Adjoint of bounded linear map is isometric isomorphism implies original map is isometric isomorphism?
$begingroup$
Suppose $X$ and $Y$ are normed spaces. Let $T$ be a bounded linear map from $X$ to $Y$. Let $T^*$ be the adjoint map from $Y^{*}$ to $X^{*}$ defined by $T^{*}(y^*) = y^* T$.
A straightforward calculation shows:
Theorem 1. If $T$ is an isometric isomorphism from $X$ onto $Y$, then $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$.
I'm trying to prove the converse.
But the best I can get is the following. (It comes by applying the above theorem with $T^*$ in place of $T$ and using that $T^{**}$ extends $T$ [if $X$ is identified with a subspace of $X^{**}$ in the natural way]).
Theorem 2. If $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$, then $T$ is an isometric isomorphism from $X$ into $Y$ and $T(X)$ is dense in $Y$.
I cannot seem to strengthen the conclusion to $T$ is surjective.
If $X$ is complete, or, more generally, if $T(X)$ is closed in $Y$, then $T$ is surjective.
But what happens if $X$ is not complete or $T(X)$ is not closed?
In the discussion of the following question, the OP claims to be able to prove that $T^{∗}$ being an isomorphism implies $T$ is surjective, but I don't see how:
$T$ is surjective if and only if the adjoint $T^*$ is an isomorphism (onto its image)
There are also some Hilbert space examples in the following links, but they don't address what I am asking about:
$T$ surjective iff $T^*$ injective in infinite-dimensional Hilbert space?
Example: operator injective, then the adjoint is NOT surjective
functional-analysis analysis operator-theory
asked Dec 17 '18 at 1:50
MichaelGaudreauMichaelGaudreau
1488
$endgroup$
add a comment |
$begingroup$
Suppose $X$ and $Y$ are normed spaces. Let $T$ be a bounded linear map from $X$ to $Y$. Let $T^*$ be the adjoint map from $Y^{*}$ to $X^{*}$ defined by $T^{*}(y^*) = y^* T$.
A straightforward calculation shows:
Theorem 1. If $T$ is an isometric isomorphism from $X$ onto $Y$, then $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$.
I'm trying to prove the converse.
But the best I can get is the following. (It comes by applying the above theorem with $T^*$ in place of $T$ and using that $T^{**}$ extends $T$ [if $X$ is identified with a subspace of $X^{**}$ in the natural way]).
Theorem 2. If $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$, then $T$ is an isometric isomorphism from $X$ into $Y$ and $T(X)$ is dense in $Y$.
I cannot seem to strengthen the conclusion to $T$ is surjective.
If $X$ is complete, or, more generally, if $T(X)$ is closed in $Y$, then $T$ is surjective.
But what happens if $X$ is not complete or $T(X)$ is not closed?
In the discussion of the following question, the OP claims to be able to prove that $T^{∗}$ being an isomorphism implies $T$ is surjective, but I don't see how:
$T$ is surjective if and only if the adjoint $T^*$ is an isomorphism (onto its image)
There are also some Hilbert space examples in the following links, but they don't address what I am asking about:
$T$ surjective iff $T^*$ injective in infinite-dimensional Hilbert space?
Example: operator injective, then the adjoint is NOT surjective
functional-analysis analysis operator-theory
asked Dec 17 '18 at 1:50
MichaelGaudreauMichaelGaudreau
1488
$endgroup$
add a comment |
$begingroup$
Suppose $X$ and $Y$ are normed spaces. Let $T$ be a bounded linear map from $X$ to $Y$. Let $T^*$ be the adjoint map from $Y^{*}$ to $X^{*}$ defined by $T^{*}(y^*) = y^* T$.
A straightforward calculation shows:
Theorem 1. If $T$ is an isometric isomorphism from $X$ onto $Y$, then $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$.
I'm trying to prove the converse.
But the best I can get is the following. (It comes by applying the above theorem with $T^*$ in place of $T$ and using that $T^{**}$ extends $T$ [if $X$ is identified with a subspace of $X^{**}$ in the natural way]).
Theorem 2. If $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$, then $T$ is an isometric isomorphism from $X$ into $Y$ and $T(X)$ is dense in $Y$.
I cannot seem to strengthen the conclusion to $T$ is surjective.
