Infinite Products and the Weierstrass Factorisation Theorem

By [F1], absolute convergence of $\prod (1+a_n)$ is exactly convergence of the real product $\prod (1+\mathrm{\mid }{a}_n\mathrm{\mid })$, and [F2] makes that equivalent to convergence of $\sum \mathrm{\mid }{a}_n\mathrm{\mid }$. This proves the equivalence of claims 1 and 2.

Assume now that $\sum \mathrm{\mid }{a}_n\mathrm{\mid }$ converges. By [F2], choose $N$ so that ${\sum }_{n\ge N}\mathrm{\mid }{a}_n\mathrm{\mid }<1\mathrm{/}2$; then $\mathrm{\mid }{a}_n\mathrm{\mid }<1\mathrm{/}2$ for every $n\ge N$, so $1+a_n\mathrm{\ne }0$ on that tail.

For $m>n\ge N$ one has $\mid {\prod }_{k=n}^m(1+a_k)-1\mid \le {\prod }_{k=n}^m(1+\mathrm{\mid }{a}_k\mathrm{\mid })-1$, and the right-hand side tends to $0$ as $n,m\to \mathrm{\infty }$ because the real tail products converge by [F2]. Hence the complex tail partial products form a Cauchy sequence, so they converge to some limit $\mathrm{\ell }\in \mathbb{C}$.

For the same tail, $\mathrm{\mid }1+a_n\mathrm{\mid }\ge 1-\mathrm{\mid }{a}_n\mathrm{\mid }>0$, so $\mid {\prod }_{k=N}^m(1+a_k)\mid \ge {\prod }_{k=N}^m(1-\mathrm{\mid }{a}_k\mathrm{\mid })$ for every $m\ge N$; by [F2], the real product ${\prod }_{k\ge N}(1-\mathrm{\mid }{a}_k\mathrm{\mid })$ converges to a positive limit because ${\sum }_{k\ge N}\mathrm{\mid }{a}_k\mathrm{\mid }$ converges and each term is in $[0,1\mathrm{/}2)$. Therefore the complex tail partial products are bounded away from $0$, so the limit $\mathrm{\ell }$ of step 2.1 is nonzero. Now [F3] makes $\prod (1+a_n)$ convergent with value $({\prod }_{k<N}(1+a_k))\mathrm{\ell }$. Since $\mathrm{\ell }\mathrm{\ne }0$, this value is nonzero exactly when the finite head has no zero factor; the tail factors are already nonzero by step 1.2.

Fix a compact set $K\subseteq \mathrm{\Omega }$. By [F1], choose $N$ so that $f_n$ has no zero on $K$ for $n\ge N$ and $M_n:={\sup }_K\mathrm{\mid }{f}_n-1\mathrm{\mid }$ satisfies ${\sum }_{n\ge N}{M}_n<\mathrm{\infty }$; enlarging $N$ if needed, assume also $M_n<1\mathrm{/}2$ for $n\ge N$.

For $m\ge \mathrm{\ell }\ge N$ and $z\in K$, one has $\mid {\prod }_{n=\mathrm{\ell }}^m{f}_n(z)-1\mid \le {\prod }_{n=\mathrm{\ell }}^m(1+M_n)-1$, while $\mathrm{\mid }{f}_n(z)\mathrm{\mid }\ge 1-M_n>1\mathrm{/}2$ for $n\ge N$. By [F2], the real products ${\prod }_{n\ge N}(1+M_n)$ and ${\prod }_{n\ge N}(1-M_n)$ converge, and the latter has positive value because all its factors are positive. Thus the tail partial products of $\prod f_n$ are uniformly bounded above and uniformly bounded away from $0$ on $K$.

The estimate of step 2.1 implies that the tail partial products are uniformly Cauchy on $K$, hence converge uniformly there to a continuous zero-free limit $Q_K$; multiplying by the finite holomorphic prefix ${\prod }_{n<N}{f}_n$ gives uniform convergence of the full partial products on $K$. Because $K$ was arbitrary, the convergence is locally uniform on $\mathrm{\Omega }$, and [F3] makes the limit function $F$ holomorphic.

