# Weak sequential continuity of certain bilinear forms on Banach algebras

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 $$\displaystyle \lim_{n\to\infty}\beta_f(x_n,x_n)\neq 0$$.

1. $$\{\beta_f:f\in A^{\ast}\}$$ 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?

• In 3. you may want to exclude $f=0$. Jun 19, 2021 at 20:21

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.
• Professor Ozawa, thank you for your answer and your detailed comment. $H^s_{\ell}(\Gamma)$ surely satisfies (1)-(3). Does $H^s_{\ell}(\Gamma)$ satisfy (4) as well? Jun 21, 2021 at 18:06
• @Onur Oktay: Your property (4) is equivalent to amenability. $H_\ell^s(\Gamma)$ is not amenable because it is dense in $C^*_\lambda\Gamma$, which is non-amenable for a non-amenable $\Gamma$. Also, it is a well-known open problem whether there exists an infinite-dimensional amenable Banach algebra whose underlying Banach space is reflexive, but there is certainly none that is isomorphic to a Hilbert space. Jun 21, 2021 at 23:04