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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.

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

2 Answers 2

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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 $c$.

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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 $F(x) = \int_0^x f(t)dt$.

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