is the sum of an increasing function and a decreasing function either monotonic or unimodal?

Question

Suppose $f(x)$ is an increasing function and $g(x)$ a decreasing function, and $h(x)=f(x)+g(x)$, what guarantees $h'(X)$ has at most one solution?

To be more specific, for the problem I have at hand, $f''(x)>0$ and $g''(x)<0$, is there at most 1 solution for $h'(x)=0$? Thanks.

Any function of bounded variation (on a closed interval) is the sum of an increasing function and a decreasing function. In particular, any $C^1$ function can be obtained this way. In case of $C^1$ functions, this can be easily made explicit: just look at integrals of the negative and positive part of the derivative. — tomasz, Mar 15 '18 at 22:16

Especially Lime · Answer 1 · 2018-03-16T08:09:45.050

1

There's no reason why this should be the case. For example, $\sin x$ is the sum of an increasing function and a decreasing function: let $f(x)=\int_0^x\max(\cos t,0)\,dt$ and $g(x)=\int_0^x\min(\cos t,0)\,dt$; then $f(x)+g(x)=\int_0^x(\max(\cos t,0)+\min(\cos t,0))\,dt=\int_0^x\cos t\,dt=\sin x$.

(You can modify this easily to get $f'(x)>0$ everywhere and $g'(x)<0$ everywhere, for example by adding $x$ to $f(x)$ and subtracting it from $g(x)$.)

edited Mar 16 '18 at 08:09

answered Mar 15 '18 at 22:07

Especially Lime

40,828

Thanks for your answer. Please take a look at my exchange with Tal-Botvinnik. – Adam Mar 16 '18 at 01:14

score 1 · Answer 2 · answered Mar 15 '18 at 22:07

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With your hypothesis only, there is no guarantee of it. It is easy to find such $f,g$ with $f+g$ having any amount of slope changes, for instance see the other answer.

There may be hypothesis of asymptotic growth of $f,g$ on $\pm \infty$ and local behavior of functions that guarantee at most one solution to the equation, but these hypothesis would be rather artificial.

Also if $f,g$ are constrained to be polynomials, there may be some assumptions on the degree and the coefficients that guarantee that too.

answered Mar 15 '18 at 22:07

Tal-Botvinnik

1,843

Okay, let's say h(0)=h(∞)=0. Would this be sufficient? To be even more specific, what I have is f(x)=(1-a^x)/x and g(x)=(a^x)ln(a), where 0<a<1. – Adam Mar 16 '18 at 01:03
Do we have any continuity hypothesis in that case? – Tal-Botvinnik Mar 16 '18 at 01:05
yes, absolutely. – Adam Mar 16 '18 at 01:06
I'm pretty sure the other answer's counterexample can be adapted to make the function converge to 0 and infinity. – Tal-Botvinnik Mar 16 '18 at 01:08
And for your special functions, why don't you just compute the derivative and equate to 0? – Tal-Botvinnik Mar 16 '18 at 01:09
Still, I don't have a proof why h'(x) has a unique solution. – Adam Mar 16 '18 at 01:11
Maybe that proof doesn't exist with your current hypothesis... I'll give it a thought – Tal-Botvinnik Mar 16 '18 at 01:12
I like this one ... $\sin x + 2x$ is increasing, $\sin x - 2x$ is decreasing, and their sum is $2 \sin x$. – GEdgar Feb 12 '24 at 18:14

is the sum of an increasing function and a decreasing function either monotonic or unimodal?

2 Answers2