Finding an exception for a seemingly universal form of functionals [closed]
How to prove that not every $fin l_infty^*$ functional is in the form of $f=f_y$, where $ y=(y_n)in l_1 $ and $$ f_y(x)=sum_n x_ny_n, , x=(x_n)in l_infty$$?
functional-analysis
closed as off-topic by The Phenotype, Saad, Leucippus, Chinnapparaj R, user10354138 Nov 26 at 8:14
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How to prove that not every $fin l_infty^*$ functional is in the form of $f=f_y$, where $ y=(y_n)in l_1 $ and $$ f_y(x)=sum_n x_ny_n, , x=(x_n)in l_infty$$?
functional-analysis
closed as off-topic by The Phenotype, Saad, Leucippus, Chinnapparaj R, user10354138 Nov 26 at 8:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – The Phenotype, Saad, Leucippus, Chinnapparaj R, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
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How to prove that not every $fin l_infty^*$ functional is in the form of $f=f_y$, where $ y=(y_n)in l_1 $ and $$ f_y(x)=sum_n x_ny_n, , x=(x_n)in l_infty$$?
functional-analysis
How to prove that not every $fin l_infty^*$ functional is in the form of $f=f_y$, where $ y=(y_n)in l_1 $ and $$ f_y(x)=sum_n x_ny_n, , x=(x_n)in l_infty$$?
functional-analysis
functional-analysis
asked Nov 25 at 19:03
kohiheku
1
1
closed as off-topic by The Phenotype, Saad, Leucippus, Chinnapparaj R, user10354138 Nov 26 at 8:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – The Phenotype, Saad, Leucippus, Chinnapparaj R, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by The Phenotype, Saad, Leucippus, Chinnapparaj R, user10354138 Nov 26 at 8:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – The Phenotype, Saad, Leucippus, Chinnapparaj R, user10354138
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1 Answer
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You can use the Hahn-Banach extension theorem on the subspace of sequences $f in l_infty$ for which $lim_{n to infty} f_n$ exists.
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can use the Hahn-Banach extension theorem on the subspace of sequences $f in l_infty$ for which $lim_{n to infty} f_n$ exists.
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You can use the Hahn-Banach extension theorem on the subspace of sequences $f in l_infty$ for which $lim_{n to infty} f_n$ exists.
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You can use the Hahn-Banach extension theorem on the subspace of sequences $f in l_infty$ for which $lim_{n to infty} f_n$ exists.
You can use the Hahn-Banach extension theorem on the subspace of sequences $f in l_infty$ for which $lim_{n to infty} f_n$ exists.
answered Nov 25 at 19:08
Bartosz Malman
8011620
8011620
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