\def\DATE{April, 11, 2020}

\line{\fchapter\hfill\the\chapterNo.\qquad Miscellanea}

\vSkipAfterTitle

\def\prevs{\the\chapterNo.\the\sectionNo.\the\statementNo}
\def\preveq{(\the\chapterNo.\the\sectionNo.\the\equationNo)}

\noindent
As its names indicates, the purpose of this chapter is to collect some
additional results. In the first section we show that for divergent solutions
whose coefficients are polynomials in $q$, the asymptotic behavior of these 
coefficients as $n$ tends to infinity is driven by their terms of highest
degree and that these terms are related by a recursion involving the \ikey{crest
polynomial}.

In the second section, we show some simple confluence results, namely that
under some suitable assumptions, the convergent solution of a $q$-algebraic 
equation in solved form, or the crest $q$-Borel transform of the solution if
that solution is divergent, converge to the solution of the limiting algebraic
equation as $q$ tends to $1$.

\bigskip

\section{Leading coefficients}

Now that we established in sections \fixedref{9.3} and \fixedref{9.4}
how we can linearize the crest except perhaps for some exceptional equations, 
we will address the following
question which underlines the results of Chapters 5 and 8: we have
seen in Theorem \RFDegree\ that the generic degree of $f_n(q)$ is given by
$H(P)n(n-h(P))/2$, and in Theorem \DSMainTh\ that when the assumptions of 
this theorem apply, $q^{-H(P)n(n-h(P))/2}f_n(q)$ remains of order $1$ as
$n$ tends to infinity, suggesting that the leading terms of $f_n(q)$ drive
its asymptotic behavior as $n$ tends to infinity. Our goal in this section
is to show that these leading terms have a particularly simple behavior
which indeed explains the asymptotics of $f_n(q)$ as $n$ tends to
infinity in the context of Theorem \DSMainTh. More precisely, we will show 
that when $\oalpha(P_0)<\oalpha(P_+)$ and $n$ is large enough, 
the top coefficients of $f_n(q)$ obey a linear
recursion driven by the crest series, and therefore can be represented as
linear combinations of algebraic/geometric series built from the roots of the 
crest series. One interesting aspect of this result is that it shows that
in some cases it is better to keep a nonlinear crest and not to attempt to
linearize it, mostly because the generic degree neglects the fact that the 
coefficients $P_A$ may depend on $q$.

To proceed, we first recall some known facts on sequences defined by a linear
recursion. We agree that any sequence $(u_n)_{n\geq 0}$ is extended to a 
sequence $u=(u_n)_{n\in\ZZ}$ by setting $u_n=0$ if $n$ is negative. 
On sequences, we define the backward shift $B$ by $(Bu)_n=u_{n-1}$. By
induction, the powers $B^k$, $k\geq 1$, are defined, and we set $B^0$ to be
the identity.

