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Why is $(0,0)$ not a local minimizer for $f (x_1,x_2)=x_1^2+x_2^3 $? Because $(0,0)$ is a critical point and Hessian matrix at $(0,0)$ is positive semi definite. Therefore isn't it a local minimum for $f(x)$? But I have to show it isn't a local minimum.

A theorem in the book stays if $x_*$ is a critical point then it is a local minimizer for $f(x)$ if $Hf(x_*)$ is positive semi definite.

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  • $\begingroup$ Fix $x_1$ and then let $x_2 \to - \infty$. $\endgroup$ – Cosmare Sep 9 '15 at 18:23
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The theorem in the book says I believe that the critical point $x_*$ is a local minimizer if $Hf(x_*)$ is positive definite. Positive semidefinite Hessian is a necessary condition for minimum (if local minimum then positive semidefinite), but not sufficient.

Compare with one dimensional case where the derivative should be strictly positive at the critical point. For example, the (critical) point $x=0$ is not a local minimizer of $f(x)=x^3$. The second derivative is zero at zero, but it is too weak to give a local minimum. In fact, the function is strictly increasing.

$(0,0)$ is not a local minimizer by the same reason because along the second variable $x_2$ the function is cubic as in the example above.

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  • $\begingroup$ yes the book says definite, but I thought difference of definite and semi definite lies with a critical point being strict or not only. So if $Hf (x*) $ is positive definite then a strict local minimum exist. But positive semi definite doesn't imply that a local minimum exist at that point. $\endgroup$ – clarkson Sep 9 '15 at 19:01
  • $\begingroup$ Right, if the second order derivative is positive definite then the local minimum is strict. If the second order derivative is not positive semidefinite then it is not a local minimum. The uncertain situation is when it is positive semidefinite. Then the second order Taylor approximation at the point is too weak to resolve whether it is loc.min. or not. One has to look at higher order Taylor terms along those directions where the second derivatives are zeros, but this is normally not done in practice, too messy. $\endgroup$ – A.Γ. Sep 9 '15 at 19:17
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Hint: Show that if $x_2 < 0$ then $f(0,x_2) < f(0,0)$. Conclude that $(0,0)$ is not a local minimum of $f$ (using the definition of a local min).

Edit: The Hessian matrix $Hf$ at $(0,0)$ is given by

$$ Hf(0,0) = \begin{pmatrix} 2 & 0 \\ 0 & 0\end{pmatrix}, $$

which is positive semidefinite but not positive definite. As @A.G. points out in his answer, this is not enough to ensure that $(0,0)$ is a local min.

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