Abstract:
In this study, we obtain Banach algebras which norm of their unit elements
is not one. These Banach algebras are subsets of
INTRODUCTION
We know normed algebras are one of the most important subjects in functional analysis. Also we know that if a normed algebra is unitary then norm of its unit is one. Most of persons who study functional analysis in a non-professional way think that norm of unit in any algebra should be one. We wish to present algebras in which their unit element’s norm is not one. We need to following definitions.
Definition 1: Kreyszig (1987) (Normed algebra):
An algebra over F (the real field
| (i) | x (yz) = (x,y)z |
| (ii) | x(y+z) = (xy)+xz, (x+y)z = xz+yz |
| (iii) | (αx)y = α(xy) = x(αy) |
Furthermore if there exists a norm ||-|| on A such that we have for any x,y∈ A,||xy||≤||x|| ||y|| then A is a normed algebra.
As is usual for normed linear spaces a normed algebra A is regarded as a metric space with the distance function d (x,y) = |x-y| (x,yεA). If A is a complete metric space with defined metric, then A is called a Banach algebra.
Definition 2: Bonsall and Duncan (1973): An element e of an algebra A is an unit element if and only if e≠0 and ex = xe = x (xεA).
Definition 3: Kreyszig (1987) (Equivalent norms): A norm ||-|| on a vector space X is said to be equivalent to a norm ||-||1 on X if there are positive numbers a and b such that for all xεX we have a||x||1≤||x||≤b||x||1.
MAIN RESULTS
In this section we introduce some subsets of
We begin with
x.y = xy |
And define norm on
| (1) |
where, c is a constant and c>1. clearly,
The defined norm in Eq. 1 is equivalent with original norm
on
| (2) |
Clearly, Ak is a subset of
| (0,…,0, xk, 0,…,0)+(0,…,0,
yk,0,...,0) = (0,…,0, xk+yk,0,…,0) α. (0, …,0, xk, 0,…,0) = (0,...,0,αxk,0,…,0) |
We define The product of Ak ‘s as follow:
| (0,…,0, xk, 0,…,0)(0,….,
0,yk,0,...,0) = (0,…,0, xkyk,0,…,0) |
And define the norm on Ak’s by:
where, c is a constant and c>1. It is easy to verify that any Ak, k≥1, induces a metric by:
It is easy to show that any Ak, k≥1, is a Banach algebra.
Since in Eq. 2 c>1 is arbitrary, thus we can obtain infinite many of Banach algebras.
Corollary 1: We can obtain infinite many of Banach algebras of
Note 1: According to last explanations We conclude that norm of unit element in an algebra depends on algebra norm’s. For example, We have:
| (3) |
where, Ds(Ak) denotes direct sum of Ak’s. If we define the product in Ak by:
| (x1, x2, …,xk).(y1,y2,…,
yk) = (x1y1, x2y2,…,xkyk) |
And norm in Ds(Ak) by:
| ||(x1, x2,..., xk)
|| = c max {|xi |: 1≤i≤k} |
where, c>1 is a constant.
It is easy to show that Ak with above definitions is a normed algebra which its unit element is (1,0,0,…….,0) and ||(1, 0, 0,..., 0)|| = c>1. Also (1, 0,0,..., 0) is the unit element in Rk. But its norm is one.
Note 2: Note 1 shows that we can write Rk as direct sum of their subsets which any of them is a Banach algebra. Although, in the left side of Eq. 3 the norm of (1, 0,……,0) is one where in right side is c(>1).