# Thread: second-order condition for convexity

1. ## second-order condition for convexity

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

2. Let $f:\Omega\subset\mathbb{R}^n\Rightarrow \mathbb{R}$ be twice differentiable and let $H(x), \ x\in \Omega$ denote the Hessian matrix.

Suppose $\langle H(x)\xi,\xi\rangle\geq 0, \ \forall \xi\in \mathbb{R}^n$.

Remember that f is convex iff
$f(x)\geq f(x_0)+\langle \nabla f(x_0),x-x_0\rangle, \ \ \ (1)$

for all $x,x_0$ in its domain $\Omega$. Now, with a simple Taylor expansion around $x_0\in \Omega$ we have

$f(x)=f(x_0)+\langle \nabla f(x_0),x-x_0\rangle +\frac{1}{2}\langle H(y)(x-x_0),x-x_0\rangle \ \ \ (2)$

for some $y$ in the line segment between $x$ and $x_0$.
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 $\{\rho_i\}=\{\rho_i(x)\}, \ x\in \Omega$ of the Hessian are all non-negative. Choose an orthonormal basis $\{\zeta_i\}$ for $\mathbb{R}^n$ consisting of eigenvectors of $H$. Then, at $x\in \Omega$ and for all $\xi=\sum_i \xi^i\zeta_i$, we have

$\langle H(x)\xi,\xi\rangle=\sum_i\sum_j\xi^i\xi^j\langle H(x)\zeta_i,\zeta_j\rangle=\sum_i\sum_j\xi^i\xi^j\ rho^i\delta_{ij}=\sum_i\rho_i(\xi^i)^2\geq 0$.