On the product of two $1$-$q$-summable series

TL;DR

This paper investigates the properties of a $q$-analog of Borel-Laplace summation, demonstrating that the product of $q$-summable solutions of certain $q$-difference equations remains $q$-summable and that the summation process acts as a ring morphism.

Contribution

It proves that the $q$-summation of the product of two $q$-summable series equals the product of their $q$-sums, supporting the conjecture that $q$-summation is a ring morphism.

Findings

The product of two $q$-summable series is $q$-summable.

The $q$-sum of the product equals the product of the $q$-sums.

The $q$-summation induces a field morphism under certain conditions.

Abstract

In this paper we consider a $q$-analog of the Borel-Laplace summation process defined by Marotte and the second author, and consider two series solutions of linear $q$-difference equations with slopes $0$ and $1$. The latter are $q$-summable and we prove that the product of the series is $q$-(multi)summable and its $q$-sum is the product of the $q$-sum of the two series. This is a first step in showing the conjecture that the $q$-summation process is a morphism of rings. We prove that the $q$-summation does induce a morphism of fields by showing that if the inverse of the $q$-Euler series is $q$-summable, then its $q$-sum is not the inverse of the $q$-sum of the $q$-Euler series.

Equations

$${\mathcal{S}}^d(f)(x):={\int }_0^{\infty e^{\mathbf{i}d}}\frac{e^{-\xi /x}}{1+\xi }d\xi .$$

$${\mathcal{S}}_q^d(g):=\frac{1}{\sqrt{2\pi \ln (q)}}{\int }_0^{\infty e^{\mathbf{i}d}}\frac{e^{-(\log (\frac{x}{\sqrt{q}\xi }){)}^2/2\log (q)}}{1+\xi }\frac{d\xi }{\xi },d\not\equiv \pi [2\pi ].$$

$${\mathcal{S}}_q^d(f_1{f}_2)={\mathcal{S}}_q^d(f_1){\mathcal{S}}_q^d(f_2).$$

$$\begin{array}[]{llll}\hat{\mathcal{B}}_{q;1}:&{\mathbb{C}}[[x]]&\to&{\mathbb{C}}[[\xi]]\\ &\displaystyle\sum_{n\geq 0}a_{n}\,x^{n}&\displaystyle\mapsto&\displaystyle\sum_{n\geq 0}a_{n}\,q^{-n(n-1)/2}\,\xi^{n}\end{array}$$

$$\begin{array}[]{llll}\hat{\mathcal{L}}_{q;1}:&{\mathbb{C}}[[\xi]]&\to&{\mathbb{C}}[[x]]\\ &\displaystyle\sum_{n\geq 0}a_{n}\,\xi^{n}&\displaystyle\mapsto&\displaystyle\sum_{n\geq 0}a_{n}\,q^{n(n-1)/2}\,x^{n}\,.\end{array}$$

$${\hat{\mathcal{B}}}_{q;1}{x}^j{\sigma }_q^m=q^{-j(j-1)/2}{\xi }^j{\sigma }_q^{m-j}{\hat{\mathcal{B}}}_{q;1},{\hat{\mathcal{L}}}_{q;1}{\xi }^j{\sigma }_q^m=q^{j(j-1)/2}{x}^j{\sigma }_q^{m+j}{\hat{\mathcal{L}}}_{q;1}.$$

$$|\xi |\ge R⟹|\varphi (\xi )|\le C{e}^{(\log |A\xi |{)}^2/(2\log (q))}.$$

$$\begin{array}[]{llll}{\mathcal{B}}_{q;1}:&{\mathbb{C}}\{\xi\}&\rightarrow&{\mathbb{E}}_{q;1}\\ &f&\mapsto&\frac{1}{\sqrt{2\pi\log(q)}\,\mathbf{i}}\,\displaystyle\int_{\partial^{+}\widetilde{D}_{r}}e^{(\log(\frac{x}{\sqrt{q}\xi}))^{2}/(2\log(q))}\,f(x)\,\frac{dx}{x}\end{array}$$

$$\begin{array}[]{llll}{\mathcal{L}}_{q;1}^{d}:&{\mathbb{E}}_{q;1}&\rightarrow&{\mathbb{C}}\{\xi\}\\ &\phi&\mapsto&\frac{1}{\sqrt{2\pi\log(q)}}\,\displaystyle\int_{0}^{\infty e^{\mathbf{i}d}}e^{-(\log(\frac{x}{\sqrt{q}\xi}))^{2}/(2\log(q))}\,\phi(\xi)\,\frac{d\xi}{\xi}\,.\end{array}$$

$${\int }_{-\infty }^{+\infty }{e}^{-a(t+b{)}^2}dt=\sqrt{\frac{\pi }{a}}.$$

$${\int }_{{\partial }^{+}{\tilde{D}}_r}{e}^{(\log (x/\sqrt{q}){)}^2/(2\log (q))}{x}^n\frac{dx}{x}=\sqrt{2\pi \log (q)}\mathbf{i}{q}^{-n(n-1)/2}$$

$${\int }_0^{+\infty }{e}^{-(\log (\sqrt{q}\xi ){)}^2/(2\log (q))}{\xi }^n\frac{d\xi }{\xi }=\sqrt{2\pi \log (q)}{q}^{n(n-1)/2}.$$

$${\tilde{D}}_R=\{x\in {\tilde{\mathbb{C}}}^{\ast }:|x|<R\},V_{\epsilon }^d=\left\{\xi \in {\tilde{\mathbb{C}}}^{\ast }:|\arg (\xi )-d|<\epsilon \right\}$$

$$\underset{d^{\prime }\in (d-\epsilon ,d+\epsilon )}{\sup }\left(\underset{(r,t)\in (0,R)\times \mathbb{R}}{\sup }\left|f(r{e}^{\mathbf{i}(t+d^{\prime })})\right|e^{-t^2/(2\log (q))}\right)<+\infty ;$$

$$\max \left(\underset{\xi \in V_{\epsilon }^d,|\xi |\le R}{\sup }|\varphi (\xi )|,\underset{\xi \in V_{\epsilon }^d,|\xi |>R}{\sup }|\varphi (\xi )|e^{-(\log |A\xi |{)}^2/(2\log (q))}\right)<+\infty .$$

$$\begin{array}[]{llll}{\mathcal{B}}_{q;1}:&{\mathbb{O}}_{q;1}^{d}&\rightarrow&{\mathbb{E}}_{q;1}^{d}\\ {\mathcal{L}}^{d}_{q;1}:&{\mathbb{E}}_{q;1}^{d}&\rightarrow&{\mathbb{O}}_{q;1}^{d}.\end{array}$$

$${\hat{\mathcal{B}}}_{q;k}={\hat{\mathcal{B}}}_{q^{1/k};1},{\hat{\mathcal{L}}}_{q;k}={\hat{\mathcal{L}}}_{q^{1/k};1}.$$

$$X_{q;k}=X_{q^{1/k};1},X\in \left\{\hat{\mathcal{B}},\hat{\mathcal{L}};\mathcal{B},{\mathcal{L}}^d,\mathbb{C}[[x]],\mathbb{E},{\mathbb{O}}^d,{\mathbb{E}}^d\right\}.$$

$${\mathcal{B}}_{q;k}\circ {\mathcal{L}}_{q;k}^d=\mathbf{Id}{|}_{{\mathbb{E}}_{q;k}^d},{\mathcal{L}}_{q;k}^d\circ {\mathcal{B}}_{q;k}=\mathbf{Id}{|}_{{\mathbb{O}}_{q;k}^d}.$$

$${\mathbb{O}}_{q;k}^d⊊{\mathbb{O}}_{q;k^{\prime }}^d,{\mathbb{E}}_{q;k}^d⊊{\mathbb{E}}_{q;k^{\prime }}^d$$

$${\mathcal{S}}_{q;\vec{s}}^d(f):={\mathcal{L}}_{q;{\tilde{s}}_r}^d\circ \cdots \circ {\mathcal{L}}_{q;{\tilde{s}}_1}^d\circ {\mathcal{C}}^d\circ {\hat{\mathcal{B}}}_{q;s_1}(f).$$

$${\mathcal{B}}_{q;{\tilde{s}}_j}\circ \cdots \circ {\mathcal{B}}_{q;{\tilde{s}}_r}^d\circ {\mathcal{S}}_{q;\vec{s}}^d(f)={\mathcal{L}}_{q;{\tilde{s}}_{j-1}}^d\circ \cdots \circ {\mathcal{L}}_{q;{\tilde{s}}_1}^d\circ {\mathcal{C}}^d\circ {\hat{\mathcal{B}}}_{q;s_1}(f)\in {\mathbb{O}}_{q;{\tilde{s}}_{j-1}}^d\cap {\mathbb{E}}_{q;{\tilde{s}}_j}^d.$$

$$L=a_n{\sigma }_q^n+\cdots +a_0\in \mathbb{C}\{x\}[{\sigma }_q],$$

$$\bigcup _{j=0}^n\left\{(j,k)\in \mathbb{N}\times [0,+\infty ):k\ge v_0(a_j)\right\}\subset {\mathbb{R}}^2,$$

