Lecture Notes in Functional Analysis

by William L. Paschke

edition 0.9

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Contents

Frontmatter

I. Hilbert Space

II. Bounded Operators

III. Compact Operators

IV. The Spectral Theorem

Index and References

E. Spectral Radius

Let preal number\( p \) be a complex polynomial preal number=equalspreal number(z)=equalsareal number0zero+plusareal number1onetimesz+plus+plusareal numbernintegertimeszninteger \( p= p\mathopen{}\left( z\right)\mathclose{}= {a}_{0}+{a}_{1}z+\dotsb+{a}_{n}{z}^{n} \). For Tlinear mapelement ofbounded linear operators(HHilbert space) \( T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \), preal number(Tlinear map)=equalsareal number0zero+plusareal number1onetimesTlinear map+plus+plusareal numbernintegertimesTlinear mapninteger \( p\mathopen{}\left( T\right)\mathclose{}= {a}_{0}+{a}_{1}T+\dotsb+{a}_{n}{T}^{n} \). If we write preal number(z)\( p\mathopen{}\left( z\right)\mathclose{} \) in factored form as preal number(z)=equalsareal numbertimesproduct (z-minusziinteger) \( p\mathopen{}\left( z\right)\mathclose{}= a\prod{} \mathopen{}\left(z-{z}_{i}\right)\mathclose{} \) then preal number(Tlinear map)=equalsareal numbertimesproduct (Tlinear map-minusziinteger) \( p\mathopen{}\left( T\right)\mathclose{}= a\prod{} \mathopen{}\left(T-{z}_{i}\right)\mathclose{} \).

Proposition II.42

For a complex polynomial preal number(z)\( p\mathopen{}\left( z\right)\mathclose{} \) as above, and Tlinear mapelement ofbounded linear operators(HHilbert space) \( T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \), σ(preal number(Tlinear map))=equals{setpreal number(λcomplex number)|such that λcomplex numberelement ofσ(Tlinear map) }set \( \mathop{\sigma}\mathopen{}\left( p\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left\{\, p\mathopen{}\left( λ\right)\mathclose{}\,\middle\vert\, , λ\in \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}, \,\right\}\mathclose{} \).

Proof. Factor preal number(z)=equalsareal numbertimes(z-minusz1one)times(z-minusz2two)timestimes(z-minuszninteger) \( p\mathopen{}\left( z\right)\mathclose{}= a\mathopen{}\left(z-{z}_{1}\right)\mathclose{}\mathopen{}\left(z-{z}_{2}\right)\mathclose{}\dotsb\mathopen{}\left(z-{z}_{n}\right)\mathclose{} \) where areal numbernot equal to0zero \( a\neq 0 \). We have 0zeronot an element ofσ(preal number(Tlinear map)) \( 0\notin \mathop{\sigma}\mathopen{}\left( p\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{} \) if and only if preal number(Tlinear map)element of bounded linear operators(Xnormed linear space) 1invertible elements \( p\mathopen{}\left( T\right)\mathclose{}\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \) if and only if Tlinear map-minusziintegerelement of bounded linear operators(Xnormed linear space) 1invertible elements \( T-{z}_{i}\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \) for 1oneless than or equal toiintegerless than or equal toninteger \( 1\leq i\leq n \) (because for Aself-adjoint operatortimesBBanach spaceelement ofTlinear mapelement ofbounded linear operators(Xnormed linear space) \( AB\in T\in \mathcal{L}\mathopen{}\left( X\right)\mathclose{} \) with Aself-adjoint operatortimesBBanach space=equalsBBanach spacetimesAself-adjoint operator \( AB= BA \), we have Aself-adjoint operatortimesBBanach spaceelement of bounded linear operators(Xnormed linear space) 1invertible elements \( AB\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \) if and only if Aself-adjoint operatorelement of bounded linear operators(Xnormed linear space) 1invertible elements \( A\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \) and BBanach spaceelement of bounded linear operators(Xnormed linear space) 1invertible elements \( B\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \)) i.e., if and only if ziintegernot an element ofσ(Tlinear map) \( {z}_{i}\notin \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{} \) for all 1oneless than or equal toiintegerless than or equal toninteger \( 1\leq i\leq n \) if and only if 0zeroelement ofpreal number(σ(Tlinear map)) \( 0\in p\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{} \). For βcomplex numberelement ofCcomplex numbers \( β\in \mathbb{C} \), βcomplex numbernot an element ofpreal number(σ(Tlinear map)) \( β\notin p\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{} \) if and only 0zeronot an element ofσ(preal number(Tlinear map)-minusβcomplex number) \( 0\notin σ\mathopen{}\left( p\mathopen{}\left( T\right)\mathclose{}-β\right)\mathclose{} \). So, apply the previous reasoning to preal number(Tlinear map)-minusβcomplex number \( p\mathopen{}\left( T\right)\mathclose{}-β \) to see that βcomplex numbernot an element ofσ(preal number(Tlinear map)) \( β\notin \mathop{\sigma}\mathopen{}\left( p\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{} \) if and only if βcomplex numbernot an element ofpreal number(σ(Tlinear map)) \( β\notin p\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{} \).

