5
$\begingroup$

Let $f$ and $g$ be real polynomials with nonnegative coefficients. Let $$ h = \frac{f}{f+g}. $$ I want to prove that the $n$-th derivative of $h$ satisfies:

There exists $C > 0$ such that $$ |h^{(n)}(x)| \le Cn!\frac{1}{x^n}, \qquad n \ge 0,\ x > 0. $$

Is it a known result? The problem seems rather simple at first glance, but I am stuck. It is easy to prove the above inequality holds with $C = C(n)$ depending on $n$. In this case, for example, we can use the Leibniz rule combined with induction.

I checked multiple cases numerically and the conjecture looks to be true.

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
8
$\begingroup$

There are counterexamples. Indeed, assume that the claimed bound holds for $x=1$. Then the Taylor series of $h(z)$ around $z=1$ converges in the disk $D=\{z:|z-1|<1\}$, hence $h(z)$ is analytic in $D$. However, the rational function $h(z)=1/(1+z^4)$ has a pole at $e^{\pi i/4}\in D$, contradiction.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Nice (counter)example! Are there any sufficient conditions on $f$ and $g$ that could guarantee the existence of such $C$? $\endgroup$
    – xen
    Commented Feb 20 at 13:36
  • 2
    $\begingroup$ @xen Sure. A sufficient condition is that $h(z)$ is bounded in the half-plane $\{z:\Re z\geq 0\}$. This follows from Cauchy's integral formula applied to the circles $\{z:|z-x|\leq x\}$. $\endgroup$
    – GH from MO
    Commented Feb 20 at 15:23
  • 4
    $\begingroup$ @GHfromMO This is almost a necessary condition too. Actually, a necessary and sufficient condition is that $h$ does not have any pole with positive real value. That it is necessary is the argument in your answer for $x \to \infty$; to check it is sufficient it is enough (Partial fraction decomposition) to check it for $h(z) = 1/(z-a)^k$ for $k\geq 0$ and $a\neq 0$ with nonnegative real part. $\endgroup$
    – Mikael de la Salle
    Commented Feb 20 at 15:38
  • 3
    $\begingroup$ @MikaeldelaSalle Nice observation! BTW your last sentence should end with "nonpositive real part". $\endgroup$
    – GH from MO
    Commented Feb 20 at 15:52
  • $\begingroup$ @GHfromMO Of course, thanks for the correction. $\endgroup$
    – Mikael de la Salle
    Commented Feb 20 at 18:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .