I have a differentiable function, $f: \mathbb{R}_{\geq 0} \mapsto \mathbb{R}_{> 0}$, and another function, $g: x \mapsto \max(a-x,0)$, for some $a > 0$. I know the following facts about $f$.
- $f$ is decreasing;
- $f$ is convex;
- $\frac{df}{dx}\geq -1$.
- $f(x) \geq g(x)$ and $f(x) > 0$ for all $x \geq 0$.
I am interested in the set $K = \{x > 0: f(x) > g(x)\}$. Denote $l = \inf K$. Obviously, $l < a$. Based on these facts, can I say that
- $f(x) = g(x)$ on $[0,l)$;
- $f(x) > g(x)$ on $[l,\infty)$?
In particular, can I rule out the scenario that $f(x) > g(x)$ everywhere except at $x = a^*$ for some $a^* < a$ and $f(a^*) = g(a^*)$? I feel like the convexity and the bounded derivative of $f$ would not permit that but so far I only have a few pictures of functions I drew and not a formal proof (please see below). If this is not the only scenario I need to rule out, what other special cases are there?
There are two cases of interest, namely if $l=0$ or $l>0$
First assume $l=0$.
$[0,l)$ is empty, so the answer is vacuously "yes".
If $f(0)>g(0)$ (e.g. $f=g+1/(x+1)$) then the answer is "Yes". If $f(0)=g(0)$ (e.g. $f=g+(x/(1+x))$) then "No".
Now assume $l>0$.
If $x<l$ then $x\not\in K$ so property 4. implies $f(x)=g(x)$, so "yes".
By continuity and non-emptyness of $[0,l)$, we have $f(l)=g(l)$, so the answer is trivially "No". A better question is whether $f>g$ on $(l,\infty)$, in which case the answer is "Yes": if $x>l$, then $l<y<x$ for some $y\in K$. By the mean value theorem and since $f'\geq g'$ (by 3.), $f(x)>g(x)$.
Note that we didn't need the facts that $f$ is decreasing or convex.