\documentclass{article}

\usepackage{amssymb}
\def\NN{\mathbb{N}}
\def\RR{\mathbb{R}}
\def\CC{\mathbb{C}}

\newtheorem{lemma}{Lemma}
\newtheorem{corollary}[lemma]{Corollary}
\newenvironment{proof}{\trivlist \item[\hskip \labelsep{\bf Proof$\,$}]}{\hfill\rlap{$\sqcap$}$\sqcup$\par}

\begin{document}

\begin{lemma}\label{lemma:ttt}
Let be given $n\in\NN_{\ge 2}$ and $\epsilon > 0$.
Then for every $a_0,\ldots,a_n \ge 0$ with $a_0 > \epsilon$ and $a_n = 1$, there exist
$t > 0$ and $k\in\{1,\ldots,n\}$ such that:
$$a_k t^k = a_0 - \epsilon$$
and such that for all $i\in\{1,\ldots,n\}$:
$$a_i t^i < a_0$$
\end{lemma}
\begin{proof}
Start with $k = n$ and $t = \root n\of{a_0 - \epsilon}$.  Then consider in turn
for $i$ the values $n-1$ down to $1$.
At each $i$ either $a_i t^i < a_0$
or $a_i t^i > a_0 - \epsilon$ (for the value of $t$ that is current at that time.)
In the first case do nothing, but in the second case
set $k$ to $i$ and $t$ to $\root i\of{(a_0 - \epsilon)/a_i}$
(in which case $t$ will decrease.)  This will give at the end a suitable $k$ and $t$.
\end{proof}

\begin{lemma}\label{lemma:max}
For $\epsilon > 0$ and $a_1,\ldots,a_n\in\RR$ there always is a $k$ such that for all $i\in\{1,\ldots,n\}$:
$$a_k > a_i - \epsilon$$
\end{lemma}
\begin{proof}
Induction with respect to $n$.
\end{proof}

\begin{lemma}\label{lemma:seq}
Let be given $n\in\NN_{\ge 2}$ and $\epsilon > 0$.
Then for every $a_0,\ldots,a_n\ge 0$ with $a_0 > \epsilon$ and $a_n = 1$, there exist:
\begin{enumerate}
\item $t > 0$
\item $k_0\ge k_1\ge k_2\ge\ldots \in\{1,\ldots,n\}$
\end{enumerate}
such that:
$$a_{k_0} t^{k_0} = a_0 - \epsilon$$
and such that for every $j$, if we let $k = k_j$ and $r = 3^{-j} t$:
$$a_k r^k > a_i r^i - \epsilon \makebox[0pt][l]{\qquad\it for all $i\in\{1,\ldots,n\}$}$$
\end{lemma}
\begin{proof}
Choose $t$ and $k_0$ according to lemma \ref{lemma:ttt}.

To get $k_{j+1}$ from $k_j$, let $k = k_j$, $r = 3^j t$
and apply lemma \ref{lemma:max} with $\epsilon/2$ to the sequence
$$a_1 (r/3),a_2 (r/3)^2,\ldots,a_k (r/3)^k$$
to get $k' = k_{j+1}$.
Then for $i\le k$ the inequality for $k_{j+1}$ directly follows,
while for $i > k$ we have:
$$a_k (r/3)^k = 3^{-k} a_k r^k > 3^{-k}\left(a_i r^i - \epsilon\right) = 3^{-k} a_i r^i - 3^{-k}\epsilon
> a_i (r/3)^i - \epsilon/2$$
and so:
$$a_{k'} (r/3)^{k'} > a_k (r/3)^k - \epsilon/2 > a_i (r/3)^i - \epsilon$$
\end{proof} 

\begin{lemma}\label{lemma:bou}
In the situation of the previous lemma, if again we take $k = k_j$ and $r = 3^{-j} t$, we have:
\begin{eqnarray*}
r^n &<& a_0\\
a_k r^k &<& a_0\\
3^{-jn} a_0 - 2\epsilon &<& a_k r^k
\end{eqnarray*}
\end{lemma}
\begin{proof}
We have $r \le t$ and so for all $i$ we have $a_i r^i \le a_i t^i < a_{k_0} t^{k_0} + \epsilon = a_0$.
Of this statement $r^n < a_0$ and $a_k r^k < a_0$ are special cases.
Finally, from $a_{k_0} r^{k_0} = 3^{-jk_0} a_{k_0} t^{k_0} \ge 3^{-jn} a_{k_0} t^{k_0} =
3^{-jn} (a_0 - \epsilon) > 3^{-jn} a_0 - \epsilon$ it follows that
$a_k r^k > a_{k_0} r^{k_0} - \epsilon > 3^{-jn} a_0 - 2\epsilon$.
\end{proof}

