# Upper bound for flows defined over normed rings

Ronald Orozco López

###### Abstract

In this paper upper bounds of flows defined over normed ring are given. For
this end, we calculate some images of the autonomous operator acting on
$\mathbb{R}^{\mathbb{N}}$.

Keywords: autonomous operator, normed rings, dynamical systems
Mathematics Subject Classification: 47H99, 46H25, 37C10.

##  1 Introduction

In [3] the autonomous operator and its properties were introduced. In this
paper we calculate upper bounds for a flow defined over a normed ring. For
this end, we calculate some images of the autonomous operator acting on
$\mathbb{R}^{\mathbb{N}}$.

This paper is organized as follows. In Section 2, some basic concept are
shown. Section 3 introduces some images of the autonomous operator. Finally,
in Section 4 the ring of convergent exponential generating functions is shown
and some upper bounds of flows defined over normed ring calculated. In all
this paper $\mathbb{N}$ will be the monoid of nonnegative integer.

##  2 Preliminaries

In this section some concept such that Bell polynomials, the ring of Hurwitz
expansion, autonomous operator and dynamical systems over integral domains are
shown.

Bell polynomials are a very useful tool in mathematics to represent the $nth$
derivative of the composite of two functions [1]. In effect, let $f$ and $g$
be two analytical functions with power series representations $\sum
a_{n}\frac{x^{n}}{n!}$ and $\sum b_{n}\frac{x^{n}}{n!}$ respectively, with
$a_{n},b_{n}\in\mathbb{C}$. Then

$f(g(x))=f(b_{0})+\sum_{n=1}^{\infty}Y_{n}(b_{1},...,b_{n};a_{1},...,a_{n})\frac{x^{n}}{n!},$
(1)

where $Y_{n}$ is the $nth$ Bell polynomial. For instance it is well known that

$\displaystyle Y_{1}(b_{1};a_{1})$ $\displaystyle=$ $\displaystyle
a_{1}b_{1},$ $\displaystyle Y_{2}(b_{1},b_{2};a_{1},a_{2})$ $\displaystyle=$
$\displaystyle a_{1}b_{2}+a_{2}b_{1}^{2},$ $\displaystyle
Y_{3}(b_{1},b_{2},b_{3};a_{1},a_{2},a_{3})$ $\displaystyle=$ $\displaystyle
a_{1}b_{3}+a_{2}(3b_{1}b_{2})+a_{3}b_{1}^{3}.$

Bell polynomials can be expressed implicitly using the Faá di Bruno formula
[2],

$Y_{n}(b_{1},...,b_{n};a_{1},...,a_{n})=\sum_{k=1}^{n}B_{n,k}a_{k},$ (2)

where

$B_{n,k}=\sum_{|p(n)|=k}\frac{n!}{j_{1}!j_{2}!\cdots
j_{n}!}\left[\frac{b_{1}}{1!}\right]^{j_{1}}\left[\frac{b_{2}}{2!}\right]^{j_{2}}\cdots\left[\frac{b_{n}}{n!}\right]^{j_{n}},$
(3)

where the sum runs over all the partitions $p(n)$ of $n$, i.e.,
$n=j_{1}+2j_{2}+\cdots+rj_{r}$, $j_{h}$ denotes the number of parts of size
$h$ and $|p(n)|=j_{1}+j_{2}+\cdots+j_{r}$ is the length of partition $p(n)$.

Let $R$ be an integral domain with identity and consider the power series of
the form $\sum_{n=0}^{\infty}a_{n}\frac{X^{n}}{n!}$ with $a_{n}\in R$, for all
$n\geq 0$ and let $\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ denote the set of
all this series. Let $\delta$ denote the derivative defined over
$\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ such that $\delta(f+g)=\delta f+\delta
g$, $\delta(fg)=f\delta g+(\delta f)g$ and
$\delta^{k}f=\delta(\delta^{k-1}f)$. Then we have the following definitions

######  Definition 1.

The sequence of derivatives of $f(X)$, denoted $\Delta f(X)$, is the sequence
$(\delta^{n}f(X))_{n\in\mathbb{N}}$. $\Delta f(X)$ is called Hurwitz expansion
of $f(X)$. Let $\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ denote the ring
of all Hurwitz expansions $\Delta f(X)$, with
$f(X)\in\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ and sum and product given by

  1. 1.

