Constant function

Let $f:[0,1]\rightarrow \mathbb{R}$ be a continuous function such that $\displaystyle \int_0^1 f(x)\, {\rm d}x =1$ and:

$$\int_0^1 \left(1-f(x) \right)e^{-f(x)}\, {\rm d}x\leq 0$$

Prove that $f(x)=1, \; \forall x \in [0, 1]$.

Solution



Consider the function $g(x)=xe^x -x , \; x \in \mathbb{R}$ . It is differentiable in $\mathbb{R}$ and after studying its monotony we easily deduce that $g$ has a minimum at $x_0=0$ that is equal to $g(0)=0$. So $g(x)\geq 0, \; \forall x \in \mathbb{R}$. Below this is presented in a monotony table:

From the assumption we have that
$$\int_{0}^{1}f(x)\, {\rm d}x =1 \Leftrightarrow \int_{0}^{1}\left ( 1-f(x) \right )\, {\rm d}x =0 \tag{1}$$

Hence:

$$\begin{aligned}
\int_{0}^{1}\left ( 1-f(x) \right )e^{-f(x)}\, {\rm d}x\leq 0 &\Rightarrow e \int_{0}^{1}\left ( 1-f(x) \right )e^{-f(x)}\, {\rm d}x\leq 0 \\
 &\Rightarrow \int_{0}^{1}\left ( 1-f(x) \right )e^{1-f(x)}\, {\rm d}x\leq 0\\
 &\overset{(1)}{\Rightarrow }\int_{0}^{1}\left ( 1-f(x) \right )e^{1-f(x)}\, {\rm d}x - \int_{0}^{1}\left ( 1-f(x) \right )\, {\rm d}x \leq 0\\
 &\Rightarrow \int_{0}^{1}\left [ \left ( 1-f(x) \right )e^{1-f(x)} - \left ( 1-f(x) \right ) \right ]\, {\rm d}x\leq 0 \\
&\Rightarrow \int_{0}^{1}g\left ( 1-f(x) \right )\, {\rm d}x\leq 0
\end{aligned}$$

Therefore $g(1-f(x))=0, \; \forall x \in [0, 1]$. Hence $$f(x)-1=0 \Leftrightarrow f(x)=1, \; \; \forall x \in [0, 1]$$ since $g$ is strictly increasing on $[0, 1]$ and the exercise is complete.

The exercise can also be found in mathematica.gr