Proof. Factor $\( p\mathopen{}\left( z\right)\mathclose{}= a\mathopen{}\left(z-{z}_{1}\right)\mathclose{}\mathopen{}\left(z-{z}_{2}\right)\mathclose{}\dotsb\mathopen{}\left(z-{z}_{n}\right)\mathclose{} \)$ where $\( a\neq 0 \)$. We have $\( 0\notin \mathop{\sigma}\mathopen{}\left( p\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{} \)$ if and only if $\( p\mathopen{}\left( T\right)\mathclose{}\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \)$ if and only if $\( T-{z}_{i}\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \)$ for $\( 1\leq i\leq n \)$ (because for $\( AB\in T\in \mathcal{L}\mathopen{}\left( X\right)\mathclose{} \)$ with $\( AB= BA \)$, we have $\( AB\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \)$ if and only if $\( A\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \)$ and $\( B\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \)$) i.e., if and only if $\( {z}_{i}\notin \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{} \)$ for all $\( 1\leq i\leq n \)$ if and only if $\( 0\in p\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{} \)$. For $\( β\in \mathbb{C} \)$, $\( β\notin p\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{} \)$ if and only $\( 0\notin σ\mathopen{}\left( p\mathopen{}\left( T\right)\mathclose{}-β\right)\mathclose{} \)$. So, apply the previous reasoning to $\( p\mathopen{}\left( T\right)\mathclose{}-β \)$ to see that $\( β\notin \mathop{\sigma}\mathopen{}\left( p\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{} \)$ if and only if $\( β\notin p\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{} \)$.

Proof. By Proposition II.42, $\( \mathop{\rho}\mathopen{}\left( {T}^{n}\right)\mathclose{}= {\mathop{\rho}\mathopen{}\left( T\right)\mathclose{}}^{n} \)$ for all $\( n \)$. On the other hand, $\( \mathop{\rho}\mathopen{}\left( T\right)\mathclose{}\leq \mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{} \)$ implies $\( {\mathop{\rho}\mathopen{}\left( T\right)\mathclose{}}^{n}\leq \mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{} \)$, so $\( \mathop{\rho}\mathopen{}\left( T\right)\mathclose{}\leq \inf_{n}{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \)$, i.e. $\( \mathop{\rho}\mathopen{}\left( T\right)\mathclose{} \)$ is less than or equal to the infimum of these critters.

Proof. Fix $\( m\in {\mathbb{Z}}^{+} \)$. For $\( n\gt m \)$ write $\( n= pm+q \)$, where $\( q\in \mathopen{}\left\{\, 1, 2, \dotsc, m-1\,\right\}\mathclose{} \)$. Notice that $\( p= p\mathopen{}\left( n\right)\mathclose{} \)$ and $\( q= q\mathopen{}\left( n\right)\mathclose{} \)$ and that now $$\[ {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}}= {\mathopen{}\left\lVert{}{T}^{pm+q}\right\rVert\mathclose{}}^{\frac{1}{n}}\leq {\mathopen{}\left\lVert{}{T}^{m}\right\rVert\mathclose{}}^{\frac{p}{n}}{\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}^{\frac{q}{n}}\text{.} \]$$ Also notice that $\( \frac{p}{n}= \frac{1-\frac{q}{n}}{m} \)$, so $$\[ {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}}\leq {\mathopen{}\left\lVert{}{T}^{m}\right\rVert\mathclose{}}^{\frac{1-\frac{q}{n}}{m}}{\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}^{\frac{q}{n}} \]$$ for all $\( n\gt m \)$. Thus $\( \limsup_{n}{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \leq {\mathopen{}\left\lVert{}{T}^{m}\right\rVert\mathclose{}}^{\frac{1}{m}} \)$ for every $\( m \)$. (Without loss of generality, $\( T\neq 0 \)$). Fine. Take the infimum on $\( m \)$: $$\[ \limsup_{n}{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \leq ν\mathopen{}\left( T\right)\mathclose{}\leq \liminf_{m}{} {\mathopen{}\left\lVert{}{T}^{m}\right\rVert\mathclose{}}^{\frac{1}{m}} \text{.} \]$$

Proof.

1. The two sides have the same degree and the same roots (both vanish at $\( {ω}_{1} \)$, $\( {ω}_{2} \)$, …, $\( {ω}_{n} \)$) and have the same value at $\( z= 0 \)$.
2. Take the derivative with respect to $\( z \)$ in part 1.

