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I have function $f(x,y)$ with $x\geq0$ and $y\geq0$. The Hessian matrix of this function has following properties.

1- $f_{xx}>0$

2- Determinant of the Hessian matrix is zero.

Can we say that the function $f(x,y)$ is jointly convex over $(x,y)$ for all of the desired domain?

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  • $\begingroup$ you want determinan be zero at all points in domain ? $\endgroup$ – Red shoes Feb 21 '18 at 0:27
  • $\begingroup$ @Redshoes I do not want. But actually for $f(x,y)$ the determinant of the Hessian matrix is zero over all the domain. $\endgroup$ – Frank Moses Feb 21 '18 at 0:30
  • $\begingroup$ @Redshoes I have also clarified in my edited post. $\endgroup$ – Frank Moses Feb 21 '18 at 0:33
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Yes. The Hessian of $f$ is semi positive definite every on the domain , because it passes the Sylvester's criterion .

So $f$ is jointly convex on its domain.

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  • $\begingroup$ shouldn't we check the (2,2) element according to Sylvester's criterion? $\endgroup$ – LinAlg Feb 21 '18 at 1:50
  • $\begingroup$ @LinAlg you don't have to. even you check it is positive because $$f_{yy} = \frac{{f_{xy}}^2 }{f_{xx}}$$ $\endgroup$ – Red shoes Feb 21 '18 at 2:45

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