Existence of constant

Let $f:[0,1] \rightarrow \mathbb{R}$ be a continuous function such that
$$\int_0^1 f(x) \, {\rm d}x = \int_0^1 xf(x) \, {\rm d}x \tag{1}$$

Prove that there exists a $c \in (0,1)$ such that $\displaystyle c f(c) = 2 \int_c^0 f(x) \, {\rm d}x$.

(Duong Viet Thong, Nam Dinh University, Vietnam)

Solution (by Paolo Perfetti)

Define the function $\displaystyle H(x)= \int_{0}^{x}\left ( \int_{0}^{y}f(z)\, {\rm d}z \right )\, {\rm d}y $.  Then trivially  $H(0)=0$. For $x=1$ we note that:

$$H(1) = \int_{0}^{1}\left ( \int_{0}^{y}f(z)\, {\rm d}z \right )\, {\rm d}y= \int_{0}^{1}\left ( \int_{z}^{1}f(z)\, {\rm d}y \right )\, {\rm d}z = \int_{0}^{1}\left ( 1-z \right )f(z)\, {\rm d}z \overset{(1)}{=}0$$

So, by Rolle's theorem , via the continuity of $f(x)$ we have the existence of $x_0 \in (0, 1)$ such that $H'(x_0)=0$, that is $\displaystyle \int_0^{x_0} f(x) \, {\rm d}x=0$. Now define the function:

$$F(x)=x^2 \int_{0}^{x}f(y)\, {\rm d}y$$

and observe that $F(0)=F(x_0)=0$. Again Rolle's theorem assures us that there exists a $c \in (0, 1)$ such that $F'(c)=0, \; c \in (0, x_0)$ that is:

$$2c \int_{0}^{c}f(x)\, {\rm d}x + c^2 f(c)=0 \overset{c \neq 0}{\iff}cf(c) = 2\int_c^0 f(x)\, {\rm d}x$$

and we are done.