Banach Algebra Counterexample 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 !

Note added by YC: this question has also been asked on MSE 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.)
 A: 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.

[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) P. G. Dixon, Radical Q-algebras.
  Glasgow Math. J. 17 (1976), no. 2, 119--126.

