Let $\Omega(A)$ be the set of (characters )non-zero homomorphic functions from a unital Banach algebra $A$ to $\mathbb{C}$.
Show $\Omega(C[0,1])$ are the pointwise evaluations.
I'm not really sure why this is true by considering $\phi \in \Omega(C[0,1])$ and taking $f\in C[0,1]$.
Thank you in advance!
Let $\phi:C([0,1])\to \mathbb{C}$ be a non-zero character. Note that $\phi$ is an algebra morphism and thus $\ker(\phi)$ is an ideal. By the first isomorphism theorem, $C([0,1])/\ker(\phi)\cong \mathbb{C}$ and thus $\ker(\phi)$ is a maximal ideal of $C([0,1])$.
For each $c\in \mathbb{C}$, let $I_c=\{f\in C([0,1])\mid f(c)=0\}$. You can easily verify that each $I_c$ is a maximal ideal. In this short note, the converse is proved as well, i.e. every maximal ideal is of this form.
Thus, there exists a $c\in \mathbb{C}$ such that $\ker(\phi)=I_c $. Note that $C([0,1])\cong \ker(\phi)\oplus \mathbb{C}1$ where $1$ denotes the constant function with value $1$, indeed, every $f\in C([0,1])$ can be written as $f=g+h$ where $h=f(c)1$ and $g=f-h$ and clearly $g\in \ker(\phi)$. Thus, $\phi(f)=\phi(g+h)=\phi(h)=\phi(f(c)1)=f(c)\phi(1)=f(c)$.