Dual spaces of uniformly convex Banach spaces
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I am interested in the dual spaces of uniformly convex Banach spaces. Given a uniformly convex Banach space $X$, can anything be said about uniform convexity of its dual space $X^*$? Or given a uniformly convex dual space $X^*$, can anything be said about the uniform convexity of $X$?
I would like links to any papers discussing these points, or any counterexamples. Thank you.
functional-analysis reference-request banach-spaces
asked Dec 16 '18 at 12:54
talfredtalfred
756
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add a comment |
$begingroup$
I am interested in the dual spaces of uniformly convex Banach spaces. Given a uniformly convex Banach space $X$, can anything be said about uniform convexity of its dual space $X^*$? Or given a uniformly convex dual space $X^*$, can anything be said about the uniform convexity of $X$?
I would like links to any papers discussing these points, or any counterexamples. Thank you.
functional-analysis reference-request banach-spaces
asked Dec 16 '18 at 12:54
talfredtalfred
756
$endgroup$
add a comment |
$begingroup$
I am interested in the dual spaces of uniformly convex Banach spaces. Given a uniformly convex Banach space $X$, can anything be said about uniform convexity of its dual space $X^*$? Or given a uniformly convex dual space $X^*$, can anything be said about the uniform convexity of $X$?
I would like links to any papers discussing these points, or any counterexamples. Thank you.
functional-analysis reference-request banach-spaces
asked Dec 16 '18 at 12:54
talfredtalfred
756
$endgroup$
I am interested in the dual spaces of uniformly convex Banach spaces. Given a uniformly convex Banach space $X$, can anything be said about uniform convexity of its dual space $X^*$? Or given a uniformly convex dual space $X^*$, can anything be said about the uniform convexity of $X$?
I would like links to any papers discussing these points, or any counterexamples. Thank you.
functional-analysis reference-request banach-spaces
functional-analysis reference-request banach-spaces
asked Dec 16 '18 at 12:54
talfredtalfred
756
asked Dec 16 '18 at 12:54
talfredtalfred
756
asked Dec 16 '18 at 12:54
talfredtalfred
756
asked Dec 16 '18 at 12:54
talfredtalfred
756
asked Dec 16 '18 at 12:54
talfredtalfred
756
756
add a comment |
add a comment |
1 Answer
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A Banach space is uniformly convex if and only if its dual is uniformly smooth. Also, since any uniformly convex Banach space is reflexive, this goes both ways, and any uniformly smooth Banach space is also reflexive. This can all be found on the Wikipedia articles for these properties, and they contain references at the bottom.
EDIT: Because uniformly convex and uniformly smooth are independent, we can find Banach spaces that are uniformly convex but which have a dual space that is not, and vise versa. Same for uniformly smooth.
Example: The norm $|cdot|_1+|cdot|_2$ is uniformly convex, but not uniformly smooth if $dim Xgeq2$. Hence its dual is uniformly smooth, but not uniformly convex.
answered Dec 16 '18 at 13:02
SmileyCraftSmileyCraft
3,401516
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1
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"Because uniformly convex and uniformly smooth are mutually exclusive..." What about Hilbert Spaces?
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– Theo Bendit
Dec 16 '18 at 13:09
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Mutually exclusive means any combination of the two can appear. Hilbert spaces are just an example where both properties are held. The norm $|cdot|_p$ for $pin{1,infty}$ is neither, and I already gave examples where either one is held, while the other is not.
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– SmileyCraft
Dec 16 '18 at 13:10
1
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I think you mean "independent" in that case. Mutually exclusive means that both cannot simultaneously occur.
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– Theo Bendit
Dec 16 '18 at 13:12
2
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Can't argue with that. I edited the post.
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– SmileyCraft
Dec 16 '18 at 13:13
add a comment |
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1 Answer
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$begingroup$
A Banach space is uniformly convex if and only if its dual is uniformly smooth. Also, since any uniformly convex Banach space is reflexive, this goes both ways, and any uniformly smooth Banach space is also reflexive. This can all be found on the Wikipedia articles for these properties, and they contain references at the bottom.
