Lecture Notes in Functional Analysis

by William L. Paschke

edition 0.9

Contents

Frontmatter

I. Hilbert Space

II. Bounded Operators

III. Compact Operators

IV. The Spectral Theorem

Index and References

Let $$$p$$$ be a complex polynomial $$$p= p\mathopen{}\left( z\right)\mathclose{}= {a}_{0}+{a}_{1}z+\dotsb+{a}_{n}{z}^{n}$$$. For $$$T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$, $$$p\mathopen{}\left( T\right)\mathclose{}= {a}_{0}+{a}_{1}T+\dotsb+{a}_{n}{T}^{n}$$$. If we write $$$p\mathopen{}\left( z\right)\mathclose{}$$$ in factored form as $$$p\mathopen{}\left( z\right)\mathclose{}= a\prod{} \mathopen{}\left(z-{z}_{i}\right)\mathclose{}$$$ then $$$p\mathopen{}\left( T\right)\mathclose{}= a\prod{} \mathopen{}\left(T-{z}_{i}\right)\mathclose{}$$$.

Proposition II.42

For a complex polynomial $$$p\mathopen{}\left( z\right)\mathclose{}$$$ as above, and $$$T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$, $$$\mathop{\sigma}\mathopen{}\left( p\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left\{\, p\mathopen{}\left( λ\right)\mathclose{}\,\middle\vert\, , λ\in \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}, \,\right\}\mathclose{}$$$.

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 $$$β\notin p\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{}$$$.

Definition II.43

For $$$T\in \mathcal{L}\mathopen{}\left( X\right)\mathclose{}$$$, the spectral radius of $$$T$$$ is denoted $$$\mathop{\rho}\mathopen{}\left( T\right)\mathclose{}$$$ and is defined to be $$$\max{} \mathopen{}\left\{\, \mathopen{}\left\lvert{}λ\right\rvert\mathclose{}\,\middle\vert\, , λ\in \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}, \,\right\}\mathclose{}$$$.

Since $$$\mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}$$$ is a subset of the disc of radius $$$\mathopen{}\left\lVert{}T\right\rVert\mathclose{}$$$, we have $$$\mathop{\rho}\mathopen{}\left( T\right)\mathclose{}\leq \mathopen{}\left\lVert{}T\right\rVert\mathclose{}$$$.

Proposition II.44

For $$$T\in \mathcal{L}\mathopen{}\left( X\right)\mathclose{}$$$, $$$\mathop{\rho}\mathopen{}\left( T\right)\mathclose{}\leq {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}}$$$ for all $$$n\in {\mathbb{Z}}^{+}$$$.

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.

Our proof of the Gelfand-Beurling formula $$$\mathop{\rho}\mathopen{}\left( T\right)\mathclose{}= \lim_{n}{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}}$$$ uses the clever and entirely elementary calculation in Rickart [9]. We begin by defining $$$ν\mathopen{}\left( T\right)\mathclose{}= \inf_{n}{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}}$$$.

Lemma II.45

$$$ν\mathopen{}\left( T\right)\mathclose{}= \lim_{n}{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}}$$$.

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 $$$m$$$: $$\limsup_{n}{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \leq ν\mathopen{}\left( T\right)\mathclose{}\leq \liminf_{m}{} {\mathopen{}\left\lVert{}{T}^{m}\right\rVert\mathclose{}}^{\frac{1}{m}} \text{.}$$

Lemma II.46

Let $$${ω}_{1}$$$, $$${ω}_{2}$$$, …, $$${ω}_{n}$$$ be the $$$n$$$th roots of unity. Then

1. $$$1-{z}^{n}= \prod_{j=1}^{n}{}\mathopen{}\left(1-{ω}_{j}z\right)\mathclose{}$$$, and
2. $$$\frac{1}{n}\sum_{j=1}^{n}{}{ω}_{j}\prod_{k\neq j}{}\mathopen{}\left(1-{ω}_{k}z\right)\mathclose{}= {z}^{n-1}$$$.

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.