$$L_{a}y=1,\mathrm{where}{L}_a=x{\sigma }_q+a.$$

$${\mathcal{S}}_q^d(E_{a,q})=\frac{1}{\sqrt{2\pi \log (q)}}{\int }_0^{\infty e^{\mathbf{i}d}}{e}^{-(\log (\frac{x}{\sqrt{q}\xi }){)}^2/(2\log (q))}\frac{1}{\xi +a}\frac{d\xi }{\xi }.$$

$$\frac{1}{\sqrt{2\pi \log (q)}}{\int }_{{\gamma }_R}{e}^{-(\log (\frac{x}{\sqrt{q}\xi }){)}^2/(2\log (q))}\frac{1}{\xi +a}\frac{d\xi }{\xi }=\frac{2\mathbf{i}\pi }{\sqrt{2\pi \log (q)}}A,$$

$$\begin{array}[]{ll}&\frac{1}{\sqrt{2\pi\log(q)}}\,\displaystyle\int_{0}^{\infty e^{\mathbf{i}d}}\left|e^{-(\log(\frac{x}{\sqrt{q}\xi}))^{2}/(2\log(q))}\,\frac{(-1)^{m}}{(\xi+a)^{m+1}}\,\frac{d\xi}{\xi}\right|\\ \leq&\frac{1}{\sqrt{2\pi\log(q)}}\,\displaystyle\int_{0}^{\infty e^{\mathbf{i}d}}\left|e^{-(\log(\frac{x}{\sqrt{q}\xi}))^{2}/(2\log(q))}\,\,\frac{d\xi}{M^{m+1}\xi}\right|\\ \leq&\infty.\end{array}$$

$$\begin{array}[]{ll}\mathcal{S}_{q}^{d}(E^{[m]}_{a,q})&=\frac{1}{\sqrt{2\pi\log(q)}}\,\displaystyle\int_{0}^{\infty e^{\mathbf{i}d}}e^{-(\log(\frac{x}{\sqrt{q}\xi}))^{2}/(2\log(q))}\,\frac{(-1)^{m}}{(\xi+a)^{m+1}}\,\frac{d\xi}{\xi}\\ &=\partial_{a}^{m}\left(\frac{1}{\sqrt{2\pi\log(q)}}\,\displaystyle\int_{0}^{\infty e^{\mathbf{i}d}}e^{-(\log(\frac{x}{\sqrt{q}\xi}))^{2}/(2\log(q))}\,\frac{1}{\xi+a}\,\frac{d\xi}{\xi}\right)\\ &=\partial_{a}^{m}\mathcal{S}_{q}^{d}(E_{a,q}).\end{array}$$

$$f=\sum _{i=0}^k{f}_i\times E_{a_i,q}^{[m_i]}$$

Full text

This project has received funding from the ANR de rerum natura ANR-19-CE40-0018.

1. 1 Notations and some prelimilary results

2. 1.1 Formal transformations

3. 1.2 Analytic transformations

4. 1.3 Transformations of arbitrary order

5. 1.4 Multisummation

6. 1.5 Linear $q$-difference equations and $q$-multissummation

7. 2 Product theorem

8. 2.1 Case of the Euler series

9. 2.2 Variant of the Euler series

10. 2.3 Product theorem

11. 3 Inverse of the $q$-Euler series

Introduction

In this paper, we are interested in the algebraic properties of a $q$-analogue of the Borel-Laplace summation process defined in [MZ00]. Before going further in the $q$-world, let us make a short overview of the theory in the setting of linear differential equations. We refer for instance to [Bal06, Bal08] for a complete description of the theory.

Consider a meromorphic linear differential equation. We have the coexistence of divergent formal power series and integral solutions. For instance, the Euler equation $x^{2}\partial_{x}y+y=1$ admits the Euler series $f:=\displaystyle\sum_{n=0}^{\infty}(-1)^{n}n!x^{n}$ as formal power series solution. On the other hand, there are integral solutions, such as

[TABLE]

The path of integration has to be understood as the half line in ${\mathbb{C}}$ of complex numbers of argument $d\in{\mathbb{R}}$. The latter integral is well defined when the path of integration does not pass through the pole $\xi=-1$, that is when $d\not\equiv\pi[2\pi]$. We may prove that the function is analytic on the sector $\arg(x)\in(d-\pi/2,d+\pi/2)$ and is asymptotic to $f$ is a certain sens. More generally, given a formal power series solution of a linear differential equation with coefficients that are germs of meromorphic functions at [math], we may for convenient $d\in{\mathbb{R}}$, construct an integral solution using Borel and Laplace transformations in direction $d$. The map $f\mapsto\mathcal{S}^{d}(f)$ that sends a formal power series to the inegral solution induces a morphism of fields. Moreover, it leaves the germs of meromorphic functions at [math] invariant and commutes with the derivation, that is $\mathcal{S}^{d}(\partial_{x}f)=\partial_{x}\mathcal{S}^{d}(f)$.

Let us now consider the case of $q$-difference equations. Let us fix $q>1$, define the $q$-difference operator $\sigma_{q}y(x):=y(qx)$, and consider the $q$-Euler equation $x\sigma_{q}y+y=1$. It admits the $q$-Euler series $g:=\displaystyle\sum_{n=0}^{\infty}(-1)^{n}q^{n(n-1)/2}x^{n}$ as divergent formal power series. An integral solution is given by

[TABLE]

We may prove that the latter is meromorphic on the Riemann surface of the logarithm and defines a multivalued complex function. Unfortunately, contrary to the differential case, there are many different $q$-summation process, due to the nonuniqueness of the $q$-analogue of the exponential. Each of the $q$-sum has advantages in certain contexts, see for instance the work of the two authors, Ramis, Sauloy, and more exhaustively, [Abd60, Abd64, Ram92, Zha99, MZ00, Zha00, Zha01, RZ02, Zha02, Zha03, Zha04, DVZ09, RSZ13, Dre15, DE16]. In this paper, we use a $q$-summation process introduced in [MZ00]. It was proved that to $f$, a formal power series solution of a linear meromorphic $q$-difference equation, we may associate, using $q$-analogues of the Borel-Laplace summation in the direction $d\in{\mathbb{R}}$ for convenient $d$, a multivalued meromorphic solution $\mathcal{S}_{q}^{d}(f)$. Let us by ${\mathbb{C}}[[x]]_{q}^{d}$ the space of series where $f\mapsto\mathcal{S}_{q}^{d}(f)$, is well defined. The authors proved that the $q$-summation process $f\mapsto\mathcal{S}_{q}^{d}(f)$, satisfies the following algebraic properties, see Proposition 1.5 below,

• •

If $f_{1},f_{2}\in{\mathbb{C}}[[x]]_{q}^{d}$, then $f_{1}+f_{2}\in{\mathbb{C}}[[x]]_{q}^{d}$ and $\mathcal{S}_{q}^{d}(f_{1}+f_{2})=\mathcal{S}_{q}^{d}(f_{1})+\mathcal{S}_{q}^{d}(f_{2})$;

• •

If $f\in{\mathbb{C}}[[x]]_{q}^{d}$, then $\sigma_{q}(f)\in{\mathbb{C}}[[x]]_{q}^{d}$ and $\mathcal{S}_{q}^{d}(\sigma_{q}(f))=\sigma_{q}\left(\mathcal{S}_{q}^{d}(f)\right)$;

• •

For all convergent series $f$, if $g\in{\mathbb{C}}[[x]]_{q}^{d}$, then $fg\textbullet in{\mathbb{C}}[[x]]_{q}^{d}$ and $\mathcal{S}_{q}^{d}(fg)=f\mathcal{S}_{q}^{d}(g)$.

In this paper we prove that under certain assumptions, the $q$-summation process commutes with the product, see Theorem 2.7 for a precise statement.

Theorem**.**

Let $f_{1},f_{2}$ be series that are solutions of linear $q$-difference equations with slopes [math] and $1$. If $f_{1}$ and $f_{2}$ belong to ${\mathbb{C}}[[x]]_{q}^{d}$, then $f_{1}f_{2}\in{\mathbb{C}}[[x]]_{q}^{d}$ and

[TABLE]

We conjecture that $f\mapsto\mathcal{S}_{q}^{d}(f)$ defines a morphism of rings and this result is a first step in that direction. Proving the latter conjecture could allows us to define the $q$-analogue of the Stokes operators in an integral way as in the differential case, and prove a $q$-analogue of the Ramis density theorem. Note that this question has been considered in another point of view in [RSZ13] and the comparison of the two approaches would be interesting.

A natural question is whether the $q$-summation process could define a morphism of fields. We answer negatively to this question in Section 3 where the inverse of the $q$-Euler series is considered. We prove that if its $q$-sum is defined, then it is not the inverse of the $q$-sum of the $q$-Euler series.

The paper is organized as follows. In Section 1, we introduce the $q$-analogues of the Borel and Laplace transformations and prove some of their basic properties. We also introduce the space of $q$-multisummable series in direction $d$, that is the set of series where the map $f\mapsto\mathcal{S}_{q}^{d}(f)$ is well defined. An example of such series is the ring of series solutions of linear $q$-difference equations. Section 2 is devoted to the proof of the product theorem. This is done by proving the result for series that are a variant of the $q$-Euler series, use the fact that series solutions of linear $q$-difference equation with slopes [math] and $1$, admit a certain decomposition into variant of $q$-Euler series, and finally, use the algebraic properties already known to conclude. In Section 3, we study the inverse of the $q$-Euler series and prove that if it $q$-multisummable, then its $q$-sum could not be the inverse of the $q$-sum of the Euler series.

