Prove that $x^4 + y^4 - 3xy = 2$ is compact

Question

The exercise consists of showing that the function $f(x,y)=x^4 + y^4$ has a global minimum and maximum under the constraint $x^4 + y^4 - 2xy = 2$.

In the solution to the exercise, it it follows that the constraint is compact if we can show that $\lim_{x^2 + y^2 \rightarrow \infty} x^4 + y^4 - 3xy - 2 \rightarrow \infty$.Why this is the case?

My intuition tells me that this is because the $x^4$ and $y^4$ terms dominates the other two terms when $x$ and $y$ gets large. This would then imply that $x$ and $y$ cannot get arbitrarily big without violating the constraint. Does this imply that if the limit of the constraint was $0$, that the domain would not be compact? Is my reasoning valid?

Many thanks,

Answer 1

If you prove that $\lim_{x^2+y^2\to\infty}(x^4+y^4-2xy)=\infty$, then there exists $K>0$ such that $x^2+y^2-2xy>2$ for $x^2+y^2>K$. Hence the solutions of $x^2+y^2-2xy-2=0$ are contained in the circle $x^2+y^2\le K$ and so they form a bounded set.

For the limit, you can use polar coordinates; given $x=r\cos\varphi$ and $y=r\sin\varphi$, you have, for $r\ge2$, \begin{align} x^4+y^4-2xy &=r^2(r^2\cos^4\varphi+r^2\sin^4\varphi-2\cos\varphi\sin\varphi) \\[6px] &\ge r^2(4\cos^4\varphi+4\sin^4\varphi-2\cos\varphi\sin\varphi) \\[6px] &\ge r^2(4-8\cos^2\varphi\sin^2\varphi-2\cos\varphi\sin\varphi) \\[6px] &=r^2(4-2\sin^22\varphi-\sin\varphi) \\[6px] &\ge r^2 \end{align} because $2\sin^22\varphi+\sin\varphi\le 3$.

Answer 2

Note that $$\eqalign{x^4+y^4-2xy&={1\over2}(x^2+y^2)^2+{1\over2}(x^2-y^2)^2-2xy\geq{1\over2}(x^2+y^2)^2-(x^2+y^2)\cr &=(x^2+y^2)^2\left({1\over2}-{1\over x^2+y^2}\right)\geq{1\over4}(x^2+y^2)^2\geq4\ ,\cr}$$ as soon as $x^2+y^2\geq4$. It follows that the constraint defines a closed and bounded, hence compact, set in the plane.