$\Delta f+\Delta g=\Delta(f+g)$,

  2. 2.

$\Delta f\ast\Delta g=\Delta(fg)$,

where $\ast$ is the convolution product in
$\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ given by

$\Delta f(X)\ast\Delta
g(X)=\left(\sum_{k=0}^{n}\binom{n}{k}\delta^{k}f(X)\delta^{n-k}g(X)\right).$

######  Definition 2.

Let $\mathop{\mbox{{Exp}}}\nolimits_{R}[X]^{\mathbb{N}}$ be the multiplicative
algebra of sequences over $\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$ with
multiplication defined term to term. The autonomous operator $\mathfrak{A}$ is
defined as the nonlinear map
$\mathfrak{A}:\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]\rightarrow\mathop{\mbox{{Exp}}}\nolimits_{R}[X]^{\mathbb{N}}$,
given by $\mathfrak{A}(\Delta f)=(A_{n}([\delta^{n-1}f]))_{n\geq 1}$, with
$[\delta^{n}f]=(f,\delta f,...,\delta^{n}f)$, in where the $A_{n}$ are
recursively defined in the next way:

$\displaystyle A_{1}([\delta^{0}f])$ $\displaystyle=$ $\displaystyle
A_{1}([f])=f,$ (4) $\displaystyle A_{n+1}([\delta^{n}f])$ $\displaystyle=$
$\displaystyle
Y_{n}(A_{1}([\delta^{0}f]),A_{2}([\delta^{1}f])...,A_{n}([\delta^{n-1}f]);\delta^{1}f,...,\delta^{n}f),$

$n\geq 1$.

######  Definition 3.

Take $\Delta f(x)$ in $\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]$. We define
the dynamical system $\Phi:R\times
R\times\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[X]\rightarrow\mathop{\mbox{{Exp}}}\nolimits_{S}[T]$,
$S=\mathop{\mbox{{Exp}}}\nolimits_{R}[X]^{\mathbb{N}}$, related to $\Delta
f(x)$ as

$\Phi(t,x,\Delta
f(x))=x+\sum_{n=1}^{\infty}A_{n}([\delta^{(n-1)}f(x)])\dfrac{t^{n}}{n!}.$ (5)

Let $R^{\sharp}$ be the ring of power series with coefficients in $R$ and let
$\epsilon_{a}:\mathop{\mbox{{Exp}}}\nolimits_{R}[X]\rightarrow R^{\sharp}$
denote the evaluation map in the element $a$ of $R$ and extend this map to
$\Phi(t,x,\Delta f(x))$ in this sense

$\epsilon_{a}\Phi(t,x,\Delta
f(x))=a+\sum_{n=1}^{\infty}\epsilon_{a}(A_{n}([\delta^{(n-1)}f(x)]))\frac{t^{n}}{n!}.$
(6)

Then $\epsilon_{a}$ maps $\mathop{\mbox{{Exp}}}\nolimits_{S}[T]$ in
$\mathop{\mbox{{Exp}}}\nolimits_{R^{\sharp}}[T]$. Finally, we define a
trajectories of $\Phi$ as

$\Gamma_{a}:=\\{\epsilon_{a}\Phi(t,y,\Delta f(y)):t\in R\\}.$ (7)

##  3 Some images of $\mathfrak{A}$

In [2] some images of $\mathfrak{A}$ were calculated for sequences in
$\mathbb{R}^{\mathbb{N}}$. We use this for to obtain the trajectories
$\Gamma_{0}$ for flow of $\delta_{t}(\Phi)=f(\Phi)$.