1. Notations and some prelimilary results

This section is devoted to review the $q$-summation theory developed in our previous papers [Zha99, MZ00]. By taking inspiration from a Phragmén–Lindelöf principle stated in [FZ99] for the space of $q$-summable functions, we shall build a $q$-Borel-Laplace summation process using two functional spaces (or saying sheaves) that will be denoted as ${\mathbb{O}}_{q;1}^{d}$ and ${\mathbb{E}}_{q;1}^{d}$; see Proposition 1.1. In most cases, we shall directly refer to the above-mentioned works [Zha99, MZ00] unless we believe some additional precisions or mentions are necessary.

1.1. Formal transformations

As usual, we will denote by ${\mathbb{C}}[[x]]$ the ${\mathbb{C}}$-vector space of all formal power series in the variable $x$ with coefficients in ${\mathbb{C}}$. On the lines of what is done in the classic Borel-Laplace summation theory (see, for example, [Bal06]), we first recall the following couple $(\hat{\mathcal{B}}_{q;1},\hat{\mathcal{L}}_{q;1})$ of formal $q$-Borel and $q$-Laplace transforms, which are really ismorphisms of ${\mathbb{C}}$-vector spaces:

[TABLE]

and

[TABLE]

By direct computations, one can find the following identities for $j$, $m\in{\mathbb{Z}}$:

[TABLE]

Let ${\mathbb{C}}\{x\}$ be the sub-space of all power series whose radius of convergence is strictly positive. Set ${\mathbb{C}}[[x]]_{q;1}=\hat{\mathcal{L}}_{q;1}\left({\mathbb{C}}\{\xi\}\right)$ the set of $q$-Gevrey series of order one.

By following [Ram92], we say that the entire function $\phi$ is called to have a $q$-exponential growth of ordre (at most) one at infinity in the following sense: for some (or any) $R>0$, one can find $C$, $A>0$ such that, for every $\xi\in{\mathbb{C}}$,

[TABLE]

The ring of such functions is called ${\mathbb{E}}_{q;1}$.

Let us denote by $\tilde{\mathbb{C}}^{*}$ the Riemann surface of the logarithm. Given $r\in(0,\infty)$, let us denote by $\partial^{+}\widetilde{D}_{r}$, the path parameterized by \left\{\begin{array}[]{lll}{\mathbb{R}}&\rightarrow&\widetilde{\mathbb{C}}^{*}\\ t&\mapsto&rq^{\mathbf{i}t}\end{array}\right.. Define the linear maps ${\mathcal{B}}_{q;1}$ and ${\mathcal{L}}_{q;1}^{d}$ as follows:

[TABLE]

and

[TABLE]

In the above, $\log$ denotes the principal branch of the logarithm on the Riemann surface $\tilde{\mathbb{C}}^{*}$, $r>0$ is chosen to be smaller than the radius of convergence of $f$, and $d\in{\mathbb{R}}$ may be arbitrary.

The integrals appeared in (1.4) and (1.5) are related with the well-konwn Gauss integral ($a>0$, $b\in{\mathbb{R}}$) :

[TABLE]

Using suitable changes of variables, it follows from the above that, for all integer $n\in{\mathbb{Z}}$ for all $r>0$:

[TABLE]

and

[TABLE]

Thus, restricting to the convergent power series spaces ${\mathbb{C}}\{x\}$ and ${\mathbb{E}}_{q;1}$, the linear maps $\hat{\mathcal{B}}_{q;1}$ and $\hat{\mathcal{L}}_{q;1}$ defined in (1.1) and (1.2) coincide with ${\mathcal{B}}_{q;1}$ and ${\mathcal{L}}^{d}_{q;1}$.

1.2. Analytic transformations

From now on, we will fix a real $d\in{\mathbb{R}}$, that will be identified with the direction of argument $d$ coming from the “origin” on the Riemann surface $\tilde{\mathbb{C}}^{*}$, namely $(0,\infty e^{\mathbf{i}d})$. For simplify, define

[TABLE]

for $R>0$ and $\varepsilon>0$. In the same time, both spaces ${\mathbb{C}}\{x\}$ and ${\mathbb{E}}_{q;1}$ are respectively extended into ${\mathbb{O}}_{q;1}^{d}$ and ${\mathbb{E}}_{q;1}^{d}$ as follows:

1. (1)

${\mathbb{O}}_{q;1}^{d}$ is composed of all analytic functions $f$ in some domain $\tilde{D}_{R}$ such that, for some suitable $\varepsilon>0$:

[TABLE] 2. (2)

${\mathbb{E}}_{q;1}^{d}$ is composed of all analytic functions $\phi$ in some domain $V_{\varepsilon}^{d}$ such that, for some suitable $A>0$ and $R>0$:

[TABLE]

As shows the following proposition, the maps ${\mathcal{B}}_{q;1}$ and ${\mathcal{L}}^{d}_{q;1}$ can be extended to ${\mathbb{O}}_{q;1}^{d}$ and ${\mathbb{E}}_{q;1}^{d}$ respectively.

Proposition 1.1**.**

The maps ${\mathcal{B}}_{q;1}$ and ${\mathcal{L}}_{q;1}^{d}$ defined by the integrals given in (1.4) and (1.5) may be extended to ${\mathbb{O}}_{q;1}^{d}$ and ${\mathbb{E}}_{q;1}^{d}$ respectively and are bijections that are each other inverses

[TABLE]

Proof.

By [MZ00], Lemmas 1.3.1 and 1.3.4 the map ${\mathcal{B}}_{q;1}$ (resp. ${\mathcal{L}}_{q;1}^{d}$) is well defined on ${\mathbb{O}}_{q;1}^{d}$ (resp. ${\mathbb{E}}_{q;1}^{d}$), and ${\mathcal{B}}_{q;1}({\mathbb{O}}_{q;1}^{d})\subset{\mathbb{E}}_{q;1}^{d}$ (resp. ${{\mathcal{L}}_{q;1}^{d}({\mathbb{E}}_{q;1}^{d})\subset{\mathbb{O}}_{q;1}^{d}}$). The fact that ${\mathcal{B}}_{q;1}\circ{\mathcal{L}}_{q;1}^{d}$ is the identity on ${\mathbb{E}}_{q;1}^{d}$ is [MZ00], Theorem 1.3.7. We may deduce also from the latter proof that ${\mathcal{L}}_{q;1}^{d}\circ{\mathcal{B}}_{q;1}$ is the identity on ${\mathbb{O}}_{q;1}^{d}$. ∎

1.3. Transformations of arbitrary order

Let $k>0$, write $q^{1/k}=e^{(\log(q))/k}$, and define the formal $q$-Borel and $q$-Laplace transforms of order $k$ in the following manner:

[TABLE]

Furthermore, by replacing $q$ with $q^{1/k}$, one sets:

[TABLE]

In this way, the maps ${\mathcal{B}}_{q;k}$ and ${\mathcal{L}}_{q;k}^{d}$ are bijections between ${\mathbb{O}}_{q;k}^{d}$ and ${\mathbb{E}}_{q;k}^{d}$ such that

[TABLE]

For $k^{\prime}>k>0$, it is easy to obtain from (1.6) that ${\mathbb{C}}[[x]]_{q;k}\supsetneq{\mathbb{C}}[[x]]_{q;k^{\prime}}$ and, further,

[TABLE]

for any $d\in{\mathbb{R}}$.

1.4. Multisummation

As in [MZ00], Section 2.3.3, we will denote by $\Omega^{+}$ the set of finite sequences of strictly increasing elements of ${\mathbb{Q}}_{>0}$. Let $\Omega^{+*}=\Omega^{+}\setminus\{\varnothing\}$. Given $(s_{1},...,s_{r})\in\Omega^{+*}$, we define $\widetilde{s}_{1},\dots,\widetilde{s}_{r}\in{\mathbb{Q}}_{>0}$ as $\widetilde{s}_{i}=\frac{1}{\frac{1}{s_{i}}-\frac{1}{s_{i+1}}}$, where we made the convention that $s_{r+1}=\infty$.

Example 1.2*.*

If we consider $(s_{1},s_{2}):=(1,2)\in\Omega^{+*}$ the associated sentence is $(\widetilde{s}_{1},\widetilde{s}_{2}):=(2,2)$.

Let ${\mathcal{C}}^{d}$ denotes the analytic continuation of any germ of analytic functions at zero along the direction of argument $d$.

