Lower bound for coercive polynomials, II

Question

This is a refinement of my earlier question (Lower bound for coercive polynomials). This time, I ask the same question but for the exponent 1. Indeed, the question is: given a coercive polynomial $f \in \mathbb{R}[x_1, \cdots, x_n]$, whether there exist constants $c_1 > 0, c_2 \in \mathbb{R}$ such that

$$\displaystyle f(x_1, \cdots, x_n) > c_1 \min\{ |x_1|, \cdots, |x_n|\} - c_2$$

The example given by Christian Remling does not exclude the exponent 1, but does exclude every exponent larger than 1.

Answer 1

Drawing up on the example of Chritian Remling, taking $$ P(x, y) = (x^2 - y^{2 k + 1})^2 + y, $$ with $k \in \mathbb{N}\setminus \{0\}$ we have $$ \lim_{t \to \infty} \frac{P (t^{2k + 1}, t^2)}{\Vert (t^{2k + 1}, t^2)\Vert^q} = \lim_{t \to \infty} \frac{t^2}{\sqrt{t^{4k + 2} + t^4}^q} > 0 $$ if and only if $$ q \le \frac{2}{2k + 1}. $$

On the other hand we have $$ \lim_{\Vert (x, y)\Vert \to \infty} P (x, y) = \infty; $$ this follows from the following facts:

• If $y^{2k + 1} \ge x^2/2$, then $y \ge 0$, $$ P(x, y) \ge y $$ and $$ \Vert (x, y)\Vert^2 \le 2 y^{2k + 1} + y^2. $$
• If $y^{2k + 1} \le x^2/2$ and $y \ge 0$, then
  $$ P(x, y) \ge \frac{x^4}{4} $$ and $$ \Vert (x, y)\Vert^2 \le x^2 + \Bigl(\frac{x^2}{2}\Bigr)^\frac{2}{2 k + 1}. $$
• If $y \le 0$, then $$ P(x, y) \ge x^4 + y^{4k + 2} + y. $$