Is the “tensoring map” from $mathcal{H}_1 times mathcal{H}_2$ to $mathcal{H}_1 otimes mathcal{H}_2$ a...
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Let $mathcal{H}_1$ and $mathcal{H_2}$ two Hilbert spaces. Construct the tensor product $mathcal{H}_1 otimes mathcal{H}_2$ as the set of all bounded antilinear operators from $mathcal{H_2}$ to $mathcal{H_1}$ such that the norm $Vert A Vert_otimes := sqrt{sum_beta Vert Av_beta Vert^2}$ is finite, where $lbrace v_beta rbrace$ is any orthonormal basis of $mathcal{H}_2$ (I followed the construction presented in Folland's A Course in Astract Harmonic Analysis).
Now for all $(u,v) in mathcal{H}_1 times mathcal{H}_2$ I can consider the antilinear operator $u otimes v : w mapsto langle v, w rangle u$, which is an element of the tensor product.
My question is: if I consider the map $bigotimes : mathcal{H}_1 times mathcal{H}_2 rightarrow mathcal{H}_1 otimes mathcal{H}_2$ defined in the natural way, is it continuous (w.r.t. the natural topologies on both spaces, i.e. the product topology on the domain and the topology induced by the norm on the codomain)? My guess is that it is continuous, but I can't prove it.
Denoting by:
$pi_1$ and $pi_2$ the two projections- $f_v = langle v,cdot rangle in mathcal{H}_2^*$
$f : v mapsto f_v$ the map which characterizes the dual space of $mathcal{H}_2$
I tried to say that $bigotimes(cdot) = f_{pi_2(cdot)}pi_1(cdot)$ but I'm confused and can't tell if this is sufficient to conclude that $bigotimes$ is continuous.
functional-analysis hilbert-spaces tensor-products
asked Jan 22 at 14:52
M. RinettiM. Rinetti
307
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add a comment |
$begingroup$
Let $mathcal{H}_1$ and $mathcal{H_2}$ two Hilbert spaces. Construct the tensor product $mathcal{H}_1 otimes mathcal{H}_2$ as the set of all bounded antilinear operators from $mathcal{H_2}$ to $mathcal{H_1}$ such that the norm $Vert A Vert_otimes := sqrt{sum_beta Vert Av_beta Vert^2}$ is finite, where $lbrace v_beta rbrace$ is any orthonormal basis of $mathcal{H}_2$ (I followed the construction presented in Folland's A Course in Astract Harmonic Analysis).
Now for all $(u,v) in mathcal{H}_1 times mathcal{H}_2$ I can consider the antilinear operator $u otimes v : w mapsto langle v, w rangle u$, which is an element of the tensor product.
My question is: if I consider the map $bigotimes : mathcal{H}_1 times mathcal{H}_2 rightarrow mathcal{H}_1 otimes mathcal{H}_2$ defined in the natural way, is it continuous (w.r.t. the natural topologies on both spaces, i.e. the product topology on the domain and the topology induced by the norm on the codomain)? My guess is that it is continuous, but I can't prove it.
Denoting by:
$pi_1$ and $pi_2$ the two projections- $f_v = langle v,cdot rangle in mathcal{H}_2^*$
$f : v mapsto f_v$ the map which characterizes the dual space of $mathcal{H}_2$
I tried to say that $bigotimes(cdot) = f_{pi_2(cdot)}pi_1(cdot)$ but I'm confused and can't tell if this is sufficient to conclude that $bigotimes$ is continuous.
functional-analysis hilbert-spaces tensor-products
asked Jan 22 at 14:52
M. RinettiM. Rinetti
307
$endgroup$
$begingroup$
You are missing squares on both sides of the definition of $|cdot|_otimes$. Furthermore, $H_1otimes H_2$ is not the set of all bounded antilinear operators, but only those for which $|cdot|_otimes$ is finite.
$endgroup$
– MaoWao
Jan 22 at 15:05
$begingroup$
Thanks, I fixed it!
$endgroup$
– M. Rinetti
Jan 22 at 15:18
add a comment |
$begingroup$
Let $mathcal{H}_1$ and $mathcal{H_2}$ two Hilbert spaces. Construct the tensor product $mathcal{H}_1 otimes mathcal{H}_2$ as the set of all bounded antilinear operators from $mathcal{H_2}$ to $mathcal{H_1}$ such that the norm $Vert A Vert_otimes := sqrt{sum_beta Vert Av_beta Vert^2}$ is finite, where $lbrace v_beta rbrace$ is any orthonormal basis of $mathcal{H}_2$ (I followed the construction presented in Folland's A Course in Astract Harmonic Analysis).
