Proving the existence of a point with a certain property for a continuous function

Let $f:[0,1]\to\mathbb{R}$ a continuous function and $\int_0^1xf(x)dx=0$. Show that there exists a point $c\in(0,1)$ so that $f(c)=(\int_c^1f(x)dx)^2$.

As a potential solution, I tried assuming that no such point exists, then the function $g(x)=f(x)-(\int_x^1f(t)dt)^2$ would have constant sign for all $x\in(0,1)$. $g(x)>0$ can't be true for all $x\in(0,1)$ since then $\int_0^1xf(x)dx>0$. But I can't figure out how to prove that $g(x)<0$ for all $x$ can't be true.

Thanks

Solution 1:

As suggested in Greg Martin's comment, let us introduce $F(x):=\int_x^1 f(t)dt$, which satisfies that $F'(x)=-f(x)$ and $F(1)=0$. Moreover, $\int_0^1 F(x)dx=\int_0^1xf(x)dx=0$, so by continuity, there exists $x_0\in (0,1)$ such that $F(x_0)=0$. Let $G(x):=e^{\int_0^xF(t)dt}F(x)$. Firstly, by definition, $$G'(x)=e^{\int_0^xF(t)dt}\big((\int_x^1f(t)dt)^2-f(x)\big).$$ Secondly, $G(x_0)=G(1)=0$, so by Rolle's theorem, there exists $c\in (x_0,1)$, such that $G'(c)=0$. The conclusion follows.