I am presently working on multipliers on Banach algebra and i found out that most result discovered so far are been establish on the fact that the algebra is faithful (without order). Please can anyone help me with the main reason why the algebra must be faithful (without order).
If A is a Banach algebra
then a mapping T defined on A (Banach algebra) is a multiplier of A if x(Ty)=(Tx)y for all
x, in A. A Banach algebra A is without
order if for all x in A, xA={O} implies x=O, or, for all x in A, Ax={O}
implies x = 0.
Most result that has been discovered so far has been on the basis that the algebra is faithful, what happen when the algebra is not faithful.
The answer is simple! I just wish i can communicate it well. From the definitions, it is clear that a Banach Algebra A is with order if every element in A is a zero divisor of every other elements of A.
Observe clearly that most result of multipliers on faithful algebra are usually concluded by applying some cancellation law i.e for example, after arriving at the equality below in a proof
gh=hTf
one concludes that g= Tf which is based on the natural assumption that h (which is non zero) cannot divide both sides to give zero (talking like a lay man) which is the same as saying Banach algebra A is without order. What I'm saying is that if A is with order then upon division of both sides by h (not equal to zero), you may get zero.
So, A has to be faithful for the cancellation law, usually applied as conclusion in most proofs I've studied so far, to hold.
Akindele, I hope this helps.
Thanks,
From your example i think you are trying to write it in the form
hg=hTf, then we can have h(g-Tf) = 0 and we apply the definition and reach a conclusion that g=Tf.
It quite clear but i was just wondering why is it that the Banach algebra must always be faithful, may be we can a result on arbitrary Banach algebra, which is not faithful.
Let A be a (Banach) algebra, possibly non-unital, and let M(A) be its multiplier algebra.
Then M(A) consists of all ordered pairs (L,R) where L, R are both linear transformations of A satisfying the following properties for all a,b\in A:
1. L(ab)=L(a)b
2. R(ab)=aR(b)
3. R(a)b=aL(b)
We can define a map \varphi : A \to M(A) by putting \varphi(x)=(L_x,R_x) for x \in A
where L_x and R_x are the linear maps defined by L_x(a)=xa resp. R_x(a)=ax. (Check that \varphi is well-defined!)
A natural question is now the following: When is \varphi injective, i.e. when can A be viewed as a subalgebra of M(A)?
The kernel of \varphi is the intersection of the left annihilator of A in A with its right annihilator in A. Therefore, \varphi is injective if and only if for every non-zero element a\in A there exists b\in A such that either ab\neq 0 or ba\neq 0.
I think that this is the main reason why you want your algebra to be "faithful". Because if it is not faithful, then the above map \varphi need not be injective and then you can not view A as a subalgebra of M(A).
(Notice that \varphi : A \to M(A) is an isomorphism if and only if A is unital.)
A few years back I had also come across the curiosity of the role of faithfulness in the study of multipliers. As a result, I prepared some detailed notes on this topic which are attached herewith. I hope these would serve to supplement the useful information already given above.
As further information to the given ones, this answer concerns some benefits that come from the hypothesis the algebra to be without order.
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