Mixing
Norm of a tensor product of operators
$begingroup$
I have two Hilbert spaces $H_1$ and $H_2$ which are subspaces of a bigger Hilbert space $H$.
I also have two bounded linear functions $T_1:H_1rightarrow H$ and $T_2:H_2rightarrow H$.
I define the tensor product space $F=H_1otimes H_2$, and a linear function on it $T=(T_1otimes T_2)$. The vector space $F$ has the induced inner product from the Hilbert spaces:
$$langlephi_1otimes psi_1, phi_2 otimes psi_2rangle = langlephi_1,phi_2ranglelanglepsi_1,psi_2rangle$$ and therefore an induced norm.
I want to show that $|T|| =|T_1|| cdot|T_2|$, but I'm stuck since I cant show that
$|T|| leq |T_1|| cdot|T_2|$ (the other inequality I've already shown).
Can anyone help?
functional-analysis
edited Apr 2 '13 at 23:15
Zev Chonoles
111k16232430
asked Apr 2 '13 at 23:12
MaxMax
36939
$endgroup$
add a comment |
$begingroup$
I have two Hilbert spaces $H_1$ and $H_2$ which are subspaces of a bigger Hilbert space $H$.
I also have two bounded linear functions $T_1:H_1rightarrow H$ and $T_2:H_2rightarrow H$.
I define the tensor product space $F=H_1otimes H_2$, and a linear function on it $T=(T_1otimes T_2)$. The vector space $F$ has the induced inner product from the Hilbert spaces:
$$langlephi_1otimes psi_1, phi_2 otimes psi_2rangle = langlephi_1,phi_2ranglelanglepsi_1,psi_2rangle$$ and therefore an induced norm.
I want to show that $|T|| =|T_1|| cdot|T_2|$, but I'm stuck since I cant show that
$|T|| leq |T_1|| cdot|T_2|$ (the other inequality I've already shown).
Can anyone help?
functional-analysis
edited Apr 2 '13 at 23:15
Zev Chonoles
111k16232430
asked Apr 2 '13 at 23:12
MaxMax
36939
$endgroup$
1
$begingroup$
LaTeX tips:<and>mean "less than" and "greater than", and produce spacing correct for that meaning only. When you want angle brackets, you need to uselangleandrangle. Also, to get proper norm symbols use|instead of||.
$endgroup$
– Zev Chonoles
Apr 2 '13 at 23:16
$begingroup$
Thanks for the correction!
$endgroup$
– Max
Apr 2 '13 at 23:19
add a comment |
$begingroup$
I have two Hilbert spaces $H_1$ and $H_2$ which are subspaces of a bigger Hilbert space $H$.
I also have two bounded linear functions $T_1:H_1rightarrow H$ and $T_2:H_2rightarrow H$.
I define the tensor product space $F=H_1otimes H_2$, and a linear function on it $T=(T_1otimes T_2)$. The vector space $F$ has the induced inner product from the Hilbert spaces:
$$langlephi_1otimes psi_1, phi_2 otimes psi_2rangle = langlephi_1,phi_2ranglelanglepsi_1,psi_2rangle$$ and therefore an induced norm.
I want to show that $|T|| =|T_1|| cdot|T_2|$, but I'm stuck since I cant show that
$|T|| leq |T_1|| cdot|T_2|$ (the other inequality I've already shown).
Can anyone help?
functional-analysis
edited Apr 2 '13 at 23:15
Zev Chonoles
111k16232430
asked Apr 2 '13 at 23:12
MaxMax
36939
$endgroup$
I have two Hilbert spaces $H_1$ and $H_2$ which are subspaces of a bigger Hilbert space $H$.
I also have two bounded linear functions $T_1:H_1rightarrow H$ and $T_2:H_2rightarrow H$.
I define the tensor product space $F=H_1otimes H_2$, and a linear function on it $T=(T_1otimes T_2)$. The vector space $F$ has the induced inner product from the Hilbert spaces:
$$langlephi_1otimes psi_1, phi_2 otimes psi_2rangle = langlephi_1,phi_2ranglelanglepsi_1,psi_2rangle$$ and therefore an induced norm.
