$\newcommand{\d}{\mathrm{d}}$I'm reading a proof of a weak version of Baker's theorem, presented here. I've managed to get almost all of the way through, but I'm struggling with section $4$, page $6$, where they prove "Baker's lemma". I want to be able to complete reading this paper since it seems to be as “elementary” a proof as one might find.
Let $f:\Bbb C\to\Bbb C$ be holomorphic, $\varepsilon>0$, and let $A,B,C,T,U$ be large real numbers. Suppose that $C\gg T/(A\log A)+UBA^{\varepsilon}$ [they do not define $\gg$] and that:
- $\left|\frac{\d^m}{\d z^m}f(z)\right|\le e^{T+U|z|}\quad\forall0\le m\le C, z\in\Bbb C$
- $\frac{\d^m}{\d z^m}f(z)=0\quad\forall0\le m\le C,\,z\in\{1,2,\cdots,[A]\}$
Then $\left|\frac{\d^m}{\d z^m}f(z)\right|\le e^{-2(T+Uz)}$ for all $0\le m\le C/2$ and all $z\in\{1,2,\cdots,[AB]\}$.
Their argument: for $0\le m\le C/2$, put $g(z)=\frac{\d^m}{\d z^m}f(z)$ and apply the maximum modulus principle to the holomorphic function $g(z)(z-1)^{-[C/2]}(z-2)^{-[C/2]}\cdots(z-[A])^{-[C/2]}$ on the disc of radius $A^{1+\varepsilon}B$, where $z\in\{1,2,\cdots,[AB]\}$: $$\begin{align}|g(z)|&\le\prod_{k=1}^{[A]}|z-k|^{[C/2]}\max_{|w|=A^{1+\varepsilon}B}|g(w)|\prod_{j=1}^{[A]}|w-k|^{-[C/2]}\\&\overset{?}{\le}e^{-(\varepsilon/2)\log A[A][C/2]}\max_{|w|=A^{1+\varepsilon}B}|g(w)|\end{align}$$ And yes, the $\log A[A][C/2]$ is a bit ambiguous. I assume from context that $\log(A)\cdot[A][C/2]$ is what is intended. I don't know how they obtained the second bound marked $?$.
Really, they're claiming that if $|w|\ge A^{\varepsilon}B|z|$ then: $$\prod_{k=1}^{[A]}\frac{|z-k|}{|w-k|}\le A^{-(\varepsilon/2)[A]}$$Or I might presume it suffices to show: $$\left|\frac{z-k}{w-k}\right|\le A^{-\varepsilon/2}$$I notice the numerator polynomial vanishes for integer $z\le[A]$ so I only need to make this bound where $|w|>|z|>k$, in which case: $$\left|\frac{z-k}{w-k}\right|\le\frac{|(z-[A])/w|}{1-A^{-\varepsilon}B^{-1}}\le\frac{A^{-\varepsilon}-B^{-1}A^{-\varepsilon}}{1-A^{-\varepsilon}B^{-1}}$$But I can't see how to improve the bound.
Moreover they claim that the bound marked $?$ completes the proof. I disagree. We know that $e^{-(\varepsilon/2)\log(A)[A][C/2]}\ll e^{-(\varepsilon/4)[T+UBA^{1+\varepsilon}\log(A)]}$ and we have a bound on $\max |g(w)|$ by assumption $(1)$, therefore they've managed to bound: $$|g(z)|\ll\exp((1-\varepsilon/4)T+UA^{1+\varepsilon}B(1-(\varepsilon/4)\log(A)))$$Which is not anywhere near the claimed bound $\le e^{-2(T+Uz)}$ (moreover, we want a proper inequality not just $\ll$)
They say their argument is taken from Baker's original work. Hopefully someone here has read and understood the original!
EDIT: I have since figured out how to make the bound by $A^{-\varepsilon/2}$. That follows from $\frac{c-1}{xc-1}\lt\frac{1}{\sqrt{x}}$ for $c>0$ and $x>\max(1/c,1)$. It is the final part, drawing the conclusion of the lemma, that I am stuck on. Oh, and I'm also really confused about how they actually intend to apply this lemma. They state a few constants: "apply to $f_{m_1,\cdots,m_{n-1}}$ with $T=h^3\log K,U=L\log K,B=h^{1/8n}$ etc." but that "etc." frustratingly hides some very important detail. Indeed, a priori we don't know that these functions $f_m$ vanish at any point whatsoever (other than when $m=0$) and the constants they mention have unclear origins. I can't guess what comes next in the "etc.", which is a shame since, otherwise, this was a nicely "accessible" proof of a difficult result.