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

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

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

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

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

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

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