Proof. We already know the limit (call it $\( ν \)$) exists, that $\( ν= \inf{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \)$, and that $\( \mathop{\rho}\mathopen{}\left( T\right)\mathclose{}\leq n \)$ by Lemma II.45 and Proposition II.44. Suppose, by way of contradiction, that $\( ν\gt \mathop{\rho}\mathopen{}\left( T\right)\mathclose{} \)$. This implies that $\( ν\gt 0 \)$. Let $\( D= \mathopen{}\left\{\, z\in \mathbb{C}\,\middle\vert\, , \mathopen{}\left\lvert{}z\right\rvert\mathclose{}\leq \frac{1}{ν}, \,\right\}\mathclose{} \)$. Define $\( f : D \to \mathcal{L}\mathopen{}\left( X\right)\mathclose{} \)$ by $\( f\mathopen{}\left( z\right)\mathclose{}= zT{ \mathopen{}\left(1-zT\right)\mathclose{} }^{-1} \)$. (Note: $\( 1-zT\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \)$. For $\( z= 0 \)$, no problem. Otherwise, $\( 0\lt \mathopen{}\left\lvert{}z\right\rvert\mathclose{}\leq \frac{1}{ν} \)$ implies $\( {z}^{-1}\geq ν\gt \mathop{\rho}\mathopen{}\left( T\right)\mathclose{} \)$ implies $\( {z}^{-1}-T\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \)$ implies $\( 1-zT\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \)$.) We remark that $\( f \)$ is uniformly continuous on $\( D \)$ because the inversion map of $\( { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \)$ to itself is continuous and $\( D \)$ is compact. Given $\( n\in {\mathbb{Z}}^{+} \)$ and $\( n \)$th roots $\( {ω}_{1} \)$, $\( {ω}_{2} \)$, …, $\( {ω}_{n} \)$ of unity, we have $\( \frac{1}{n}\sum_{j=1}^{n}{} {ω}_{j} \prod_{k\neq j}{}\mathopen{}\left(1-{ω}_{k}z\right)\mathclose{}= {z}^{n-1} \)$. (This is a polynomial identity.) Replace $\( z \)$ by $\( zT \)$ to get $\( \frac{1}{n}\sum_{j=1}^{n}{} {ω}_{j} \prod_{k\neq j}{}\mathopen{}\left(1-{ω}_{k}zT\right)\mathclose{}= {z}^{n-1}{T}^{n-1} \)$. Multiply by $\( zT \)$ to get $\( \frac{1}{n}\sum_{j=1}^{n}{} {ω}_{j} z T \prod_{k\neq j}{}\mathopen{}\left(1-{ω}_{k}zT\right)\mathclose{}= {z}^{n}{T}^{n} \)$. By Lemma II.46, $\( 1-{z}^{n}{T}^{n}= \prod_{k=1}^{n}{}\mathopen{}\left(1-{ω}_{k}zT\right)\mathclose{} \)$. All the factors commute, so for $\( z\in D \)$, we can multiply both sides by $\( { \mathopen{}\left(1-{z}^{n}{T}^{n}\right)\mathclose{} }^{-1} \)$ to obtain $$\[ \frac{1}{n}\sum_{j=1}^{n}{} f\mathopen{}\left( z{ω}_{j}\right)\mathclose{} = {z}^{n}{T}^{n}{\mathopen{}\left(1-{z}^{n}{T}^{n}\right)\mathclose{}}^{-1} \text{.} \]$$

Given $\( ε\gt 0 \)$ there exists $\( δ\gt 0 \)$ such that $\( \mathopen{}\left\lVert{}f\mathopen{}\left( {z}_{1}\right)\mathclose{}-f\mathopen{}\left( {z}_{2}\right)\mathclose{}\right\rVert\mathclose{}\lt ε \)$ whenever $\( \mathopen{}\left\lvert{}{z}_{1}-{z}_{2}\right\rvert\mathclose{}\lt δ \)$. Take $\( μ \)$ with $\( μ\gt ν \)$ and $\( \frac{1}{μ}-\frac{1}{ν}\lt δ \)$. We have $\( \mathopen{}\left\lvert{}{ν}^{-1}{ω}_{j}-{μ}^{-1}{ω}_{j}\right\rvert\mathclose{}\lt δ \)$, and hence $\( \mathopen{}\left\lVert{}f\mathopen{}\left( {ν}^{-1}{ω}_{j}\right)\mathclose{}-f\mathopen{}\left( {μ}^{-1}{ω}_{j}\right)\mathclose{}\right\rVert\mathclose{}\lt ε \)$, for $\( 1\leq j\leq n \)$. It follows from Equation II.B that $$\[ \mathopen{}\left\lVert{}{ν}^{{-}n}{T}^{n}{\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1}-{μ}^{{-}n}{T}^{n}{\mathopen{}\left(1-{μ}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1}\right\rVert\mathclose{}\lt ε \text{.} \]$$

