Mixing
Is the natural embedding from $X$ to $X^{**}$ a homeomorphism with respect to the weak topologies if $X$ is...
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If $X$ is a reflexive Banach space, is it true that the natural embedding $Lambda$ of $X$ into its double dual is a homeomorphism? Here we equip $X$ with the weak topology, and $X^{**}$ with the weak-* topology, i.e. $sigma(X^{**},X^*)$. I think that this is true. To start with I have shown that it is at least continuous by using nets. I also wanted to show that the inverse is continuous (but I could not figure this out using nets), but here's how far I got: So assume that the net $f_a rightarrow f$ weakly (here the $f_a$ lie in $X^{**}$). We want to show $Lambda^{-1}(f_a) rightarrow Lambda^{-1}(f)$ weakly. Note that $f_a rightarrow f$ weakly is equivalent to $f_a(x) rightarrow f(x)$ for all $x in X^{*}$. And that the conclusion is equivalent to $x(Lambda^{-1}(f_a)) rightarrow x(Lambda^{-1}(f))$ weakly for all $x in X^*$. This is where I'm getting a bit confused: Is $x(Lambda^{-1}(f_a))=f_a(x)$?
functional-analysis banach-spaces
asked Jan 6 at 15:23
VercingetorixVercingetorix
285
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add a comment |
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If $X$ is a reflexive Banach space, is it true that the natural embedding $Lambda$ of $X$ into its double dual is a homeomorphism? Here we equip $X$ with the weak topology, and $X^{**}$ with the weak-* topology, i.e. $sigma(X^{**},X^*)$. I think that this is true. To start with I have shown that it is at least continuous by using nets. I also wanted to show that the inverse is continuous (but I could not figure this out using nets), but here's how far I got: So assume that the net $f_a rightarrow f$ weakly (here the $f_a$ lie in $X^{**}$). We want to show $Lambda^{-1}(f_a) rightarrow Lambda^{-1}(f)$ weakly. Note that $f_a rightarrow f$ weakly is equivalent to $f_a(x) rightarrow f(x)$ for all $x in X^{*}$. And that the conclusion is equivalent to $x(Lambda^{-1}(f_a)) rightarrow x(Lambda^{-1}(f))$ weakly for all $x in X^*$. This is where I'm getting a bit confused: Is $x(Lambda^{-1}(f_a))=f_a(x)$?
functional-analysis banach-spaces
asked Jan 6 at 15:23
VercingetorixVercingetorix
285
$endgroup$
add a comment |
$begingroup$
If $X$ is a reflexive Banach space, is it true that the natural embedding $Lambda$ of $X$ into its double dual is a homeomorphism? Here we equip $X$ with the weak topology, and $X^{**}$ with the weak-* topology, i.e. $sigma(X^{**},X^*)$. I think that this is true. To start with I have shown that it is at least continuous by using nets. I also wanted to show that the inverse is continuous (but I could not figure this out using nets), but here's how far I got: So assume that the net $f_a rightarrow f$ weakly (here the $f_a$ lie in $X^{**}$). We want to show $Lambda^{-1}(f_a) rightarrow Lambda^{-1}(f)$ weakly. Note that $f_a rightarrow f$ weakly is equivalent to $f_a(x) rightarrow f(x)$ for all $x in X^{*}$. And that the conclusion is equivalent to $x(Lambda^{-1}(f_a)) rightarrow x(Lambda^{-1}(f))$ weakly for all $x in X^*$. This is where I'm getting a bit confused: Is $x(Lambda^{-1}(f_a))=f_a(x)$?
functional-analysis banach-spaces
asked Jan 6 at 15:23
VercingetorixVercingetorix
285
$endgroup$
If $X$ is a reflexive Banach space, is it true that the natural embedding $Lambda$ of $X$ into its double dual is a homeomorphism? Here we equip $X$ with the weak topology, and $X^{**}$ with the weak-* topology, i.e. $sigma(X^{**},X^*)$. I think that this is true. To start with I have shown that it is at least continuous by using nets. I also wanted to show that the inverse is continuous (but I could not figure this out using nets), but here's how far I got: So assume that the net $f_a rightarrow f$ weakly (here the $f_a$ lie in $X^{**}$). We want to show $Lambda^{-1}(f_a) rightarrow Lambda^{-1}(f)$ weakly. Note that $f_a rightarrow f$ weakly is equivalent to $f_a(x) rightarrow f(x)$ for all $x in X^{*}$. And that the conclusion is equivalent to $x(Lambda^{-1}(f_a)) rightarrow x(Lambda^{-1}(f))$ weakly for all $x in X^*$. This is where I'm getting a bit confused: Is $x(Lambda^{-1}(f_a))=f_a(x)$?
functional-analysis banach-spaces
functional-analysis banach-spaces
asked Jan 6 at 15:23
VercingetorixVercingetorix
285
asked Jan 6 at 15:23
VercingetorixVercingetorix
285
asked Jan 6 at 15:23
VercingetorixVercingetorix
285
asked Jan 6 at 15:23
VercingetorixVercingetorix
285
asked Jan 6 at 15:23
VercingetorixVercingetorix
285
285
add a comment |
add a comment |
1 Answer
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Yes, but don't use nets, as that is very cumbersome. Since $X$ is isometric to $X^{**}$ under the canonical embedding, that means they are isomorphic under the weak topologies. But since $X^{**}$ is reflexive, its weak and weak-* topologies coincide.
answered Jan 6 at 15:47
Ben WBen W
2,234615
$endgroup$
$begingroup$
Why must they be isomorphic under the weak topologies if $X$ is isometric to $X^{**}$ under the canonical embedding?