If $X$ is complete, or, more generally, if $T(X)$ is closed in $Y$, then $T$ is surjective.
But what happens if $X$ is not complete or $T(X)$ is not closed?
In the discussion of the following question, the OP claims to be able to prove that $T^{∗}$ being an isomorphism implies $T$ is surjective, but I don't see how:
$T$ is surjective if and only if the adjoint $T^*$ is an isomorphism (onto its image)
There are also some Hilbert space examples in the following links, but they don't address what I am asking about:
$T$ surjective iff $T^*$ injective in infinite-dimensional Hilbert space?
Example: operator injective, then the adjoint is NOT surjective
functional-analysis analysis operator-theory
asked Dec 17 '18 at 1:50
MichaelGaudreauMichaelGaudreau
1488
$endgroup$
Suppose $X$ and $Y$ are normed spaces. Let $T$ be a bounded linear map from $X$ to $Y$. Let $T^*$ be the adjoint map from $Y^{*}$ to $X^{*}$ defined by $T^{*}(y^*) = y^* T$.
A straightforward calculation shows:
Theorem 1. If $T$ is an isometric isomorphism from $X$ onto $Y$, then $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$.
I'm trying to prove the converse.
But the best I can get is the following. (It comes by applying the above theorem with $T^*$ in place of $T$ and using that $T^{**}$ extends $T$ [if $X$ is identified with a subspace of $X^{**}$ in the natural way]).
Theorem 2. If $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$, then $T$ is an isometric isomorphism from $X$ into $Y$ and $T(X)$ is dense in $Y$.
I cannot seem to strengthen the conclusion to $T$ is surjective.
If $X$ is complete, or, more generally, if $T(X)$ is closed in $Y$, then $T$ is surjective.
But what happens if $X$ is not complete or $T(X)$ is not closed?
In the discussion of the following question, the OP claims to be able to prove that $T^{∗}$ being an isomorphism implies $T$ is surjective, but I don't see how:
$T$ is surjective if and only if the adjoint $T^*$ is an isomorphism (onto its image)
There are also some Hilbert space examples in the following links, but they don't address what I am asking about:
$T$ surjective iff $T^*$ injective in infinite-dimensional Hilbert space?
Example: operator injective, then the adjoint is NOT surjective
functional-analysis analysis operator-theory
functional-analysis analysis operator-theory
asked Dec 17 '18 at 1:50
MichaelGaudreauMichaelGaudreau
1488
asked Dec 17 '18 at 1:50
MichaelGaudreauMichaelGaudreau
1488
edited Dec 18 '18 at 20:12
MichaelGaudreau
asked Dec 17 '18 at 1:50
MichaelGaudreauMichaelGaudreau
1488
asked Dec 17 '18 at 1:50
MichaelGaudreauMichaelGaudreau
1488
asked Dec 17 '18 at 1:50
MichaelGaudreauMichaelGaudreau
1488
1488
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The statement does not hold unless you assume both spaces are Banach spaces. This is because the dual space is the dual space of a normed space is (naturally isomorphic to) the dual of its completion.
Indeed, let $X$ be an infinite-dimensional, not complete normed space, let $tilde X$ denote its completion, and let $T:Xto tilde X$ be the canonical inclusion. Then $T^*$ is an isometric isomorphism, but $T$ is not a surjective isometry.
answered Dec 17 '18 at 4:25
AweyganAweygan
14.5k21442
$endgroup$
$begingroup$
This example clearly shows that completeness of $X$ is necessary to get $T$ surjective in Theorem 2. But I am having trouble understanding why $Y$ needs to be complete. I am pretty sure Theorem 2 is true as stated. Suppose we add the hypothesis that $X$ is complete and $Y$ is not complete. Then $T(X)$ will be complete because of the conclusion that $T$ is an isometric isomorphism. Hence $overline{T(X)} = T(X)$. Combining this with the conclusion that $T(X)$ is dense in $Y$, we get that $Y = overline{T(X)} = T(X)$. But this will mean that $Y$ is complete. Contradiction. Continued in next...
$endgroup$
– MichaelGaudreau
Dec 18 '18 at 20:25
$begingroup$
So is this saying that when $X$ is complete and $Y$ is not complete, then $T^{*}:Y^{*} to X^{*}$ cannot be an isometric isomorphism? This is confusing because the duals $Y^{*}$ and $X^*$ cannot detect whether $X$ and $Y$ are complete. What is going on here?
