The extension of the tensor product of two operators

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Let $E$, $F$ be two complex Hilbert spaces and $mathcalL(E)$ (resp. $mathcalL(F)$) be the algebra of all bounded linear operators on $E$ (resp. $F$).

The algebraic tensor product of $E$ and $F$ is given by
$$E otimes F:=leftxi=sum_i=1^dv_iotimes w_i:;din mathbbN^*,;;v_iin E,;;w_iin F
right.$$

In $E otimes F$, we define
$$
langle xi,etarangle=sum_i=1^nsum_j=1^m langle x_i,z_jrangle_1langle y_i ,t_jrangle_2,
$$
for $xi=displaystylesum_i=1^nx_iotimes y_iin E otimes F$ and $eta=displaystylesum_j=1^mz_jotimes w_jin E otimes F$.

The above sesquilinear form is an inner product in $E otimes F$.

It is well known that $(E otimes F,langlecdot,cdotrangle)$ is not a complete space. Let $E widehatotimes F$ be the completion of $E otimes F$ under the inner product $langlecdot,cdotrangle$.

If $Tin mathcalL(E)$ and $Sin mathcalL(F)$, then the tensor product of $T$ and $S$ is denoted $Totimes S$ and defined as
$$big(Totimes Sbig)bigg(sum_k=1^d x_kotimes y_kbigg)=sum_k=1^dTx_k otimes Sy_k,;;forall,sum_k=1^d x_kotimes y_kin E otimes F,$$
which lies in $mathcalL(E otimes F)$. The extension of $Totimes S$ over the Hilbert space $E widehatotimes F$, denoted by $T widehatotimes S$, is the tensor product of $T$ and $S$ on the tensor product space, which lies in $mathcalL(EwidehatotimesF)$.

Let $operatornameIm X$ and $overlineoperatornameIm X$ denote respectively the range of an operator $X$ and the closure of its range.

Let $T,Min mathcalL(E)$ and $S,Nin mathcalL(F)$. If $operatornameIm (T)subseteq overlineoperatornameIm (M)$ and $operatornameIm SsubseteqoverlineoperatornameIm (N)$. I want to show that
$$operatornameIm(T widehatotimes S)subseteq overlineoperatornameIm(M widehatotimes N).$$

Note that I show that
$$overlineoperatornameIm MotimesoverlineoperatornameIm NsubseteqoverlineoperatornameIm(M otimes N).$$

functional-analysis operator-theory hilbert-spaces

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Let $E$, $F$ be two complex Hilbert spaces and $mathcalL(E)$ (resp. $mathcalL(F)$) be the algebra of all bounded linear operators on $E$ (resp. $F$).

The algebraic tensor product of $E$ and $F$ is given by
$$E otimes F:=leftxi=sum_i=1^dv_iotimes w_i:;din mathbbN^*,;;v_iin E,;;w_iin F
right.$$

In $E otimes F$, we define
$$
langle xi,etarangle=sum_i=1^nsum_j=1^m langle x_i,z_jrangle_1langle y_i ,t_jrangle_2,
$$
for $xi=displaystylesum_i=1^nx_iotimes y_iin E otimes F$ and $eta=displaystylesum_j=1^mz_jotimes w_jin E otimes F$.

The above sesquilinear form is an inner product in $E otimes F$.

It is well known that $(E otimes F,langlecdot,cdotrangle)$ is not a complete space. Let $E widehatotimes F$ be the completion of $E otimes F$ under the inner product $langlecdot,cdotrangle$.

If $Tin mathcalL(E)$ and $Sin mathcalL(F)$, then the tensor product of $T$ and $S$ is denoted $Totimes S$ and defined as
$$big(Totimes Sbig)bigg(sum_k=1^d x_kotimes y_kbigg)=sum_k=1^dTx_k otimes Sy_k,;;forall,sum_k=1^d x_kotimes y_kin E otimes F,$$
which lies in $mathcalL(E otimes F)$. The extension of $Totimes S$ over the Hilbert space $E widehatotimes F$, denoted by $T widehatotimes S$, is the tensor product of $T$ and $S$ on the tensor product space, which lies in $mathcalL(EwidehatotimesF)$.

Let $operatornameIm X$ and $overlineoperatornameIm X$ denote respectively the range of an operator $X$ and the closure of its range.

Let $T,Min mathcalL(E)$ and $S,Nin mathcalL(F)$. If $operatornameIm (T)subseteq overlineoperatornameIm (M)$ and $operatornameIm SsubseteqoverlineoperatornameIm (N)$. I want to show that
$$operatornameIm(T widehatotimes S)subseteq overlineoperatornameIm(M widehatotimes N).$$

Note that I show that
$$overlineoperatornameIm MotimesoverlineoperatornameIm NsubseteqoverlineoperatornameIm(M otimes N).$$

functional-analysis operator-theory hilbert-spaces

share|cite|improve this question

asked yesterday

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up vote
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down vote

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Let $E$, $F$ be two complex Hilbert spaces and $mathcalL(E)$ (resp. $mathcalL(F)$) be the algebra of all bounded linear operators on $E$ (resp. $F$).