I want to show that $|T|| =|T_1|| cdot|T_2|$, but I'm stuck since I cant show that
$|T|| leq |T_1|| cdot|T_2|$ (the other inequality I've already shown).
Can anyone help?
functional-analysis
functional-analysis
edited Apr 2 '13 at 23:15
Zev Chonoles
111k16232430
asked Apr 2 '13 at 23:12
MaxMax
36939
edited Apr 2 '13 at 23:15
Zev Chonoles
111k16232430
asked Apr 2 '13 at 23:12
MaxMax
36939
edited Apr 2 '13 at 23:15
Zev Chonoles
111k16232430
edited Apr 2 '13 at 23:15
Zev Chonoles
111k16232430
edited Apr 2 '13 at 23:15
Zev Chonoles
111k16232430
111k16232430
asked Apr 2 '13 at 23:12
MaxMax
36939
asked Apr 2 '13 at 23:12
MaxMax
36939
asked Apr 2 '13 at 23:12
MaxMax
36939
36939
1
$begingroup$
LaTeX tips:<and>mean "less than" and "greater than", and produce spacing correct for that meaning only. When you want angle brackets, you need to uselangleandrangle. Also, to get proper norm symbols use|instead of||.
$endgroup$
– Zev Chonoles
Apr 2 '13 at 23:16
$begingroup$
Thanks for the correction!
$endgroup$
– Max
Apr 2 '13 at 23:19
add a comment |
1
$begingroup$
LaTeX tips:<and>mean "less than" and "greater than", and produce spacing correct for that meaning only. When you want angle brackets, you need to uselangleandrangle. Also, to get proper norm symbols use|instead of||.
$endgroup$
– Zev Chonoles
Apr 2 '13 at 23:16
$begingroup$
Thanks for the correction!
$endgroup$
– Max
Apr 2 '13 at 23:19
1
1
$begingroup$
LaTeX tips:
< and > mean "less than" and "greater than", and produce spacing correct for that meaning only. When you want angle brackets, you need to use langle and rangle. Also, to get proper norm symbols use | instead of ||.$endgroup$
– Zev Chonoles
Apr 2 '13 at 23:16
$begingroup$
LaTeX tips:
< and > mean "less than" and "greater than", and produce spacing correct for that meaning only. When you want angle brackets, you need to use langle and rangle. Also, to get proper norm symbols use | instead of ||.$endgroup$
– Zev Chonoles
Apr 2 '13 at 23:16
$begingroup$
Thanks for the correction!
$endgroup$
– Max
Apr 2 '13 at 23:19
$begingroup$
Thanks for the correction!
$endgroup$
– Max
Apr 2 '13 at 23:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$newcommand{norm}[1]{left|{#1}right|}
newcommand{ip}[1]{leftlangle{#1}rightrangle}
newcommand{abs}[1]{left|{#1}right|}$Let $S in B(H_1)$, $T in B(H_2)$; the claim is that $norm{S otimes T} leq norm{S}norm{T}$.
Let me first show that $S otimes I$ is bounded with $norm{S otimes I} leq norm{S}$; the same proof, mutatis mutandis, will show that $I otimes T$ is bounded with $norm{I otimes T} leq norm{T}$. Since the algebraic tensor product $H_1 odot H_2$ is dense in $H_1 otimes H_2$, it suffices to show that $norm{(S otimes I)v} leq norm{S}norm{v}$ for any $v in H_1 odot H_2$.
So, let $v = sum_{k=1}^N x_k otimes y_k in H_1 odot H_2$; by performing Gram--Schmidt orthogonalisation on ${y_k}$ and expressing the $y_k$ in terms of the resulting orthonormal basis for $operatorname{span}{y_k}$, we may assume without loss of generality that ${y_k}$ is orthonormal. On the one hand, it follows that ${x_k otimes y_k}$ is orthogonal, so that
$$
norm{v}^2 = norm{sum_{k=1}^N x_k otimes y_k}^2 = sum_{k=1}^N norm{x_k otimes y_k}^2 = sum_{k=1}^N norm{x_k}^2.
$$
On the other hand, since $(S otimes I)(x_k otimes y_k) = Sx_k otimes y_k$, it follows that ${Sx_k otimes y_k}$ is also orthogonal, so that by the same computation, mutatis mutandis,
$$
norm{(S otimes I)v}^2 = sum_{k=1}^N norm{S x_k}^2 leq sum_{k=1}^N norm{S}^2 norm{x_k}^2 = norm{S}^2 sum_{k=1}^N norm{x_k}^2 = norm{S}^2norm{v}^2.
$$
Thus, $norm{(S otimes I)v} leq norm{S}norm{v}$, as required.