On the fixed compact set $K$, write $F=\left({\prod }_{n<N}{f}_n\right)Q_K$ with $Q_K$ holomorphic and zero-free by step 3.1. If no factor vanishes identically on any connected component of $\mathrm{\Omega }$, the finitely many prefix factors have only isolated zeros on their respective components, and [F4] shows that every zero of $F$ on $K$, with its multiplicity, comes from that finite prefix and no tail factor contributes a new zero.

Fix a compact set $K\subseteq \mathrm{\Omega }$ disjoint from the zero set of $F$. By [F1], choose $N$ so that $f_n$ has no zero on $K$ for $n\ge N$, and write $P_m={\prod }_{n=0}^m{f}_n$. Then $P_m\to F$ uniformly on $K$, and for $m\ge N$ the functions $P_m$ are zero-free on $K$.

By [F2], the derivatives $P_m^{\mathrm{\prime }}$ converge uniformly on $K$ to $F^{\mathrm{\prime }}$. Since $F$ has no zero on the compact set $K$, the uniform convergence of $P_m$ to $F$ makes $P_m$ uniformly bounded away from $0$ on $K$ for all large $m$, so $P_m^{\mathrm{\prime }}\mathrm{/}{P}_m\to F^{\mathrm{\prime }}\mathrm{/}F$ uniformly on $K$.

For each finite product, ordinary differentiation gives $P_m^{\mathrm{\prime }}\mathrm{/}{P}_m={\sum }_{n=0}^m{f}_n^{\mathrm{\prime }}\mathrm{/}{f}_n$ on $K$. Passing to the uniform limit from step 2.1 yields the stated series identity and uniform convergence.

Write $S_p(u)=u+u^2\mathrm{/}2+\cdots +u^p\mathrm{/}p$, with the empty sum $0$ when $p=0$. Using [F1], [F2], [F3], and [F4], differentiate $E_p(u)=(1-u)e^{S_p(u)}$ to obtain $E_p^{\mathrm{\prime }}(u)=-e^{S_p(u)}+(1-u)e^{S_p(u)}(1+u+\cdots +u^{p-1})=-u^p{e}^{S_p(u)}\mathrm{.}$

Fix $w$ with $\mathrm{\mid }w\mathrm{\mid }\le 1$. By step 1.1, $\frac{d}{dt}{E}_p(tw)=w{E}_p^{\mathrm{\prime }}(tw)=-w^{p+1}{t}^p{e}^{S_p(tw)}(0\le t\le 1)\mathrm{.}$ Integrating from $0$ to $1$ and using $E_p(0)=1$ gives $1-E_p(w)=w^{p+1}{\int }_0^1{t}^p{e}^{S_p(tw)}dt\mathrm{.}$

For $0\le t\le 1$ and $1\le k\le p$, one has $\mathrm{Re}\text{\hspace{0.17em}⁣}\left(\frac{(tw{)}^k}{k}\right)\le \frac{\mathrm{\mid }tw{\mathrm{\mid }}^k}{k}\le \frac{t^k}{k},$ so $\mathrm{\mid }{e}^{S_p(tw)}\mathrm{\mid }=e^{\mathrm{Re}{S}_p(tw)}\le e^{S_p(t)}\mathrm{.}$ Taking absolute values in step 2.1 therefore yields $\mathrm{\mid }1-E_p(w)\mathrm{\mid }\le \mathrm{\mid }w{\mathrm{\mid }}^{p+1}{\int }_0^1{t}^p{e}^{S_p(t)}dt\mathrm{.}$

For real $t\in [0,1]$, step 1.1 gives $\frac{d}{dt}{E}_p(t)=-t^p{e}^{S_p(t)}\mathrm{.}$ Hence ${\int }_0^1{t}^p{e}^{S_p(t)}dt=E_p(0)-E_p(1)=1,$ because $E_p(0)=1$ and $E_p(1)=0$ by [F1]. Substituting this into step 3.1 proves $\mathrm{\mid }1-E_p(w)\mathrm{\mid }\le \mathrm{\mid }w{\mathrm{\mid }}^{p+1},$ and the displayed bound with $C=1$ follows.