Let $C$ be an arbitrary polynomial which does not vanish at $0$. Then
$1/C(z)$ is a power series. Substituting $B^k$ for $z^k$ in that power
series, we define $1/C(B)$. If $(u_n)$ is a sequence, 
then $t_n=\bigl(1/C(B)\bigr)u_n$ is well defined and vanishes on the negative
integers. Furthermore, $C(B)t_n=u_n$. This last identity shows that
$(t_n)$ obeys a linear recursion, writing $C_i$ for the coefficient of $z^i$
in $C(z)$,
$$
  C_0t_n = u_n-\sum_{1\leq i\leq \deg C} C_i t_{n-i} \, .
$$
Such a sequence can be described differently. Up to multiplying $t_n$ by $C(0)$
we may assume that $C(0)$ is $1$. Let $\rho_i$, $1\leq i\leq k$, 
be the distinct roots of $C$, with $\mu_i$ the multiplicity of $\rho_i$.
As in section \fixedref{6.2}, we write the partial fractions expansion 
of $1/C(z)$, asserting that there exist some $\lambda_{i,j}$ such that
$$
  {1\over C(z)}
  = \sum_{1\leq i\leq k} \sum_{1\leq j\leq \mu_i}
  {\lambda_{i,j}\over (1-z/\rho_i)^j} \, .
$$
Expanding $1/(1-z/\rho_i)^j$, this implies that the relation $C(B)t_n=u_n$
forces
$$
  t_n=\sum_{1\leq i\leq k} \sum_{1\leq j\leq\mu_i}
  \lambda_{i,j} \sum_{m\geq 0}  {(m+j-1)_{j-1}\over (j-1)!} 
  {u_{n-m}\over \rho_i^m} \, .\eqno{\equa{LCA}}
$$
in particular, if we know that $C(B)t_n=0$ for $n$ large enough, that is
$u_n=0$ for $n$ larger than some $N$ then \LCA\ implies
$$\eqalign{
  t_n
  &{}=\sum_{1\leq i\leq k} \sum_{0\leq j< \mu_i} \lambda_{i,j}
    \sum_{n-N\leq m\leq n}  {(m+j-1)_{j-1}\over (j-1)!} {u_{n-m}\over \rho_i^m} \cr
  &{}=\sum_{1\leq i\leq k} \sum_{0\leq j<\mu_i} \lambda_{i,j}
    \sum_{0\leq m\leq N} {(n-m+j-1)_{j-1}\over (j-1)!} {u_m\over \rho_i^n} 
    \rho_i^m \, .\cr
}
$$
Thus, we may express $t_n$ as a linear combination of the
more elementary sequences $(n-m+j-1)_{j-1}/\rho_i^n$, or even simpler ones,
$n^p/\rho_i^n$ for $0\leq p\leq \max_{1\leq i\leq k}\mu_i$. 

We are now ready to 
describe the top coefficients of $f_n(q)$. Recall that 
once $\oalpha(P_0)<\oalpha(P_+)$, there is no loss of generality in assuming
that $\oalpha(P_0)=0$, and following the results of sections \fixedref{9.3} 
and \fixedref{9.4}, we may also assume that the crest is linear.


\Theorem{\label{LCoeff}
  Let $P$ be in solved form, satisfying the uniqueness condition, and such 
  that $0=\oalpha(P_0)<\oalpha(P_+)$. Assume
  furthermore that $\overline a(P)$ is finite and that the coefficients of $P$
  do not depend on $q$. Then for
  any fixed integer $k$ in $\ZZ$, there exists $N_k$ such that for any
  $n$ at least $N_k$,
  $$
    [q^k]\inot{\calC_{P,q,f_0}}(B)[z^n]\widehat\inot{B_{P,q}f(z)} = 0 \, .
  $$
}

The meaning of the conclusion is the following. By definition of the 
crest $q$-Borel transform,
$$
  [z^n]\widehat B_{P,q,f_0}f(z) = q^{-H(P)n(n-h(P))/2}f_n(q) \, .
$$
Since $[q^k]$ and $B$ commute here, the conclusion of Theorem \LCoeff\
asserts that
$$
  \calC_{P,q,f_0}(B)[q^k]\bigl(q^{-H(P)n(n-h(P))/2}f_n(q)\bigr) = 0 \, ,
$$
that is
$$
  \calC_{P,q,f_0}(B) [q^{H(P)n(n-h(P))/2+k}] f_n(q) = 0 \, .
  \eqno{\equa{LCB}}
$$

Also, if the crest polynomial has a unique root $\rho$ of multiplicity
$\mu$, then Theorem \LCoeff\ implies that for any $k$ in $\ZZ$ there exists
$u_{m,k}$ such that, provided $n$
is large enough,
$$\displaylines{\qquad
  [q^{H(P)n(n-h(P))/2+k}] f_n(q)
  \hfill\cr\hfill 
  {}= \sum_{0\leq j\leq \mu} \lambda_{1,j} \sum_{0\leq m\leq N} 
    {(n+m+j-1)_{j-1}\over (j-1)!} {u_{m,k}\over \rho^n} \rho^m\, .
  \qquad\cr}
$$
Arranging the terms, this means that there is a polynomial $\Pi_{\mu,k}(n)$ 
in $n$, of degree $\mu$, such that for any $n$ large enough,
$$
  [q^{H(P)n(n-h(P))/2+k}] f_n(q) 
  = {\Pi_{\mu,k}(n) \over \rho^n}\, .
$$