EDIT: Because uniformly convex and uniformly smooth are independent, we can find Banach spaces that are uniformly convex but which have a dual space that is not, and vise versa. Same for uniformly smooth.
Example: The norm $|cdot|_1+|cdot|_2$ is uniformly convex, but not uniformly smooth if $dim Xgeq2$. Hence its dual is uniformly smooth, but not uniformly convex.
answered Dec 16 '18 at 13:02
SmileyCraftSmileyCraft
3,401516
$endgroup$
1
$begingroup$
"Because uniformly convex and uniformly smooth are mutually exclusive..." What about Hilbert Spaces?
$endgroup$
– Theo Bendit
Dec 16 '18 at 13:09
$begingroup$
Mutually exclusive means any combination of the two can appear. Hilbert spaces are just an example where both properties are held. The norm $|cdot|_p$ for $pin{1,infty}$ is neither, and I already gave examples where either one is held, while the other is not.
$endgroup$
– SmileyCraft
Dec 16 '18 at 13:10
1
$begingroup$
I think you mean "independent" in that case. Mutually exclusive means that both cannot simultaneously occur.
$endgroup$
– Theo Bendit
Dec 16 '18 at 13:12
2
$begingroup$
Can't argue with that. I edited the post.
$endgroup$
– SmileyCraft
Dec 16 '18 at 13:13
add a comment |
$begingroup$
A Banach space is uniformly convex if and only if its dual is uniformly smooth. Also, since any uniformly convex Banach space is reflexive, this goes both ways, and any uniformly smooth Banach space is also reflexive. This can all be found on the Wikipedia articles for these properties, and they contain references at the bottom.
EDIT: Because uniformly convex and uniformly smooth are independent, we can find Banach spaces that are uniformly convex but which have a dual space that is not, and vise versa. Same for uniformly smooth.
Example: The norm $|cdot|_1+|cdot|_2$ is uniformly convex, but not uniformly smooth if $dim Xgeq2$. Hence its dual is uniformly smooth, but not uniformly convex.
answered Dec 16 '18 at 13:02
SmileyCraftSmileyCraft
3,401516
$endgroup$
1
$begingroup$
"Because uniformly convex and uniformly smooth are mutually exclusive..." What about Hilbert Spaces?
$endgroup$
– Theo Bendit
Dec 16 '18 at 13:09
$begingroup$
Mutually exclusive means any combination of the two can appear. Hilbert spaces are just an example where both properties are held. The norm $|cdot|_p$ for $pin{1,infty}$ is neither, and I already gave examples where either one is held, while the other is not.
$endgroup$
– SmileyCraft
Dec 16 '18 at 13:10
1
$begingroup$
I think you mean "independent" in that case. Mutually exclusive means that both cannot simultaneously occur.
$endgroup$
– Theo Bendit
Dec 16 '18 at 13:12
2
$begingroup$
Can't argue with that. I edited the post.
$endgroup$
– SmileyCraft
Dec 16 '18 at 13:13
add a comment |
$begingroup$
A Banach space is uniformly convex if and only if its dual is uniformly smooth. Also, since any uniformly convex Banach space is reflexive, this goes both ways, and any uniformly smooth Banach space is also reflexive. This can all be found on the Wikipedia articles for these properties, and they contain references at the bottom.
EDIT: Because uniformly convex and uniformly smooth are independent, we can find Banach spaces that are uniformly convex but which have a dual space that is not, and vise versa. Same for uniformly smooth.
Example: The norm $|cdot|_1+|cdot|_2$ is uniformly convex, but not uniformly smooth if $dim Xgeq2$. Hence its dual is uniformly smooth, but not uniformly convex.
answered Dec 16 '18 at 13:02
SmileyCraftSmileyCraft
3,401516
$endgroup$
A Banach space is uniformly convex if and only if its dual is uniformly smooth. Also, since any uniformly convex Banach space is reflexive, this goes both ways, and any uniformly smooth Banach space is also reflexive. This can all be found on the Wikipedia articles for these properties, and they contain references at the bottom.
EDIT: Because uniformly convex and uniformly smooth are independent, we can find Banach spaces that are uniformly convex but which have a dual space that is not, and vise versa. Same for uniformly smooth.
