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Benefits of choosing a Hamel bases for $L^p$ including a specific linearly independent subset
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According to the book I am studying (Royden & Fitzpatrick):
We can infer from Zorn's Lemma that every linear space possesses a Hamel basis.
A Hamel Basis is defined as a subset $mathcalB $ of a linear space $X$ provided each vector in $X$ is expressible as a unique $bffinite$ linear combination of vectors in $mathcalB$.
When I first read about the concept of Hamel bases (compared to other sets referred to as basis), I recognized that bases of finite spaces are Hamel basis, and that, for example, $x^k_k=0^infty$ is a Hamel basis of $mathbbR[x]$, the infinite-dimensional space of polynomials with real coefficients.
As comments have pointed out, there might not be a Hamel basis that is explicit to state for other common infinite-dimensional linear spaces ($ell^p$ or $L^p$) - their Hamel bases would need to be uncountable.
However, even if we cannot explicitly characterize the whole basis, could we choose one that includes a specific (countable) set of vectors as a subset, given they are 'independent'? (For example, include $ x^k_k=0^infty$ in the Hamel basis for $L^p([a,b])$). If so, what are the applications of this fact?
functional-analysis vector-spaces normed-spaces hamel-basis
edited Jul 29 at 7:43
Asaf Karagila
291k31402732
asked Jul 29 at 5:23
ertl
427110
add a comment |Â
up vote
1
down vote
favorite
According to the book I am studying (Royden & Fitzpatrick):
We can infer from Zorn's Lemma that every linear space possesses a Hamel basis.
A Hamel Basis is defined as a subset $mathcalB $ of a linear space $X$ provided each vector in $X$ is expressible as a unique $bffinite$ linear combination of vectors in $mathcalB$.
When I first read about the concept of Hamel bases (compared to other sets referred to as basis), I recognized that bases of finite spaces are Hamel basis, and that, for example, $x^k_k=0^infty$ is a Hamel basis of $mathbbR[x]$, the infinite-dimensional space of polynomials with real coefficients.
As comments have pointed out, there might not be a Hamel basis that is explicit to state for other common infinite-dimensional linear spaces ($ell^p$ or $L^p$) - their Hamel bases would need to be uncountable.
However, even if we cannot explicitly characterize the whole basis, could we choose one that includes a specific (countable) set of vectors as a subset, given they are 'independent'? (For example, include $ x^k_k=0^infty$ in the Hamel basis for $L^p([a,b])$). If so, what are the applications of this fact?
functional-analysis vector-spaces normed-spaces hamel-basis
edited Jul 29 at 7:43
Asaf Karagila
291k31402732
asked Jul 29 at 5:23
ertl
427110
1
When the dimension is continuum, I think there is almost no hope for writing such explicit basis. Choice of one that includes a specific countable linearly independent set is possible though.
– i707107
Jul 29 at 5:32
Possible duplicate of A Hamel basis for $ell^p$?
– adfriedman
Jul 29 at 6:02
Thank you for the link! I changed the question to ask what (if any) 'benefits' there might be to being able to choose a Hamel basis including a certain linearly independent set of vectors. This is not a duplicate as far as I can tell.
– ertl
Jul 29 at 6:26
Yes, an arbitrary linearly independent subset $S$ can be extended to a Hamel basis. Just apply Zorn's lemma to the family of all linearly independent sets which contain $S$, ordered by inclusion. The maximal element will be a Hamel basis.
– mechanodroid
Jul 29 at 12:38
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
According to the book I am studying (Royden & Fitzpatrick):
We can infer from Zorn's Lemma that every linear space possesses a Hamel basis.
A Hamel Basis is defined as a subset $mathcalB $ of a linear space $X$ provided each vector in $X$ is expressible as a unique $bffinite$ linear combination of vectors in $mathcalB$.
When I first read about the concept of Hamel bases (compared to other sets referred to as basis), I recognized that bases of finite spaces are Hamel basis, and that, for example, $x^k_k=0^infty$ is a Hamel basis of $mathbbR[x]$, the infinite-dimensional space of polynomials with real coefficients.
As comments have pointed out, there might not be a Hamel basis that is explicit to state for other common infinite-dimensional linear spaces ($ell^p$ or $L^p$) - their Hamel bases would need to be uncountable.