Since Theorem \RFDegree\ asserts that the generic degree of $f_n(q)$
is $H(P)n\bigl(n-h(P)\bigr)/2+O(1)$, identity \LCB\ shows that the top
coefficients of $f_n(q)$ obey a linear recursion, and can therefore be
represented as linear combinations of terms
$n^i/\rho_j^n$, $1\leq i\leq \mu_j$, built from the crest
polynomial. In particular, their order of magnitude is that of
$n^{\mu_1}/\rho_1^n$ where $\rho_1$ is a root of smallest modulus of
the crest polynomial of maximal multiplicty $\mu_1$.  Coupled with
Theorem \DSMainTh, this shows that when $|q|$ is greater than $1$, the
asymptotic behavior of $f_n(q)$ is in fact driven by its terms of
largest degree in $q$.

In some sense, Theorem \LCoeff\ systematizes the empirical observation made
by Drake (2009) for some specific equations that the top 
coefficients of $f_n(q)$ seem to remain constant for $n$ large enough.

The following elementary example illustrates this point and more generally, 
our discussion.

\bigskip

\Example Consider the \ikey{$q$-Catalan equation}
$$
  f(z)=1+zf(z)f(qz) \, .\eqno{\equa{LCqCatalanA}}
$$
It is in solved form, satisfies the uniqueness condition, and for this
equation $0=\oalpha(P_0)<\oalpha(P_+)=1$. We have $f_0=1$ and the crest 
polynomial is $1-z$. It has a unique root, $1$. Theorem \LCoeff\ asserts
that for any fixed $k$ and any $n$ large enough,
$$
  (1-B)[q^{n(n-1)/2+k}]f_n(q)=0 \, .
  \eqno{\equa{LCqCatalanB}}
$$
Thus, if we write $f_n(q)=\sum_{0\leq j\leq \delta_n} f_{n,i}q^i$, we have
for any $k$ in $\ZZ$ and any $n$ large enough,
$$
  f_{n,n(n-1)/2+k}=f_{n-1,(n-1)(n-2)/2+k} \, .
  \eqno{\equa{LCqCatalanBa}}
$$

This can be seen directly, in particular when $k$ vanishes. Indeed, applying
$[z^n]$ to \LCqCatalanA, we obtain
$$
  f_n(q)=\One\{\, n=0\,\} +\sum_{0\leq i\leq n-1}  q^i f_i(q) f_{n-1-i}(q) \, .
$$
Therefore, 
$$
  \deg_q f_n(q) 
  = \max_{0\leq i\leq n-1} \bigl( i+\deg_q f_i(q) + \deg_q f_{n-1-i}(q)\bigr) 
  \, ,
  \eqno{\equa{LCqCatalanC}}
$$
with $\deg_q f_0(q)=0$. The sequence $n(n-1)/2$ is $0$ when $n=0$ and
obeys the same recursion as $\deg_q f_n(q)$. Therefore, 
$\deg_q f_n(q)=n(n-1)/2$. This implies that the maximum in \LCqCatalanC\ is 
achieved only when $i=n-1$. Thus, $[q^{n(n-1)/2}]q^i f_i f_{n-1-i}=0$ if
$i\leq n-2$. Therefore,
$$\eqalign{
  [q^{n(n-1)/2}]f_n 
  &{}=[q^{n(n-1)/2}] q^{n-1} f_{n-1}f_0 \cr
  &{}=[q^{n(n-1)/2 - (n-1)}] f_{n-1} \cr
  &{}=[q^{(n-1)(n-2)/2}] f_{n-1} \, .\cr}
$$
This is what \LCqCatalanB\ asserts when $k$ is $0$, meaning that the top
coefficient of $f_n(q)$ is constant whenever $n$ is large enough.

