Existence of constant

Let $f:[0, 1] \rightarrow \mathbb{R}$ be a continuous function such that $f(0)=0$ and

$$\int_0^1 f(x) \, {\rm d}x = \int_0^1 x f(x) \, {\rm d}x \tag{1}$$

Prove that there exists a $c \in (0, 1)$ such that

$$\int_0^c x f(x) \, {\rm d}x = \frac{c}{2} \int_0^c f(x) \, {\rm d}x$$

Solution

Let $F$ be a primitive of the function $f$ and $G$ be a primitive of $F$ such that $G(0)=G'(0)=0$. Thus $G''(0)=f(0)=0$. We can now conclude $G(1)$. Indeed using $(1)$ we have that:

\begin{align*}
\int_{0}^{1} f(x) \, {\rm d}x  = \int_{0}^{1} x f(x) \, {\rm d}x&\Leftrightarrow G'(1) - G'(0) = G'(1) - G(1) + G(0) \\
 &\Leftrightarrow G(1)=0
\end{align*}

Let us consider the function $\displaystyle H(x) = \left\{\begin{matrix}
\dfrac{G(x)}{x^2} & , & x \neq 0 \\\\
 0& , &  x =0
\end{matrix}\right.$ which is continuous in $[0, 1]$ because

$$H(0)= \lim_{x\rightarrow 0} \frac{G(x)}{x^2} = \lim_{x\rightarrow 0} \frac{G'(x)}{2x} = \lim_{x\rightarrow 0} \frac{G''(x)}{2} = 0$$

So, Rolle's conditions are satisfied for $H$ therefore there exists a $c \in (0, 1)$ such that $H'(c)=0$. This in return means that:

\begin{align*}
H'(a)=0 &\Leftrightarrow c G'(c) - 2G(c) =0  \\
 &\Leftrightarrow c G'(c) - G(c) + G(0)= \frac{c}{2}\left ( G'(c) - G'(0) \right ) \\
 &\Leftrightarrow \int_{0}^{c} x f(x) \, {\rm d}x = \frac{c}{2}\int_{0}^{c} f(x) \,{\rm d}x
\end{align*}

which is what we desire.