Hi,
I am wondering how I can prove that a twice differentiable function F is convex if and only if its domain is convex and the Hessian of F is positive definitive, i.e. ∇Fxx is greater or equal to zero for all x∈ domain F
Thanks in advance
Hi,
I am wondering how I can prove that a twice differentiable function F is convex if and only if its domain is convex and the Hessian of F is positive definitive, i.e. ∇Fxx is greater or equal to zero for all x∈ domain F
Thanks in advance
Let be twice differentiable and let denote the Hessian matrix.
Suppose .
Remember that f is convex iff
for all in its domain . Now, with a simple Taylor expansion around we have
for some in the line segment between and .
Deduce that (2) implies (1).
For the converse, if f is convex, then certain subdeterminants of the Hessian matrix satisfy certain conditions. These imply that the eigenvalues of the Hessian are all non-negative. Choose an orthonormal basis for consisting of eigenvectors of . Then, at and for all , we have
.