The algebraic tensor product of $E$ and $F$ is given by
$$E otimes F:=leftxi=sum_i=1^dv_iotimes w_i:;din mathbbN^*,;;v_iin E,;;w_iin F
right.$$

In $E otimes F$, we define
$$
langle xi,etarangle=sum_i=1^nsum_j=1^m langle x_i,z_jrangle_1langle y_i ,t_jrangle_2,
$$
for $xi=displaystylesum_i=1^nx_iotimes y_iin E otimes F$ and $eta=displaystylesum_j=1^mz_jotimes w_jin E otimes F$.

The above sesquilinear form is an inner product in $E otimes F$.

It is well known that $(E otimes F,langlecdot,cdotrangle)$ is not a complete space. Let $E widehatotimes F$ be the completion of $E otimes F$ under the inner product $langlecdot,cdotrangle$.

If $Tin mathcalL(E)$ and $Sin mathcalL(F)$, then the tensor product of $T$ and $S$ is denoted $Totimes S$ and defined as
$$big(Totimes Sbig)bigg(sum_k=1^d x_kotimes y_kbigg)=sum_k=1^dTx_k otimes Sy_k,;;forall,sum_k=1^d x_kotimes y_kin E otimes F,$$
which lies in $mathcalL(E otimes F)$. The extension of $Totimes S$ over the Hilbert space $E widehatotimes F$, denoted by $T widehatotimes S$, is the tensor product of $T$ and $S$ on the tensor product space, which lies in $mathcalL(EwidehatotimesF)$.

Let $operatornameIm X$ and $overlineoperatornameIm X$ denote respectively the range of an operator $X$ and the closure of its range.

Let $T,Min mathcalL(E)$ and $S,Nin mathcalL(F)$. If $operatornameIm (T)subseteq overlineoperatornameIm (M)$ and $operatornameIm SsubseteqoverlineoperatornameIm (N)$. I want to show that
$$operatornameIm(T widehatotimes S)subseteq overlineoperatornameIm(M widehatotimes N).$$

Note that I show that
$$overlineoperatornameIm MotimesoverlineoperatornameIm NsubseteqoverlineoperatornameIm(M otimes N).$$

functional-analysis operator-theory hilbert-spaces

share|cite|improve this question

asked yesterday

Student

2,4802524

Let $E$, $F$ be two complex Hilbert spaces and $mathcalL(E)$ (resp. $mathcalL(F)$) be the algebra of all bounded linear operators on $E$ (resp. $F$).

The algebraic tensor product of $E$ and $F$ is given by
$$E otimes F:=leftxi=sum_i=1^dv_iotimes w_i:;din mathbbN^*,;;v_iin E,;;w_iin F
right.$$

In $E otimes F$, we define
$$
langle xi,etarangle=sum_i=1^nsum_j=1^m langle x_i,z_jrangle_1langle y_i ,t_jrangle_2,
$$
for $xi=displaystylesum_i=1^nx_iotimes y_iin E otimes F$ and $eta=displaystylesum_j=1^mz_jotimes w_jin E otimes F$.

The above sesquilinear form is an inner product in $E otimes F$.

It is well known that $(E otimes F,langlecdot,cdotrangle)$ is not a complete space. Let $E widehatotimes F$ be the completion of $E otimes F$ under the inner product $langlecdot,cdotrangle$.

If $Tin mathcalL(E)$ and $Sin mathcalL(F)$, then the tensor product of $T$ and $S$ is denoted $Totimes S$ and defined as
$$big(Totimes Sbig)bigg(sum_k=1^d x_kotimes y_kbigg)=sum_k=1^dTx_k otimes Sy_k,;;forall,sum_k=1^d x_kotimes y_kin E otimes F,$$
which lies in $mathcalL(E otimes F)$. The extension of $Totimes S$ over the Hilbert space $E widehatotimes F$, denoted by $T widehatotimes S$, is the tensor product of $T$ and $S$ on the tensor product space, which lies in $mathcalL(EwidehatotimesF)$.

Let $operatornameIm X$ and $overlineoperatornameIm X$ denote respectively the range of an operator $X$ and the closure of its range.

Let $T,Min mathcalL(E)$ and $S,Nin mathcalL(F)$. If $operatornameIm (T)subseteq overlineoperatornameIm (M)$ and $operatornameIm SsubseteqoverlineoperatornameIm (N)$. I want to show that
$$operatornameIm(T widehatotimes S)subseteq overlineoperatornameIm(M widehatotimes N).$$

Note that I show that
$$overlineoperatornameIm MotimesoverlineoperatornameIm NsubseteqoverlineoperatornameIm(M otimes N).$$

functional-analysis operator-theory hilbert-spaces

share|cite|improve this question

asked yesterday

Student

2,4802524

share|cite|improve this question

share|cite|improve this question

asked yesterday

Student

2,4802524

asked yesterday

Student

2,4802524

asked yesterday

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2,4802524

2,4802524

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Firstly, it is immediate that $$operatornameIm(T otimes S) = operatornameIm(T) otimes operatornameIm(S) subseteqoverlineoperatornameIm(M)otimes overlineoperatornameIm(N) subseteq overlineoperatornameIm(Motimes N)$$
and so $overlineoperatornameIm(T otimes S) subseteq overlineoperatornameIm(Motimes N)$

