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Is injective co-isometry an isometry?
Clash Royale CLAN TAG#URR8PPP
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Let $E$ be a normed space. Let $T:Eto E$ be an injective continuous linear map, such that $T^*$ is an isometry. Does it follows that $T$ itself is an isometry (in fact it is then an isometric isomorphism)?
This is true if $E$ is reflexive. Indeed, if $T$ is injective, $T^*$ has a dense range. An isometry with a dense range must be an isometric isomorphism, and so $T=T^**$ is also an isometric isomorphism. However, if $E$ is not reflexive we cannot conclude that $T^*$ has a dense image, only weak* dense, and so I expect that there is a counterexample.
functional-analysis banach-spaces
asked Jul 25 at 2:42
erz
433212
add a comment |Â
up vote
3
down vote
favorite
Let $E$ be a normed space. Let $T:Eto E$ be an injective continuous linear map, such that $T^*$ is an isometry. Does it follows that $T$ itself is an isometry (in fact it is then an isometric isomorphism)?
This is true if $E$ is reflexive. Indeed, if $T$ is injective, $T^*$ has a dense range. An isometry with a dense range must be an isometric isomorphism, and so $T=T^**$ is also an isometric isomorphism. However, if $E$ is not reflexive we cannot conclude that $T^*$ has a dense image, only weak* dense, and so I expect that there is a counterexample.
functional-analysis banach-spaces
asked Jul 25 at 2:42
erz
433212
1
You should look at this question
– Norbert
Jul 27 at 10:01
@Norbert my question follows from yours. Thank you!
– erz
Jul 29 at 4:13
Then you should post your answer. It is ok to do so.
– Norbert
Jul 29 at 13:02
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $E$ be a normed space. Let $T:Eto E$ be an injective continuous linear map, such that $T^*$ is an isometry. Does it follows that $T$ itself is an isometry (in fact it is then an isometric isomorphism)?
This is true if $E$ is reflexive. Indeed, if $T$ is injective, $T^*$ has a dense range. An isometry with a dense range must be an isometric isomorphism, and so $T=T^**$ is also an isometric isomorphism. However, if $E$ is not reflexive we cannot conclude that $T^*$ has a dense image, only weak* dense, and so I expect that there is a counterexample.
functional-analysis banach-spaces
asked Jul 25 at 2:42
erz
433212
Let $E$ be a normed space. Let $T:Eto E$ be an injective continuous linear map, such that $T^*$ is an isometry. Does it follows that $T$ itself is an isometry (in fact it is then an isometric isomorphism)?
This is true if $E$ is reflexive. Indeed, if $T$ is injective, $T^*$ has a dense range. An isometry with a dense range must be an isometric isomorphism, and so $T=T^**$ is also an isometric isomorphism. However, if $E$ is not reflexive we cannot conclude that $T^*$ has a dense image, only weak* dense, and so I expect that there is a counterexample.
functional-analysis banach-spaces
asked Jul 25 at 2:42
erz
433212
edited Jul 25 at 3:04
asked Jul 25 at 2:42
erz
433212
asked Jul 25 at 2:42
erz
433212
asked Jul 25 at 2:42
erz
433212
433212
1
You should look at this question
– Norbert
Jul 27 at 10:01
@Norbert my question follows from yours. Thank you!
– erz
Jul 29 at 4:13
Then you should post your answer. It is ok to do so.
– Norbert
Jul 29 at 13:02
add a comment |Â
1
You should look at this question
– Norbert
Jul 27 at 10:01
@Norbert my question follows from yours. Thank you!
– erz
Jul 29 at 4:13
Then you should post your answer. It is ok to do so.
– Norbert
Jul 29 at 13:02
1
1
You should look at this question
– Norbert
Jul 27 at 10:01
You should look at this question
– Norbert
Jul 27 at 10:01
@Norbert my question follows from yours. Thank you!
– erz
Jul 29 at 4:13
@Norbert my question follows from yours. Thank you!
– erz
Jul 29 at 4:13
Then you should post your answer. It is ok to do so.
– Norbert
Jul 29 at 13:02
Then you should post your answer. It is ok to do so.
– Norbert
Jul 29 at 13:02
add a comment |Â
1 Answer
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1
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Edit: This answer was for an older version of the question where $T$ was not assumed injective.
Use $E = c_0$ with sup norm and $T$ as the left-shift map, so $E^* = ell^1$. Then, if $y=(y_1, y_2, dots) in ell^1$, we have $T^*(y) = (0,y_1,y_2, dotsc)$ which is clearly an isometry on $ell^1$, but $T$ is not an isometry.
answered Jul 25 at 2:57
B. Mehta
11.7k21944
I am very sorry, I forgot to mention the most crucial part that $T$ is assumed to be injective. I wrote this in the second half of the question, and in the title, but failed to state the question itself correctly.
