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?

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

1 Answer 1


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

You must log in to answer this question.

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