Now, observe that since $(S otimes T) = (S otimes I)(I otimes T)$ on $H_1 odot H_2$, it follows by the boundedness of $S otimes I$ and $I otimes T$ that $S otimes T$ is also bounded with norm
$$
norm{S otimes T} leq norm{S otimes I}norm{I otimes T} leq norm{S}norm{T},
$$
as required.
answered Apr 3 '13 at 1:38
Branimir ĆaćićBranimir Ćaćić
10.2k22047
$endgroup$
$begingroup$
But what about the other inequality? How do I prove that $|T_1 otimes T_2| leq |T_1| |T_2|$
$endgroup$
– Max
Apr 3 '13 at 1:50
$begingroup$
@Max Sorry about that, I clearly misread your question! Now I've answered the question you actually asked.
$endgroup$
– Branimir Ćaćić
Apr 3 '13 at 7:23
add a comment |
$begingroup$
Let me add the other direction. Suppose $varepsilon >0$. Then from the boundedness of operators $T_{1}$ and $T_{2}$ we know that there exist unit vectors in the Hilbert spaces $psi in H_{1}$, $xi in H_{2}$ such that:
$left | T_{1} psi right |> left | T_{1} right | - epsilon$, $left | T_{2} xi right |> left | T_{2} right | - epsilon$
Using this we can use the definition of the tensor product to see that:
$left | left (T_{1}otimes T_{2} right )left ( psiotimes xi right ) right |_{H_{1}bigotimes H_{2}}=left | left ( T_{1}psiotimes T_{2}xi right ) right |_{H_{1}bigotimes H_{2}}=left |T_{1}psi right |_{H_{1}} left | T_{2}xi right |_{H_{2}}geq$
$ geqleft | T_{1} right |left | T_{2} right |-epsilon left | T_{1} right | - epsilon left | T_{2} right | + epsilon^{2}geqleft | T_{1} right |left | T_{2} right |-epsilon left | T_{1} right | - epsilon left | T_{2} right |$
Sending $epsilon$ to 0 shows the desired result.
answered Jan 30 at 12:03
Alex EastAlex East
1
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$newcommand{norm}[1]{left|{#1}right|}
newcommand{ip}[1]{leftlangle{#1}rightrangle}
newcommand{abs}[1]{left|{#1}right|}$Let $S in B(H_1)$, $T in B(H_2)$; the claim is that $norm{S otimes T} leq norm{S}norm{T}$.
Let me first show that $S otimes I$ is bounded with $norm{S otimes I} leq norm{S}$; the same proof, mutatis mutandis, will show that $I otimes T$ is bounded with $norm{I otimes T} leq norm{T}$. Since the algebraic tensor product $H_1 odot H_2$ is dense in $H_1 otimes H_2$, it suffices to show that $norm{(S otimes I)v} leq norm{S}norm{v}$ for any $v in H_1 odot H_2$.
So, let $v = sum_{k=1}^N x_k otimes y_k in H_1 odot H_2$; by performing Gram--Schmidt orthogonalisation on ${y_k}$ and expressing the $y_k$ in terms of the resulting orthonormal basis for $operatorname{span}{y_k}$, we may assume without loss of generality that ${y_k}$ is orthonormal. On the one hand, it follows that ${x_k otimes y_k}$ is orthogonal, so that
$$
norm{v}^2 = norm{sum_{k=1}^N x_k otimes y_k}^2 = sum_{k=1}^N norm{x_k otimes y_k}^2 = sum_{k=1}^N norm{x_k}^2.
$$
On the other hand, since $(S otimes I)(x_k otimes y_k) = Sx_k otimes y_k$, it follows that ${Sx_k otimes y_k}$ is also orthogonal, so that by the same computation, mutatis mutandis,
$$
norm{(S otimes I)v}^2 = sum_{k=1}^N norm{S x_k}^2 leq sum_{k=1}^N norm{S}^2 norm{x_k}^2 = norm{S}^2 sum_{k=1}^N norm{x_k}^2 = norm{S}^2norm{v}^2.
$$
Thus, $norm{(S otimes I)v} leq norm{S}norm{v}$, as required.
