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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)

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Answer the question B (here)

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  • $f$ 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

1 Answers1

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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 $x$.

Harald Hanche-Olsen
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