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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?
The Next CEO of Stack OverflowCoordinate functions of Schauder basisLinearly independentSchauder basis for a separable Banach spaceWhat is the difference between a Hamel basis and a Schauder basis?Hamel basis for subspacesExistence of weak Schauder-basis for concrete example.Isomorphisms with invariant linearly independent dense subset.Linear independence and Schauder basisWhy isn't every Hamel basis a Schauder basis?Schauder basis that is not Hilbert basis
$begingroup$
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 59 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 59 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 59 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 59 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
asked 59 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
asked 59 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
asked 59 mins ago
Keshav SrinivasanKeshav Srinivasan
2,38621446
asked 59 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 55 mins ago
GEdgarGEdgar
63.3k268172
$endgroup$
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1 Answer
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1 Answer
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oldest
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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 55 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 55 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 55 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 55 mins ago
GEdgarGEdgar
63.3k268172
answered 55 mins ago
GEdgarGEdgar
63.3k268172
answered 55 mins ago
GEdgarGEdgar
63.3k268172
answered 55 mins ago
GEdgarGEdgar
63.3k268172
63.3k268172
add a comment |
add a comment |
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