integral inequality with derivative

Question

Let a function $f:[0,1]→\mathbb{R}$ have a continuous derivative and $$\int_{0}^{1}f(x)dx=0$$ Show that for every $\alpha \in [0,1]$, $$\left|\int_{0}^{\alpha}f(x)dx\right|\le\frac{1}{8} \mathbb{sup} \{|f'(x)|:x\in[0,1]\}$$

Answer 1

This problem can be reformulated as follows by letting $F(a)=\int_0^a f(x) dx$. Then, for $F\in C^2([0,1])$ with $F(0)=F(1)$, show that $|F(x)| \leq \frac{M}{8}$, where $M=\sup_{x\in [0,1]} |F''(x)|$.

Let $c$ be a critical point of $F$ such that $F'(c)=0$. Then Taylor expanding, we get $$0 = F(0) = F(c)+ \frac{1}{2} F''(\xi_0) (c)^2,$$ $$0 = F(1) = F(c)+ \frac{1}{2} F''(\xi_1) (c-1)^2.$$

Rearranging and estimating, we have $|F(c)|\leq \frac{1}{2} M c^2$ and $|F(c)|\leq \frac{1}{2} M (c-1)^2$. Since $c\in [0,1]$, either $c\leq 1/2$ or $(1-c) \leq 1/2$. Thus, using the appropriate inequality, we get $|F(c)|\leq \frac{M}{8}$.

The maximum of $|F|$ is obtained at a critical point $c$. (We ignore the endpoints since $F(0)=F(1)=0$.) Hence, $|F(x)|\leq |F(c)| \leq \frac{M}{8}$, which is what we wished to show.