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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?

MJD
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Peter Pan
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1 Answers1

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I would say it is the Hessian ...

Which shall be equal to:

\begin{matrix} 12x^2 +2 & -6 \\ -6 & 6 & . \end{matrix}

Which is positive definite for the solution at hand, right?

Peter Pan
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  • 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 07 '14 at 21:21
  • Willt, thanks for your help. So a definite hessian is also semi definite? – Peter Pan Aug 08 '14 at 16:17