Definition 1.3* ([MZ00], Definition 2.3.4).*

Let $d\in{\mathbb{R}}$, $\vec{s}=(s_{1},...,s_{r})\in\Omega^{+*}$. The power series $f\in{\mathbb{C}}[[x]]$ is $q$-multisummable of order $\vec{s}$ in the direction of argument $d$ if and only if, the following conditions are fulfilled:

1. (1)

$\hat{\mathcal{B}}_{q;s_{1}}(f)\in{\mathbb{C}}\{\xi\}$ and ${\mathcal{C}}^{d}\circ\hat{\mathcal{B}}_{q;s_{1}}(f)\in{\mathbb{E}}_{q;\widetilde{s}_{1}}^{d}$. 2. (2)

for $1\leq j<r$, ${\mathcal{L}}_{q;\widetilde{s}_{j}}^{d}\circ\dots\circ{\mathcal{L}}_{q;\widetilde{s}_{1}}^{d}\circ{\mathcal{C}}^{d}\circ\hat{\mathcal{B}}_{q;s_{1}}(f)\in{\mathbb{E}}_{q;\widetilde{s}_{j+1}}^{d}$.

Let ${\mathbb{C}}[[x]]_{q;\vec{s}}^{d}$ be the set of all $q$-multisummable power series of order $\vec{s}$ in the direction of argument $d$. For $f\in{\mathbb{C}}[[x]]_{q;\vec{s}}^{d}$, define its (multi-)sum function ${\mathcal{S}}_{q;\vec{s}}^{d}(f)$ in the direction $d$:

[TABLE]

We list first properties of the intermediate $q$-sum that will be very important in the sequel. It is a straightforward consequence of the definition combined with Proposition 1.1.

Proposition 1.4**.**

1. Let $d\in{\mathbb{R}}$, $\vec{s}=(s_{1},\dots,s_{r})\in\Omega^{+*}$ and $f\in{\mathbb{C}}[[x]]_{q;\vec{s}}^{d}$.
2. (1)

One has ${\mathcal{S}}_{q;\vec{s}}^{d}(f)\in{\mathbb{O}}_{q;\widetilde{s}_{r}}^{d}$. 3. (2)

${\mathcal{B}}_{q;\widetilde{s}_{1}}\circ\dots\circ{\mathcal{B}}_{q;\widetilde{s}_{r}}\circ\mathcal{S}_{q;\vec{s}}^{d}(f)=\widehat{{\mathcal{B}}}_{q;s_{1}}(f)$. 4. (3)

For $j=2,\dots,r$,

[TABLE]

Let $\vec{s}\subset\vec{t}\in\Omega^{+*}$, $d\in{\mathbb{R}}$, and assume that $\hat{f}\in{\mathbb{C}}[[x]]_{q;\vec{s}}^{d}$. Then by [MZ00], Lemma 2.4.1, $\hat{f}\in{\mathbb{C}}[[x]]_{q;\vec{t}}^{d}$ and $\mathcal{S}_{q;\vec{s}}^{d}(\hat{f})=\mathcal{S}_{q;\vec{t}}^{d}(\hat{f})$. Then, we may omit $\vec{s}$ and write $\mathcal{S}_{q}^{d}$ instead of ${\mathcal{S}}_{q;\vec{s}}^{d}$. For $d\in{\mathbb{R}}$, we let ${\mathbb{C}}[[x]]_{q}^{d}=\displaystyle\bigcup_{\vec{s}\in\Omega^{+}}{\mathbb{C}}[[x]]_{q;\vec{s}}^{d}$. We say that $d\in{\mathbb{R}}$ is a singular direction of $f\in{\mathbb{C}}[[x]]$, if $f\notin{\mathbb{C}}[[x]]_{q}^{d}$.

We say that $f\in{\mathbb{C}}[[x]]$ is a $q$-multisummable series, and we write $f\in\mathcal{MS}_{q}$, if the set of its singular direction is finite modulo $2\pi{\mathbb{Z}}$. The set of $q$-multisummable series, form a ${\mathbb{C}}\{x\}$-module and the $q$-summation process is a morphism, as shows the following proposition. Let ${\mathbb{O}}_{q}^{d}:=\displaystyle\bigcup_{s\in{\mathbb{Q}}_{>0}}{\mathbb{O}}_{q;s}^{d}$.

Proposition 1.5**.**

Let $d\in{\mathbb{R}}$. The $q$-summation process \left\{\begin{array}[]{lll}{\mathbb{C}}[[x]]_{q}^{d}&\rightarrow&{\mathbb{O}}_{q}^{d}\\ f&\mapsto&{\mathcal{S}}_{q}^{d}(f)\end{array}\right. satisfies the following algebraic properties:

• •

For all $f_{1},f_{2}\in{\mathbb{C}}[[x]]_{q}^{d}$, we have $f_{1}+f_{2}\in{\mathbb{C}}[[x]]_{q}^{d}$, and $\mathcal{S}_{q}^{d}(f_{1}+f_{2})=\mathcal{S}_{q}^{d}(f_{1})+\mathcal{S}_{q}^{d}(f_{2})$;

• •

For all $f\in{\mathbb{C}}[[x]]_{q}^{d}$, we have $\sigma_{q}(f)\in{\mathbb{C}}[[x]]_{q}^{d}$, and $\mathcal{S}_{q}^{d}(\sigma_{q}(f))=\sigma_{q}\left(\mathcal{S}_{q}^{d}(f)\right)$;

• •

For all $f\in{\mathbb{C}}\{x\}$, $g\in{\mathbb{C}}[[x]]_{q}^{d}$, we have $fg\in{\mathbb{C}}[[x]]_{q}^{d}$ and $\mathcal{S}_{q}^{d}(fg)=f\,\mathcal{S}_{q}^{d}(g)$.

1.5. Linear $q$-difference equations and $q$-multissummation

Let $L$ be a $q$-difference operator of the following form:

[TABLE]

where $a_{n}a_{0}\neq 0$, $n>0$. The Newton polygon associated with $L$, denoted by $\mathcal{N\!P}(L)$, is the convex hull, in the plane ${\mathbb{R}}^{2}$, of the finite set of ascending half-lines

[TABLE]

where $v_{0}$ denotes the $x$-adic valuation – it is worth recalling that $v_{0}(a_{j})=+\infty$ when $a_{j}=0$. Let $(d_{1},n_{1}),\dots,(d_{r+1},n_{r+1})$ with ${d_{1}<\dots<d_{r+1}}$, be a minimal subset of ${\mathbb{N}}\times{\mathbb{N}}$ for the inclusion, such that the lower part of the boundary of $\mathcal{N\!P}(L)$ is the convex hull of the finite set of the points having as their coordinates $(d_{1},n_{1}),\dots,(d_{r+1},n_{r+1})$. Letting $\displaystyle s_{j}=\frac{n_{j+1}-n_{j}}{d_{j+1}-d_{j}}$, one gets the (finite) slopes of $\mathcal{N\!P}(L)$. Note that, by construction, the sequence $(s_{j})_{1\leq j\leq r}$ is strictly increasing.

With regard to the summability of the formal power series solutions of linear $q$-difference equations, one can quote the following result.

Theorem 1.6** ([MZ00], Theorem 3.3.5).**

Let $L$ be as in (1.9), and let $f\in{\mathbb{C}}[[x]]$. Suppose that the associated Newton polygon $\mathcal{N\!P}(L)$ has the integers $s_{1}<...<s_{r}$ as all its positive slopes. If $Lf\in{\mathbb{C}}\{x\}$, then $f\in\mathcal{MS}_{q}$. More precisely, for all $d\in{\mathbb{R}}$ that is not a singular direction, one has $f\in{\mathbb{C}}[[x]]_{q;(s_{1},\dots,s_{r})}^{d}$. Furthermore, for all $d\in{\mathbb{R}}$ that is not a singular direction, $\mathcal{S}_{q}^{d}(f)$ is solution of (1.9).

Example 1.7*.*

Let $a\in{\mathbb{C}}^{*}$ and let $E_{a,q}$ be the unique series that is solution of the following first order linear $q$-difference equation:

[TABLE]

When $a=1$ we recover the $q$-Euler equation and $E_{a,q}$ is the $q$-Euler series. With (1.3) we find that $(\xi+a)\widehat{{\mathcal{B}}}_{q;1}(E_{a,q})=1$. Then, $\widehat{{\mathcal{B}}}_{q;s_{r}}(E_{a,q})\in{\mathbb{C}}\{x\}\cap{\mathbb{E}}_{q;1}^{d}$, for all $d\in{\mathbb{R}}$ such that $\arg(-a)\neq d$. We therefore obtain

[TABLE]

By (1.3), $\mathcal{S}_{q}^{d}(E_{a,q})$ is solution of (1.10). In virtue of Proposition 1.4, we find that $\mathcal{S}_{q}^{d}(E_{a,q})\in{\mathbb{O}}_{q;1}^{d}$, and is therefore analytic on some domain of the form $\tilde{D}_{R}$ for some $R>0$. Since $\mathcal{S}_{q}^{d}(E_{a,q})$ is solution of (1.10) and $q>1$, we deduce that $\mathcal{S}_{q}^{d}(E_{a,q})$ is meromorphic on the Riemann surface of the logarithm.

