Is the Banach space $M_{ntimes n} (mathbb{C})$ with normal structure?
Is the Banach space $M_{ntimes n} (mathbb{C})$ with normal structure?
I know the Banach space $oplus_{1}^{n} mathbb{C}$ is with normal structure but I can't fine a subspace of $M_n(mathbb{C})$ without nondiametral point.
banach-spaces
asked Nov 18 at 16:24
Dadrahm
441112
add a comment |
Is the Banach space $M_{ntimes n} (mathbb{C})$ with normal structure?
I know the Banach space $oplus_{1}^{n} mathbb{C}$ is with normal structure but I can't fine a subspace of $M_n(mathbb{C})$ without nondiametral point.
banach-spaces
asked Nov 18 at 16:24
Dadrahm
441112
add a comment |
Is the Banach space $M_{ntimes n} (mathbb{C})$ with normal structure?
I know the Banach space $oplus_{1}^{n} mathbb{C}$ is with normal structure but I can't fine a subspace of $M_n(mathbb{C})$ without nondiametral point.
banach-spaces
asked Nov 18 at 16:24
Dadrahm
441112
Is the Banach space $M_{ntimes n} (mathbb{C})$ with normal structure?
I know the Banach space $oplus_{1}^{n} mathbb{C}$ is with normal structure but I can't fine a subspace of $M_n(mathbb{C})$ without nondiametral point.
banach-spaces
banach-spaces
asked Nov 18 at 16:24
Dadrahm
441112
asked Nov 18 at 16:24
Dadrahm
441112
asked Nov 18 at 16:24
Dadrahm
441112
asked Nov 18 at 16:24
Dadrahm
441112
asked Nov 18 at 16:24
Dadrahm
441112
441112
add a comment |
add a comment |
1 Answer
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Every finite-dimensional Banach space has a normal structure because every non-trivial compact convex set contains a non-diametral point. See also Lemma 4.1 in Goebel & Kirk's Topics in Metric Fixed Point Theory.
More generally, there is an easy criterion for a reflexive space to not have a normal structure.
Suppose that $X$ is a reflexive Banach space. If $X$ fails to have a normal structure then you will find in $X$ a sequence $(x_n)_{n=1}^infty$ of unit vectors that converges to 0 weakly and such that ${rm diam}{x_1, x_2, ldots} leqslant 1$.
answered Nov 18 at 22:19
Tomek Kania
12.1k11943
Thank you verry much.
– Dadrahm
Nov 19 at 7:26
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
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active
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votes
Every finite-dimensional Banach space has a normal structure because every non-trivial compact convex set contains a non-diametral point. See also Lemma 4.1 in Goebel & Kirk's Topics in Metric Fixed Point Theory.
More generally, there is an easy criterion for a reflexive space to not have a normal structure.
Suppose that $X$ is a reflexive Banach space. If $X$ fails to have a normal structure then you will find in $X$ a sequence $(x_n)_{n=1}^infty$ of unit vectors that converges to 0 weakly and such that ${rm diam}{x_1, x_2, ldots} leqslant 1$.
answered Nov 18 at 22:19
Tomek Kania
12.1k11943
Thank you verry much.
– Dadrahm
Nov 19 at 7:26
add a comment |
Every finite-dimensional Banach space has a normal structure because every non-trivial compact convex set contains a non-diametral point. See also Lemma 4.1 in Goebel & Kirk's Topics in Metric Fixed Point Theory.
More generally, there is an easy criterion for a reflexive space to not have a normal structure.
Suppose that $X$ is a reflexive Banach space. If $X$ fails to have a normal structure then you will find in $X$ a sequence $(x_n)_{n=1}^infty$ of unit vectors that converges to 0 weakly and such that ${rm diam}{x_1, x_2, ldots} leqslant 1$.
answered Nov 18 at 22:19
Tomek Kania
12.1k11943
Thank you verry much.
– Dadrahm
Nov 19 at 7:26
add a comment |
Every finite-dimensional Banach space has a normal structure because every non-trivial compact convex set contains a non-diametral point. See also Lemma 4.1 in Goebel & Kirk's Topics in Metric Fixed Point Theory.
More generally, there is an easy criterion for a reflexive space to not have a normal structure.
Suppose that $X$ is a reflexive Banach space. If $X$ fails to have a normal structure then you will find in $X$ a sequence $(x_n)_{n=1}^infty$ of unit vectors that converges to 0 weakly and such that ${rm diam}{x_1, x_2, ldots} leqslant 1$.
answered Nov 18 at 22:19
Tomek Kania
12.1k11943
Every finite-dimensional Banach space has a normal structure because every non-trivial compact convex set contains a non-diametral point. See also Lemma 4.1 in Goebel & Kirk's Topics in Metric Fixed Point Theory.
More generally, there is an easy criterion for a reflexive space to not have a normal structure.
Suppose that $X$ is a reflexive Banach space. If $X$ fails to have a normal structure then you will find in $X$ a sequence $(x_n)_{n=1}^infty$ of unit vectors that converges to 0 weakly and such that ${rm diam}{x_1, x_2, ldots} leqslant 1$.
answered Nov 18 at 22:19
Tomek Kania
12.1k11943
answered Nov 18 at 22:19
Tomek Kania
12.1k11943
answered Nov 18 at 22:19
Tomek Kania
12.1k11943
answered Nov 18 at 22:19
Tomek Kania
12.1k11943
12.1k11943
Thank you verry much.
– Dadrahm
Nov 19 at 7:26
add a comment |
Thank you verry much.
– Dadrahm
Nov 19 at 7:26
Thank you verry much.
– Dadrahm
Nov 19 at 7:26
Thank you verry much.
– Dadrahm
Nov 19 at 7:26
add a comment |
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