Fix a compact disc $K_R=\{z:\mathrm{\mid }z\mathrm{\mid }\le R\}$. Since $(a_n)$ has no finite accumulation point, $\mathrm{\mid }{a}_n\mathrm{\mid }\to \mathrm{\infty }$, so there is $N$ with $\mathrm{\mid }{a}_n\mathrm{\mid }>R$ for $n\ge N$; then $\mathrm{\mid }z\mathrm{/}{a}_n\mathrm{\mid }\le 1$ on $K_R$ for every $n\ge N$.

For $z\in K_R$ and $n\ge N$, [F1] gives $\mathrm{\mid }1-E_p(z\mathrm{/}{a}_n)\mathrm{\mid }\le \mathrm{\mid }z\mathrm{/}{a}_n{\mathrm{\mid }}^{p+1}\le R^{p+1}\mathrm{\mid }{a}_n{\mathrm{\mid }}^{-(p+1)}$. Because the reciprocal power series converges, the Weierstrass $M$-test makes ${\sum }_{n\ge N}{\sup }_{z\in K_R}\mathrm{\mid }1-E_p(z\mathrm{/}{a}_n)\mathrm{\mid }$ convergent.

By [F2], the product ${\prod }_{n\ge 1}{E}_p(z\mathrm{/}{a}_n)$ is a canonical product, and step 2.1 is exactly the normal-convergence condition on the arbitrary compact disc $K_R$. Hence [F3] makes the product entire, with zeros exactly at the points $a_n$ and with their multiplicities.

Set $p_n:=n$ for every $n\ge 1$. Fix $R>0$. Because $(a_n)$ has no finite accumulation point, only finitely many terms satisfy $\mathrm{\mid }{a}_n\mathrm{\mid }\le 2R$; choose $N$ so that $\mathrm{\mid }{a}_n\mathrm{\mid }>2R$ for all $n\ge N$.

For $\mathrm{\mid }z\mathrm{\mid }\le R$ and $n\ge N$, one has $\mathrm{\mid }z\mathrm{/}{a}_n\mathrm{\mid }\le R\mathrm{/}\mathrm{\mid }{a}_n\mathrm{\mid }<1\mathrm{/}2\le 1$, and $E_n(z\mathrm{/}{a}_n)\mathrm{\ne }0$ because its unique zero occurs at $z=a_n$, outside the closed disc $\mathrm{\mid }z\mathrm{\mid }\le R$. Using [F2] with $p=n$ gives $\mathrm{\mid }1-E_n(z\mathrm{/}{a}_n)\mathrm{\mid }\le \mathrm{\mid }z\mathrm{/}{a}_n{\mathrm{\mid }}^{n+1}\le (R\mathrm{/}\mathrm{\mid }{a}_n\mathrm{\mid }{)}^{n+1}\le 2^{-n-1}\mathrm{.}$ Therefore ${\sum }_{n\ge N}{\sup }_{\mathrm{\mid }z\mathrm{\mid }\le R}\mathrm{\mid }1-E_n(z\mathrm{/}{a}_n)\mathrm{\mid }$ converges.

Step 2.1 is exactly the normal-convergence condition on the closed disc $\mathrm{\mid }z\mathrm{\mid }\le R$, and $R$ was arbitrary. Hence [F3] makes ${\prod }_{n\ge 1}{E}_n(z\mathrm{/}{a}_n)$ an entire function whose zeros are exactly the points $a_n$, counted with multiplicity. Multiplying by $z^m$ contributes precisely the order-$m$ zero at $0$ and no other zeros, so $P(z)=z^m{\prod }_{n\ge 1}{E}_n(z\mathrm{/}{a}_n)$ has exactly the stated zero divisor. Since $p_n=n\ge 0$, this is a Weierstrass product in the sense of [F1].

Let $m$ be the order of the zero of $f$ at $0$, with $m=0$ if $f(0)\mathrm{\ne }0$. If the nonzero zero multiset is infinite, [F2] gives an entire function $P(z)=z^m{\prod }_{n\ge 1}{E}_{p_n}(z\mathrm{/}{a}_n)$ with exactly the same zeros as $f$, counted with multiplicity. If it is finite, list it as $a_1,\dots ,a_N$ and put $P(z)=z^m{\prod }_{j=1}^N{E}_0(z\mathrm{/}{a}_j)$, with the empty product equal to $1$; [F5] gives the same zero-divisor conclusion.

