What ways are there to prove that a function with more than 2 variable is concave?ٍ
I know we can check that the associated Hessian matrix is negative (semi)-definite, but are there other ways?
1 Answer
$\begingroup$This list isn't exhaustive but here are a few more ways. Note that a function $f:\operatorname{dom}(f)\subset\Bbb R^n \to \Bbb R^n$ is concave if and only if $-f$ is convex. So I'll give some tricks to use in order to prove that a functions is convex.
You can always check directly the definition.
$f$ is differentiable and $\operatorname{dom}(f)$ is an open set, then $f$ is convex if and only if $\operatorname{dom}(f)$ is convex and $$ f(y)\geq f(x)+\langle \nabla f(x),y-x\rangle \qquad \forall x,y \in \operatorname{dom}(f)$$
If $f\in C^2(\operatorname{dom}(f))$ and $\operatorname{dom}(f)$ is open, then $f$ is convex if and only if $\operatorname{dom}(f)$ is convex and the Hessian of $f$ is positive semi-definite.
$f$ is convex if and only if the epigraph of $f$ is convex.
If you can prove that $f$ has two distinct local minima, then $f$ is not convex.(see)
the composition of nondecreasing convex functions is convex. (see)