It is instructive to see what happens if we linearize the crest. We translate 
and simplify the equation, in effect setting $f(z)=1+zg(z)$, to obtain
$$
  g(z)=1+zg(z)+qzg(qz)+qz^2g(z)g(qz) \, .
$$
This equation has a linear crest, $(0;0)-q(1;1)$, but the coefficients of
the associated $q$-operator depend on $q$. To satisfy the assumptions of 
Theorem \LCoeff, we may embed the equation into a parametric family, say
$$
  g(z)=1+zg(z)+rzg(qz)+rz^2g(z)g(qz) \, .
  \eqno{\equa{LCqCatalanBb}}
$$
The coefficients of $g$ are now some polynomials in $r$ and $q$, 
say $g_n(r,q)$. This equation has height $1/2$ and co-height $1$. 
The crest polynomial is $1-rz$. It has a unique root, $1/r$. For this 
new equation, Theorem \LCoeff\ asserts that if $n$ is at least some $N$,
then
$$
  (1-rB)[q^{n(n-1)/2+k}]g_n(r,q)=0 \, .
  \eqno{\equa{LCqCatalanEmbedB}}
$$
Thus, if we write $g_n(q)=\sum_{0\leq j\leq \delta_n} g_{n,i}(r)q^i$, we have for 
any integer $k$ and any $n$ large enough,
$$
  g_{n,n(n-1)/2+k}(r)=r g_{n-1,(n-1)(n-2)/2+k}(r) \, .
  \eqno{\equa{LCqCatalanD}}
$$

Since $f(z)=1+zg(z)$, we have $g_n(q,q)=f_{n+1}(q)$, so that the 
sequence $f_n(q)$ is related to the sequence $g_n(r,q)$ on the diagonal $r=q$.
It is unclear if some useful information on $f_n(q)$ can then be recovered from
that we have on $g_n(r,q)$.\hfill\qed

\bigskip

\noindent{\bf Proof of Theorem \LCoeff.} As in section \fixedref{8.2}, let
$$
  \inot{v_n}=H(P)n\bigl(n-h(P)\bigr)/2 \, .
$$ 
For a $q$-factor $A=(a;\alpha_1,\ldots,\alpha_\ell)$ and any nonnegative 
integers $n_1,\ldots,n_\ell$ with $n_1+\cdots+n_\ell=n-a$, set, as in 
section \fixedref{8.2},
$$
  \inot{\Delta_A(n_1,\ldots,n_\ell)}
  = v_n -\sum_{1\leq i\leq \ell} v_{n_i} -\sum_{1\leq i\leq\ell} \alpha_i n_i
  \, .
$$
Defining $g_n$ by $f_n(q)=q^{v_n}g_n(1/q)$, we rewrite \DSgnRecursion\ as
$$\displaylines{\qquad
  \sum_{A\in P_0} P_A q^{\alpha_1 n} g_n(1/q)
  +\sum_{A\in P_+} P_A \sum_{n_1+\cdots+n_\ell=n-a} 
  q^{-\Delta_A(n_1,\ldots,n_\ell)}
  \hfill\cr\hfill
  {}\times g_{n_1}(1/q)\cdots g_{n_\ell}(1/q)
  + q^{-v_n} P_{(n;\sp)} = 0 \, .
  \qquad \equa{LCoeffA} \cr}
$$
Note that since the coefficients of $P$ do not depend on $q$, 
Theorem \RFDegree\ implies $\delta_n=v_n+O(1)$ as $n$ tends to infinity.
Consequently, 
$$
  \max_{n\geq 0} \deg_q g_n(1/q) < \infty \, .
  \eqno{\equa{LCoeffC}}
$$ 
Lemma \DSDeltaAMinLoc\
implies that for any of the cases (ii), (iii) and (iv) of that lemma, there
exists some positive $\Theta$ such that
$$
  \Delta_A(n_1,\ldots,n_\ell) \geq (n-a)\Theta +H(P)a\bigl(a-h(P)\bigr)/2 \, .
  \eqno{\equa{LCoeffB}}
$$
In particular, since $\Theta$ is positive and $\overline a(P)$ is finite, since
\LCoeffC\ holds, for $n$ large enough
$$
  [q^k] q^{-\Delta_A(n_1,\ldots,n_\ell)} g_{n_1}(1/q)\cdots g_{n_\ell}(1/q)
  = 0 \, .
$$
Thus, applying $[q^k]$ to \LCoeffA, we obtain for $n$ large enough,
$$
  P_{(0;0)} [q^k] g_n(1/q) 
  + \sum_{A\in \widehat P_+} P_A [q^k] 
  \bigl( q^{H(P)a(a-h(P))/2} g_{n-a}(1/q)\bigr) =0\, ,
$$
that is
$$
  [q^k] P_{(0;0)}g_n(1/q)
  + [q^k] \sum_{A\in \widehat P_+} P_A q^{H(P)a(a-h(P))/2}B^a g_n(1/q) =0\, ,
$$
or, equivalently,
$$
  [q^k]\calC_{P,q}(B) g_n(1/q)=0 \, .
$$
This is the result since the series $\sum_{n\geq 0} g_n z^n$ is 
$\widehat\calB_{P,q}f(z)$.\hfill\qed

\bigskip

The proof of Theorem \LCoeff\ shows that it is possible to refine the result,
showing that it holds as long as $k\leq \epsilon n$ for some 
positive $\epsilon$ which depends on $\Theta$ in \LCoeffB.


\section{A simple \ikey{confluence} result}

Consider a polynomial $q$-algebraic equation in solved form, $P_qf(z)=0$. 
The solution $f$ is a power series in $z$ which depends also on $q$, and 
can be written, as in Chapter \fixedref{5},
$$
  f(z;q)=\sum_{n\geq 0} \inot{f_n(q)}z^n \, . 
$$
The confluence problem is to assess
the continuity of $f$ with respect to $q$. A considerable variety of 
situations may occur, as we will show with some examples. Our goal in this
section is not to develop an extensive theory, but, on the contrary, to
prove two specific and simple results, with the precise aim of enlightening
the numerical calculations which we will carry out in Chapter \fixedref{13}.

We will be only interested in results as $q$ tends to $1$. Since we need to 
distinguish according to the position of $|q|$ with respect to $1$, we 
write $\lim_{q\to 1-}$ for $\lim_{\ss q\to 1\atop\ss |q|<1}$ and $\lim_{q\to 1+}$ 
for $\lim_{\ss q\to 1\atop\ss |q|>1}$.

\bigskip

\Examples (i) Consider the equation
$$
  f(z)=1+(1-q)zf(qz) \, .
$$
Applying $[z^n]$, we obtain the recursion
$$
  f_n=\One\{\, n=0\,\} +(1-q)q^{n-1}f_{n-1} \, .
$$
Therefore, $f_n=(1-q)^n q^{n(n-1)/2}$ and
$$
  f(z;q)=\sum_{n\geq 0} q^{n(n-1)/2} \bigl((1-q)z\bigr)^n\, .
$$
Thus, $f(z;q)$ is a function for $|q|<1$, which is defined on the complex
plane. We have $\lim_{q\to 1} f_n(q)=\One\{\, n=0\,\}=f_n(1)$. 
Consequently,
$$
  \lim_{\ss q\to 1-}f(z;q)=1=f(z;1)
$$
and this convergence holds locally uniformly on $\CC$. 

If $|q|>1$, then $f(z;q)$ is a divergent series, but
$$
  \calB_{q,1/2}f(z;q)
  =\sum_{n\geq 0} \bigl((1-q)z\bigr)^n
  = {1\over 1-(1-q)z} \, .
$$
This $q$-Borel transform is defined on the 
disk 
$$
  \{\, z\in \CC \,:\, |z|<1/|1-q|\,\}\, ,
$$
and
$$
  \lim_{q\to 1+} \calB_{q,1/2}f(z;q)=f(z;1)\,,
$$
and this convergence is locally uniform.

\noindent (ii) Consider the equation
$$
  (1-q)f(z)=1+zf(z)-zf(qz) \, .
$$
When $q$ is $1$, this equation gives the formal identity $0=1$, and, therefore,
has no solution. But $f(z;q)=1/(1-q)$ is a solution of the equation. Clearly,
this solution has no limit as $q$ tends to $1$.

However, if one is interested in describing the behavior of $f(z;q)$
for $q$ near $1$, one could consider $g(z;q)=(1-q)f(z;q)$, which satisfies 
the equation
$$
  g(z)=1+z{g(z)-g(zq)\over 1-q}
$$
and assert that $\lim_{q\to 1} g(z;q)=1$.

