Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The square form is $H:=x^T\nabla^2 f(x) x= 2 a b$ where $x=[a,b]$. Now $f(x_1,x_2)=x_1x_2$ in $\mathbb R^2_{++}$ (problem b).

I am perplexed:

  • I think my teacher means that this not positively semidefinite because $H>0$ -condition is not satisfied when $x\in \mathbb R^2$.

  • I think the question restrict the domain to $\mathbb R^2_{++}$ so $H>0$ so positively semidefinite.

  • my teacher says that there is only one definition for positively-semi-definiteness and they match.

  • By different domains with each pos.semi-definiteness -definition, I got different answers so not matching definitions. I think I tried the the determinant rule -thing.

Now is this function positively-definite and when? What is called definiteness when you restrict the domain? I feel it quite stupid if definites is really limited to $\mathbb R^2$ or $\mathbb C^2$.

Question B (source here)

enter image description here

Answer the question B (here)

enter image description here

share|improve this question
$f$ is certainly positive-definite. The definition of positive definite is $x\neq 0\implies f(x,x)>0$. This is true for all $x$ in the domain of $f$. –  Alex Becker Sep 27 '12 at 18:03
@AlexBecker: One doesn't usually use the term positive definite about arbitrary functions just about quadratic forms. And in that case, one looks at the entire domain of the form. $f$ just happens to be the restriction of a quadratic form to the first quadrant. But due to the restriction, I would not call it a quadratic form. –  Harald Hanche-Olsen Sep 27 '12 at 18:08
@HaraldHanche-Olsen But since the question is being asked at all, presumably either a yes or not answer is wanted. And the answer certainly isn't no. –  Alex Becker Sep 27 '12 at 18:10
@AlexBecker But the problems quoted in the question don't ask if $f$ is positive-definite or not! That the question is asked at all, indicates confusion, not the need for a yes or no answer. –  Harald Hanche-Olsen Sep 27 '12 at 18:43
add comment

1 Answer

up vote 1 down vote accepted

Your confusion appears to arise from looking at $x^T\nabla^2 f(x) x$. This has no relation to the positive definiteness or convexity of $f$! What you need to look at is the form $u^T\nabla^2 f(x) u$ as a function of the vector $u$, for any fixed $x$. And as a quadratic form of $u$, it is not positive definite (it is $u_1u_2$), hence $f$ is neither convex nor concave near $x$, for any $x$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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