Hadamard factorization for finite-order entire functions

Since $p+1>\rho$, [F2] gives ${\sum }_{n\ge 1}\mathrm{\mid }{a}_n{\mathrm{\mid }}^{-(p+1)}<\mathrm{\infty }$. Therefore [F3] constructs the canonical product $P(z):={\prod }_{n\ge 1}{E}_p(z\mathrm{/}{a}_n)$, and $P$ has exactly the nonzero zeros of $f$, with multiplicity.

Fix a real number $\sigma$ with $\rho <\sigma \le p+1$. By [F1], for all sufficiently large $R$ one has $\log M_f(R)\le R^{\sigma }$, and [F2] gives a finite sum $S_{\sigma }:={\sum }_{n\ge 1}\mathrm{\mid }{a}_n{\mathrm{\mid }}^{-\sigma }<\mathrm{\infty }$.

The quotient $H(z):=f(z)\mathrm{/}(z^{m}P(z))$ is therefore entire and zero-free: the factor $z^m$ removes the zero at $0$, and step 1.1 removes every other zero of $f$ with the correct multiplicity.

There is a constant $A_{\sigma }\ge 0$ such that $\frac{1}{\mathrm{\mid }{E}_p(w)\mathrm{\mid }}\le \exp (A_{\sigma }\mathrm{\mid }w{\mathrm{\mid }}^{\sigma })$ whenever $\mathrm{\mid }w\mathrm{\mid }\ge 2$ or $\mathrm{\mid }w\mathrm{\mid }\le 1\mathrm{/}2$. Indeed, if $\mathrm{\mid }w\mathrm{\mid }\ge 2$, then [F4] gives $\frac{1}{\mathrm{\mid }{E}_p(w)\mathrm{\mid }}=\frac{1}{\mathrm{\mid }1-w\mathrm{\mid }}\exp \text{\hspace{0.17em}⁣}(-\mathrm{Re}\text{\hspace{0.17em}⁣}(w+\frac{w^2}{2}+\cdots +\frac{w^p}{p}))\le \exp \text{\hspace{0.17em}⁣}\left({\sum }_{k=1}^p\frac{\mathrm{\mid }w{\mathrm{\mid }}^k}{k}\right)\le \exp (A_{\sigma }\mathrm{\mid }w{\mathrm{\mid }}^{\sigma }),$ because $\mathrm{\mid }1-w\mathrm{\mid }\ge 1$ and $k\le p<\sigma +1$. If $\mathrm{\mid }w\mathrm{\mid }\le 1\mathrm{/}2$, then [F4] gives $\mathrm{\mid }1-E_p(w)\mathrm{\mid }\le \mathrm{\mid }w{\mathrm{\mid }}^{p+1}\le \mathrm{\mid }w{\mathrm{\mid }}^{\sigma }\le 1\mathrm{/}2$, so $\frac{1}{\mathrm{\mid }{E}_p(w)\mathrm{\mid }}\le \frac{1}{1-\mathrm{\mid }1-E_p(w)\mathrm{\mid }}\le \exp (2\mathrm{\mid }w{\mathrm{\mid }}^{\sigma })\mathrm{.}$ Enlarge the constant once to cover both cases.

Fix $r\ge 1$ such that $r$ is not one of the moduli $\mathrm{\mid }{a}_n\mathrm{\mid }$, and put $H_1(z):=\frac{f(z)}{z^m{\prod }_{\mathrm{\mid }{a}_n\mathrm{\mid }\le 2r}{E}_p(z\mathrm{/}{a}_n)},H_2(z):={\prod }_{\mathrm{\mid }{a}_n\mathrm{\mid }>2r}{E}_p(z\mathrm{/}{a}_n{)}^{-1}\mathrm{.}$ Then $H=H_1{H}_2$. If $\mathrm{\mid }z\mathrm{\mid }=4r$, every factor of $H_1$ satisfies $\mathrm{\mid }z\mathrm{/}{a}_n\mathrm{\mid }\ge 2$, so steps 1.2 and 2.2 give $\mathrm{\mid }{H}_1(z)\mathrm{\mid }\le \mathrm{\mid }f(z)\mathrm{\mid }\exp \text{\hspace{0.17em}⁣}\left(A_{\sigma }\mathrm{\mid }z{\mathrm{\mid }}^{\sigma }{\sum }_{\mathrm{\mid }{a}_n\mathrm{\mid }\le 2r}\mathrm{\mid }{a}_n{\mathrm{\mid }}^{-\sigma }\right)\le \exp ((1+A_{\sigma }{S}_{\sigma })\mathrm{\mid }z{\mathrm{\mid }}^{\sigma })$ for all sufficiently large $r$. The function $H_1$ is entire by step 2.1, so [F7] applies on the disc $\mathrm{\mid }z\mathrm{\mid }<4r$ and gives the same bound for $\mathrm{\mid }z\mathrm{\mid }=r$. On that circle every factor of $H_2$ satisfies $\mathrm{\mid }z\mathrm{/}{a}_n\mathrm{\mid }<1\mathrm{/}2$, so step 2.2 gives $\mathrm{\mid }{H}_2(z)\mathrm{\mid }\le \exp \text{\hspace{0.17em}⁣}\left(A_{\sigma }\mathrm{\mid }z{\mathrm{\mid }}^{\sigma }{\sum }_{\mathrm{\mid }{a}_n\mathrm{\mid }>2r}\mathrm{\mid }{a}_n{\mathrm{\mid }}^{-\sigma }\right)\le \exp (A_{\sigma }{S}_{\sigma }\mathrm{\mid }z{\mathrm{\mid }}^{\sigma })\mathrm{.}$ Therefore $\mathrm{\mid }H(z)\mathrm{\mid }\le \exp (B_{\sigma }\mathrm{\mid }z{\mathrm{\mid }}^{\sigma })$ on $\mathrm{\mid }z\mathrm{\mid }=r$ for all sufficiently large admissible $r$, with $B_{\sigma }:=1+2A_{\sigma }{S}_{\sigma }$. Since such radii occur arbitrarily large, $H$ has order at most $\rho$.

