Spectral radius formula

First, if $a=0$, then $r(a)=0$ by $0\le r(a)\le \mathrm{\parallel }a\mathrm{\parallel }$ in [L10], and every positive power and root norm is zero, proving the formula. In the remainder assume $a\mathrm{\ne }0$, so $\mathrm{\parallel }a\mathrm{\parallel }>0$ and divisions by this number are legitimate. For $z\mathrm{\ne }0$ one has $1-za=-z(a-z^{-1}1)$; hence $1-za$ is invertible exactly when $a-z^{-1}1$ is, that is exactly when $z^{-1}\in {\rho }_A(a)$.

For every $n\ge 1$ one has $\mathrm{\parallel }{a}^{m+n}\mathrm{\parallel }\le \mathrm{\parallel }{a}^m\mathrm{\parallel }\mathrm{\parallel }{a}^n\mathrm{\parallel }$ for all $m,n\ge 1$, so the positive-indexed family $u_n:=\mathrm{\parallel }{a}^n\mathrm{\parallel }$ is submultiplicative and nonnegative, and its root sequence is $v_j=u_{j+1}^{1\mathrm{/}(j+1)}$ for $j\ge 0$.

For every $n\ge 1$, [L9] gives ${\sigma }_A(a^n)=\{{\lambda }^n:\lambda \in {\sigma }_A(a)\}$, hence by [L10] applied to $a^n$ and the multiplicativity of the modulus, $r(a^n)=\max \{\mathrm{\mid }\lambda {\mathrm{\mid }}^n:\lambda \in {\sigma }_A(a)\}=r(a{)}^n$; and $r(a^n)\le \mathrm{\parallel }{a}^n\mathrm{\parallel }$ by the last clause of [L10], so $r(a{)}^n\le \mathrm{\parallel }{a}^n\mathrm{\parallel }$.

For $\mathrm{\mid }z\mathrm{\mid }<1\mathrm{/}\mathrm{\parallel }a\mathrm{\parallel }$ one has $\mathrm{\parallel }za\mathrm{\parallel }=\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }a\mathrm{\parallel }<1$, so [L2] applies to $y:=za$ and gives $(1-za{)}^{-1}={\sum }_{n\ge 0}{a}^n{z}^n$; comparing with the identity of [step 1.1] this is a power series in $z$ whose value at $z=0$ is $1$ and whose linear coefficient is $a$.

Fix a real $R>r(a)$ and put $D_R:=\{z\in \mathbb{C}:\mathrm{\mid }z\mathrm{\mid }<1\mathrm{/}R\}$. For $z\in D_R$ with $z\mathrm{\ne }0$ one has $\mathrm{\mid }{z}^{-1}\mathrm{\mid }>R>r(a)$, so $z^{-1}\mathrm{\notin }{\sigma }_A(a)$ because every spectral point has modulus at most $r(a)$; by [step 1.1] the element $1-za$ is invertible. Hence the function $h(z):=(1-za{)}^{-1}$ for $z\mathrm{\ne }0$, $h(0):=1$, is a well-defined map $D_R\to A$.

$h$ is complex differentiable at every $z_0\in D_R$, $z_0\mathrm{\ne }0$: by [step 1.1] one has $h(z)=z^{-1}R(z^{-1},a)$ for $z\mathrm{\ne }0$, so with $u:=z^{-1}$, $v:=z_0^{-1}$ the difference is $h(z)-h(z_0)=uR(u)-vR(v)=(v-u)aR(u)R(v)$, the identity following from the resolvent identity [L3] in the form $R(u)-R(v)=(v-u)R(u)R(v)$ together with the rearrangement $1-vR(v)=-aR(v)$ of $vR(v)=1+aR(v)$; since $v-u=(z-z_0)\mathrm{/}(z{z}_0)$, the difference quotient is $\frac{h(z)-h(z_0)}{z-z_0}=a\frac{R(z^{-1})R(z_0^{-1})}{z{z}_0}$, which converges to $aR(z_0^{-1}{)}^2\mathrm{/}{z}_0^2$ as $z\to z_0$ by continuity of the resolvent [L4] and of $1\mathrm{/}z$.

At $z_0=0$ the map $h$ is complex differentiable with $h^{\mathrm{\prime }}(0)=a$: by [step 2.1], $h(z)=1+za+{\sum }_{n\ge 2}{a}^n{z}^n$ for $\mathrm{\mid }z\mathrm{\mid }<1\mathrm{/}\mathrm{\parallel }a\mathrm{\parallel }$, and the remainder is bounded by ${\sum }_{n\ge 2}\mathrm{\parallel }a{\mathrm{\parallel }}^n\mathrm{\mid }z{\mathrm{\mid }}^n=\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{\parallel }a{\mathrm{\parallel }}^2\mathrm{/}(1-\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }a\mathrm{\parallel })=o(\mathrm{\mid }z\mathrm{\mid })$. Combined with [step 3.1] this shows that $h$ is holomorphic on $D_R$.

Let $\phi \in A^{\ast }$ be a bounded linear functional and let $g:=\phi \circ h:D_R\to \mathbb{C}$. Since $h$ is holomorphic by [step 4.1] and $\phi$ is continuous linear, $g$ is holomorphic on $D_R$ with $g^{\mathrm{\prime }}(z)=\phi (h^{\mathrm{\prime }}(z))$.

