Proving an integral identity

Question

Let $ f : \left[ 0, 1 \right] \rightarrow \mathbb{R} $ be a continuous function. Knowing that: $$ \int_0^1 (2x-1)f(x)dx = 0 $$ Show that $ \exists c \in \left(0,1\right)$ such that: $$ \int_0^c (x-c)(f(x)-f(c))dx = 0 $$

Let $F$ be a primitive of $f$ such that $F(0) = 0$. ( i.e. $F(x)=\int_0^xf(t)dt$ and $F'(x)=f(x)$ )

I have attempted to solve this and managed to show that there exists $ c, t \in [0,1] $ such that $ F(c) = F(1)/2 = f(t)/2 $ using the mean value theorem: $$ 2\int_0^1 x f(x)dx = \int_0^1 f(x) dx = \frac{F(1)-F(0)}{1-0} = f(t) \\ $$ Doing the first integral by parts we get: $$ \int_0^1 xf(x)dx = xF(x)\vert_0^1 \; - \int_0^1 F(x)dx = \\ = F(1) -\int_0^1F(x)dx $$

Using the mean value theorem, we get that: $$ \exists c\in [0,1] \text{ such that } \int_0^1F(x)dx = F(c) $$ Plugging everything in the first equation gives us: $$ 2( F(1) - F(c) ) = F(1) = f(t) \\ \Rightarrow F(1) = 2 F(c) \\ \Rightarrow F(c) = \frac{F(1)}{2} = \frac{f(t)}{2} $$ I do not know if these results will help solving this exercise or not.

I have already posted the question here but got no answer.

Answer 1

Let $H(c)=\int_0^c (x-c)\left (f(x)-f(c)\right) \, dx$. Then $$ \int_0^1 H(c)\, dc=\int_0^1 \, dx \int_x^1 (x-c)(f(x)-f(c))\, dc. $$ The innermost integral is $$-\frac {(x-1)^2}{2}f(x)- \int_x^1 (x-c) f(c)\, dc dc=(x-\frac 12 -\frac 12 x^2)f(x)-x\int_x^1 f(c)\, dc+\int_x^1 cf(c)\, dc.$$ Integrating by parts $$ -\int_0^1 x \int_x^1 f(c)\, dc=-\frac 12 \int_0^1 x^2 f(x)\, dx, \qquad \int_0^1dx \int_x^1 cf(c)\, dc=\int_0^1 x^2 f(x)\, dx $$ and summing all terms and using the assumption $\int_0^1 H(c)\, dc=0$.