# Global Maximums and Minimums

My book states:

"It is also true that if $x^*$ is an interior point and:

• a global maximum of $f$ , then $d^2f(x^*)$ is negative semi-definite.
• a global minimum of $f$ , then $d^2f(x^*)$ is positive semi-definite.

But, it is not true that if $x^*$ is a critical point and $d^2f(x^*)$ is negative (positive) semidefinite, then $x^*$ is a local maximum (minimum).

However consider the function: $x^4+x^2-6xy+3y^2$ .

It has a global minimum on $(x,y)=(1,1)$ and $(x,y)=(-1,-1)$ .

However the Hessian at this points, does not seem to be semi-definite.

Can anyone help me?

• what do you think is the matrix of second partial derivatives of your function? – Will Jagy Aug 7 '14 at 20:59
• I see your deleted second "answer." Yes, a definite hessian is also semi-definite. – Will Jagy Aug 7 '14 at 22:47

• yes, positive definite at $(1,1)$ and $(-1,-1)$ which are local (and global) minima, and indefinite at $(0,0)$ which is a saddle point. Maybe your trouble is the prefix "semi." The word semidefinite includes definite. – Will Jagy Aug 7 '14 at 21:21