Apply [F5] to the zero-free entire function $H$. Since $H$ has no zeros at all, its Weierstrass product part is empty, so there is an entire function $g$ with $H=e^g$. Hence $f(z)=z^m{e}^{g(z)}P(z)$.

For $R>0$, let $A_R:=1+{\max }_{\mathrm{\mid }\zeta \mathrm{\mid }=R}\mathrm{Re}(g(\zeta )-g(0))$. Because $\mathrm{\mid }H\mathrm{\mid }=e^{\mathrm{Re}g}$ and step 3.1 bounds $M_H(R)$ by $\exp (C_{\sigma }^{\mathrm{\prime }\mathrm{\prime }}(1+R^{\sigma }))$, one has $A_R\le C_{\sigma }^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(1+R^{\sigma })$. On the disc $\mathrm{\mid }z\mathrm{\mid }<R$, define $F_R(z):=(g(z)-g(0))\mathrm{/}(2A_R-(g(z)-g(0)))$. If $\mathrm{\mid }z\mathrm{\mid }=R$, then $\mathrm{Re}(g(z)-g(0))\le A_R-1$, so $\mathrm{\mid }2A_R-(g(z)-g(0)){\mathrm{\mid }}^2-\mathrm{\mid }g(z)-g(0){\mathrm{\mid }}^2=4A_R(A_R-\mathrm{Re}(g(z)-g(0)))>0$, hence $\mathrm{\mid }{F}_R(z)\mathrm{\mid }<1$ on the boundary circle. Also $F_R(0)=0$, so $F_R(z)\mathrm{/}z$ extends holomorphically across $0$. If $\mathrm{\mid }{F}_R(z)\mathrm{/}z\mathrm{\mid }$ had an interior local maximum larger than $1\mathrm{/}R$, then multiplying by the constant $R$ would give an interior local modulus maximum for a nonconstant holomorphic function, contradicting [F8]. Therefore $\mathrm{\mid }{F}_R(z)\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }\mathrm{/}R$ for $\mathrm{\mid }z\mathrm{\mid }<R$.

For $\mathrm{\mid }z\mathrm{\mid }\le R\mathrm{/}2$, step 4.2 gives $\mathrm{\mid }{F}_R(z)\mathrm{\mid }\le 1\mathrm{/}2$, so $\mathrm{\mid }g(z)-g(0)\mathrm{\mid }=2A_R\mathrm{\mid }{F}_R(z)\mathrm{\mid }\mathrm{/}\mathrm{\mid }1+F_R(z)\mathrm{\mid }\le 2A_R$. Together with the bound on $A_R$, this yields $\mathrm{\mid }g(z)\mathrm{\mid }\le C_{\sigma }^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(1+R^{\sigma })$ for $\mathrm{\mid }z\mathrm{\mid }\le R\mathrm{/}2$. Taking $R:=2(1+\mathrm{\mid }z\mathrm{\mid })$ gives a global growth estimate $\mathrm{\mid }g(z)\mathrm{\mid }\le C_{\sigma }^{(5)}(1+\mathrm{\mid }z\mathrm{\mid }{)}^{\sigma }$ on $\mathbb{C}$.

Step 5.1 holds for every $\sigma >\rho$. Applying [F6] to any one such $\sigma$ makes $g$ a polynomial; because the polynomial degree is an integer and the bound is available for every $\sigma >\rho$, the degree of $g$ is at most $p=\lfloor \rho \rfloor$. Put $Q:=g$. Then step 4.1 becomes $f(z)=z^m{e}^{Q(z)}{\prod }_{n\ge 1}{E}_p(z\mathrm{/}{a}_n)$ with $\mathrm{deg}Q\le p$, exactly as claimed.