Suppose that $A$ is a Banach algebra satisfying the condition $x\in A, x\cdot A=\{0\}\Rightarrow x=0.$ Given a nonempty subset $C\subset A$ we define the left annihilator $\ell(C):=\{x\in A : x\cdot C=\{0\}\}$.

Let $B$ be a closed subalgebra $B$ of $A$. Is it true that $\ell(B)=\{0\}\Rightarrow B=A$?

ADDENDUM: I am interested in the cases $A=L_1(G)^{**}$ and $A=L_1(G)^{**}/L_1(G)$, where $G$ is a compact abelian group, $L_1(G)$ is the convolution algebra, and $L_1(G)^{**}$ is endowed with the first Arens product. In this case $L_1(G)$ is an ideal in $L_1(G)^{**}$, and both examples satisfy the required condition because $L_1(G)^{**}$ admits a right identity $E$ ($x\cdot E= x$ for each $x\in A=L_1(G)^{**}$).

  • $\begingroup$ If you assume A is unital and take B to be any proper, unital subalgebra then this immediately gives you a counterexample. Perhaps you are thinking of a different situation with extra hypotheses? $\endgroup$ – Yemon Choi Sep 14 '16 at 12:21
  • $\begingroup$ Yes. After asking the question I realized that taking as $A$ the bounded operators on a Banach space $X$, and as B the compact operators on $X$, I can get a counterexample. The problem is that I do not know what kind of conditions could help. $\endgroup$ – M.González Sep 14 '16 at 14:21
  • $\begingroup$ The case in which I am interested is $A=L_1(G)^{**}$ with one of the Arens products induced by the convolution algebra $L_1(G)$, where $G$ is a LCA group. $\endgroup$ – M.González Sep 14 '16 at 14:22
  • $\begingroup$ I guess $L^1(G)<L^1(G)^{**}$ is a unital subalgebra in case $G$ is discrete, and you can use approximate identity in the general case to give a counter example. $\endgroup$ – Uri Bader Sep 14 '16 at 16:39
  • $\begingroup$ @UriBader I agree discrete G works but due to the failure of Arens regularity I think the question could be harder for for non-discrete G... (Maybe this is related to elements of $L^\infty(G)^*$ which annihilates $LUC(G)$ ?) $\endgroup$ – Yemon Choi Sep 14 '16 at 16:43

No. Let $A=C([0,1])$ and $B=\{f\mid f(0)=0\}$. $A$ is unital, so it certainly satisfies your condition, but for every $f\neq 0$, $fB\neq \{0\}$.

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  • $\begingroup$ @Yemon's answer would be to take $B$ to be the constant functions, which I guess is even easier... $\endgroup$ – Uri Bader Sep 14 '16 at 12:31
  • $\begingroup$ In this case, any closed subalgebra of functions that are not all $0$ on any interval of positive length will do. $\endgroup$ – Robert Israel Sep 14 '16 at 15:42

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