TeXmacs export test

hbghlyj

$\newcommand{\mathd}{\mathrm{d}}$

Lemma 1

Let $f : [a, b] \rightarrow \mathbb{R}$ be any continuous function such that $f (x) \geqslant 0$ for all $x \in [a, b]$ and \[ \int_a^b f (x) \mathd x = 0 . \] Then $f (x) = 0$ for all $x \in [a, b]$.

Problem

Find all continuous function $f : [0, 1] \rightarrow \mathbb{R}$ that satisfy \[ \int_0^1 f (x) (1 - f (x)) \mathd x = \frac{1}{4} . \]

Solution

Since \[ f (x) (1 - f (x)) = - \left( f (x) - \frac{1}{2} \right)^2 + \frac{1}{4} . \] So, \begin{eqnarray*} \int_0^1 f (x) (1 - f (x)) & = & \int_0^1 \left( - \left( f (x) - \frac{1}{2} \right)^2 + \frac{1}{4} \right) \mathd x\\ & = & - \int_0^1 \left( f (x) - \frac{1}{2} \right)^2 \mathd x + \int_0^1 \frac{1}{4} \mathd x\\ & = & - \int_0^1 \left( f (x) - \frac{1}{2} \right)^2 \mathd x + \frac{1}{4} \end{eqnarray*} But $\int_0^1 f (x) (1 - f (x)) \mathd x = \frac{1}{4}$ Therefore, \[ \frac{1}{4} = - \int_0^1 \left( f (x) - \frac{1}{2} \right)^2 \mathd x + \frac{1}{4} . \] i.e. \[ - \int_0^1 \left( f (x) - \frac{1}{2} \right)^2 \mathd x = 0 \quad \Rightarrow \quad \int_0^1 \left( f (x) - \frac{1}{2} \right)^2 \mathd x = 0 . \] But $\left( f (x) - \frac{1}{2} \right)^2 \geqslant 0$ for all $x \in [0, 1]$. From Lemma 1, we have \[ \left( f (x) - \frac{1}{2} \right)^2 = 0 \Rightarrow f (x) - \frac{1}{2} = 0 \] i.e. \[ f (x) = \frac{1}{2} \quad \forall x \in [0, 1] . \] Hence, exactly one function satisfy the condition.