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Let $f$ is positive-valued concave function and $g$ is another positive-valued concave function, such that $f:\mathbb{R}_+ \mapsto \mathbb{R}_+$ and $g:\mathbb{R}_+ \mapsto \mathbb{R}_+$. Additionally, $f\ge g$ over all domain $\mathbb{R}_+$.

Do their difference $f-g$ is concave function too?

This question is different from this.

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3 Answers 3

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No, let $f(x) = 1-e^{-x}$ and $g(x) = 1 - e^{-x/2}$.

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  • $\begingroup$ I have edited a question. When it is sure that their difference is always positive too. $\endgroup$
    – kaka
    Commented Dec 6, 2016 at 6:33
  • $\begingroup$ @kaka Let $f(x) = \sqrt{x}$ and $g(x) = \frac{1}{2} + \frac{1}{4}\log x$. $\endgroup$
    – Erik M
    Commented Dec 6, 2016 at 6:42
  • $\begingroup$ $(f-g)(x+0.5)$ is concave. $\endgroup$
    – kaka
    Commented Dec 6, 2016 at 6:52
  • $\begingroup$ It is not clear what you are asking for. The original question asked if the difference of two concave functions is concave, the counterexample shows this is not true in general. $\endgroup$
    – Erik M
    Commented Dec 6, 2016 at 6:55
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No. Take:

$f(x)=1-2^{-x}$

$g(x)=\frac x2,\ x\in[0,1]$ and $g(x)=f(x),\ x\in(1,+\infty)$

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I was trying to find out necessary conditions for it to be true. I was wondering about $f>g$. In fact, a slight modification of the above counter-example is sufficient to get another counter-example for the case $f>g$.

I came up with another counter-example. In fact the difference between two concave functions such that $f>g$ can be convex. Let $f$ be defined over $\mathbb{R}$, $$ f(x)= \left\{ \begin{array}{l} -\vert x \vert^3 \quad\text{if}\quad x\in [-0.25,0.25] \\ -3/16\vert x\vert +1/32 \quad\text{else}, \end{array} \right. $$ and $g(x)= -x^2$. Then $f$ and $g$ are concave and $f>g$ but $h = f-g$ is convex. You can see an illustration below. I have also illustrated the counter-examples given by Erik M and GLay.

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