Can someone give me an example of a Banach Algebra which does not have an isometric representation in a Hilbert Space ?


(if possible, can you add a proof or a reference ? )


Thank you very much !




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**Note added by YC**: this question has also been [asked on MSE](https://math.stackexchange.com/questions/423702/banach-algebra-counterexample) where someone has given a much better, elementary example to the OP's question. (However, my example also works for the question of topologically isomorphic representations, not just the isometric ones.)



Banach Algebra Counterexample
Any Banach algebra which is not Arens regular cannot be embedded as a closed subalgebra of B(H), even if you allow for isomorphic embeddings that have closed range yet are not isometric.


If you are only interested in Banach $\ast$-algebras and isometric $\ast$-homomorphic embeddings, then it is easier to find examples, as Owen Sizemore has indicated.




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**[Not directly relevant, but perhaps of background interest to the OP]**
By the way, although the question asks about isometric embeddings, there are interesting and slightly unexpected examples of Banach algebras $A$ for which there is an injective homomorphism $A\to B(\ell^2)$ that has closed range, showing that non-selfadjoint operator algebra theory has to be a lot wilder than the self-adjoint case. Examples include: $\ell^p$ for $1\leq p <\infty$ with pointwise product; and the algebras $C^k([0,1]^m)$ of $k$-times continuously differentiable functions on the $m$-cube. You can even get radical commutative Banach algebras embedded into $B(\ell^2)$ in this way, see



> 
> [MR0410386 (53 #14136)](http://www.ams.org/mathscinet-getitem?mr=410386) P. G. Dixon, *Radical Q-algebras.*
>  Glasgow Math. J. 17 (1976), no. 2, 119--126.
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