Mixing
If $E$ is Banach and $E^*$ is its dual, is every $T:E^*rightarrow E^*$ an adjoint?
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If $E$ is a (complex) Banach and $E^*$ is it's dual, is every bounded $T:E^*rightarrow E^*$ an adjoint? I am mostly interested in when $T$ is an automorphism.
If not, is ${T: E^*rightarrow E^* | mbox{T is an adjoint}}$ dense in $B(E^*)$?
Thanks in advance!
real-analysis functional-analysis
asked Jan 19 at 21:24
RandomWalkerRandomWalker
1919
$endgroup$
add a comment |
$begingroup$
If $E$ is a (complex) Banach and $E^*$ is it's dual, is every bounded $T:E^*rightarrow E^*$ an adjoint? I am mostly interested in when $T$ is an automorphism.
If not, is ${T: E^*rightarrow E^* | mbox{T is an adjoint}}$ dense in $B(E^*)$?
Thanks in advance!
real-analysis functional-analysis
asked Jan 19 at 21:24
RandomWalkerRandomWalker
1919
$endgroup$
$begingroup$
Can you define what means to be an adjoint map?
$endgroup$
– alexp9
Jan 19 at 21:30
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@Rhcpy99 Presumbaly $T$ is an adjoint if $T=S^*$ for some $Sin B(E)$.
$endgroup$
– Aweygan
Jan 19 at 21:33
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@Aweygan Yes, that's what I meant. Sorry for the confusion
$endgroup$
– RandomWalker
Jan 19 at 21:34
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What is easy: if $E$ is reflexive, then every $Tin B(E^*)$ is an adjoint.
$endgroup$
– amsmath
Jan 19 at 21:55
add a comment |
$begingroup$
If $E$ is a (complex) Banach and $E^*$ is it's dual, is every bounded $T:E^*rightarrow E^*$ an adjoint? I am mostly interested in when $T$ is an automorphism.
If not, is ${T: E^*rightarrow E^* | mbox{T is an adjoint}}$ dense in $B(E^*)$?
Thanks in advance!
real-analysis functional-analysis
asked Jan 19 at 21:24
RandomWalkerRandomWalker
1919
$endgroup$
If $E$ is a (complex) Banach and $E^*$ is it's dual, is every bounded $T:E^*rightarrow E^*$ an adjoint? I am mostly interested in when $T$ is an automorphism.
If not, is ${T: E^*rightarrow E^* | mbox{T is an adjoint}}$ dense in $B(E^*)$?
Thanks in advance!
real-analysis functional-analysis
real-analysis functional-analysis
asked Jan 19 at 21:24
RandomWalkerRandomWalker
1919
asked Jan 19 at 21:24
RandomWalkerRandomWalker
1919
asked Jan 19 at 21:24
RandomWalkerRandomWalker
1919
asked Jan 19 at 21:24
RandomWalkerRandomWalker
1919
asked Jan 19 at 21:24
RandomWalkerRandomWalker
1919
1919
$begingroup$
Can you define what means to be an adjoint map?
$endgroup$
– alexp9
Jan 19 at 21:30
$begingroup$
@Rhcpy99 Presumbaly $T$ is an adjoint if $T=S^*$ for some $Sin B(E)$.
$endgroup$
– Aweygan
Jan 19 at 21:33
$begingroup$
@Aweygan Yes, that's what I meant. Sorry for the confusion
$endgroup$
– RandomWalker
Jan 19 at 21:34
$begingroup$
What is easy: if $E$ is reflexive, then every $Tin B(E^*)$ is an adjoint.
$endgroup$
– amsmath
Jan 19 at 21:55
add a comment |
$begingroup$
Can you define what means to be an adjoint map?
$endgroup$
– alexp9
Jan 19 at 21:30
$begingroup$
@Rhcpy99 Presumbaly $T$ is an adjoint if $T=S^*$ for some $Sin B(E)$.
$endgroup$
– Aweygan
Jan 19 at 21:33
$begingroup$
@Aweygan Yes, that's what I meant. Sorry for the confusion
$endgroup$
– RandomWalker
Jan 19 at 21:34
$begingroup$
What is easy: if $E$ is reflexive, then every $Tin B(E^*)$ is an adjoint.
$endgroup$
– amsmath
Jan 19 at 21:55
$begingroup$
Can you define what means to be an adjoint map?