$endgroup$
– MichaelGaudreau
Dec 18 '18 at 20:35
1
$begingroup$
I don't really know how to answer. This may help: even though $X^*$ and $Y^*$ can't detect whether or not $X$ and $Y$ are complete, the map $T^*:Y^*to X^*$ is constructed from the map $T:Xto Y$.
$endgroup$
– Aweygan
Dec 18 '18 at 23:52
$begingroup$
Yes, I've come to the same understanding. Thanks for your excellent example.
$endgroup$
– MichaelGaudreau
Dec 19 '18 at 1:33
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Dec 19 '18 at 1:35
add a comment |
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$begingroup$
The statement does not hold unless you assume both spaces are Banach spaces. This is because the dual space is the dual space of a normed space is (naturally isomorphic to) the dual of its completion.
Indeed, let $X$ be an infinite-dimensional, not complete normed space, let $tilde X$ denote its completion, and let $T:Xto tilde X$ be the canonical inclusion. Then $T^*$ is an isometric isomorphism, but $T$ is not a surjective isometry.
answered Dec 17 '18 at 4:25
AweyganAweygan
14.5k21442
$endgroup$
$begingroup$
This example clearly shows that completeness of $X$ is necessary to get $T$ surjective in Theorem 2. But I am having trouble understanding why $Y$ needs to be complete. I am pretty sure Theorem 2 is true as stated. Suppose we add the hypothesis that $X$ is complete and $Y$ is not complete. Then $T(X)$ will be complete because of the conclusion that $T$ is an isometric isomorphism. Hence $overline{T(X)} = T(X)$. Combining this with the conclusion that $T(X)$ is dense in $Y$, we get that $Y = overline{T(X)} = T(X)$. But this will mean that $Y$ is complete. Contradiction. Continued in next...
$endgroup$
– MichaelGaudreau
Dec 18 '18 at 20:25
$begingroup$
So is this saying that when $X$ is complete and $Y$ is not complete, then $T^{*}:Y^{*} to X^{*}$ cannot be an isometric isomorphism? This is confusing because the duals $Y^{*}$ and $X^*$ cannot detect whether $X$ and $Y$ are complete. What is going on here?
$endgroup$
– MichaelGaudreau
Dec 18 '18 at 20:35
1
$begingroup$
I don't really know how to answer. This may help: even though $X^*$ and $Y^*$ can't detect whether or not $X$ and $Y$ are complete, the map $T^*:Y^*to X^*$ is constructed from the map $T:Xto Y$.
$endgroup$
– Aweygan
Dec 18 '18 at 23:52
$begingroup$
Yes, I've come to the same understanding. Thanks for your excellent example.
$endgroup$
– MichaelGaudreau
Dec 19 '18 at 1:33
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Dec 19 '18 at 1:35
add a comment |
$begingroup$
The statement does not hold unless you assume both spaces are Banach spaces. This is because the dual space is the dual space of a normed space is (naturally isomorphic to) the dual of its completion.
Indeed, let $X$ be an infinite-dimensional, not complete normed space, let $tilde X$ denote its completion, and let $T:Xto tilde X$ be the canonical inclusion. Then $T^*$ is an isometric isomorphism, but $T$ is not a surjective isometry.
answered Dec 17 '18 at 4:25
AweyganAweygan
14.5k21442
$endgroup$
$begingroup$
This example clearly shows that completeness of $X$ is necessary to get $T$ surjective in Theorem 2. But I am having trouble understanding why $Y$ needs to be complete. I am pretty sure Theorem 2 is true as stated. Suppose we add the hypothesis that $X$ is complete and $Y$ is not complete. Then $T(X)$ will be complete because of the conclusion that $T$ is an isometric isomorphism. Hence $overline{T(X)} = T(X)$. Combining this with the conclusion that $T(X)$ is dense in $Y$, we get that $Y = overline{T(X)} = T(X)$. But this will mean that $Y$ is complete. Contradiction. Continued in next...
$endgroup$
– MichaelGaudreau
Dec 18 '18 at 20:25
$begingroup$
So is this saying that when $X$ is complete and $Y$ is not complete, then $T^{*}:Y^{*} to X^{*}$ cannot be an isometric isomorphism? This is confusing because the duals $Y^{*}$ and $X^*$ cannot detect whether $X$ and $Y$ are complete. What is going on here?
