8
$\begingroup$

Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, $f \in C^2$. Show that $f$ is convex function iff Hessian matrix is nonnegative-definite.

$f(x,y)$ is convex if $f( \lambda x + (1-\lambda )y) \le \lambda f(x) + (1- \lambda)f(y)$ for any $x,y \in \mathbb{R}^2$.

Hessian matrix is nonnegative-definite if $f_{xx}'' x^2 + f_{x,y}(x+y) + f_{yy}''y^2 \ge 0$

I know the definition but I have no idea how prove the If and only if condition or first and second implication?

$\endgroup$
  • $\begingroup$ Hint: use Taylor's theorem. $\endgroup$ – Paul Siegel Mar 20 '14 at 20:07
4
$\begingroup$

I would use restrictions to lines: $\phi_{a,b}(t) = f(a+tb)$ where $t\in\mathbb R$ and $a,b\in\mathbb R^2$ and $b\ne 0$. The key points are:

  1. $f$ is convex if and only if $\phi_{a,b}$ is convex for every $a,b$. This follows from the fact that definition of convexity involves points on the same line.

  2. The Hessian of $f$ is nonnegative definite (aka positive semidefinite) if and only if $\phi_{a,b}''\ge 0$ for all $a,b$. Indeed, Hessian is a symmetric matrix and for such matrices being nonnegative is equivalent to $b^THb\ge 0$ for all $b\in\mathbb R^2$. By the multivariable chain rule, $b^THb$ gives $\phi_{a,b}''$.

$\endgroup$
  • $\begingroup$ How do you get from part 1 and 2 that $f(\lambda x +(1-\lambda)y)\leq (1-\lambda)f(x) + \lambda f(y)$? How do you extract this 'line manipulating inequality' from the second derivative being non negative? $\endgroup$ – user162520 Aug 13 '14 at 4:42
  • 1
    $\begingroup$ @user162520 I took the one-dimensional result as known: a function on real line is convex if and only if its second derivative is nonnegative. See math.stackexchange.com/q/279750 $\endgroup$ – user147263 Aug 13 '14 at 4:45
  • $\begingroup$ Cool, thanks brother! $\endgroup$ – user162520 Aug 13 '14 at 4:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.