Definition II.43

For Tlinear mapelement ofbounded linear operators(Xnormed linear space) \( T\in \mathcal{L}\mathopen{}\left( X\right)\mathclose{} \), the spectral radius of Tlinear map\( T \) is denoted ρspectral radius(Tlinear map) \( \mathop{\rho}\mathopen{}\left( T\right)\mathclose{} \) and is defined to be maxmaximum {set|modulusλcomplex number|modulus|such that λcomplex numberelement ofσ(Tlinear map) }set \( \max{} \mathopen{}\left\{\, \mathopen{}\left\lvert{}λ\right\rvert\mathclose{}\,\middle\vert\, , λ\in \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}, \,\right\}\mathclose{} \).

Since σ(Tlinear map) \( \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{} \) is a subset of the disc of radius Tlinear map\( \mathopen{}\left\lVert{}T\right\rVert\mathclose{} \), we have ρ(Tlinear map)less than or equal toTlinear map \( \mathop{\rho}\mathopen{}\left( T\right)\mathclose{}\leq \mathopen{}\left\lVert{}T\right\rVert\mathclose{} \).

Proposition II.44

For Tlinear mapelement ofbounded linear operators(Xnormed linear space) \( T\in \mathcal{L}\mathopen{}\left( X\right)\mathclose{} \), ρ(Tlinear map)less than or equal to Tlinear mapninteger 1oneninteger \( \mathop{\rho}\mathopen{}\left( T\right)\mathclose{}\leq {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \) for all nintegerelement ofZintegers+positive elements \( n\in {\mathbb{Z}}^{+} \).

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

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

Lemma II.45

ν(Tlinear map)=equalslimlimitninteger Tlinear mapninteger 1oneninteger \( ν\mathopen{}\left( T\right)\mathclose{}= \lim_{n}{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \).

