Justify why there can't be a strict local extremum at $\|x\|$ = $\sqrt{1 \over 2}$ for $f(x) = \|x\|^4 - \|x\|^2$ 
Given $f: \Bbb R^2 \to \Bbb R$ defined by $$f(x) = \|x\|^4 - \|x\|^2$$ with $x := (x_1, x_2)$, justify why there can't be a strict local extremum at $\|x\|$ = $\sqrt{1 \over 2}$.


Approach
Well, I would guess that there can't be a strict local extremum if I could justify that the gradient of the function isn't $0$ at the given point.
But in this thread I received answers on where the gradient of the function is $0$. Rodrigo de Azevedo expressed his solution in a way that might be helpful in this case: he found that the gradient of the function vanishes at $\|x\|_2 = \sqrt{1 \over 2}$. Am I supposed to go ahead from there and argue why the gradient can't be identical at the point given in the exercise?
 A: The given function is
$$
f(x) = \Vert x \Vert^4 - \Vert x \Vert^2 = \left(x_1^2 + x_2^2 \right)^2 - (x_1^2 + x_2^2) = x_1^4 + 2x_1^2x_2^2 + x_2^4 -x_1^2 -x_2^2
$$
We find that
$$
\nabla f(x) = \left[ \begin{array}{c} \frac{\partial f}{\partial x_1}  \\[2mm]
\frac{\partial f}{\partial x_2} \\[2mm]
\end{array} \right] = 
\left[ \begin{array}{c} 4x_1^3 + 4x_1 x_2^2 - 2x_1  \\[2mm]
 4x_2^3 + 4x_1^2 x_2 - 2x_2\\[2mm]
\end{array} \right]
$$
The critical points of $f$ are obtained by solving
$$
\Delta f(x) = 0
$$
or equivalently, the system of equations
$$
\left. \begin{array}{ccc}
4x_1^3 + 4x_1 x_2^2 - 2x_1 & = & 0 \\[2mm]
 4x_2^3 + 4x_1^2 x_2 - 2x_2 & = & 0 \\[2mm]
\end{array} \right. \tag{1}
$$
Clearly, $(x_1, x_2) = (0, 0)$ is a solution of (1).
Hence, the origin is a critical point of the function $f(x)$.
Next, we suppose that $x_1 \neq 0$ and $x_2 \neq 0$.
Dividing the first equation of (1) by $x_1$ and the second equation of (1)
by $x_2$, we get:
$$
4x_1^2 + 4x_2^2 - 2 = 0\\
4x_2^2 + 4x_1^1 - 2 = 0
$$
or equivalently, the equation
$$
x_1^2 + x_2^2 = {1 \over 2}
$$
Thus, we conclude that the critical points of $f$ consist of the origin and
all points on the circle with centre at the origin and radius ${1 \over 2}$,
which can be also represented compactly as
$$
\Vert \mathbf{x} \Vert = \sqrt{1 \over 2}
$$
Calculating the Hessian of $f$, we obtain
$$
Hf(x) = \left[ \begin{array}{cc}
 12 x_1^2 + 4 x_2^2 -2  & 8 x_1 x_2 \\[2mm]
8 x_1 x_2 & 12 x_2^2 + 4 x_1^2 - 2 \\[2mm]
\end{array} \right]
$$
Along the circle $\Vert \mathbf{x} \Vert = \sqrt{1 \over 2}$ or
$ 4 x_1^2 + 4 x_2^2 - 2 = 0$, we get
$$
Hf(x) = \left[ \begin{array}{cc}
  8 x_1^2  & 8 x_1 x_2 \\[2mm]
8 x_1 x_2 & 8 x_2^2 \\[2mm]
\end{array} \right]
$$
If we calculate the principal minors of $Hf(x)$, we get
$$
\Delta_1 = 8 x_1^2 > 0, \ \ \Delta_2 = 0
$$
Using the Sylvester's test, we conclude that $Hf(x)$ is not positive definite.
However, using the Sylvester's test, we can conclude that $Hf(x)$ is
positive semi-definite.
Hence, any point on the circle $C$ defined by $x_1^2 + x_2^2 =  {1 \over 2}$ is a local
minimizer for $f$ but not a strict local minimizer. This is because all points on $C$ have the same value of $f$ and the definition for strict local minimizer requires that in a small neighborhood $N$ of a point $\xi \in C$, $f(\xi)$
is smaller than $f(x), x \neq \xi, x \in N$. Obviously, any small neighborhood
$N$ of $\xi \in C$ will include points in $C$.