Now, observe that since $(S otimes T) = (S otimes I)(I otimes T)$ on $H_1 odot H_2$, it follows by the boundedness of $S otimes I$ and $I otimes T$ that $S otimes T$ is also bounded with norm
$$
norm{S otimes T} leq norm{S otimes I}norm{I otimes T} leq norm{S}norm{T},
$$
as required.
answered Apr 3 '13 at 1:38
Branimir ĆaćićBranimir Ćaćić
10.2k22047
$endgroup$
$begingroup$
But what about the other inequality? How do I prove that $|T_1 otimes T_2| leq |T_1| |T_2|$
$endgroup$
– Max
Apr 3 '13 at 1:50
$begingroup$
@Max Sorry about that, I clearly misread your question! Now I've answered the question you actually asked.
$endgroup$
– Branimir Ćaćić
Apr 3 '13 at 7:23
add a comment |
$begingroup$
$newcommand{norm}[1]{left|{#1}right|}
newcommand{ip}[1]{leftlangle{#1}rightrangle}
newcommand{abs}[1]{left|{#1}right|}$Let $S in B(H_1)$, $T in B(H_2)$; the claim is that $norm{S otimes T} leq norm{S}norm{T}$.
Let me first show that $S otimes I$ is bounded with $norm{S otimes I} leq norm{S}$; the same proof, mutatis mutandis, will show that $I otimes T$ is bounded with $norm{I otimes T} leq norm{T}$. Since the algebraic tensor product $H_1 odot H_2$ is dense in $H_1 otimes H_2$, it suffices to show that $norm{(S otimes I)v} leq norm{S}norm{v}$ for any $v in H_1 odot H_2$.
So, let $v = sum_{k=1}^N x_k otimes y_k in H_1 odot H_2$; by performing Gram--Schmidt orthogonalisation on ${y_k}$ and expressing the $y_k$ in terms of the resulting orthonormal basis for $operatorname{span}{y_k}$, we may assume without loss of generality that ${y_k}$ is orthonormal. On the one hand, it follows that ${x_k otimes y_k}$ is orthogonal, so that
$$
norm{v}^2 = norm{sum_{k=1}^N x_k otimes y_k}^2 = sum_{k=1}^N norm{x_k otimes y_k}^2 = sum_{k=1}^N norm{x_k}^2.
$$
On the other hand, since $(S otimes I)(x_k otimes y_k) = Sx_k otimes y_k$, it follows that ${Sx_k otimes y_k}$ is also orthogonal, so that by the same computation, mutatis mutandis,
$$
norm{(S otimes I)v}^2 = sum_{k=1}^N norm{S x_k}^2 leq sum_{k=1}^N norm{S}^2 norm{x_k}^2 = norm{S}^2 sum_{k=1}^N norm{x_k}^2 = norm{S}^2norm{v}^2.
$$
Thus, $norm{(S otimes I)v} leq norm{S}norm{v}$, as required.
Now, observe that since $(S otimes T) = (S otimes I)(I otimes T)$ on $H_1 odot H_2$, it follows by the boundedness of $S otimes I$ and $I otimes T$ that $S otimes T$ is also bounded with norm
$$
norm{S otimes T} leq norm{S otimes I}norm{I otimes T} leq norm{S}norm{T},
$$
as required.
answered Apr 3 '13 at 1:38
Branimir ĆaćićBranimir Ćaćić
10.2k22047
$endgroup$
$begingroup$
But what about the other inequality? How do I prove that $|T_1 otimes T_2| leq |T_1| |T_2|$
$endgroup$
– Max
Apr 3 '13 at 1:50
$begingroup$
@Max Sorry about that, I clearly misread your question! Now I've answered the question you actually asked.
