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Let's consider two non-negative concave functions $f_1(x)$ and $f_2(x)$ such that $f_1(x) \geq f_2(x)$ always, $f_1(x)= f_2(x) = 0$ at $x=0$ and both attain their maxima at $x\rightarrow \infty$, i.e., $f'_1(x)=f'_2(x)=0$ for $x\rightarrow \infty$. Moreover, $\lim_{x \rightarrow \infty} f_1(x) - f_2(x) = 1$.

Is it possible to state that $f_1(x)-f_2(x)$ also attains maxima at $x\rightarrow \infty$ without doing any calculations with the specific form of functions or the only option is to find the first derivative of $f_1(x)-f_2(x)$ and show that it's always positive??

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A simple counterexample is $$ f_1(x) = \min(4x, 2) , \\ f_2(x) = \min(x, 1) $$ which satisfy the given conditions, but the maximum of $f_1(x) - f_2(x) $ is $\frac 32$ at $x = \frac 12$.

A counterexample with differentiable functions is $$ f_1(x) = \frac{4}{\pi} \arctan(4x), \\ f_2(x) = \frac{2}{\pi} \arctan(x) $$ which satisfy the given conditions, but $$ f_1'(x) - f_2'(x) = \frac{2(7 - 8x^2)}{\pi(x^2+1)(16x^2+1)} $$ shows that the maximum of the difference is attained at $x = \sqrt{7/8}$.

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  • $\begingroup$ Thanks @Martin... I guess then the only option is to check the first derivative. $\endgroup$ – kphy Jul 16 '17 at 7:37

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