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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$begingroup$
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 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
$endgroup$
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 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
$endgroup$
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 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
$endgroup$
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 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
asked 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
asked 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
asked 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
asked 1 hour ago
Keshav SrinivasanKeshav Srinivasan
2,39121446
2,39121446
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 1 hour 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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
GEdgarGEdgar
63.3k268172
answered 1 hour ago
GEdgarGEdgar
63.3k268172
answered 1 hour ago
GEdgarGEdgar
63.3k268172
answered 1 hour ago
GEdgarGEdgar
63.3k268172
63.3k268172
add a comment |
add a comment |
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