\noindent (iii) Consider the equation 
$$
  f(z)-f(qz)=(1-q)zf(z) \, .
$$
Applying $[z^n]$ yields 
$$
  (1-q^n)f_n=(1-q)f_{n-1} \, ,
$$
and, therefore,
$$
  f_n=\prod_{1\leq k\leq n} {1-q\over 1-q^k} f_0 \, ,
$$
with $f_0$ a free parameter. This yields
$$
  f(z;q)=f_0+f_0\sum_{n\geq 1} \prod_{1\leq k\leq n} {1-q\over 1-q^k}z^n \, .
$$
We have $\lim_{q\to 1} f_n(q)=f_0/n!$. To study the convergence of $f(z;q)$ as
$q$ tends to $1-$ say, we restrict ourselves to $q$ being real, namely $q$ in
$(0,1)$. Then
$$
  0\leq {1-q\over 1-q^n}\leq 1 \, ,
$$
so that
$$
  \Bigl| f(z;q)-f_0-f_0\sum_{1\leq n\leq N} \prod_{1\leq k\leq n} 
         {1-q\over 1-q^n}z^n \Bigr| 
  \leq |f_0| \sum_{n>N} |z|^n \, .
$$
Therefore, for any $z$ of modulus less than $1$, for any positive $\epsilon$, 
provided $N$ is large enough,
$$
  \limsup_{\ss q\to 1-\atop\ss q\in (0,1)} 
  \Bigl| f(z;q)-f_0-\sum_{1\leq n\leq N} 
  \prod_{1\leq k\leq n} {1-q\over 1-q^n}z^n\Bigr| 
  \leq \epsilon
$$
and, consequently,
$$
  \lim_{\ss q\to 1-\atop\ss q\in (0,1)} f(z;q)=f_0e^z \,,
$$
and this convergence is locally uniform in the 
disk $\{\, z\in \CC\,:\, |z|<1\,\}$.

Formally, the limiting equation for $q=1$ reads $0=0$, and any power series
is a solution. However, upon rewriting the equation as
$$
  {f(z)-f(qz)\over 1-q}=zf(z)
$$
and assuming that $f$ is differentiable, taking the limit as $q$ tends to $1$ 
yields
the differential equation $zf'(z)=zf(z)$, to which $f_0e^z$ are the only 
solutions. In that sense, $f(z;q)$ converges to the limiting solution as
$q$ tends to $1-$.\hfill\qed

\bigskip

For linear equations, the confluence problem with differential equations as
limit has been studied, among others, 
by Sauloy (2000), Dreyfus (2015), Le Stum and Quir\'os (2015), and 
by Andr\'e and Di Vizzio (2004) in the $p$-adic setting. Our purpose being 
quite specific, we will instead consider the much simpler confluence with 
an algebraic equation as a limit.

Formally, the confluence problem is somewhat complicated to state in full
generality, because
a $q$-operator is a fomal weighted sum of $q$-factors, so that both the 
coefficients and the $q$-factors may depend on $q$. So, for the purpose of this
section, we consider $q$-operators of the form
$$
  P_q=\sum_{A\in P} P_A(q)(a;\alpha_1,\ldots,\alpha_\ell)_q \, ,
  \eqno{\equa{MiscPqDef}}
$$
the important point being here that the summation is over a fixed set $P$,
which does not depend on $q$, and with the assumption that
$$
  \vcenter{\hsize=230pt\noindent $P_A(q)\not=0$ for any $A$ in $P$ and any $q$ in a 
  punctured neighborhood of $1$.} 
  \eqno{\equa{MiscBasicAssumption}}
$$
So the support of $P_q$ is $P$ when $q\not =1$, but may be smaller when $q=1$,
as for example in $P_q=(1-q)(0;0)$.

Our first result deals with the convergence of the coefficients of the 
solution as $q$ tends to $1$.

