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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?

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  • $\begingroup$ In 3. you may want to exclude $f=0$. $\endgroup$ Jun 19, 2021 at 20:21
  • $\begingroup$ @DirkWerner I edited the post to correct the oversight. Professor Werner, I'm glad to see that you read this post. I need a new point of view, and perhaps an expert in the field like yourself could lend me some insight. I'll be grateful for your comments. $\endgroup$
    – Onur Oktay
    Jun 20, 2021 at 1:50

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Onur Oktay
    Jun 21, 2021 at 18:06
  • $\begingroup$ Thank you for the link to Lafforgue's article. In his proof, G is a discrete group and $\ell$ is a length function for which G satisfies RD property w.r.t. $\ell$. Is there a relationship between the amenability of the algebra $H^s_{\ell}(G)$ and that of the group $G$? I am curious if there is a relationship similar to $C^{\ast}(G) = C^{\ast}_{\lambda}(G)$ iff $G$ is amenable, or $L^1(G)$ is amenable iff $G$ is amenable. $\endgroup$
    – Onur Oktay
    Jun 21, 2021 at 18:38
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    $\begingroup$ @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. $\endgroup$ Jun 21, 2021 at 23:04
  • $\begingroup$ Thank you again, this small exchange has been beneficial for me. Since (4)=amenable and $(2), (4)\Rightarrow (1)$, then question 2 is same as asking if there exists a reflexive amenable Banach algebra satisfying (3). Perhaps it is not reasonable anymore to ask for a counterexample that disproves an open problem. However, do you think the extra assumption (3) would be of some use? Would it yield a partial result on the affirmative or negative way? $\endgroup$
    – Onur Oktay
    Jun 22, 2021 at 9:20
  • $\begingroup$ Looking at $H^s_{\ell}(F_{\infty})$ from my oversimplified point of view of (1)-(4), it is as though one first constructs two weakly null sequences, then defines an algebra of functions on the free group generated by the terms of this sequence, making sure that the sequence is indeed weakly null. The norm on $H^s_{\ell}(F_{\infty})$ clearly meet this need. Perhaps weighted Orlicz algebras (with the weight $w(x)=(1+\ell(x))^s$) which are reflexive are other examples that satisfy (1)-(3). $\endgroup$
    – Onur Oktay
    Jun 22, 2021 at 9:20

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