I need to prove that $f(x)=(x_2 - x_1^2)^2 + x_1^5 $ has one stationary point which is neither a local max nor min
The stationary point of $f(x)$ is found by $\nabla f(x)=0$ which gives $x_1 = x_2 = 0$
In order to be a local minimiser it has to be positive definite and so $s^T \nabla f(x^*) s > 0$ ,where $x^*$ is my stationary point and $s\neq 0$.
I find $\nabla^2 f(0,0) = \begin{bmatrix} 0 & 0 \\ 0 &2 \end{bmatrix}$
Am I able to prove this by just taking $s = \begin{bmatrix} 1 \\0 \end{bmatrix}$ ?
Every neighborhood of $(0,0)$ contains points where $f$ is positive and points where $f$ is negative: just take $x_1$ small and positive and $x_2=x_1^{2}$ to make $f$ positive and $x_1$ small and negative and $x_2=x_1^{2}$ to make $f$ negative. Hence $(0,0)$ is neither a local minimum nor a local maximum.