A function with positive Hessian at a critical point, without having a minimum there

Question

I have a problem with a little instance:

$f(x,y) = \begin{cases} (x^4-3x^2y^2+y^2)/(x^2+y^2) & otherwise \\ 0 & \text{(x,y)=(0,0)} \end{cases}$

This is a example of a function which gradient at $(0,0)$ is $(0,0)$ and the Hessian Matrix is positiv definite at $(0,0)$, but still it has no minimum at $(0,0)$.

I can show that it has no minimum because of $f(-y,y)$ and $f(x,0)$ (furthermore a look at the graph) have different signs, but can someone explain why the hessian is still positive at this point?

Answer 1

I get the Hessian at $(0,0)$ to be $$\begin{pmatrix}2&0 \\ 0&0\end{pmatrix}$$ which is positive semidefinite, but not positive definite. Indeed:

• the second partial derivative in $x$ is $2$, because $f(x,0)\equiv x^2$
• the second partial derivative in $y$ is $0$, because $f(0,y)\equiv 1$.
• to find that the mixed partial is $0$, you could calculate $f_x(x,y)$ and set $x=0$ there, obtaining $f_x(0,y)\equiv 0$.