######  Theorem 1.

Let $(a^{n})_{n\geq 0}$, $(n!)_{n\geq 0}$, $((a)_{n})_{n\geq 0}$ be sequences
in $\mathbb{R}^{\mathbb{N}}$. Then

$\displaystyle\mathfrak{A}\left((a^{n})_{n\geq 0}\right)$ $\displaystyle=$
$\displaystyle\left(a^{n-1}(n-1)!\right)_{n\geq 1},$ (8)
$\displaystyle\mathfrak{A}\left((n!)_{n\geq 0}\right)$ $\displaystyle=$
$\displaystyle\left((-1)^{n+1}2^{n}\binom{1/2}{n}n!\right)_{n\geq 1},$ (9)
$\displaystyle\mathfrak{A}\left(((a)_{n})_{n\geq 0}\right)$ $\displaystyle=$
$\displaystyle\left(\binom{\frac{1}{1-a}}{n}(1-a)^{n}n!\right)_{n\geq 1},$
(10)

where $(a)_{k}=a(a+1)\cdots(a+k-1)$ is the Pochhammer symbol.

###### Proof.

Let $y^{\prime}=f(y)$ be autonmous differential equation with value initial
$y(0)=0$ and $f$ analytic and invertible in a subset $\Omega$ of $\mathbb{R}$.
We suppose that $y=f^{-1}(g)$, where $g$ is a differential function in
$\Omega$. Then

$(f^{-1})^{\prime}(g)g^{\prime}=g$

and therefore

$\int\frac{(f^{-1})^{\prime}(g)dg}{g}=x+c.$ (11)

The $i$th derivatives of the functions $f(z)=e^{az}$, $f(z)=\frac{1}{1-z}$ and
$f(z)=(1+z)^{a}$ valuates in $0$ are
$f^{i}\left(0\right)=a^{i},~{}f^{i}\left(0\right)=i!$ and
$f^{i}\left(0\right)=\left(a\right)_{i}=$ $a(a-1)\cdots(a-i+1),$
$i>0,~{}f\left(0\right)=1,$ respectively. Applying $(\ref{solucion_auto_1})$
for to obtain a solution to the equation $y^{\prime}=f\left(y\right)$, where
$f$ denote some of the given functions, we have respectively

$\displaystyle\frac{1}{a}\log\left|\frac{1}{1-at}\right|$ $\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}A_{n}\left(1,a,a^{2},...,a^{n-1}\right)\frac{t^{n}}{n!}=\sum_{n=1}^{\infty}a^{n-1}\frac{t^{n}}{n},$
$\displaystyle 1-\sqrt{1-2t}$ $\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}A_{n}\left(0!,1!,2!,...,(n-1)!\right)\frac{t^{n}}{n!}$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}2^{n}\binom{1/2}{n}t^{n},$
$\displaystyle\left[1+(1-a)t\right]^{1/(1-a)}-1$ $\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}A_{n}\left(1,(a)_{1},(a)_{2},...,(a)_{n-1}\right)\frac{t^{n}}{n!}$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}\binom{\frac{1}{1-a}}{n}(1-a)^{n}t^{n}.$

∎

######  Corollary 1.

From above theorem we have that

$\displaystyle\epsilon_{0}\Phi(t,x,\Delta(e^{\alpha x}))$ $\displaystyle=$
$\displaystyle\Phi(t,0,(\alpha^{n})_{n\geq 0})$ $\displaystyle=$
$\displaystyle\frac{1}{a}\log\left|\dfrac{1}{1-\alpha t}\right|$
$\displaystyle\epsilon_{0}\Phi\left(t,x,\Delta\left(\frac{1}{1-x}\right)\right)$
$\displaystyle=$ $\displaystyle\Phi\left(t,0,(n!)_{n\geq 0}\right)$
$\displaystyle=$ $\displaystyle 1-\sqrt{1-2t}$
$\displaystyle\epsilon_{0}\Phi(t,x,\Delta(1+x)^{\alpha})$ $\displaystyle=$
$\displaystyle\Phi\left(t,0,((\alpha)_{n})_{n\geq 0}\right)$ $\displaystyle=$
$\displaystyle[1+(1-\alpha)t]^{1/(1-\alpha)}-1.$