When $\arg(-a)\neq d,d^{\prime}$, $\mathcal{S}_{q}^{d}(E_{a,q})$ and $\mathcal{S}_{q}^{d^{\prime}}(E_{a,q})$ are two meromorphic solutions, and we might compare the two functions. Assume that $d<d^{\prime}$. For $R>0$ let $\gamma_{R}$ be the path that goes from [math] to $Re^{\mathbf{i}d}$ in straight line, from $Re^{\mathbf{i}d}$ to $Re^{\mathbf{i}d^{\prime}}$ following positively the circle of center [math] and radius $R$, and coming from $Re^{\mathbf{i}d^{\prime}}$ to [math] in straight line. When $R$ is sufficiently big, residue theorem yields that

[TABLE]

where $A=\mathrm{res}(\xi^{-1}e^{-(\log(\frac{x}{\sqrt{q}\xi}))^{2}/(2\log(q))},-a)$ denotes the residue at $\xi=-a$. The latter is zero if and only if there exists $k\in{\mathbb{Z}}$ such that $\arg(-a)<d<d^{\prime}<\arg(-a)+2k\pi$. When $R$ goes to infinity, the integral from $Re^{\mathbf{i}d}$ to $Re^{\mathbf{i}d^{\prime}}$ tends to [math]. Then, when $R$ goes to infinity the left hand side of (1.12) tends to $\mathcal{S}_{q}^{d}(E_{a,q})-\mathcal{S}_{q}^{d^{\prime}}(E_{a,q})$ while the right hand side stay equal to $\frac{2\mathbf{i}\pi}{\sqrt{2\pi\log(q)}}A$. This shows that if for all $k\in{\mathbb{Z}}$, $\arg(-a)+2k\pi\notin(d,d^{\prime})$, then $\mathcal{S}_{q}^{d}(E_{a,q})$ and $\mathcal{S}_{q}^{d^{\prime}}(E_{a,q})$ are equal.

Example 1.8*.*

Let us see $E_{a,q}$ as a function of $a$ and for $m\in{\mathbb{N}}$, let $E^{[m]}_{a,q}=\partial_{a}^{m}E_{a,q}$ be the formal derivative. Obviously, the formal Borel transformation commutes with $\partial_{a}$ and we find for all $m$, $\hat{\mathcal{B}}_{q;1}(E^{[m]}_{a,q})=\partial_{a}^{m}\hat{\mathcal{B}}_{q;1}(E_{a,q})=\partial_{a}^{m}\frac{1}{\xi+a}=\frac{(-1)^{m}}{(\xi+a)^{m+1}}$. Consider $\mathcal{S}_{q}^{d}(E_{a,q})$ as an integral depending upon the parameter $a$ and let us study its differentialbility. Let $f(a,m,\xi,x):=\xi^{-1}e^{-(\log(\frac{x}{\sqrt{q}\xi}))^{2}/(2\log(q))}\,\frac{(-1)^{m}}{(\xi+a)^{m+1}}$. Let $d\in{\mathbb{R}}$, let us fix a compact $K\subset{\mathbb{C}}^{*}\setminus{\mathbb{R}}_{>0}e^{\mathbf{i}(d+\pi)}$ . Let $M:=\displaystyle\min_{\begin{subarray}{c}a\in K,\\ \xi\in{\mathbb{R}}_{>0}e^{\mathbf{i}d}\end{subarray}}(\xi+a)$. Note that $M>0$. Then, for all $m$, let us set $g(m,\xi,x):=\left|\xi^{-1}e^{-(\log(\frac{x}{\sqrt{q}\xi}))^{2}/(2\log(q))}\,M^{-m-1}\right|$. Then, we may locally dominate the following integral depending upon $a$

[TABLE]

Then the integral depending upon $a$ may be differentiate and we find, $E^{[m]}_{a,q}\in{\mathbb{C}}[[x]]_{q;1}^{d}$ and

[TABLE]

The following result, whose prove may be deduced from [DVZ09], Theorem 4.20 will be used toward the proof of our main result. Let us make the convention that $E^{0}_{a,q}=1$ for all $a\in{\mathbb{C}}^{*}$.

Theorem 1.9**.**

Let $L$ be as in (1.9), and let $f\in{\mathbb{C}}[[x]]$ be a solution of (1.9). Suppose that the associated Newton polygon $\mathcal{N\!P}(L)$ has slopes [math] and $1$. Then, there exists a decomposition of the form

[TABLE]

such that $f_{0},\dots,f_{k}\in{\mathbb{C}}\{x\}$, and $a_{i}\in{\mathbb{C}}^{*}$, $m_{i}\in{\mathbb{N}}$.

2. Product theorem

The goal of this section is to prove the main result of the paper, that is the product of two series solution of linear $q$-difference equations with slopes [math] and $1$ is $q$-multisummable, and the $q$-sum of the product is the product of the $q$-sums. By Theorem 1.9 two such series admits a decomposition involving convergent series and variants of $q$-Euler series. The strategy of the proof is to show the product theorem for series of the form $E^{[m]}_{a,q}$, $E^{[n]}_{b,q}$ where $a,b\in{\mathbb{C}}^{*}$, $m,n\in{\mathbb{N}}^{*}$, and then use the Theorem 1.9, together with the fact that the $q$-summation process is a morphism of ${\mathbb{C}}\{x\}$-modules, see Proposition 1.5, to deduce the result.

2.1. Case of the Euler series

We are first going to consider the particular case where series are of the form $E_{a,q}$, $E_{b,q}$ where $a,b\in{\mathbb{C}}^{*}$ are fixed complex numbers. So let us study the summability of the product $E_{a,q}E_{b,q}$. Now, the equation into the form $x\,\sigma_{q}E_{a,q}=1-aE_{a,q}$, $x\,\sigma_{q}E_{b,q}=1-bE_{b,q}$, consider the product of both sides of, and observe then that

[TABLE]

Furthermore, as $L_{a}L_{b}=L_{b}L_{a}$, we find $L_{a}L_{b}E_{a,q}=L_{b}1=x+b$ and $L_{a}L_{b}E_{b,q}=L_{a}1=x+a$. Since and $L_{a}L_{b}1=(x\sigma_{q}+a)(x+b)=qx^{2}+(a+b)x+ab$, it follows that $E_{a,q}E_{b,q}$ is solution to the following functional equation:

[TABLE]

It is obvious that the associated Newton polygon for $L_{a},L_{b}$ have a unique slope, that equals 1. One sees that the slopes of the Newton polygon of

[TABLE]

are $1$ and $2$. Consequently, by Theorem 1.6, one may expect to have $E_{a,q}E_{b,q}\in\mathcal{MS}_{q}$. Let

[TABLE]

Then for all $d\in{\mathbb{R}}\setminus S$, one has $E_{a,q}\in{\mathbb{C}}[[x]]_{q;1}^{d}$ and $E_{b,q}\in{\mathbb{C}}[[x]]_{q;1}^{d}$. The goal of that subsection is to prove:

Theorem 2.1**.**

Given $d\in{\mathbb{R}}\setminus S$, $E_{a,q}E_{b,q}\in{\mathbb{C}}[[x]]_{q;(1,2)}^{d}$ and for all $x\in\widetilde{{\mathbb{C}}}^{*}$, we have

[TABLE]

Since $E_{\star,q}=\displaystyle\sum_{n\geq 0}(-1)^{n}(\star)^{n}\,q^{n(n-1)/2}\,x^{n}$, it is straightforward to check that we have ${f_{1}=\hat{{\mathcal{B}}}_{q;1}(E_{a,q}E_{b,q})\in{\mathbb{C}}\{\zeta\}}$.

Lemma 2.2**.**

The power series $f_{1}$ represents the only analytic function on ${\mathbb{C}}\setminus(-aq^{\mathbb{N}},-bq^{\mathbb{N}})$ that satisfies the $q$-difference equation

[TABLE]

Furthermore, for any given $d\in{\mathbb{R}}\setminus S$, we have $f_{1}\in{\mathbb{E}}_{q;2}^{d}$.

Proof.

By considering the first relation in (1.3), one gets that

[TABLE]

By definition, $\hat{{\mathcal{B}}}_{q;1}(qx^{2}-ab)=x^{2}-ab$. Thus, applying $\hat{{\mathcal{B}}}_{q;1}$ in both sides of (2.1) yields the functional equation (2.2) for $f_{1}$.

By putting (2.2) into the following form:

[TABLE]

iterating this last relation shows that

[TABLE]

Thus, $f_{1}$ is analytic on the domain ${\mathbb{C}}\setminus(-aq^{\mathbb{N}},-bq^{\mathbb{N}})$. The uniqueness of $f_{1}$ comes from the fact that the associated homogeneous equation with (2.2), $(q^{-1}\zeta^{2}\sigma_{q}^{-1}-ab)y=0$, has not nontrivial analytic solution at $\zeta=0$.

To prove that $f_{1}\in{\mathbb{E}}_{q;2}^{d}$ for all $d\in{\mathbb{R}}\setminus S$, we may use the same reasoning as in [Dre15], Proposition 2.13, (3). ∎

Given any $d\in{\mathbb{R}}\setminus S$, let $f_{2}^{d}:={\mathcal{L}}_{q;2}^{d}(f_{1})$. Let $S^{\prime}=S\cup\{\arg(\sqrt{ab})+\pi{\mathbb{Z}}\}$.

Lemma 2.3**.**

If $d\in{\mathbb{R}}\setminus S$, one has for all $\zeta\in\tilde{\mathbb{C}}^{*}$

[TABLE]

Furthermore, when $d\in{\mathbb{R}}\setminus S^{\prime}$, we find $f_{2}^{d}\in{\mathbb{E}}_{q;2}^{d}$.

