(Weak) basis for the space of bounded sequences
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The canonical basis is not a Schauder basis of the space of bounded sequences, but in some way, it uniquely determines every element in the space. Is it a basis in a weaker sense? How is it called?
Thanks a lot.
sequences-and-series banach-spaces schauder-basis
asked Jan 10 at 15:40
SergioSergio
1
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add a comment |
$begingroup$
The canonical basis is not a Schauder basis of the space of bounded sequences, but in some way, it uniquely determines every element in the space. Is it a basis in a weaker sense? How is it called?
Thanks a lot.
sequences-and-series banach-spaces schauder-basis
asked Jan 10 at 15:40
SergioSergio
1
$endgroup$
$begingroup$
It's a Schauder basis of $c_0$, the sequences with limit $0$. I don't see in what sense it "uniquely determines" anything not in $c_0$. Maybe you're thinking of a basis of $ell^1$, whose dual is $ell^infty$?
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– Robert Israel
Jan 10 at 15:46
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That makes sense, any sequence in $ell^infty$ can be understood as a operator of $ell^1$ sequences. Then we can take a Schauder basis of $ell^1$ and describe the behavior of the $ell^infty$ operator through the images of the basis elements. Thanks! I will post it as solution :-)
$endgroup$
– Sergio
Jan 11 at 10:27
add a comment |
$begingroup$
The canonical basis is not a Schauder basis of the space of bounded sequences, but in some way, it uniquely determines every element in the space. Is it a basis in a weaker sense? How is it called?
Thanks a lot.
sequences-and-series banach-spaces schauder-basis
asked Jan 10 at 15:40
SergioSergio
1
$endgroup$
The canonical basis is not a Schauder basis of the space of bounded sequences, but in some way, it uniquely determines every element in the space. Is it a basis in a weaker sense? How is it called?
Thanks a lot.
sequences-and-series banach-spaces schauder-basis
sequences-and-series banach-spaces schauder-basis
asked Jan 10 at 15:40
SergioSergio
1
asked Jan 10 at 15:40
SergioSergio
1
asked Jan 10 at 15:40
SergioSergio
1
asked Jan 10 at 15:40
SergioSergio
1
asked Jan 10 at 15:40
SergioSergio
1
1
$begingroup$
It's a Schauder basis of $c_0$, the sequences with limit $0$. I don't see in what sense it "uniquely determines" anything not in $c_0$. Maybe you're thinking of a basis of $ell^1$, whose dual is $ell^infty$?
$endgroup$
– Robert Israel
Jan 10 at 15:46
$begingroup$
That makes sense, any sequence in $ell^infty$ can be understood as a operator of $ell^1$ sequences. Then we can take a Schauder basis of $ell^1$ and describe the behavior of the $ell^infty$ operator through the images of the basis elements. Thanks! I will post it as solution :-)
$endgroup$
– Sergio
Jan 11 at 10:27
add a comment |
$begingroup$
It's a Schauder basis of $c_0$, the sequences with limit $0$. I don't see in what sense it "uniquely determines" anything not in $c_0$. Maybe you're thinking of a basis of $ell^1$, whose dual is $ell^infty$?
$endgroup$
– Robert Israel
Jan 10 at 15:46
$begingroup$
That makes sense, any sequence in $ell^infty$ can be understood as a operator of $ell^1$ sequences. Then we can take a Schauder basis of $ell^1$ and describe the behavior of the $ell^infty$ operator through the images of the basis elements. Thanks! I will post it as solution :-)
$endgroup$
– Sergio
Jan 11 at 10:27
$begingroup$
It's a Schauder basis of $c_0$, the sequences with limit $0$. I don't see in what sense it "uniquely determines" anything not in $c_0$. Maybe you're thinking of a basis of $ell^1$, whose dual is $ell^infty$?
$endgroup$
– Robert Israel
Jan 10 at 15:46
$begingroup$
It's a Schauder basis of $c_0$, the sequences with limit $0$. I don't see in what sense it "uniquely determines" anything not in $c_0$. Maybe you're thinking of a basis of $ell^1$, whose dual is $ell^infty$?
