The Riemann xi function is entire of order one, real on the real axis, and symmetric under $s\mapsto 1-s$

By [L3], $\mathrm{\Lambda }$ has simple poles at $0$ and $1$. Multiplying by $\frac{1}{2}s(s-1)$ in [L1] cancels exactly those poles, so $\xi$ is entire. The same two facts give $\xi (1-s)=\frac{1}{2}(1-s)(-s)\mathrm{\Lambda }(1-s)=\frac{1}{2}s(s-1)\mathrm{\Lambda }(s)=\xi (s)\mathrm{.}$

For real $x>1$, the Dirichlet series in [L5] is a sum of positive real terms, so $\zeta (x)\in \mathbb{R}$. The remaining factors in [L1] are also real there, hence $\xi (x)\in \mathbb{R}$ for all $x>1$. Therefore the entire functions $s\mapsto \xi (s)$ and $s\mapsto \overline{\xi (\stackrel{‾}{s})}$ agree on $(1,\mathrm{\infty })$, so [A1] makes them equal on all of $\mathbb{C}$. In particular $\xi (x)$ is real for every real $x$.

On the half-plane $\mathrm{Re}s\ge 2$, [L5] gives $\mathrm{\mid }\zeta (s)\mathrm{\mid }\le {\sum }_{n\ge 1}{n}^{-2}$. Applying [L4] to $z=s\mathrm{/}2$ on that sector shows $\mathrm{\mid }\mathrm{\Gamma }(s\mathrm{/}2)\mathrm{\mid }\le \exp (C\mathrm{\mid }s\mathrm{\mid }\log (2+\mathrm{\mid }s\mathrm{\mid }))$ for some constant $C$, hence [L1] gives $\mathrm{\mid }\xi (s)\mathrm{\mid }\le \exp (C_1\mathrm{\mid }s\mathrm{\mid }\log (2+\mathrm{\mid }s\mathrm{\mid }))(\mathrm{Re}s\ge 2)\mathrm{.}$ By the symmetry from step 1.1, the same bound holds on $\mathrm{Re}s\le -1$. On the strip $-1\le \mathrm{Re}s\le 2$, the explicit split formula in [L3] gives $\xi (s)=\frac{1}{2}+\frac{1}{4}s(s-1){\int }_1^{\mathrm{\infty }}(\theta (t)-1)(t^{s\mathrm{/}2-1}+t^{-s\mathrm{/}2-1\mathrm{/}2})dt\mathrm{.}$ For $t\ge 1$ and $-1\le \mathrm{Re}s\le 2$, both powers of $t$ have modulus at most $1$, while $\theta (t)-1$ decays exponentially in $t$. Hence the integral is uniformly bounded on the strip, so $\mathrm{\mid }\xi (s)\mathrm{\mid }\le C_2(1+\mathrm{\mid }s{\mathrm{\mid }}^2)$ there. Therefore the same exponential bound holds on all of $\mathbb{C}$, and $\xi$ has order at most $1$.

Along the positive real axis, $\zeta (r)\to 1$ as $r\to \mathrm{\infty }$ by [L5], so [L1] and [L4] give $\log \mathrm{\mid }\xi (r)\mathrm{\mid }=\frac{r}{2}\log r+O(r)(r\to \mathrm{\infty })\mathrm{.}$ Thus $\log \log M_{\xi }(r)\ge \log \log \mathrm{\mid }\xi (r)\mathrm{\mid }=\log r+o(\log r)$, which rules out order smaller than $1$. Combining this with step 2.1 shows that $\xi$ has order exactly $1$.