Proof. Fix mintegerelement ofZintegers+positive elements \( m\in {\mathbb{Z}}^{+} \). For ninteger>greater thanminteger \( n\gt m \) write ninteger=equalspintegertimesminteger+plusqinteger \( n= pm+q \), where qintegerelement of{set1one2twominteger-minus1one}set \( q\in \mathopen{}\left\{\, 1, 2, \dotsc, m-1\,\right\}\mathclose{} \). Notice that pinteger=equalspinteger(ninteger) \( p= p\mathopen{}\left( n\right)\mathclose{} \) and qinteger=equalsqinteger(ninteger) \( q= q\mathopen{}\left( n\right)\mathclose{} \) and that now Tlinear mapninteger 1oneninteger =equals Tlinear mappintegertimesminteger+plusqinteger 1oneninteger less than or equal to Tlinear mapminteger pintegerninteger times Tlinear map qintegerninteger . \[ {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}}= {\mathopen{}\left\lVert{}{T}^{pm+q}\right\rVert\mathclose{}}^{\frac{1}{n}}\leq {\mathopen{}\left\lVert{}{T}^{m}\right\rVert\mathclose{}}^{\frac{p}{n}}{\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}^{\frac{q}{n}}\text{.} \] Also notice that pintegerninteger=equals 1one-minusqintegerninteger minteger \( \frac{p}{n}= \frac{1-\frac{q}{n}}{m} \), so Tlinear mapninteger 1oneninteger less than or equal to Tlinear mapminteger 1one-minusqintegerninteger minteger times Tlinear map qintegerninteger \[ {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}}\leq {\mathopen{}\left\lVert{}{T}^{m}\right\rVert\mathclose{}}^{\frac{1-\frac{q}{n}}{m}}{\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}^{\frac{q}{n}} \] for all ninteger>greater thanminteger\( n\gt m \). Thus lim suplimit supremumninteger Tlinear mapninteger 1oneninteger less than or equal to Tlinear mapminteger 1oneminteger \( \limsup_{n}{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \leq {\mathopen{}\left\lVert{}{T}^{m}\right\rVert\mathclose{}}^{\frac{1}{m}} \) for every minteger\( m \). (Without loss of generality, Tlinear mapnot equal to0zero \( T\neq 0 \)). Fine. Take the infimum on minteger\( m \): lim suplimit supremumninteger Tlinear mapninteger 1oneninteger less than or equal toν(Tlinear map)less than or equal tolim inflimit infimumminteger Tlinear mapminteger 1oneminteger . \[ \limsup_{n}{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \leq ν\mathopen{}\left( T\right)\mathclose{}\leq \liminf_{m}{} {\mathopen{}\left\lVert{}{T}^{m}\right\rVert\mathclose{}}^{\frac{1}{m}} \text{.} \]

Lemma II.46

Let ωroot of unity1one\( {ω}_{1} \), ωroot of unity2two\( {ω}_{2} \), …, ωroot of unityninteger\( {ω}_{n} \) be the ninteger\( n \)th roots of unity. Then

  1. 1one-minuszninteger=equalsproductjinteger=1oneninteger(1one-minusωroot of unityjintegertimesz) \( 1-{z}^{n}= \prod_{j=1}^{n}{}\mathopen{}\left(1-{ω}_{j}z\right)\mathclose{} \), and
  2. 1onenintegertimessummationjinteger=1onenintegerωroot of unityjintegertimesproductkintegerjinteger(1one-minusωroot of unitykintegertimesz)=equalszninteger-minus1one \( \frac{1}{n}\sum_{j=1}^{n}{}{ω}_{j}\prod_{k\neq j}{}\mathopen{}\left(1-{ω}_{k}z\right)\mathclose{}= {z}^{n-1} \).

Proof.

  1. The two sides have the same degree and the same roots (both vanish at ωroot of unity1one\( {ω}_{1} \), ωroot of unity2two\( {ω}_{2} \), …, ωroot of unityninteger\( {ω}_{n} \)) and have the same value at z=equals0zero \( z= 0 \).
  2. Take the derivative with respect to z\( z \) in part 1.

Here is a preview of the argument that ν\( ν \) is the same as ρ\( \mathop{\rho} \): If ν(Tlinear map)=equals0zero \( ν\mathopen{}\left( T\right)\mathclose{}= 0 \), then ρ(Tlinear map)=equals0zero \( \mathop{\rho}\mathopen{}\left( T\right)\mathclose{}= 0 \) because ρ(Tlinear map)less than or equal toν(Tlinear map) \( \mathop{\rho}\mathopen{}\left( T\right)\mathclose{}\leq ν\mathopen{}\left( T\right)\mathclose{} \). Henceforth, assume ν=equalsν(Tlinear map)>greater than0zero \( ν= ν\mathopen{}\left( T\right)\mathclose{}\gt 0 \) for fixed Tlinear mapelement ofbounded linear operators(Xnormed linear space) \( T\in \mathcal{L}\mathopen{}\left( X\right)\mathclose{} \). We have ρless than or equal toν \( \mathop{\rho}\leq ν \) so 1oneνless than or equal to1oneρ \( \frac{1}{ν}\leq \frac{1}{\mathop{\rho}} \). If ρ<less thanν \( \mathop{\rho}\lt ν \) we would have λcomplex number-minusTlinear mapelement of bounded linear operators(Xnormed linear space) 1invertible elements \( λ-T\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \) for every λcomplex number\( λ \) with |modulusλcomplex number|modulusgreater than or equal toν \( \mathopen{}\left\lvert{}λ\right\rvert\mathclose{}\geq ν \) which would make 1one-minusztimesTlinear mapelement of bounded linear operators(Xnormed linear space) 1invertible elements \( 1-zT\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \) for all |modulusz|modulusless than or equal to1oneν \( \mathopen{}\left\lvert{}z\right\rvert\mathclose{}\leq \frac{1}{ν} \). The thing we want to look at is the function ffunction\( f \) defined by ffunction(z)=equalsztimesTlinear maptimes (1one-minusztimesTlinear map) 1inverse \( f\mathopen{}\left( z\right)\mathclose{}= zT{ \mathopen{}\left(1-zT\right)\mathclose{} }^{-1} \) which is continuous (hence uniformly continuous) on the closed disc of radius 1oneν \( \frac{1}{ν} \). The strategy is to express 1onenintegertimessummationjinteger=1oneninteger ffunction(ztimesωjinteger) \( \frac{1}{n}\sum_{j=1}^{n}{} f\mathopen{}\left( z{ω}_{j}\right)\mathclose{} \) in terms of ffunction\( f \) using Lemma II.46.
Theorem II.47 (Gelfand-Beurling)

