Mixing
When are the topological duals $A^*$ and $B^*$ isomorphic?
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If I have to normed vector spaces $A$ and $B$, I was wondering when the topological duals are isomorphic (i.e. $A^* cong B^*)$ . Is it sufficient that $A cong B$? Or that $A$ has to be isometric to $B$?
If it is sufficient for $A$ isometric to $B$, given an isometry $f$, what would be the isomorphism mapping between the spaces $A^*$ and $B^*$?
functional-analysis dual-spaces
asked Jan 12 at 20:38
roi_saumonroi_saumon
56438
$endgroup$
add a comment |
$begingroup$
If I have to normed vector spaces $A$ and $B$, I was wondering when the topological duals are isomorphic (i.e. $A^* cong B^*)$ . Is it sufficient that $A cong B$? Or that $A$ has to be isometric to $B$?
If it is sufficient for $A$ isometric to $B$, given an isometry $f$, what would be the isomorphism mapping between the spaces $A^*$ and $B^*$?
functional-analysis dual-spaces
asked Jan 12 at 20:38
roi_saumonroi_saumon
56438
$endgroup$
add a comment |
$begingroup$
If I have to normed vector spaces $A$ and $B$, I was wondering when the topological duals are isomorphic (i.e. $A^* cong B^*)$ . Is it sufficient that $A cong B$? Or that $A$ has to be isometric to $B$?
If it is sufficient for $A$ isometric to $B$, given an isometry $f$, what would be the isomorphism mapping between the spaces $A^*$ and $B^*$?
functional-analysis dual-spaces
asked Jan 12 at 20:38
roi_saumonroi_saumon
56438
$endgroup$
If I have to normed vector spaces $A$ and $B$, I was wondering when the topological duals are isomorphic (i.e. $A^* cong B^*)$ . Is it sufficient that $A cong B$? Or that $A$ has to be isometric to $B$?
If it is sufficient for $A$ isometric to $B$, given an isometry $f$, what would be the isomorphism mapping between the spaces $A^*$ and $B^*$?
functional-analysis dual-spaces
functional-analysis dual-spaces
asked Jan 12 at 20:38
roi_saumonroi_saumon
56438
asked Jan 12 at 20:38
roi_saumonroi_saumon
56438
asked Jan 12 at 20:38
roi_saumonroi_saumon
56438
asked Jan 12 at 20:38
roi_saumonroi_saumon
56438
asked Jan 12 at 20:38
roi_saumonroi_saumon
56438
56438
add a comment |
add a comment |
1 Answer
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$begingroup$
You just need to have $A simeq B$.
Consider $f$ an isomorphism between $A$ and $B$. By definition, $f$ is linear and continuous and it turns out that linearity and continuity are preserved through composition. If you consider the application that associates to each linear $l$ form in $A^*$ the linear form $l circ f^{-1}$ in $B^*$ then you get an isomorphism from $A^*$ to $B^*$ (with inverse $l rightarrow l circ f$). (You can easily verify that this application between $A^*$ and $B^*$ is also linear and continuous.)
answered Jan 12 at 21:37
Paul CottalordaPaul Cottalorda
3865
$endgroup$
$begingroup$
Nice explanation, thanks!
$endgroup$
– roi_saumon
Jan 12 at 22:01
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You just need to have $A simeq B$.
Consider $f$ an isomorphism between $A$ and $B$. By definition, $f$ is linear and continuous and it turns out that linearity and continuity are preserved through composition. If you consider the application that associates to each linear $l$ form in $A^*$ the linear form $l circ f^{-1}$ in $B^*$ then you get an isomorphism from $A^*$ to $B^*$ (with inverse $l rightarrow l circ f$). (You can easily verify that this application between $A^*$ and $B^*$ is also linear and continuous.)
answered Jan 12 at 21:37
Paul CottalordaPaul Cottalorda
3865
$endgroup$
$begingroup$
Nice explanation, thanks!
$endgroup$
– roi_saumon
Jan 12 at 22:01
add a comment |
$begingroup$
You just need to have $A simeq B$.
Consider $f$ an isomorphism between $A$ and $B$. By definition, $f$ is linear and continuous and it turns out that linearity and continuity are preserved through composition. If you consider the application that associates to each linear $l$ form in $A^*$ the linear form $l circ f^{-1}$ in $B^*$ then you get an isomorphism from $A^*$ to $B^*$ (with inverse $l rightarrow l circ f$). (You can easily verify that this application between $A^*$ and $B^*$ is also linear and continuous.)
answered Jan 12 at 21:37
Paul CottalordaPaul Cottalorda
3865
$endgroup$
$begingroup$
Nice explanation, thanks!
$endgroup$
– roi_saumon
Jan 12 at 22:01
add a comment |
$begingroup$
You just need to have $A simeq B$.
Consider $f$ an isomorphism between $A$ and $B$. By definition, $f$ is linear and continuous and it turns out that linearity and continuity are preserved through composition. If you consider the application that associates to each linear $l$ form in $A^*$ the linear form $l circ f^{-1}$ in $B^*$ then you get an isomorphism from $A^*$ to $B^*$ (with inverse $l rightarrow l circ f$). (You can easily verify that this application between $A^*$ and $B^*$ is also linear and continuous.)
answered Jan 12 at 21:37
Paul CottalordaPaul Cottalorda
3865
$endgroup$
You just need to have $A simeq B$.
Consider $f$ an isomorphism between $A$ and $B$. By definition, $f$ is linear and continuous and it turns out that linearity and continuity are preserved through composition. If you consider the application that associates to each linear $l$ form in $A^*$ the linear form $l circ f^{-1}$ in $B^*$ then you get an isomorphism from $A^*$ to $B^*$ (with inverse $l rightarrow l circ f$). (You can easily verify that this application between $A^*$ and $B^*$ is also linear and continuous.)
answered Jan 12 at 21:37
Paul CottalordaPaul Cottalorda
3865
answered Jan 12 at 21:37
Paul CottalordaPaul Cottalorda
3865
answered Jan 12 at 21:37
Paul CottalordaPaul Cottalorda
3865
answered Jan 12 at 21:37
Paul CottalordaPaul Cottalorda
3865
3865
$begingroup$
Nice explanation, thanks!
$endgroup$
– roi_saumon
Jan 12 at 22:01
add a comment |
$begingroup$
Nice explanation, thanks!
$endgroup$
– roi_saumon
Jan 12 at 22:01
$begingroup$
Nice explanation, thanks!
$endgroup$
– roi_saumon
Jan 12 at 22:01
$begingroup$
Nice explanation, thanks!
$endgroup$
– roi_saumon
Jan 12 at 22:01
add a comment |
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