# What can we say about the function which has following kind of Hessian matrix?

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?

• you want determinan be zero at all points in domain ? – Red shoes Feb 21 '18 at 0:27
• @Redshoes I do not want. But actually for $f(x,y)$ the determinant of the Hessian matrix is zero over all the domain. – Frank Moses Feb 21 '18 at 0:30
• @Redshoes I have also clarified in my edited post. – Frank Moses Feb 21 '18 at 0:33

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.
• @LinAlg you don't have to. even you check it is positive because $$f_{yy} = \frac{{f_{xy}}^2 }{f_{xx}}$$ – Red shoes Feb 21 '18 at 2:45