Existence of constant (4)

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

(1) $\begin{equation*} \int_0^1 f(x) \, \mathrm{d}x = \int_0^1 x f(x) \, \mathrm{d}x \end{equation*}$

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

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

Solution

For starters let us consider the function $G(x) = \int_0^x t f(t) \, \mathrm{d}t$ and $F(x) = \int_0^x f(t)\, \mathrm{d}t$ . Trivially , it is $F(0)=G(0)=0$ and we note that:

$\int_{0}^{1} f(x) \, \mathrm{d}t = \int_{0}^{1} t f(t) \, \mathrm{d}t \Rightarrow F(1) = G(1)$

Consider now the function

$H(x) = \left\{\begin{matrix} \dfrac{G(x)}{x} - F(x) & , & x \in (0, 1] \\\\ 0 & , & x =0 \end{matrix}\right.$

Clearly $H$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$ with derivative

$H'(x) = \frac{x^2 f(x)-G(x)}{x^2} - f(x)$

Furthermore , $H(1)=G(1)-F(1)=0=H(0)$ . Hence , by Rolle’s theorem there exists a $x_0 \in (0, 1)$ such that $H'(x_0)=0$ ; that is $G(x_0)=0$ . Finally, let us consider the function

$\Phi(x) = \left\{\begin{matrix} \dfrac{G(x)}{x} & , & x \in (0, x_0] \\\\ 0& , & x=0 \end{matrix}\right.$

which satisfies Rolle’s conditions on $[0, x_0]$ . Hence, there exists a $c \in (0, x_0) \subseteq (0, 1)$ such that $\Phi'(c)=0$ . However,

$\begin{align*} \Phi'(c) =0 &\Leftrightarrow \frac{c G'(c) - G(c)}{c^2} =0 \\ &\Leftrightarrow \frac{c \cdot c f(c) - G(c)}{c^2} =0 \\ &\Leftrightarrow c^2 f(c) = G(c) \\ &\Leftrightarrow c^2 f(c) = \int_0^c x f(x) \, \mathrm{d}x \end{align*}$

Existence of constant (4)

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