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

## H. Operators on Tensor Products

Proposition II.69

Let $$$V$$$ and $$$W$$$ be vector spaces and $$${φ}_{1}\in V^{*}$$$ and $$${φ}_{2}\in W^{*}$$$ linear functionals. Then there is a unique linear functional $$${φ}_{1}\otimes {φ}_{2}$$$ on $$$V\otimes W$$$ such that $$\mathopen{}\left({φ}_{1}\otimes {φ}_{2}\right)\mathclose{}\mathopen{}\left( x\otimes y\right)\mathclose{}= {φ}_{1}\mathopen{}\left( x\right)\mathclose{}{φ}_{2}\mathopen{}\left( y\right)\mathclose{} \text{.}$$ for all $$$x\in V$$$ and $$$y\in W$$$.

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

Proposition II.70

If $$$\sum_{j=1}^{n}{} {x}_{j}\otimes {y}_{j} = 0$$$ where $$${x}_{j}\in V$$$ and the $$${y}_{j}\in W$$$ are linearly independent, then $$${x}_{1}= \dotsb= {x}_{n}= 0$$$.

Proof. Since the $$${y}_{j}$$$ are linearly independent, there exist functionals $$${φ}_{i}\in W^{*}$$$ such that $$${φ}_{i}\mathopen{}\left( {y}_{j}\right)\mathclose{}= \delta_{ij}$$$ where $$$\delta_{ij}$$$ is the Kronecker $$$δ$$$. Let $$$τ\in V^{*}$$$. Then by Proposition II.69 we have $$0= \mathopen{}\left(τ\otimes {φ}_{j}\right)\mathclose{}\mathopen{}\left( \sum_{i=1}^{n}{}{x}_{i}\otimes {y}_{i}\right)\mathclose{}= \sum_{i=1}^{n}{}τ\mathopen{}\left( {x}_{i}\right)\mathclose{}{φ}_{j}\mathopen{}\left( {y}_{i}\right)\mathclose{}= \sum_{i=1}^{n}{}τ\mathopen{}\left( {x}_{i}\right)\mathclose{}\delta_{ij}= τ\mathopen{}\left( {x}_{j}\right)\mathclose{} \text{.}$$ Thus $$$τ\mathopen{}\left( {x}_{j}\right)\mathclose{}= 0$$$ for arbitrary $$$τ$$$, and $$${x}_{j}= 0$$$.

If $$${H}_{1}$$$ and $$${H}_{2}$$$ are Hilbert spaces and $$$γ$$$ is a bounded bilinear map from $$${H}_{1}\times {H}_{2}$$$ to a Hilbert space $$$K$$$, then the norm $$$\mathopen{}\left\lVert{}γ\mathopen{}\left( \mathopen{}\left(x, y\right)\mathclose{}\right)\mathclose{}\right\rVert\mathclose{}\leq c\mathopen{}\left\lVert{}x\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{}$$$ where $$$c$$$ is a constant. We know from our definition of the tensor product that $$$γ$$$ may be associated with a bounded linear operator $$$T : {H}_{1}\otimes {H}_{2} \to K$$$. It so happens that $$$\mathopen{}\left\lVert{}T\right\rVert\mathclose{}= \mathopen{}\left\lVert{}γ\right\rVert\mathclose{}_{2}$$$.

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 $$$H\otimes K$$$.

Theorem II.71

If $$$T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ and $$$S\in \mathcal{L}\mathopen{}\left( K\right)\mathclose{}$$$ are bounded, then the operator $$\mathopen{}\left(T\otimes S\right)\mathclose{}\mathopen{}\left( \mathopen{}\left(x\otimes y\right)\mathclose{}\right)\mathclose{}= T\mathopen{}\left( x\right)\mathclose{}\otimes S\mathopen{}\left( y\right)\mathclose{}$$ is in $$$\mathcal{L}\mathopen{}\left( H\otimes K\right)\mathclose{}$$$ and is bounded.

Proof. If $$$\mathrm{I}$$$ is the identity operator, then it is easy to show $$$T\otimes \mathrm{I}$$$ and $$$\mathrm{I}\otimes S$$$ are bounded. Then $$$T\otimes S= \mathopen{}\left(T\otimes \mathrm{I}\right)\mathclose{}\mathopen{}\left(\mathrm{I}\otimes S\right)\mathclose{}$$$ is bounded. The characterization of the tensor product shows $$$T\otimes S\in \mathcal{L}\mathopen{}\left( H\otimes K\right)\mathclose{}$$$.

Exercise II.72

For $$${T}_{1}$$$ and $$${T}_{2}$$$ in $$$\mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ and $$${S}_{1}$$$ and $$${S}_{2}$$$ in $$$\mathcal{L}\mathopen{}\left( K\right)\mathclose{}$$$, bounded, and $$$λ\in \mathbb{C}$$$, the following hold:

1. $$$\mathopen{}\left(λ{T}_{1}+{T}_{2}\right)\mathclose{}\otimes {S}_{1}= λ\mathopen{}\left({T}_{1}\otimes {S}_{1}\right)\mathclose{}+{T}_{2}\otimes {S}_{1}$$$,
2. $$${T}_{1}\otimes \mathopen{}\left({S}_{1}+{S}_{2}\right)\mathclose{}= {T}_{1}\otimes {S}_{1}+{T}_{1}\otimes {S}_{2}$$$,
3. $$$\mathopen{}\left({T}_{1}\otimes {S}_{1}\right)\mathclose{}\mathopen{}\left({T}_{2}\otimes {S}_{2}\right)\mathclose{}= {T}_{1}{T}_{2}\otimes {S}_{1}{S}_{2}$$$,
4. $$$\mathopen{}\left\lVert{}{T}_{1}\otimes {S}_{1}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}{T}_{1}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{S}_{1}\right\rVert\mathclose{}$$$,
5. $$$\mathopen{}\left({T}_{1}\otimes {T}_{2}\right)\mathclose{}^{*}= \mathopen{}\left({T}_{1}\right)\mathclose{}^{*}\otimes \mathopen{}\left({T}_{2}\right)\mathclose{}^{*}$$$.