Now, $operatornameIm(T otimes S) = T hat otimes S (E otimes F)$. But $T hat otimes S$ is a continuous map on $E hat otimes F$ and so $$T hat otimes S(E hat otimes F) = T hat otimes S( overlineE otimes F) subseteq overlineT hat otimes S(E otimes F).$$
Therefore $operatornameIm(T hat otimes S) subseteq overlineoperatornameIm(Totimes S) subseteq overlineoperatornameIm(Motimes N)$

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answered 6 hours ago

Rhys Steele

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1 Answer
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1 Answer
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active

oldest

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up vote
1
down vote



accepted

Firstly, it is immediate that $$operatornameIm(T otimes S) = operatornameIm(T) otimes operatornameIm(S) subseteqoverlineoperatornameIm(M)otimes overlineoperatornameIm(N) subseteq overlineoperatornameIm(Motimes N)$$
and so $overlineoperatornameIm(T otimes S) subseteq overlineoperatornameIm(Motimes N)$

Now, $operatornameIm(T otimes S) = T hat otimes S (E otimes F)$. But $T hat otimes S$ is a continuous map on $E hat otimes F$ and so $$T hat otimes S(E hat otimes F) = T hat otimes S( overlineE otimes F) subseteq overlineT hat otimes S(E otimes F).$$
Therefore $operatornameIm(T hat otimes S) subseteq overlineoperatornameIm(Totimes S) subseteq overlineoperatornameIm(Motimes N)$

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answered 6 hours ago

Rhys Steele

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up vote
1
down vote



accepted

Firstly, it is immediate that $$operatornameIm(T otimes S) = operatornameIm(T) otimes operatornameIm(S) subseteqoverlineoperatornameIm(M)otimes overlineoperatornameIm(N) subseteq overlineoperatornameIm(Motimes N)$$
and so $overlineoperatornameIm(T otimes S) subseteq overlineoperatornameIm(Motimes N)$

Now, $operatornameIm(T otimes S) = T hat otimes S (E otimes F)$. But $T hat otimes S$ is a continuous map on $E hat otimes F$ and so $$T hat otimes S(E hat otimes F) = T hat otimes S( overlineE otimes F) subseteq overlineT hat otimes S(E otimes F).$$
Therefore $operatornameIm(T hat otimes S) subseteq overlineoperatornameIm(Totimes S) subseteq overlineoperatornameIm(Motimes N)$

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answered 6 hours ago

Rhys Steele

5,4151727

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Firstly, it is immediate that $$operatornameIm(T otimes S) = operatornameIm(T) otimes operatornameIm(S) subseteqoverlineoperatornameIm(M)otimes overlineoperatornameIm(N) subseteq overlineoperatornameIm(Motimes N)$$
and so $overlineoperatornameIm(T otimes S) subseteq overlineoperatornameIm(Motimes N)$

Now, $operatornameIm(T otimes S) = T hat otimes S (E otimes F)$. But $T hat otimes S$ is a continuous map on $E hat otimes F$ and so $$T hat otimes S(E hat otimes F) = T hat otimes S( overlineE otimes F) subseteq overlineT hat otimes S(E otimes F).$$
Therefore $operatornameIm(T hat otimes S) subseteq overlineoperatornameIm(Totimes S) subseteq overlineoperatornameIm(Motimes N)$

share|cite|improve this answer

answered 6 hours ago

Rhys Steele

5,4151727

Firstly, it is immediate that $$operatornameIm(T otimes S) = operatornameIm(T) otimes operatornameIm(S) subseteqoverlineoperatornameIm(M)otimes overlineoperatornameIm(N) subseteq overlineoperatornameIm(Motimes N)$$
and so $overlineoperatornameIm(T otimes S) subseteq overlineoperatornameIm(Motimes N)$

Now, $operatornameIm(T otimes S) = T hat otimes S (E otimes F)$. But $T hat otimes S$ is a continuous map on $E hat otimes F$ and so $$T hat otimes S(E hat otimes F) = T hat otimes S( overlineE otimes F) subseteq overlineT hat otimes S(E otimes F).$$
Therefore $operatornameIm(T hat otimes S) subseteq overlineoperatornameIm(Totimes S) subseteq overlineoperatornameIm(Motimes N)$

share|cite|improve this answer

answered 6 hours ago

Rhys Steele

5,4151727

share|cite|improve this answer

share|cite|improve this answer

answered 6 hours ago

Rhys Steele

5,4151727

answered 6 hours ago

Rhys Steele

5,4151727

answered 6 hours ago

Rhys Steele

5,4151727

5,4151727

add a comment | 

add a comment | 

 

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