###### Proof.

Let $f(x)$ be the exponential generating function for some of the sequences of
previous theorem. Then $\epsilon_{0}\Phi(t,x,\Delta
f(x))=\sum_{n=0}^{\infty}\epsilon_{0}A_{n}([\delta^{n-1}f(x)])\frac{t^{n}}{n!}$
is the exponential generating function of such sequence. ∎

##  4 Upper bounds for $\Phi$ when the ring $R$ is a normed ring

We shall use results of the above section to find an upper bound for the flow
of a one-dimensional dynamical system defined over a normed ring $R$. In all
this section $R$ is a normed integral domain.

######  Definition 4.

([4]) A normed ring $R$ is a ring $R$ which can be endowed with a non-negative
real-valued function $\parallel\cdot\parallel:R\rightarrow\mathbb{R}_{\geq
0}$, namely a norm, satisfying for all $x,y\in R$:

  1. 1.

$\parallel x\parallel=0$ if and only if $x=0$,

  2. 2.

$\parallel x+y\parallel\leq\parallel x\parallel+\parallel y\parallel$,

  3. 3.

$\parallel-a\parallel=\parallel a\parallel$

  4. 4.

$\parallel xy\parallel\leq\parallel x\parallel\parallel y\parallel$,

######  Definition 5.

A sequence $a_{1},a_{1},...$ of elements of $R$ is Cauchy if for each
$\epsilon>0$ there exists $k_{0}$ such that $\lVert
a_{n}-a_{m}\rVert<\epsilon$ for all $n,m\geq k_{0}$. A normed ring $R$ is
complete if every Cauchy sequence converges.

Let $R$ be a complete normed ring. Consider the following subset of
$\mathop{\mbox{{Exp}}}\nolimits_{R}[x]$:

$\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}=\left\\{\sum_{n=0}^{\infty}a_{n}\frac{x^{n}}{n!}:a_{n}\in
R,\
\lim_{n\rightarrow\infty}\bigg{\lVert}\frac{a_{n}}{n!}\bigg{\rVert}=0\right\\}.$

From definition of $\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}$ it followed
that $\lVert\frac{a_{n}}{n!}\rVert$ is bounded. Then we can define $\lVert
f\rVert=\max\left(\lVert\frac{a_{n}}{n!}\rVert\right)_{n\in\mathbb{N}}$ for
each $f\in\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}$. On the other hand, it is
clear that $\delta^{k}f(x)\in\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}$ for
all $f$ in $\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}$ and all $k\geq 1$ and
thus
$\lVert\delta^{k}f\rVert=\max\left(\lVert\frac{a_{n+k}}{n!}\rVert\right)$. We
define the subset of Hurwitz expansion of
$\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}$ as
$\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}=\\{\Delta f(x):\
f(x)\in\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}\\}$.

######  Lemma 1.

:

  1. 1.

$\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}$ is a subring of
$\mathop{\mbox{{Exp}}}\nolimits_{R}[x]$ containing $R$.

  2. 2.

$\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}$ is a subring of
$\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}[x]$.

###### Proof.

:

  1. 1.

