Is every Banach space embedded in a reflexive space?Is every Banach space densely embedded in a Hilbert space?reflexive Banach spaceIs there a non-reflexive Banach space with every proper closed subspace reflexive?A Banach space is reflexive if a closed subspace and its quotient space are both reflexiveClosed subspace of a reflexive Banach space is reflexiveEquivalent characterizations for Reflexive spacesIs any closed and bounded subset of a reflexive Banach space compact in the weak topology?Is every Banach space densely embedded in a Hilbert space?Is a reflexive space necessarily an L-embedded space?Reflexive closure of Banach spaceDoes the norm $mathcalH := | w(x,y)log^+ (w(x,y)) |_L^1$ define a reflexive Banach space?
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Is every Banach space embedded in a reflexive space?
Is every Banach space densely embedded in a Hilbert space?reflexive Banach spaceIs there a non-reflexive Banach space with every proper closed subspace reflexive?A Banach space is reflexive if a closed subspace and its quotient space are both reflexiveClosed subspace of a reflexive Banach space is reflexiveEquivalent characterizations for Reflexive spacesIs any closed and bounded subset of a reflexive Banach space compact in the weak topology?Is every Banach space densely embedded in a Hilbert space?Is a reflexive space necessarily an L-embedded space?Reflexive closure of Banach spaceDoes the norm $mathcalH := | w(x,y)log^+ (w(x,y)) |_L^1$ define a reflexive Banach space?
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Can every Banach space be continuously embedded in a reflexive Banach space? As for example for $ L^infty(Omega)$ and $L^2(Omega)$, where $ Omegasubset R^n$ is a bounded open set.
functional-analysis banach-spaces reflexive-space
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|
show 1 more comment
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Can every Banach space be continuously embedded in a reflexive Banach space? As for example for $ L^infty(Omega)$ and $L^2(Omega)$, where $ Omegasubset R^n$ is a bounded open set.
functional-analysis banach-spaces reflexive-space
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The second example you ask about is a Hilbert space, and is therefore a reflexive Banach space.
$endgroup$
– Keen-ameteur
Mar 12 at 10:43
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I think you can modify Tomek's answer here to show that the answer is "no" (reflexive spaces have equivalent strictly convex renormings).
$endgroup$
– David Mitra
Mar 12 at 12:01
$begingroup$
What is "embedding"? So, for example, is the set-theoretic inclusion $L^infty[0,1] to L^2[0,1]$ a "continuous embedding"? You do not require the image to be closed?
$endgroup$
– GEdgar
Mar 12 at 12:49
$begingroup$
By the continuous embedding, I mean that : $|y|_L^2(Omega)le C |y|_L^infty(Omega), forall yin L^infty(Omega) $ for some constant $C>0,$ without any constraint on the image.
$endgroup$
– Rabat
Mar 12 at 14:38
$begingroup$
I have seen the Tomek's answer. But the strict convexity property is not preserved by equivalence of norm.
$endgroup$
– Rabat
Mar 12 at 14:49
|
show 1 more comment
$begingroup$
Can every Banach space be continuously embedded in a reflexive Banach space? As for example for $ L^infty(Omega)$ and $L^2(Omega)$, where $ Omegasubset R^n$ is a bounded open set.
functional-analysis banach-spaces reflexive-space
$endgroup$
Can every Banach space be continuously embedded in a reflexive Banach space? As for example for $ L^infty(Omega)$ and $L^2(Omega)$, where $ Omegasubset R^n$ is a bounded open set.
functional-analysis banach-spaces reflexive-space
functional-analysis banach-spaces reflexive-space
asked Mar 12 at 10:29
RabatRabat
505
505
$begingroup$
The second example you ask about is a Hilbert space, and is therefore a reflexive Banach space.
$endgroup$
– Keen-ameteur
Mar 12 at 10:43
$begingroup$
I think you can modify Tomek's answer here to show that the answer is "no" (reflexive spaces have equivalent strictly convex renormings).
$endgroup$
– David Mitra
Mar 12 at 12:01
$begingroup$
What is "embedding"? So, for example, is the set-theoretic inclusion $L^infty[0,1] to L^2[0,1]$ a "continuous embedding"? You do not require the image to be closed?
$endgroup$
– GEdgar
Mar 12 at 12:49
$begingroup$
By the continuous embedding, I mean that : $|y|_L^2(Omega)le C |y|_L^infty(Omega), forall yin L^infty(Omega) $ for some constant $C>0,$ without any constraint on the image.