For Tlinear mapelement ofbounded linear operators(Xnormed linear space) \( T\in \mathcal{L}\mathopen{}\left( X\right)\mathclose{} \), we have ρ(Tlinear map)=equalsmaxmaximum|modulusσ(Tlinear map)|modulus=equalslimlimitninteger Tlinear mapninteger 1oneninteger . \[ \mathop{\rho}\mathopen{}\left( T\right)\mathclose{}= \max{}\mathopen{}\left\lvert{}\mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}\right\rvert\mathclose{}= \lim_{n}{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \text{.} \]

Proof. We already know the limit (call it ν\( ν \)) exists, that ν=equalsinfinfimum Tlinear mapninteger 1oneninteger \( ν= \inf{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \), and that ρ(Tlinear map)less than or equal toninteger \( \mathop{\rho}\mathopen{}\left( T\right)\mathclose{}\leq n \) by Lemma II.45 and Proposition II.44. Suppose, by way of contradiction, that ν>greater thanρ(Tlinear map) \( ν\gt \mathop{\rho}\mathopen{}\left( T\right)\mathclose{} \). This implies that ν>greater than0zero \( ν\gt 0 \). Let Ddiagonal operator=equals{setzelement ofCcomplex numbers|such that |modulusz|modulusless than or equal to1oneν }set \( D= \mathopen{}\left\{\, z\in \mathbb{C}\,\middle\vert\, , \mathopen{}\left\lvert{}z\right\rvert\mathclose{}\leq \frac{1}{ν}, \,\right\}\mathclose{} \). Define ffunction:mapsDdiagonal operatortobounded linear operators(Xnormed linear space) \( f : D \to \mathcal{L}\mathopen{}\left( X\right)\mathclose{} \) by ffunction(z)=equalsztimesTlinear maptimes (1one-minusztimesTlinear map) 1inverse \( f\mathopen{}\left( z\right)\mathclose{}= zT{ \mathopen{}\left(1-zT\right)\mathclose{} }^{-1} \). (Note: 1one-minusztimesTlinear mapelement of bounded linear operators(Xnormed linear space) 1invertible elements \( 1-zT\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \). For z=equals0zero \( z= 0 \), no problem. Otherwise, 0zero<less than|modulusz|modulusless than or equal to1oneν \( 0\lt \mathopen{}\left\lvert{}z\right\rvert\mathclose{}\leq \frac{1}{ν} \) implies z1inversegreater than or equal toν>greater thanρ(Tlinear map) \( {z}^{-1}\geq ν\gt \mathop{\rho}\mathopen{}\left( T\right)\mathclose{} \) implies z1inverse-minusTlinear mapelement of bounded linear operators(Xnormed linear space) 1invertible elements \( {z}^{-1}-T\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \) implies 1one-minusztimesTlinear mapelement of bounded linear operators(Xnormed linear space) 1invertible elements \( 1-zT\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \).) We remark that ffunction\( f \) is uniformly continuous on Ddiagonal operator\( D \) because the inversion map of bounded linear operators(Xnormed linear space) 1invertible elements \( { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \) to itself is continuous and Ddiagonal operator\( D \) is compact. Given nintegerelement ofZintegers+positive elements \( n\in {\mathbb{Z}}^{+} \) and ninteger\( n \)th roots ω1one\( {ω}_{1} \), ω2two\( {ω}_{2} \), …, ωninteger\( {ω}_{n} \) of unity, we have 1onenintegertimessummationjinteger=1oneninteger ωjinteger timesproductkintegerjinteger(1one-minusωkintegertimesz)=equalszninteger-minus1one \( \frac{1}{n}\sum_{j=1}^{n}{} {ω}_{j} \prod_{k\neq j}{}\mathopen{}\left(1-{ω}_{k}z\right)\mathclose{}= {z}^{n-1} \). (This is a polynomial identity.) Replace z\( z \) by ztimesTlinear map\( zT \) to get 1onenintegertimessummationjinteger=1oneninteger ωjinteger timesproductkintegerjinteger(1one-minusωkintegertimesztimesTlinear map)=equalszninteger-minus1onetimesTlinear mapninteger-minus1one \( \frac{1}{n}\sum_{j=1}^{n}{} {ω}_{j} \prod_{k\neq j}{}\mathopen{}\left(1-{ω}_{k}zT\right)\mathclose{}= {z}^{n-1}{T}^{n-1} \). Multiply by ztimesTlinear map\( zT \) to get 1onenintegertimessummationjinteger=1oneninteger ωjinteger z Tlinear map timesproductkintegerjinteger(1one-minusωkintegertimesztimesTlinear map)=equalsznintegertimesTlinear mapninteger \( \frac{1}{n}\sum_{j=1}^{n}{} {ω}_{j} z T \prod_{k\neq j}{}\mathopen{}\left(1-{ω}_{k}zT\right)\mathclose{}= {z}^{n}{T}^{n} \). By Lemma II.46, 1one-minusznintegertimesTlinear mapninteger=equalsproductkinteger=1oneninteger(1one-minusωkintegertimesztimesTlinear map) \( 1-{z}^{n}{T}^{n}= \prod_{k=1}^{n}{}\mathopen{}\left(1-{ω}_{k}zT\right)\mathclose{} \). All the factors commute, so for zelement ofDdiagonal operator \( z\in D \), we can multiply both sides by (1one-minusznintegertimesTlinear mapninteger) 1inverse \( { \mathopen{}\left(1-{z}^{n}{T}^{n}\right)\mathclose{} }^{-1} \) to obtain (II.B) 1onenintegertimessummationjinteger=1oneninteger ffunction(ztimesωjinteger) =equalsznintegertimesTlinear mapnintegertimes(1one-minusznintegertimesTlinear mapninteger)1inverse . \[ \frac{1}{n}\sum_{j=1}^{n}{} f\mathopen{}\left( z{ω}_{j}\right)\mathclose{} = {z}^{n}{T}^{n}{\mathopen{}\left(1-{z}^{n}{T}^{n}\right)\mathclose{}}^{-1} \text{.} \]

