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

H. Operators on Tensor Products

Proposition II.69

Let Vvector space\( V \) and Wvector space\( W \) be vector spaces and φfunction1oneelement ofVvector space* \( {φ}_{1}\in V^{*} \) and φfunction2twoelement ofWvector space* \( {φ}_{2}\in W^{*} \) linear functionals. Then there is a unique linear functional φfunction1oneφfunction2two \( {φ}_{1}\otimes {φ}_{2} \) on Vvector spaceWvector space \( V\otimes W \) such that (φfunction1oneφfunction2two)(xvectoryvector)=equalsφfunction1one(xvector)timesφfunction2two(yvector) . \[ \mathopen{}\left({φ}_{1}\otimes {φ}_{2}\right)\mathclose{}\mathopen{}\left( x\otimes y\right)\mathclose{}= {φ}_{1}\mathopen{}\left( x\right)\mathclose{}{φ}_{2}\mathopen{}\left( y\right)\mathclose{} \text{.} \] for all xvectorelement ofVvector space\( x\in V \) and yvectorelement ofWvector space\( y\in W \).

Proof. The function defined by (xvector, yvector)is mapped toφfunction1one(xvector)timesφfunction2two(yvector) \( \mathopen{}\left(x, y\right)\mathclose{}\mapsto {φ}_{1}\mathopen{}\left( x\right)\mathclose{}{φ}_{2}\mathopen{}\left( y\right)\mathclose{} \) is bilinear.

Proposition II.70

If summationjinteger=1oneninteger xvectorjintegeryvectorjinteger =equals0zero \( \sum_{j=1}^{n}{} {x}_{j}\otimes {y}_{j} = 0 \) where xvectorjintegerelement ofVvector space \( {x}_{j}\in V \) and the yvectorjintegerelement ofWvector space \( {y}_{j}\in W \) are linearly independent, then xvector1one=equals=equalsxvectorninteger=equals0zero \( {x}_{1}= \dotsb= {x}_{n}= 0 \).

Proof. Since the yvectorjinteger\( {y}_{j} \) are linearly independent, there exist functionals φlinear functionaliintegerelement ofWvector space* \( {φ}_{i}\in W^{*} \) such that φlinear functionaliinteger(yvectorjinteger)=equalsδiintegerjintegerKronecker delta function \( {φ}_{i}\mathopen{}\left( {y}_{j}\right)\mathclose{}= \delta_{ij} \) where δiintegerjintegerKronecker delta function \( \delta_{ij} \) is the Kronecker δpositive real number\( δ \). Let τlinear functionalelement ofVvector space* \( τ\in V^{*} \). Then by Proposition II.69 we have 0zero=equals(τlinear functionalφlinear functionaljinteger)(summationiinteger=1onenintegerxvectoriintegeryvectoriinteger)=equalssummationiinteger=1onenintegerτlinear functional(xvectoriinteger)timesφlinear functionaljinteger(yvectoriinteger)=equalssummationiinteger=1onenintegerτlinear functional(xvectoriinteger)timesδiintegerjintegerKronecker delta function=equalsτlinear functional(xvectorjinteger) . \[ 0= \mathopen{}\left(τ\otimes {φ}_{j}\right)\mathclose{}\mathopen{}\left( \sum_{i=1}^{n}{}{x}_{i}\otimes {y}_{i}\right)\mathclose{}= \sum_{i=1}^{n}{}τ\mathopen{}\left( {x}_{i}\right)\mathclose{}{φ}_{j}\mathopen{}\left( {y}_{i}\right)\mathclose{}= \sum_{i=1}^{n}{}τ\mathopen{}\left( {x}_{i}\right)\mathclose{}\delta_{ij}= τ\mathopen{}\left( {x}_{j}\right)\mathclose{} \text{.} \] Thus τlinear functional(xvectorjinteger)=equals0zero \( τ\mathopen{}\left( {x}_{j}\right)\mathclose{}= 0 \) for arbitrary τlinear functional\( τ \), and xvectorjinteger=equals0zero \( {x}_{j}= 0 \).

If HHilbert space1one\( {H}_{1} \) and HHilbert space2two\( {H}_{2} \) are Hilbert spaces and γmapping\( γ \) is a bounded bilinear map from HHilbert space1one×Cartesian productHHilbert space2two \( {H}_{1}\times {H}_{2} \) to a Hilbert space KHilbert Space\( K \), then the norm γmapping((tuplexvector, yvector)tuple)less than or equal tocreal numbertimesxvectortimesyvector \( \mathopen{}\left\lVert{}γ\mathopen{}\left( \mathopen{}\left(x, y\right)\mathclose{}\right)\mathclose{}\right\rVert\mathclose{}\leq c\mathopen{}\left\lVert{}x\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{} \) where creal number\( c \) is a constant. We know from our definition of the tensor product that γmapping\( γ \) may be associated with a bounded linear operator Tlinear map:mapsHHilbert space1oneHHilbert space2twotoKHilbert Space \( T : {H}_{1}\otimes {H}_{2} \to K \). It so happens that Tlinear map=equalsγmapping2two \( \mathopen{}\left\lVert{}T\right\rVert\mathclose{}= \mathopen{}\left\lVert{}γ\right\rVert\mathclose{}_{2} \).

