# Difference of concave functions

Suppose that there are two concave functions $f_1(x)$ and $f_2(x)$ defined on $x\geq0$. In addition, the functions are positive, smooth, bounded ($|f_2|\leq b_2,|f_1|\leq b_1$ such that $b_2 = b_1<\infty$), monotone increasing and $f_1\geq f_2$ where $f_1(0)=f_2(0)=0$. I'm trying to be as much specific as I can because my question is a bit vague:

Is there something general that can be said about the difference $f_1-f_2$ without the further knowledge of the behavior of their derivatives? Namely, is there something to say about the number of local extrema or stationary points the difference can have?

To elaborate: clearly, $f'_1(0)>f'_2(0)$ but for $x>0$ the derivative of $f_2$ can become greater than that of $f_1$ at least once. In that case there is one stationary point. Can there be more? What changes if $b_2< b_1<\infty$ or if both functions are unbounded?

Feel free to point me to books, web sites or lecture notes where the answers are. This is not a homework.

EDIT: Loosely speaking, can a difference of two functions satisfying the assumptions above have many ($>1$) humps and dips? If yes provide an example or argument.

What would the two functions have to satisfy for their difference to have just one stationary point (that is one function approaches the other, crosses it and departs)?

## 2 Answers

I think the sufficient condition to have at most two stationary points is to show that the first derivatives itself are strictly convex/concave increasing/decreasing (monotonic convex or monotonic concave).

If for example, we have $f_1'> 0$ and $f_2' >0$ and furthermore, $f_1''' > 0$ and $f_2'''<0$, then $f_1'$ is convex and $-f_2'$ is also convex.

$f_1' - f_2'$ will have at most two zeros (since sum of convex is convex) and hence at most two critical points of $f_1 -f_2$.

The humps and dips will go away in this case.

If I understand your question correctly, nothing prevents $f'_2$ from being larger than $f'_1$ as many times as you like, for example, qualitatively

Here, $f'_2$ and $f'_1$ are plotted. If you'd like to make $f_1$, $f_2$ bounded, increase $y$-values on this plot by, say, 20 and set $f'_2(x)$ and $f'_1(x)$ to $0$ for $x \ge 20$

EDIT: more precisely, $f'_1(x) = 21.9 - x$ and $f'_2(x) = 21 - x + \sin x$ , for $x \le 20$, and both are $0$ for $x > 20$.

• What is what on your plot? Do your functions satisfy my assumptions? – leprotic May 9 '14 at 17:18
• Ok, it is not the whole positive line but thx anyway. – leprotic May 9 '14 at 17:56