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Let $(f_n)_{n \in \mathbb{N}} \subset C([0,1], \mathbb{R})$ be a sequence of functions such that $\int_0^1 f_n (t) dt =0$ and $\vert f_n'(x) \vert < x$ then there is a subsequence that converge uniformly.

By Arzela-Ascoli theorem all I need to do is to prove that $(f_n)_{n \in \mathbb{N}}$ is uniformly bounded and equicontinuous, I already proved that is equicontinuous but I don't know how to use the hypothesis to prove that is uniformly bounded.

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  • $\begingroup$ You can't have $|f'(0)| < |0|=0.$ $\endgroup$ – zhw. Dec 5 '17 at 20:35
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From the mean value theorem for integrals it follows that $f_n(x_n) = \int_0^1 f_n (t) dt = 0$ for some $x_n$.

From $-x \le f_n'(x) \le x$ it follows that $f_n(x) - \frac 12 x^2$ is decreasing, and $f_n(x) + \frac 12 x^2$ is increasing, so that $$ \vert f_n(x) - f_n(y) \vert \le \vert \frac 12 x^2 - \frac 12 y^2 \vert $$ for all $x, y \in [0, 1]$.

Combining these facts we get $$ \vert f_n(x) \vert \le \vert f_n(x_n) \vert + \vert f_n(x) - f_n(x_n) \vert \le 0 + \vert \frac 12 x^2 - \frac 12 x_n^2 \vert \le \frac 12 \, . $$

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  • $\begingroup$ Very minor point, but how do you get equality in the second equation without assuming something like $f' \in L_1[0,1]$? (The result follows from the ordinary MVT, so the integral is not necessary.) $\endgroup$ – copper.hat Dec 5 '17 at 21:58
  • $\begingroup$ @copper.hat: You are right. Hopefully OK now. $\endgroup$ – Martin R Dec 6 '17 at 9:31
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If $|f'(x)| \le 1$ on $[0,1]$ and $|f(a)| > 2$ for some $a \in [0,1]$, then $$|f(x)| \ge |f(a)| - \left|\int_a^x f'(t)\; dt \right| > 1$$ Since $f$ is continuous, either $f(x) > 1$ everywhere or $f(x) < -1$ everywhere; in either case it is impossible to have $\int_0^1 f(x)\; dx = 0$.

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