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This is almost surely an Arzela-Ascoli question, since it comes from an old exam of which such problems are quite common. Unfortunately, I can't seem to get it though.

$\{ f_n \}$ is a sequence of functions $[0,1] \to \mathbb{R}$ satisfying $|f_n'(x)| \leq \frac{1 + |\ln (x)|}{\sqrt{x}}$ and $-10 \leq \int_0^1 f_n(x) dx \leq 10$ for every $n$. The question is to show the existence of a uniformly convergent subsequence.

In these kinds of situations, the bounds given usually turn into something nice to force uniform boundedness. Maybe I could hope to establish equicontinuity using the integral somehow, but I'm not told that the $f_n$ are always positive, so the absolute values would probably screw things up anyway.

Is it an easy problem I'm missing or is this totally the wrong track? If so, what's the right track?

A. Thomas Yerger
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1 Answers1

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I assume the $f_n$ are differentiable on $[0,1]$ and are continuously differentiable on $(0,1].$

Here's a sketch that should help you along (ask if you have questions). First, the function

$$x\to \int_0^x \frac{1+|\ln t|}{\sqrt t}\,dt$$

is continuous, hence uniformly continuous, on $[0,1].$ Thus given $\epsilon>0,$ we will have

$$ \int_x^y \frac{1+|\ln t|}{\sqrt t}\, dt < \epsilon$$

if $x,y$ are close to each other.

It follows that for such $x,y$ we have

$$|f_n(y) - f_n(x)| = |\int_x^y f_n'(t)\,dt| \le \int_x^y |f_n'(t)|\,dt \le \int_x^y \frac{1+|\ln t|}{\sqrt t}\, dt < \epsilon$$

for every $n.$ This shows the $f_n$ are equicontinuous on $[0,1].$ By Arzela-Ascoli, we will have the desired conclusion if we show the $f_n$ are uniformly bounded.

Now the given condition on the integrals implies that for each $n,$ there is $x_n\in [0,1]$ such that $|f(x_n)|\le 10.$ Thus for each $n$ we have

$$|f_n(x)| = |f_n(x_n) + \int_{x_n}^x f_n'(t)\,dt\,|\le |f_n(x_n)| + \int_{x_n}^x |f_n'(t)|\,dt$$ $$ \le 10 + \int_{x_n}^x \frac{1+|\ln t|}{\sqrt t}\, dt \le 10 + \int_{0}^1 \frac{1+|\ln t|}{\sqrt t}\, dt <\infty. $$

This gives uniform boundedness and we're done.

zhw.
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