Spectrum is nonempty compact and norm bounded

If $\mathrm{\mid }z\mathrm{\mid }>\mathrm{\parallel }a\mathrm{\parallel }$ then $\mathrm{\parallel }a\mathrm{/}z\mathrm{\parallel }=\mathrm{\parallel }a\mathrm{\parallel }\mathrm{/}\mathrm{\mid }z\mathrm{\mid }<1$, so by [L3] the element $1-a\mathrm{/}z$ is invertible and hence $z1-a=z(1-a\mathrm{/}z)$ is invertible with $R(z,a)=z^{-1}(1-a\mathrm{/}z{)}^{-1}={\sum }_{n\ge 0}{z}^{-n-1}{a}^n\mathrm{.}$ The Neumann-series norm estimate gives $\mathrm{\parallel }R(z,a)\mathrm{\parallel }\le \mathrm{\mid }z{\mathrm{\mid }}^{-1}\mathrm{/}(1-\mathrm{\parallel }a\mathrm{\parallel }\mathrm{/}\mathrm{\mid }z\mathrm{\mid })=1\mathrm{/}(\mathrm{\mid }z\mathrm{\mid }-\mathrm{\parallel }a\mathrm{\parallel })$. Therefore $z\in {\rho }_A(a)$ and ${\sigma }_A(a)\subseteq \{\mathrm{\mid }z\mathrm{\mid }\le \mathrm{\parallel }a\mathrm{\parallel }\}$.

The set ${\rho }_A(a)$ is open by [L4], so its complement ${\sigma }_A(a)$ is closed; combined with the boundedness of [step 1.1] this makes ${\sigma }_A(a)$ a closed bounded subset of $\mathbb{C}$, hence compact, which is claim 1.

Suppose, for contradiction, that ${\sigma }_A(a)=\mathrm{\varnothing }$, so that ${\rho }_A(a)=\mathbb{C}$ and $R(z,a)$ is defined for every $z\in \mathbb{C}$.

The element $R(0,a)=(0-a{)}^{-1}=-a^{-1}$ is nonzero: if $a^{-1}=0$ then $1=a{a}^{-1}=0$, contradicting [L1]; here $a$ is invertible because $0\in {\rho }_A(a)$.

By [L6], applied to the distinct points $R(0,a)$ and $0$ in $A$, there is a bounded linear functional $\phi :A\to \mathbb{C}$ with $\phi (R(0,a))\mathrm{\ne }0$.

Define $g:\mathbb{C}\to \mathbb{C}$ by $g(z):=\phi (R(z,a))$. Then $g$ is holomorphic on $\mathbb{C}$: at each $z_0$ the resolvent is complex differentiable with $R^{\mathrm{\prime }}(z_0,a)=-R(z_0,a{)}^2$ by [L4], and a bounded linear functional is complex differentiable with ${\phi }^{\mathrm{\prime }}(x)=\phi$ for $x\in A$, so the chain rule gives $g^{\mathrm{\prime }}(z_0)=-\phi (R(z_0,a{)}^2)$; thus $g$ is entire.

The function $g$ is bounded: on the compact set $\{\mathrm{\mid }z\mathrm{\mid }\le \mathrm{\parallel }a\mathrm{\parallel }+1\}$ the norm $\mathrm{\parallel }R(z,a)\mathrm{\parallel }$ is bounded by some $C_1<\mathrm{\infty }$ because $z\mapsto R(z,a)$ is norm continuous by [L4], and for $\mathrm{\mid }z\mathrm{\mid }>\mathrm{\parallel }a\mathrm{\parallel }+1$ the estimate in [step 1.1] gives $\mathrm{\parallel }R(z,a)\mathrm{\parallel }\le 1\mathrm{/}(\mathrm{\mid }z\mathrm{\mid }-\mathrm{\parallel }a\mathrm{\parallel })\le 1$; hence $\mathrm{\mid }g(z)\mathrm{\mid }\le \mathrm{\parallel }\phi \mathrm{\parallel }\max (C_1,1)$ for every $z\in \mathbb{C}$.

By [L5] the bounded entire function $g$ is constant; since the estimate in [step 1.1] gives $\mathrm{\parallel }R(z,a)\mathrm{\parallel }\le 1\mathrm{/}(\mathrm{\mid }z\mathrm{\mid }-\mathrm{\parallel }a\mathrm{\parallel })\to 0$ as $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$ and $\phi$ is continuous, $g(z)\to 0$ along $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$, so the constant value is $0$ and $g\equiv 0$.

But $g(0)=\phi (R(0,a))\mathrm{\ne }0$ by the choice of $\phi$ in [step 5.1], contradicting $g\equiv 0$; hence ${\sigma }_A(a)\mathrm{\ne }\mathrm{\varnothing }$, which is claim 2.

Claim 1 was proved in [step 2.1] and claim 2 in [step 9.1], so the spectrum of $a$ is a nonempty compact subset of the disc of radius $\mathrm{\parallel }a\mathrm{\parallel }$.