Let $\mathfrak A$ be a Banach algebra with a bounded approximate identity (BAI), and let $\square$ and $\lozenge$ denote, resp., the first and the second Arens products of $\mathfrak A''$. Consider the first topological center of $\mathfrak A''$; $$\mathfrak Z^{(1)}_t(\mathfrak A''):=\{\Phi\in\mathfrak A'': \Phi\square\Psi=\Phi\lozenge\Psi, \text{for all $\Psi\in\mathfrak A''$}\}$$ When do we have the inclusion ${\mathfrak A}\mathfrak Z^{(1)}_t(\mathfrak A'')\subset {\mathfrak A}$? in particular, does it hold when $\mathfrak A$ is weakly sequentially complete?

$\begingroup$ An obvious sufficient condition is that the 1st topological center coincides with $A$ itself  this is either left SAI or right SAI, I forget which $\endgroup$– Yemon ChoiDec 10, 2017 at 19:31

1$\begingroup$ It might help attract better responses if you (a) provide some context for why you are interested in the question (b) say something about what you have tried so far in order to find cases where the inclusion holds and where it fails $\endgroup$– Yemon ChoiDec 10, 2017 at 19:32
1 Answer
To addres your last question: recall that $A(G)$ is weakly sequentially complete, since it is the isometric predual of a von Neumann algebra. Moreover, if $G$ is compact then $A(G)$ is unital, and so $A(G)\cdot Z_t(A(G)^{**}) = Z_t(A(G)^{**})$.
Therefore, to get a "counterexample" to your claim/question, it suffices to find $G$ compact such that $Z_t(A(G)^{**})$ is strictly larger than $A(G)$. According to an unpublished announcement/calculation of V. Losert, we can achieve this by taking $G=SU(3)$.
Update 20171231: would whoever left a downvote care to explain, in a comment, what is unsatisfactory or deficient with this answer?

$\begingroup$ Thank you for informing me of this example. Also $\mathfrak A:=\ell^1(\mathbb Z^k,\omega)$ with $\omega(\mathbf n)=1+\\mathbf n\$ can serve as an counterexample of my last claim; indeed this is an unital Arens regular Banach algebra. But all these examples are unital. How about nonunital ones? If $\mathfrak A$ is a nonunital WSC Banach algebra, does the inclusion hold? $\endgroup$ Dec 13, 2017 at 12:18

$\begingroup$ Why not try a maximal ideal in $A(SU(3))$? Such an ideal always has a bounded approximate identity and is nonunital; and its bidual is a complemented 2sided codimension1 ideal in $A(SU(3))^{**}$. I have not thought about the details, but it seems likely to me that this ideal will also be WSC and you could try to prove it is a counterexample to your original question by using Losert's result. $\endgroup$ Dec 13, 2017 at 13:05

$\begingroup$ Can you please make me known about the references of facts used in your answer and comment; 1Is a predual of a von Neumann algebra always WSC? 2Where do I find the calculation of V. Losert or How can I refer to it? 3 Why a maximal ideal in $A(SU(3))$ has BAI, and why its bidual is a complemented 2sided codimension1 ideal in $A(SU(3))^{∗∗}$? Of course this maximal ideal should be WSC because it is norm closed. $\endgroup$ Dec 20, 2017 at 5:38

$\begingroup$ @MeisamSoleimaniMalekan I saw your request, but unfortunately I am busy with other tasks. The predual of any von Neumann algebra is always WSC, this result is used often in the work of people such as Lau, Granirer, Neufang... and I think you can find it in volume 2 of Takesaki's book(s) on Operator Algebras. Losert's calculation was never published, unfortunately. $\endgroup$ Dec 20, 2017 at 15:05

$\begingroup$ @MeisamSoleimaniMalekan The fact that every maximal ideal in A(G) has a BAI when G is amenable follows from the fact that for any G, and any $p\in G$ and any compact neighbourhood $K\ni p$, we can always find $f\in A(G)$ which has norm $1$, which is supported inside $K$, and which satisfies $f(p)=1$. This gives a net $(f_K)$ and if $A(G)$ has a b.a.i. $(u_i)$ then you just consider the net $(u_if_K)$ with a suitable partial ordering of the indexing sets, to obtain a BAI for the maximal ideal corresponding to the point $p$ $\endgroup$ Dec 20, 2017 at 15:06