On submultiplicative constants of an algebra

Abstract

If $\mathcal{A}$ is a finite-dimensional commutative associative real algebra with norm $\| \cdot\|$ then we say that the rth submultiplicative constant of $\mathcal{A}$ is the smallest constant $m_{r}(\mathcal{A})$ for which $\| x_{1} x_{2} \cdots x_{r} \| \leq m_{r}(\mathcal {A}) \| x_{1} \| \| x_{2} \| \cdots \| x_{r} \|$. For a product algebra, we show that there exist zero divisors where equality is attained in the inequality defining $m_{r}( \mathcal{A})$. We also study $\rho_{\mathcal{A}} = \limsup_{r \rightarrow\infty} \sqrt[r]{m_{r}(\mathcal{A})}$. We explain how $\rho_{\mathcal{A}}$ appears in the generalization of the Cauchy–Hadamard criterion for hypercomplex power series. We find the submultiplicative constants and $\rho_{\mathcal{A}}$ for the real group algebra of the cyclic group of order n as well as the complicated numbers $\mathcal{C}_{n} = \{ a_{1}+a_{2}k+ \cdots+ a_{n} k^{n-1} | a_{i} \in\mathbb{R}, k^{n} = -1 \}$ with Euclidean norm. Submultiplicative constants for the n-dual numbers $\Delta_{n}$ with Euclidean norm are also calculated or conjectured for $n \leq6$. We show, for $n \geq2$, $\rho_{\Delta_{n}} = 1$ for $\Delta_{n}$ given the p-norm.

1 Introduction

Suppose $V_{1}, \dots, V_{r}$ are finite-dimensional spaces over either $\mathbb {R}$ or $\mathbb {C}$, and X is a normed space. It is known that any bounded multilinear map $T: V_{1} \times V_{2} \times\cdots\times V_{r} \rightarrow X$ is continuous. Furthermore, finite-dimensional $V_{1}, \dots, V_{r}$ have compact unit ball whose Cartesian products are likewise compact. Therefore, the extreme value theorem yields that the norm $\lVert T \rVert_{\mathrm{op}} = \sup\{ \lVert T(v_{1}, v_{2}, \ldots, v_{r}) \rVert : \lVert v_{i} \rVert \le1, 1 \le i \le r \}$ is attained by evaluation of T at some point in the Cartesian product of the unit-balls of $V_{1}, \dots, V_{r}$. Since r-fold multiplication on an algebra $\mathcal {A}$ provides an $\mathcal {A}$-valued r-linear map over $\mathcal {A}$, we have the following result:

Theorem 1.1

If $\mathcal {A}$ is a finite-dimensional unital associative algebra (with multiplication denoted as ⋆) over $\mathbb{R}$ or $\mathbb{C}$ with a norm $\lVert\cdot \rVert$ then there exists a smallest constant $m_{r}(\mathcal {A}) \in\mathbb{R}$ for each $r \in\mathbb{N}, r \le2$ such that $\lVert x_{1} \star x_{2} \star\cdots\star x_{r} \rVert \le m_{r}(\mathcal {A}) \lVert x_{1} \rVert \lVert x_{2} \rVert \cdots \lVert x_{r} \rVert$ for all $x_{1}, x_{2}, \ldots, x_{r} \in \mathcal {A}$. Furthermore, there exist $y_{1},\dots, y_{r} \in \mathcal {A}$ for which equality is attained; $\lVert y_{1} \star y_{2} \star\cdots\star y_{r} \rVert = m_{r}(\mathcal {A}) \lVert y_{1} \rVert \lVert y_{2} \rVert \cdots \lVert y_{r} \rVert$.

We call $m_{r}(\mathcal {A})$ the rth submultiplicative constant of $\mathcal {A}$ (with respect to the norm $\lVert\cdot \rVert$). Usually, we denote $m_{2}(\mathcal {A}) = m_{\mathcal {A}}$ and name it the submultiplicative constant. This improves the less sharp use of $m_{\mathcal {A}}$ in [1].

Proposition 1.2

If $m_{\mathcal {A}}$ is the submultiplicative constant for $\mathcal {A}$ then $m_{r}(\mathcal {A}) \leq m_{\mathcal {A}}^{r-1}$. Also, if there exists $x \in \mathcal {A}$ for which $\| x^{r} \| = m_{\mathcal {A}}^{r-1} \| x\|^{r}$ for each $r \in \mathbb {N}$ then $m_{r}(\mathcal {A}) = m_{\mathcal {A}}^{r-1}$.

Proof

If $\| x \star y \| \leq m_{\mathcal {A}}\| x \| \| y \|$ for all $x,y \in \mathcal {A}$ then by repeated application of the inequality for the norm of the product of $x_{1},x_{2}, \dots, x_{r}$ we find

$$ \Vert x_{1} \star x_{2} \star\cdots\star x_{r} \Vert \leq m_{\mathcal {A}}^{r-1} \Vert x_{1} \Vert \Vert x_{2} \Vert \cdots \Vert x_{r} \Vert . $$

(1)

Therefore, $m_{r}(\mathcal {A}) \leq m_{\mathcal {A}}^{r-1}$. Suppose there exists $x \in \mathcal {A}$ for which $\| x^{r} \| = m_{\mathcal {A}}^{r-1} \| x \|^{r}$. Setting $x_{1} = x_{2} = \cdots= x_{r} = x$ shows (1) is sharp, hence $m_{r}(\mathcal {A}) = m_{\mathcal {A}}^{r-1}$. □

From these constants, we denote another constant related to the normed algebra $\mathcal {A}$

$$ \rho_{\mathcal {A}}= \limsup_{r \rightarrow\infty} \sqrt[r]{ m_{r}( \mathcal {A}) }. $$

(2)

Observe that $\rho_{\mathcal {A}}$ exists since $m_{r}(\mathcal {A}) \leq m_{\mathcal {A}}^{r - 1}$, so $\rho_{\mathcal {A}}\leq m_{\mathcal {A}}$. In another direction, substitute $x_{1} = x_{2} = \cdots= x_{r} = 1$ to obtain $\lVert1^{r} \rVert \le m_{r}(\mathcal {A}) \lVert1 \rVert^{r}$, which implies $m_{r}(\mathcal {A}) \ge \lVert1 \rVert^{-(r - 1)}$ and

$$ \lVert1 \rVert^{-1} \leq\rho_{\mathcal {A}}\leq m_{\mathcal {A}}. $$

(3)

For a finite-dimensional associative real algebra $\mathcal {A}$, the radius of convergence for the geometric series $1+z+z^{2}+ \cdots$ is shown to be at least $1/m_{\mathcal {A}}$ in Theorem 4.5 of [2]. We offer an improved radius in what follows:

Theorem 1.3

If $\sum_{n = 0}^{\infty}{c_{n} (z - z_{0})^{n}}$ has $0< C = \limsup_{n \rightarrow\infty} \sqrt[n]{ \lVert c_{n} \rVert} < \infty$ then the power series converges absolutely for $z \in \mathcal {A}$ such that $\lVert z - z_{0} \rVert < (C \rho_{A})^{-1}$.