The quotient $h=f\mathrm{/}P$ is holomorphic on $\mathbb{C}$: away from the common zeros this is immediate, and at each zero the matching multiplicities from step 1.1 and [F1] remove the singularity. Moreover $h$ has no zero anywhere, because every zero of $f$ was already cancelled by $P$.

By [F3], the whole plane is homologically simply connected, so [F4] gives an entire function $g$ with $e^g=h$. Substituting the definition of $h$ from step 2.1 yields the required infinite or finite factorization of $f$.

Let $(p_n)$ be the poles of $f$, listed with multiplicity equal to the order of the pole. By [F2], there is an entire function $q$ whose zeros are exactly the points $p_n$ with those multiplicities and with no other zeros.

On $\mathbb{C}\setminus \{p_n\}$ define $g=fq$. Near a pole $p_n$ of order $m$, the zero of $q$ has the same order $m$, so the product $g$ is locally bounded on the punctured neighbourhood of $p_n$. By [F3], each such singularity is removable, hence $g$ extends to an entire function on $\mathbb{C}$.

Away from the poles, $f=g\mathrm{/}q$ by construction, and both sides are meromorphic with the same removed singularities at the poles. Therefore $f$ is the quotient of the two entire functions $g$ and $q$.

For $z=x+iy$, [F4] gives $\sin (\pi z)=(e^{i\pi z}-e^{-i\pi z})\mathrm{/}(2i)$, so $\mathrm{\mid }\sin (\pi z)\mathrm{\mid }\le \frac{1}{2}(e^{-\pi y}+e^{\pi y})\le e^{\pi \mathrm{\mid }z\mathrm{\mid }}$. Along the imaginary axis one has $\mathrm{\mid }\sin (\pi ir)\mathrm{\mid }=(e^{\pi r}-e^{-\pi r})\mathrm{/}2$ for $r>0$. Therefore [F3] makes $f(z)=\sin (\pi z)$ an entire function of order $1$.

By [F1], the zero at $0$ has order $1$ and the nonzero zeros are exactly $\pm 1,\pm 2,\dots$. Applying [F2] with $\rho =1$ yields a polynomial $Q$ of degree at most $1$ such that $\sin (\pi z)=z{e}^{Q(z)}{\prod }_{n\ge 1}{E}_1(z\mathrm{/}n)E_1(-z\mathrm{/}n)$. Since $E_1(w)E_1(-w)=(1-w)e^w(1+w)e^{-w}=1-w^2$, this becomes $\sin (\pi z)=z{e}^{Q(z)}{\prod }_{n\ge 1}(1-z^2\mathrm{/}{n}^2)$.

By [F4], the function $\sin (\pi z)$ is odd, while ${\prod }_{n\ge 1}(1-z^2\mathrm{/}{n}^2)$ is even. Therefore the quotient $e^{Q(z)}=\frac{\sin (\pi z)}{z{\prod }_{n\ge 1}(1-z^2\mathrm{/}{n}^2)}$ is even. Writing $Q(z)=az+b$, this means $e^{az+b}=e^{-az+b}$ for every $z$, so $e^{2az}=1$ on $\mathbb{C}$. Therefore $a=0$, and $Q$ is constant.

Dividing the power series in [F4] by $\pi z$ gives $\sin (\pi z)\mathrm{/}(\pi z)=1-{\pi }^2{z}^2\mathrm{/}6+O(z^4)$. Step 2.1 with step 3.1 gives $\sin (\pi z)\mathrm{/}z=e^b{\prod }_{n\ge 1}(1-z^2\mathrm{/}{n}^2)$, and substituting $z=0$ shows $e^b=\pi$. Therefore $\sin (\pi z)=\pi z{\prod }_{n\ge 1}(1-z^2\mathrm{/}{n}^2)$. The convergence is locally uniform because step 2.1 is a normally convergent canonical-product factorization.

Assume first that $f$ has no zero on $\mathrm{\mid }z\mathrm{\mid }=R$. Because the closed disc is compact, $f$ has only finitely many zeros in $\mathrm{\mid }z\mathrm{\mid }<R$; applying [F2] repeatedly gives $f(z)={\prod }_{k=1}^N(z-a_k)g(z)$ on a neighbourhood of the closed disc, where $g$ is holomorphic and zero-free there.