We prove that $\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}$ is closed under
addition and multiplication. To that end take
$f=\sum_{n=0}^{\infty}a_{n}\frac{x^{n}}{n!}$ and
$g=\sum_{n=0}^{\infty}b_{n}\frac{x^{n}}{n!}$ in
$\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}$. It is easy to see that
$\lVert\frac{a_{n}}{n!}+\frac{b_{n}}{n!}\rVert\rightarrow 0$. Now, consider
$\epsilon>0$ and let $n_{0}$ be a positive number such that $\lVert
a_{k}\rVert<\frac{\epsilon}{2^{k}}$ if $k\geq\frac{n_{0}}{2}$ and $\lVert
b_{l}\rVert<\frac{\epsilon}{2^{l}}$ if $l\geq\frac{n_{0}}{2}$. Now let $n\geq
n_{0}$ and define $l=n-k$. Then $k\geq\frac{n_{0}}{2}$ or
$l\geq\frac{n_{0}}{2}$. Therefore
$\lVert\sum_{k=0}^{n}\binom{n}{k}a_{k}b_{n-k}\rVert\leq\sum_{k=0}^{n}\binom{n}{k}\lVert
a_{k}\rVert\lVert b_{n-k}\rVert<\epsilon^{2}$ and thus
$fg=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\binom{n}{k}a_{k}b_{n-k}\frac{x^{n}}{n!}$
belongs to $\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}$, as claimed.

  2. 2.

Follows from previous and that $\delta^{k}(f+g)$ and $\delta^{k}(fg)$ belong
to $\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}$.

∎

Let $R^{\prime}$ be a complete normed ring extension of $R$. Then each $c\in
R^{\prime}$ with $\lVert c\rVert\leq 1$ defines an evaluation homomorphism
$\epsilon_{c}:\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}\rightarrow R^{\prime}$
given by

$f(x)=\sum_{n=0}^{\infty}a_{n}\frac{x^{n}}{n!}\mapsto
f(c)=\sum_{n=0}^{\infty}a_{n}\frac{c^{n}}{n!}.$

######  Theorem 2.

Let $\Delta f(x)\in\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}$ and
$g(x)\in\mathop{\mbox{{Exp}}}\nolimits_{\mathbb{R}}\\{x\\}$ with
$[x^{n}]g(x)=a_{n}$ such that $\lVert\epsilon_{c}(\delta^{n}f(x))\rVert\leq
a_{n}$, for some $c$ in $R^{\prime}$ and such that $\Phi(t,0,\Delta
g(0))\in\mathop{\mbox{{Exp}}}\nolimits_{\mathbb{R}}\\{t\\}$. Then

$\lVert\epsilon_{c}(A_{n}([\delta^{n-1}f(x)])\rVert\leq A_{n}([a_{n-1}])$

and

$\lVert\Phi(t,c,\Delta f(c))-c\rVert\leq\Phi(t,0,\Delta g(0)).$

###### Proof.

The proof is by induction. This is true for $n=1$, since
$\lVert\epsilon_{c}(A_{1}([\delta^{0}f])\rVert\leq A_{1}(a_{0})$. Suppose that
the statement it holds for all positive number smaller than $n$ and let’s
consider the case $n+1$. Then

$\displaystyle\lVert\epsilon_{c}(A_{n+1}([\delta^{n}f]))\rVert$
$\displaystyle\leq$
$\displaystyle\sum_{k=1}^{n}\sum_{|p(n)|=k}\frac{n!}{j_{1}!\cdots
j_{n}!}\left[\frac{\lVert\epsilon_{c}(A_{1})\rVert}{1!}\right]^{j_{1}}\cdots\left[\frac{\lVert\epsilon_{c}(A_{n})\rVert}{n!}\right]^{j_{n}}\lVert\epsilon_{c}(\delta^{k}f)\rVert$
$\displaystyle\leq$
$\displaystyle\sum_{k=1}^{n}\sum_{|p(n)|=k}\frac{n!}{j_{1}!\cdots
j_{n}!}\left[\frac{A_{1}([a_{0}])}{1!}\right]^{j_{1}}\cdots\left[\frac{A_{n}([a_{n-1}])}{n!}\right]^{j_{n}}a_{k}$
$\displaystyle=$ $\displaystyle A_{n+1}([a_{n}])$