Given εpositive real number>greater than0zero \( ε\gt 0 \) there exists δpositive real number>greater than0zero \( δ\gt 0 \) such that ffunction(z1one)-minusffunction(z2two)<less thanεpositive real number \( \mathopen{}\left\lVert{}f\mathopen{}\left( {z}_{1}\right)\mathclose{}-f\mathopen{}\left( {z}_{2}\right)\mathclose{}\right\rVert\mathclose{}\lt ε \) whenever |modulusz1one-minusz2two|modulus<less thanδpositive real number \( \mathopen{}\left\lvert{}{z}_{1}-{z}_{2}\right\rvert\mathclose{}\lt δ \). Take μpositive real number\( μ \) with μpositive real number>greater thanν \( μ\gt ν \) and 1oneμpositive real number-minus1oneν<less thanδpositive real number \( \frac{1}{μ}-\frac{1}{ν}\lt δ \). We have |modulusν1inversetimesωjinteger-minusμpositive real number1inversetimesωjinteger|modulus<less thanδpositive real number \( \mathopen{}\left\lvert{}{ν}^{-1}{ω}_{j}-{μ}^{-1}{ω}_{j}\right\rvert\mathclose{}\lt δ \), and hence ffunction(ν1inversetimesωjinteger)-minusffunction(μpositive real number1inversetimesωjinteger)<less thanεpositive real number \( \mathopen{}\left\lVert{}f\mathopen{}\left( {ν}^{-1}{ω}_{j}\right)\mathclose{}-f\mathopen{}\left( {μ}^{-1}{ω}_{j}\right)\mathclose{}\right\rVert\mathclose{}\lt ε \), for 1oneless than or equal tojintegerless than or equal toninteger \( 1\leq j\leq n \). It follows from Equation II.B that (II.C) normνnintegertimesTlinear mapnintegertimes(1one-minusνnintegertimesTlinear mapninteger)1inverse-minusμpositive real numbernintegertimesTlinear mapnintegertimes(1one-minusμpositive real numbernintegertimesTlinear mapninteger)1inversenorm<less thanεpositive real number . \[ \mathopen{}\left\lVert{}{ν}^{{-}n}{T}^{n}{\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1}-{μ}^{{-}n}{T}^{n}{\mathopen{}\left(1-{μ}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1}\right\rVert\mathclose{}\lt ε \text{.} \]