By [F3], choose a holomorphic logarithm $L$ of $g$ on $\mathrm{\mid }z\mathrm{\mid }<R$. Applying [F1] to $L$ on the circle $\mathrm{\mid }z\mathrm{\mid }=R$ and taking real parts yields $\log \mathrm{\mid }g(0)\mathrm{\mid }=\frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }g(R{e}^{it})\mathrm{\mid }dt$.

For each zero $a_k$ with $\mathrm{\mid }{a}_k\mathrm{\mid }<R$, write $R{e}^{it}-a_k=R{e}^{it}(1-a_k{R}^{-1}{e}^{-it})$. The factor $1-a_k{R}^{-1}z$ is zero-free on the closed unit disc, so the same argument as in step 2.1 shows $\frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }1-a_k{R}^{-1}{e}^{-it}\mathrm{\mid }dt=0$; hence $\frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }R{e}^{it}-a_k\mathrm{\mid }dt=\log R$.

Taking logarithms of the factorization in step 1.1 on the boundary circle and averaging, step 2.1 gives the mean for $g$ and step 3.1 contributes one $\log R$ for each zero. Rearranging yields $\log \mathrm{\mid }f(0)\mathrm{\mid }=\frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }f(R{e}^{it})\mathrm{\mid }dt-{\sum }_{k=1}^N\log (R\mathrm{/}\mathrm{\mid }{a}_k\mathrm{\mid })$.

If $f$ has zeros on $\mathrm{\mid }z\mathrm{\mid }=R$, apply step 4.1 to radii $r<R$ with no zero on $\mathrm{\mid }z\mathrm{\mid }=r$; as $r\uparrow R$, the zero list inside $\mathrm{\mid }z\mathrm{\mid }<r$ stabilizes except when $r$ crosses one of finitely many zero moduli, and the boundary integral converges to the stated radial limit.

If $a_k$ is a zero with $\mathrm{\mid }{a}_k\mathrm{\mid }\le r$, then $\log (R\mathrm{/}\mathrm{\mid }{a}_k\mathrm{\mid })\ge \log (R\mathrm{/}r)$. There are exactly $n(r)$ such zeros, counted with multiplicity.

Therefore the Jensen sum in [F1] satisfies ${\sum }_{k=1}^N\log (R\mathrm{/}\mathrm{\mid }{a}_k\mathrm{\mid })\ge n(r)\log (R\mathrm{/}r)$. Substituting this lower bound into [F1] and rearranging gives the stated inequality.

Let $m$ be the order of the zero of $f$ at $0$, with $m=0$ if $f(0)\mathrm{\ne }0$. By [F4], there is an entire function $g$ with $g(0)\mathrm{\ne }0$ and $f(z)=z^{m}g(z),$ so $g$ has exactly the same nonzero zeros as $f$, with the same multiplicities.

If that nonzero zero multiset is finite, every reciprocal power sum over it is finite and its exponent is $0\le \rho$, so the conclusion holds. Hence assume from now on that it is infinite and enumerate it as $(a_n{)}_{n\ge 1}$.

For $r\ge 1$ and $\mathrm{\mid }z\mathrm{\mid }=r$, step 1.1 gives $\mathrm{\mid }g(z)\mathrm{\mid }=\mathrm{\mid }f(z)\mathrm{\mid }\mathrm{/}{r}^m\le \mathrm{\mid }f(z)\mathrm{\mid }$, hence $M_g(r)\le M_f(r)$. Therefore $g$ has order at most $\rho$ by [F1].

Fix real numbers $\sigma ,s$ with $\rho <\sigma <s$. By [F1] and step 3.1, for all sufficiently large $r$ one has $\log M_g(2r)\le (2r{)}^{\sigma }$. Applying [F2] to $g$, whose value at $0$ is nonzero by step 1.1, yields $n(r)\log 2\le \frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }g(2r{e}^{it})\mathrm{\mid }dt-\log \mathrm{\mid }g(0)\mathrm{\mid }\le (2r{)}^{\sigma }-\log \mathrm{\mid }g(0)\mathrm{\mid },$ where $n(r)$ counts the nonzero zeros of $f$ in $\mathrm{\mid }z\mathrm{\mid }\le r$. Thus $n(r)\le C{r}^{\sigma }$ for all large $r$.