\Proposition{\label{MiscfnqToOne}
  Let $P_q$ be a $q$-operator as in \MiscPqDef, 
  satisfying \MiscBasicAssumption, in solved form, satisfying the
  uniqueness condition for any $q$ in a neighborhood of $1$, with 
  finite support, such that $\lim_{q\to 1} P_A(q)=P_A(1)$ for any $A$ in $P$,
  and  $\lim_{q\to 1} \sum_{A \in P} P_A(q)\not=0$. Then
  $$
    \lim_{q\to 1} \inot{f_n(q)}=f_n(1)\, .
  $$
}

\Proof Consider the basic recursion \basicRecursion. The result follows by
induction on $n$.\hfill\qed

\bigskip

We now strengthen this result by showing that not only do the coefficients
converge but, when $f(z;q)$ is analytic, this solution converges, and when 
$f(z;q)$ is divergent, its crest $q$-Borel transform converges.

\Theorem{\label{MiscfqToOne}
  Under the assumptions of Proposition \MiscfnqToOne,
  
  \medskip
  \noindent(i) if $0=\ualpha(P_0)\leq \ualpha(P_+)$ then $\lim_{q\to 1-}f(z;q)
  = f(z;1)$ locally uniformly in a disk centered at $0$;

  \medskip
  \noindent(ii) if $0=\oalpha(P_0)<\oalpha(P_+)$, 
  then $\lim_{q\to 1+} \inot{\widehat\calB_{P,q}f}(z;q)=f(z;1)$ locally uniformly in 
  a disk centered at $0$.
}

\bigskip

\Proof  
For any $A$ in $P$, set $Q_A=1+|P_A(1)|$. Since $P$ has a finite
support, $|P_A(q)|\leq Q_A$ for any $q$ in some neighborhood $\calU$ of $1$ in
the unit disk $\{\, q\in \CC\,:\, |q|\leq 1\,\}$.

\noindent (i) Since we are interested in a limit as $q$ tends to $1$ with
the modulus of $q$ being less than $1$, we choose $\calU$ in the unit disk 
centered at $0$ and small enough so that
$$
  \lambda=\min_{q\in \calU} \Bigl|\sum_{A\in P_0} P_A(q)\Bigr|\wedge 1
$$ 
does not vanish, which is posible under the assumptions of 
Proposition \MiscfnqToOne.

Using Puiseux's theorem, define $h(z)$ as the power series solution of
$$
  \lambda h(z)
  = \sum_{A\in P_+} Q_A z^a h(z)^\ell + \sum_{n\in \NN} Q_{(n;\sp)}z^n \, .
$$
The coefficients $h_n=[z^n]h$ satisfy the recursion
$$
  \lambda h_n
  =\sum_{A\in P_+} Q_A \sum_{n_1+\cdots+n_\ell=n-a} h_{n_1}\cdots h_{n_\ell}
  + Q_{(n;\sp)} \, .
$$
The power series $h$ is analytic in a neighborhood of the origin. Given the
basic recursion \basicRecursion, for any $q$ in $\calU$ we have
$$
  \lambda h_0=Q_{(0;\sp)} \geq |P_{(0;\sp)}|=|f_0(q)| \, . 
$$
In particular, since $\lambda$ is at most $1$, we have $|f_0(q)|\leq h_0$
for any $q$ in $\calU$.
Assume that we have proved that for any $q$ in $\calU$ and any $0\leq i<n$
the inequality $f_i(q)\leq h_i$ holds. Since any $q$ is in $\calU$ is of 
modulus less than $1$ and since $\ualpha(P_+)$ is nonnegative, the basic 
recursion \basicRecursion\ yields
$$\eqalign{
  \Bigl| \sum_{A\in P_0} P_A q^{\alpha_1 n} f_n(q)\Bigr|
  &{}\leq \sum_{A\in P_+} Q_A \sum_{n_1+\cdots+n_\ell=n-a} 
    h_{n_1}\cdots h_{n_\ell} +Q_{(n;\sp)} \cr
  &{}=\lambda h_n \, .\cr}
$$
Given how $\lambda$ is defined, this implies $|f_n(q)|\leq h_n$, and this
now holds for any nonnegative integer $n$.