However, even if we cannot explicitly characterize the whole basis, could we choose one that includes a specific (countable) set of vectors as a subset, given they are 'independent'? (For example, include $ x^k_k=0^infty$ in the Hamel basis for $L^p([a,b])$). If so, what are the applications of this fact?
functional-analysis vector-spaces normed-spaces hamel-basis
edited Jul 29 at 7:43
Asaf Karagila
291k31402732
asked Jul 29 at 5:23
ertl
427110
According to the book I am studying (Royden & Fitzpatrick):
We can infer from Zorn's Lemma that every linear space possesses a Hamel basis.
A Hamel Basis is defined as a subset $mathcalB $ of a linear space $X$ provided each vector in $X$ is expressible as a unique $bffinite$ linear combination of vectors in $mathcalB$.
When I first read about the concept of Hamel bases (compared to other sets referred to as basis), I recognized that bases of finite spaces are Hamel basis, and that, for example, $x^k_k=0^infty$ is a Hamel basis of $mathbbR[x]$, the infinite-dimensional space of polynomials with real coefficients.
As comments have pointed out, there might not be a Hamel basis that is explicit to state for other common infinite-dimensional linear spaces ($ell^p$ or $L^p$) - their Hamel bases would need to be uncountable.
However, even if we cannot explicitly characterize the whole basis, could we choose one that includes a specific (countable) set of vectors as a subset, given they are 'independent'? (For example, include $ x^k_k=0^infty$ in the Hamel basis for $L^p([a,b])$). If so, what are the applications of this fact?
functional-analysis vector-spaces normed-spaces hamel-basis
edited Jul 29 at 7:43
Asaf Karagila
291k31402732
asked Jul 29 at 5:23
ertl
427110
edited Jul 29 at 7:43
Asaf Karagila
291k31402732
edited Jul 29 at 7:43
Asaf Karagila
291k31402732
edited Jul 29 at 7:43
Asaf Karagila
291k31402732
291k31402732
asked Jul 29 at 5:23
ertl
427110
asked Jul 29 at 5:23
ertl
427110
asked Jul 29 at 5:23
ertl
427110
427110
1
When the dimension is continuum, I think there is almost no hope for writing such explicit basis. Choice of one that includes a specific countable linearly independent set is possible though.
– i707107
Jul 29 at 5:32
Possible duplicate of A Hamel basis for $ell^p$?
– adfriedman
Jul 29 at 6:02
Thank you for the link! I changed the question to ask what (if any) 'benefits' there might be to being able to choose a Hamel basis including a certain linearly independent set of vectors. This is not a duplicate as far as I can tell.
– ertl
Jul 29 at 6:26
Yes, an arbitrary linearly independent subset $S$ can be extended to a Hamel basis. Just apply Zorn's lemma to the family of all linearly independent sets which contain $S$, ordered by inclusion. The maximal element will be a Hamel basis.
– mechanodroid
Jul 29 at 12:38
add a comment |Â
1
When the dimension is continuum, I think there is almost no hope for writing such explicit basis. Choice of one that includes a specific countable linearly independent set is possible though.
– i707107
Jul 29 at 5:32
Possible duplicate of A Hamel basis for $ell^p$?
– adfriedman
Jul 29 at 6:02
Thank you for the link! I changed the question to ask what (if any) 'benefits' there might be to being able to choose a Hamel basis including a certain linearly independent set of vectors. This is not a duplicate as far as I can tell.
– ertl
Jul 29 at 6:26
Yes, an arbitrary linearly independent subset $S$ can be extended to a Hamel basis. Just apply Zorn's lemma to the family of all linearly independent sets which contain $S$, ordered by inclusion. The maximal element will be a Hamel basis.
– mechanodroid
Jul 29 at 12:38
1
1
When the dimension is continuum, I think there is almost no hope for writing such explicit basis. Choice of one that includes a specific countable linearly independent set is possible though.
– i707107
Jul 29 at 5:32
When the dimension is continuum, I think there is almost no hope for writing such explicit basis. Choice of one that includes a specific countable linearly independent set is possible though.
– i707107
Jul 29 at 5:32
Possible duplicate of A Hamel basis for $ell^p$?
– adfriedman
Jul 29 at 6:02
Possible duplicate of A Hamel basis for $ell^p$?
