The Riemann zeta function has no zeros on the closed half-plane $\mathrm{Re}s\ge 1$, except for its pole at $1$

By [L1], only the boundary line $\mathrm{Re}s=1$ remains to be checked. Suppose $t\mathrm{\ne }0$ and $\zeta (1+it)=0$. Since [L2] makes zeta holomorphic at $1+it$, there are $\delta >0$ and $C>0$ such that $\mathrm{\mid }\zeta (\sigma +it)\mathrm{\mid }\le C(\sigma -1)(1<\sigma <1+\delta )\mathrm{.}$

For $\sigma >1$ and real $u$, absolute convergence in [L3] and the power series $-\log (1-z)={\sum }_{m\ge 1}{z}^m\mathrm{/}m$ give $\log \mathrm{\mid }\zeta (\sigma +iu)\mathrm{\mid }={\sum }_p{\sum }_{m\ge 1}\frac{\cos (mu\log p)}{m{p}^{m\sigma }}\mathrm{.}$ Consequently $\log \text{\hspace{0.17em}⁣}\mid \zeta (\sigma {)}^3\zeta (\sigma +it{)}^4\zeta (\sigma +2it)\mid ={\sum }_p{\sum }_{m\ge 1}\frac{3+4\cos (mt\log p)+\cos (2mt\log p)}{m{p}^{m\sigma }}\ge 0,$ because $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2\ge 0$. Exponentiating gives $\mid \zeta (\sigma {)}^3\zeta (\sigma +it{)}^4\zeta (\sigma +2it)\mid \ge 1.$ On the other hand, [L2] gives $\zeta (\sigma )=1\mathrm{/}(\sigma -1)+O(1)$ as $\sigma \downarrow 1$, so $\zeta (\sigma {)}^3=O((\sigma -1{)}^{-3})$. The point $1+2it$ is not $1$, so [L2] also makes $\zeta (\sigma +2it)$ bounded as $\sigma \downarrow 1$. Combining these bounds with step 1.1 yields $\mid \zeta (\sigma {)}^3\zeta (\sigma +it{)}^4\zeta (\sigma +2it)\mid =O(\sigma -1)\to 0,$ contradicting the lower bound above.

Therefore $\zeta (1+it)\mathrm{\ne }0$ for every $t\mathrm{\ne }0$. At $t=0$, [L2] says $s=1$ is a simple pole, not a zero. Together with [L1], this proves that zeta has no zeros on $\mathrm{Re}s\ge 1$.