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We know a function $f \in C^2(R)$ has a Local Maximum in the origin $(0,0)$. What can you say about the differential: $d_{(0,0)}^2f(1,-1)<0$?

I've recently got this on a test and I'm not sure if I said the right thing. Can someone give me a correct answer please? Thank you in advance!

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When $f$ has a local extremum at ${\bf 0}:=(0,0)$ then in any case ${\bf 0}$ is a critical point of $f$, i.e. $\nabla f({\bf 0})={\bf 0}$.

Let $$ H:=\left[\matrix{f_{xx} & f_{xy} \cr f_{yx} & f_{yy}\cr}\right]_{\bf 0}$$ be the matrix of second partials of $f$ at ${\bf 0}$. The origin is a nondegenerate critical point if $\det H\ne0$. When this is the case then a local maximum at ${\bf 0}$ is only possible if $H$ is negative definite. This means that $$d^2f({\bf 0}).(X,Y)=f_{xx} X^2 + 2 f_{xy}XY + f_{yy} Y^2<0$$ for all $(X,Y)\ne(0,0)\ .$

But we can also have a local maximum at ${\bf 0}$ when ${\bf 0}$ is a degenerate critical point. Consider the function $$f(x,y):=-x^4-y^4\ .$$ In this case $d^2f({\bf 0}).(X,Y)=0$ for all $(X,Y)$.

But we never can have $d^2f({\bf 0}).(X,Y)>0$ for some $(X,Y)$ when the origin is a local maximum.

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Amazing! Thank you very much! –  Grozav Alex Ioan Jan 22 '13 at 19:42

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