Example: The norm $|cdot|_1+|cdot|_2$ is uniformly convex, but not uniformly smooth if $dim Xgeq2$. Hence its dual is uniformly smooth, but not uniformly convex.
answered Dec 16 '18 at 13:02
SmileyCraftSmileyCraft
3,401516
edited Dec 16 '18 at 13:12
answered Dec 16 '18 at 13:02
SmileyCraftSmileyCraft
3,401516
answered Dec 16 '18 at 13:02
SmileyCraftSmileyCraft
3,401516
answered Dec 16 '18 at 13:02
SmileyCraftSmileyCraft
3,401516
3,401516
1
$begingroup$
"Because uniformly convex and uniformly smooth are mutually exclusive..." What about Hilbert Spaces?
$endgroup$
– Theo Bendit
Dec 16 '18 at 13:09
$begingroup$
Mutually exclusive means any combination of the two can appear. Hilbert spaces are just an example where both properties are held. The norm $|cdot|_p$ for $pin{1,infty}$ is neither, and I already gave examples where either one is held, while the other is not.
$endgroup$
– SmileyCraft
Dec 16 '18 at 13:10
1
$begingroup$
I think you mean "independent" in that case. Mutually exclusive means that both cannot simultaneously occur.
$endgroup$
– Theo Bendit
Dec 16 '18 at 13:12
2
$begingroup$
Can't argue with that. I edited the post.
$endgroup$
– SmileyCraft
Dec 16 '18 at 13:13
add a comment |
1
$begingroup$
"Because uniformly convex and uniformly smooth are mutually exclusive..." What about Hilbert Spaces?
$endgroup$
– Theo Bendit
Dec 16 '18 at 13:09
$begingroup$
Mutually exclusive means any combination of the two can appear. Hilbert spaces are just an example where both properties are held. The norm $|cdot|_p$ for $pin{1,infty}$ is neither, and I already gave examples where either one is held, while the other is not.
$endgroup$
– SmileyCraft
Dec 16 '18 at 13:10
1
$begingroup$
I think you mean "independent" in that case. Mutually exclusive means that both cannot simultaneously occur.
$endgroup$
– Theo Bendit
Dec 16 '18 at 13:12
2
$begingroup$
Can't argue with that. I edited the post.
$endgroup$
– SmileyCraft
Dec 16 '18 at 13:13
1
1
$begingroup$
"Because uniformly convex and uniformly smooth are mutually exclusive..." What about Hilbert Spaces?
$endgroup$
– Theo Bendit
Dec 16 '18 at 13:09
$begingroup$
"Because uniformly convex and uniformly smooth are mutually exclusive..." What about Hilbert Spaces?
$endgroup$
– Theo Bendit
Dec 16 '18 at 13:09
$begingroup$
Mutually exclusive means any combination of the two can appear. Hilbert spaces are just an example where both properties are held. The norm $|cdot|_p$ for $pin{1,infty}$ is neither, and I already gave examples where either one is held, while the other is not.
$endgroup$
– SmileyCraft
Dec 16 '18 at 13:10
$begingroup$
Mutually exclusive means any combination of the two can appear. Hilbert spaces are just an example where both properties are held. The norm $|cdot|_p$ for $pin{1,infty}$ is neither, and I already gave examples where either one is held, while the other is not.
$endgroup$
– SmileyCraft
Dec 16 '18 at 13:10
1
1
$begingroup$
I think you mean "independent" in that case. Mutually exclusive means that both cannot simultaneously occur.
$endgroup$
– Theo Bendit
Dec 16 '18 at 13:12
$begingroup$
I think you mean "independent" in that case. Mutually exclusive means that both cannot simultaneously occur.
$endgroup$
– Theo Bendit
Dec 16 '18 at 13:12
2
2
$begingroup$
Can't argue with that. I edited the post.
$endgroup$
– SmileyCraft
Dec 16 '18 at 13:13
$begingroup$
Can't argue with that. I edited the post.
$endgroup$
– SmileyCraft
Dec 16 '18 at 13:13
add a comment |
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