Lower bound of $x^4+y^4-2(x-y)^2$

Question

Let's consider the function $f:\mathbb{R}^2\to\mathbb{R}$ with $f(x,y)=x^4+y^4-2(x-y)^2$.

I want to show that it is bounded below.

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First, I manipulated $f$ as follows: $$ x^4+y^4-2(x-y)^2=x^4-2x^2+y^4-2y^2+4xy+x^2-x^2+4y^2-4y^2=\\ x^4-3x^2+y^4-6y^2+(x+2y)^2=x^2(x^2-3)+y^2(y^2-6)+(x+2y)^2. $$ If $f$ had no lower bound then there must exist a sequence $(x_k,y_k)_{k\in\mathbb{N}}$ such that $\lim\limits_{k\to\infty}f(x_k,y_k)=-\infty$. The limit of $(x_k,y_k)_{k\in\mathbb{N}}$ must attain one of the following four values:

$\lim\limits_{k\to\infty}(x_k,y_k)=(-\infty,+\infty)$,$\lim\limits_{k\to\infty}(x_k,y_k)=(+\infty,-\infty)$,$\lim\limits_{k\to\infty}(x_k,y_k)=(c,-\infty)$,$\lim\limits_{k\to\infty}(x_k,y_k)=(-\infty,c)$,

where $c\in\mathbb{R}$. In fact, if $(x_k,y_k)_{k\in\mathbb{N}}$ had a finite limit, e.g. $\lim\limits_{k\to\infty}(x_k,y_k)=(c,d)$ where $c,d\in\mathbb{R}$ then by continuity $\lim\limits_{k\to\infty}f(x_k,y_k)$ would be finite which we had ruled out.

However, we see that in each of the four cases $\lim\limits_{k\to\infty}f(x_k,y_k)=\infty$. So $\lim\limits_{k\to\infty}f(x_k,y_k)=-\infty$ is impossible which shows that $f$ is bounded below.

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Is this correct? Do you have a quicker or more elegant approach? This seems very complicated.

Answer 1

You are right that if $f$ is not bounded below then $f(x_k, y_k) \to -\infty$ for some sequence $(x_k, y_k)$, and that sequence can not have a finite limit point. However, that does not imply that one of the four cases

• $\lim_{k\to\infty}(x_k,y_k)=(-\infty,+\infty)$,
• $\lim_{k\to\infty}(x_k,y_k)=(+\infty,-\infty)$,
• $\lim_{k\to\infty}(x_k,y_k)=(c,-\infty)$,
• $\lim_{k\to\infty}(x_k,y_k)=(-\infty,c)$

must occur. You can only conclude that $\Vert (x_k, y_k) \Vert = \sqrt{x_k^2 + y_k^2} \to \infty$ (which implies $|x_k| \to \infty$ or $|y_k| \to \infty$, or both).

But your representation of $f$ as $$ f(x, y) = x^2(x^2-3)+y^2(y^2-6)+(x+2y)^2 $$ can be used as a starting point for a proof, you only have to show that all three terms on the right-hand side are bounded below. The last term is non-negative, and the first two terms are monic polynomials in a single variable of even degree, and therefore bounded below.

Side note: a more symmetric representation is $$ f(x, y) = x^2(x^2-4)+y^2(y^2-4)+2(x+y)^2 \\ = (x^2-2)^2 + (y^2-2)^2 +2(x+y)^2 -8 $$ which leads to the sharp lower bound $f(x, y) \ge -8$. Equality holds if $x^2=y^2=2$ and $x=-y$.

Answer 2

I am not an expert for check your analysis. Therefore, I just want to deal with the algebraic part.

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Your function is given as follows:

$$f(x,y):=x^4+y^4-2(x-y)^2$$

If you take $x=a+b,~ y=a-b$, then we have

$$\begin{align}g(a,b):&=2a^4 + 12 a^2 b^2 + 2 b^4 - 8 b^2\\ &=2a^2(a^2+6b^2)+2b^4-8b^2\\ &=2a^2(a^2+6b^2)+2\left(b^2-2\right)^2-8\\ &≥-8\end{align}$$

We conclude that,

$$\min \left\{g(a,b)\right\}=-8\\ \text{at}~a=0,~b=\pm\sqrt 2$$

This means,

$$\min\left\{f(x,y)\right\}=-8\\ \text{at}~x=\pm\sqrt 2,~y=\mp\sqrt 2.$$