Given $f:[a,b] \to R$ is continuously differentiable on $[a,b]$.
Supposed that $\int_a^b f(x)dx=0$; $f(a)=f(b)=0$
Prove $|\int_a^x f(t)dt|\le \frac {(b-a)^2}{16} sup_{a\le x\le b} |f'(x)|$ with $\forall x\in [a,b]$
I tried to use Taylor expansion and integral by parts but only reach to $\frac 18$ in the coefficient.
The stationary points for the function $$ F(x)=\int_{a}^{x}f(t)\,dt $$ occur when $f(x)=0$. Up to replacing $f(x)$ with $f(a+b-x)$, we can assume that $|F(x)|$ attains its maximum in a point $x_0\in(a,b)$ such that $x_0-a\leq\frac{b-a}{2}$. Integration by parts gives: $$\int_{a}^{x_0}f(t)\,dx = -\int_{a}^{x_0} t\,f'(t)\,dt =\int_{a}^{x_0}\left(\frac{a+x_0}{2}-t\right)\,f'(t)\,dt,$$ hence: $$\left|\int_{a}^{x}f(t)\,dt\right|\leq \max_{x\in[a,b]}|f'(x)|\cdot\int_{a}^{x_0}\left|\frac{a+x_0}{2}-t\right|\,dt = \max_{x\in[a,b]}|f'(x)|\cdot\frac{(x_0-a)^2}{4}$$ so: $$\left|\int_{a}^{x}f(t)\,dt\right|\leq\max_{x\in[a,b]}|f'(x)|\cdot\frac{(b-a)^2}{16}$$ as wanted.