Indeed, now that we have a norm structure, we might rephrase the original definition in terms of bounded linear mappings (see Kadison [6]). Here, we continue on to investigate bounded operators on HHilbert spaceKHilbert Space \( H\otimes K \).

Theorem II.71

If Tbounded linear operatorelement ofbounded linear operators(HHilbert space) \( T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \) and Sbounded linear operatorelement ofbounded linear operators(KHilbert Space) \( S\in \mathcal{L}\mathopen{}\left( K\right)\mathclose{} \) are bounded, then the operator (Tbounded linear operatorSbounded linear operator)((xvectoryvector))=equalsTbounded linear operator(xvector)Sbounded linear operator(yvector) \[ \mathopen{}\left(T\otimes S\right)\mathclose{}\mathopen{}\left( \mathopen{}\left(x\otimes y\right)\mathclose{}\right)\mathclose{}= T\mathopen{}\left( x\right)\mathclose{}\otimes S\mathopen{}\left( y\right)\mathclose{} \] is in bounded linear operators(HHilbert spaceKHilbert Space) \( \mathcal{L}\mathopen{}\left( H\otimes K\right)\mathclose{} \) and is bounded.

Proof. If Iidentity operator\( \mathrm{I} \) is the identity operator, then it is easy to show Tbounded linear operatorIidentity operator \( T\otimes \mathrm{I} \) and Iidentity operatorSbounded linear operator \( \mathrm{I}\otimes S \) are bounded. Then Tbounded linear operatorSbounded linear operator=equals(Tbounded linear operatorIidentity operator)times(Iidentity operatorSbounded linear operator) \( T\otimes S= \mathopen{}\left(T\otimes \mathrm{I}\right)\mathclose{}\mathopen{}\left(\mathrm{I}\otimes S\right)\mathclose{} \) is bounded. The characterization of the tensor product shows Tbounded linear operatorSbounded linear operatorelement ofbounded linear operators(HHilbert spaceKHilbert Space) \( T\otimes S\in \mathcal{L}\mathopen{}\left( H\otimes K\right)\mathclose{} \).

Exercise II.72

For Tbounded linear operator1one \( {T}_{1} \) and Tbounded linear operator2two \( {T}_{2} \) in bounded linear operators(HHilbert space) \( \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \) and Sbounded linear operator1one \( {S}_{1} \) and Sbounded linear operator2two \( {S}_{2} \) in bounded linear operators(KHilbert Space) \( \mathcal{L}\mathopen{}\left( K\right)\mathclose{} \), bounded, and λcomplex numberelement ofCcomplex numbers \( λ\in \mathbb{C} \), the following hold:

  1. (λcomplex numbertimesTbounded linear operator1one+plusTbounded linear operator2two)Sbounded linear operator1one=equalsλcomplex numbertimes(Tbounded linear operator1oneSbounded linear operator1one)+plusTbounded linear operator2twoSbounded linear operator1one \( \mathopen{}\left(λ{T}_{1}+{T}_{2}\right)\mathclose{}\otimes {S}_{1}= λ\mathopen{}\left({T}_{1}\otimes {S}_{1}\right)\mathclose{}+{T}_{2}\otimes {S}_{1} \),
  2. Tbounded linear operator1one(Sbounded linear operator1one+plusSbounded linear operator2two)=equalsTbounded linear operator1oneSbounded linear operator1one+plusTbounded linear operator1oneSbounded linear operator2two \( {T}_{1}\otimes \mathopen{}\left({S}_{1}+{S}_{2}\right)\mathclose{}= {T}_{1}\otimes {S}_{1}+{T}_{1}\otimes {S}_{2} \),
  3. (Tbounded linear operator1oneSbounded linear operator1one)times(Tbounded linear operator2twoSbounded linear operator2two)=equalsTbounded linear operator1onetimesTbounded linear operator2twoSbounded linear operator1onetimesSbounded linear operator2two \( \mathopen{}\left({T}_{1}\otimes {S}_{1}\right)\mathclose{}\mathopen{}\left({T}_{2}\otimes {S}_{2}\right)\mathclose{}= {T}_{1}{T}_{2}\otimes {S}_{1}{S}_{2} \),
  4. Tbounded linear operator1oneSbounded linear operator1one=equalsTbounded linear operator1onetimesSbounded linear operator1one \( \mathopen{}\left\lVert{}{T}_{1}\otimes {S}_{1}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}{T}_{1}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{S}_{1}\right\rVert\mathclose{} \),
  5. (Tbounded linear operator1oneTbounded linear operator2two)*=equals(Tbounded linear operator1one)*(Tbounded linear operator2two)* \( \mathopen{}\left({T}_{1}\otimes {T}_{2}\right)\mathclose{}^{*}= \mathopen{}\left({T}_{1}\right)\mathclose{}^{*}\otimes \mathopen{}\left({T}_{2}\right)\mathclose{}^{*} \).


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