Remark 2.4*.*

We will prove in the sequel that for all $d\in{\mathbb{R}}\setminus S$, we have $f_{2}^{d}\in{\mathbb{E}}_{q;2}^{d}$.

Proof.

Let $q^{\prime}=q^{1/2}$, $d\in{\mathbb{R}}\setminus S$, and transform (2.2) into one $q^{\prime}$-difference equation as follows:

[TABLE]

With the help of (1.3), in which $q$ is replaced by $q^{\prime}$, we deduce that

[TABLE]

As $\displaystyle\frac{\xi^{2}-ab}{(\xi+a)(\xi+b)}=1-\frac{a}{\xi+a}-\frac{b}{\xi+b}$ we obtain that

[TABLE]

This shows (2.4) for all $d\in{\mathbb{R}}\setminus S$. Furthermore, by Proposition 1.4, $\mathcal{S}_{q;2}^{d}(E_{a,q^{1/2}})\in{\mathbb{O}}_{q;2}^{d}$, so it is defined in the neighborhood of [math] in the Riemann surface $\tilde{\mathbb{C}}^{*}$. By Theorem 1.6, $\mathcal{S}_{q;2}^{d}(E_{a,q^{1/2}})$ is solution of the same equation as $E_{a,q^{1/2}}$, which implies that we have $(x\sigma_{q^{\prime}}+a)\mathcal{S}_{q;2}^{d}(E_{a,q^{1/2}})=1$. We deduce similarly to the proof of [Dre15], Proposition 2.13, (3), that $\mathcal{S}_{q;2}^{d}(E_{a,q^{1/2}})\in{\mathbb{E}}_{q;2}^{d}$. Then, the right hand side of (2.4) belongs to ${\mathbb{E}}_{q;2}^{d}$ for all $d\in{\mathbb{R}}\setminus S$.

For all $\zeta\in\tilde{\mathbb{C}}^{*}$ such that $\zeta^{2}\not=q^{1/2}ab$, i.e. $\zeta\not=q^{1/4}\,e^{k\pi\mathbf{i}}\sqrt{ab}$ for any $k\in{\mathbb{Z}}$, we deduce that $\frac{1}{q^{\prime-1}\zeta^{2}-ab}\in{\mathbb{E}}_{q;2}^{d}$. Then for all $d\in{\mathbb{R}}\setminus S^{\prime}$, we find

[TABLE]

∎

We are now ready the prove Theorem 2.1.

Proof of Theorem 2.1.

Let us begin with the case where $d\in{\mathbb{R}}\setminus S^{\prime}$, and then consider the general case.

Case $d\in{\mathbb{R}}\setminus S^{\prime}$.

Let us first consider the situation where $d\in{\mathbb{R}}\setminus S^{\prime}$. If one defines

[TABLE]

for all $x\in\tilde{\mathbb{C}}^{*}$, one can write

[TABLE]

Let $\Phi$ : $(\xi_{1},\xi_{2})\mapsto(\xi,\zeta)$ be the homeomorphism from ${\mathbb{R}}_{>0}\times{\mathbb{R}}_{>0}$ onto itself defined by

[TABLE]

It is a bijection of ${\mathbb{R}}_{>0}\times{\mathbb{R}}_{>0}$ with inverse

[TABLE]

Furthermore, the Jacobian is given by

[TABLE]

Then,

[TABLE]

Let us prove the following technical lemma.

Lemma 2.5**.**

We have the following equality

[TABLE]

Proof of Lemma 2.5.

We have to prove that $f(x):=\frac{e_{q}(\xi_{1}/x)\,e_{q}(\xi_{2}/x)}{e_{q^{2}}(\xi/q)\,e_{q^{1/2}}(\zeta/x)}$ equals to $1$. The following holds

[TABLE]

Let us expand the expression of $f(x)$ that will be of the form $e^{\alpha/2\log(q)}e^{\log(\beta)\log(x)/\log(q)}$ where

[TABLE]

and

[TABLE]

Replacing $\xi$ and $\zeta$ by their expression in $\xi_{1},\xi_{2}$ gives

[TABLE]

and

[TABLE]

This completes the proof of the lemma. ∎

Let us continue the proof of Theorem 2.1. If

[TABLE]

and

[TABLE]

making the change of variables $\Phi$ in (2.5) yields that

[TABLE]

Let $f_{2}^{d}$ be the function considered in Lemma 2.3. By Lemma 2.3, $E_{a,q}E_{b,q}\in{\mathbb{C}}[[x]]_{q;(1,2)}^{d}$ and

[TABLE]

In virtue of Lemma 2.3, we have to prove

[TABLE]

In what follows, we will prove (2.6) for all $\zeta>0$, and by analytic continuation principle this permits us to get (2.6) for all $\zeta\in\tilde{\mathbb{C}}^{*}$.

Let $\xi=u^{2}$ with $u>0$, write $\sqrt{\xi}=u$, and note that

[TABLE]

Define for every convenient $\zeta\in{\mathbb{C}}$ such that the denominator does not vanish on the path of integration:

[TABLE]

and

[TABLE]

A straightforward computation shows that

[TABLE]

In view of the relation $\displaystyle\frac{d\xi}{\xi}=2\,\frac{du}{u}$ one finds that

[TABLE]

By using appropriate changes of variables, it follows that, for all convenient $\zeta\in{\mathbb{C}}$ such that the denominator does not vanish on the path of integration

[TABLE]

and

[TABLE]

If we put $w=1/v$, we find

[TABLE]

and with $e_{q^{1/2}}(x)=e_{q^{1/2}}(q^{-1/2}x^{-1})$, we obtain that the latter is equal to

[TABLE]

Therefore, one deduces from the above the expression expected in (2.6) for $\psi^{*}$.

General case.

It remains to prove the result for $d\in{\mathbb{R}}\setminus S$. Recall that $S^{\prime}=S\cup\{\arg(\sqrt{ab})+\pi{\mathbb{Z}}\}$ and for all $d\in{\mathbb{R}}\setminus S^{\prime}$, $E_{a,q}E_{b,q}\in{\mathbb{C}}[[x]]_{q;(1,2)}^{d}$ and for all $x\in\widetilde{{\mathbb{C}}}^{*}$,

[TABLE]

If $S=S^{\prime}$ there is nothing to prove. Assume that this is not the case. Let $d_{0}\in\{\arg(\sqrt{ab})+\pi{\mathbb{Z}}\}$ that does not belong to $S$. Let $d^{\prime}<d_{0}<d^{\prime\prime}$ such that $(d^{\prime},d^{\prime\prime})\cap S^{\prime}=\{d_{0}\}$. Then, $(d^{\prime},d^{\prime\prime})\cap S=\varnothing$. By Example 1.7, we deduce that for $\star\in\{a,b\}$, $\mathcal{S}_{q;1}^{d}(E_{\star,q})$ is independent of $d\in(d^{\prime},d^{\prime\prime})$. Then, the right hand side of (2.7), seen as a function of $d$, is independent of $d$ in $(d^{\prime},d^{\prime\prime})$. By Proposition 1.4, $\mathcal{S}_{q;1}^{d}(E_{a,q})$ and $\mathcal{S}_{q;1}^{d}(E_{b,q})$ belong to ${\mathbb{O}}_{q;1}^{d}$ and it follows by definition that for all $d\in(d^{\prime},d^{\prime\prime})$, $\mathcal{S}_{q;1}^{d}(E_{a,q})\times\mathcal{S}_{q;1}^{d}(E_{b,q})\in{\mathbb{O}}_{q;2}^{d}$ and we may apply ${\mathcal{B}}_{q;2}$ to it. By Proposition 1.4, and the proof of the theorem in the case $d\in{\mathbb{R}}\setminus S^{\prime}$, for all $d\in(d^{\prime},d^{\prime\prime})\setminus\{d_{0}\}$, we have

[TABLE]

By Lemma 2.3, for all $d\in(d^{\prime},d^{\prime\prime})\setminus\{d_{0}\}$

[TABLE]

Let $f_{2}^{d_{0}}:={\mathcal{B}}_{q;2}(\mathcal{S}_{q;1}^{d_{0}}(E_{a,q})\mathcal{S}_{q;1}^{d_{0}}(E_{b,q}))$. Since $\mathcal{S}_{q;1}^{d}(E_{a,q})\mathcal{S}_{q;1}^{d}(E_{b,q})$ is independent of $d\in(d^{\prime},d^{\prime\prime})$, it follows that $f_{2}^{d}$ is independent of $d\in(d^{\prime},d^{\prime\prime})$ and we deduce that

[TABLE]

Lemma 2.3 shows that $f_{2}^{d_{0}}(\zeta)={\mathcal{L}}_{q;2}^{d_{0}}(f_{1})$, proving with Proposition 1.1 that

[TABLE]

Then $E_{a,q}E_{b,q}\in{\mathbb{C}}[[x]]_{q;(1,2)}^{d_{0}}$ and $\mathcal{S}_{q;(1,2)}^{d_{0}}(E_{a,q}E_{b,q})=\mathcal{S}_{q;1}^{d_{0}}(E_{a,q})\mathcal{S}_{q;1}^{d_{0}}(E_{b,q})$. ∎

2.2. Variant of the Euler series

Consider now the situation where the two series are of the form $E^{[m]}_{a,q}$, $E^{[n]}_{b,q}$ where $a,b\in{\mathbb{C}}^{*}$, $m,n\in{\mathbb{N}}^{*}$. Recall that we have set ${S:=\{\arg(-a)+2\pi{\mathbb{Z}}\}\cup\{\arg(-b)+2\pi{\mathbb{Z}}\}}$. More precisely, let us prove the following.

