First isomorphism theorem for Banach algebra

Question

Let $A$ and $B$ be Banach algebras and $\phi:A\to B$ be a continuous homomorphism. Can we see $$A/\ker(\phi)\cong \phi(A)$$ as a Banach algebra isomorphism?

Answer 1

If the norm on $\phi(A)$ is the norm inherited as a subspace of $B$ then the answer is no. Of course $A/\ker(\phi)$ is a Banach algebra, since $\ker(\phi)$ is a closed ideal in $A$, but $\phi(A)$ need not even be a Banach algebra in the first place.

If we define a norm on $C^1(\Bbb T)$ by $$||f||=||f||_\infty+||f'||_\infty$$ then $C^1(\Bbb T)$ is a Banach algebra. The only bit that might not be clear is that the norm is submultiplicative, but this follows from the product rule:$$||fg||\le||f||_\infty||g||_\infty+||f||_\infty||g'||_\infty+||f'||_\infty||g||_\infty\le||f||\,||g||.$$

Let $A=C^1(\Bbb T)$ and $B=C(\Bbb T)$. Define $\phi:A\to B$ by $\phi(f)=f$. Then $\phi(A)$ is not complete, in fact it's dense in $B$.