If $z$ is in the disk of convergence of $h$ and $q$ is in $\calU$, then
$$
  \Bigl|f(z;q)-\sum_{0\leq n<N}f_n(q) z^n\Bigr|
  \leq \sum_{n\geq N} |z|^n h_n \, .
$$
Since Proposition \MiscfnqToOne\ implies $\lim_{q\to 1-} f_n(q)=f_n(1)$, it 
follows that
$$
  \limsup_{q\to 1-} \Bigl| f(z;q)-\sum_{0\leq n<N} f_n(1)z^n\Bigr|
  \leq \sum_{n\geq N} |z|^n h_n \, .
$$
This implies that $\lim_{q\to 1-}f(z;q)=f(z;1)$ and that $f(z;1)$ is
well defined. Moreover, by Puiseux's
theorem, $f(z;1)$ is analytic in a neighborhood of $0$.

\noindent(ii) We now take $\calU$ contained in the annulus defined 
by $1\leq |q|\leq 2$. 
As in the proof of Theorem \DSMainTh, let $H_P=H(P)$ and 
$h_P=h(P)$. Define the power series $g(z;p)=\sum_{n\geq 0} g_n(q)z^n$ by
$g_n(q)=q^{-H_Pn(n-h_P)/2}f_n$ so that $g(z;q)=\widehat\calB_{P,q}f(z)$. Similarly
to what we did to prove Theorem \DSMainTh, define the power series $h(z)$ by
$$
  h(z)=\sum_{A\in P_+} Q_A z^ah(z)^\ell+Q_\sp(z) \, .
$$
By Puiseux's theorem, $h$ is holomorphic in a neighborhood of the origin.
Set
$$
  C=1\vee \sup_{q\in \calU} \sup_{n\geq 0} 
  {1\over |\sum_{A\in P_0} P_A(q)q^{\alpha_1n}|} \, .
$$
Given $L=\max_{A\in P} \ell$, define also $c=C^L 2^{H_Ph_P^2}$. Clearly, for any
$q$ in $\calU$, the inequality $|g_0(q)|\leq h_0$ holds. Assume that we have 
proved
$|g_i(q)|\leq C c^i h_i$ for any $0\leq i<n$. Then, \DSgnRecursion\ and
Lemma \DSDeltaAMin\ imply that $|\sum_{A\in P_0} P_A q^{\alpha_1 n}g_n(q)|$ 
is at most
$$\displaylines{\quad
  c^{n-1} C^L \sum_{A\in P_+} |Q_A| |q|^{-H_Pa(a-h_P)/2}
  \sum_{n_1+\cdots+n_\ell=n-a} h_{n_1}\cdots h_{n_\ell} 
  \hfill\cr\hfill
  {}+ |q|^{-H_Pn(n-h_P)/2}Q_{(n;\sp)}
  \, .\quad\cr}
$$
Since $|q|$ is at most $2$ in $\calU$, we 
have $|q|^{-H_Pn(n-h_P)/2}\leq 2^{H_Ph_P^2/2}$ because $n(n-h_P)\geq -h_P^2/4$. 
Therefore, $|g_n(q)|$ is at most
$$\displaylines{\qquad
  c^{n-1}C^{L+1} 2^{H_Ph_P^2}
  \Bigl( \sum_{A\in P_+} |Q_A| \sum_{n_1+\cdots+n_\ell=n-a} 
  h_{n_1}\cdots h_{n_\ell}+Q_{(n;\sp)}\Bigr)
  \hfill\cr\hfill
  {}\leq Cc^n h_n \, . \qquad\cr}
$$
Thus, if $z$ is such that $cz$ is in the disk of convergence of $h$, 
we have
$$
  \Bigl| \widehat\calB_{P,q}f(z;q)-\sum_{0\leq i<N} g_i(q)z^i\Bigr|
  \leq C\sum_{n\geq N} |cz|^n h_n \, . 
$$
Since Proposition \MiscfnqToOne\ yields $\lim_{q\to 1+} g_i(q)=f_i(1)$, we
proved that $\lim_{q\to 1+}\widehat\calB_{P,q}f(z;q)=f(z;1)$.\hfill\qed
{"config": "arxiv", "file": "2006.09527/miscellanea1.tex"}