$endgroup$
– MichaelGaudreau
Dec 18 '18 at 20:35
1
$begingroup$
I don't really know how to answer. This may help: even though $X^*$ and $Y^*$ can't detect whether or not $X$ and $Y$ are complete, the map $T^*:Y^*to X^*$ is constructed from the map $T:Xto Y$.
$endgroup$
– Aweygan
Dec 18 '18 at 23:52
$begingroup$
Yes, I've come to the same understanding. Thanks for your excellent example.
$endgroup$
– MichaelGaudreau
Dec 19 '18 at 1:33
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Dec 19 '18 at 1:35
add a comment |
$begingroup$
The statement does not hold unless you assume both spaces are Banach spaces. This is because the dual space is the dual space of a normed space is (naturally isomorphic to) the dual of its completion.
Indeed, let $X$ be an infinite-dimensional, not complete normed space, let $tilde X$ denote its completion, and let $T:Xto tilde X$ be the canonical inclusion. Then $T^*$ is an isometric isomorphism, but $T$ is not a surjective isometry.
answered Dec 17 '18 at 4:25
AweyganAweygan
14.5k21442
$endgroup$
The statement does not hold unless you assume both spaces are Banach spaces. This is because the dual space is the dual space of a normed space is (naturally isomorphic to) the dual of its completion.
Indeed, let $X$ be an infinite-dimensional, not complete normed space, let $tilde X$ denote its completion, and let $T:Xto tilde X$ be the canonical inclusion. Then $T^*$ is an isometric isomorphism, but $T$ is not a surjective isometry.
answered Dec 17 '18 at 4:25
AweyganAweygan
14.5k21442
answered Dec 17 '18 at 4:25
AweyganAweygan
14.5k21442
answered Dec 17 '18 at 4:25
AweyganAweygan
14.5k21442
answered Dec 17 '18 at 4:25
AweyganAweygan
14.5k21442
14.5k21442
$begingroup$
This example clearly shows that completeness of $X$ is necessary to get $T$ surjective in Theorem 2. But I am having trouble understanding why $Y$ needs to be complete. I am pretty sure Theorem 2 is true as stated. Suppose we add the hypothesis that $X$ is complete and $Y$ is not complete. Then $T(X)$ will be complete because of the conclusion that $T$ is an isometric isomorphism. Hence $overline{T(X)} = T(X)$. Combining this with the conclusion that $T(X)$ is dense in $Y$, we get that $Y = overline{T(X)} = T(X)$. But this will mean that $Y$ is complete. Contradiction. Continued in next...
$endgroup$
– MichaelGaudreau
Dec 18 '18 at 20:25
$begingroup$
So is this saying that when $X$ is complete and $Y$ is not complete, then $T^{*}:Y^{*} to X^{*}$ cannot be an isometric isomorphism? This is confusing because the duals $Y^{*}$ and $X^*$ cannot detect whether $X$ and $Y$ are complete. What is going on here?
$endgroup$
– MichaelGaudreau
Dec 18 '18 at 20:35
1
$begingroup$
I don't really know how to answer. This may help: even though $X^*$ and $Y^*$ can't detect whether or not $X$ and $Y$ are complete, the map $T^*:Y^*to X^*$ is constructed from the map $T:Xto Y$.
$endgroup$
– Aweygan
Dec 18 '18 at 23:52
$begingroup$
Yes, I've come to the same understanding. Thanks for your excellent example.
$endgroup$
– MichaelGaudreau
Dec 19 '18 at 1:33
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Dec 19 '18 at 1:35
add a comment |
$begingroup$
This example clearly shows that completeness of $X$ is necessary to get $T$ surjective in Theorem 2. But I am having trouble understanding why $Y$ needs to be complete. I am pretty sure Theorem 2 is true as stated. Suppose we add the hypothesis that $X$ is complete and $Y$ is not complete. Then $T(X)$ will be complete because of the conclusion that $T$ is an isometric isomorphism. Hence $overline{T(X)} = T(X)$. Combining this with the conclusion that $T(X)$ is dense in $Y$, we get that $Y = overline{T(X)} = T(X)$. But this will mean that $Y$ is complete. Contradiction. Continued in next...
