Is a linearly independent set whose span is dense a Schauder basis? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Coordinate 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
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Is a linearly independent set whose span is dense a Schauder basis?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Coordinate 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 Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,47521547
$endgroup$
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 Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,47521547
$endgroup$
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 Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,47521547
$endgroup$
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 Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,47521547
asked Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,47521547
asked Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,47521547
asked Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,47521547
asked Apr 1 at 21:28
Keshav SrinivasanKeshav Srinivasan
2,47521547
2,47521547
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 $C[0,1]$ by the Weierstrass approximation theorem. But it is not a Schauder basis of that space, since 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 subset) that have no Schauder basis at all.
answered Apr 1 at 21:33
GEdgarGEdgar
63.6k269175
$endgroup$
add a comment |
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$begingroup$
No, certainly not. The linearly independent set $1, x, x^2, x^3, dots$ has span dense in $C[0,1]$ by the Weierstrass approximation theorem. But it is not a Schauder basis of that space, since 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 subset) that have no Schauder basis at all.
answered Apr 1 at 21:33
GEdgarGEdgar
63.6k269175
$endgroup$
add a comment |
$begingroup$
No, certainly not. The linearly independent set $1, x, x^2, x^3, dots$ has span dense in $C[0,1]$ by the Weierstrass approximation theorem. But it is not a Schauder basis of that space, since 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 subset) that have no Schauder basis at all.
answered Apr 1 at 21:33
GEdgarGEdgar
63.6k269175
$endgroup$
add a comment |
$begingroup$
No, certainly not. The linearly independent set $1, x, x^2, x^3, dots$ has span dense in $C[0,1]$ by the Weierstrass approximation theorem. But it is not a Schauder basis of that space, since 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 subset) that have no Schauder basis at all.
answered Apr 1 at 21:33
GEdgarGEdgar
63.6k269175
$endgroup$
No, certainly not. The linearly independent set $1, x, x^2, x^3, dots$ has span dense in $C[0,1]$ by the Weierstrass approximation theorem. But it is not a Schauder basis of that space, since 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 subset) that have no Schauder basis at all.
answered Apr 1 at 21:33
GEdgarGEdgar
63.6k269175
edited Apr 5 at 0:27
answered Apr 1 at 21:33
GEdgarGEdgar
63.6k269175
answered Apr 1 at 21:33
GEdgarGEdgar
63.6k269175
answered Apr 1 at 21:33
GEdgarGEdgar
63.6k269175
63.6k269175
add a comment |
add a comment |
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