Does $f(x,y)=x^4+y^4$ have an extremum at $(0,0)$?

Question

Let us consider the following function $$f(x,y)=x^4+y^4.$$

I want to find the local extrema of $f$.

We can verify that $$(\frac{\partial f}{\partial x}(x,y),\frac{\partial f}{\partial y}(x,y) )=(0,0)\Longleftrightarrow (x,y)=(0,0)$$

Further the hessian matrix of $f$ is given by $$H_f(x,y)=\begin{pmatrix} 12x^2 &0\\ 0 &12y^2 \end{pmatrix}.$$ So the determinant of the matrix $H_f(0,0)$ is equal to $0$.

Answer 1

$f(x,y)=x^4+y^4 \ge 0=f(0,0)$ for all $(x,y)\in\Bbb R^2$.

So YES, it is a minimum (in fact, at $(0,0)$ the function attains its global minimum).

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MOREOVER

For all $n\in\Bbb N$, the same argument says that $f_n(x):=x^{2n}+y^{2n}$ as a (global) minimum at $(0,0)$.

For odd exponents, namely, for functions $g_n(x,y):=x^{2n+1}+y^{2n+1}$, the poinbt $(0,0)$ is a Saddle point: Given $\varepsilon>0$, $$g_n(\varepsilon,0)=\varepsilon^{2n+1}>0>-\varepsilon^{2n+1}=g_n(-\varepsilon,0)$$

Answer 2

Interestingly, even if the determinant is $0$ at $(x,y) = (0,0)$ we can say some things. The matrix is positive semidefinite at this point but is positive definite at any point in a neighborhood of $(0,0)$. Hence it's a local minimum.

However if you want to prove that it's a global minimum, you should consider Tito Eliatron's answer.

Answer 3

You have looked for the critical points of $f$. There is only $(0,0)$, and the Hessian there is $0$. This means that you have to study qualitatively what happens at $(0,0)$; it does not mean that things cannot be decided.

We have $f(0,0)=0$, and when $x\ne0$ or $y\ne0$ then $f(x,y)>0$. This already proves that $f$ has a global (a fortiori: a local) minimum at $(0,0)$. Furthermore $$\nabla f(x,y)=(4x^3,4y^3)\ne(0,0)\qquad\bigl((x,y)\ne(0,0)\bigr)$$ shows that $f$ has no other local minima or maxima in ${\mathbb R}^2$.