Show that $A^o$ is convex, balanced, closed in $A'$
0
$begingroup$
Let $E$ is a normed space and $A subset E$. Define
$$A^o := {y in A': |y(x)| leq 1, forall x in A}.$$
in which, $A'$ is the dual space of $A$.
With this definition, how to show that $A^o$ is convex, balanced, closed in $A'$?
linear-algebra functional-analysis convex-analysis dual-spaces
asked Jan 15 at 13:46
MinhMinh
31119
$endgroup$
add a comment |
0
$begingroup$
Let $E$ is a normed space and $A subset E$. Define
$$A^o := {y in A': |y(x)| leq 1, forall x in A}.$$
in which, $A'$ is the dual space of $A$.
With this definition, how to show that $A^o$ is convex, balanced, closed in $A'$?
linear-algebra functional-analysis convex-analysis dual-spaces
asked Jan 15 at 13:46
MinhMinh
31119
$endgroup$
$begingroup$
Are you using the weak-$ast$ of the norm topology on $A'$? In any case, each of these properties follows pretty immediately from the definitions. Have you tried to prove it yourself?
$endgroup$
– MaoWao
Jan 15 at 14:05
$begingroup$
$forall y_1,y_2 in A^o; forall alpha,beta in mathbb{K}, |alpha| + |beta| leq 1 Rightarrow alpha y_1 + beta y_2 in A$ i.e $forall x in A, |(alpha y_1 + beta y_2)(x)| = |alpha y_1 (x) + beta y_2 (x)| le |alpha||y_1 (x)| + |beta||y_2 (x)| le |alpha|.1 + |beta|.1 le 1$. $Rightarrow sup_{x in A} (alpha y_1 + beta y_2) le 1 Rightarrow alpha y_1 + beta y_2 in A^o$. So, $A^o$ is convex, balanced. Is that right?
$endgroup$
– Minh
Jan 15 at 14:36
$begingroup$
Yes, that's correct.
$endgroup$
– MaoWao
Jan 15 at 14:38
$begingroup$
Thank you! I have proved it.
$endgroup$
– Minh
Jan 15 at 15:05
add a comment |
0
0
0
$begingroup$
Let $E$ is a normed space and $A subset E$. Define
$$A^o := {y in A': |y(x)| leq 1, forall x in A}.$$
in which, $A'$ is the dual space of $A$.
With this definition, how to show that $A^o$ is convex, balanced, closed in $A'$?
linear-algebra functional-analysis convex-analysis dual-spaces
asked Jan 15 at 13:46
MinhMinh
31119
$endgroup$
Let $E$ is a normed space and $A subset E$. Define
$$A^o := {y in A': |y(x)| leq 1, forall x in A}.$$
in which, $A'$ is the dual space of $A$.
With this definition, how to show that $A^o$ is convex, balanced, closed in $A'$?
linear-algebra functional-analysis convex-analysis dual-spaces
linear-algebra functional-analysis convex-analysis dual-spaces
asked Jan 15 at 13:46
MinhMinh
31119
asked Jan 15 at 13:46
MinhMinh
31119
asked Jan 15 at 13:46
MinhMinh
31119
asked Jan 15 at 13:46
MinhMinh
31119
asked Jan 15 at 13:46
MinhMinh
31119
31119
$begingroup$
Are you using the weak-$ast$ of the norm topology on $A'$? In any case, each of these properties follows pretty immediately from the definitions. Have you tried to prove it yourself?
$endgroup$
– MaoWao
Jan 15 at 14:05
$begingroup$
$forall y_1,y_2 in A^o; forall alpha,beta in mathbb{K}, |alpha| + |beta| leq 1 Rightarrow alpha y_1 + beta y_2 in A$ i.e $forall x in A, |(alpha y_1 + beta y_2)(x)| = |alpha y_1 (x) + beta y_2 (x)| le |alpha||y_1 (x)| + |beta||y_2 (x)| le |alpha|.1 + |beta|.1 le 1$. $Rightarrow sup_{x in A} (alpha y_1 + beta y_2) le 1 Rightarrow alpha y_1 + beta y_2 in A^o$. So, $A^o$ is convex, balanced. Is that right?
$endgroup$
– Minh
Jan 15 at 14:36
$begingroup$
Yes, that's correct.
$endgroup$
– MaoWao
Jan 15 at 14:38
$begingroup$
Thank you! I have proved it.
$endgroup$
– Minh
Jan 15 at 15:05
add a comment |
$begingroup$
Are you using the weak-$ast$ of the norm topology on $A'$? In any case, each of these properties follows pretty immediately from the definitions. Have you tried to prove it yourself?
