1
$\begingroup$

Let $f \in \mathbb{R}[x,y]$. I want to understand when $f$ has the following property: for all sufficiently large (positive) $k$, the level curves defined by

$$\displaystyle f(x,y) = k$$

consist of a finite number of bounded components. Clearly it is sufficient to assume that $f$ is positive definite, or when the Hessian of $f$ is strictly positive definite for all $(x,y) \in \mathbb{R}^2$.

Is there a relatively simple necessary and sufficient condition for this phenomenon?

Improve this question
$\endgroup$
6
  • 3
    $\begingroup$ The necessary and sufficient condition is that $|f(x,y)|\to+\infty$ as $x^2+y^2\to +\infty$. $\endgroup$
    – Alexandre Eremenko
    Commented Nov 25, 2022 at 2:55
  • $\begingroup$ For a Zariski closed subset $X$ of $\mathbf{R}^2$, the property "$X$ consists of a finite number of bounded components" is equivalent to "$X$ is bounded". So you require that for every $k$ (real? integer?) large enough, $f^{-1}(\{k\})$ is bounded. Or I misunderstand your formulation? $\endgroup$
    – YCor
    Commented Nov 25, 2022 at 10:20
  • $\begingroup$ @AlexandreEremenko Well, constant maps also satisfy the condition. $\endgroup$
    – YCor
    Commented Nov 25, 2022 at 10:23
  • 1
    $\begingroup$ @YCor: I disagree: for a constant polynomial, one level set is unbounded. $\endgroup$
    – Alexandre Eremenko
    Commented Nov 25, 2022 at 14:06
  • 1
    $\begingroup$ @AlexandreEremenko I know, but the question specifies "all sufficiently large (positive) $k$" $\endgroup$
    – YCor
    Commented Nov 25, 2022 at 15:20

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .