Consider a function $g(x): [0,\infty) \rightarrow \mathbb{R}$ that is
- strictly decreasing
- concave
Does such a function always need to hit zero eventually?
If not, are we able to say something about the slope (bound it maybe?)
Also, if the answer is "no", what if we require strict concavity instead?
This may be an obvious question, but my thinking is as follows
– Reasons I feel it should hit zero: Suppose it has a slope of $m$ at $x=0$, and that $g(0)=c$ Then for $x=\frac{c}{m}$ we would have $g(x)=0$, even if the slope never became steeper
I think that proof (sketch) is valid?
Now, if instead the slope could decrease (say convex instead of concave), then its possible for the function to never reach zero? Sigmoids, for example.
Best Answer
Your sketch is roughly correct assuming that $g$ is differentiable. But the statement is still true without that assumption.
By concavity, for $s>1$, $$\dfrac{g(s)-g(0)}{s-0}\le\dfrac{g(1)-g(0)}{1-0}\Rightarrow g(s)\le s(g(1)-g(0))+g(0),$$ which is eventually negative since $g(1)<g(0)$.