Theorem: Let $A$ be a unital Banach algebra. Then for $a \in A$ the spectrum $\sigma (a) \neq \varnothing$.
Consider the following proof:
The first step that seems unnecessary to me:
Let's say we have shown that
$$f: \mathbb C \to A, \lambda \mapsto (a-\lambda)^{-1}$$
is bounded entire. My next step would be to conclude: by Liouville $f$ is constant therefore $a = a-1$. However, this proof continues by arguing that for every $\tau \in A^\ast$ the map $\tau \circ f$ is bounded entire and hence $a = a-1$.
Why is this step using $A^\ast$ done? Is it necessary? To me it seems to be enough that $f$ is differentiable and bounded on $\mathbb C$ to derive the desired contradiction.
The other step that is not clear to me are the two lines of inequalities before the ''Consequently, ...''. After $1-|\lambda^{-1}| \|a\| > {1\over 2}$ my next step would be to take the inverse on both sides to get ${1 \over 1- |\lambda^{-1}| \|a\| } < 2$. What am I missing?
Thank you for clarification.