Proof

For $C = \limsup_{n \rightarrow\infty} \sqrt[n]{ \lVert c_{n} \rVert}$ and $\lVert z - z_{0} \rVert < (C \rho_{A})^{-1}$ observe

$$ \limsup_{n \rightarrow\infty} \sqrt[n]{ \bigl\lVert c_{n} (z - z_{0})^{n} \bigr\rVert } \le\limsup_{n \rightarrow\infty} \sqrt[n]{m_{n}(\mathcal {A}) \lVert c_{n} \rVert \lVert z - z_{0} \rVert^{n}} = \rho_{A} C \lVert z-z_{0} \rVert < 1. $$

(4)

Thus $\limsup_{n \rightarrow\infty} \sqrt[n]{ \lVert c_{n} (z - z_{0})^{n} \rVert} < 1$, and we find $\sum_{n = 0}^{\infty}{c_{n} (z - z_{0})^{n}}$ converges absolutely by applying the Root Test for a series over an algebra (Theorem 4.2 of [2]). □

The theorem above shows that convergence of a power series $\sum c_{n} (z-z_{o})^{n}$ over an algebra $\mathcal {A}$ is governed both by the usual Cauchy–Hadamard criterion involving $\limsup_{n \rightarrow\infty} \sqrt[n]{\| c_{n} \|}$ and the constant $\rho_{\mathcal {A}}$. We note $\rho _{\mathcal {A}}$ depends on both $\mathcal {A}$ and the norm which $\mathcal {A}$ is given. In the remaining sections of this paper, we endeavor to calculate both $m_{r}(\mathcal {A})$ and $\rho_{\mathcal {A}}$ for several interesting algebras.

2 Transfer principle

Theorem 2.1

(Transfer Principle)

If $\mathcal {A}, \mathcal {B}$ are finite-dimensional, unital, associative, normed algebras over $\mathbb {R}$ or $\mathbb {C}$ for which there exists an algebra isomorphism $\varphi: \mathcal {A}\rightarrow \mathcal {B}$ which is also a dilation (meaning there exists a constant $\nu> 0$ so that $\lVert\varphi(x) \rVert = \nu \lVert x \rVert$ for all $x \in \mathcal {A}$). Then, for $r \ge2$, $\nu^{r - 1} \cdot m_{r}(\mathcal {B}) = m_{r}(\mathcal {A})$ and $\rho _{\mathcal {A}}= \nu\cdot\rho_{\mathcal {B}}$.

Proof

Observe $\lVert\varphi(x_{1} \cdots x_{r}) \rVert = \lVert\varphi(x_{1}) \cdots\varphi(x_{r}) \rVert \le m_{r}(\mathcal {B}) \lVert\varphi(x_{1}) \rVert \cdots \lVert\varphi(x_{r}) \rVert$. Then as φ is a dilation with constant ν, it follows $\| x_{1} \cdots x_{r} \| \leq m_{r}(\mathcal {B}) \nu ^{r-1} \lVert x_{1} \rVert \cdots \lVert x_{r} \rVert$. Recall $m_{r}(\mathcal {A})$ is the smallest constant for which $\| x_{1} \cdots x_{r} \| \leq m_{r}(\mathcal {A}) \lVert x_{1} \rVert \cdots \lVert x_{r} \rVert$ for all $x_{1}, \dots, x_{r} \in \mathcal {A}$. Therefore, $m_{r}(\mathcal {A}) \leq m_{r}(\mathcal {B}) \nu^{r-1}$. Observe $\varphi^{-1}: \mathcal {B}\rightarrow \mathcal {A}$ is a dilation with constant $\frac{1}{\nu}$ hence $m_{r}(\mathcal {B}) \leq m_{r}(\mathcal {A}) \frac{1}{\nu ^{r-1}}$, from which we find $m_{r}(\mathcal {A}) \geq\nu^{r-1}m_{r}(\mathcal {B})$. Hence $\nu^{r-1}m_{r}(\mathcal {B}) = m_{r}(\mathcal {A})$ and thus $\rho_{\mathcal {A}}= \limsup_{r \rightarrow\infty} \sqrt[r]{ \nu^{r-1}m_{r}(\mathcal {B})} = \nu\limsup_{r \rightarrow\infty} \sqrt[r]{m_{r}(\mathcal {B})} = \nu\rho _{\mathcal {B}}$. □

Theorem 2.2

(Product Algebra with Weighted p-norm)

Let $\mathcal {A}_{1}, \dots, \mathcal {A}_{s}$, be finite- dimensional associative normed algebras with rth submultiplicative constant $m_{i,r}$ for $r\geq2$ and with $\rho_{i} = \limsup_{r \rightarrow\infty} \sqrt [r]{m_{i,r}}$ for $1 \leq i \leq s$. If $\mathcal {A}= \mathcal {A}_{1} \times \mathcal {A}_{2} \times\cdots\times \mathcal {A}_{s}$ then define weighted p-norm with weights $a_{1},\dots, a_{s} >0$ by $\lVert(x_{1}, x_{2}, \ldots, x_{s}) \rVert = \max_{1 \le i \le s} {a_{i} \lVert x_{i} \rVert}$ for $p = \infty$ and

$$\bigl\lVert (x_{1}, x_{2}, \ldots, x_{s}) \bigr\rVert = \Biggl( \sum_{i = 1}^{s} a_{i}^{p} \lVert x_{i} \rVert^{p} \Biggr)^{\frac{1}{p}} $$

for $p< \infty$. Then $\mathcal {A}$ with the weighted p-norm defines a normed associative algebra with $m_{r}(\mathcal {A}) = \max_{1 \le i \le s} \{ m_{i,r}/a_{i}^{r - 1} \}$ and $\rho_{\mathcal {A}}= \max_{1 \le i \le s} \frac{\rho_{i}}{a_{i}}$.

Proof

Given $x_{i, j} \in A_{i}$ for $1 \leq i \leq s, 1 \leq j \leq r$. Let $\mu_{r} = m_{r}(\mathcal {A}) = \max_{1 \le i \le s} \{ m_{i,r}/a_{i}^{r - 1} \}$. Then

$$\begin{aligned} \mu_{r}^{p} \prod_{j = 1}^{r} \bigl\lVert (x_{1, j}, \ldots, x_{s, j}) \bigr\rVert ^{p} & = \mu_{r}^{p} \prod_{j = 1}^{r} \sum_{i = 1}^{s} a_{i}^{p} \lVert x_{i, j} \rVert^{p} \\ & \ge\mu_{r}^{p} \sum_{i = 1}^{s} \prod_{j = 1}^{r} a_{i}^{p} \lVert x_{i, j} \rVert^{p} \\ &= \mu_{r}^{p} \sum_{i = 1}^{s} \bigl(a_{i}^{p}\bigr)^{r} \Biggl( \prod _{j = 1}^{r} \lVert x_{i, j} \rVert \Biggr)^{p} \\ & \ge\max_{1 \le i \le s} \biggl\{ \frac{m_{i,r}^{p}}{a_{i}^{(r - 1)p}} \biggr\} \sum_{i = 1}^{s} \bigl(a_{i}^{p} \bigr)^{r} \Biggl( \frac{1}{m_{i, r}} \Biggl\lVert \prod _{j = 1}^{r} x_{i, j} \Biggr\rVert \Biggr)^{p} \\ & \ge\sum_{i = 1}^{s} \frac{m_{i,r}^{p}}{a_{i}^{(r - 1) p}} \cdot a_{i}^{rp} \cdot\frac{1}{m_{i, r}^{p}} \Biggl\lVert \prod_{j = 1}^{r} x_{i, j} \Biggr\rVert ^{p} \\ &= \sum_{i = 1}^{s} a_{i}^{p} \Biggl\lVert \prod_{j = 1}^{r} x_{i, j} \Biggr\rVert ^{p} \\ & = \Biggl\lVert \prod_{j = 1}^{r} ( x_{1, j}, \ldots, x_{s, j} ) \Biggr\rVert ^{p}. \end{aligned}$$

