I currently have a problem understanding a proof:
Let $A$ be a commutative Banach Algebra with unit and norm $||.||$. Let $M$ be a subspace of $A$ with codimension 1, which contains no invertable elements.
Now let $\phi$ be a continous linear functional on $A$, such that $M = ker \ \phi$.
How can I show that $|\phi(x^n)| \leq ||\phi||\ ||x||^n$?
I assume that $||\phi|| := \sup\limits_{||y||=1} |\phi(y)|$ as the normal norm of a functional or operator but that was nowhere defined.
(It was also given that $\phi(e) = 1$, but I think this is not important here.)
There is nothing to do with invertible elements and unit (you don't even need a unital algebra). Just consider inequality $||ab|| \le ||a||\; ||b||$ from which follows that $||x^n|| \le ||x||^n$. And therefore we obtain $$|\phi(x^n)| \le ||\phi||\;||x^n|| \le ||\phi||\;||x||^n$$