0
$\begingroup$

Sometimes it is claimed that $\zeta$ cannot be equal to zero for Re$(z)>1$ because every term in the Euler product to which $\zeta$ absolutely converges is non-zero. This can be easily refuted with

$\dfrac{1}{\zeta(1)}=0=\left( \dfrac{1}{2} \right)\left( \dfrac{2}{3} \right)\left( \dfrac{4}{5} \right)...$

Instead, it is said that Hurwitz' theorem gives the rigorous proof that Re$(z)>1\implies\zeta(z)\neq0$. The theorem is: "Let $\{f_k\}$ be a sequence of holomorphic functions on a connected open set $G$ that converge uniformly on compact subsets of $G$ to a holomorphic function $f$ which is not constantly zero on $G$. If $f$ has a zero of order $m$ at $z_0$ then for every small enough $\rho>0$ and for sufficiently large $n\in\mathbb{N}$, $f_k$ has exactly $m$ zeros in the disk defined by $|z-z_0|<\rho$, including multiplicity. These zeros converge to $z_0$ as $k\to\infty$."

I think the proof I'm looking for depends on the corollary: "Let $G$ be a connected, open set and $\{f_n\}$ a sequence of holomorphic functions which converge uniformly on compact subsets of $G$ to a holomorphic function $f$. If each $f_n$ is nonzero everywhere in $G$, then $f$ is either identically zero or also is nowhere zero."

$\endgroup$
4
  • 1
    $\begingroup$ Your example is of an infinite product that diverges to zero. $\endgroup$ Commented Jun 5, 2019 at 21:30
  • $\begingroup$ Yes, I see that this diverges to zero and I can show that by taking the log of the product. Can you say how I can prove that zeta has no zeros real part greater than one? $\endgroup$ Commented Jun 5, 2019 at 21:45
  • 2
    $\begingroup$ @hodop: it relies on the following result: if $a_n$ is a sequence of complex numbers, none being equal to $-1$, such that $\sum_n{|a_n|}=S< \infty$, then $P_n=\prod_{k=1}^n{(1+a_k)}$ converges to a nonzero complex number. To prove it, show that $P_n$ is bounded by $e^S$ and then that if $n>m$, $|P_n-P_m| \leq e^S\left(e^{\sum_m^{\infty}{|a_k|}}-1\right)=o(1)$ as $m \rightarrow \infty$. Thus $P_n$ is Cauchy hence converges. Now apply the same result with the sequence $a’_k=\frac{-a_k}{1+a_k}$, so that $(1+a_k)(1+a’_k)=1$: it follows that $1/P_n$ converges, hence the limit of $P_n$ is nonzero. $\endgroup$
    – Aphelli
    Commented Jun 5, 2019 at 22:35
  • 2
    $\begingroup$ Read about convergent products of holomorphic functions - one result is that their zeros are precisely the zeros of the factors, so in particular here the factors have no zeros, so zeta has no zeros when it is expressed as a product (which happens precisely for $\Re(s)>1$) $\endgroup$
    – Conrad
    Commented Jun 5, 2019 at 23:29

0

You must log in to answer this question.

Browse other questions tagged .