$endgroup$
– Robert Israel
Jan 10 at 15:46
$begingroup$
That makes sense, any sequence in $ell^infty$ can be understood as a operator of $ell^1$ sequences. Then we can take a Schauder basis of $ell^1$ and describe the behavior of the $ell^infty$ operator through the images of the basis elements. Thanks! I will post it as solution :-)
$endgroup$
– Sergio
Jan 11 at 10:27
$begingroup$
That makes sense, any sequence in $ell^infty$ can be understood as a operator of $ell^1$ sequences. Then we can take a Schauder basis of $ell^1$ and describe the behavior of the $ell^infty$ operator through the images of the basis elements. Thanks! I will post it as solution :-)
$endgroup$
– Sergio
Jan 11 at 10:27
add a comment |
2 Answers
2
active
oldest
votes
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A subspace of a normed vector space is closed if and only if it is weakly closed Closed $iff$ weakly closed subspace. Hence, a set is a Schauder basis if and only if it is a "basis in the weaker sense".
answered Jan 10 at 15:50
SmileyCraftSmileyCraft
3,776519
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add a comment |
$begingroup$
Any sequence in $ell^infty$ can be understood as a operator of $ell^1$ sequences. Then we can take a Schauder basis of $ell^1$ and describe the behavior of the $ell^infty$ operator through the images of the basis elements.
answered Jan 11 at 10:29
SergioSergio
1
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
A subspace of a normed vector space is closed if and only if it is weakly closed Closed $iff$ weakly closed subspace. Hence, a set is a Schauder basis if and only if it is a "basis in the weaker sense".
answered Jan 10 at 15:50
SmileyCraftSmileyCraft
3,776519
$endgroup$
add a comment |
$begingroup$
A subspace of a normed vector space is closed if and only if it is weakly closed Closed $iff$ weakly closed subspace. Hence, a set is a Schauder basis if and only if it is a "basis in the weaker sense".
answered Jan 10 at 15:50
SmileyCraftSmileyCraft
3,776519
$endgroup$
add a comment |
$begingroup$
A subspace of a normed vector space is closed if and only if it is weakly closed Closed $iff$ weakly closed subspace. Hence, a set is a Schauder basis if and only if it is a "basis in the weaker sense".
answered Jan 10 at 15:50
SmileyCraftSmileyCraft
3,776519
$endgroup$
A subspace of a normed vector space is closed if and only if it is weakly closed Closed $iff$ weakly closed subspace. Hence, a set is a Schauder basis if and only if it is a "basis in the weaker sense".
answered Jan 10 at 15:50
SmileyCraftSmileyCraft
3,776519
answered Jan 10 at 15:50
SmileyCraftSmileyCraft
3,776519
answered Jan 10 at 15:50
SmileyCraftSmileyCraft
3,776519
answered Jan 10 at 15:50
SmileyCraftSmileyCraft
3,776519
3,776519
add a comment |
add a comment |
$begingroup$
Any sequence in $ell^infty$ can be understood as a operator of $ell^1$ sequences. Then we can take a Schauder basis of $ell^1$ and describe the behavior of the $ell^infty$ operator through the images of the basis elements.
answered Jan 11 at 10:29
SergioSergio
1
$endgroup$
add a comment |
$begingroup$
Any sequence in $ell^infty$ can be understood as a operator of $ell^1$ sequences. Then we can take a Schauder basis of $ell^1$ and describe the behavior of the $ell^infty$ operator through the images of the basis elements.
answered Jan 11 at 10:29
SergioSergio
1
$endgroup$
add a comment |
$begingroup$
Any sequence in $ell^infty$ can be understood as a operator of $ell^1$ sequences. Then we can take a Schauder basis of $ell^1$ and describe the behavior of the $ell^infty$ operator through the images of the basis elements.
answered Jan 11 at 10:29
SergioSergio
1
$endgroup$
Any sequence in $ell^infty$ can be understood as a operator of $ell^1$ sequences. Then we can take a Schauder basis of $ell^1$ and describe the behavior of the $ell^infty$ operator through the images of the basis elements.
answered Jan 11 at 10:29
SergioSergio
1
answered Jan 11 at 10:29
SergioSergio
1
answered Jan 11 at 10:29
SergioSergio
1
answered Jan 11 at 10:29
SergioSergio
1
1
add a comment |
add a comment |
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$begingroup$
It's a Schauder basis of $c_0$, the sequences with limit $0$. I don't see in what sense it "uniquely determines" anything not in $c_0$. Maybe you're thinking of a basis of $ell^1$, whose dual is $ell^infty$?
$endgroup$
– Robert Israel
Jan 10 at 15:46
$begingroup$
That makes sense, any sequence in $ell^infty$ can be understood as a operator of $ell^1$ sequences. Then we can take a Schauder basis of $ell^1$ and describe the behavior of the $ell^infty$ operator through the images of the basis elements. Thanks! I will post it as solution :-)
$endgroup$
– Sergio
Jan 11 at 10:27