Equation II.C holds for every $\( n \)$. Because $\( μ\gt \lim_{n}{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \)$ we have $\( {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}}\leq r\lt μ \)$ for sufficiently large $\( n \)$. Thus $\( \mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}\leq {r}^{n} \)$, $\( \mathopen{}\left\lVert{}{μ}^{{-}n}{T}^{n}\right\rVert\mathclose{}\leq {\mathopen{}\left(\frac{r}{μ}\right)\mathclose{}}^{n} \to 0 \)$. So $\( {μ}^{{-}n}{T}^{n} \to 0 \)$ and $\( {μ}^{{-}n}{T}^{n}{\mathopen{}\left(1-{μ}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1} \to 0 \)$. By Equation II.C, $\( \mathopen{}\left\lVert{}{ν}^{{-}n}{T}^{n}{\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1}\right\rVert\mathclose{}\lt ε \)$ for sufficiently large $\( n \)$, so $\( {ν}^{{-}n}{T}^{n}{\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1} \to 0 \)$. Notice $\( 1= \mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}{\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1}= {\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1}-\mathopen{}\left({ν}^{{-}n}{T}^{n}\right)\mathclose{}{\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1} \)$. But $\( \mathopen{}\left({ν}^{{-}n}{T}^{n}\right)\mathclose{}{\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1} \to 0 \)$ so $\( {\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1} \to 1 \)$ implies $\( {ν}^{{-}n}{T}^{n} \to 0 \)$. However, $\( ν\leq {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \)$ for all $\( n \)$, i.e. $\( 1\leq {\mathopen{}\left\lVert{}{ν}^{{-}n}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \)$ for all $\( n \)$, which is a contradiction.

Proof. $\( \mathopen{}\left\lVert{} T^{*}T\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{} T^{*}\right\rVert\mathclose{}\mathopen{}\left\lVert{}T\right\rVert\mathclose{}= \mathopen{}\left\lVert{}{T}^{2}\right\rVert\mathclose{} \)$. Moreover, $$\[ \mathopen{}\left\lVert{}{T}^{2}\right\rVert\mathclose{}= \sup_{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1}{}{\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}= \sup_{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1}{}\mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}, T\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}= \sup_{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1}{}\mathopen{}\left\langle{} T^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x\right\rangle\mathclose{}\leq \sup_{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1}{}\mathopen{}\left\lVert{} T^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{}= \mathopen{}\left\lVert{} T^{*}\mathopen{}\left( T\right)\mathclose{}\right\rVert\mathclose{} \text{.} \]$$

Proof. Notice that $$\[ {\mathopen{}\left\lVert{}{T}^{2}\right\rVert\mathclose{}}^{2}= \mathopen{}\left\lVert{} \mathopen{}\left({T}^{2}\right)\mathclose{}^{*}{T}^{2}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\mathopen{}\left( T^{*} T^{*}\right)\mathclose{}\mathopen{}\left(TT\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\mathopen{}\left( T^{*}T\right)\mathclose{}\mathopen{}\left( T^{*}T\right)\mathclose{}\right\rVert\mathclose{}= {\mathopen{}\left\lVert{} T^{*}T\right\rVert\mathclose{}}^{2}= {\mathopen{}\left({\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}^{2}\right)\mathclose{}}^{2} \]$$ so $\( \mathopen{}\left\lVert{}{T}^{2}\right\rVert\mathclose{}= {\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}^{2} \)$ for all normal operators. $\( T \)$ normal implies $\( {T}^{2} \)$ normal implies $\( \mathopen{}\left\lVert{}{T}^{4}\right\rVert\mathclose{}= {\mathopen{}\left\lVert{}{T}^{2}\right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}^{4} \)$ etc., for all $\( n \)$. $\( {\mathopen{}\left\lVert{}{T}^{2n}\right\rVert\mathclose{}}^{\frac{1}{2n}}= \mathopen{}\left\lVert{}T\right\rVert\mathclose{} \)$. Let $\( n \to \infty \)$, to get $\( \mathop{\rho}\mathopen{}\left( T\right)\mathclose{}= \mathopen{}\left\lVert{}T\right\rVert\mathclose{} \)$.