Zero free entire function of exponential type is an exponential

A disk estimate including zero growth. Suppose $H$ is entire, $H(0)=0$, and $\mathrm{Re}H(w)\le A+B\mathrm{\mid }w\mathrm{\mid }$ with $A,B\ge 0$. Fix $w\mathrm{\ne }0$ and $0<t<1$, put $R=\mathrm{\mid }w\mathrm{\mid }\mathrm{/}t$ and $K=A+BR+1>0$. For $\mathrm{\mid }\zeta \mathrm{\mid }<1$, the denominator $2K-H(R\zeta )$ has real part at least $K+1>0$. Thus $G(\zeta )=H(R\zeta )\mathrm{/}(2K-H(R\zeta ))$ is holomorphic by [F2] and satisfies $G(0)=0$. For $\eta =H(R\zeta )$, $\mathrm{\mid }2K-\eta {\mathrm{\mid }}^2-\mathrm{\mid }\eta {\mathrm{\mid }}^2=4K(K-\mathrm{Re}\eta )>0$, so $\mathrm{\mid }G(\zeta )\mathrm{\mid }<1$. The local power series at zero shows that $G(\zeta )\mathrm{/}\zeta$ extends holomorphically through zero. For $0<r<1$, [F5] applied on $\mathrm{\mid }\zeta \mathrm{\mid }\le r$ gives $\mathrm{\mid }G(\zeta )\mathrm{/}\zeta \mathrm{\mid }\le 1\mathrm{/}r$ there; fixing $\zeta$ and letting $r$ increase to one proves $\mathrm{\mid }G(\zeta )\mathrm{\mid }\le \mathrm{\mid }\zeta \mathrm{\mid }$. Hence $\mathrm{\mid }H(R\zeta )\mathrm{\mid }\le \mathrm{\mid }\zeta \mathrm{\mid }(2K+\mathrm{\mid }H(R\zeta )\mathrm{\mid })$. At $\zeta =tw\mathrm{/}\mathrm{\mid }w\mathrm{\mid }$ this becomes $\mathrm{\mid }H(w)\mathrm{\mid }\le [2t(A+1)+2B\mathrm{\mid }w\mathrm{\mid }]\mathrm{/}(1-t)$. Let $t$ decrease to zero to conclude $\mathrm{\mid }H(w)\mathrm{\mid }\le 2B\mathrm{\mid }w\mathrm{\mid }$; at $w=0$ the same holds. In particular $A=B=0$ never produces division by zero, and $B=0$ forces $H=0$.

Construct the entire logarithm. By the local power series and derivative statements of [F2], $f^{\mathrm{\prime }}$ is holomorphic, and because $f$ is zero-free, $g=f^{\mathrm{\prime }}\mathrm{/}f$ is holomorphic. On the convex plane [F1] gives a primitive $h$; subtract its value at zero so $h(0)=0$ and $h^{\mathrm{\prime }}=f^{\mathrm{\prime }}\mathrm{/}f$. A primitive is holomorphic by definition, so [F2] also supplies its local power series.

By [F3] and [F4], $(f{e}^{-h}{)}^{\mathrm{\prime }}=f^{\mathrm{\prime }}{e}^{-h}-f{h}^{\mathrm{\prime }}{e}^{-h}=0$. Hence $f{e}^{-h}$ is constant with value $f(0)$, and the exponential addition formula gives $f=f(0)e^h$. Here $f(0)\mathrm{\ne }0$ by zero-freeness.

The growth bound at zero gives $M\mathrm{/}\mathrm{\mid }f(0)\mathrm{\mid }\ge 1$. Let $A$ be the unique real number with $e^A=M\mathrm{/}\mathrm{\mid }f(0)\mathrm{\mid }$, supplied by [F4]. Since the exponential is increasing and $e^0=1$, $A\ge 0$. Taking moduli in step 2.1 and using [F4] gives $e^{\mathrm{Re}h(z)}\le e^{A+C\mathrm{\mid }z\mathrm{\mid }}$, so $\mathrm{Re}h(z)\le A+C\mathrm{\mid }z\mathrm{\mid }$. The disk estimate of step 1.1, applied directly to $h$, yields $\mathrm{\mid }h(z)\mathrm{\mid }\le 2C\mathrm{\mid }z\mathrm{\mid }$ for all $z$.

Near zero write the convergent power series $h(z)={\sum }_{k\ge 1}{b}_k{z}^k$, with no constant term because $h(0)=0$. Then $h(z)\mathrm{/}z={\sum }_{k\ge 1}{b}_k{z}^{k-1}$ extends analytically through zero, with value $b_1=h^{\mathrm{\prime }}(0)$; the shifted series converges on the same disk by comparison on any smaller radius. Away from zero the quotient is holomorphic by [F2]. This defines an entire function $q$, bounded by $2C$ off zero by step 3.1 and at zero by continuity. By [F5], $q$ is constant and equals $h^{\mathrm{\prime }}(0)=f^{\mathrm{\prime }}(0)\mathrm{/}f(0)$ from step 1.2. Consequently $h(z)=az$ with the stated $a$, and step 2.1 gives $f(z)=f(0)e^{az}$.

If $C=0$ the bound in step 3.1 forces $h=0$ and hence $a=0$ and $f=f(0)$, including $f=1,M=1$. If $f(0)=1$, step 4.1 gives the advertised normalized formula. The assumptions exclude $M=0$ and $f(0)=0$; the removable value at zero and the open parameter limit $t\downarrow 0$ have both been checked. All constructions use uniquely determined analytic operations or one primitive, not any simultaneous choice of arbitrary witnesses.