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I'm struggling with a problem which has been asked a lot of times but always in different versions so I'm still not quote sure about my answer. I have a function l:R→R:x↦l(x) which is the result of different smaller functions namely: $$l(x)=f(x)g(x)-g(x)h(x)$$ and $0<x<1$, with $f(x)>0$;$g(x)>0$ and $h(x)>0$.

Let $f(x)$ and $g(x)$ be convex as well as decreasing over the range of $x$ and let $h(x)$ be concave and also decreasing. Is the following statement true?

  1. The product of $f(x)$ and $g(x)$ is concave and the product of $g(x)$ and $h(x)$ is concave.

  2. But because of $-g(x)h(x)$ the second term of $l(x)$ is convex

  3. We are only interested in positive values for $l(x)$ we can say that $f(x)g(x)>g(x)h(x)$, $l(x)$ is concave for all $l(x)>0$

Thank you.

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  • $\begingroup$ Did you mean "Let $f(x)$ and $g(x)$..."? $\endgroup$ Mar 11, 2017 at 13:41
  • $\begingroup$ oh yes. I will edit the post, thank you $\endgroup$
    – PAS
    Mar 11, 2017 at 13:43
  • $\begingroup$ We need to investigate the second derivative: $$(fg)'' = f'' + 2f'g' + g''$$ Now, $f,g$ are convex so $f'' ,g'' >0 $. Further more, $f,g$ are decreasing so $f' , g' < 0 \Rightarrow f'g'>0 $, meaning that $(fg)''>0$ and the product is convex. About the product of $gh$, the statement feels wrong. I think taking $g(x)=e^{-x}$ and $h$ as some sort of $\arctan$ after a translation will do the job $\endgroup$ Mar 11, 2017 at 14:05
  • $\begingroup$ Thank you for your answer and you are totally right with your remark on $f$ and $g$. To be honest I'm not quite sure about the second derivative of $f''$ so I might be wrong about this. The problem is the following: I know that $l(x)$ definitely concave but the extent depends on a lot of different parameters $(a, b, c, d...)$. The second derivative for $l(x)$ is not an option and I can't really apply any other concept like $l(1-\alpha)x+\alpha y)>(1-\alpha) l(x) +\alpha l(y)$ Are there any options left? $\endgroup$
    – PAS
    Mar 11, 2017 at 14:37
  • $\begingroup$ the second derivative of f(x)g(x) is wrong. It should be: gf" + 2f'g' + fg". Since f > 0 ang g >0, (fg)' is still > 0 $\endgroup$
    – Hass
    Sep 14, 2017 at 12:22

2 Answers 2

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The claim is false in general.

However, Exercise 3.32.(a) on p. 119 in the book quoted by Convexity of the product of two functions in higher dimensions provides the following sufficient condition to obtain your result: $f,g: \mathbf{R} \rightarrow \mathbf{R}$ are convex, both nondecreasing (or non increasing) and positive on an interval.

A condition (stronger than convexity) that directly addresses your issue is log-convexity. The set of log-convex functions is closed under product, sum and positive scaling.

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  • $\begingroup$ What about an "inverse" problem when we know that a product of two functions is convex and that one of the functions in the product is also convex (and positive on some interval in R)? What can we say about the other function in the product? $\endgroup$
    – Confounded
    Jul 3, 2018 at 12:20
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Thank you for your answer. I was checking my formula again for different parameter values and it looks pretty bad for my original claim that $l(x)$ is concave, which is off the table now. In general $l(x)$ can look like this (again for different parameter values): https://ibb.co/eKY16F

The original problem (maybe I should make another thread for this) is the following: There is a value for $x$ that maximizes and minimizes $l(x)$ in the first quadrant (so for non-negative values). The range for $x$ is a compact set with $x[0,x°]$ where $x°$ is the zero of the function $l(x)$ which again depends on different parameter values. Using the theorem of Weierstrass I can say that: (I) That there exists a maximum and a minimum for $l(x)$ with the relevant $x$ values in the compact set $x[0, x°]$.

The Problem is: I can't really say something about the uniqueness of the maximum or the minimum of $l(x)$. I can't use the Karush-Kuhn-Tucker-Theorem because $l(x)$ is not concave. Maybe I should check for quasi-concavity?

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  • $\begingroup$ Monotone=quasiconcave. Any chance that $l$ is either concave or monotone? $\endgroup$
    – mlc
    Mar 12, 2017 at 13:58
  • $\begingroup$ Well for some parameter values $l$ is monotone decreasing (and looks convex) but for other values it is hump-shaped (and concave). I hope this answers your question. Both functions in the link are $l$ for different parameter values. $\endgroup$
    – PAS
    Mar 12, 2017 at 14:06
  • $\begingroup$ If your $l(x)$ is always concave or monotone (of either kind), the function is quasiconcave. There is a version of the KKT theorem that holds for a quasiconcave objective function. $\endgroup$
    – mlc
    Mar 12, 2017 at 16:42
  • $\begingroup$ Thank you for your answer. Yes, it's the Arrow-Enthoven-Theorem. But I'm still struggling with the sufficient condition for a quasiconcave objective function, especially the proof for a pretty complex function. But as you were saying "a function is quasiconcave if it is monotone" and although the dotted line is convex (see the link) it is always monotone decreasing. So my function is either hump-shaped or monotone decreasing (depending on the parameter values) and therefore it is quasiconcave? (sry if this is a kinda dump question). $\endgroup$
    – PAS
    Mar 13, 2017 at 13:53
  • $\begingroup$ After looking at it I think my function is quasiconcave. But I don't think the proof for this is quite difficult for my case. $\endgroup$
    – PAS
    Mar 13, 2017 at 14:18

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