Now for all $(u,v) in mathcal{H}_1 times mathcal{H}_2$ I can consider the antilinear operator $u otimes v : w mapsto langle v, w rangle u$, which is an element of the tensor product.
My question is: if I consider the map $bigotimes : mathcal{H}_1 times mathcal{H}_2 rightarrow mathcal{H}_1 otimes mathcal{H}_2$ defined in the natural way, is it continuous (w.r.t. the natural topologies on both spaces, i.e. the product topology on the domain and the topology induced by the norm on the codomain)? My guess is that it is continuous, but I can't prove it.
Denoting by:
$pi_1$ and $pi_2$ the two projections- $f_v = langle v,cdot rangle in mathcal{H}_2^*$
$f : v mapsto f_v$ the map which characterizes the dual space of $mathcal{H}_2$
I tried to say that $bigotimes(cdot) = f_{pi_2(cdot)}pi_1(cdot)$ but I'm confused and can't tell if this is sufficient to conclude that $bigotimes$ is continuous.
functional-analysis hilbert-spaces tensor-products
asked Jan 22 at 14:52
M. RinettiM. Rinetti
307
$endgroup$
Let $mathcal{H}_1$ and $mathcal{H_2}$ two Hilbert spaces. Construct the tensor product $mathcal{H}_1 otimes mathcal{H}_2$ as the set of all bounded antilinear operators from $mathcal{H_2}$ to $mathcal{H_1}$ such that the norm $Vert A Vert_otimes := sqrt{sum_beta Vert Av_beta Vert^2}$ is finite, where $lbrace v_beta rbrace$ is any orthonormal basis of $mathcal{H}_2$ (I followed the construction presented in Folland's A Course in Astract Harmonic Analysis).
Now for all $(u,v) in mathcal{H}_1 times mathcal{H}_2$ I can consider the antilinear operator $u otimes v : w mapsto langle v, w rangle u$, which is an element of the tensor product.
My question is: if I consider the map $bigotimes : mathcal{H}_1 times mathcal{H}_2 rightarrow mathcal{H}_1 otimes mathcal{H}_2$ defined in the natural way, is it continuous (w.r.t. the natural topologies on both spaces, i.e. the product topology on the domain and the topology induced by the norm on the codomain)? My guess is that it is continuous, but I can't prove it.
Denoting by:
$pi_1$ and $pi_2$ the two projections- $f_v = langle v,cdot rangle in mathcal{H}_2^*$
$f : v mapsto f_v$ the map which characterizes the dual space of $mathcal{H}_2$
I tried to say that $bigotimes(cdot) = f_{pi_2(cdot)}pi_1(cdot)$ but I'm confused and can't tell if this is sufficient to conclude that $bigotimes$ is continuous.
functional-analysis hilbert-spaces tensor-products
functional-analysis hilbert-spaces tensor-products
asked Jan 22 at 14:52
M. RinettiM. Rinetti
307
asked Jan 22 at 14:52
M. RinettiM. Rinetti
307
edited Jan 22 at 15:18
M. Rinetti
asked Jan 22 at 14:52
M. RinettiM. Rinetti
307
asked Jan 22 at 14:52
M. RinettiM. Rinetti
307
asked Jan 22 at 14:52
M. RinettiM. Rinetti
307
307
$begingroup$
You are missing squares on both sides of the definition of $|cdot|_otimes$. Furthermore, $H_1otimes H_2$ is not the set of all bounded antilinear operators, but only those for which $|cdot|_otimes$ is finite.
$endgroup$
– MaoWao
Jan 22 at 15:05
$begingroup$
Thanks, I fixed it!
$endgroup$
– M. Rinetti
Jan 22 at 15:18
add a comment |
$begingroup$
You are missing squares on both sides of the definition of $|cdot|_otimes$. Furthermore, $H_1otimes H_2$ is not the set of all bounded antilinear operators, but only those for which $|cdot|_otimes$ is finite.
$endgroup$
– MaoWao
Jan 22 at 15:05
$begingroup$
Thanks, I fixed it!
$endgroup$
– M. Rinetti
Jan 22 at 15:18
$begingroup$
You are missing squares on both sides of the definition of $|cdot|_otimes$. Furthermore, $H_1otimes H_2$ is not the set of all bounded antilinear operators, but only those for which $|cdot|_otimes$ is finite.
$endgroup$
– MaoWao
Jan 22 at 15:05
$begingroup$
You are missing squares on both sides of the definition of $|cdot|_otimes$. Furthermore, $H_1otimes H_2$ is not the set of all bounded antilinear operators, but only those for which $|cdot|_otimes$ is finite.
$endgroup$
– MaoWao
Jan 22 at 15:05
$begingroup$
Thanks, I fixed it!