$endgroup$
– Branimir Ćaćić
Apr 3 '13 at 7:23
add a comment |
$begingroup$
$newcommand{norm}[1]{left|{#1}right|}
newcommand{ip}[1]{leftlangle{#1}rightrangle}
newcommand{abs}[1]{left|{#1}right|}$Let $S in B(H_1)$, $T in B(H_2)$; the claim is that $norm{S otimes T} leq norm{S}norm{T}$.
Let me first show that $S otimes I$ is bounded with $norm{S otimes I} leq norm{S}$; the same proof, mutatis mutandis, will show that $I otimes T$ is bounded with $norm{I otimes T} leq norm{T}$. Since the algebraic tensor product $H_1 odot H_2$ is dense in $H_1 otimes H_2$, it suffices to show that $norm{(S otimes I)v} leq norm{S}norm{v}$ for any $v in H_1 odot H_2$.
So, let $v = sum_{k=1}^N x_k otimes y_k in H_1 odot H_2$; by performing Gram--Schmidt orthogonalisation on ${y_k}$ and expressing the $y_k$ in terms of the resulting orthonormal basis for $operatorname{span}{y_k}$, we may assume without loss of generality that ${y_k}$ is orthonormal. On the one hand, it follows that ${x_k otimes y_k}$ is orthogonal, so that
$$
norm{v}^2 = norm{sum_{k=1}^N x_k otimes y_k}^2 = sum_{k=1}^N norm{x_k otimes y_k}^2 = sum_{k=1}^N norm{x_k}^2.
$$
On the other hand, since $(S otimes I)(x_k otimes y_k) = Sx_k otimes y_k$, it follows that ${Sx_k otimes y_k}$ is also orthogonal, so that by the same computation, mutatis mutandis,
$$
norm{(S otimes I)v}^2 = sum_{k=1}^N norm{S x_k}^2 leq sum_{k=1}^N norm{S}^2 norm{x_k}^2 = norm{S}^2 sum_{k=1}^N norm{x_k}^2 = norm{S}^2norm{v}^2.
$$
Thus, $norm{(S otimes I)v} leq norm{S}norm{v}$, as required.
Now, observe that since $(S otimes T) = (S otimes I)(I otimes T)$ on $H_1 odot H_2$, it follows by the boundedness of $S otimes I$ and $I otimes T$ that $S otimes T$ is also bounded with norm
$$
norm{S otimes T} leq norm{S otimes I}norm{I otimes T} leq norm{S}norm{T},
$$
as required.
answered Apr 3 '13 at 1:38
Branimir ĆaćićBranimir Ćaćić
10.2k22047
$endgroup$
$newcommand{norm}[1]{left|{#1}right|}
newcommand{ip}[1]{leftlangle{#1}rightrangle}
newcommand{abs}[1]{left|{#1}right|}$Let $S in B(H_1)$, $T in B(H_2)$; the claim is that $norm{S otimes T} leq norm{S}norm{T}$.
Let me first show that $S otimes I$ is bounded with $norm{S otimes I} leq norm{S}$; the same proof, mutatis mutandis, will show that $I otimes T$ is bounded with $norm{I otimes T} leq norm{T}$. Since the algebraic tensor product $H_1 odot H_2$ is dense in $H_1 otimes H_2$, it suffices to show that $norm{(S otimes I)v} leq norm{S}norm{v}$ for any $v in H_1 odot H_2$.
So, let $v = sum_{k=1}^N x_k otimes y_k in H_1 odot H_2$; by performing Gram--Schmidt orthogonalisation on ${y_k}$ and expressing the $y_k$ in terms of the resulting orthonormal basis for $operatorname{span}{y_k}$, we may assume without loss of generality that ${y_k}$ is orthonormal. On the one hand, it follows that ${x_k otimes y_k}$ is orthogonal, so that
$$
norm{v}^2 = norm{sum_{k=1}^N x_k otimes y_k}^2 = sum_{k=1}^N norm{x_k otimes y_k}^2 = sum_{k=1}^N norm{x_k}^2.
$$
On the other hand, since $(S otimes I)(x_k otimes y_k) = Sx_k otimes y_k$, it follows that ${Sx_k otimes y_k}$ is also orthogonal, so that by the same computation, mutatis mutandis,
$$
norm{(S otimes I)v}^2 = sum_{k=1}^N norm{S x_k}^2 leq sum_{k=1}^N norm{S}^2 norm{x_k}^2 = norm{S}^2 sum_{k=1}^N norm{x_k}^2 = norm{S}^2norm{v}^2.
$$
Thus, $norm{(S otimes I)v} leq norm{S}norm{v}$, as required.
