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 $B(H)$.

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