\begin{lemma}\label{lemma:sel}
For each sequence $k_0 \ge k_1 \ge k_2 \ge \ldots\in\{1,\ldots,n\}$ there is
a $j\in\NN$ with $j < 2n$ such that $k_{j-1} = k_j = k_{j+1}$.
\end{lemma}
\begin{proof}
Induction with respect to $n$.
\end{proof}

\begin{lemma}\label{lemma:est}
For every $\epsilon > 0$, $a_1,\ldots,a_n \ge 0$, $k\in\{1,\ldots,n\}$ and $r > 0$ such that for all $i\in\{1,\ldots,n\}$:
\begin{eqnarray*}
a_k (r/3)^k &>& a_i (r/3)^i - \epsilon\\
a_k (3r)^k &>& a_i (3r)^i - \epsilon
\end{eqnarray*}
holds:
$$\sum_{i=1\atop i\ne k}^n a_i r^i < (1 - 3^{-n}) a_k r^k + 3^n\epsilon$$
\end{lemma}
\begin{proof}
From the first assumption follows that
\begin{eqnarray*}
a_i r^i &=& 3^i a_i (r/3)^i\\
&<& 3^i (a_k (r/3)^k + \epsilon)\\
&<& 3^{i-k} a_k r^k + 3^i\epsilon
\end{eqnarray*}
and therefore
\begin{eqnarray*}
\sum_{i=1}^{k-1} a_i r^i &<& \sum_{i=1}^{k-1} \left(3^{i-k} a_k r^k + 3^n\epsilon\right)\\
&=& \big(\sum_{i=1}^{k-1} 3^{i-k}\big) a_k r^k + \big(\sum_{i=1}^{k-1} 3^i\big)\epsilon\\
&=& {1\over 2}(1-3^{1-k}) a_k r^k + {1\over 2}(3^k-3^1)\epsilon\\
&<& {1\over 2}(1 - 3^{-n}) a_k r^k + {1\over 2}3^n\epsilon
\end{eqnarray*}
%
In exactly the same way it follows from the other assumption that
$$a_i r^i < 3^{k-i} a_k r^k + 3^{-i}\epsilon$$
and
$$\sum_{i=k+1}^n a_i r^i < {1\over 2}(1 - 3^{-n}) a_k r^k + {1\over 2}3^n\epsilon$$
%
Together this gives
\begin{eqnarray*}
\sum_{i=1\atop i\ne k}^n a_i r^i &=& \sum_{i=1}^{k-1} a_i r^i + \sum_{i=k+1}^n a_i r^i\\
&<& {1\over 2}(1 - 3^{-n}) a_k r^k + {1\over 2}3^n\epsilon + {1\over 2}(1 - 3^{-n}) a_k r^k + {1\over 2}3^n\epsilon\\
&=& (1 - 3^{-n}) a_k r^k + 3^n\epsilon
\end{eqnarray*}
as required.
\end{proof}

\begin{lemma}\label{lemma:est1}
For every $\epsilon > 0$ and $a_0,\ldots,a_n \ge 0$ with $a_0 > 0$, there
exist $r > 0$ and $k\in\{1,\ldots,n\}$ such that:
\begin{eqnarray}
r^n &<& a_0 \label{ineq:est1}\\
a_k r^k &<& a_0 \label{ineq:est2}\\
3^{-2n^2} a_0 - 2\epsilon &<& a_k r^k \label{ineq:est3}\\
\sum_{i=1\atop i\ne k}^n a_i r^i &<& (1 - 3^{-n}) a_k r^k + 3^n\epsilon \label{ineq:est4}
\end{eqnarray}
\end{lemma}
\begin{proof}
Take $t$ and $k_0, k_1, \ldots$ as in lemma \ref{lemma:seq}.  According to
lemma \ref{lemma:sel} there is a $j < 2n$ such that $k_{j-1} = k_j = k_{j+1}$.
Take $k = k_j$ and $r = 3^{-j} t$.  Then inequalities (\ref{ineq:est1}), (\ref{ineq:est2}) and
(\ref{ineq:est3}) are given by lemma \ref{lemma:bou}
(the inequality $3^{-2n^2} a_0 < 3^{-jn} a_0$ holds because $j < 2n$),
and inequality (\ref{ineq:est4}) is given by lemma \ref{lemma:est}.
\end{proof}

