Calculus: $$|f_n(x)|=\left|f(0)+\int_0^xf_n'(t)dt\right|\leq 0+\int_0^x|f_n(t)|dt\leq \int_0^xMdt=Mx\leq M$$

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    $\begingroup$ The usual hypothesis of the Arzela-Ascoli theorem, besides equicontinuity, is that $\sup_n |f_n(x)| < \infty$ for each $x$, which is usually expressed as "the sequence is pointwise bounded". I'd expect "uniformly bounded" to mean $\sup_{n, x} |f_n(x)| < \infty$. Of course, it follows from the theorem that assuming equicontinuity, the two conditions are equivalent, but the weaker condition of pointwise boundedness should be easier to check. $\endgroup$ – Nate Eldredge Feb 7 '12 at 14:41
  • $\begingroup$ @NateEldredge Ah, thank you for the clarification. $\endgroup$ – Alex Becker Feb 7 '12 at 15:07

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