Now, observe that since $(S otimes T) = (S otimes I)(I otimes T)$ on $H_1 odot H_2$, it follows by the boundedness of $S otimes I$ and $I otimes T$ that $S otimes T$ is also bounded with norm
$$
norm{S otimes T} leq norm{S otimes I}norm{I otimes T} leq norm{S}norm{T},
$$
as required.
answered Apr 3 '13 at 1:38
Branimir ĆaćićBranimir Ćaćić
10.2k22047
edited Apr 3 '13 at 18:33
answered Apr 3 '13 at 1:38
Branimir ĆaćićBranimir Ćaćić
10.2k22047
answered Apr 3 '13 at 1:38
Branimir ĆaćićBranimir Ćaćić
10.2k22047
answered Apr 3 '13 at 1:38
Branimir ĆaćićBranimir Ćaćić
10.2k22047
10.2k22047
$begingroup$
But what about the other inequality? How do I prove that $|T_1 otimes T_2| leq |T_1| |T_2|$
$endgroup$
– Max
Apr 3 '13 at 1:50
$begingroup$
@Max Sorry about that, I clearly misread your question! Now I've answered the question you actually asked.
$endgroup$
– Branimir Ćaćić
Apr 3 '13 at 7:23
add a comment |
$begingroup$
But what about the other inequality? How do I prove that $|T_1 otimes T_2| leq |T_1| |T_2|$
$endgroup$
– Max
Apr 3 '13 at 1:50
$begingroup$
@Max Sorry about that, I clearly misread your question! Now I've answered the question you actually asked.
$endgroup$
– Branimir Ćaćić
Apr 3 '13 at 7:23
$begingroup$
But what about the other inequality? How do I prove that $|T_1 otimes T_2| leq |T_1| |T_2|$
$endgroup$
– Max
Apr 3 '13 at 1:50
$begingroup$
But what about the other inequality? How do I prove that $|T_1 otimes T_2| leq |T_1| |T_2|$
$endgroup$
– Max
Apr 3 '13 at 1:50
$begingroup$
@Max Sorry about that, I clearly misread your question! Now I've answered the question you actually asked.
$endgroup$
– Branimir Ćaćić
Apr 3 '13 at 7:23
$begingroup$
@Max Sorry about that, I clearly misread your question! Now I've answered the question you actually asked.
$endgroup$
– Branimir Ćaćić
Apr 3 '13 at 7:23
add a comment |
$begingroup$
Let me add the other direction. Suppose $varepsilon >0$. Then from the boundedness of operators $T_{1}$ and $T_{2}$ we know that there exist unit vectors in the Hilbert spaces $psi in H_{1}$, $xi in H_{2}$ such that:
$left | T_{1} psi right |> left | T_{1} right | - epsilon$, $left | T_{2} xi right |> left | T_{2} right | - epsilon$
Using this we can use the definition of the tensor product to see that:
$left | left (T_{1}otimes T_{2} right )left ( psiotimes xi right ) right |_{H_{1}bigotimes H_{2}}=left | left ( T_{1}psiotimes T_{2}xi right ) right |_{H_{1}bigotimes H_{2}}=left |T_{1}psi right |_{H_{1}} left | T_{2}xi right |_{H_{2}}geq$
$ geqleft | T_{1} right |left | T_{2} right |-epsilon left | T_{1} right | - epsilon left | T_{2} right | + epsilon^{2}geqleft | T_{1} right |left | T_{2} right |-epsilon left | T_{1} right | - epsilon left | T_{2} right |$
Sending $epsilon$ to 0 shows the desired result.
