# Product of a linear and a monotonically decreasing function

I have the following function $$f(x)>0$$: $$f(x)=g(x)(1-k(x))$$ $$\text{with } g(x)>0, 0 I know that $$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 $$x_1. We know that $$g(x_1) 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.

• 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. Mar 15, 2019 at 9:29
• Also, if $k'>0$ how come $k$ is decreasing? Mar 15, 2019 at 9:33

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

• 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?
– PAS
Mar 12, 2019 at 16:05
• 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. Mar 12, 2019 at 16:12
• 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 $k''(x)>0$
– PAS
Mar 12, 2019 at 16:56

I am suggesting the following example: $$k(x) = (1+x)^{-\frac{1}{2}}.$$ Then, $$0 Moreover $$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.

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