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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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    @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.2012-09-27

1 Answers 1

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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 $x$.