Is a linearly independent set whose span is dense a Schauder basis? The Next CEO of Stack...
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Is a linearly independent set whose span is dense a Schauder basis?
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Is a linearly independent set whose span is dense a Schauder basis?
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$begingroup$
If $X$ is a Banach space, then a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$. My question is, if $A$ is a linearly independent subset of $X$ such that the closure of the span of $A$ equals $X$, then is $A$ necessarily a Schauder basis of $X$?
If not, does anyone know of any counterexamples?
linear-algebra functional-analysis banach-spaces normed-spaces schauder-basis
asked 58 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
$endgroup$
add a comment |
$begingroup$
If $X$ is a Banach space, then a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$. My question is, if $A$ is a linearly independent subset of $X$ such that the closure of the span of $A$ equals $X$, then is $A$ necessarily a Schauder basis of $X$?
If not, does anyone know of any counterexamples?
linear-algebra functional-analysis banach-spaces normed-spaces schauder-basis
asked 58 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
$endgroup$
add a comment |
$begingroup$
If $X$ is a Banach space, then a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$. My question is, if $A$ is a linearly independent subset of $X$ such that the closure of the span of $A$ equals $X$, then is $A$ necessarily a Schauder basis of $X$?
If not, does anyone know of any counterexamples?
linear-algebra functional-analysis banach-spaces normed-spaces schauder-basis
asked 58 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
$endgroup$
If $X$ is a Banach space, then a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$. My question is, if $A$ is a linearly independent subset of $X$ such that the closure of the span of $A$ equals $X$, then is $A$ necessarily a Schauder basis of $X$?
If not, does anyone know of any counterexamples?
linear-algebra functional-analysis banach-spaces normed-spaces schauder-basis
linear-algebra functional-analysis banach-spaces normed-spaces schauder-basis
asked 58 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
asked 58 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
asked 58 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
asked 58 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
asked 58 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
2,38621446
add a comment |
add a comment |
1 Answer
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$begingroup$
No, certainly not. The linearly independent set ${1, x, x^2, x^3, dots}$ has span dense in $C[0,1]$, but is not a Schauder basis of that space. (Not every continuous function is given by a power series.)
A Schauder basis is, in general, much harder to construct than a set with dense span.
Since Enflo we know that there are separable Banach spaces (hence they have countable, dense, linearly independent set) that have no Schauder basis at all.
answered 53 mins ago
GEdgarGEdgar
63.3k268172
$endgroup$
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1 Answer
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1 Answer
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$begingroup$
No, certainly not. The linearly independent set ${1, x, x^2, x^3, dots}$ has span dense in $C[0,1]$, but is not a Schauder basis of that space. (Not every continuous function is given by a power series.)
A Schauder basis is, in general, much harder to construct than a set with dense span.
Since Enflo we know that there are separable Banach spaces (hence they have countable, dense, linearly independent set) that have no Schauder basis at all.
answered 53 mins ago
GEdgarGEdgar
63.3k268172
$endgroup$
add a comment |
$begingroup$
No, certainly not. The linearly independent set ${1, x, x^2, x^3, dots}$ has span dense in $C[0,1]$, but is not a Schauder basis of that space. (Not every continuous function is given by a power series.)
A Schauder basis is, in general, much harder to construct than a set with dense span.
Since Enflo we know that there are separable Banach spaces (hence they have countable, dense, linearly independent set) that have no Schauder basis at all.
answered 53 mins ago
GEdgarGEdgar
63.3k268172
$endgroup$
add a comment |
$begingroup$
No, certainly not. The linearly independent set ${1, x, x^2, x^3, dots}$ has span dense in $C[0,1]$, but is not a Schauder basis of that space. (Not every continuous function is given by a power series.)
A Schauder basis is, in general, much harder to construct than a set with dense span.
Since Enflo we know that there are separable Banach spaces (hence they have countable, dense, linearly independent set) that have no Schauder basis at all.
answered 53 mins ago
GEdgarGEdgar
63.3k268172
$endgroup$
No, certainly not. The linearly independent set ${1, x, x^2, x^3, dots}$ has span dense in $C[0,1]$, but is not a Schauder basis of that space. (Not every continuous function is given by a power series.)
A Schauder basis is, in general, much harder to construct than a set with dense span.
Since Enflo we know that there are separable Banach spaces (hence they have countable, dense, linearly independent set) that have no Schauder basis at all.
answered 53 mins ago
GEdgarGEdgar
63.3k268172
answered 53 mins ago
GEdgarGEdgar
63.3k268172
answered 53 mins ago
GEdgarGEdgar
63.3k268172
answered 53 mins ago
GEdgarGEdgar
63.3k268172
63.3k268172
add a comment |
add a comment |
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