$endgroup$
– M. Rinetti
Jan 22 at 15:18
$begingroup$
Thanks, I fixed it!
$endgroup$
– M. Rinetti
Jan 22 at 15:18
add a comment |
1 Answer
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The map $otimes$ is obviously bilinear. A bilinear map $betacolon H_1times H_2to K$ is continuous if and only if there exists a constant $C>0$ such that $|beta(u,v)|leq C|u||v|$ for all $uin H_1$, $vin H_2$. The proof is very similar to the characterization of continuous linear maps as bounded operators.
In your case one has $|uotimes v|_otimes=|langle v,frac{v}{|v|}vrangle u|=|u||v|$, which shows that $otimes$ is continuous.
answered Jan 22 at 15:06
MaoWaoMaoWao
3,533617
$endgroup$
$begingroup$
Thank you, it is all clear now!
$endgroup$
– M. Rinetti
Jan 22 at 15:27
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The map $otimes$ is obviously bilinear. A bilinear map $betacolon H_1times H_2to K$ is continuous if and only if there exists a constant $C>0$ such that $|beta(u,v)|leq C|u||v|$ for all $uin H_1$, $vin H_2$. The proof is very similar to the characterization of continuous linear maps as bounded operators.
In your case one has $|uotimes v|_otimes=|langle v,frac{v}{|v|}vrangle u|=|u||v|$, which shows that $otimes$ is continuous.
answered Jan 22 at 15:06
MaoWaoMaoWao
3,533617
$endgroup$
$begingroup$
Thank you, it is all clear now!
$endgroup$
– M. Rinetti
Jan 22 at 15:27
add a comment |
$begingroup$
The map $otimes$ is obviously bilinear. A bilinear map $betacolon H_1times H_2to K$ is continuous if and only if there exists a constant $C>0$ such that $|beta(u,v)|leq C|u||v|$ for all $uin H_1$, $vin H_2$. The proof is very similar to the characterization of continuous linear maps as bounded operators.
In your case one has $|uotimes v|_otimes=|langle v,frac{v}{|v|}vrangle u|=|u||v|$, which shows that $otimes$ is continuous.
answered Jan 22 at 15:06
MaoWaoMaoWao
3,533617
$endgroup$
$begingroup$
Thank you, it is all clear now!
$endgroup$
– M. Rinetti
Jan 22 at 15:27
add a comment |
$begingroup$
The map $otimes$ is obviously bilinear. A bilinear map $betacolon H_1times H_2to K$ is continuous if and only if there exists a constant $C>0$ such that $|beta(u,v)|leq C|u||v|$ for all $uin H_1$, $vin H_2$. The proof is very similar to the characterization of continuous linear maps as bounded operators.
In your case one has $|uotimes v|_otimes=|langle v,frac{v}{|v|}vrangle u|=|u||v|$, which shows that $otimes$ is continuous.
answered Jan 22 at 15:06
MaoWaoMaoWao
3,533617
$endgroup$
The map $otimes$ is obviously bilinear. A bilinear map $betacolon H_1times H_2to K$ is continuous if and only if there exists a constant $C>0$ such that $|beta(u,v)|leq C|u||v|$ for all $uin H_1$, $vin H_2$. The proof is very similar to the characterization of continuous linear maps as bounded operators.
In your case one has $|uotimes v|_otimes=|langle v,frac{v}{|v|}vrangle u|=|u||v|$, which shows that $otimes$ is continuous.
answered Jan 22 at 15:06
MaoWaoMaoWao
3,533617
answered Jan 22 at 15:06
MaoWaoMaoWao
3,533617
answered Jan 22 at 15:06
MaoWaoMaoWao
3,533617
answered Jan 22 at 15:06
MaoWaoMaoWao
3,533617
3,533617
$begingroup$
Thank you, it is all clear now!
$endgroup$
– M. Rinetti
Jan 22 at 15:27
add a comment |
$begingroup$
Thank you, it is all clear now!
$endgroup$
– M. Rinetti
Jan 22 at 15:27
$begingroup$
Thank you, it is all clear now!
$endgroup$
– M. Rinetti
Jan 22 at 15:27
$begingroup$
Thank you, it is all clear now!
$endgroup$
– M. Rinetti
Jan 22 at 15:27
add a comment |
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$begingroup$
You are missing squares on both sides of the definition of $|cdot|_otimes$. Furthermore, $H_1otimes H_2$ is not the set of all bounded antilinear operators, but only those for which $|cdot|_otimes$ is finite.
$endgroup$
– MaoWao
Jan 22 at 15:05
$begingroup$
Thanks, I fixed it!
$endgroup$
– M. Rinetti
Jan 22 at 15:18