For which values of $\alpha , \beta \in \mathbb{R}$ is $f(x,y) = x^2 + y^2 + 2\alpha xy + \beta$ coercive?

Question

I know that a function is coercive if $lim_{||x|| \to \infty}f(x) = \infty$, but I don't know how to find the values of $\alpha$ and $\beta$ that would make this function coercive. I figured that $\beta$ didn't matter since it wouldn't make a difference as the function goes to infinity. $\alpha$ I couldn't figure out though. I know that $\alpha$ can't be -1 since then if $x=y$ the function would not be coercive, but other than that I'm not sure. What should I do to find the values needed?

Thanks

Answer 1

Note that $\beta$ is irrelevant to the behavior of the function at infinity. Hence let us focus only on $x^2+y^2+2\alpha xy$. By completing the square, we have \begin{align} x^2+y^2+2\alpha xy = (1-\alpha)(x^2+y^2) +\alpha(x+ y)^2\geq (1-\alpha)(x^2+y^2) \end{align} whenever $\alpha<1$. In the case $\alpha>1$, we see that \begin{align} x^2+y^2+2\alpha xy= (x+y)^2+2(\alpha-1)xy \end{align} then take $y=-x$ gives $-2(\alpha-1)x^2\rightarrow -\infty$ as $x^2+y^2\rightarrow \infty$ which means the function is not coercive.