\begin{lemma}\label{lemma:est2}
Let $a_0,\ldots,a_n\ge 0$ and $b_0,\ldots,b_n\in\CC$ with $a_i = |b_i|$ for
$i=1,\ldots,n$.
Furthermore, let $k\in\{1,\ldots,n\}$, $r > 0$ and  $z\in\CC$ with $r = |z|$.  Then:
$$\big|\sum_{i=0}^n b_i z^i\big| < \left|b_0 + b_k z^k\right| + \sum_{i=1\atop i\ne k}^n a_i r^i$$
\end{lemma}
\begin{proof}
Repeated application of the triangle inquality for the complex numbers.
\end{proof}

\begin{lemma}\label{lemma:est3}
Given $a_0,\,a_k > 0$, $b_0,\,b_0',\,b_k\in\CC$, $k\in\{1,\ldots,n\}$, $r > 0$ and $\eta > 0$
such that:
\begin{eqnarray*}
|b_0'| &=& a_0 \\
|b_k| &=& a_k \\
|b_0 - b_0'| &<& \eta\\
a_k r^k &<& a_0
\end{eqnarray*}
then there exists a $z\in\CC$ such that $|z| = r$ and:
$$\left|b_0 + b_k z^k\right| < a_0 - a_k r^k + \eta$$
\end{lemma}
\begin{proof}
Take
$$z = r\,\root k\of{-{a_k\over a_0}{b_0'\over b_k}}$$
Then we have:
$$\Big|{-{a_k\over a_0}{b_0'\over b_k}}\Big| = {a_k\left|b_0'\right|\over a_0\left|b_k\right|}
= {a_k a_0\over a_0 a_k} = 1$$
so
$$\Bigg|\root k\of{-{a_k\over a_0}{b_0'\over b_k}}\Bigg| = 1$$
and so $|z| = r$.

Because $a_k r^k < a_0$ we get $\big|a_0 - a_k r^k\big| = a_0 - a_k r^k$ and therefore
\begin{eqnarray*}
|b_0' + b_k z^k| &=& \Big|b_0' + b_k r^k \big(-{a_k\over a_0}{b_0'\over b_k}\big)\Big|\\
&=& \big|{b_0'\over a_0}(a_0 - a_k r^k)\big|\\
&=& {|b_0'|\over a_0}|a_0 - a_k r^k|\\
&=& a_0 - a_k r^k
\end{eqnarray*}
From this it follows that
$\left|b_0 + b_k z^k\right| \le \left|b_0' + b_k z^k\right| + |b_0 - b_0'| < a_0 - a_k r^k + \eta$.
\end{proof}

\begin{lemma}\label{lemma:nzc}
For $\eta > 0$ and $z\in\CC$ there is a $z'\in\CC$ with $z' \mathrel{\#} 0$ and $|z' - z| < \eta$.
\end{lemma}
\begin{proof}
Because $z + \eta/2 \mathrel{\#} z - \eta/2$, either $z + \eta/2 \mathrel{\#} 0$ or
$z - \eta/2 \mathrel{\#} 0$.  For both choices $|z' - z| = \eta/2 < \eta$.
\end{proof}

\begin{lemma}\label{lemma:eps0}
Given a finite list of inequalities
\begin{eqnarray*}
p_0\epsilon &<& q_0\\
p_1\epsilon &<& q_1\\
p_2\epsilon &<& q_2\\
\ldots
\end{eqnarray*}
with $p_i,q_i > 0$, there is an $\epsilon > 0$ that satisfies it.
\end{lemma}
\begin{proof}
Induction with respect to the length of the list.
\end{proof}

