Convex optimisation and non-globally convex functions

Question

Suppose I am maximising some function $f(x)$ such that $f'(x)=0 \iff x=c$. However $f''(x)<0 \iff c-k<x<c+k$ for some positive real number $k$. Even though the second order condition holds locally, is the inference that $x=c$ is a maximum still permitted?

Answer 1

Yes, if $f$ is sufficiently smooth (i.e. $C^2$), you can conclude that

1) either $f$ attains its maximum at a local maximum, or never attains its maximum but keeps increasing towards the boundary of its domain (in this case $\pm\infty$);

2) $x=c$ is the unique local maximum of $f$;

3) $f$ is decreasing in both directions away from $c$. This is because $f'<0$ at $x=c+k$ since $$f'(c+k) = f'(c) + \int_c^{c+k} f''(x)\,dx < 0$$ and $f'(x)$ must remain negative for $x\in (c+k,\infty)$ since otherwise by the intermediate value theorem $f'(x) = 0$ for some $x\in (c+k,\infty)$, violating the condition that $x=c$ is the unique critical point of $f$. An exact same argument applies when $x<c$.