Number of local minima of the product of a decreasing function and a linear function

Question

Let $f(x)=ax+b$ where $a,b>0$. Suppose that $g(x)>0, \forall x>1$ is a strictly decreasing function with $g''(x)>0$ and $\lim_{x\rightarrow\infty}g(x)=C>0$ where $C$ is a constant number. Is that possible to prove that $r(x)=f(x)g(x)$ has at most one local minimum on $x\in(1,+\infty)$? If not, may I have a counterexample $f(x)$, $g(x)$?

Answer 1

The differential equation $$ y'=-\frac{ay}{ax+b},\qquad y(1)=2C$$ has a solution with $y>C$ in some interval $[1,u]$. Note that $$ y''=-\frac{a(ax+b)y'-a^2y}{(ax+b)^2}=\frac{2ay^2}{(ax+b)^2}>\underbrace{\frac{2aC^2}{(au+b)^2}}_{=:D}>0$$ and $$ y'<-\frac{aC}{au+b}$$ for all $x\in[a,u]$.

If we picked $g(x)=y(x)$ (with a suitable extension beyond $u$), we'd have $$ r'(x)=f'(x)g(x)+f(x)g'(x)=ay+(ax+b)y'=0$$ for all $x\in(1,u)$. But we want strict local minima, don't we? To this, some tiny perturbation helps:

For arbitrary natural $n$ and for some yet to be specified constants $s>0$, let $$g(x)=\begin{cases}y(x)+s\sin \frac{n\pi x}{u}&\text{if }1\le x\le u\\\text{left as exercise}&\text{if }x>u\end{cases}$$ It is straightforward that the lower branch can be picked suitably to guarantee $g\in C^2[1,\infty)$, $g''(x)>0$ for all $x\in [1,\infty)$, and $\lim_{x\to\infty}g(x)=C$, provided the upper branch gives us $g(u)>C$, $g'(u)<0$, and $g''(x)>0$ for all $x\in [1,u]$. And indeed, we have $$ g'(x)=y'(x)+\frac{sn\pi}{u}\cos\frac{n\pi x}{u}<0$$ for all $x\in[1,u]$, provided we pick $s<\frac{aCu}{(au+b)n\pi}$. Also, $$ g''(x)=y''(x)-\frac{sn^2\pi^2}{u^2}\sin\frac{n\pi x}{u}>0$$ for all $x\in[1,u]$, provided we pick $s<\frac{2aC^2u^2}{(au+b)^2n^2\pi^2}$.

Now what about the extrema of $r$? For $1<x<u$, we have $$\begin{align}r'(x)&=f'(x)g(x)+f(x)g'(x)\\&=f'(x)(g(x)-y(x))+f(x)(g'(x)-y'(x))\\ &=s\cdot\left(a\sin\frac{n\pi x}{u}+(ax+b)\frac{n\pi }{u}\cos\frac{n\pi x}{u}\right)\end{align}$$ which has several isolated simple zeroes, about half of which lead to strict local minima of $r(x)$.