Lecture Notes in Functional Analysis
by William L. Paschke
edition 0.9
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Proposition II.69
Let V vector space \( V \) and W vector space \( W \) be vector spaces and
φ function 1 one ∈ element of V vector space *
\(
{φ}_{1}\in V^{*}
\)
and
φ function 2 two ∈ element of W vector space *
\(
{φ}_{2}\in W^{*}
\)
linear functionals. Then there is a unique linear functional
φ function 1 one ⊗ φ function 2 two
\(
{φ}_{1}\otimes {φ}_{2}
\)
on
V vector space ⊗ W vector space
\(
V\otimes W
\)
such that
( φ function 1 one ⊗ φ function 2 two ) ( x vector ⊗ y vector ) = equals φ function 1 one ( x vector ) times φ function 2 two ( y vector )
.
\[
\mathopen{}\left({φ}_{1}\otimes {φ}_{2}\right)\mathclose{}\mathopen{}\left( x\otimes y\right)\mathclose{}= {φ}_{1}\mathopen{}\left( x\right)\mathclose{}{φ}_{2}\mathopen{}\left( y\right)\mathclose{}
\text{.}
\] for all x vector ∈ element of V vector space \( x\in V \) and y vector ∈ element of W vector space \( y\in W \) .
The function defined by
( x vector , y vector ) ↦ is mapped to φ function 1 one ( x vector ) times φ function 2 two ( y vector )
\(
\mathopen{}\left(x, y\right)\mathclose{}\mapsto {φ}_{1}\mathopen{}\left( x\right)\mathclose{}{φ}_{2}\mathopen{}\left( y\right)\mathclose{}
\)
is bilinear.
Proposition II.70
If
∑ summation j integer = 1 one n integer
x vector j integer ⊗ y vector j integer
= equals 0 zero
\(
\sum_{j=1}^{n}{}
{x}_{j}\otimes {y}_{j}
= 0
\)
where
x vector j integer ∈ element of V vector space
\(
{x}_{j}\in V
\)
and the
y vector j integer ∈ element of W vector space
\(
{y}_{j}\in W
\)
are linearly independent, then
x vector 1 one = equals ⋯ = equals x vector n integer = equals 0 zero
\(
{x}_{1}= \dotsb= {x}_{n}= 0
\) .
Since the y vector j integer \( {y}_{j} \) are linearly independent, there exist functionals
φ linear functional i integer ∈ element of W vector space *
\(
{φ}_{i}\in W^{*}
\)
such that
φ linear functional i integer ( y vector j integer ) = equals δ i integer j integer Kronecker delta function
\(
{φ}_{i}\mathopen{}\left( {y}_{j}\right)\mathclose{}= \delta_{ij}
\)
where
δ i integer j integer Kronecker delta function
\(
\delta_{ij}
\)
is the Kronecker δ positive real number \( δ \) . Let
τ linear functional ∈ element of V vector space *
\(
τ\in V^{*}
\) .
Then by Proposition II.69 we have
0 zero = equals ( τ linear functional ⊗ φ linear functional j integer ) ( ∑ summation i integer = 1 one n integer x vector i integer ⊗ y vector i integer ) = equals ∑ summation i integer = 1 one n integer τ linear functional ( x vector i integer ) times φ linear functional j integer ( y vector i integer ) = equals ∑ summation i integer = 1 one n integer τ linear functional ( x vector i integer ) times δ i integer j integer Kronecker delta function = equals τ linear functional ( x vector j integer )
.
\[
0= \mathopen{}\left(τ\otimes {φ}_{j}\right)\mathclose{}\mathopen{}\left( \sum_{i=1}^{n}{}{x}_{i}\otimes {y}_{i}\right)\mathclose{}= \sum_{i=1}^{n}{}τ\mathopen{}\left( {x}_{i}\right)\mathclose{}{φ}_{j}\mathopen{}\left( {y}_{i}\right)\mathclose{}= \sum_{i=1}^{n}{}τ\mathopen{}\left( {x}_{i}\right)\mathclose{}\delta_{ij}= τ\mathopen{}\left( {x}_{j}\right)\mathclose{}
\text{.}
\]
Thus
τ linear functional ( x vector j integer ) = equals 0 zero
\(
τ\mathopen{}\left( {x}_{j}\right)\mathclose{}= 0
\)
for arbitrary τ linear functional \( τ \) , and
x vector j integer = equals 0 zero
\(
{x}_{j}= 0
\) .
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 Hilbert space ⊗ K Hilbert Space
\(
H\otimes K
\) .
Theorem II.71
If
T bounded linear operator ∈ element of ℒ bounded linear operators ( H Hilbert space )
\(
T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}
\)
and
S bounded linear operator ∈ element of ℒ bounded linear operators ( K Hilbert Space )
\(
S\in \mathcal{L}\mathopen{}\left( K\right)\mathclose{}
\)
are bounded, then the
operator
( T bounded linear operator ⊗ S bounded linear operator ) ( ( x vector ⊗ y vector ) ) = equals T bounded linear operator ( x vector ) ⊗ S bounded linear operator ( y vector )
\[
\mathopen{}\left(T\otimes S\right)\mathclose{}\mathopen{}\left( \mathopen{}\left(x\otimes y\right)\mathclose{}\right)\mathclose{}= T\mathopen{}\left( x\right)\mathclose{}\otimes S\mathopen{}\left( y\right)\mathclose{}
\]
is in
ℒ bounded linear operators ( H Hilbert space ⊗ K Hilbert Space )
\(
\mathcal{L}\mathopen{}\left( H\otimes K\right)\mathclose{}
\)
and is bounded.
If I identity operator \( \mathrm{I} \) is the identity operator, then it is easy to show
T bounded linear operator ⊗ I identity operator
\(
T\otimes \mathrm{I}
\)
and
I identity operator ⊗ S bounded linear operator
\(
\mathrm{I}\otimes S
\)
are bounded. Then
T bounded linear operator ⊗ S bounded linear operator = equals ( T bounded linear operator ⊗ I identity operator ) times ( I identity operator ⊗ S bounded 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
T bounded linear operator ⊗ S bounded linear operator ∈ element of ℒ bounded linear operators ( H Hilbert space ⊗ K Hilbert Space )
\(
T\otimes S\in \mathcal{L}\mathopen{}\left( H\otimes K\right)\mathclose{}
\) .