(5)

From the above we derive

$$ \mu_{r} \prod_{j = 1}^{r} \bigl\lVert (x_{1, j}, \ldots, x_{s, j}) \bigr\rVert \geq \Biggl\lVert \prod_{j = 1}^{r} ( x_{1, j}, \ldots, x_{s, j} ) \Biggr\rVert . $$

(6)

Let k be an index between 1 and s such that $\frac {m_{k,r}}{a_{k}^{r - 1}} = \mu_{r}$ and, following the last sentence in Theorem 1.1, let $y_{k, j} \in \mathcal {A}_{k}$ ($1 \le j \le r$) be such that $m_{k, r} \prod_{j = 1}^{r} \lVert y_{k, j} \rVert = \lVert\prod_{j = 1}^{r} y_{k, j} \rVert$. Let $e_{k} \in \mathcal {A}$ denote $(0,\dots, 1, \dots, 0)$ where 1 is in the kth entry. Then calculate

$$\begin{aligned} \mu_{r}^{p} \prod_{j = 1}^{r} \lVert y_{k, j} e_{k} \rVert^{p} &= \Biggl( a_{k} m_{k,r} \prod_{j = 1}^{r} \lVert y_{k, j} \rVert \Biggr)^{p} \\ &= \Biggl( a_{k} \Biggl\lVert \prod _{j = 1}^{r} y_{k, j} \Biggr\rVert \Biggr)^{p} \\ &= \Biggl\lVert \prod_{j = 1}^{r} (y_{k, j} e_{k}) \Biggr\rVert ^{p}, \end{aligned}$$

(7)

which indicates (6) is sharp, as equality is attained by setting $(x_{1, j}, \ldots, x_{s, j}) = y_{k, j} e_{k}$ for $j=1,\dots, r$. Thus $m_{r}(\mathcal {A}) = \max_{1 \le i \le s} \{ m_{i,r}/a_{i}^{r - 1} \}$. Finally, observe

$$\begin{aligned} \rho_{\mathcal {A}} &= \limsup_{r \rightarrow\infty} \sqrt[r]{ \max _{1 \le i \le s} \bigl\{ m_{i,r}/a_{i}^{r - 1} \bigr\} } \\ &= \limsup_{r \rightarrow\infty} \max_{1 \le i \le s} \biggl\{ \frac{\sqrt[r]{m_{i,r}}}{\sqrt[r]{a_{i}^{r - 1}}} \biggr\} \\ &= \max_{1 \le i \le s} \biggl\{ \limsup_{r \rightarrow\infty} \frac{\sqrt[r]{m_{i,r}}}{\sqrt[r]{a_{i}^{r - 1}}} \biggr\} \\ &= \max_{1 \le i \le s} \biggl\{ \frac{\rho_{i}}{a_{i}} \biggr\} . \end{aligned}$$

(8)

 □

Remark 2.3

If $\varphi: \mathcal {B}\rightarrow \mathcal {A}= \mathcal {A}_{1} \times\cdots\times \mathcal {A}_{s}$ is an isomorphism which is also a dilation then the proof given suggests a calculational method for finding an element $x \in \mathcal {B}$ for which $\| x^{2} \| = m_{\mathcal {B}} \| x \|^{2}$. In particular, select the element $y_{k, j} e_{k}$ for $r=2$ as in the proof and set $x=\varphi^{-1}(y_{k, j} e_{k})$. In this construction we will find that x is a zero-divisor. However, if we consider algebras which are not product algebras then the situation is different. For example, in Theorem 5.1(1) we find that the unit $x = \sqrt {2}+\varepsilon$ gives $\| x \| = m_{\mathcal {A}}\| x\|^{2}$.

3 Submultiplicative constants for cyclic group algebra

The n-hyperbolic numbers are the algebra $\mathcal {H}_{n} = \mathbb{R}[x] / \langle x^{n} - 1 \rangle$ which is usually represented as

$$\mathcal {H}_{n} = \bigl\{ a_{0} + a_{1} j + \cdots+ a_{n - 1} j^{n - 1} | a_{i} \in \mathbb{R} \bigr\} , $$

where we identify j with $x + \langle x^{n} - 1 \rangle$ and 1 with $1 + \langle x^{n} - 1 \rangle$. One may note that $\mathcal {H}_{n}$ is the real group algebra of the cyclic group of order n. In this section we derive the submultiplicative constants for hyperbolic numbers provided that $\mathcal {H}_{n}$ is given the norm

$$\bigl\Vert a_{0} + a_{1} j + \cdots+ a_{n - 1} j^{n - 1} \bigr\Vert = \sqrt{a_{0}^{2} + a_{1}^{2} + \cdots+ a_{n - 1}^{2}}. $$

3.1 $\mathcal {H}_{n}$ for even n

Notice that if $n=2m$ then we have the following factorization of $x^{n}-1$:

$$ x^{n}-1 = (x - 1) (x + 1) \prod_{k = 1}^{m - 1} \bigl( x - e^{i \frac{k \pi}{m}} \bigr) \bigl( x - e^{-i \frac{k \pi}{m}} \bigr). $$

(9)

The Chinese remainder theorem and the factorization above suggests we form the isomorphism $\varphi: \mathcal {H}_{2m} \rightarrow\mathbb{R}^{2} \times\mathbb{C}^{m - 1}$ we describe next. If $p(j) = a_{0}+ a_{1}j + \cdots+a_{n}j^{n-1} \in \mathcal {H}_{n}$ then $p(c) = a_{0}+a_{1}c+ \cdots+ a_{n} c^{n-1}$ for $c \in \mathbb {C}$, thus for $p(j) \in \mathcal {H}_{n}$ define

$$ \varphi\bigl(p(j)\bigr) = \bigl(p(1),p(-1),p\bigl(e^{i\pi/m} \bigr),p\bigl(e^{2i\pi/m}\bigr), \dots, p\bigl(e^{(m-1)i\pi/m}\bigr)\bigr). $$

(10)

Observe $\varphi: \mathcal {H}_{n} \rightarrow \mathbb {R} \times \mathbb {R} \times \mathbb {C} ^{m-1}$ is an isomorphism of real associative algebras. Our goal is to use Theorem 2.1 by assigning $\mathcal{B} = \mathbb {R} \times \mathbb {R} \times \mathbb {C} ^{m-1}$ the appropriate weighted norm as to make (10) a dilation. After some experimentation, one finds that

$$ \bigl\lVert (a, b, z_{1}, z_{2}, \ldots, z_{m - 1}) \bigr\rVert = \sqrt {a^{2} + b^{2} + 2 \bigl( \vert z_{1} \vert ^{2} + \cdots+ \vert z_{m - 1} \vert ^{2} \bigr)} $$

