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This is a follow up from this question.

I have a polynomial function $f(x,y)$ that is unbounded in every direction. In other words, if we choose a direction $(a,b)\in S^1$ and keep moving along the curve $(ta,tb)$, then $$\lim_{t\to\infty}f(ta,tb)=\infty.$$

Can we conclude that there exists some disc of radius $R$ such that outside of this disc, the function $f(x,y)> 1$ (alternatively, $f(x,y)\geq 1$; see below)?

Note that I had earlier asked this question for a smooth function, but Iosif Pinelis gave a wonderful counterexample of the form $f(x,y)=x^2y^2+(x^2+y^2)e^{-x^2y^2(x^2+y^2)}$. I wonder if this is true for a polynomial function though.

Note: The "obvious" way to approach this problem is to define a function $g(\theta)$ on $S^1$, where $$g(\theta)=\inf\limits_{t_0} \{t_0:f(t\theta)\geq 1\text{ for all }t\geq t_0\}.$$ If this is a continuous function, then using the compactness of $S^1$, we can conclude that the function attains a maximum at some point on $S^1$, and then use that maximum value to define $R$.

However, based on a couple of examples that I've constructed, this function is not turning out to be continuous. However, changing the requirement from $f(x,y)>1$ to $f(x,y)\geq 1$ seems to do the trick. Is this true in general?

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For instance $f(x,y):= (y-x^2)^2 $.

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    $\begingroup$ Thanks. If I understand this correctly, along every line of the form $y=ax$, it is unbounded, as can be deduced from the formula. However, if we keep moving along the parabola $y=x^2$, the function remains zero. Neat! $\endgroup$
    – Ryan Hendricks
    Commented Jul 18 at 21:22

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