Can a representation of a Banach algebra be unbounded?


To clarify, I'm not asking about a representation as unbounded operators, but
rather a homomorphism $\pi: A \to B(H)$ for some Hilbert space $H$, with the property
that $\sup\_{a \neq 0} \frac{\|\pi(a)\|}{\|a\|} = \infty$.


This question is inspired by chapter 2.5 of Arveson's *Spectral Theory*, which proves that every representation of a Banach \*-algebra is contractive. This raises the question of what happens when a Banach algebra does not have an (isometric) involution, or when a homomorphism does not respect it.


I know of examples of unbounded homomorphisms of Banach algebras (the simplest: if $E$ is an infinite-dimensional Banach space, turn both $E$ and $\mathbb{C}$ into Banach algebras by defining all products to be zero, and any unbounded functional on $E$ will become an unbounded homomorphism), but none in which the codomain is $B(H)$.



Unbounded representations of Banach algebras
As Dima Shlyakhtenko indicated, there are examples, obtained for example by modifying your zero product example to $x\mapsto\begin{bmatrix}0&f(x)\\\ 0&0\end{bmatrix}$ for some unbounded linear functional $f$. 


A generalization of the result you mentioned is found in Theorem 4.1.20 of Rickart's *General theory of Banach algebras*. A special case of that theorem says that if the range of the representation is a $\*$-subalgebra of $B(H)$, then the representation is automatically continuous.


What looks like a good reference is Dales's *[Banach algebras and automatic continuity](http://books.google.com/books?id=WZ6DQgAACAAJ)*, but I don't currently have a copy. It is referenced in an *Encyclopaedia of mathematics* article by the same author called "[Automatic continuity for Banach algebras](http://books.google.com/books?id=47YC2h295JUC&lpg=PP1&pg=PA48#v=onepage&q&f=false)," which also gives some other references that might help. According to the article, there are some Banach spaces $E$ such that every homomorphism from $B(E)$ to another Banach algebra is continuous. (It doesn't say which ones.)