(11)

serves our purpose. Considering $p(j) = \sum_{k = 0}^{2m - 1} a_{k} j^{k}$,

$$\begin{aligned} \bigl\lVert \varphi \bigl( p(j) \bigr) \bigr\rVert ^{2} ={}& \Biggl( \sum_{k = 0}^{2m - 1} a_{k} \Biggr)^{2} + \Biggl( \sum_{k = 0}^{2m - 1} (-1)^{k} a_{k} \Biggr)^{2} + 2 \sum _{s = 1}^{m - 1} \Biggl\vert \sum _{k = 0}^{2m - 1} a_{k} e^{\frac {iks\pi}{m}} \Biggr\vert ^{2} \\ ={}& 2 \sum_{k = 0}^{2m - 1} a_{k}^{2} + 2 \sum_{p < q} a_{p} a_{q} \bigl( 1 + (-1)^{p + q} \bigr) \\ & + 2 \sum_{s = 1}^{m - 1} \Biggl[ \Biggl( \sum_{k = 0}^{2m - 1} a_{k} \cos \biggl( \frac{ks \pi}{m} \biggr) \Biggr)^{2} + \Biggl( \sum _{k = 0}^{2m - 1} a_{k} \sin \biggl( \frac{ks \pi}{m} \biggr) \Biggr)^{2} \Biggr] \\ ={}& 2m \sum_{k = 0}^{2m - 1} a_{k}^{2} + 2 \sum_{p < q} a_{p} a_{q} \bigl( 1 + (-1)^{p + q} \bigr) \\ &{} + 4 \sum_{s = 1}^{m - 1} \sum _{p < q} a_{p} a_{q} \biggl[ \cos \biggl( \frac{ps \pi}{m} \biggr) \cos \biggl( \frac{qs \pi}{m} \biggr) + \sin \biggl( \frac{ps \pi}{m} \biggr) \sin \biggl( \frac {qs \pi}{m} \biggr) \biggr] \\ ={}& 2m \bigl\Vert p(j) \bigr\Vert ^{2} + 2 \sum _{p < q} a_{p} a_{q} \Biggl[ 1 + (-1)^{p + q} + 2 \sum_{s = 1}^{m - 1} \cos \biggl( \frac{(p - q)s\pi}{m} \biggr) \Biggr] \\ ={}& 2m \bigl\Vert p(j) \bigr\Vert ^{2} + 2 \sum _{p < q} a_{p} a_{q} \sum _{s = 0}^{2m - 1} \cos \biggl(\frac{(p - q)s\pi}{m} \biggr). \end{aligned}$$

(12)

Lemma 3.1

If $p,q,m$ are integers and $p< q$ then $\sum_{s = 0}^{2m - 1} \cos (\frac{(p - q)s\pi}{m} )=0$.

Proof

If $p-q$ is not a multiple of m then let $\theta= \pi/m$ and note the points $1,e^{i\theta},\dots, e^{i(2m-1)\theta}$ form a set of 2m-symmetrically arranged points on the unit circle whose centroid is the origin. Since $e^{i\theta}=\cos\theta+i\sin\theta$, we identify $\frac{1}{n}\sum_{s = 0}^{2m - 1} \cos (\frac{(p - q)s\pi}{m} )$ as the real coordinate of the centroid. If $p-q$ is a multiple of m then $\cos (\frac{(p - q)s\pi}{m} ) = (-1)^{s}$ and the lemma follows. □

In view of Lemma 3.1 and (12), we find $\lVert\varphi( z) \rVert^{2} = 2m \| z \|^{2}$ for each $z \in \mathcal {H}_{n}$. Thus φ is a dilation with constant $\nu= \sqrt{2m}$. Let $\mathcal{B} = \mathbb {R} ^{2} \times \mathbb {C} ^{m-1}$ have norm as in (11). Then Theorem 2.2 indicates the submultiplicative constant of $\mathcal{B}$ is $m_{\mathcal{B}}=1$. Therefore, Theorem 2.1 provides $m_{\mathcal {H}_{n}} = \nu m_{\mathcal{B}}$, thus $m_{\mathcal {H}_{n}} = \sqrt{2m} = \sqrt{n}$.

3.2 $\mathcal {H}_{n}$ for odd n

Notice that if $n=2m+1$ then we have the following factorization of $x^{n}-1$:

$$ x^{n}-1 = (x - 1)\prod_{k = 1}^{m} \bigl( x - e^{\frac{2\pi k i}{2m+1}} \bigr) \bigl( x - e^{-\frac{2\pi k i}{2m+1}} \bigr). $$

(13)

As in (10) we define an isomorphism $\varphi: \mathcal {H}_{2m+1} \rightarrow \mathbb {R} \times \mathbb {C} ^{m}$ for each $p(j) \in \mathcal {H}_{2m+1}$ by

$$ \varphi\bigl(p(j)\bigr) = \bigl( p(1), p \bigl( e^{\frac{ 2 \pi i}{2m + 1}} \bigr), p \bigl( e^{\frac{ 4 \pi i}{2m + 1}} \bigr), \ldots, p \bigl( e^{ \frac{2m \pi i }{2m + 1}} \bigr) \bigr). $$

(14)

Let $\mathcal{B} = \mathbb {R} \times \mathbb {C} ^{m}$ have norm

$$ \bigl\lVert (a, z_{1}, \ldots, z_{m}) \bigr\rVert = \sqrt{a^{2} + 2 \bigl( \vert z_{1} \vert ^{2} + \cdots+ \vert z_{m} \vert ^{2} \bigr)}. $$

(15)

Thus for $p(j) = \sum_{k=0}^{2m}a_{k} j^{k}$,

$$\begin{aligned} \bigl\lVert \varphi \bigl( p(j) \bigr) \bigr\rVert ^{2} ={} &\Biggl( \sum_{k = 0}^{2m} a_{k} \Biggr)^{2} + 2 \sum_{s = 1}^{m} \Biggl\vert \sum_{k = 0}^{2m} a_{k} e^{i\frac{i2k s\pi}{2m + 1}} \Biggr\vert ^{2} \\ ={}& \sum_{k = 0}^{2m} a_{k}^{2} + 2 \sum_{p < q} a_{p} a_{q}\\ &{}+ 2 \sum_{s = 1}^{m} \Biggl[ \Biggl( \sum_{k = 0}^{2m} a_{k} \cos \biggl( \frac{2ks \pi}{2m + 1} \biggr) \Biggr)^{2} + \Biggl( \sum_{k = 0}^{2m} a_{k} \sin \biggl( \frac{2ks \pi}{2m + 1} \biggr) \Biggr)^{2} \Biggr] \\ ={}& (2m + 1) \sum_{k = 0}^{2m} a_{k}^{2} + 2 \sum_{p < q} a_{p} a_{q} \\ &{} + 4 \sum_{s = 1}^{m} \sum _{p < q} a_{p} a_{q} \biggl[ \cos \biggl( \frac{ 2sp \pi}{2m + 1} \biggr) \cos \biggl( \frac{2sq \pi}{2m + 1} \biggr)\\ &{} + \sin \biggl( \frac{2sp \pi}{2m + 1} \biggr) \sin \biggl( \frac{2sq \pi}{2m + 1} \biggr) \biggr] \\ ={}& (2m + 1) \bigl\Vert p(j) \bigr\Vert ^{2} + 2 \sum _{p < q } a_{p} a_{q} \Biggl[1 + 2 \sum_{s = 1}^{m} \cos \biggl( \frac{2s(p - q) \pi}{2m + 1} \biggr) \Biggr] \\ ={}& (2m + 1) \bigl\Vert p(j) \bigr\Vert ^{2} + 2 \sum _{p < q} a_{p} a_{q} \sum _{s = 0}^{2m} \cos \biggl( \frac{2s(p - q) \pi}{2m + 1} \biggr). \end{aligned}$$

(16)

The proof of Lemma 3.1 is easily adapted to show the following result:

Lemma 3.2

If $p,q,m$ are integers and $p< q$ then $\sum_{s = 0}^{2m} \cos (\frac{2s(p - q)\pi}{2m+1} )=0$.