and the statement follows. Then

$\displaystyle\lVert\Phi(t,c,\Delta f(c))-c\rVert$ $\displaystyle=$
$\displaystyle\bigg{\lVert}\sum_{n=1}^{\infty}A_{n}([\delta^{n-1}f(c)])\frac{t^{n}}{n!}\bigg{\rVert}$
$\displaystyle\leq$ $\displaystyle\sum_{n=1}^{\infty}\parallel
A_{n}([\delta^{n-1}f(c)]\parallel\frac{t^{n}}{n!}$ $\displaystyle\leq$
$\displaystyle\sum_{n=1}^{\infty}A_{n}([a_{n-1}])\frac{t^{n}}{n!}=\Phi(t,0,\Delta
g(0)).$

$\Phi(t,0,\Delta g(0))\in\mathop{\mbox{{Exp}}}\nolimits_{\mathbb{R}}\\{t\\}$
implies that $\lVert A_{n}(\delta^{n-1}g(0))\rVert/n!\rightarrow 0$ and
therefore $\Phi(t,c,\Delta
f(c))\in\mathop{\mbox{{Exp}}}\nolimits_{R^{\prime}}\\{t\\}$. ∎

Now, we will use corollary 1 for get upper bound for flow defined over normed
ring.

######  Corollary 2.

Let $\Delta f(x)\in\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}$ such that
$\lVert\epsilon_{c}(\delta^{n}f(x))\rVert\leq a^{n}$, for some $c$ in
$R^{\prime}$, $a\in\mathbb{R}$. Then

$\lVert\epsilon_{c}(A_{n}([\delta^{n-1}f(x)]))\rVert\leq a^{n}(n-1)!$

and

$\lVert\Phi(t,c,\Delta
f(c)-c\rVert\leq\frac{1}{a}\log\left|\frac{1}{1-at}\right|$

with $|at|<1$ and $a>0$.

######  Corollary 3.

Let $\Delta f(x)\in\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}$ such that
$\lVert\epsilon_{c}(\delta^{n}f(x))\rVert\leq n!$, for some $c$ in
$R^{\prime}$. Then

$\lVert\epsilon_{c}(A_{n}([\delta^{n}f(x)]))\rVert\leq(-1)^{n+1}2^{n}n!\binom{1/2}{n}$

and

$\lVert\Phi(t,c,\Delta f(c)-c\rVert\leq 1-\sqrt{1-2t}$

with $|2t|<1$.

######  Corollary 4.

Let $\Delta f(x)\in\Delta\mathop{\mbox{{Exp}}}\nolimits_{R}\\{x\\}$ such that
$\lVert\epsilon_{c}(f(x))\rVert\leq 1$ and
$\lVert\epsilon_{c}(\delta^{n}f(x))\rVert\leq(a)_{n}$, $n\geq 1$, for some $c$
in $R^{\prime}$, $1\neq a\in\mathbb{R}$. Then

$\lVert\epsilon_{c}(A_{n}([\delta^{n}f(x)]))\rVert\leq\binom{\frac{1}{1-a}}{n}(1-a)n!$

and

$\lVert\Phi(t,c,\Delta f(c))-c\rVert\leq\left(1+(1-a)t\right)^{1/(1-a)}-1$

with $|t|<1$.

## References

  * [1] Comtent, L.Analyse Combinatoire, Tome Premier, Presses Universitaires de France (1970).
  * [2] Orozco, R. Solución Analítica de la Ecuación Diferencial Autónoma de orden $k$, Revista De Matemática Teoría y Aplicaciones vol(23) fasc.1: 63–84.
  * [3] Orozco, R. Autonomous Operator and Differential Systems of order 1 over Integral Domains, arXiv:1812.11518.
  * [4] Gelfan, I., Raikov, M.D.A., Shilov, G.E. Commutative normed rings, Chelsea Publishing Company: New York, NY, USA, 1964.

E-mail address, R. Orozco<EMAIL_ADDRESS>