4
$\begingroup$

Let $\{z_j\}$ be the sequence of zeros on an entire function $f$. We define the convergence exponent of $\{z_j\}$ as $$b=\inf\left\{\lambda>0\ \text{s.t.}\ \sum_{j=1}^{+\infty}\frac{1}{|z_j|^{\lambda}}<+\infty\right\}$$ Let $n(r)$ be the number of $z_j$'s with $|z_j|\leq r$. Then the following identity holds: $$b=\limsup_{r\rightarrow +\infty}\frac{\log{\ n(r)}}{\log{r}}$$ Do you think i should use Jensen formula to prove this?

$\endgroup$
4
  • $\begingroup$ I think this should be possible to show straightforwardly. If $n(r)>r^\lambda$ infinitely often, then ... $\endgroup$ Jan 10 '13 at 20:51
  • $\begingroup$ Is this a homework problem? $\endgroup$ Jan 10 '13 at 21:34
  • $\begingroup$ @haigen von eitzen suppose i know $\lim_{r\rightarrow +\infty}\frac{n(r)}{r^{\lambda}}=0$. What could i deduce from this? $\endgroup$ Jan 10 '13 at 21:35
  • $\begingroup$ @greg martin a part of it $\endgroup$ Jan 10 '13 at 21:36
0
$\begingroup$

Hint: note that $$ \sum_{j\colon U\le |z_j|<2U} \frac1{|z_j|^\lambda} \le \sum_{j\colon U\le |z_j|<2U} \frac1{U^\lambda} \le \frac{n(2U)}{U^\lambda}. $$ Therefore \begin{align*} \sum_{j=1}^\infty \frac1{|z_j|^\lambda} &\le \sum_{j\colon |z_j|<1} \frac1{|z_j|^\lambda} + \sum_{k=1}^\infty \sum_{j\colon 2^{k-1}\le |z_j|<2^k} \frac1{|z_j|^\lambda} \\ &\le \text{(finite number of terms)} + \sum_{k=1}^\infty \frac{n(2^k)}{(2^{k-1})^\lambda}. \end{align*} In this way, knowledge about the growth of $n(r)$ can be converted into bounds on the sum in question. Lower bounds can be handled similarly.

$\endgroup$
2
  • $\begingroup$ thank you very much, but i can't understand the second line in your answer. we have $\sum_{j=1}^{+\infty}\frac{1}{|z_j|^{\lambda}}=$ sum of finite number of terms +$\sum_{U=1}^{+\infty}(\displaystyle\sum_{U\leq|z_j|<2U}\frac{1}{|z_j|^{\lambda}})\leq\sum_{U=1}^{+\infty}\frac{n(2U)}{U^{\lambda}}$ and how can i get your second line? $\endgroup$ Jan 12 '13 at 13:45
  • $\begingroup$ sorry, I had a typo in there - and I added another step while I was at it, hopefully better now $\endgroup$ Jan 13 '13 at 7:55

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.