In view of the Lemma 3.2 and (16), we find $\lVert\varphi( z) \rVert^{2} = (2m+1) \| z \|^{2}$ for each $z \in \mathcal {H}_{n}$ where $n=2m+1$. Thus φ is a dilation with constant $\nu= \sqrt{2m+1}$. Let $\mathcal{B} = \mathbb {R} \times \mathbb {C} ^{m}$ have norm as in (15). Then Theorem 2.2 indicates that the submultiplicative constant of $\mathcal{B}$ is $m_{\mathcal {B}}=1$. Therefore, Theorem 2.1 provides $m_{\mathcal {H}_{n}} = \nu m_{\mathcal{B}}$ and thus $m_{\mathcal {H}_{n}} = \sqrt{2m+1} = \sqrt{n}$.

Theorem 3.3

If $\mathcal {H}_{n}$ is given norm $\| a_{0}+a_{1}j+ \cdots +a_{n}j^{n-1} \| = \sqrt{ a_{0}^{2}+a_{2}^{2}+ \cdots +a_{n}^{2}}$ then $m_{\mathcal {H}_{n}} = \sqrt{n}$. Moreover, for $r \geq2$ we find $m_{r}( \mathcal {H}_{n}) = n^{(r-1)/2}$ and $\rho_{\mathcal {H}_{n}} = \sqrt{n}$.

Proof

We have already proved $m_{\mathcal {H}_{n}} = \sqrt{n}$. Consider $x = 1+j+j^{2}+ \cdots+ j^{n-1}$ and observe $x^{2}=nx$, hence $x^{r} = n^{r-1}x$ for $r \in \mathbb {N}$. Moreover, $\| x\| = \sqrt{n}$, thus $\| x^{r} \| = m_{\mathcal {H}_{n}}^{r-1} \| x \|^{r}$ and Proposition 1.2 provides $m_{r}(\mathcal {H}_{n}) = n^{(r-1)/2}$. Finally, calculate

$$ \rho_{\mathcal {H}_{n}} = \limsup_{r \rightarrow\infty} \sqrt[r]{ m_{r}( \mathcal {H}_{n}) } = \limsup_{r \rightarrow\infty} \sqrt[r]{ n^{(r-1)/2} } = \sqrt{n}. $$

(17)

 □

It is interesting to note that the inequality $\| xy \| \leq\sqrt{n} \| x\| \| y \|$ is sharp for $x=y = 1+j+j^{2}+ \cdots+ j^{n-1}$ since $\| x \| = \sqrt{n}$ and as

$$ x^{2} = \bigl(1+j+j^{2}+ \cdots+ j^{n-1} \bigr)^{2} = n\bigl(1+j+j^{2}+ \cdots+ j^{n-1}\bigr) = nx. $$

(18)

Thus $\| xy \| = n \| x \| = n \sqrt{n} = \sqrt{n} \| x\| \| y\|$. The same x has $x(x-n)=0$, which shows x and $x-n$ are zero-divisors. This illustrates Remark 2.3.

4 Submultiplicative constant of the complicated numbers

We follow [3] and say $\mathcal{C}_{n} = \{ a_{0}+a_{1}k+ \cdots +a_{n-1}k^{n-1} | a_{i} \in \mathbb {R} , k^{n}=-1 \}$ is the set of complicated numbers. Let us assume the norm on $\mathcal{C}_{n}$ is given by

$$ \bigl\Vert a_{0}+a_{1}k+ \cdots+a_{n-1}k^{n-1} \bigr\Vert = \sqrt{a_{0}^{2}+ \cdots+a_{n-1}^{2}} $$

(19)

throughout this section.

Theorem 4.1

The submultiplicative constant

$$m_{\mathcal{C}_{n}} = \textstyle\begin{cases} \sqrt{n/2} & \textit{if }n\textit{ is even,} \\ \sqrt{n} & \textit{if}n\textit{ is odd.} \end{cases} $$

Proof

If n is odd then $\varPsi(k) = -j$ defines an isomorphism of $\mathcal {C}_{n}$ and $\mathcal{H}_{n}$ since $\varPsi(k)^{n} = (-j)^{n} = (-1)^{n}j^{n}=-1$. Extending Ψ to arbitrary elements,

$$\varPsi \bigl( a_{0}+a_{1}k+ \cdots+a_{n-1}k^{n-1} \bigr) = a_{0}-a_{1}j+a_{2}j^{2}+ \cdots+a_{n-1}j^{n-1}. $$

If $\| a_{0}+a_{1}k+ \cdots+a_{n-1}k^{n-1} \| = \sqrt{a_{0}^{2}+ \cdots +a_{n-1}^{2}}$ then using the norm before discussed for $\mathcal{H}_{n}$ observe $\| \varPsi(z) \| = \| z \|$. Therefore, from Theorem 2.1 we find $m_{\mathcal{C}_{n}} = m_{\mathcal{H}_{n}}$.

Consider $\mathcal{C}_{n}$ for $n=2m$ with $m \geq1$. Let $\theta_{j} = \frac{\pi}{n}(2j+1)$ and $x_{j} = \exp ( i\theta_{j} )$ for $j=0,1, \dots, n/2-1$. Define $\varphi: \mathcal{C}_{n} \rightarrow \mathbb {C} ^{n/2}$ by

$$ \varphi\bigl( f(k)\bigr) = \bigl(f(x_{0}), \dots, f(x_{n/2-1}) \bigr). $$

(20)

If $f(k) = a_{0}+a_{1}k+ \cdots+ a_{n-1}k^{n-1} \in\mathcal{C}_{n}$ then calculate

$$\begin{aligned} \bigl\vert f(x_{j}) \bigr\vert ^{2} &= \Biggl\vert \sum_{q=1}^{n-1} a_{q}\exp ( iq\theta_{j} ) \Biggr\vert ^{2} \\ &= \Biggl\vert \sum_{q=1}^{n-1} a_{q} \bigl[ \cos ( q\theta_{j} )+i\sin ( q \theta_{j} ) \bigr] \Biggr\vert ^{2} \\ & = \Biggl[ \sum_{q=1}^{n-1} a_{q} \cos ( q\theta_{j} ) \Biggr]^{2}+ \Biggl[ \sum_{q=1}^{n-1} a_{q}\sin ( q \theta_{j} ) \Biggr]^{2} \\ & = \sum_{q=1}^{n-1} a_{q}^{2}+ \sum_{p< q} a_{p}a_{q} \bigl[\cos ( p\theta_{j} )\cos ( q \theta_{j} )+\sin ( p\theta_{j} )\sin ( q\theta_{j} ) \bigr] \\ & = \bigl\Vert f(k) \bigr\Vert ^{2}+ \sum _{p< q} a_{p}a_{q} \cos \bigl( (p-q) \theta_{j} \bigr). \end{aligned}$$

(21)

Observe that, much for the same reasons as given for Lemma 3.1,

$$ \sum_{j=0}^{n/2-1}\cos \bigl( (p-q)\theta_{j} \bigr) = \sum_{j=0}^{n/2-1} \cos \biggl( \frac{\pi(p-q)(2j+1)}{2m} \biggr)=0. $$

(22)

