Spectrum of complex normed space bounded operator This is a problem from Functional Analysis by B.V Limaye which I am finding difficult to deal with:
Let $X$ be a normed space over complex field and let  $$   A: X\rightarrow X $$ be a bounded linear operator. Then show that spectrum of A is non-empty.
 A: Suppose by contradiction that $\sigma(A)=\emptyset$, then $\rho(A)=\mathbb{C}$ and $:\mathbb{C}\ni \lambda \mapsto R_{\lambda}(A)$ (where $R_{\lambda}(A)$ is the resolvent of $A$ in $\lambda$) is analytic on the whole complex plane, take any $f'\in X'$ and $x\in X$, define $g(\lambda):= f(R_{\lambda}(A)x) $. By expanding the analytic series of the resolvent, using the continuity of $f'$ and the fact that $\sum_{n \in \mathbb{N}}(\lambda-\mu)^nR_{\mu}(A)^n$ converges in the operator topology,
$$
g(\lambda)=\sum_{n \in \mathbb{N}}(\lambda-\mu)^nf(R_{\mu}(A)^{n+1}x)
$$
Therefore $g$ is analytic on the whole $\mathbb{C}$. Now when $\lambda>1+ \vert\vert A\vert\vert $ one can also expand the resolvent as (you just need $\lambda> \vert\vert A\vert\vert $ for the expression below, but the above relation is usefull when making the estimate for $g$)
$$
R_{\lambda}(A)=-\sum_{n \in \mathbb{N}}\lambda^{-n-1}A^n
$$
so using the absolute convergence of $g(\lambda)$ one has the estimate 
$$
\vert g(\lambda)\vert \leq \sum_{n \in \mathbb{N}}\vert \lambda\vert ^{-n-1} \vert f(A^nx)\vert \leq \frac{\vert\vert f \vert\vert \vert\vert x\vert\vert}{\vert\lambda\vert}\frac{1}{1-\vert \lambda \vert^{-1} \vert\vert A\vert\vert}\leq \frac{c}{\vert \lambda \vert}
$$
thus $g$ is bounded for $\lambda>1+ \vert\vert A\vert\vert $. It is also bounded when $\lambda \leq 1+ \vert\vert A\vert\vert $ (it is continuous, being holomorphic, in a compact), so by Liouville's Theorem it must be constant, but the estimate given implies that the only constant possible is $0$. Using the arbitrariness of $f$ and $x$ one has $R_{\lambda}(A)=0$, which is a contradiction.
