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I'm wondering what the hessian matrix of a function tells me about its critical points.

Quick Example: Let $f(x,y) = \frac{x^2}{2}+\frac{y^2}{2}$ and $M:=\{(x,y)\in\mathbb R^2 | \frac{x^2}{2}+y^2\leq1\}$

We get a minima in $p_1=(0,0)$ and two maxima at $p_{2,3}=(\pm\sqrt{2},0)$

The Hessian Matrix of $f$ is $Hess(f)=\begin{pmatrix}1&0\\0&1\end{pmatrix}$

I always though, that:

$Hess(f(p_i))$ positive definit $\Rightarrow p_i$ is minima

$Hess(f(p_i))$ negativedefinit $\Rightarrow p_i$ is maxima

$Hess(f(p_i))$ indefinit $\Rightarrow p_i$ is saddle point

But here we have a positive definit Hessian Matrix for any Point. Thus all of them would be maxima. But they can't be, since there must be a minima between two maxima.

Question 1: So whats going on here? Where's my mistake?

Question 2: What exactly tells me, if my hessian matrix is (pos/neg) semi-definit? In which cases can I follow what? How do I handle these cases?

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    The maxima at the boundary are not critical points of the function. – 2017-01-19
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    What do you call "between" in two dimensions ? – 2017-01-19
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    Yves Daoust: You just can't have two maxima without a minima. amd: True, nevertheless are $p_{2,3}$ global extremas. But you just hinted me to something: The Hessian Works only for critical points, true. So only one is left. $p_1$ – 2017-01-19

2 Answers 2

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The only critical point is $p_1=(0,0)$, which is indeed a minimum. $p_2,p_3$ are ordinary points. There is no rule invalidated.

On the boundary, we can switch to a parametric representation,

$$x=\sqrt2\cos t,y=\sin t.$$

Then

$$f(t)=2\cos^2t+\sin^2t=\cos^2t+1$$ has critical points where

$$f'(t)=2\cos t\sin t=0.$$

The sign of the Hessian

$$f''(t)=2\sin^2t-2\cos^2t$$ will tell you if they are minima or maxima. (On a full turn, there are two minima and two maxima.)

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    Yeah I think I see my mistake. I basically look at $f|_M$. But the hessian is not defined for the borders, thus I can't use it to tell anything about the global extrem at the border but rather have to evaluate them by putting them into f. Thanks for the reminder. – 2017-01-19
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    the question "What does it tell me if the hessian is semi-definit" still remians :) – 2017-01-19
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    @xotix: you can't conclude, third order criteria are required. Compare $x^3$ and $x^4$. – 2017-01-19
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    okay thanks you – 2017-01-19
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If the hessian is positive (negative) semidefinite everywhere, the function is convex (concave). That means that a stationary point is a global minimum (maximum).

When the Hessian is indefinite, the function is neither convex nor concave. A stationary point could be a local optimum.

Since your objective and constraints are convex, you could use KKT conditions to find a global minimum if the stationary points are infeasible.

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    Never heard of KKT so I doub't we have to use it. Thanks though – 2017-01-19