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Let $f:\mathbb{R}^n\to \mathbb{R}^n$ be a $C^2$-function and let $H=\left(\frac{\partial^2f}{\partial x_i \partial x_j}\right)_{1\le i,j\le n}$ be its Hessian matrix. Suppose I know that $ \det H(x_1,\ldots,x_n)\ge 0$ for all $x=(x_1,\ldots,x_n)$.

Does this have any geometric meaning for $f$?

e.g., when $n=1$, this means that $f$ is convex. This no longer holds for $n\ge 2$, but when $f$ is convex the Hessian determinant is certainly positive, so perhaps one could wonder if a weaker property holds.

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    The title and the last paragraph have "positive", the first paragraph has $\ge0$.2011-09-07
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    @trony: when judging the convexity of a function, we need to check if $H$ is positive definite, but not if $\det H$ is positive.2011-09-07
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    @Shiyu: Please read my post again. I am not asking whether H is positive definite, I even said this is wrong, I wondered whether it had some other implications to the geometry of $f$.2011-09-07
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    @trony: I've read your post again, but I still didn't see much meaning from$\det H$. There are many situations in which $\det H>0$. I don't think $\det H>0$ can imply anything. In stead, checking the eigenvalues of $H$ might be meaningful.2011-09-07

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