– erz
Jul 25 at 3:03
Ah, thank you for the correction.
– B. Mehta
Jul 25 at 3:05
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Edit: This answer was for an older version of the question where $T$ was not assumed injective.
Use $E = c_0$ with sup norm and $T$ as the left-shift map, so $E^* = ell^1$. Then, if $y=(y_1, y_2, dots) in ell^1$, we have $T^*(y) = (0,y_1,y_2, dotsc)$ which is clearly an isometry on $ell^1$, but $T$ is not an isometry.
answered Jul 25 at 2:57
B. Mehta
11.7k21944
I am very sorry, I forgot to mention the most crucial part that $T$ is assumed to be injective. I wrote this in the second half of the question, and in the title, but failed to state the question itself correctly.
– erz
Jul 25 at 3:03
Ah, thank you for the correction.
– B. Mehta
Jul 25 at 3:05
add a comment |Â
up vote
1
down vote
Edit: This answer was for an older version of the question where $T$ was not assumed injective.
Use $E = c_0$ with sup norm and $T$ as the left-shift map, so $E^* = ell^1$. Then, if $y=(y_1, y_2, dots) in ell^1$, we have $T^*(y) = (0,y_1,y_2, dotsc)$ which is clearly an isometry on $ell^1$, but $T$ is not an isometry.
answered Jul 25 at 2:57
B. Mehta
11.7k21944
I am very sorry, I forgot to mention the most crucial part that $T$ is assumed to be injective. I wrote this in the second half of the question, and in the title, but failed to state the question itself correctly.
– erz
Jul 25 at 3:03
Ah, thank you for the correction.
– B. Mehta
Jul 25 at 3:05
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Edit: This answer was for an older version of the question where $T$ was not assumed injective.
Use $E = c_0$ with sup norm and $T$ as the left-shift map, so $E^* = ell^1$. Then, if $y=(y_1, y_2, dots) in ell^1$, we have $T^*(y) = (0,y_1,y_2, dotsc)$ which is clearly an isometry on $ell^1$, but $T$ is not an isometry.
answered Jul 25 at 2:57
B. Mehta
11.7k21944
Edit: This answer was for an older version of the question where $T$ was not assumed injective.
Use $E = c_0$ with sup norm and $T$ as the left-shift map, so $E^* = ell^1$. Then, if $y=(y_1, y_2, dots) in ell^1$, we have $T^*(y) = (0,y_1,y_2, dotsc)$ which is clearly an isometry on $ell^1$, but $T$ is not an isometry.
answered Jul 25 at 2:57
B. Mehta
11.7k21944
edited Jul 25 at 3:05
answered Jul 25 at 2:57
B. Mehta
11.7k21944
answered Jul 25 at 2:57
B. Mehta
11.7k21944
answered Jul 25 at 2:57
B. Mehta
11.7k21944
11.7k21944
I am very sorry, I forgot to mention the most crucial part that $T$ is assumed to be injective. I wrote this in the second half of the question, and in the title, but failed to state the question itself correctly.
– erz
Jul 25 at 3:03
Ah, thank you for the correction.
– B. Mehta
Jul 25 at 3:05
add a comment |Â
I am very sorry, I forgot to mention the most crucial part that $T$ is assumed to be injective. I wrote this in the second half of the question, and in the title, but failed to state the question itself correctly.
– erz
Jul 25 at 3:03
Ah, thank you for the correction.
– B. Mehta
Jul 25 at 3:05
I am very sorry, I forgot to mention the most crucial part that $T$ is assumed to be injective. I wrote this in the second half of the question, and in the title, but failed to state the question itself correctly.
– erz
Jul 25 at 3:03
I am very sorry, I forgot to mention the most crucial part that $T$ is assumed to be injective. I wrote this in the second half of the question, and in the title, but failed to state the question itself correctly.
– erz
Jul 25 at 3:03
Ah, thank you for the correction.
– B. Mehta
Jul 25 at 3:05
Ah, thank you for the correction.
– B. Mehta
Jul 25 at 3:05
add a comment |Â
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1
You should look at this question
– Norbert
Jul 27 at 10:01
@Norbert my question follows from yours. Thank you!
– erz
Jul 29 at 4:13
Then you should post your answer. It is ok to do so.
– Norbert
Jul 29 at 13:02