3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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