\begin{lemma}\label{lemma:eps}
Let be given $b_0,\ldots,b_n\in\CC$ with $b_n = 1$ and $c\in\RR$ with $|b_0| < c$.
Then there are $b_0'\in\CC$, $a_0$ and $\eta > 0$ such that:
\begin{eqnarray}
|b_0 - b_0'| &<& \eta\\
|b_0'| &=& a_0\\
a_0 &>& 0\\
a_0 + 3\eta &<& c  \label{ineq:eta}
\end{eqnarray}
and an $\epsilon > 0$ such that:
\begin{eqnarray}
2(3^n + 1)\epsilon &<& \eta  \label{ineq:epsilon}\\
2\epsilon &<& 3^{-2n^2} a_0\\
\epsilon &<& a_0
\end{eqnarray}
\end{lemma}
\begin{proof}
Take
$$\eta = {1\over 4}(c - |b_0|)$$
so $|b_0| = c - 4\eta$.
Then choose an arbitrary $b_0' \mathrel{\#} 0$ with $|b_0' - b_0| < \eta$ and take $a_0 = |b_0'|$.
To see that (\ref{ineq:eta}) is satisfied, calculate:
$$a_0 = |b_0'| \le |b_0' - b_0| + |b_0| < \eta + c - 4\eta = c - 3\eta$$
%
The existence of a suitable $\epsilon$ then follows from lemma \ref{lemma:eps0}.
\end{proof}

\begin{lemma}\label{lemma:eps1}
For:
$$q = 1 - 3^{-2n^2-n}$$
we have that $q > {1\over 2}$ and because of that
inequalities (\ref{ineq:eta}) and (\ref{ineq:epsilon}) imply:
$$q a_0 + 3^n \epsilon + \epsilon + \eta < qc$$
\end{lemma}
\begin{proof}
We get
$$a_0 + 2\cdot 3^n\epsilon + 2\epsilon + 2\eta = a_0 + 2(3^n + 1)\epsilon + 2\eta < a_0 + \eta + 2\eta < c$$
Using that $1 < 2q$, this gives
$$q a_0 + 3^n\epsilon + \epsilon + \eta < q a_0 + 2q 3^n\epsilon + 2q \epsilon + 2q \eta\
= q (a_0 + 2\cdot 3^n\epsilon + 2\epsilon + 2\eta) < qc $$
\end{proof}

\begin{lemma}\label{lemma:key}
Let be given $b_0,\ldots,b_n\in\CC$ with $b_n = 1$ and $c\in\RR$ with $|b_0| < c$.
Let $q$ be as in the previous lemma.  Then there is a $z\in\CC$ with
$$|z| < c^{1/n}$$
and:
$$\big|\sum_{i=0}^n b_i z^i\big| < qc$$
\end{lemma}
\begin{proof}
Take $b_0'$, $a_0$, $\eta$ and $\epsilon$ as in lemma \ref{lemma:eps}.  Take $a_i = |b_i|$
for $i\in\{1,\ldots,n\}$.
Take $r$ and $k$ as in lemma \ref{lemma:est1}.  Finally take $z$ as in lemma \ref{lemma:est3}.

Then plugging all conditions and results of lemmas \ref{lemma:est1}, \ref{lemma:est2}, \ref{lemma:est3},
\ref{lemma:eps} and \ref{lemma:eps1} together we get
$$r^n < a_0 < c - 3\eta < c$$
so
$$|z| = r < c^{1/n}$$
and
\begin{eqnarray*}
\big|\sum_{i=0}^n b_i z^i\big| &<& \left|b_0 + b_k z^k\right| + \sum_{i=1\atop i\ne k}^n a_i r^i\\
&<& \left(a_0 - a_k r^k + \eta\right) + \left((1 - 3^{-n}) a_k r^k + 3^n\epsilon\right)\\
&=& a_0 - 3^{-n} a_k r^k + 3^n\epsilon + \eta\\
&<& a_0 - 3^{-n} (3^{-2n^2} a_0 - 2\epsilon) + 3^n\epsilon + \eta\\
&=& (1 - 3^{-2n^2-n}) a_0 + 3^n\epsilon + 3^{-n} 2\epsilon +\eta\\
&<& (1 - 3^{-2n^2-n}) a_0 + 3^n\epsilon + \epsilon + \eta\\
&=& q a_0 + 3^n\epsilon + \epsilon + \eta\\
&<& q c
\end{eqnarray*}
\end{proof}

\end{document}