Corollary 2.6**.**

Given $d\in{\mathbb{R}}\setminus S$, $E^{[m]}_{a,q}E^{[n]}_{b,q}\in{\mathbb{C}}[[x]]_{q;(1,2)}^{d}$ and for all $x\in\widetilde{{\mathbb{C}}}^{*}$, we have

[TABLE]

Proof.

Let us fix $d\in{\mathbb{R}}\setminus S$. As we can see in Example 1.8, we may differentiate $\mathcal{S}_{q;1}^{d}(E_{a,q})$ with respect to $a$ and find $\mathcal{S}_{q;1}^{d}(E^{[m]}_{a,q})=\partial_{a}^{m}\mathcal{S}_{q;1}^{d}(E_{a,q})$. Therefore we may differentiate $\mathcal{S}_{q;1}^{d}(E_{a,q})\mathcal{S}_{q;1}^{d}(E_{b,q})$ with respect to $a$ and find $\partial_{a}^{m}\mathcal{S}_{q;1}^{d}(E_{a,q})\mathcal{S}_{q;1}^{d}(E_{b,q})=\mathcal{S}_{q;1}^{d}(E^{[m]}_{a,q})\mathcal{S}_{q;1}^{d}(E_{b,q})(x)$. If we show that we may differentiate $\mathcal{S}_{q;(1,2)}^{d}(E_{a,q}E_{b,q})$ with respect to $a$ with $\partial_{a}^{m}\mathcal{S}_{q;(1,2)}^{d}(E_{a,q}E_{b,q})=\mathcal{S}_{q;(1,2)}^{d}(E^{[m]}_{a,q}E_{b,q})$, we will deduce with Theorem 2.1 that $\mathcal{S}_{q;(1,2)}^{d}(E^{[m]}_{a,q}E_{b,q})=\mathcal{S}_{q;1}^{d}(E^{[m]}_{a,q})\mathcal{S}_{q;1}^{d}(E_{b,q})$. If we proceed similarly with the $b$ derivation we will complete the proof. So it suffices to show that we may differentiate $\mathcal{S}_{q;(1,2)}^{d}(E_{a,q}E_{b,q})$ with respect to $a$ with $\partial_{a}^{m}\mathcal{S}_{q;(1,2)}^{d}(E_{a,q}E_{b,q})=\mathcal{S}_{q;(1,2)}^{d}(E^{[m]}_{a,q}E_{b,q})$.

By definition, the derivation in $a$ commutes with $\hat{{\mathcal{B}}}_{q;1}$ so that $\partial_{a}^{m}f_{1}=\partial_{a}^{m}\hat{{\mathcal{B}}}_{q;1}(E_{a,q}E_{b,q})=\hat{{\mathcal{B}}}_{q;1}(E^{[m]}_{a,q}E_{b,q})$. Let us fix a compact $K\subset{\mathbb{C}}^{*}\setminus{\mathbb{R}}_{>0}e^{\mathbf{i}(d+\pi)}$. Recall, see (2.3) , that we have the expression ${f_{1}=\sum_{n\geq 0}\frac{\zeta^{2n}\,(abq^{n}-\zeta^{2})}{(ab)^{n}q^{n^{2}}\,(aq^{n}+\zeta)(bq^{n}+\zeta)}}$. Then, there exists $M>0$ such that for all $\zeta\in e^{\mathbf{i}d}$, for all $n\geq 0$, for all $a\in K$, $\left|\frac{\zeta^{2n}\,(abq^{n}-\zeta^{2})}{(ab)^{n}q^{n^{2}}\,(aq^{n}+\zeta)(bq^{n}+\zeta)}\right|<\left|\frac{M\zeta^{2n+2}}{b^{n}q^{n^{2}}(bq^{n}+\zeta)}\right|$. Similarly to the proof of [Dre15], Proposition 2.13, (3), we find that $\sum_{n\geq 0}\left|\frac{M\zeta^{2n+2}}{b^{n}q^{n^{2}}(bq^{n}+\zeta)}\right|\in{\mathbb{E}}_{q;2}^{d}$. Furthermore, we deduce as in Example 1.8 that $\partial_{a}^{m}f_{1}\in{\mathbb{E}}_{q;2}^{d}$ and ${\mathcal{L}}_{q;2}^{d}(\partial_{a}^{m}f_{1})=\partial_{a}^{m}{\mathcal{L}}_{q;2}^{d}(f_{1})=\partial_{a}^{m}f_{2}^{d}$. We now use (2.4) to deal with $f_{2}^{d}$. We need to bound $\mathcal{S}_{q;2}^{d}(E_{a,q^{1/2}})$ and $\frac{1}{(q^{\prime-1}\zeta^{2}-ab)}$ uniformly in $a$. Let us begin by $\mathcal{S}_{q;2}^{d}(E_{a,q^{1/2}})$. As we can see in Example 1.8, there exists $M>0$ such that for all $a\in K$,

[TABLE]

Setting $x=re^{\mathbf{i}t}$ we find that the latter expression is bounded by

[TABLE]

Now consider $\frac{1}{(q^{\prime-1}\zeta^{2}-ab)}$ in the case $d\in{\mathbb{R}}\setminus S^{\prime}$. Let $M^{\prime}>0$ such that for all $a\in K$, for all $\zeta\in{\mathbb{R}}_{>0}e^{\mathbf{i}d}$, $\left|(q^{\prime-1}\zeta^{2}-ab)\right|>M^{\prime}$. Then, we may bound the function ${f_{2}^{d}(\zeta)=\frac{-a\,\mathcal{S}_{q;2}^{d}(E_{a,q^{1/2}})(\zeta)-b\,\mathcal{S}_{q;2}^{d}(E_{b,q^{1/2}})(\zeta)}{(q^{\prime-1}\zeta^{2}-ab)}}$ uniformly in $a\in K$ by a function where ${\mathcal{L}}_{q;2}^{d}$ may be applied. This shows that we may differentiate and $\partial_{a}^{m}{\mathcal{L}}_{q;2}^{d}(f_{2}^{d})={\mathcal{L}}_{q;2}^{d}(\partial_{a}^{m}f_{2}^{d})$. For the remaining cases $d\notin S$, we may have a problem to bound when $q^{\prime-1}\zeta^{2}=ab$. By Remark 2.4 in that case, $f_{2}^{d}\in{\mathbb{E}}_{q;2}^{d}$ proving that $f_{2}^{d}$ is analytic at $q^{\prime-1}\zeta^{2}=ab$ and thus may be correctly bounded. We have proved that

[TABLE]

Then, $E^{[m]}_{a,q}E_{b,q}\in{\mathbb{C}}[[x]]_{q;(1,2)}^{d}$ and $\mathcal{S}_{q;(1,2)}^{d}(E^{[m]}_{a,q}E_{b,q})=\partial_{a}^{m}\mathcal{S}_{q;(1,2)}^{d}(E_{a,q}E_{b,q})$. This was the sufficient fact to conclude the proof. ∎

2.3. Product theorem

Let us now state and prove the main result of the paper.

Theorem 2.7**.**

Let $f$ (resp. $g$) be a series solution of a linear $q$ difference equation with slopes [math] and $1$. Then, $fg\in\mathcal{MS}_{q}$. More precisely, let $d\in{\mathbb{R}}$, such that $f\in{\mathbb{C}}[[x]]_{q;1}^{d}$ and $g\in{\mathbb{C}}[[x]]_{q;1}^{d}$. Then $f\times g\in{\mathbb{C}}[[x]]_{q;(1,2)}^{d}$ and for all $x\in\widetilde{{\mathbb{C}}}^{*}$,

[TABLE]

Proof.