$endgroup$
– alexp9
Jan 19 at 21:30
$begingroup$
Can you define what means to be an adjoint map?
$endgroup$
– alexp9
Jan 19 at 21:30
$begingroup$
@Rhcpy99 Presumbaly $T$ is an adjoint if $T=S^*$ for some $Sin B(E)$.
$endgroup$
– Aweygan
Jan 19 at 21:33
$begingroup$
@Rhcpy99 Presumbaly $T$ is an adjoint if $T=S^*$ for some $Sin B(E)$.
$endgroup$
– Aweygan
Jan 19 at 21:33
$begingroup$
@Aweygan Yes, that's what I meant. Sorry for the confusion
$endgroup$
– RandomWalker
Jan 19 at 21:34
$begingroup$
@Aweygan Yes, that's what I meant. Sorry for the confusion
$endgroup$
– RandomWalker
Jan 19 at 21:34
$begingroup$
What is easy: if $E$ is reflexive, then every $Tin B(E^*)$ is an adjoint.
$endgroup$
– amsmath
Jan 19 at 21:55
$begingroup$
What is easy: if $E$ is reflexive, then every $Tin B(E^*)$ is an adjoint.
$endgroup$
– amsmath
Jan 19 at 21:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A linear map $Tin B(E^*)$ is of the form $T=S^*$ for some $Sin B(E)$ if an only if $T$ is weak$^*$-to-weak$^*$ continuous.
Indeed, the reverse implication is easy (take a weak$^*$-convergent net $(x_gamma^*)$, and show that $(Tx_gamma^*)$ is weak$^*$-convergent). For the forward implication, for each $xin X$, the map $E^*tomathbb C$ given by $x^*mapsto langle Tx^*,xrangle$ is a weak$^*$-continuous linear functional, whence there is some $Sxin E$ such that $langle Tx^*,xrangle=langle x^*,Sxrangle$ for all $x^*in E^*$. Showing that the map $S:xmapsto Sx$ is in $B(E)$ and $S^*=T$ isn't terribly difficult (the former follows from the closed graph theorem, and the latter is by construction).
As for the question about density, I'm not aware of any general results. If $E$ is reflexive, then ${T: E^*rightarrow E^* mid Tmbox{ is an adjoint}}=B(E^*)$, but there are some weird Banach spaces, and I imagine counterexamples exist.
EDIT Note that the adjoint map $B(E)to B(E^*)$, $Tmapsto T^*$ is an isometry, whence the image is closed. Thus to show that the image of $B(E)$ under the adjoint map is not dense, it suffices to show that there is some element of $ B(E^*)$ which is not an adjoint. An example of such an operator is provided the comments following the answer on this question.
answered Jan 19 at 21:53
AweyganAweygan
14.4k21441
$endgroup$
$begingroup$
Extremely thorough and helpful. Thank you very much.
$endgroup$
– RandomWalker
Jan 19 at 22:12
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 19 at 22:14
$begingroup$
Follow up: is the subspace of adjoints complemented in $B(E^*)$?
$endgroup$
– RandomWalker
Jan 19 at 22:14
$begingroup$
Interesting question. I can't say anything concrete, but I doubt it.
$endgroup$
– Aweygan
Jan 19 at 22:24
add a comment |
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1 Answer
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active
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1 Answer
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oldest
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active
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active
oldest
votes
$begingroup$
A linear map $Tin B(E^*)$ is of the form $T=S^*$ for some $Sin B(E)$ if an only if $T$ is weak$^*$-to-weak$^*$ continuous.
Indeed, the reverse implication is easy (take a weak$^*$-convergent net $(x_gamma^*)$, and show that $(Tx_gamma^*)$ is weak$^*$-convergent). For the forward implication, for each $xin X$, the map $E^*tomathbb C$ given by $x^*mapsto langle Tx^*,xrangle$ is a weak$^*$-continuous linear functional, whence there is some $Sxin E$ such that $langle Tx^*,xrangle=langle x^*,Sxrangle$ for all $x^*in E^*$. Showing that the map $S:xmapsto Sx$ is in $B(E)$ and $S^*=T$ isn't terribly difficult (the former follows from the closed graph theorem, and the latter is by construction).
As for the question about density, I'm not aware of any general results. If $E$ is reflexive, then ${T: E^*rightarrow E^* mid Tmbox{ is an adjoint}}=B(E^*)$, but there are some weird Banach spaces, and I imagine counterexamples exist.
