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Research Article
 

On Unit Element’s Norm in Some Banach Spaces



Abdolmohammad Forouzanfar, Sajad Khorshidvandpour and Mohsen Moosavi
 
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ABSTRACT

In this study, we obtain Banach algebras which norm of their unit elements is not one. These Banach algebras are subsets of . Also, we present some interesting properties.

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  How to cite this article:

Abdolmohammad Forouzanfar, Sajad Khorshidvandpour and Mohsen Moosavi, 2012. On Unit Element’s Norm in Some Banach Spaces. Asian Journal of Mathematics & Statistics, 5: 167-169.

DOI: 10.3923/ajms.2012.167.169

URL: https://scialert.net/abstract/?doi=ajms.2012.167.169
 
Received: January 18, 2012; Accepted: February 29, 2012; Published: May 18, 2012



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 Image for - On Unit Element’s Norm in Some Banach Spaces or the complex field C) is a linear space A over F together with a mapping (x, y)→xy of AxA into A that satisfies the following axioms (for all x,y,zεA, αεF):

(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 Image for - On Unit Element’s Norm in Some Banach Spacesk which are Banach algebras with defined norm. In these algebras the norm of unit element isn’t one.

We begin with Image for - On Unit Element’s Norm in Some Banach Spaces (real numbers set). Consider Image for - On Unit Element’s Norm in Some Banach Spaces with ordinary addition and scalar product. Image for - On Unit Element’s Norm in Some Banach Spaces with defined operations is a vector space. We define the product of Image for - On Unit Element’s Norm in Some Banach Spaces as follow:

x.y = xy

And define norm on Image for - On Unit Element’s Norm in Some Banach Spaces by:

Image for - On Unit Element’s Norm in Some Banach Spaces
(1)

where, c is a constant and c>1. clearly, Image for - On Unit Element’s Norm in Some Banach Spaces is a normed algebra with 1 as unit element, but ||1|| = c|1| = c>1.

The defined norm in Eq. 1 is equivalent with original norm on Image for - On Unit Element’s Norm in Some Banach Spaces. The norm of 1 with original norm is one, whereas with (1) isn’t one. We define, In general, for kεN:

Image for - On Unit Element’s Norm in Some Banach Spaces
(2)

Clearly, Ak is a subset of Image for - On Unit Element’s Norm in Some Banach Spacesk for k≥1. Ak ‘s with following operations are vector spaces on Image for - On Unit Element’s Norm in Some Banach Spaces:

(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:

Image for - On Unit Element’s Norm in Some Banach Spaces

where, c is a constant and c>1. It is easy to verify that any Ak, k≥1, induces a metric by:

Image for - On Unit Element’s Norm in Some Banach Spaces

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 Image for - On Unit Element’s Norm in Some Banach Spacesk which norm of their unit element’s norm is not one.

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:

Image for - On Unit Element’s Norm in Some Banach Spaces
(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).

REFERENCES

1:  Kreyszig, E., 1987. Introductory Functional Analysis with Application. 1st Edn., John Wiley and Sons, Canada, ISBN-13: 978-0471504597, pp: 49-76

2:  Bonsall, F.F. and J. Duncan, 1973. Complete Normed Algebras. Springer-Verlag, New York, USA., ISBN-13: 9780387063867, pp: 1-20

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