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I have the following function $f(x)>0$: $$f(x)=g(x)(1-k(x))$$ $$\text{with } g(x)>0, 0<k(x)≤1 \text{ and } x \in \Bbb R^+.$$ I know that $g(x)$ is a linear and $k(x)$ a decreasing function with the following properties: $$g'>0 \text{ and } g''=0$$ $$k'>0 \text{ and } k''>0 \text{ or } k''<0 $$

My question is: Is it possible to prove that $f(x)$ has either no maximum or only one maximum (which is global)?


For this problem it seems quite intuitive to me there has to be only 1 maximum because $g(x)$ is increasing at a constant rate while $(1-k(x))$ is decreasing. Based on the fact that $(1-h(x))$ must be zero at some value for $x$, $f(x)$ can not increase forever and after it reaches a maximum it has to decrease; it is also possible that $f(x)$ is decreasing all the time.

Therefore I came up with another way of thinking about this problem. Based on the values for $g(x)$ and $k(x)$, $f(x)$ is either decreasing or is strictly concave over $x$. Therefore $f(x)$ is strictly-quasi-concave. This means that the following statement must be true: $$g(\lambda x_1 +(1-\lambda)x_2 )(1-k(\lambda x_1 +(1-\lambda)x_2 ))>\min(g(x_1)(1-k(x_1)); g(x_2)(1-k(x_2))$$ for $x_1<x_2$. We know that $g(x_1)<g(\lambda x_1+(1-\lambda)x_2)<g(x_2)$ because $g$ is constantly increasing with $x$. The same argument is also true $(1-k(x_1))<(1-k(\lambda x_1 +(1-\lambda)x_2 )<(1-k(x_2))$ because $(1-k(x))$ is monotonically decreasing with $x$. Therefore the product of $g(\lambda x_1 +(1-\lambda)x_2 )(1-k(\lambda x_1 +(1-\lambda)x_2 ))$ must be between $g(x_1)(1-k(x_1))$ and $g(x_2)(1-k(x_2)$. This means that the above statement must be true.

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    $\begingroup$ Do you mean $(k'>0)∧((k''>0)∨(k''<0))$ or $((k'>0)∧(k''>0))∨(k''<0)$? Also, if the question has been significantly altered, in your case the conditions are changed, it would be better post it as a new question. $\endgroup$ Mar 15, 2019 at 9:29
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    $\begingroup$ Also, if $k'>0$ how come $k$ is decreasing? $\endgroup$ Mar 15, 2019 at 9:33

2 Answers 2

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It holds $$f''(x) = -2g'(x)k'(x) - g(x)k''(x) < 0.$$ Hence $f$ is concave and allows only one maximum.

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  • $\begingroup$ Thank you for your answer. Do you mean strictly concave? Because this would only allow one maximum. Regarding the proof: is it sufficient to say that the first derivative is unclear because $f'(x)=g'(x)-g'(x)k(x)-g(x)k(x)$ is either increasing or decreasing based on the values for $g(x)$ and $k(x)$ but that the second derivative is clearly negative and therefore $f(x)$ is either decreasing or strictly concave? $\endgroup$
    – PAS
    Mar 12, 2019 at 16:05
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    $\begingroup$ Yes, it is strictly concave since $f''(x) < 0$. You do not need to argue with the first derivative since the second derivative already implies that $f$ is strictly concave. $\endgroup$
    – Klaus
    Mar 12, 2019 at 16:12
  • $\begingroup$ Unfortunately I made a mistake in one of my assumptions. I'm very sorry. I edited my post above. In short: I can no longer assume that $k''(x)>0$ $\endgroup$
    – PAS
    Mar 12, 2019 at 16:56
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I am suggesting the following example: $$k(x) = (1+x)^{-\frac{1}{2}}.$$ Then, $0<k(x)≤1 \text{ and } x \in \Bbb R^+.$ Moreover $$k'(x) = -\frac{1}{2}(1+x)^{-\frac{3}{2}} < 0.$$ Finally, $$k''(x) = \frac{3}{4}(1+x)^{-\frac{5}{2}} > 0.$$ Take for $g(x) = 1+x$. Then, $$f(x) = (1+x) (1-(1+x)^{-\frac{1}{2}}) = (1+x)-(1+x)^{\frac{1}{2}},$$ which does not have a global maximum.

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  • $\begingroup$ You are totally right. What if I change my statement and argue that the above function as either no maximum or only one maximum? $\endgroup$
    – PAS
    Mar 13, 2019 at 6:36

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