Question about norm of homomorphism on Banachalgebras I'm currently reading a book where it is stated that if $A$ is a unital banach algebra, then every homomorphism $\phi: A \to \mathbb{C}$ has either norm $1$ or is $0$, i.e $\|\phi\|=1$ or $\phi=0$.
I know what a homomorphism is but could someone explain to me what exactly the norm of a homomorphism is? 
and how can it be proven that it's always either $1$ or $0$?
 A: If $\phi:A \to \mathbb{C}$ is continuous, its norm is defined as the smallest number $C > 0$ such that
$$
\lvert \phi(a) \rvert \leq C \lVert a \rVert.
$$
This is the same definition as for any linear operator between Banach spaces. Equivalently, you also have
$$
\lVert \phi \rVert = \sup \{ \lvert \phi(a) \rvert : a \in A, \lVert a \rVert =1\}.
$$
Now we prove that $\phi$ has norm one by contradiction. Suppose there's some $a \in A$ such that $\lvert \phi(x) \rvert > \lVert x \rVert$. Then we have that $\phi(a) I - a$ is invertible in $A$. This is due to the following lemma

Let $A$ be a unital Banach algebra and $x \in A$. If $\lambda \in \mathbb{C}$ with $\lvert \lambda \rvert > \lVert x \rVert$, then $\lambda$ isn't contained in the spectrum of $x$. Furthermore we have
  $$ (\lambda I -x)^{-1} = \sum_{k \geq 0} \lambda^{-k-1} x^k. $$

But we have $\phi \left( \phi(a)I -a \right) = 0$, which is in contradiction with the fact that $\phi(a) I - a$ is invertible. We conclude that $\lvert \phi(x) \rvert \leq \lVert x \rVert$ and so $\lVert \phi \rVert \leq 1$. At last since $\lvert \phi(I) \rvert = 1$ we must have $\lVert \phi \rVert = 1$.
