Jensen's Formula application $f(z)$ has no zeros on $|z|=r, r<1$

Question

I am reading the Balazard Saias and Yor paper where he defines, $s=s(z)=\frac{1}{1-z}$ and $$f(z)=(s-1)\zeta(s)$$ where $\zeta$ is the Riemann zeta function.Then by Jensen's formula we have for $f(0)\neq 0$ and $r<1$, $$\frac{1}{2\pi}\int_{-\pi}^{\pi}log\ |f(re^{i\theta})|d\theta=log\ |f(0)|+\sum_{|\alpha|<r,f(\alpha)=0} log \ \frac{r}{|\alpha|}$$

For applying Jensen's formula we need that $f(z)$ has no zeros on $|z|=r$.

Question Show that $f(z)$ has no zeros on $|z|=r$, $r<1$.

Attempt $$f(z)=(s-1)\zeta(s)$$ $$f(z)= \frac{z}{1-z}\zeta(\frac{1}{1-z})$$ Since $|z|=r$ so write $z=re^{i\theta}$ $$\zeta(\frac{1}{1-z})= \zeta(\frac{1}{1-re^{i\theta}}) $$ $$\zeta(\frac{1}{1-z})= \zeta(\frac{1}{1-rcos\theta-irsin\theta}) $$ $$\zeta(\frac{1}{1-z})= \zeta(\frac{1-rcos\theta+irsin\theta}{(1-rcos\theta)^2+r^2sin^2\theta}) $$ $$\zeta(\frac{1}{1-z})= \zeta(\frac{1-rcos\theta+irsin\theta}{r^2-2rcos\theta+1}) $$ How to show that $f(re^{i\theta})\neq 0$ on $|z|=r, r<1$?.

Answer 1

Jensen's theorem is valid even if $f$ has zeroes on the circle $|z|=r$ (in practice since such circles are discrete, one avoids them easily in the usual applications of Jensen which is to count zeros in open discs)

The proof is standard assuming the result first without zeroes; then the number of zeroes on the circle $|z|=r$ is finite ($f$ is analytic in a neighborhood of $|z|=r$ in Jensen), hence let them be $a_1,..a_n, |a_k|=r$ counted with multiplicity.

Hence $f=(z-a_1)..(z-a_n)g$ with $g$ same zeroes as $f$ except those on the circle and $\log |f|=\log |g|+\log |z-a_1|+...\log |z-a_k|$

But now applying Jensen for $g$, using the easy to prove fact that $\int_0^{2\pi}\log |r(e^{i\theta}-e^{i\theta_0})|d\theta=2\pi \log r$, the fact that $|\log f(0)|=n\log r+|\log g(0)|$ and the observation that $\log r/|a_k|=0$ gives the result.

Answer 2

The question is equivalent to showing that $\zeta(s)\ne0$ for $|s|<1$.

• When $\Re s\le 0$, we know that $\zeta(s)$ has only trivial zeros at $s=-2,-4,-6,\dots$, by the functional equation for $\zeta(s)$.
• When $\Re s>0$, we can use the fact that $$ \sum_{\substack{\rho \\ \zeta(\rho)=0 \\ 0<\Re\rho<1}} \Re\frac1\rho = \frac{C_0}2+1-\frac12\log4\pi\approx0.0231 $$ to show that $\zeta(s)$ has no nontrivial zeros $\rho$ with $|\Im\rho|\le6.5$. (See for example Exercise 2 in Section 10.2 of Montgomery and Vaughan's Multiplicative Number Theory I. Classical Theory.)

Together these two results prove the desired statement and then some.