EDIT Note that the adjoint map $B(E)to B(E^*)$, $Tmapsto T^*$ is an isometry, whence the image is closed. Thus to show that the image of $B(E)$ under the adjoint map is not dense, it suffices to show that there is some element of $ B(E^*)$ which is not an adjoint. An example of such an operator is provided the comments following the answer on this question.
answered Jan 19 at 21:53
AweyganAweygan
14.4k21441
$endgroup$
$begingroup$
Extremely thorough and helpful. Thank you very much.
$endgroup$
– RandomWalker
Jan 19 at 22:12
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 19 at 22:14
$begingroup$
Follow up: is the subspace of adjoints complemented in $B(E^*)$?
$endgroup$
– RandomWalker
Jan 19 at 22:14
$begingroup$
Interesting question. I can't say anything concrete, but I doubt it.
$endgroup$
– Aweygan
Jan 19 at 22:24
add a comment |
$begingroup$
A linear map $Tin B(E^*)$ is of the form $T=S^*$ for some $Sin B(E)$ if an only if $T$ is weak$^*$-to-weak$^*$ continuous.
Indeed, the reverse implication is easy (take a weak$^*$-convergent net $(x_gamma^*)$, and show that $(Tx_gamma^*)$ is weak$^*$-convergent). For the forward implication, for each $xin X$, the map $E^*tomathbb C$ given by $x^*mapsto langle Tx^*,xrangle$ is a weak$^*$-continuous linear functional, whence there is some $Sxin E$ such that $langle Tx^*,xrangle=langle x^*,Sxrangle$ for all $x^*in E^*$. Showing that the map $S:xmapsto Sx$ is in $B(E)$ and $S^*=T$ isn't terribly difficult (the former follows from the closed graph theorem, and the latter is by construction).
As for the question about density, I'm not aware of any general results. If $E$ is reflexive, then ${T: E^*rightarrow E^* mid Tmbox{ is an adjoint}}=B(E^*)$, but there are some weird Banach spaces, and I imagine counterexamples exist.
EDIT Note that the adjoint map $B(E)to B(E^*)$, $Tmapsto T^*$ is an isometry, whence the image is closed. Thus to show that the image of $B(E)$ under the adjoint map is not dense, it suffices to show that there is some element of $ B(E^*)$ which is not an adjoint. An example of such an operator is provided the comments following the answer on this question.
answered Jan 19 at 21:53
AweyganAweygan
14.4k21441
$endgroup$
$begingroup$
Extremely thorough and helpful. Thank you very much.
$endgroup$
– RandomWalker
Jan 19 at 22:12
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 19 at 22:14
$begingroup$
Follow up: is the subspace of adjoints complemented in $B(E^*)$?
$endgroup$
– RandomWalker
Jan 19 at 22:14
$begingroup$
Interesting question. I can't say anything concrete, but I doubt it.
$endgroup$
– Aweygan
Jan 19 at 22:24
add a comment |
$begingroup$
A linear map $Tin B(E^*)$ is of the form $T=S^*$ for some $Sin B(E)$ if an only if $T$ is weak$^*$-to-weak$^*$ continuous.
Indeed, the reverse implication is easy (take a weak$^*$-convergent net $(x_gamma^*)$, and show that $(Tx_gamma^*)$ is weak$^*$-convergent). For the forward implication, for each $xin X$, the map $E^*tomathbb C$ given by $x^*mapsto langle Tx^*,xrangle$ is a weak$^*$-continuous linear functional, whence there is some $Sxin E$ such that $langle Tx^*,xrangle=langle x^*,Sxrangle$ for all $x^*in E^*$. Showing that the map $S:xmapsto Sx$ is in $B(E)$ and $S^*=T$ isn't terribly difficult (the former follows from the closed graph theorem, and the latter is by construction).
As for the question about density, I'm not aware of any general results. If $E$ is reflexive, then ${T: E^*rightarrow E^* mid Tmbox{ is an adjoint}}=B(E^*)$, but there are some weird Banach spaces, and I imagine counterexamples exist.
EDIT Note that the adjoint map $B(E)to B(E^*)$, $Tmapsto T^*$ is an isometry, whence the image is closed. Thus to show that the image of $B(E)$ under the adjoint map is not dense, it suffices to show that there is some element of $ B(E^*)$ which is not an adjoint. An example of such an operator is provided the comments following the answer on this question.
answered Jan 19 at 21:53
AweyganAweygan
14.4k21441
$endgroup$
A linear map $Tin B(E^*)$ is of the form $T=S^*$ for some $Sin B(E)$ if an only if $T$ is weak$^*$-to-weak$^*$ continuous.