$endgroup$
– Rabat
Mar 12 at 14:38
$begingroup$
I have seen the Tomek's answer. But the strict convexity property is not preserved by equivalence of norm.
$endgroup$
– Rabat
Mar 12 at 14:49
|
show 1 more comment
$begingroup$
The second example you ask about is a Hilbert space, and is therefore a reflexive Banach space.
$endgroup$
– Keen-ameteur
Mar 12 at 10:43
$begingroup$
I think you can modify Tomek's answer here to show that the answer is "no" (reflexive spaces have equivalent strictly convex renormings).
$endgroup$
– David Mitra
Mar 12 at 12:01
$begingroup$
What is "embedding"? So, for example, is the set-theoretic inclusion $L^infty[0,1] to L^2[0,1]$ a "continuous embedding"? You do not require the image to be closed?
$endgroup$
– GEdgar
Mar 12 at 12:49
$begingroup$
By the continuous embedding, I mean that : $|y|_L^2(Omega)le C |y|_L^infty(Omega), forall yin L^infty(Omega) $ for some constant $C>0,$ without any constraint on the image.
$endgroup$
– Rabat
Mar 12 at 14:38
$begingroup$
I have seen the Tomek's answer. But the strict convexity property is not preserved by equivalence of norm.
$endgroup$
– Rabat
Mar 12 at 14:49
$begingroup$
The second example you ask about is a Hilbert space, and is therefore a reflexive Banach space.
$endgroup$
– Keen-ameteur
Mar 12 at 10:43
$begingroup$
The second example you ask about is a Hilbert space, and is therefore a reflexive Banach space.
$endgroup$
– Keen-ameteur
Mar 12 at 10:43
$begingroup$
I think you can modify Tomek's answer here to show that the answer is "no" (reflexive spaces have equivalent strictly convex renormings).
$endgroup$
– David Mitra
Mar 12 at 12:01
$begingroup$
I think you can modify Tomek's answer here to show that the answer is "no" (reflexive spaces have equivalent strictly convex renormings).
$endgroup$
– David Mitra
Mar 12 at 12:01
$begingroup$
What is "embedding"? So, for example, is the set-theoretic inclusion $L^infty[0,1] to L^2[0,1]$ a "continuous embedding"? You do not require the image to be closed?
$endgroup$
– GEdgar
Mar 12 at 12:49
$begingroup$
What is "embedding"? So, for example, is the set-theoretic inclusion $L^infty[0,1] to L^2[0,1]$ a "continuous embedding"? You do not require the image to be closed?
$endgroup$
– GEdgar
Mar 12 at 12:49
$begingroup$
By the continuous embedding, I mean that : $|y|_L^2(Omega)le C |y|_L^infty(Omega), forall yin L^infty(Omega) $ for some constant $C>0,$ without any constraint on the image.
$endgroup$
– Rabat
Mar 12 at 14:38
$begingroup$
By the continuous embedding, I mean that : $|y|_L^2(Omega)le C |y|_L^infty(Omega), forall yin L^infty(Omega) $ for some constant $C>0,$ without any constraint on the image.
$endgroup$
– Rabat
Mar 12 at 14:38
$begingroup$
I have seen the Tomek's answer. But the strict convexity property is not preserved by equivalence of norm.
$endgroup$
– Rabat
Mar 12 at 14:49
$begingroup$
I have seen the Tomek's answer. But the strict convexity property is not preserved by equivalence of norm.
$endgroup$
– Rabat
Mar 12 at 14:49
|
show 1 more comment
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$begingroup$
The second example you ask about is a Hilbert space, and is therefore a reflexive Banach space.
$endgroup$
– Keen-ameteur
Mar 12 at 10:43
$begingroup$
I think you can modify Tomek's answer here to show that the answer is "no" (reflexive spaces have equivalent strictly convex renormings).
$endgroup$
– David Mitra
Mar 12 at 12:01
$begingroup$
What is "embedding"? So, for example, is the set-theoretic inclusion $L^infty[0,1] to L^2[0,1]$ a "continuous embedding"? You do not require the image to be closed?
$endgroup$
– GEdgar
Mar 12 at 12:49
$begingroup$
By the continuous embedding, I mean that : $|y|_L^2(Omega)le C |y|_L^infty(Omega), forall yin L^infty(Omega) $ for some constant $C>0,$ without any constraint on the image.
$endgroup$
– Rabat
Mar 12 at 14:38
$begingroup$
I have seen the Tomek's answer. But the strict convexity property is not preserved by equivalence of norm.
$endgroup$
– Rabat
Mar 12 at 14:49