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Thread: second-order condition for convexity

  1. #1
    Feb 2010

    second-order condition for convexity

    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
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  2. #2
    Super Member Rebesques's Avatar
    Jul 2005
    My house.
    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.
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