By Theorem 1.9, there exists a decomposition

[TABLE]

such that $f_{0},\dots,f_{k}\in{\mathbb{C}}\{x\}$, and $a_{i}\in{\mathbb{C}}^{*}$, $m_{i}\in{\mathbb{N}}$. The same holds for $g$

[TABLE]

with $g_{0},\dots,g_{k}\in{\mathbb{C}}\{x\}$, and $b_{j}\in{\mathbb{C}}^{*}$, $n_{j}\in{\mathbb{N}}$. Then we find

[TABLE]

Let $d\in{\mathbb{R}}$, such that $f\in{\mathbb{C}}[[x]]_{q;1}^{d}$ and $g\in{\mathbb{C}}[[x]]_{q;1}^{d}$. The singular directions of $f$ (resp. $g$) correspond to the directions $\arg(-a_{i})+2\pi{\mathbb{Z}}$ with $m_{i}\neq 0$ (resp. $\arg(-b_{j})+2\pi{\mathbb{Z}}$ with $n_{j}\neq 0$). Let $S\subset{\mathbb{R}}$ be the union of the singular directions of $f$ and $g$. By Corollary 2.6, when $m_{i},n_{j}\in{\mathbb{N}}^{*}$, $E^{[m_{i}]}_{a_{i},q}E^{[n_{j}]}_{b_{j},q}\in{\mathbb{C}}[[x]]_{q;(1,2)}^{d}$ and for $x\in\widetilde{{\mathbb{C}}}^{*}$, $\mathcal{S}_{q;(1,2)}^{d}(E^{[m_{i}]}_{a_{i},q}E^{[n_{j}]}_{b_{j},q})(x)=\mathcal{S}_{q;1}^{d}(E^{[m_{i}]}_{a_{i},q})\mathcal{S}_{q;1}^{d}(E^{[n_{j}]}_{b_{j},q})(x)$. By Proposition 1.5, the map $f\mapsto\mathcal{S}_{q}^{d}(f)$ is a morphism of ${\mathbb{C}}\{x\}$-modules so that we have $f_{i}g_{j}\mathcal{S}_{q;(1,2)}^{d}\left(E^{[m_{i}]}_{a_{i},q}E^{[n_{j}]}_{b_{j},q}\right)=\mathcal{S}_{q;(1,2)}^{d}\left(f_{i}g_{j}E^{[m_{i}]}_{a_{i},q}E^{[n_{j}]}_{b_{j},q}\right)$, $\mathcal{S}_{q;1}^{d}(f)=\displaystyle\sum_{i=0}^{k}f_{i}\mathcal{S}_{q;1}^{d}\left(E^{[m_{i}]}_{a_{i},q}\right)$, and $\mathcal{S}_{q;1}^{d}(g)=\displaystyle\sum_{j=0}^{\ell}g_{j}\mathcal{S}_{q;1}^{d}\left(E^{[n_{j}]}_{b_{j},q}\right)$. Then,

[TABLE]

which shows that $fg\in{\mathbb{C}}[[x]]_{q;(1,2)}^{d}$. This completes the proof. ∎

3. Inverse of the $q$-Euler series

In this section we study the inverse of the $q$-Euler series $E_{1,q}$. More precisely, we prove that if it is $q$-multisumable in the direction $d=0$, then ${\mathcal{S}}_{q}^{0}(E_{1,q}^{-1})\neq{\mathcal{S}}_{q}^{0}(E_{1,q})^{-1}$. In particular, this means that we have no hope to define a morphism of fields.

One can observe that

[TABLE]

This implies that

[TABLE]

Furthermore, applying the residue theorem as in Example 1.7 yields that

[TABLE]

Iterating (3.2) several times implies that

[TABLE]

Thus, combining this together with (3.1) gives the following identity for all $x\in\tilde{\mathbb{C}}^{*}$:

[TABLE]

Furthermore, replacing $x$ with $x\,e^{2\pi\mathbf{i}N}$ in the above in (3.3) implies immediately that

[TABLE]

The following lemma, will permit us to control the sums in (3.3) and (3.4).

Lemma 3.1**.**

Let $(N,r_{0},\lambda)\in{\mathbb{N}}^{*}\times{\mathbb{C}}\times{\mathbb{R}}_{>0}$ and define

[TABLE]

For all $r_{1}<r_{2}$, one has the existence of $C,C^{\prime}>0$ such that for all $\Re(r_{0})\in[r_{1},r_{2}]$

[TABLE]

Proof.

Let us fix $r_{1}<r_{2}$. Consider $e^{-\lambda(N+r_{0})^{2}}f(N;r_{0},\lambda)$ with $\Re(r_{0})\in[r_{1},r_{2}]$. On has

[TABLE]

The degree two polynomial $x\mapsto\lambda x(x-2N-2\Re(r_{0}))$ has a minimum at $N+\Re(r_{0})$. Therefore, there exists $C^{\prime}>0$ such that for all $\Re(r_{0})\in[r_{1},r_{2}]$, for all $N\in{\mathbb{N}}^{*}$

[TABLE]

Then, $\sum_{k=0}^{N-1}e^{k\lambda(k-2N-2r_{0})}$ converges, when $N$ goes to infinity proving the upper bound.

On has

[TABLE]

Again, $\left|\sum_{k=1}^{N-1}e^{k\lambda(k-2N-2r_{0})}\right|\leq N\max(e^{\lambda(1-2N-2\Re(r_{0}))},e^{(N-1)\lambda(-N-1-2\Re(r_{0}))})$, proving that there exists $C>0$, such that for all $\Re(r_{0})\in[r_{1},r_{2}]$, such that when $N$ is sufficiently big ${\left|1-\left|\sum_{k=1}^{N-1}e^{k\lambda(k-2N-2r_{0})}\right|\;\right|\geq C}$. Up to take a smaller $C$ we find that for all ${\Re(r_{0})\in[r_{1},r_{2}]}$, for all $N\in{\mathbb{N}}^{*}$, $C\left|e^{\lambda(N+r_{0})^{2}}\right|\leq\left|f(N;r_{0},\lambda)\right|$. ∎

Let us now use (3.3) and (3.4) to give an estimate for ${\mathcal{S}}_{q;1}^{0}(E_{1,q})(x)$.

Lemma 3.2**.**

We have the existence of $C(r),R>0$ such that

[TABLE]

Proof.

First, we will consider the case where $t\to+\infty$. Choose $N\in{\mathbb{N}}$ such that $t\displaystyle\in[N,N+1)$, and set

[TABLE]

One can notice that, for any integer $k$:

[TABLE]

By combining (3.5) together with (3.3), it follows that we have the existence of $C>0$ such that for all $N\in{\mathbb{N}}^{*}$ for all $t>0$

[TABLE]

When $t$ goes to $+\infty$, $e^{-(\log(\frac{re^{\mathbf{i}t}}{\sqrt{q}}e^{-\pi\mathbf{i}}))^{2}/(2\log(q))}=O(e^{\frac{t^{2}}{2\log(q)}})$. Taking the inverse in the latter expression gives the result for $t>0$ sufficiently big. Up to take a bigger constant we deduce the result for all $t>0$.

Finally, one can notice the case of $\arg(x)\to-\infty$ may be treated in a similar way, using (3.4) instead of (3.3). ∎

Let us prove now prove that the Borel transformation of a function with a negative $q$-exponential angular growth is still a function with a negative $q$-exponential angular growth.

Lemma 3.3**.**

Let $s,s^{\prime}\in{\mathbb{Q}}_{>0}$. Let $f$ be a function meromorphic on the neighborhood of [math] in the Riemann surface of the logarithm. Assume that there exist $C(R),R>0$ such that

[TABLE]

Then, $f\in{\mathbb{O}}_{q;s}^{0}$ and we may consider ${\mathcal{B}}_{q;s}^{0}(f)$. Furthermore, there exist $C^{\prime}(R)\in(0,\infty)$ such that

[TABLE]

Proof.

By definition, we may apply ${\mathcal{B}}_{q;s}^{0}$ to $f$. Let $g={\mathcal{B}}_{q;s}^{0}(f)$. We have

[TABLE]

To simplify, let us write $C$ instead of $C(R)$. If we write $x=re^{\mathbf{i}\theta}$, we find $\frac{dx}{x}=\mathbf{i}d\theta$. Then,

[TABLE]

We now remark that $\frac{1}{s}-\frac{1}{s^{2}(s^{-1}+s^{\prime-1})}=\frac{1}{s}-\frac{1}{s+s^{2}/s^{\prime}}=\frac{s+s^{2}/s^{\prime}-s}{s^{2}+s^{3}/s^{\prime}}=\frac{s^{2}/s^{\prime}}{s^{2}+s^{3}/s^{\prime}}=\frac{1}{s+s^{\prime}}$. This proves the result. ∎

We are now ready to prove the main result of the section.

Theorem 3.4**.**

For all finite strictly increasing sequence $\vec{s}$ of positive rational numbers, there exists no power series $f\in{\mathbb{C}}[[x]]_{q;\vec{s}}^{0}$ such that $\mathcal{S}_{q;\vec{s}}^{0}(f)=\left(\mathcal{S}_{q;1}^{0}(E_{1,q})\right)^{-1}$. Therefore, if $E_{1,q}^{-1}\in{\mathbb{C}}[[x]]_{q}^{0}$, then ${\mathcal{S}}_{q}^{0}(E_{1,q}^{-1})\neq{\mathcal{S}}_{q}^{0}(E_{1,q})^{-1}$.

Proof.

To the contrary, assume that $\mathcal{S}_{q;\vec{s}}^{d}(f)=\left(\mathcal{S}_{q;1}^{d}(\hat{E}_{q})\right)^{-1}$ for some suitable increasing sequence $\vec{s}=(s_{1},...,s_{r})$. Let $(\widetilde{s}_{1},...,\widetilde{s}_{r})$ its associated sequence as in (1.4). By Proposition 1.4, one has:

[TABLE]

and

[TABLE]

Furthermore, Lemma 3.2 states that

[TABLE]

Thus, using several times Lemma 3.3 implies that

[TABLE]

where $\tilde{s}$ is a convenient positive constant. Since $\hat{\mathcal{B}}_{q;s_{1}}(f)$ is a convergent power series, we find that the only possibility is that ${\hat{\mathcal{B}}_{q;s_{1}}(f)=0}$. This implies $f=0$, which is a contradiction. ∎