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Constant function

Let f:[0, 1] \rightarrow \mathbb{R} be a continous function such that \bigintsss_0^1 f(x) \, {\rm d}x=1 and

(1)   \begin{equation*}\int_0^1 \left(1-f(x) \right)e^{-f(x)}\, {\rm d}x\leq 0 \end{equation*}

Prove that f(x)=1 forall x\in \mathbb{R}.


Consider the function g(x)=xe^x-x , \; x \in \mathbb{R}  which is differentiable in \mathbb{R}. We can easily see that g has a global minimum at x_0=0 that is equal to g(0)=0. Visually we have that:

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Clearly as we can see it holds that g(x) \geq 0 forall x \in \mathbb{R}. Also:

(2)   \begin{equation*} \int_{0}^{1}f(x)\, {\rm d}x =1 \Leftrightarrow \int_{0}^{1}\left ( 1-f(x) \right )\, {\rm d}x =0\end{equation*}

Thus (1) gives us:

\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{(2)}{\Rightarrow }\int_{0}^{1}\left ( 1-f(x) \right )e^{1-f(x)}\, {\rm d}x - \\ &\quad \quad \quad - \int_{0}^{1}\left ( 1-f(x) \right )\, {\rm d}x \leq 0\\ &\Rightarrow \int_{0}^{1}\bigg [ \left ( 1-f(x) \right )e^{1-f(x)} - \\&\quad \quad \quad -\left ( 1-f(x) \right )  \bigg]\, {\rm d}x\leq 0 \\ &\Rightarrow \int_{0}^{1}g\left ( 1-f(x) \right )\, {\rm d}x\leq 0 \end{aligned}

Therefore g \left( 1- f(x) \right)=0 . Thus f(x)=1 forall  x \in \mathbb{R}.

The exercise can also be found in mathematica.gr

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