# Continuity of double centralizers in Banach algebras

I had some problems with a certain exercise, came up with a solution, but I'm not sure it is correct.

Exercise ("MURPHY, C*-Algebras and Operator Theory", Chapter 2, exercise 1) Let $A$ be a Banach algebra such that for all $a\in A$, the implication $$Aa=0\text{ or }aA=0\Rightarrow a=0\qquad (1)$$ holds. Let $L,R$ be linear mappings from $A$ to itself such that for all $a,b\in A$, $$L(ab)=L(a)b,\quad R(ab)=aR(b),\qquad (2)$$ $$\text{and } R(a)b=aL(b).\qquad (3)$$ Show that $L$ and $R$ are necessarily continuous.

A double centralizer in a Banach algebra is defined to be a pair $(L,R)$ of bounded linear maps from $A$ to itself satisfying conditions (2) and (3). This exercise shows that if the Banach algebra satisfies condition (1) (for example, if it is a C*-algebra), then boundedness is consequence of (2) and (3).

I spent quite some time trying to solve this. Then,

Solution: By the Closed Graph Theorem, to show that $L$ is continuous, it suffices to show that for every sequence $\left\{x_n\right\}\subseteq A$ such that $x_n\rightarrow 0$ and such that $\left\{L(x_n)\right\}$ also converges, then $\lim L(x_n)=0$.

Let $x_n\rightarrow 0$ and $L(x_n)\rightarrow y$ in $A$. Then for every $n\in\mathbb{N}$ and for every $b\in A$, $$bL(x_n)=R(b)x_n,\quad\text{ by (3)}.$$ Taking $n\rightarrow\infty$, we have, for every $b\in A$, $$by=R(b)0=0,$$ that is, $Ay=0$. By (1), this implies that $y=0$.

Continuity of $R$ is proved similarly. Q.E.D.

Now, the problem is that I didn't use condition (2), and that seems too strong. Is there any problem with my reasoning? Thank you.

EDIT (see Martin's answer): I just found out that (1) and (3) imply (2). This explains why (2) was not used above. Let's show this:

Suppose $A$ is an algebra (not necessarily normed) satisfying (1) and let $L,R$ be linear maps from $A$ to $A$ satisfying (3).

Let $a,b\in A$. Then for every $c\in A$, $$c(L(ab)-L(a)b)=cL(ab)-cL(a)b\overset{(3)}{=}R(c)(ab)-(R(c)a)b=R(c)ab-R(c)ab=0.$$ Then $A(L(ab)-L(a)b)=0$, hence, by (1), $L(ab)=L(a)b$, for every $a,b\in A$. Similarly, $R(ab)=aR(b)$ for every $a,b\in A$. Therefore, (2) holds.

-
+1. Your argument is correct. You only need (3). –  1015 Dec 25 '13 at 18:05
Your argument is correct. Note that (3) almost implies (2). Indeed, $$cL(ab)=R(c)ab=cL(a)b.$$ So, if $A$ is unital or has a approximate identity, the implication holds.