If $\mathbb {C} ^{n/2}$ has norm $\| (z_{1}, \dots, z_{n/2}) \|^{2} = |z_{1}|^{2}+ \cdots+ |z_{n/2}|^{2}$ then Theorem 2.2 indicates $m_{ \mathbb {C} ^{n/2}}=1$. Notice (21) and (22) yield

$$ \bigl\Vert \varphi\bigl( f(k)\bigr) \bigr\Vert ^{2} = \sum _{j=0}^{n/2-1} \bigl\vert f(x_{j}) \bigr\vert ^{2} = \frac {n}{2} \bigl\Vert f(k) \bigr\Vert ^{2}. $$

(23)

Consequently, φ is a dilation with $\nu= \sqrt{n/2}$, and Theorem 2.1 provides $m_{\mathcal{C}_{n}} = \sqrt{n/2}$. □

In the special case $n=2$, we note $\mathcal{C}_{2}$ has a multiplicative norm. However, for other $n >2$ the norm is not multiplicative. Furthermore, $\| xy \| = m_{\mathcal{C}_{n}} \| x \| \| y \|$ only for select $x,y$. In particular, if n is odd then $x = 1-k+k^{2}-\cdots+k^{n-1} \in\mathcal{C}_{n}$ has $x^{2} = nx$, thus $\| x^{2} \| = \sqrt{n} \| x \|^{2}$. In contrast, for even $n \neq2$ notice $x = \sum_{j=0}^{n-1}\cos ( \frac{j \pi}{n} ) k^{j}$ is a nonzero zero-divisor with $x^{2} = \frac{n}{2}x$ and $\| x \| = \sqrt {\frac{n}{2}}$. Thus $\| x^{2} \| = \sqrt{\frac{n}{2}} \| x \|^{2}$. Once again, Remark 2.3 is illustrated.

5 Submultiplicative constant of the dual numbers

The n-dual number is defined as an element of $\Delta_{n} = \mathbb {R}[x] / \langle x^{n} \rangle$, or equivalently, $a_{0} + a_{1} \varepsilon + \cdots+ a_{n - 1} \varepsilon^{n - 1}$ where $\varepsilon^{n} = 0$. We study the norm on $\Delta_{n}$ defined by

$$ \bigl\lVert a_{0} + a_{1} \varepsilon+ \cdots+ a_{n - 1} \varepsilon^{n - 1} \bigr\rVert = \sqrt {a_{0}^{2} + a_{1}^{2} + \cdots+ a_{n - 1}^{2}}. $$

(24)

Theorem 5.1

The submultiplicative constant for $\Delta_{n}$ is

• (1) for $n=2$, $m_{2}(\Delta_{2}) = \frac{2}{\sqrt{3}}$;

• (2) for $n=3$, $m_{2}(\Delta_{3}) = \frac{4}{3}$.

Proof

In all of the cases, we mainly use the AM–GM inequality. Begin with claim (1). Suppose $x = a_{0} + a_{1} \varepsilon, y = b_{0} + b_{1} \varepsilon$ and observe:

$$\begin{aligned} 4 \lVert x \rVert^{2} \cdot \lVert y \rVert^{2} & = 4 \bigl(a_{0}^{2} + a_{1}^{2}\bigr) \bigl(b_{0}^{2} + b_{1}^{2}\bigr) \\ & = 3 \bigl(a_{0}^{2} b_{0}^{2} + a_{1}^{2} b_{0}^{2} + a_{0}^{2} b_{1}^{2}\bigr) + \bigl(a_{0}^{2} b_{0}^{2} + 4a_{1}^{2} b_{1}^{2}\bigr) + \bigl(a_{1}^{2} b_{0}^{2} + a_{0}^{2} b_{1}^{2}\bigr) \\ & \ge3 \bigl(a_{0}^{2} b_{0}^{2} + a_{1}^{2} b_{0}^{2} + a_{0}^{2} b_{1}^{2}\bigr) + 4 a_{0} b_{0} a_{1} b_{1} + 2 a_{1} b_{0} a_{0} b_{1} \\ & = 3 \bigl[ (a_{0} b_{0})^{2} + (a_{1} b_{0} + a_{0} b_{1})^{2} \bigr] = 3 \bigl\lVert a_{0} b_{0} + (a_{1} b_{0} + a_{0} b_{1}) \varepsilon \bigr\rVert ^{2} \\ & = 3 \lVert x y \rVert^{2}. \end{aligned}$$

(25)

If we set $a_{0} = b_{0} = \sqrt{2}$ and $a_{1} = b_{1} = 1$ then equality is attained. Next, prove claim (2). We suppose $x = a_{0} + a_{1} \varepsilon + a_{2} \varepsilon^{2}$ and $y = b_{0} + b_{1} \varepsilon+ b_{2} \varepsilon ^{2}$. Calculate:

$$\begin{aligned} 16 \lVert x \rVert^{2} \cdot \lVert y \rVert^{2} ={}& 16 \bigl(a_{0}^{2} + a_{1}^{2} + a_{2}^{2}\bigr) \bigl(b_{0}^{2} + b_{1}^{2} + b_{2}^{2}\bigr) \\ ={}& 9 \bigl( a_{0}^{2} b_{0}^{2} + a_{1}^{2} b_{0}^{2} + a_{0}^{2} b_{1}^{2} + a_{0}^{2} b_{2}^{2} + a_{1}^{2} b_{1}^{2} + a_{2}^{2} b_{0}^{2} \bigr) + 16 a_{1}^{2} b_{2}^{2} + 4 a_{0}^{2} b_{1}^{2} \\ &{} + 16 a_{2}^{2} b_{1}^{2} + 4 a_{1}^{2} b_{0}^{2} + 16 a_{2}^{2} b_{2}^{2} + a_{0}^{2} b_{0}^{2} + 3 \bigl( a_{0}^{2} b_{1}^{2} + a_{1}^{2} b_{0}^{2} \bigr) \\ &{} + 6 \bigl( a_{0}^{2} b_{0}^{2} + a_{1}^{2} b_{1}^{2} \bigr) + \frac{1}{2} a_{1}^{2} b_{1}^{2} + 2 a_{2}^{2} b_{0}^{2} + \frac{1}{2} a_{1}^{2} b_{1}^{2} + 2 a_{0}^{2} b_{2}^{2} + 5 \bigl( a_{0}^{2} b_{2}^{2} + a_{2}^{2} b_{0}^{2} \bigr) \\ \ge{}&9 \bigl( a_{0}^{2} b_{0}^{2} + a_{1}^{2} b_{0}^{2} + a_{0}^{2} b_{1}^{2} + a_{0}^{2} b_{2}^{2} + a_{1}^{2} b_{1}^{2} + a_{2}^{2} b_{0}^{2} \bigr) \\ &{} + 16 a_{1} b_{2} a_{0} b_{1} + 16 a_{2} b_{1} a_{1} b_{0} + 8 a_{2} b_{2} a_{0} b_{0} + 6 a_{0} b_{1} a_{1} b_{0} \\ &{} + 12 a_{0} b_{0} a_{1} b_{1} + 2 a_{1} b_{1} a_{2} b_{0} + 2 a_{1} b_{1} a_{0} b_{2} + 10 a_{0} b_{2} a_{2} b_{0} \\ ={}& 9 \bigl[ (a_{0} b_{0})^{2} + (a_{1} b_{0} + a_{0} b_{1})^{2} + (a_{0} b_{2} + a_{1} b_{1} + a_{2} b_{0})^{2} \bigr] \\ ={}& 9 \bigl\lVert a_{0} b_{0} + (a_{1} b_{0} + a_{0} b_{1}) \varepsilon+ (a_{0} b_{2} + a_{1} b_{1} + a_{2} b_{0}) \varepsilon^{2} \bigr\rVert ^{2} \\ ={}& 9 \lVert xy \rVert^{2}. \end{aligned}$$