$endgroup$
– MichaelGaudreau
Dec 18 '18 at 20:25
$begingroup$
So is this saying that when $X$ is complete and $Y$ is not complete, then $T^{*}:Y^{*} to X^{*}$ cannot be an isometric isomorphism? This is confusing because the duals $Y^{*}$ and $X^*$ cannot detect whether $X$ and $Y$ are complete. What is going on here?
$endgroup$
– MichaelGaudreau
Dec 18 '18 at 20:35
1
$begingroup$
I don't really know how to answer. This may help: even though $X^*$ and $Y^*$ can't detect whether or not $X$ and $Y$ are complete, the map $T^*:Y^*to X^*$ is constructed from the map $T:Xto Y$.
$endgroup$
– Aweygan
Dec 18 '18 at 23:52
$begingroup$
Yes, I've come to the same understanding. Thanks for your excellent example.
$endgroup$
– MichaelGaudreau
Dec 19 '18 at 1:33
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Dec 19 '18 at 1:35
$begingroup$
This example clearly shows that completeness of $X$ is necessary to get $T$ surjective in Theorem 2. But I am having trouble understanding why $Y$ needs to be complete. I am pretty sure Theorem 2 is true as stated. Suppose we add the hypothesis that $X$ is complete and $Y$ is not complete. Then $T(X)$ will be complete because of the conclusion that $T$ is an isometric isomorphism. Hence $overline{T(X)} = T(X)$. Combining this with the conclusion that $T(X)$ is dense in $Y$, we get that $Y = overline{T(X)} = T(X)$. But this will mean that $Y$ is complete. Contradiction. Continued in next...
$endgroup$
– MichaelGaudreau
Dec 18 '18 at 20:25
$begingroup$
This example clearly shows that completeness of $X$ is necessary to get $T$ surjective in Theorem 2. But I am having trouble understanding why $Y$ needs to be complete. I am pretty sure Theorem 2 is true as stated. Suppose we add the hypothesis that $X$ is complete and $Y$ is not complete. Then $T(X)$ will be complete because of the conclusion that $T$ is an isometric isomorphism. Hence $overline{T(X)} = T(X)$. Combining this with the conclusion that $T(X)$ is dense in $Y$, we get that $Y = overline{T(X)} = T(X)$. But this will mean that $Y$ is complete. Contradiction. Continued in next...
$endgroup$
– MichaelGaudreau
Dec 18 '18 at 20:25
$begingroup$
So is this saying that when $X$ is complete and $Y$ is not complete, then $T^{*}:Y^{*} to X^{*}$ cannot be an isometric isomorphism? This is confusing because the duals $Y^{*}$ and $X^*$ cannot detect whether $X$ and $Y$ are complete. What is going on here?
$endgroup$
– MichaelGaudreau
Dec 18 '18 at 20:35
$begingroup$
So is this saying that when $X$ is complete and $Y$ is not complete, then $T^{*}:Y^{*} to X^{*}$ cannot be an isometric isomorphism? This is confusing because the duals $Y^{*}$ and $X^*$ cannot detect whether $X$ and $Y$ are complete. What is going on here?
$endgroup$
– MichaelGaudreau
Dec 18 '18 at 20:35
1
1
$begingroup$
I don't really know how to answer. This may help: even though $X^*$ and $Y^*$ can't detect whether or not $X$ and $Y$ are complete, the map $T^*:Y^*to X^*$ is constructed from the map $T:Xto Y$.
$endgroup$
– Aweygan
Dec 18 '18 at 23:52
$begingroup$
I don't really know how to answer. This may help: even though $X^*$ and $Y^*$ can't detect whether or not $X$ and $Y$ are complete, the map $T^*:Y^*to X^*$ is constructed from the map $T:Xto Y$.
$endgroup$
– Aweygan
Dec 18 '18 at 23:52
$begingroup$
Yes, I've come to the same understanding. Thanks for your excellent example.
$endgroup$
– MichaelGaudreau
Dec 19 '18 at 1:33
$begingroup$
Yes, I've come to the same understanding. Thanks for your excellent example.
$endgroup$
– MichaelGaudreau
Dec 19 '18 at 1:33
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Dec 19 '18 at 1:35
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Dec 19 '18 at 1:35
add a comment |
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