Unital homomorphism to semisimple Banach algebra is automatically continuous

Question

I need to prove that any unital homomorphism $\phi: A \to B$, where $A$ is unital Banach algebra and $B$ is semisimple Banach algebra is continuous.

The definition of "semisimple" I know is that the kernel of Gelfand transform which is the same as $ \{b \in B: \sigma(b) = 0 \}$ equals $\{0 \}$.

I am asking for some hint. Also may be I am not aware of some result needed.

Aren't Banach algebra homomorphisms often assumed to be continuous? What is your definition? — Cameron L. Williams, Jun 20 '18 at 18:55
@Cameron Williams, just homorphism as algebras (I didn't mean morphism in category Banach algebras) — Invincible, Jun 20 '18 at 19:00
See chapter 6, section 2 in Banach and Locally Convex Algebras by Helemskii, A.Ya. There is a lot on these matters at the end of that section. — Norbert, Jul 12 '18 at 16:57

score 1 · Answer 1 · answered Jul 27 '18 at 15:09

This is not quite right as stated.

Johnson proved that a surjective homomorphism onto a semi-simple Banach algebra is automatically continuous. (For the proof see, e.g., Theorem 5.1.5 in Dales' Banach Algebras and Automatic Continuity.)

Apparently, it is an open problem whether it is enough to assume that the homomorphism considered has dense range (Question 5.1.A in Dales' book).

Unital homomorphism to semisimple Banach algebra is automatically continuous

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