Indeed, the reverse implication is easy (take a weak$^*$-convergent net $(x_gamma^*)$, and show that $(Tx_gamma^*)$ is weak$^*$-convergent). For the forward implication, for each $xin X$, the map $E^*tomathbb C$ given by $x^*mapsto langle Tx^*,xrangle$ is a weak$^*$-continuous linear functional, whence there is some $Sxin E$ such that $langle Tx^*,xrangle=langle x^*,Sxrangle$ for all $x^*in E^*$. Showing that the map $S:xmapsto Sx$ is in $B(E)$ and $S^*=T$ isn't terribly difficult (the former follows from the closed graph theorem, and the latter is by construction).
As for the question about density, I'm not aware of any general results. If $E$ is reflexive, then ${T: E^*rightarrow E^* mid Tmbox{ is an adjoint}}=B(E^*)$, but there are some weird Banach spaces, and I imagine counterexamples exist.
EDIT Note that the adjoint map $B(E)to B(E^*)$, $Tmapsto T^*$ is an isometry, whence the image is closed. Thus to show that the image of $B(E)$ under the adjoint map is not dense, it suffices to show that there is some element of $ B(E^*)$ which is not an adjoint. An example of such an operator is provided the comments following the answer on this question.
answered Jan 19 at 21:53
AweyganAweygan
14.4k21441
edited Jan 19 at 22:07
answered Jan 19 at 21:53
AweyganAweygan
14.4k21441
answered Jan 19 at 21:53
AweyganAweygan
14.4k21441
answered Jan 19 at 21:53
AweyganAweygan
14.4k21441
14.4k21441
$begingroup$
Extremely thorough and helpful. Thank you very much.
$endgroup$
– RandomWalker
Jan 19 at 22:12
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 19 at 22:14
$begingroup$
Follow up: is the subspace of adjoints complemented in $B(E^*)$?
$endgroup$
– RandomWalker
Jan 19 at 22:14
$begingroup$
Interesting question. I can't say anything concrete, but I doubt it.
$endgroup$
– Aweygan
Jan 19 at 22:24
add a comment |
$begingroup$
Extremely thorough and helpful. Thank you very much.
$endgroup$
– RandomWalker
Jan 19 at 22:12
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 19 at 22:14
$begingroup$
Follow up: is the subspace of adjoints complemented in $B(E^*)$?
$endgroup$
– RandomWalker
Jan 19 at 22:14
$begingroup$
Interesting question. I can't say anything concrete, but I doubt it.
$endgroup$
– Aweygan
Jan 19 at 22:24
$begingroup$
Extremely thorough and helpful. Thank you very much.
$endgroup$
– RandomWalker
Jan 19 at 22:12
$begingroup$
Extremely thorough and helpful. Thank you very much.
$endgroup$
– RandomWalker
Jan 19 at 22:12
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 19 at 22:14
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 19 at 22:14
$begingroup$
Follow up: is the subspace of adjoints complemented in $B(E^*)$?
$endgroup$
– RandomWalker
Jan 19 at 22:14
$begingroup$
Follow up: is the subspace of adjoints complemented in $B(E^*)$?
$endgroup$
– RandomWalker
Jan 19 at 22:14
$begingroup$
Interesting question. I can't say anything concrete, but I doubt it.
$endgroup$
– Aweygan
Jan 19 at 22:24
$begingroup$
Interesting question. I can't say anything concrete, but I doubt it.
$endgroup$
– Aweygan
Jan 19 at 22:24
add a comment |
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$begingroup$
Can you define what means to be an adjoint map?
$endgroup$
– alexp9
Jan 19 at 21:30
$begingroup$
@Rhcpy99 Presumbaly $T$ is an adjoint if $T=S^*$ for some $Sin B(E)$.
$endgroup$
– Aweygan
Jan 19 at 21:33
$begingroup$
@Aweygan Yes, that's what I meant. Sorry for the confusion
$endgroup$
– RandomWalker
Jan 19 at 21:34
$begingroup$
What is easy: if $E$ is reflexive, then every $Tin B(E^*)$ is an adjoint.
$endgroup$
– amsmath
Jan 19 at 21:55