I have the function $g(x,y) = x^6 -y^6x^2$ and want to prove that the origin is a saddle point.
I know that a critical point with an indefinite Hessian matrix is a saddle point, but this is only a sufficient condition.
$(0,0)$ is indeed a critical point of $g$, but the Hessian matrix is everywhere $0$ and hence it is positive and negative semi-definite and so not indefinite.
How would I go about concluding that the origin is indeed a saddle point here? Unfortunately, the only definitions of saddle points that I could find gave the usual sufficient condition of an indefinite Hessian.
Many thanks.
Best Answer
On the line $y=0$, $g(x,y)=x^6$, which is concave up.
On the curve $x=y^2$, $g(x,y)=y^{12}-y^{10}=y^{10}(y^2-1)$, which is concave down.
More details, as requested:
A saddle point is stationary, but neither a local max nor a local min. $g(x,y)$ is stationary at the origin, because both partials are zero. $(0,0)$ is not a local max by the first observation above, it is not a local min by the second one.