Let $A$ be a Banach algebra. Some textbooks define a (left ) multiplier as a map $T:A\rightarrow A$ satisfying $T(ab)=T(a)b$ for all $a,b\in A$ and assume that $A$ needs to be a without order Banach algebra, that is, if $xA=(0)$ for some $x\in A$, then $x=0$. (right multiplier is defined in a similar way). However, some other define a (left) multiplier as a linear map (even continuous) on a (general) Banach algebra satisfying the above condition. Am I missing something here? I know the first definition does not assume that $T$ is linear however it follows from that $A$ is without order. Is a multiplier assumed to be continuous by definition?


I would be interested to know exactly what sources you are using, as any standard textbook does prove the following.

I actually don't know what happens for just a left multiplier (though also do not know any counter-examples off the top of my head), but suppose we have a double multiplier $(L,R)$, that is, $L$ is a left multiplier, and $R$ is a right multiplier, and $R(a)b = aL(b)$ for all $a,b\in A$.

Notice that being "without order" is equivalent to:

For $a,b\in A$, if $ac=bc$ for all $c\in A$, then $a=b$.

There is an obvious equivalent definition on the other side. I will assume both these conditions hold for $A$.

Using this, we find that for $a,b\in A$ and $\lambda$ a scalar, $$ c(L(a)+\lambda L(b)) = cL(a) + c (\lambda L(b)) = R(c)a + R(c) (\lambda b) = R(c)(a+\lambda b) = cL(a+\lambda b), $$ for any $c\in A$, and thus $L$ is linear. Similarly $R$ is linear. In fact, it is enough that $R(a)b=aL(b)$ for $a,b\in A$; then a similar argument shows that $L$ is automatically a left multiplier, and $R$ a right multiplier.

We don't assume that $L$ and $R$ are bounded. However, we can use the closed graph theorem in the following way: if $(a_n)\rightarrow 0$ and $L(a_n)\rightarrow a$, then $$ ba = \lim_n bL(a_n) = \lim_n R(b) a_n = 0 $$ for any $b$, and so $a=0$, and thus $L$ is continuous; similarly $R$ is continuous.

  • $\begingroup$ Question about the literature: it seems from a quick search online that the terminology "without order" is mostly used in the commutative setting, and the phrase doesn't seem to appear in the Big Book of Dales. Do you know any sources which use "without order" in the 2-sided noncommutative setting? $\endgroup$ – Yemon Choi 2 days ago
  • $\begingroup$ @YemonChoi: I think you are quite correct. I was merely copying the terminology the Questioner used; I have not seen "without order" myself. Barry Johnson used the term "faithful" which seems reasonable; I think I've also seen "non-degenerate" used (both these terms are of course a little overused in Mathematics). $\endgroup$ – Matthew Daws 2 days ago

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