If $A$ is a Banach agebra and $M$ is a Banach $A$-bimodule then a linear map $T:A\to M$ is called an $A$-module homomorphism if $$T(ab)=aT(b),\quad T(ab)=T(a)b,\qquad a,b\in A.$$ Also $A\hat{\otimes} A$ is a two sided $A$-module with the following actions $$c(a\otimes b)=ca\otimes b,\quad (a\otimes b)c=a\otimes bc.$$ (Here $\hat{\otimes}$ deotes the projective tensor product.)

Does there exist a Banach algebra $A$ and a net of linear $A$-module homomorphisms $\{\rho_\alpha:A\to A\hat{\otimes} A\}_{\alpha\in I}$ such that for the linear product map $\pi:A\hat{\otimes} A\to A,$ $\pi(a\otimes b)=ab$ we have:

  1. $\pi\circ\rho_\alpha(a)\to a$ in norm for all $a\in A$;

  2. the set $\left\{\frac{\|\pi\circ\rho_\alpha(a)b-ab\|}{\|ab\|}:a,b\in A,\alpha\in I\right\}$ is bounded;

  3. the set $\left\{\frac{\|\pi\circ\rho_\alpha(a)-a\|}{\|a\|}:a\in A,\alpha\in I\right\}$ is not bounded.

  • 1
    $\begingroup$ (I just rolled back an edit which replaced imperfect English with slightly worse English) $\endgroup$
    – Yemon Choi
    Oct 9, 2016 at 2:56
  • $\begingroup$ Just to get the background straight in my head: is this related to questions about approximate biprojectivity? $\endgroup$
    – Yemon Choi
    Oct 9, 2016 at 2:57
  • $\begingroup$ Yes! It is related to approximate biprojectivity. Thanks for edit. $\endgroup$ Oct 9, 2016 at 15:11

1 Answer 1


The following is not a complete answer, but suggests that it will be difficult to find such an $A$ (and it leads me to conjecture that no such $A$ exists).

Let $A^2$ denote $\{ ab \colon a,b\in A\}$. (This notation is not entirely standard: many authors use $A^2$ to denote the linear span of this set, and some use $A^2$ to denote the closed linear span of this set. If $A$ has a bounded approximate identity then all of these sets are the same, by Cohen's factorization theorem.) Then your condition 2 is equivalent to

2'. the set $\left\{\frac{\|\pi\circ\rho_\alpha(x)-x\|}{\|x\|} \colon x\in A^2,\alpha\in I\right\}$ is bounded.

So in the case where $A^2$ is dense in $A$, a straightforward argument shows that condition 2' contradicts condition 3.

On the other hand, condition 1 clearly implies that ${\rm lin} A^2$ is dense in $A$.

Therefore, if a Banach algebra $A$ exists which satisfies 1, 2 and 3, it must have the property that $A^2$ is not dense in $A$ but ${\rm lin} A^2$ is dense in $A$. This seems quite difficult to me; most examples I can think of where ${\rm lin} A^2$ is dense in $A$ have the property that $A^2$ is dense in $A$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .