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For each positive integer $n$ let $f_n : [0,1] \to R$ be a continuous function, differentiable on $(0,1]$, such that $$|f_n^{~'} (x)| \leq\frac{1 + |\ln x|}{\sqrt{x}} $$for $0 < x \leq 1$. and such that $$−10 \leq \int^1_0 f_n (x)~\mathrm{d}x \leq 10.$$ Prove that $\{f_n \}$ has a uniformly convergent subsequence on $[0,1]$.

I try to use the Arzelà–Ascoli theorem, which requires me to prove $\{f_n\}$ is uniformly bounded and equicontinuous.

From the second inequality, we conclude $\{f_n\}$ is uniformly bounded.

However, I stuck with proving they are equicontinuous.

I try to use the Lagrange theorem so that it suffices to show the $\{f_n^{~'}\}$ is uniformly bounded.

However, the first inequality cannot give us what we want since the function on the right side is not bounded on $(0,1]$. So what should I do?

Aolong Li
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    Hello fellow IU student. I posted this question a while back, also while studying there. I'll link you my question momentarily. – A. Thomas Yerger Aug 01 '17 at 20:47
  • The second inequality alone doesn't imply uniform boundedness. You need the equicontinuity (which you get from the first). – Daniel Fischer Aug 01 '17 at 20:48
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    https://math.stackexchange.com/questions/2281164/uniformly-convergent-subsequence – A. Thomas Yerger Aug 01 '17 at 20:48
  • @AlfredYerger Thank you very much! – Aolong Li Aug 01 '17 at 21:07
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    @AolongLi No problem, and hopefully luck will be with us both in a few weeks! – A. Thomas Yerger Aug 01 '17 at 21:34
  • @DanielFischer Could you make it more precisely on why the second inequality doesn't imply the uniform boundness? – Aolong Li Aug 01 '17 at 21:50
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    When you integrate, large values of a function at some points can be compensated by small values or large values of the opposite sign. For example, if $g_n(x) = n \cdot (2x-1)$, then $g_n(1) = n$, but $\int_0^1 g_n(x),dx = 0$ for all $n$. So to conclude the uniform boundedness, you need something beyond the bound on the integrals. The bound on the derivatives gives that. We have $$\lvert f_n(y) - f_n(x)\rvert \leqslant \int_x^y \lvert f_n'(t)\rvert,dt \leqslant \int_x^y \frac{1+\lvert\ln t\rvert}{\sqrt{t}},dt\leqslant \int_0^1\frac{1+\lvert\ln t\rvert}{\sqrt{t}},dt.$$ – Daniel Fischer Aug 02 '17 at 10:09
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    If we call that last integral $A$, then we have the inequality $\lvert f_n(y) - f_n(x)\rvert \leqslant A$ for all $x,y$ and all $n$, so in particular $f_n(0) - A \leqslant f_n(x) \leqslant f_n(0) + A$ for all $x$ and all $n$, which gives $f_n(0) - A \leqslant \int_0^1 f_n(x),dx \leqslant 10$ and $-10 \leqslant \int_0^1 f_n(x),dx \leqslant f_n(0) + A$. Together, $-(10+A) \leqslant f_n(0) \leqslant 10+A$, and therefore $\lvert f_n(x)\rvert \leqslant 10 + 2A$ for all $x$. (We get the tighter bound $\lvert f_n(x)\rvert \leqslant 10+A$ – Daniel Fischer Aug 02 '17 at 10:09
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    if instead of using $f_n(0)$ we use $f_n(y)$ for an arbitrary but fixed $y$, and with a bit more work, we could get even tighter bounds, but all that matters is a uniform bound, so $10 + 2A$ is enough.) For the equicontinuity, note that since $\frac{1+\lvert\ln x\rvert}{\sqrt{x}}$ is decreasing, for $x < y$ we have $$\lvert f_n(y) - f_n(x)\rvert \leqslant \int_x^y \lvert f_n'(t)\rvert,dt \leqslant \int_x^y \frac{1+\lvert\ln t\rvert}{\sqrt{t}},dt \leqslant \int_0^{y-x} \frac{1+\lvert\ln t\rvert}{\sqrt{t}},dt.$$ – Daniel Fischer Aug 02 '17 at 10:09
  • @DanielFischer Thank you!!! – Aolong Li Aug 02 '17 at 16:48
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    This is not an exact duplicate actually. I wrote the old solution, but there I assumed the $f_n$ were continuously differentiable on $(0,1]$ so we could use the elementary FTC. We are not assuming that here! One can make the old solution work, but we need to appeal to the Lebesgue version of the FTC. – zhw. Aug 03 '17 at 10:37

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