0
$\begingroup$

Expanding on a previous post I made recently, let $$ \mathscr{P}:= \{ p(x) \in \mathbb{R}[x] \mid p(x) \ge 0,~\forall x\ge 0\}. $$ The Pòlya-Szegö theorem (see Theorem 3.21 here) asserts that $p \in \mathscr{P}$ if and only if $$ p(x) = f(x) + xg(x), $$ in which $f$ and $g$ are polynomial sum-of-squares.

Is anything known about polynomials $p$ such that $p,~p' \in \mathscr{P}$? Obviously, these polynomials must enjoy many desirable properties (for instance, $p$ must be monotonically increasing on $[0,\infty)$).

Is it possible that if $p$, $p' \in \mathscr{P}$, then $p^{(k)} \in \mathscr{P}$ for $2 \le k \le \deg{p}-1$?

In particular, I am interested in proving that $|p(z)| \le p(|z|)$ for every $z \in \mathbb{C}$ or knowing whether this property is independent of a polynomial and its derivative belonging to $\mathscr{P}$.

Any references/ideas/partial results are greatly appreciated.

$\endgroup$

1 Answer 1

6
$\begingroup$

The equivalent condition is $p(0)\geqslant 0$, $p'(x) \geqslant 0$ for $x\geqslant 0$.Thus this class reduces to $\mathscr {P} $ by taking antiderivative.

The inequality $|p(z)|\leqslant p(|z|)$ fails for $p(z) =z^2 - z^3 +z^4$ and $z=-1$. Also $p'''\notin \mathscr{P}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.