If a $C^\ast$algebra is reflexive (as a Banach space) then it is finite dimensional. Can anyone provide (or give a reference to) a nice example of an infinite dimensional noncommutative Banach algebra which is reflexive as a Banach space (perhaps even uniformly convex)?

11$\begingroup$ I don't know how nice it is, but the algebra of HilbertSchmidt operators is a Hilbert space as a Banach space. $\endgroup$ – Narutaka OZAWA Feb 6 '14 at 8:56

2$\begingroup$ Just take a reflexive Banach space and turn it into a Banach algebra by setting 1) $xy=0$ for all $x,y$, 2) $xy=\phi(x)y$ with some functional $\phi$ with $\phi\le 1$, 3) add a unit to example 2). I guess none of these would be "nice"... and 1) isn't even noncommutative. $\endgroup$ – UwF Feb 6 '14 at 9:50

$\begingroup$ Was there any reason why you wanted it noncommutative? There are semisimple commutative Banach algebras with no nontrivial idempotents that are isomorphic to Hilbert spaces. $\endgroup$ – Yemon Choi Feb 27 '14 at 5:03

$\begingroup$ Yes, there is a reason :) I'm doing some potential theory on normed spaces, and wanted a couple of concrete examples of spaces which are NOT algebras of functions. $\endgroup$ – Joakim Arnlind Feb 28 '14 at 6:49
If $G$ is a compact group, and we choose the normalized Haar measure $\mu$ on $G$, then $L^2(G)$ is a Banach algebra with convolution. Indeed, if $f$ and $h$ are in $L^2(G)$, then we use Hölder's inequality at the second step, together with the fact that the constant function 1 has $L^2$norm equal to 1, to compute \begin{align*} f\ast h(x)\leq \int_Gf(y)h(xy)1\ d\mu(y)\leq \f\_2\h\_2. \end{align*} Thus $$\f\ast h\_2\leq \f\ast h\_{\infty}\leq \f\_2 \h\_2.$$ This Banach $\ast$algebra is not only reflexive, but it is also a Hilbert space.
EDIT: I was short of time yesterday when I wrote this, and in particular didn't see Ozawa's comment. I want to add that our examples are closely related. In fact, and this is based on a conversation with Chris Phillips, write $L^2(G)=\bigoplus_{\pi\in\widehat{G}}H_\pi$ according to PeterWeyl Theorem, and give the HilbertSchmidt operators on $H_\pi$ the HilbertSchdmidt norm: $\T\=tr(T^*T)^{1/2}$, with composition as product, and the usual adjoint of operators on Hilbert spaces. Then I think that $L^2(G)$, with convolution as product and complex conjugation as involution, should be isomorphic to the direct sum (over $\pi$ in $\widehat{G}$) of the HilbertSchmidt operators on $H_\pi$.
The example I gave is, in some sense, a sum of a bunch of Ozawa's examples.