# Does $\int_0^1 f(x)=\int_0^1 xf(x)$ imply $\int_0^x f(t)$ has a root

My question is whether or not the following is true:

If $$f:[0,1]\to \mathbb{R}$$ is a continous function such that $$\int_0^1 f(x)dx=\int_0^1 xf(x)dx$$ then there exist $$c\in(0,1)$$ such that $$\int_0^c f(x)dx=0$$

It is quite clear that in the interval $$(0,1)$$ there must be some points $$a$$ such that $$f(1)f(a)<0$$ but I can't say from this if the statement above is true or not.

• Do you have an example of a non-zero function that satisfies the first equality? Jan 5, 2019 at 12:27
• @Yanko $\operatorname{sinc}(2\pi (1-x) )$ works Jan 5, 2019 at 12:35
• @Yanko Surley there are such functions. There exist $a,b$ such that $ax+b$ satisfies that equation. Jan 5, 2019 at 12:35
• @KaviRamaMurthy Right, for $a=6,b=-2$. Thanks. Jan 5, 2019 at 12:37

Let $$F(x)=\int_0^{x} f(t)\, dt$$. Then $$F(1)=\int_0^{1} f(t)\, dt=\int_0^{1} tf(t)\, dt=tF(t)|_0^{1}-\int_0^{1}F(t)\, dt$$. This gives $$\int_0^{1}F(t)\, dt=0$$. If $$F$$ has no zeros then it is strictly positive or strictly negative throughout and its integral cannot be $$0$$. Hence $$F(c)=0$$ for some $$c$$.
Hint: Note that $$\int_0^1 F(x)dx =\int_0^1\left(\int_0^x f(t)dt\right)dx=\int_0^1\left(\int_t^1dx\right)f(t)dt= \int_0^1 (1-t)f(t)dt$$ where $$F(x) = \int_0^x f(t)dt$$.