Number of itersection points between two monotonously decreasing functions Let $f(x)$ and $g(x)$ be two different functions, where:


*

*$f(x)$ is infinitely differentiable for every $x$

*$g(x)$ is infinitely differentiable for every $x$

*$f'(x)$ is negative for every $x$ (i.e., $f(x)$ is monotonously decreasing)

*$g'(x)$ is negative for every $x$ (i.e., $g(x)$ is monotonously decreasing)


I am suspecting that the possible number of intersection points for any such pair of functions is:


*

*$0$

*$1$

*$2$

*$\infty$ (if the functions "merge" at some point; not even sure if that's possible)


How can we prove or refute this?
I emphasized infinitely differentiable in order to avoid function definitions such as:
$
 f(x)=
 \begin{cases}
  \dots & \text{$ x \leq K$}\\
  \dots & \text{$ x \geq K$}\\
 \end{cases}
$
Thanks
 A: The property you are looking for isn't going to work as stated. By choosing a suitable value of $c$ near $-2$, the functions $f(x)=\sin x-2x$ and $g(x)=cx$ can be made to intersect $n$ times for any integer $n$, or $\infty$ times for $c=2$, even though they both have strictly negative derivatives.
Even if you demand that all derivatives are negative, you still can't limit the number of intersection points. Consider the function $h(x)=p(2^x)=\sum_{k=0}^na_k2^{kx}$, where $p(t)$ is a polynomial. There is a unique polynomial of degree $n$ passing through any $n+1$ points; choose $p(t)$ such that $p(0)=1$ and $p(2^x)=0$ for $x\in\{1,2,\dots,n\}$. Then $h(x)$ is a linear combination of exponentials that has roots at $\{1,2,\dots,n\}$. Now let $f(x)$ be the sum of all positive terms and $g(x)$ be the sum of all negative terms in $h(x)$. Then $f(x)+g(x)=h(x)$, and $f(x)$ has entirely positive derivatives, and $g(x)$ has entirely negative derivatives. (Note that neither $f(x)$ or $g(x)$ is zero, because $h$ has a root and $f,g$ don't.) Thus $-f$ and $g$ are functions with all negative derivatives, crossing at $\{1,2,\dots,n\}$.
As an example for $n=4$:
$$f(x)=1+\frac{35}{32}2^{2x}+\frac1{64}2^{4x};\qquad g(x)=-\frac{15}82^x-\frac{15}{64}2^{3x}$$
A challenge to readers: can you generalize this construction to a countably infinite number of intersection points?

But your original idea is not too far off the mark; there is a property of the sort you are looking for, namely convexity. A function with a positive second derivative is convex, and convex functions can intersect a line at most twice. (If you allow the derivative to be zero, it may coincide with the line on an interval, hence your "$\infty$ intersections" case.) Thus, a revised theorem statement:

If $f(x)$ and $g(x)$ are twice-differentiable functions such that $f''(x)<g''(x)$ for all $x$, then $f$ intersects $g$ at most twice.

A: Let $g(x) = -x^3 - x$ and let $f(x) = -2x$. Both have a negative first derivative, but they intersect at three points.
Edit: to follow up on Mario Carneiro's comment below: try to convince yourself that for any $n \in \mathbb N$ there exists a smooth function $h_n : \mathbb R \to \mathbb R$ such that 1) $h_n'(x) \le 1$ for all $x$, and 2) $h_n$ has exactly $n$ zeroes. Then $f(x) = -2x$ and $g(x) = -2x + h_n(x)$ both have derivatives less than or equal to $-1$ but intersect at $n$ points.
