Question: Let $f(x):[0,\infty)\to \mathbb R $ be a twice differentiable function such that $f(0)=0$ and $f''(x)<0$ $\forall x\in (0,\infty)$ and $f(1)=\alpha,\alpha>0.$ Which of the following will hold true:
- $f$ is necessarily bounded above.
- $lim_{x \to \infty}f'(x)=0$
- $f(x)$ can have atmost one root.
$1$. Since $f''(x)<0$, therefore $f'(x)$ is strictly decreasing, so $f'(x)$ can have atmost one root (no comments on $f(x)$ from here). Also $f(x)$ is convex, so choose $f(x)=-x^2+2x$, satisfies the given conditions at the same time it has $2$ roots so 3. is false, also $f'(x)=-2x+2$ does not have a $0$ limiting value. It is not bounded. Is this a right counter example ?
$2.$ We are given $f(1)=\alpha>0,$ if we define $f:[0,1]\subset[0,\infty)\to \mathbb R$ and given that $f(x)$ is a twice differentiable function, so we can use Mean Value Theorem on $[0,1]$ there exist $c\in(0,1)$ such that $f'(c)=\alpha>0$, that is $f(x)$ is increasing at $c$. Whenver we have $f'(x)>0$ at a point in the domain, then $f(x)$ will have an extremum at a point so it will be a local minimum. Can I use it somewhere to get a different conclusion. Answer says 1. is true. How to do that?
I think none of these is true. Please correct if wrong.
Best Answer
For an unbounded above counterexample, we can take:
For any constants $C>0$, $a>0$. Motivation: although $f’’(x)=-e^{-x-a}<0$, $f’’$ rapidly decays and $f’$ quickly becomes stationary, roughly speaking (I was thinking about how slowly $\ln\ln x$ grows!). Therefore it’s possible for $f’$ to stay positive, and for $f$ to grow arbitrarily large. You can easily check $\lim_{x\to\infty}f(x)=\infty$ despite it satisfying the given conditions.
This would also work without the condition $f(1)>0,f(0)=0$, up to a different choice of constants.