Let $f(z)=e^{\ -\beta \ z^2}g(z)$ where $\beta \geq 0$ and $g(z)$ is a real entire function of genus $p\leq 1$ with real zeros. Prove that the zeros of the derivative $f^{\prime}(z)$ are also real and interlace with the of $f(z)$

suggestion: use hadamard theorem and note that $$\dfrac{f^{\prime}(z)}{f(z)}=-2\beta z+\dfrac{g^{\prime}(z)}{g(z)}$$

But I do not see how to take advantage of this, I hope you can guide me.

  • $\begingroup$ @ hrelmonio: what do you mean by the zeroes interlace? $\endgroup$ – Robert Lewis Dec 18 '13 at 6:16
  • $\begingroup$ There's a particular hint I'd like to give, but it would be best if you would say what you've tried so far so that I don't say something unhelpful. $\endgroup$ – Antonio Vargas Dec 18 '13 at 6:17
  • $\begingroup$ How do you define the genius of the function $g$? $\endgroup$ – Mercy King Dec 18 '13 at 22:42
  • $\begingroup$ @Mercy: en.wikipedia.org/wiki/Entire_function :) $\endgroup$ – helmonio Dec 19 '13 at 18:04

This is only a partial answer for the case $p=0$

In this case $g(z)$ takes the form $$Cz^m\prod_{n=1}^\infty \left(1-\frac{z}{x_n} \right) $$ where $x_n \in \mathbb R$ are the zeros (multiple zeros being repeated), which satisfy $$\sum_{n=1}^\infty \frac{1}{|x_n|}<\infty. $$ We then have $$f(z)=Ce^{- \beta z^2} z^m \prod_{n=1}^\infty \left(1-\frac{z}{x_n} \right), $$ and as you said $$\frac{f'(z)}{f(z)}=-2 \beta z+ \frac{m}{z}+ \sum_{n=1}^\infty \frac{1}{z-x_n}. $$ Taking the imaginary part of this we find $$\text{Im} \frac{f'(z)}{f(z)}= \left[-2\beta-\frac{m}{x^2+y^2}-\sum_{n=1}^\infty \frac{1}{(x-x_n)^2+y^2} \right] y, $$ where $z=x+iy$. Thus zeros of the derivative $f'(z)$ can only occur when $y=0$, that is when $z$ is real. I will try to develop this solution further, but that's all I've got for now.


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