answered Jan 30 at 12:03
Alex EastAlex East
1
$endgroup$
add a comment |
$begingroup$
Let me add the other direction. Suppose $varepsilon >0$. Then from the boundedness of operators $T_{1}$ and $T_{2}$ we know that there exist unit vectors in the Hilbert spaces $psi in H_{1}$, $xi in H_{2}$ such that:
$left | T_{1} psi right |> left | T_{1} right | - epsilon$, $left | T_{2} xi right |> left | T_{2} right | - epsilon$
Using this we can use the definition of the tensor product to see that:
$left | left (T_{1}otimes T_{2} right )left ( psiotimes xi right ) right |_{H_{1}bigotimes H_{2}}=left | left ( T_{1}psiotimes T_{2}xi right ) right |_{H_{1}bigotimes H_{2}}=left |T_{1}psi right |_{H_{1}} left | T_{2}xi right |_{H_{2}}geq$
$ geqleft | T_{1} right |left | T_{2} right |-epsilon left | T_{1} right | - epsilon left | T_{2} right | + epsilon^{2}geqleft | T_{1} right |left | T_{2} right |-epsilon left | T_{1} right | - epsilon left | T_{2} right |$
Sending $epsilon$ to 0 shows the desired result.
answered Jan 30 at 12:03
Alex EastAlex East
1
$endgroup$
add a comment |
$begingroup$
Let me add the other direction. Suppose $varepsilon >0$. Then from the boundedness of operators $T_{1}$ and $T_{2}$ we know that there exist unit vectors in the Hilbert spaces $psi in H_{1}$, $xi in H_{2}$ such that:
$left | T_{1} psi right |> left | T_{1} right | - epsilon$, $left | T_{2} xi right |> left | T_{2} right | - epsilon$
Using this we can use the definition of the tensor product to see that:
$left | left (T_{1}otimes T_{2} right )left ( psiotimes xi right ) right |_{H_{1}bigotimes H_{2}}=left | left ( T_{1}psiotimes T_{2}xi right ) right |_{H_{1}bigotimes H_{2}}=left |T_{1}psi right |_{H_{1}} left | T_{2}xi right |_{H_{2}}geq$
$ geqleft | T_{1} right |left | T_{2} right |-epsilon left | T_{1} right | - epsilon left | T_{2} right | + epsilon^{2}geqleft | T_{1} right |left | T_{2} right |-epsilon left | T_{1} right | - epsilon left | T_{2} right |$
Sending $epsilon$ to 0 shows the desired result.
answered Jan 30 at 12:03
Alex EastAlex East
1
$endgroup$
Let me add the other direction. Suppose $varepsilon >0$. Then from the boundedness of operators $T_{1}$ and $T_{2}$ we know that there exist unit vectors in the Hilbert spaces $psi in H_{1}$, $xi in H_{2}$ such that:
$left | T_{1} psi right |> left | T_{1} right | - epsilon$, $left | T_{2} xi right |> left | T_{2} right | - epsilon$
Using this we can use the definition of the tensor product to see that:
$left | left (T_{1}otimes T_{2} right )left ( psiotimes xi right ) right |_{H_{1}bigotimes H_{2}}=left | left ( T_{1}psiotimes T_{2}xi right ) right |_{H_{1}bigotimes H_{2}}=left |T_{1}psi right |_{H_{1}} left | T_{2}xi right |_{H_{2}}geq$
$ geqleft | T_{1} right |left | T_{2} right |-epsilon left | T_{1} right | - epsilon left | T_{2} right | + epsilon^{2}geqleft | T_{1} right |left | T_{2} right |-epsilon left | T_{1} right | - epsilon left | T_{2} right |$
Sending $epsilon$ to 0 shows the desired result.
answered Jan 30 at 12:03
Alex EastAlex East
1
answered Jan 30 at 12:03
Alex EastAlex East
1
answered Jan 30 at 12:03
Alex EastAlex East
1
answered Jan 30 at 12:03
Alex EastAlex East
1
1
add a comment |
add a comment |
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LaTeX tips:
<and>mean "less than" and "greater than", and produce spacing correct for that meaning only. When you want angle brackets, you need to uselangleandrangle. Also, to get proper norm symbols use|instead of||.$endgroup$
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