Let $A$ be a Banach algebra and $Bil(A)$ denote the space of bounded bilinear forms on $A$.
$Bil(A)$ is a Banach $A$-bimodule with the module operations
\begin{eqnarray\*}
\beta a(x,y) &:=& \beta(ax,y) \\
a \beta(x,y) &:=& \beta(x,ya)
\end{eqnarray\*}
for each $\beta\in Bil(A)$ and each $a,x,y\in A$.
Further, for each $f\in A^{\ast}$, define $$\beta\_f(x,y) := f(xy) \hspace{4mm} \forall x,y\in A.$$
The set $\{\beta\_f:f\in A^{\ast}\}$ is a bi-submodule of $Bil(A)$.


Consider the following hypotheses:


1. $A$ is unital.
2. $A$ is reflexive as Banach space.
3. $\beta\_f$ is *not* weakly sequentially continuous (wsc) for all $f\in A^{\ast}\backslash\{0\}$.


$\beta \in Bil(A)$ is wsc if $\beta(x\_n,y\_n)\to 0$ whenever $(x\_n)$ and $(y\_n)$ are weakly null sequences.


3'. Same as 3 with $x\_n=y\_n$, i.e., for each $f\in A^{\ast}\backslash\{0\}$ there exists a weakly null $(x\_n)$ such that $\displaystyle \lim\_{n\to\infty}\beta\_f(x\_n,x\_n)\neq 0$.


4. $\{\beta\_f:f\in A^{\ast}\}$ is a direct bimodule summand of $Bil(A)$, i.e., there exists another bi-submodule $K$ of $Bil(A)$ such that $Bil(A) = K\oplus \{\beta\_f:f\in A^{\ast}\}$.


**Question 1.** Does there exist an infinite dimensional Banach algebra that satisfies *1, 2, 3 (or 3')*?


**Question 2.** Does there exist an infinite dimensional Banach algebra that satisfies *1, 2, 3 (or 3'), 4*?



Weak sequential continuity of certain bilinear forms on Banach algebras
YES.
Consider the Jolissaint--Lafforgue Sobolev algebra $H\_\ell^s(\Gamma)$.
(I don't know the common name for it.)
Here we take $\Gamma=F\_\infty$ to be the free group of countably infinite rank, $\ell$ the standard word length, and $s>2$.
It is the completion of the complex group algebra ${\mathbb C}\Gamma$ under
the Sobolev norm
$$\| f \|= (\sum\_ x |f(x)|^2(1+\ell(x))^{2s})^{1/2}.$$
V. Lafforgue (<https://mathscinet.ams.org/mathscinet-getitem?mr=1774859>) has proved that
$H\_\ell^s(\Gamma)$ is a "Banach algebra" (see the comment below) which is embedded densely in the reduced group $\mathrm{C}^\*$-algebra $\mathrm{C}^\*\_\lambda(\Gamma)$ and is closed under the holomorphic functional calculus there. (Property RD for $F\_\infty$ is due to Haagerup.)
From the latter property, we see that $H\_\ell^s(\Gamma)$ is simple, because the $\mathrm{C}^\*$-algebra
$\mathrm{C}^\*\_\lambda(F\_\infty)$ is simple (Powers).
The Banach algebra $H\_\ell^s(\Gamma)$ is unital and isomorphic to a Hilbert space as a Banach space.
For the standard free basis $\{s\_n\}$ of $\Gamma=F\_\infty$, the corresponding "unitary"
elements and their inverses are uniformly bounded in $H\_\ell^s(\Gamma)$. Property (3) follows from this.


Comment: Note that the above Sobolev norm only satisfies $\|f \* g\|\le C\|f\|\|g\|$ for some universal constant $C$,
but one can renorm it via $H\_\ell^s(\Gamma)\hookrightarrow B(H\_\ell^s(\Gamma))$ to make
it satisfies $\|f \* g\|'\le \|f\|'\|g\|'$ and $\|1\|'=1$. Note that by Lumer's theorem, a unital infinite-dimensional Banach algebra cannot be isometric to a Hilbert space.