(26)

Equality occurs if, for instance, $a_{0} = b_{0} = a_{1} = b_{1} = 2, a_{2} = b_{2} = 1$. □

It is interesting to note that in both cases ($n = 2, 3$) we find equality occurs when

$$ \frac{a_{0}}{b_{0}} = \frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}}. $$

(27)

Assuming the above trend continues, we conjecture that:

• (1) for $n = 4$, $m_{2}(\Delta_{4}) = \sqrt{\frac{2 (1103 + 33\sqrt {33})}{1153}} \approx1.49736$. We note equality in $\| xy \| \leq m_{2}(\Delta_{4}) \| x \| \|y \|$ is attained for $x = y = \sqrt{5 + \sqrt{\frac{11}{3}}} + ( 1 + \sqrt{\frac {11}{3}} ) \varepsilon+ \sqrt{3 + \sqrt{\frac{11}{3}}} \varepsilon^{2} + \varepsilon^{3} \approx2.62961 + 2.91485 \varepsilon+ 2.21695 \varepsilon^{2} + \varepsilon^{3}$.

• (2) for $n = 5$, the closed form expression is too lengthy for us to display here, however, an approximation of the constant is $m_{2}(\Delta _{5}) \approx1.64748$. Equality in $\| xy \| \leq m_{2}(\Delta_{5}) \| x \| \|y \|$ is attained when $x = y \approx1 - 1.14862 \varepsilon+ 1.03046 \varepsilon^{2} - 0.700308 \varepsilon^{3} + 0.304849 \varepsilon^{4}$.

• (3) for $n = 6$, the submultiplicative constant is $m_{2}(\Delta_{6}) \approx1.78611$ and equality in $\| xy \| \leq m_{2}(\Delta_{6}) \| x \| \|y \|$ is attained at $x = y \approx1 - 1.16152 \varepsilon+ 1.12963 \varepsilon^{2} - 0.915093 \varepsilon^{3} + 0.589126 \varepsilon^{4} - 0.253601 \varepsilon^{5}$.

Theorem 5.2

For $r \in\mathbb{N}, r \ge2$, we have $m_{r}(\Delta_{2}) = \sqrt{\frac{r + 1}{ ( 1 + \frac{1}{r} )^{r}}}$.

Proof

Considering $a_{i}, b_{i} \in\mathbb{R}$ for $1 \le i \le r$, since $\varepsilon^{2}=0$, we find

$$ \prod_{i = 1}^{r} (a_{i} + b_{i} \varepsilon) = \prod_{i = 1}^{r} a_{i} + \varepsilon\sum_{i = 1}^{r} \biggl( \prod_{j \neq i} a_{j} \biggr) b_{i}. $$

(28)

Let $g^{2}$ be the smallest constant such that the inequality below holds for arbitrary $a_{i}, b_{i}$,

$$ g^{2} \prod_{i = 1}^{r} \bigl(a_{i}^{2} + b_{i}^{2}\bigr) \ge \prod_{i = 1}^{r} a_{i}^{2} + \Biggl[ \sum_{i = 1}^{r} \biggl( \prod _{j \neq i} a_{j} \biggr) b_{i} \Biggr]^{2}. $$

(29)

Substituting $a_{i} = 1$ and $b_{i} = 0$, we have $g^{2} \ge1$. If there is at least one $a_{i} = 0$ then, without loss of generality, we may suppose $a_{r} = 0$, and thus for all $a_{i}, b_{i} \in\mathbb{R}$

$$ g^{2} \prod_{i = 1}^{r - 1} \bigl(a_{i}^{2} + b_{i}^{2}\bigr) b_{r}^{2} \ge\prod_{i = 1}^{r - 1} \bigl(a_{i}^{2} + b_{i}^{2}\bigr) b_{r}^{2} \ge a_{1}^{2} a_{2}^{2} \cdots a_{r - 1}^{2} b_{r}^{2}. $$

(30)

Thus (29) holds trivially. Therefore, let us suppose all $a_{i}$ are non-zero. Then we can normalize the above inequality to (setting $x_{i} = b_{i} / a_{i}$)

$$ g^{2} \prod_{i = 1}^{r} \bigl(1 + x_{i}^{2}\bigr) \ge1 + \Biggl( \sum _{i = 1}^{r} x_{i} \Biggr)^{2}. $$

(31)

Define $f(x_{1}, \ldots, x_{r}) = \frac{1 + (x_{1} + \cdots+ x_{r})^{2}}{(1 + x_{1}^{2}) \cdots(1 + x_{r}^{2})}$ and note by construction $g^{2} \geq f(x_{1}, \ldots, x_{r})$ for all $x_{1}, \dots, x_{r}$. We seek solutions to $\nabla f = 0$ in the interest of finding the global maximum of f. Set $S = x_{1} + x_{2} + \cdots+ x_{n}$ and calculate

$$ \frac{\partial f}{\partial x_{i}} = -2 \frac{(S - x_{i})(x_{i} S - 1)}{(1 + x_{i}^{2}) \prod_{k = 1}^{r} (1 + x_{k}^{2})}. $$

(32)

Hence, $\nabla f = 0$ if and only if $(S - x_{i})(x_{i} S - 1) = 0$ for all $1 \le i \le r$.

• (1) If $S = 0$ then $(S - x_{i})(x_{i} S - 1) = 0$ implies $x_{1}=0, x_{2}=0, \dots, x_{r}=0$.

• (2) Suppose $x_{1}, x_{2}, \ldots, x_{r} \neq0$. Then $(S - x_{i})(x_{i} S - 1) = 0$ yields

  $$S + \frac{1}{S} = x_{1} + \frac{1}{x_{1}} = x_{2} + \frac{1}{x_{2}} = \cdots = x_{r} + \frac{1}{x_{r}}, $$

  and we find solution $x_{1} = x_{2} = \cdots= x_{r} = \pm\frac{1}{\sqrt{r}}$.

Observe $f(0, \ldots, 0) = 1$ whereas $f ( \frac{1}{\sqrt{r}}, \ldots, \frac{1}{\sqrt{r}} ) = f ( -\frac{1}{\sqrt{r}}, \ldots, -\frac{1}{\sqrt{r}} ) = \frac{r + 1}{ ( 1 + \frac{1}{r} )^{r}} \ge1$ for all $r \geq2$. Therefore we conclude $m_{r} (\Delta_{2}) = \sqrt{g} = \sqrt{\frac{r + 1}{ ( 1 + \frac{1}{r} )^{r}}}$. □

Let $\Delta_{n}$ have p-norm defined for $p< \infty$ by

$$ \bigl\Vert a_{1}+a_{2}\varepsilon+ \cdots+a_{n} \varepsilon^{n-1} \bigr\Vert = \sqrt [p]{a_{1}^{p}+a_{2}^{p}+ \cdots+a_{n}^{p}}. $$

(33)

Theorem 5.3

Suppose $p \in(1, \infty)$. If $\Delta_{n}$ is endowed with p-norm then $\rho_{\Delta_{n}} = 1$.

Proof

Let it be understood that $m_{r}(\Delta_{n})$ is defined with respect to the p-norm throughout this proof. Suppose $a_{i,j} > 0$. We will attempt to bound $m_{r} (\Delta_{n})$ by an appropriate constant. We have

$$ \prod_{i = 1}^{r} \sum _{j = 0}^{n - 1} a_{i,j} \varepsilon^{j} = \sum_{k = 0}^{n - 1} \Biggl( \sum _{d_{1} + \cdots+ d_{r} = k} \prod_{i = 1}^{r} a_{i, d_{i}} \Biggr) \varepsilon^{k}. $$

(34)

Therefore,

$$ m_{r}(\Delta_{n})^{p} \prod _{i = 1}^{r} \sum_{j = 0}^{n - 1} a_{i, j}^{p} \ge \sum_{k = 0}^{n - 1} \Biggl( \sum_{d_{1} + \cdots+ d_{r} = k} \prod _{i = 1}^{r} a_{i, d_{i}} \Biggr)^{p}. $$

(35)

A few observations about (35) are as follows:

• (1) There are $\binom{k + r - 1}{k}$ distinct terms $\prod_{i = 1}^{r} a_{i, d_{i}}$ in the sum

  $$\sum_{d_{1} + \cdots+ d_{r} = k} \prod_{i = 1}^{r} a_{i, d_{i}}. $$

• (2) Each term $\prod_{i = 1}^{r} a_{i, d_{i}}$ to the power p occurs in the product

  $$\prod_{i = 1}^{r} \sum _{j = 0}^{n - 1} a_{i, j}^{p}. $$

• (3) $\binom{(k + 1) + r - 1}{k + 1} = \binom{k + r - 1}{k} + \binom {k + r - 1}{k + 1} \ge\binom{k + r - 1}{k}$, so the sequence $\binom {k + r - 1}{k}$ is increasing in variable k and thus $\max_{0 \le k < n} \bigl\{ \binom{k + r - 1}{k}\bigr\} = \binom{n + r - 2}{n - 1} = O(r^{n - 1})$.

• (4) By Hölder’s inequality with $q = \frac{p}{p - 1}$,

  $$x_{1} \cdot1 + x_{2} \cdot1 + \cdots+ x_{s} \cdot1 \le \bigl( x_{1}^{p} + x_{2}^{p} + \cdots+ x_{s}^{p} \bigr)^{\frac{1}{p}} \bigl( 1^{q} + 1^{q} + \cdots + 1^{q} \bigr)^{\frac{1}{q}}. $$

  Consequently, $( x_{1} + x_{2} + \cdots+ x_{s} )^{p} \le s^{p - 1} ( x_{1}^{p} + x_{2}^{p} + \cdots+ x_{s}^{p} )$.

Combining the observations above, we find:

$$\begin{aligned} \sum_{k = 0}^{n - 1} \Biggl( \sum _{d_{1} + \cdots+ d_{r} = k} \prod_{i = 1}^{r} a_{i, d_{i}} \Biggr)^{p} & \le\sum_{k = 0}^{n - 1} \binom{k + r - 1}{k}^{p - 1} \Biggl( \sum_{d_{1} + \cdots+ d_{r} = k} \prod_{i = 1}^{r} a_{i, d_{i}}^{p} \Biggr) \\ & \le\binom{n + r - 2}{n - 1}^{p - 1} \sum_{k = 0}^{n - 1} \Biggl( \sum_{d_{1} + \cdots+ d_{r} = k} \prod _{i = 1}^{r} a_{i, d_{i}}^{p} \Biggr) \\ & \le\binom{n + r - 2}{n - 1}^{p} \prod_{i = 1}^{r} \sum_{j = 0}^{n - 1} a_{i, j}^{p}. \end{aligned}$$

(36)

Thus,

$$ m_{r}(\Delta_{n}) \le\binom{n + r - 2}{n - 1} = O\bigl( r^{n-1}\bigr). $$

(37)

Since n is fixed, $\rho_{\Delta_{n}} \le\limsup_{r \rightarrow \infty} \sqrt[r]{ r^{n-1}} = 1$. However, we found in (3) that $\rho_{\Delta_{n}} \ge \lVert 1 \rVert^{-1} = 1$. Thus, $\rho_{\Delta_{n}} = \lVert1 \rVert^{-1} = 1$. □

6 Submultiplicative constants for the dualplex numbers

An n-dualplex number is defined as an element of $\mathcal {A}=\mathbb {R}[x] / \langle(x^{2}+1)^{n} \rangle$, or equivalently, $a_{0} + a_{1} q + \cdots+ a_{2n-1} q^{2n-1}$ where $a_{i} \in\mathbb{R}$ and $(q^{2} + 1)^{n} = 0$. If we use norm

$$\bigl\lVert a_{0} + a_{1} q + \cdots+ a_{2n-1} q^{2n-1} \bigr\rVert = \sqrt{a_{0}^{2} + a_{1}^{2} + \cdots+ a_{2n - 1}^{2}} $$

then for $n=1$ we have complex numbers with the usual multiplicative norm and multiplicative constant 1. However, for $n=2$ we conjecture that the submultiplicative constant is given by $m_{\mathcal {A}}= \sqrt{\frac{71 + 17 \sqrt{17}}{6}}$ where $\| xy \| = m_{\mathcal {A}}\| x \| \| y\|$ if we set $x = y = (3 - \sqrt{17}) q + 2 q^{3}$.

7 Convergence of geometric series over some algebras

In this section we use Theorem 1.3 in conjunction with the results of the previous sections to study the geometric series over an algebra.

Example 7.1

Consider the geometric series $\sum_{n = 0}^{\infty}z^{n}$ in the N-hyperbolic numbers $\mathcal {H}_{N}$. Observe $c_{n}=1$ and $C = \limsup_{n \rightarrow\infty} \sqrt[n]{ \lVert c_{n} \rVert} = 1$. We found $\rho _{\mathcal {H}_{N}} = \sqrt{N}$ in Theorem 3.3. Theorem 1.3 gives that the geometric series converges for $\| z \| < \frac{1}{\sqrt{N}}$. Suppose $R > \frac {1}{\sqrt{N}}$, pick $z = \frac{R}{\sqrt{N}} ( 1 + j + j^{2} + \cdots+ j^{N - 1} )$, and notice $\lVert z \rVert = R$ and that the geometric series diverges at z by the Root Test as

$$\begin{aligned} \limsup_{n \rightarrow\infty} \sqrt[n]{ \bigl\lVert z^{n} \bigr\rVert } & = \limsup_{n \rightarrow\infty} \sqrt[n]{\frac{R^{n}}{\sqrt{N}^{n}} \bigl\lVert N^{n - 1} \bigl( 1 + j + j^{2} + \cdots+ j^{N - 1} \bigr) \bigr\rVert } \\ & = \limsup_{n \rightarrow\infty} \sqrt[n]{R^{n} \sqrt{N}^{n - 1}} = R \sqrt{N} > 1. \end{aligned}$$

(38)

We see the radius of convergence $1/\sqrt{N}$ is maximal.

Example 7.2

Consider the Dual Numbers $\Delta_{N}$ for some $N\geq2$. Apply Theorems 5.3 and 1.3 to see why the geometric series $\sum_{n = 0}^{\infty}z^{n}$ converges for $\| z \| < 1$. Also, if $R>1$ then setting $z=R$ gives a point with $\| z \| > 1$ where the geometric series diverges.