Fix $R^{\mathrm{\prime }}$ with $r(a)<R^{\mathrm{\prime }}<R$ and put $D_{R^{\mathrm{\prime }}}:=\{z:\mathrm{\mid }z\mathrm{\mid }<1\mathrm{/}{R}^{\mathrm{\prime }}\}$. The argument of steps 2.2-4.1 with $R^{\mathrm{\prime }}$ in place of $R$ shows that $h$ is holomorphic on $D_{R^{\mathrm{\prime }}}$; since $1\mathrm{/}R<1\mathrm{/}{R}^{\mathrm{\prime }}$, the disc $D_{R^{\mathrm{\prime }}}$ contains the closed disc of radius $1\mathrm{/}R$ around $0$, the same inverse formula extends $h$ consistently, and $g=\phi \circ h$ extends by that formula as well. Thus [L5] applies to this extended $g$ with $\rho =1\mathrm{/}R$ and gives $g^{(n)}(0)=\frac{n!}{2\pi i}{\int }_{\mathrm{\mid }\zeta \mathrm{\mid }=1\mathrm{/}R}g(\zeta ){\zeta }^{-n-1}d\zeta$ for every $n\ge 0$.

For $0<\rho <\min (1\mathrm{/}R,1\mathrm{/}\mathrm{\parallel }a\mathrm{\parallel })$ the series of [step 2.1] converges uniformly on the circle $\mathrm{\mid }\zeta \mathrm{\mid }=\rho$, so $g(\zeta )=\phi ((1-\zeta a{)}^{-1})={\sum }_{k\ge 0}\phi (a^k){\zeta }^k$ uniformly there, and [L5] also applies on this smaller circle. For fixed $n$, the uniform remainder after multiplying by ${\zeta }^{-n-1}$ is bounded by $\mathrm{\parallel }\phi \mathrm{\parallel }{\rho }^{-n-1}(\rho \mathrm{\parallel }a\mathrm{\parallel }{)}^{N+1}\mathrm{/}(1-\rho \mathrm{\parallel }a\mathrm{\parallel })$, which tends to zero. By [L12] its integral tends to zero. Integrating the finite sums and using [L6] therefore gives $g^{(n)}(0)=n!\phi (a^n)$ for every $n\ge 0$.

Norm estimate for the coefficients, using [L12] on the circle of length $2\pi \mathrm{/}R$: for $n\ge 0$, by [step 7.1] and the integral formula of [step 6.1], $\mathrm{\mid }\phi (a^n)\mathrm{\mid }=\frac{1}{2\pi }\mid {\int }_{\mathrm{\mid }\zeta \mathrm{\mid }=1\mathrm{/}R}g(\zeta ){\zeta }^{-n-1}d\zeta \mid \le ({\sup }_{\mathrm{\mid }\zeta \mathrm{\mid }=1\mathrm{/}R}\mathrm{\parallel }h(\zeta )\mathrm{\parallel })\mathrm{\parallel }\phi \mathrm{\parallel }{R}^n$; the supremum is finite because [step 6.1] places the circle $\mathrm{\mid }\zeta \mathrm{\mid }=1\mathrm{/}R$ as a compact subset of the larger disc $D_{R^{\mathrm{\prime }}}$, on which the argument of [step 4.1] makes $h$ holomorphic and hence continuous.

Put $C_R:={\sup }_{\mathrm{\mid }\zeta \mathrm{\mid }=1\mathrm{/}R}\mathrm{\parallel }h(\zeta )\mathrm{\parallel }<\mathrm{\infty }$. This constant is positive because $h(1\mathrm{/}R)$ is invertible and therefore nonzero. Taking the supremum in [step 8.1] over all $\phi$ with $\mathrm{\parallel }\phi \mathrm{\parallel }\le 1$ and using [L7] gives $\mathrm{\parallel }{a}^n\mathrm{\parallel }\le C_R{R}^n$ for every $n\ge 0$; hence $v_j\le R{C}_R^{1\mathrm{/}(j+1)}$ for every $j\ge 0$. By [L8] and step 1.2, $v_j$ has a real limit $Q$ equal to the stated infimum. By [L11], $C_R^{1\mathrm{/}(j+1)}\to 1$, and passing to these real limits gives $Q\le R$. (If $Q>R$, convergence of both sequences would contradict their termwise inequality.) Since this holds for every $R>r(a)$, $Q\le r(a)$: otherwise choose $R=(Q+r(a))\mathrm{/}2$.

By [L8] applied to the submultiplicative family $u_n=\mathrm{\parallel }{a}^n\mathrm{\parallel }$ of [step 1.2], the limit $Q:={\lim }_j{v}_j={\inf }_n\mathrm{\parallel }{a}^n{\mathrm{\parallel }}^{1\mathrm{/}n}$ exists; [step 9.1] gives $Q\le r(a)$, while [step 1.3] gives $\mathrm{\parallel }{a}^n{\mathrm{\parallel }}^{1\mathrm{/}n}\ge r(a)$ for every $n$, hence $Q\ge r(a)$. Therefore $Q=r(a)$ and the formula holds for $a\mathrm{\ne }0$; step 1.1 already proved the zero case.