Weak Closure, clarification
$begingroup$
I'm working on the following question:
Show that the closed unit ball $B(0,1)$ in a normed space is also weakly closed.
I think I want to show that $$lim_{nrightarrow infty}f(x_n) in f(B(0,1)) subset mathbb{R}$$ for all $f in X^*$
This is strange to me though because for each $f$, I will get a different image in $mathbb{R}$. This correct though right? I'm checking that the limit lives in the image, so long as the image is created by an element of $X^*$.
EDIT:
This question is not a duplicate of the linked question. In fact, I have an answer already. I wanted to get input on this idea that "I'm checking if the limit lives in ALL images formed from elements of $X^*$" is a way to think about this problem because I've never heard it posed that way before.
functional-analysis soft-question
asked Dec 4 '18 at 15:43
yoshiyoshi
1,186817
$endgroup$
add a comment |
$begingroup$
I'm working on the following question:
Show that the closed unit ball $B(0,1)$ in a normed space is also weakly closed.
I think I want to show that $$lim_{nrightarrow infty}f(x_n) in f(B(0,1)) subset mathbb{R}$$ for all $f in X^*$
This is strange to me though because for each $f$, I will get a different image in $mathbb{R}$. This correct though right? I'm checking that the limit lives in the image, so long as the image is created by an element of $X^*$.
EDIT:
This question is not a duplicate of the linked question. In fact, I have an answer already. I wanted to get input on this idea that "I'm checking if the limit lives in ALL images formed from elements of $X^*$" is a way to think about this problem because I've never heard it posed that way before.
functional-analysis soft-question
asked Dec 4 '18 at 15:43
yoshiyoshi
1,186817
$endgroup$
2
$begingroup$
Possible duplicate of closed unit ball in a Banach space is closed in the weak topology
$endgroup$
– aduh
Dec 4 '18 at 15:50
$begingroup$
You cannot show weak closedness by checking closedness against weakly converging sequences.
$endgroup$
– daw
Dec 5 '18 at 7:22
add a comment |
$begingroup$
I'm working on the following question:
Show that the closed unit ball $B(0,1)$ in a normed space is also weakly closed.
I think I want to show that $$lim_{nrightarrow infty}f(x_n) in f(B(0,1)) subset mathbb{R}$$ for all $f in X^*$
This is strange to me though because for each $f$, I will get a different image in $mathbb{R}$. This correct though right? I'm checking that the limit lives in the image, so long as the image is created by an element of $X^*$.
EDIT:
This question is not a duplicate of the linked question. In fact, I have an answer already. I wanted to get input on this idea that "I'm checking if the limit lives in ALL images formed from elements of $X^*$" is a way to think about this problem because I've never heard it posed that way before.
functional-analysis soft-question
asked Dec 4 '18 at 15:43
yoshiyoshi
1,186817
$endgroup$
I'm working on the following question:
Show that the closed unit ball $B(0,1)$ in a normed space is also weakly closed.
I think I want to show that $$lim_{nrightarrow infty}f(x_n) in f(B(0,1)) subset mathbb{R}$$ for all $f in X^*$
This is strange to me though because for each $f$, I will get a different image in $mathbb{R}$. This correct though right? I'm checking that the limit lives in the image, so long as the image is created by an element of $X^*$.
EDIT:
This question is not a duplicate of the linked question. In fact, I have an answer already. I wanted to get input on this idea that "I'm checking if the limit lives in ALL images formed from elements of $X^*$" is a way to think about this problem because I've never heard it posed that way before.
functional-analysis soft-question
functional-analysis soft-question
asked Dec 4 '18 at 15:43
yoshiyoshi
1,186817
asked Dec 4 '18 at 15:43
yoshiyoshi
1,186817
edited Dec 4 '18 at 15:53
yoshi
asked Dec 4 '18 at 15:43
yoshiyoshi
1,186817
asked Dec 4 '18 at 15:43
yoshiyoshi
1,186817
asked Dec 4 '18 at 15:43
yoshiyoshi
1,186817
1,186817
2
$begingroup$
Possible duplicate of closed unit ball in a Banach space is closed in the weak topology
$endgroup$
– aduh
Dec 4 '18 at 15:50
$begingroup$
You cannot show weak closedness by checking closedness against weakly converging sequences.
$endgroup$
– daw
Dec 5 '18 at 7:22
add a comment |
2
$begingroup$
Possible duplicate of closed unit ball in a Banach space is closed in the weak topology
$endgroup$
– aduh
Dec 4 '18 at 15:50
$begingroup$
You cannot show weak closedness by checking closedness against weakly converging sequences.
$endgroup$
– daw
Dec 5 '18 at 7:22
2
2
$begingroup$
Possible duplicate of closed unit ball in a Banach space is closed in the weak topology
$endgroup$
– aduh
Dec 4 '18 at 15:50
$begingroup$
Possible duplicate of closed unit ball in a Banach space is closed in the weak topology
$endgroup$
– aduh
Dec 4 '18 at 15:50
$begingroup$
You cannot show weak closedness by checking closedness against weakly converging sequences.
$endgroup$
– daw
Dec 5 '18 at 7:22
$begingroup$
You cannot show weak closedness by checking closedness against weakly converging sequences.
$endgroup$
– daw
Dec 5 '18 at 7:22
add a comment |
1 Answer
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$begingroup$
First, you cannot show weak closedness by weakly converging sequences.
Second, (sequential) closedness usually means that limits of converging sequences of elements from the closed set are in the closed set as well. For weak convergence, this has nothing to do with any functionals. So you idea does not work.
answered Dec 5 '18 at 7:24
dawdaw
24.1k1544
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add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, you cannot show weak closedness by weakly converging sequences.
Second, (sequential) closedness usually means that limits of converging sequences of elements from the closed set are in the closed set as well. For weak convergence, this has nothing to do with any functionals. So you idea does not work.
answered Dec 5 '18 at 7:24
dawdaw
24.1k1544
$endgroup$
add a comment |
$begingroup$
First, you cannot show weak closedness by weakly converging sequences.
Second, (sequential) closedness usually means that limits of converging sequences of elements from the closed set are in the closed set as well. For weak convergence, this has nothing to do with any functionals. So you idea does not work.
answered Dec 5 '18 at 7:24
dawdaw
24.1k1544
$endgroup$
add a comment |
$begingroup$
First, you cannot show weak closedness by weakly converging sequences.
Second, (sequential) closedness usually means that limits of converging sequences of elements from the closed set are in the closed set as well. For weak convergence, this has nothing to do with any functionals. So you idea does not work.
answered Dec 5 '18 at 7:24
dawdaw
24.1k1544
$endgroup$
First, you cannot show weak closedness by weakly converging sequences.
Second, (sequential) closedness usually means that limits of converging sequences of elements from the closed set are in the closed set as well. For weak convergence, this has nothing to do with any functionals. So you idea does not work.
answered Dec 5 '18 at 7:24
dawdaw
24.1k1544
answered Dec 5 '18 at 7:24
dawdaw
24.1k1544
answered Dec 5 '18 at 7:24
dawdaw
24.1k1544
answered Dec 5 '18 at 7:24
dawdaw
24.1k1544
24.1k1544
add a comment |
add a comment |
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2
$begingroup$
Possible duplicate of closed unit ball in a Banach space is closed in the weak topology
$endgroup$
– aduh
Dec 4 '18 at 15:50
$begingroup$
You cannot show weak closedness by checking closedness against weakly converging sequences.
$endgroup$
– daw
Dec 5 '18 at 7:22