Are these two steps in this proof necessary? 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. 
 A: Liouville's theorem (the one about bounded entire functions) is in complex analysis only proved for complex-valued functions. The generalisation to $\mathbb{C}^n$-valued functions is immediate, but for general Banach-space-valued (or Banach-algebra-valued) functions, it needs to be proved before it is used. The proof in that general case is by reducing it to the $\mathbb{C}$-valued case via continuous linear forms, and then by Hahn-Banach the conclusion that $f$ itself is constant is reached.
If the general form is available, that step is not necessary, but usually, it isn't, since the non-emptiness of spectra is - in books on functional analysis - typically the only place where it is used.
The other step is necessary, because we don't have $\lVert a^{-1}\rVert = \lVert a\rVert^{-1}$ in general. Consider the operator $M$ on $\ell^2$ that multiplies every even-indexed term with $2$, and every odd-indexed term with $\frac12$. Then $\lVert M\rVert = \lVert M^{-1}\rVert = 2$, so from $1-\lvert \lambda^{-1}\rvert\lVert a\rVert < \frac12$ we cannot directly conclude $\lVert (1-\lambda^{-1}a)^{-1}\rVert < 2$, the special structure of the inverse is needed for that conclusion.
