-1
$\begingroup$

We know that functions in $\mathcal{LP}$-class, and only these, are uniform limits, on compact subsets of $\mathbb{C}$, of polynomials with only real zeros.

Question:

  1. Does it means that these function in $\mathcal{LP}$-class has only real zeroes?

  2. Let $\phi(x)$ be an entire function but not in $\mathcal{LP}$-class. Does it means that $\phi(x)$ have a nonreal zero?

$\endgroup$
2
  • 1
    $\begingroup$ Up to Wiki en.wikipedia.org/wiki/Laguerre-Pólya_class , the answer to Q1 is affirmative. $\endgroup$
    – user64494
    Aug 8, 2017 at 4:41
  • 1
    $\begingroup$ The answer to Q2 is no. One may take an entire function of infinite order with "many" real zeroes, say the Hadamard product $$ \prod_{n=1}^\infty(1-x/\log(n+2))\exp(x/\log(n+2)+\cdots +1/n(x/\log(n+2))^n). $$The questions are not at the research level. $\endgroup$
    – user64494
    Aug 8, 2017 at 6:33

1 Answer 1

1
$\begingroup$

To the first question the answer is yes: the limit of functions with all zeros on a closed set has zeros on the same closed set, if this limit is not identically zero.

To the second question the answer is no: $e^{z^2}$ and $ze^{z^2}$ are not in LP, but have only real zeros.

$\endgroup$
2
  • $\begingroup$ Thanks for your answer! Here I have another question. If $\phi(x)$ be a real entire function of the form $e^{\alpha x^2}\phi_1(x)$, where $\alpha \geq 0$ and $\phi_1(x)$ has genus 0 or 1. If $\phi(x)$ does not belong to $\mathcal{LP}$, then $\phi(x)$ has a nonreal zero. Is it true? This comes from the proof of Theorem 2.2 in Iterated Laguerre and Turan inequalities (emis.de/journals/JIPAM/article191.html). $\endgroup$
    – Dennis Jia
    Aug 10, 2017 at 1:19
  • $\begingroup$ The function you wrote never belongs to LP when $\alpha>0$. It belongs to LP if and only if $\alpha\leq 0$ and all zeros of $g$ are real. $\endgroup$ Aug 10, 2017 at 11:33

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.