Existence of constant

[Tolaso J Kos]

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

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

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

[Papapetros Vaggelis]

If there exists \(\displaystyle{c\in\left(0,1\right)}\) such that \(\displaystyle{c\,f(c)=2\,\int_{c}^{0}f(x)\,\mathrm{d}x}\) , then :

\(\displaystyle{\begin{aligned} c\,f(c)+2\,\int_{0}^{c}f(x)\,\mathrm{d}x=0&\iff c^2\,f(c)+2\,c\,\int_{0}^{c}f(x)\,\mathrm{d}x=0\\&\iff c^2\,\left[\dfrac{\mathrm{d}}{\mathrm{d}x}\int_{0}^{x}f(t)\,\mathrm{d}t\right]_{x=c}+\left[\dfrac{\mathrm{d}}{\mathrm{d}x}x^2\right]_{x=c}\,\int_{0}^{c}f(t)\,\mathrm{d}t=0\\&\iff \left[\dfrac{\mathrm{d}}{\mathrm{d}x}\,x^2\,\int_{0}^{x}f(t)\,\mathrm{d}t\right]_{x=c}=0\end{aligned}}\)

So, in order to prove that there exists such a \(\displaystyle{c\in\left(0,1\right)}\) , it is sufficient to prove that the function

\(\displaystyle{g:\left[0,1\right]\longrightarrow \mathbb{R}\,\,,g(x)=x^2\,\int_{0}^{x}f(t)\,\mathrm{d}t}\) satisfies the conditions of

\(\displaystyle{\rm{Rolle's}}\) theorem on \(\displaystyle{\left[0,1\right]}\) or on \(\displaystyle{\left[0,x_0\right]\subseteq \left[0,1\right]}\) .

The function \(\displaystyle{g}\) is continuous at \(\displaystyle{\left[0,1\right]}\) and differentiable at \(\displaystyle{\left(0,1\right)}\)

with \(\displaystyle{g(0)=0}\) . What about \(\displaystyle{g(1)}\) ? We are not sure if \(\displaystyle{g(1)=0}\) . From the hypothesis,

\(\displaystyle{\int_{0}^{1}\left(1-x\right)\,f(x)\,\mathrm{d}x=0\iff \int_{0}^{1}\,\int_{x}^{1}f(x)\,\mathrm{d}y\,\mathrm{d}x\,=0 \; (I)}\) .

We define \(\displaystyle{K:\left[0,1\right]\longrightarrow \mathbb{R}}\) by \(\displaystyle{K(x)=\int_{0}^{x}\,\left(\int_{0}^{y}f(t)\,\mathrm{d}t\right)\,\mathrm{d}y}\) .

The function \(\displaystyle{K}\) is continuous at \(\displaystyle{\left[0,1\right]}\) , differentiable at \(\displaystyle{\left(0,1\right)}\) with


\(\displaystyle{K^\prime(x)=\int_{0}^{x}f(t)\,\mathrm{d}t\, , \,0<x<1\,\,,}\)

\( \displaystyle K(0)=0 \)

and

\(\displaystyle{\begin{aligned} K(1)&=\int_{0}^{1}\,\int_{0}^{y}f(t)\,\mathrm{d}t\,\mathrm{d}y\\&=\left[y\,\int_{0}^{y}f(t)\,\mathrm{d}t\right]_{0}^{1}-\int_{0}^{1}y\,f(y)\,\mathrm{d}y\\&=\int_{0}^{1}f(y)\,\mathrm{d}y-\int_{0}^{1}y\,f(y)\,\mathrm{d}y\\&=0\end{aligned}}\)

According to \(\displaystyle{\rm{Rolle's}}\) theorem, \(\displaystyle{K^\prime(x_0)=\int_{0}^{x_0}f(t)\,\mathrm{d}t=0}\) for some \(\displaystyle{x_0\in\left(0,1\right)}\).

Now, \(\displaystyle{g(0)=g(x_0)=0}\) and the exercise comes to an end.