I'm studying a book, which wants to prove that for some constants $C,\kappa$, we have $$\left|\frac{\zeta ''(s)\zeta (s) -2\zeta '(s)}{\zeta (s)^2}\right|\le C*|t|^\kappa~~~~,\text{for }s=\sigma+it,~|t|\ge 1, \sigma >1.$$

It has proved so far, that for every $m\in \mathbb{N}_0,$ we have constants $C_m$ with $$|\zeta^{(m)}(s)|\le C_m|t|~~~\text{, for }|t|\ge 1,\sigma>1$$ and for some $\delta>0$ we have: $$|\zeta(s)|\ge\delta|t|^{-4}~~~\text{, for }|t|\ge 1,\sigma>1.$$

It's obvious that the first inequality holds for e.g. $\kappa=9$. But my book says that the first inequality is true for $\kappa=5,$ too and I fail to see how we can prove this using these inequalities.

How can I prove the first inequality for $\kappa =5$? (is it even possible?)

  • $\begingroup$ What book are you reading? $\endgroup$ – davidlowryduda Jun 19 '15 at 8:42
  • $\begingroup$ It's a german book by E. Freitag and R. Busam: "Funktionentheorie 1" (beginning of chapter 6 in the 4th edition of the book). The result is needed for the proof of the prime number theorem. The exact value of $\kappa$ is not relevant for the proof, but it's stated that the inequality holds for e.g. $\kappa=5$. As is I missed no obvious solution apparently, I guess it's just a mistake in the book as they stated that the first inequality follows directly from the other two. $\endgroup$ – Scooby Jul 24 '15 at 17:29

I think we can get a better approximation of $1/\zeta^{2}\left(s\right) $. Take $\epsilon>0 $ and $\sigma\geq1+\epsilon>1 $. We recall that if $\sigma>1 $ we have $$\frac{1}{\zeta\left(s\right)}=\sum_{n\geq1}\frac{\mu\left(s\right)}{n^{s}} \tag{1} $$ where $\mu\left(s\right) $ is the Mobius function. Now if we consider the partial sum of $(1) $ we have, by partial summation, $$ \sum_{n\leq N}\frac{\mu\left(n\right)}{n^{s}}=M\left(N\right)N^{-s}+s\int_{1}^{N}M\left(u\right)u^{-s-1}du $$ where $M\left(N\right)=\sum_{n\leq N}\mu\left(n\right) $ is the Mertens function. By PNT, we know that $M\left(N\right)=o\left(N\right) $ as $N\rightarrow\infty $, then $$\left|\frac{1}{\zeta\left(s\right)}\right|=O\left(\left|s\right|\int_{1}^{\infty}u^{-\sigma}du\right)=O\left(\frac{\left|s\right|}{1-\sigma}\right) $$ then we have that exists some $C>0 $ such that $$\left|\frac{1}{\zeta\left(s\right)}\right|\leq C\left|t\right| $$ and so $$\left|\frac{\zeta''\left(s\right)\zeta\left(s\right)-2\zeta'\left(s\right)}{\zeta^{2}\left(s\right)}\right|\leq C_{1}\left|t\right|^{3}. $$

  • $\begingroup$ cool but this is useful only for $s = 1+it$ because for $s > 1$ : $\frac{1}{\zeta(s)} = \sum \mu_n n^{-s}$, $ |1/\zeta(s)| < \sum n^{-\sigma} = \zeta(\sigma) = \mathcal{O}(1/(\sigma-1))$ $\endgroup$ – reuns Jul 18 '15 at 8:35
  • $\begingroup$ $\mathcal{O}( |t|^0)$ ? it's better than $\mathcal{O}(|t|)$, but again yours is useful for $\Re(s) = 1$ (and $\Re(s) > \sigma_0$) $\endgroup$ – reuns Jul 18 '15 at 8:50
  • $\begingroup$ Yes, you're right :) $\endgroup$ – Marco Cantarini Jul 18 '15 at 8:59
  • $\begingroup$ and I'm wondering if just with the PNT we can say that $1/|\zeta(1+it)| = \mathcal{O}(|t|^\epsilon)$ $\endgroup$ – reuns Jul 18 '15 at 9:01

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.