Prove that $\exists a, b\in(0,1)$ such that $\int_0^{a} xf(x)dx=0\text{ and }\int_0^bxf(x)dx=\frac{b^2f(b)}{2}.$

Solution 1:

The first part is answered in Olympiad calculus problem. We can build on Christian Blatter's answer to solve the second part.

Define $F(x) = \int_0^x f(t) \, dt$ and $\phi(x) = \frac 1x \int_0^x F(t)\, dt$, $\phi(0) = 0$. In Christian's answer it is demonstrated that $$ \phi'(x) = \frac{1}{x^2} \int_0^x tf(t) \, dt $$ has a zero $a \in (0, 1)$, so that $\int_0^a tf(t) \, dt = 0$.

In our case we additionally have $f(0) = 0$, which implies that $\phi$ is differentiable at $x=0$ with $\phi'(0) = 0$. Applying the mean-value theorem to $\phi'$ gives that $$ \phi''(x) = -\frac{2}{x^3} \int_0^x tf(t) \, dt + \frac{f(x)}{x} $$ has a zero $b \in (0, a)$, so that $\int_0^b tf(t) \, dt = \frac{b^2f(b)}{2}$.