Deduce stability from strict convexity of gradient systems

Question

Given a twice differentiable function $f(x):\mathbb R^n\rightarrow \mathbb R$, and its corresponding gradient system $$\dot x=-\nabla f(x)$$

My question is:

If $f(x)$ is strictly convex, can we conclude that all trajectories will converge to the unique global minimum (i.e. the unique global minimum is globally asymptotically equilibrium)?

Answer 1

Strict convexity is not enough as it may not guarantee the existence of am equilibrium point. For instance $f(x)=e^{-x}$ is strictly convex.

So, you will need

• strict convexity and the existence of a minimum for $f$, or
• strong convexity.

Assume that the minimum is 0 and is attained at $x^*=0$, then $f$ is positive definite. We can therefore consider the function

$$V(x)=f(x)$$

which is obviously positive definite. Then, we have that

$$\dot{V}(x)=-||\nabla f(x)||_2^2$$

which is negative definite since $\nabla f(x)=0$ if and only if $x=0$. So, we have the local convergence to the equilibrium point.

To have the global convergence, we need the radial unboundedness. This is automatically achieved $f$ is strongly convex or $f$ is strictly convex under the assumption of positive definiteness of $f$.