Existence of constants

Let $f:\mathbb{R} \rightarrow (0, +\infty)$ be a continuous function such that:

$$\int_0^1 f(x)\, {\rm d}x = \int_0^1 f^2(x)\, {\rm d}x $$

Prove that there exist $a, b \in \mathbb{R}$ such that $\displaystyle \int_a^b \frac{{\rm d}x}{f(x)}=1$.

Solution

Since $f(x)>0 , \; x \in [0, 1]$ we have that:

\begin{align*}
-\left ( f(x)-1 \right )^2\leq 0 &\Rightarrow -f(x)\left ( f(x)-1 \right )+\left ( f(x)-1 \right )\leq 0 \\
 &\Rightarrow f(x)-1\leq f(x)\left ( f(x)-1 \right ) \\
 &\overset{f(x)>0}{=\!=\!\Rightarrow} \frac{f(x)-1}{f(x)}\leq f(x)-1 \\
 &\Rightarrow 1- \frac{1}{f(x)}\leq f(x)-1
\end{align*}

Hence

$$ \int_{0}^{1}\left ( 1-\frac{1}{f(x)} \right )\, {\rm d}x \leq \int_{0}^{1}\left ( f(x)-1 \right )\,{\rm d}x \Rightarrow \int_{0}^{1}f(x)\, {\rm d}x + \int_{0}^{1}\frac{{\rm d}x}{f(x)}\geq 2 \tag{*}$$

However,

\begin{align*}
\int_{0}^{1}\left ( f(x) -1\right )^2\, {\rm d}x\geq 0 &\Rightarrow \int_{0}^{1}f^2(x)-2\int_{0}^{1}f(x)\, {\rm d}x + \int_{0}^{1}\, {\rm d}x \geq 0 \\
 &\Rightarrow \int_{0}^{1}f(x)\, {\rm d}x \leq 1 \quad \quad (**)
\end{align*}


Now, equations $(*), \; (**)$ imply that $\displaystyle \int_{0}^{1}\frac{{\rm d}x}{f(x)}\geq 1$.

Now, let us consider the function

$$g(x)=\int_{1/2-x}^{1/2+x}\frac{{\rm d}x}{f(x)}$$

that is defined in $\left [ -\frac{1}{2}, \frac{1}{2} \right ]$ and continuous in it. It also holds that:

$$g\left ( -\frac{1}{2} \right )=\int_{1}^{0}\frac{{\rm d}x}{f(x)}\leq 0<1$$

and similarly $g\left ( \frac{1}{2} \right )=\int_{0}^{1}\frac{{\rm d}x}{f(x)}\geq 1$. Using the intermediate value theorem we have the existence of an $x_0 \in (-1/2, 1/2]$ such that $g(x_0)=1$ that is

$$\int_{1/2-x_0}^{1/2+x_0}\frac{{\rm d}x}{f(x)}=1$$

which is what we want.