$endgroup$
– MaoWao
Jan 15 at 14:05
$begingroup$
$forall y_1,y_2 in A^o; forall alpha,beta in mathbb{K}, |alpha| + |beta| leq 1 Rightarrow alpha y_1 + beta y_2 in A$ i.e $forall x in A, |(alpha y_1 + beta y_2)(x)| = |alpha y_1 (x) + beta y_2 (x)| le |alpha||y_1 (x)| + |beta||y_2 (x)| le |alpha|.1 + |beta|.1 le 1$. $Rightarrow sup_{x in A} (alpha y_1 + beta y_2) le 1 Rightarrow alpha y_1 + beta y_2 in A^o$. So, $A^o$ is convex, balanced. Is that right?
$endgroup$
– Minh
Jan 15 at 14:36
$begingroup$
Yes, that's correct.
$endgroup$
– MaoWao
Jan 15 at 14:38
$begingroup$
Thank you! I have proved it.
$endgroup$
– Minh
Jan 15 at 15:05
$begingroup$
Are you using the weak-$ast$ of the norm topology on $A'$? In any case, each of these properties follows pretty immediately from the definitions. Have you tried to prove it yourself?
$endgroup$
– MaoWao
Jan 15 at 14:05
$begingroup$
Are you using the weak-$ast$ of the norm topology on $A'$? In any case, each of these properties follows pretty immediately from the definitions. Have you tried to prove it yourself?
$endgroup$
– MaoWao
Jan 15 at 14:05
$begingroup$
$forall y_1,y_2 in A^o; forall alpha,beta in mathbb{K}, |alpha| + |beta| leq 1 Rightarrow alpha y_1 + beta y_2 in A$ i.e $forall x in A, |(alpha y_1 + beta y_2)(x)| = |alpha y_1 (x) + beta y_2 (x)| le |alpha||y_1 (x)| + |beta||y_2 (x)| le |alpha|.1 + |beta|.1 le 1$. $Rightarrow sup_{x in A} (alpha y_1 + beta y_2) le 1 Rightarrow alpha y_1 + beta y_2 in A^o$. So, $A^o$ is convex, balanced. Is that right?
$endgroup$
– Minh
Jan 15 at 14:36
$begingroup$
$forall y_1,y_2 in A^o; forall alpha,beta in mathbb{K}, |alpha| + |beta| leq 1 Rightarrow alpha y_1 + beta y_2 in A$ i.e $forall x in A, |(alpha y_1 + beta y_2)(x)| = |alpha y_1 (x) + beta y_2 (x)| le |alpha||y_1 (x)| + |beta||y_2 (x)| le |alpha|.1 + |beta|.1 le 1$. $Rightarrow sup_{x in A} (alpha y_1 + beta y_2) le 1 Rightarrow alpha y_1 + beta y_2 in A^o$. So, $A^o$ is convex, balanced. Is that right?
$endgroup$
– Minh
Jan 15 at 14:36
$begingroup$
Yes, that's correct.
$endgroup$
– MaoWao
Jan 15 at 14:38
$begingroup$
Yes, that's correct.
$endgroup$
– MaoWao
Jan 15 at 14:38
$begingroup$
Thank you! I have proved it.
$endgroup$
– Minh
Jan 15 at 15:05
$begingroup$
Thank you! I have proved it.
$endgroup$
– Minh
Jan 15 at 15:05
add a comment |
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$begingroup$
Are you using the weak-$ast$ of the norm topology on $A'$? In any case, each of these properties follows pretty immediately from the definitions. Have you tried to prove it yourself?
$endgroup$
– MaoWao
Jan 15 at 14:05
$begingroup$
$forall y_1,y_2 in A^o; forall alpha,beta in mathbb{K}, |alpha| + |beta| leq 1 Rightarrow alpha y_1 + beta y_2 in A$ i.e $forall x in A, |(alpha y_1 + beta y_2)(x)| = |alpha y_1 (x) + beta y_2 (x)| le |alpha||y_1 (x)| + |beta||y_2 (x)| le |alpha|.1 + |beta|.1 le 1$. $Rightarrow sup_{x in A} (alpha y_1 + beta y_2) le 1 Rightarrow alpha y_1 + beta y_2 in A^o$. So, $A^o$ is convex, balanced. Is that right?
$endgroup$
– Minh
Jan 15 at 14:36
$begingroup$
Yes, that's correct.
$endgroup$
– MaoWao
Jan 15 at 14:38
$begingroup$
Thank you! I have proved it.
$endgroup$
– Minh
Jan 15 at 15:05