Given a 1-dimensional gradient system which has one equilibrium point $x=0$
$$\frac{dx}{dt}=-f'(x),x\in R$$
where $f'(x)$ denotes $\frac{df(x)}{dx}$.
Let $f(x)$ be a convex function, i.e. $f''(x)>0$.
(1) How to prove $x=0$ is locally asymptotic stable?
(2) on the other hand, does globally asymptotic stability deduces convexity of $f$?
My attempt to (1) is as below and I am wondering whether it is correct:
Since $f'(x)|_{x=0}=0$ and $f''(x)| _{x=0}>0$, then $f'(x)|_{x\in U_-(0)}<0$ and $f'(x)|_{x\in U_+(0)}>0$.
Thus $\frac{dx}{dt}\big |_{x\in U_-(0)}>0$, which means $x(t)$ will increase and approach to $x=0$ as $t$ increases.
While $\frac{dx}{dt}\big |_{x\in U_+(0)}<0$, which means $x(t)$ will decrease and approach to $x=0$ as $t$ increases.
Thus, $x=0$ is a locally asymptotically stable equilibrium point.
A general method to show locally asymptotic stablity is the Lyapunov method. The intuition behind is to define a Lyapunov function / energy function, such that the dynamics of the system makes the energy decay along trajectory. Then if the energy function is lower bounded, we will be stable.
More formally, we just need to show $h(x)=f(x)-f(0)$ is a valid Lyapnov function for the dynamics you described.
To proof the local positive definite of $h(x)$ we can show it with the convexity of $f(x)$: the whole function resides on top of the tangent line at $x=0$ $$ f(x)\geq f(0) + f'(0)(x-0)\\ h(x) = f(x)-f(0)\geq x f'(0) = 0 $$ For convex functions, the local minimum is also global minimum, thus we can say $h(x)\geq 0$ and equality holds if.f. $x=0$.
Then we verify $h(s)$ is a lyapnov function, and by the theorem above the system is locally asymptotic stable around 0.