Equation II.C holds for every ninteger\( n \). Because μpositive real number>greater thanlimlimitninteger Tlinear mapninteger 1oneninteger \( μ\gt \lim_{n}{} {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \) we have Tlinear mapninteger 1oneninteger less than or equal torreal number<less thanμpositive real number \( {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}}\leq r\lt μ \) for sufficiently large ninteger\( n \). Thus Tlinear mapnintegerless than or equal torreal numberninteger \( \mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}\leq {r}^{n} \), μpositive real numbernintegertimesTlinear mapnintegerless than or equal to(rreal numberμpositive real number)nintegerconverges to0zero \( \mathopen{}\left\lVert{}{μ}^{{-}n}{T}^{n}\right\rVert\mathclose{}\leq {\mathopen{}\left(\frac{r}{μ}\right)\mathclose{}}^{n} \to 0 \). So μpositive real numbernintegertimesTlinear mapnintegerconverges to0zero \( {μ}^{{-}n}{T}^{n} \to 0 \) and μpositive real numbernintegertimesTlinear mapnintegertimes(1one-minusμpositive real numbernintegertimesTlinear mapninteger)1inverseconverges to0zero \( {μ}^{{-}n}{T}^{n}{\mathopen{}\left(1-{μ}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1} \to 0 \). By Equation II.C, νnintegertimesTlinear mapnintegertimes(1one-minusνnintegertimesTlinear mapninteger)1inverse<less thanεpositive real number \( \mathopen{}\left\lVert{}{ν}^{{-}n}{T}^{n}{\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1}\right\rVert\mathclose{}\lt ε \) for sufficiently large ninteger\( n \), so νnintegertimesTlinear mapnintegertimes(1one-minusνnintegertimesTlinear mapninteger)1inverseconverges to0zero \( {ν}^{{-}n}{T}^{n}{\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1} \to 0 \). Notice 1one=equals(1one-minusνnintegertimesTlinear mapninteger)times(1one-minusνnintegertimesTlinear mapninteger)1inverse=equals(1one-minusνnintegertimesTlinear mapninteger)1inverse-minus(νnintegertimesTlinear mapninteger)times(1one-minusνnintegertimesTlinear mapninteger)1inverse \( 1= \mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}{\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1}= {\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1}-\mathopen{}\left({ν}^{{-}n}{T}^{n}\right)\mathclose{}{\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1} \). But (νnintegertimesTlinear mapninteger)times(1one-minusνnintegertimesTlinear mapninteger)1inverseconverges to0zero \( \mathopen{}\left({ν}^{{-}n}{T}^{n}\right)\mathclose{}{\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1} \to 0 \) so (1one-minusνnintegertimesTlinear mapninteger)1inverseconverges to1one \( {\mathopen{}\left(1-{ν}^{{-}n}{T}^{n}\right)\mathclose{}}^{-1} \to 1 \) implies νnintegertimesTlinear mapnintegerconverges to0zero \( {ν}^{{-}n}{T}^{n} \to 0 \). However, νless than or equal to Tlinear mapninteger 1oneninteger \( ν\leq {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \) for all ninteger\( n \), i.e. 1oneless than or equal to νnintegertimesTlinear mapninteger 1oneninteger \( 1\leq {\mathopen{}\left\lVert{}{ν}^{{-}n}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}} \) for all ninteger\( n \), 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 σ(Tlinear map)not equal toempty set \( \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}\neq \emptyset \). Here's how. Begin by defining ν\( ν \) as an infimum and then proving Lemma II.46, which identifies ν\( ν \) as the corresponding limit. No knowledge of σ(Tlinear map) \( \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{} \) is needed for this. It is easy to see that if |modulusλcomplex number|modulus>greater thanν \( \mathopen{}\left\lvert{}λ\right\rvert\mathclose{}\gt ν \), then λcomplex number-minusTlinear map \( λ-T \) is invertible. (Indeed, let Aself-adjoint operator=equals1oneλcomplex number \( A= \frac{1}{λ} \). Then Aself-adjoint operatorninteger<less than1one \( \mathopen{}\left\lVert{}{A}^{n}\right\rVert\mathclose{}\lt 1 \) which makes 1one-minusAself-adjoint operatorninteger \( 1-{A}^{n} \) invertible. Since 1one-minusAself-adjoint operatorninteger=equals(1one-minusAself-adjoint operator)times(1one+plusAself-adjoint operator+plus+plusAself-adjoint operatorninteger-minus1one) \( 1-{A}^{n}= \mathopen{}\left(1-A\right)\mathclose{}\mathopen{}\left(1+A+\dotsb+{A}^{n-1}\right)\mathclose{} \), we conclude that 1one-minusAself-adjoint operator \( 1-A \) is invertible.) Observe as well that if Tlinear map\( T \) is invertible, then for all ninteger\( n \) 1one=equalsI=equals Tlinear mapnintegertimesTlinear mapninteger 1oneninteger less than or equal to Tlinear mapninteger 1oneninteger times Tlinear mapninteger 1oneninteger , \[ 1= \mathopen{}\left\lVert{}I\right\rVert\mathclose{}= {\mathopen{}\left\lVert{}{T}^{n}{T}^{{-}n}\right\rVert\mathclose{}}^{\frac{1}{n}}\leq {\mathopen{}\left\lVert{}{T}^{n}\right\rVert\mathclose{}}^{\frac{1}{n}}{\mathopen{}\left\lVert{}{T}^{{-}n}\right\rVert\mathclose{}}^{\frac{1}{n}} \text{,} \] so ν(Tlinear map)timesν(Tlinear map1inverse)greater than or equal to1one \( ν\mathopen{}\left( T\right)\mathclose{}ν\mathopen{}\left( {T}^{-1}\right)\mathclose{}\geq 1 \). It follows that if ν(Tlinear map)=equals0zero \( ν\mathopen{}\left( T\right)\mathclose{}= 0 \), then Tlinear map\( T \) is not invertible, i.e., 0zeroelement ofσ(Tlinear map) \( 0\in \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{} \). Stripped of all mention of ρ(Tlinear map) \( \mathop{\rho}\mathopen{}\left( T\right)\mathclose{} \) or σ(Tlinear map) \( \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{} \), the calculations in the proof of 2.20 now show that if ν>greater than0zero \( ν\gt 0 \), then there must be a z\( z \) of modulus 1oneν \( \frac{1}{ν} \) such that 1one-minusztimesTlinear map \( 1-zT \) is noninvertible, so 1onez \( \frac{1}{z} \) belongs to σ(Tlinear map) \( \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{} \).

Back to our Hilbert space HHilbert space\( H \).

Proposition II.49

Tlinear map2two=equalsTlinear map*timesTlinear map \( {\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}^{2}= \mathopen{}\left\lVert{} T^{*}T\right\rVert\mathclose{} \) for all Tlinear mapelement ofbounded linear operators(HHilbert space) \( T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \).

