If the function $f$ is convex, the global optimum can be find, by solving a equation system?

Question

Is this statement in terms of a convex function $f:R^n \to R$ correct?

If the function $f$ is convex, the global optimum can be find, by solving a equation system.

I would say yes, because the local optimum of an convex function is the global optimum. Could this statement be true or Am I wrong?

Answer 1

You need more than that - at least, you need to ensure that local optimum exists.

Consider $f(x) = e^x$. It is a strictly convex function $\Bbb R \mapsto \Bbb R$, yet it does not have a global minimum on $\Bbb R$.

The standard criterion "If $f$ is $C^2$, if Hessian is positive definite at a point $x$ and gradient is zero at $x$, then $x$ is a local minimum" indeed can yield a system of equations, but you need a $C^2$ function in the first place.

Answer 2

The validity of this statement depends on details.

• If $f$ is differentiable, convex and everywhere finite, then this is true: $x^*$ is a global minimizer of $f$ if and only if $\nabla f(x^*) = 0$.

• If $f$ is convex (and not differentiable), then this is false in general: $x^*$ is a global minimizer of $f$ if and only if $0\in \partial f(x^*)$ where $\partial f(x^*)$ is the subdifferential of $f$ at $x^*$. In general this can be a set, and hence, you do not have a system of equations.

The case where $f$ is convex and differentiable but only defined on a convex subset of $R^n$, is somehow a subcase of the previous one (by allowing $f$ to assume the value $+\infty$ since the definition of the subdifferential also works in this case).