$endgroup$
– Vercingetorix
Jan 6 at 16:01
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The weak topologies are generated by the respective norm topologies. Since the latter are identical, so are the former.
$endgroup$
– Ben W
Jan 6 at 16:04
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See here for more info: math.stackexchange.com/questions/900903/…
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– Ben W
Jan 6 at 16:06
add a comment |
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1 Answer
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1 Answer
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Yes, but don't use nets, as that is very cumbersome. Since $X$ is isometric to $X^{**}$ under the canonical embedding, that means they are isomorphic under the weak topologies. But since $X^{**}$ is reflexive, its weak and weak-* topologies coincide.
answered Jan 6 at 15:47
Ben WBen W
2,234615
$endgroup$
$begingroup$
Why must they be isomorphic under the weak topologies if $X$ is isometric to $X^{**}$ under the canonical embedding?
$endgroup$
– Vercingetorix
Jan 6 at 16:01
$begingroup$
The weak topologies are generated by the respective norm topologies. Since the latter are identical, so are the former.
$endgroup$
– Ben W
Jan 6 at 16:04
$begingroup$
See here for more info: math.stackexchange.com/questions/900903/…
$endgroup$
– Ben W
Jan 6 at 16:06
add a comment |
$begingroup$
Yes, but don't use nets, as that is very cumbersome. Since $X$ is isometric to $X^{**}$ under the canonical embedding, that means they are isomorphic under the weak topologies. But since $X^{**}$ is reflexive, its weak and weak-* topologies coincide.
answered Jan 6 at 15:47
Ben WBen W
2,234615
$endgroup$
$begingroup$
Why must they be isomorphic under the weak topologies if $X$ is isometric to $X^{**}$ under the canonical embedding?
$endgroup$
– Vercingetorix
Jan 6 at 16:01
$begingroup$
The weak topologies are generated by the respective norm topologies. Since the latter are identical, so are the former.
$endgroup$
– Ben W
Jan 6 at 16:04
$begingroup$
See here for more info: math.stackexchange.com/questions/900903/…
$endgroup$
– Ben W
Jan 6 at 16:06
add a comment |
$begingroup$
Yes, but don't use nets, as that is very cumbersome. Since $X$ is isometric to $X^{**}$ under the canonical embedding, that means they are isomorphic under the weak topologies. But since $X^{**}$ is reflexive, its weak and weak-* topologies coincide.
answered Jan 6 at 15:47
Ben WBen W
2,234615
$endgroup$
Yes, but don't use nets, as that is very cumbersome. Since $X$ is isometric to $X^{**}$ under the canonical embedding, that means they are isomorphic under the weak topologies. But since $X^{**}$ is reflexive, its weak and weak-* topologies coincide.
answered Jan 6 at 15:47
Ben WBen W
2,234615
answered Jan 6 at 15:47
Ben WBen W
2,234615
answered Jan 6 at 15:47
Ben WBen W
2,234615
answered Jan 6 at 15:47
Ben WBen W
2,234615
2,234615
$begingroup$
Why must they be isomorphic under the weak topologies if $X$ is isometric to $X^{**}$ under the canonical embedding?
$endgroup$
– Vercingetorix
Jan 6 at 16:01
$begingroup$
The weak topologies are generated by the respective norm topologies. Since the latter are identical, so are the former.
$endgroup$
– Ben W
Jan 6 at 16:04
$begingroup$
See here for more info: math.stackexchange.com/questions/900903/…
$endgroup$
– Ben W
Jan 6 at 16:06
add a comment |
$begingroup$
Why must they be isomorphic under the weak topologies if $X$ is isometric to $X^{**}$ under the canonical embedding?
$endgroup$
– Vercingetorix
Jan 6 at 16:01
$begingroup$
The weak topologies are generated by the respective norm topologies. Since the latter are identical, so are the former.
$endgroup$
– Ben W
Jan 6 at 16:04
$begingroup$
See here for more info: math.stackexchange.com/questions/900903/…
$endgroup$
– Ben W
Jan 6 at 16:06
$begingroup$
Why must they be isomorphic under the weak topologies if $X$ is isometric to $X^{**}$ under the canonical embedding?
$endgroup$
– Vercingetorix
Jan 6 at 16:01
$begingroup$
Why must they be isomorphic under the weak topologies if $X$ is isometric to $X^{**}$ under the canonical embedding?
$endgroup$
– Vercingetorix
Jan 6 at 16:01
$begingroup$
The weak topologies are generated by the respective norm topologies. Since the latter are identical, so are the former.
$endgroup$
– Ben W
Jan 6 at 16:04
$begingroup$
The weak topologies are generated by the respective norm topologies. Since the latter are identical, so are the former.
$endgroup$
– Ben W
Jan 6 at 16:04
$begingroup$
See here for more info: math.stackexchange.com/questions/900903/…
$endgroup$
– Ben W
Jan 6 at 16:06
$begingroup$
See here for more info: math.stackexchange.com/questions/900903/…
$endgroup$
– Ben W
Jan 6 at 16:06
add a comment |
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