Proof. Tlinear map*timesTlinear mapless than or equal toTlinear map*timesTlinear map=equalsTlinear map2two \( \mathopen{}\left\lVert{} T^{*}T\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{} T^{*}\right\rVert\mathclose{}\mathopen{}\left\lVert{}T\right\rVert\mathclose{}= \mathopen{}\left\lVert{}{T}^{2}\right\rVert\mathclose{} \). Moreover, Tlinear map2two=equalssupsupremumxvector=equals1oneTlinear map(xvector)2two=equalssupsupremumxvector=equals1oneTlinear map(xvector), Tlinear map(xvector)=equalssupsupremumxvector=equals1oneTlinear map*(Tlinear map(xvector)), xvectorless than or equal tosupsupremumxvector=equals1oneTlinear map*(Tlinear map(xvector))timesxvector=equalsTlinear map*(Tlinear map) . \[ \mathopen{}\left\lVert{}{T}^{2}\right\rVert\mathclose{}= \sup_{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1}{}{\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}= \sup_{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1}{}\mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}, T\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}= \sup_{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1}{}\mathopen{}\left\langle{} T^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, x\right\rangle\mathclose{}\leq \sup_{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1}{}\mathopen{}\left\lVert{} T^{*}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{}= \mathopen{}\left\lVert{} T^{*}\mathopen{}\left( T\right)\mathclose{}\right\rVert\mathclose{} \text{.} \]

Definition II.50

An operator Tlinear mapelement ofbounded linear operators(HHilbert space) \( T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \) is called normal if Tlinear map*timesTlinear map=equalsTlinear maptimesTlinear map* \( T^{*}T= T T^{*} \).

Example II.51

The multiplication operator by a bounded measurable function on L2Lebesgue space(Ωmeasure space) \( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( Ω\right)\mathclose{} \) is a basic model.

Proposition II.52

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

Proof. Notice that Tlinear map2two2two=equals(Tlinear map2two)*timesTlinear map2two=equals(Tlinear map*timesTlinear map*)times(Tlinear maptimesTlinear map)=equals(Tlinear map*timesTlinear map)times(Tlinear map*timesTlinear map)=equals Tlinear map*timesTlinear map 2two =equals (Tlinear map2two) 2two \[ {\mathopen{}\left\lVert{}{T}^{2}\right\rVert\mathclose{}}^{2}= \mathopen{}\left\lVert{} \mathopen{}\left({T}^{2}\right)\mathclose{}^{*}{T}^{2}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\mathopen{}\left( T^{*} T^{*}\right)\mathclose{}\mathopen{}\left(TT\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\mathopen{}\left( T^{*}T\right)\mathclose{}\mathopen{}\left( T^{*}T\right)\mathclose{}\right\rVert\mathclose{}= {\mathopen{}\left\lVert{} T^{*}T\right\rVert\mathclose{}}^{2}= {\mathopen{}\left({\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}^{2}\right)\mathclose{}}^{2} \] so Tlinear map2two=equalsTlinear map2two \( \mathopen{}\left\lVert{}{T}^{2}\right\rVert\mathclose{}= {\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}^{2} \) for all normal operators. Tlinear map\( T \) normal implies Tlinear map2two \( {T}^{2} \) normal implies Tlinear map4four=equalsTlinear map2two2two=equalsTlinear map4four \( \mathopen{}\left\lVert{}{T}^{4}\right\rVert\mathclose{}= {\mathopen{}\left\lVert{}{T}^{2}\right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}^{4} \) etc., for all ninteger\( n \). Tlinear map2twotimesninteger 1one2twotimesninteger =equalsTlinear map \( {\mathopen{}\left\lVert{}{T}^{2n}\right\rVert\mathclose{}}^{\frac{1}{2n}}= \mathopen{}\left\lVert{}T\right\rVert\mathclose{} \). Let nintegerconverges toinfinity \( n \to \infty \), to get ρ(Tlinear map)=equalsTlinear map \( \mathop{\rho}\mathopen{}\left( T\right)\mathclose{}= \mathopen{}\left\lVert{}T\right\rVert\mathclose{} \).


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