Here is a preview of the argument that $$$ν$$$ is the same as $$$\mathop{\rho}$$$: If $$$ν\mathopen{}\left( T\right)\mathclose{}= 0$$$, then $$$\mathop{\rho}\mathopen{}\left( T\right)\mathclose{}= 0$$$ because $$$\mathop{\rho}\mathopen{}\left( T\right)\mathclose{}\leq ν\mathopen{}\left( T\right)\mathclose{}$$$. Henceforth, assume $$$ν= ν\mathopen{}\left( T\right)\mathclose{}\gt 0$$$ for fixed $$$T\in \mathcal{L}\mathopen{}\left( X\right)\mathclose{}$$$. We have $$$\mathop{\rho}\leq ν$$$ so $$$\frac{1}{ν}\leq \frac{1}{\mathop{\rho}}$$$. If $$$\mathop{\rho}\lt ν$$$ we would have $$$λ-T\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1}$$$ for every $$$λ$$$ with $$$\mathopen{}\left\lvert{}λ\right\rvert\mathclose{}\geq ν$$$ which would make $$$1-zT\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1}$$$ for all $$$\mathopen{}\left\lvert{}z\right\rvert\mathclose{}\leq \frac{1}{ν}$$$. The thing we want to look at is the function $$$f$$$ defined by $$$f\mathopen{}\left( z\right)\mathclose{}= zT{ \mathopen{}\left(1-zT\right)\mathclose{} }^{-1}$$$ which is continuous (hence uniformly continuous) on the closed disc of radius $$$\frac{1}{ν}$$$. The strategy is to express $$$\frac{1}{n}\sum_{j=1}^{n}{} f\mathopen{}\left( z{ω}_{j}\right)\mathclose{}$$$ in terms of $$$f$$$ using Lemma II.46.
Theorem II.47 (Gelfand-Beurling)

For $$$T\in \mathcal{L}\mathopen{}\left( X\right)\mathclose{}$$$, we have $$\mathop{\rho}\mathopen{}\left( T\right)\mathclose{}= \max{}\mathopen{}\left\lvert{}\mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}\right\rvert\mathclose{}= \lim_{n}{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \text{.}$$

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.

Remark II.48

The argument above, which uses only algebraic manipulations and the rudiments of higher analysis, actually yields another proof that $$$\mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}\neq \emptyset$$$. Here's how. Begin by defining $$$ν$$$ as an infimum and then proving Lemma II.46, which identifies $$$ν$$$ as the corresponding limit. No knowledge of $$$\mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}$$$ is needed for this. It is easy to see that if $$$\mathopen{}\left\lvert{}λ\right\rvert\mathclose{}\gt ν$$$, then $$$λ-T$$$ is invertible. (Indeed, let $$$A= \frac{1}{λ}$$$. Then $$$\mathopen{}\left\lVert{}{A}^{n}\right\rVert\mathclose{}\lt 1$$$ which makes $$$1-{A}^{n}$$$ invertible. Since $$$1-{A}^{n}= \mathopen{}\left(1-A\right)\mathclose{}\mathopen{}\left(1+A+\dotsb+{A}^{n-1}\right)\mathclose{}$$$, we conclude that $$$1-A$$$ is invertible.) Observe as well that if $$$T$$$ is invertible, then for all $$$n$$$ $$1= \mathopen{}\left\lVert{}I\right\rVert\mathclose{}= {\mathopen{}\left\lVert{}{T}^{n}{T}^{{-}n}\right\rVert\mathclose{}}^{\frac{1}{n}}\leq {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}}{\mathopen{}\left\lVert{}{T}^{{-}n}\right\rVert\mathclose{}}^{\frac{1}{n}} \text{,}$$ so $$$ν\mathopen{}\left( T\right)\mathclose{}ν\mathopen{}\left( {T}^{-1}\right)\mathclose{}\geq 1$$$. It follows that if $$$ν\mathopen{}\left( T\right)\mathclose{}= 0$$$, then $$$T$$$ is not invertible, i.e., $$$0\in \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}$$$. Stripped of all mention of $$$\mathop{\rho}\mathopen{}\left( T\right)\mathclose{}$$$ or $$$\mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}$$$, the calculations in the proof of 2.20 now show that if $$$ν\gt 0$$$, then there must be a $$$z$$$ of modulus $$$\frac{1}{ν}$$$ such that $$$1-zT$$$ is noninvertible, so $$$\frac{1}{z}$$$ belongs to $$$\mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}$$$.

Back to our Hilbert space $$$H$$$.

Proposition II.49

$$${\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}^{2}= \mathopen{}\left\lVert{} T^{*}T\right\rVert\mathclose{}$$$ for all $$$T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$.

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

Definition II.50

An operator $$$T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ is called normal if $$$T^{*}T= T T^{*}$$$.

Example II.51

The multiplication operator by a bounded measurable function on $$$\mathrm{L}^{\mathrm{2}}\mathopen{}\left( Ω\right)\mathclose{}$$$ is a basic model.

Proposition II.52

If $$$T$$$ is normal then $$$\mathopen{}\left\lVert{}T\right\rVert\mathclose{}= \mathop{\rho}\mathopen{}\left( T\right)\mathclose{}$$$.

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 $$$\mathop{\rho}\mathopen{}\left( T\right)\mathclose{}= \mathopen{}\left\lVert{}T\right\rVert\mathclose{}$$$.