– adfriedman
Jul 29 at 6:02
Thank you for the link! I changed the question to ask what (if any) 'benefits' there might be to being able to choose a Hamel basis including a certain linearly independent set of vectors. This is not a duplicate as far as I can tell.
– ertl
Jul 29 at 6:26
Thank you for the link! I changed the question to ask what (if any) 'benefits' there might be to being able to choose a Hamel basis including a certain linearly independent set of vectors. This is not a duplicate as far as I can tell.
– ertl
Jul 29 at 6:26
Yes, an arbitrary linearly independent subset $S$ can be extended to a Hamel basis. Just apply Zorn's lemma to the family of all linearly independent sets which contain $S$, ordered by inclusion. The maximal element will be a Hamel basis.
– mechanodroid
Jul 29 at 12:38
Yes, an arbitrary linearly independent subset $S$ can be extended to a Hamel basis. Just apply Zorn's lemma to the family of all linearly independent sets which contain $S$, ordered by inclusion. The maximal element will be a Hamel basis.
– mechanodroid
Jul 29 at 12:38
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
could we choose one that includes a specific (countable) set of vectors as a subset, given they are 'independent'?
Yes, we can.
If so, what are the applications of this fact?
None.
answered Jul 29 at 15:18
user357151
13.8k31140
1
Care to explain the "None"?
– T_M
Jul 29 at 15:26
1
I think "none" is at least an exaggeration, if not just false. For example, a simple application of the fact that we can always extend a linearly independent set to a basis (assuming AC) is the existence of a discontinuous linear functional on any infinite dimensional normed space.
– Rhys Steele
Aug 1 at 20:20
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
could we choose one that includes a specific (countable) set of vectors as a subset, given they are 'independent'?
Yes, we can.
If so, what are the applications of this fact?
None.
answered Jul 29 at 15:18
user357151
13.8k31140
1
Care to explain the "None"?
– T_M
Jul 29 at 15:26
1
I think "none" is at least an exaggeration, if not just false. For example, a simple application of the fact that we can always extend a linearly independent set to a basis (assuming AC) is the existence of a discontinuous linear functional on any infinite dimensional normed space.
– Rhys Steele
Aug 1 at 20:20
add a comment |Â
up vote
0
down vote
could we choose one that includes a specific (countable) set of vectors as a subset, given they are 'independent'?
Yes, we can.
If so, what are the applications of this fact?
None.
answered Jul 29 at 15:18
user357151
13.8k31140
1
Care to explain the "None"?
– T_M
Jul 29 at 15:26
1
I think "none" is at least an exaggeration, if not just false. For example, a simple application of the fact that we can always extend a linearly independent set to a basis (assuming AC) is the existence of a discontinuous linear functional on any infinite dimensional normed space.
– Rhys Steele
Aug 1 at 20:20
add a comment |Â
up vote
0
down vote
up vote
0
down vote
could we choose one that includes a specific (countable) set of vectors as a subset, given they are 'independent'?
Yes, we can.
If so, what are the applications of this fact?
None.
answered Jul 29 at 15:18
user357151
13.8k31140
could we choose one that includes a specific (countable) set of vectors as a subset, given they are 'independent'?
Yes, we can.
If so, what are the applications of this fact?
None.
answered Jul 29 at 15:18
user357151
13.8k31140
answered Jul 29 at 15:18
user357151
13.8k31140
answered Jul 29 at 15:18
user357151
13.8k31140
answered Jul 29 at 15:18
user357151
13.8k31140
13.8k31140
1
Care to explain the "None"?
– T_M
Jul 29 at 15:26
1
I think "none" is at least an exaggeration, if not just false. For example, a simple application of the fact that we can always extend a linearly independent set to a basis (assuming AC) is the existence of a discontinuous linear functional on any infinite dimensional normed space.
– Rhys Steele
Aug 1 at 20:20
add a comment |Â
1
Care to explain the "None"?
– T_M
Jul 29 at 15:26
1
I think "none" is at least an exaggeration, if not just false. For example, a simple application of the fact that we can always extend a linearly independent set to a basis (assuming AC) is the existence of a discontinuous linear functional on any infinite dimensional normed space.
– Rhys Steele
Aug 1 at 20:20
1
1
Care to explain the "None"?
– T_M
Jul 29 at 15:26
Care to explain the "None"?
– T_M
Jul 29 at 15:26
1
1
I think "none" is at least an exaggeration, if not just false. For example, a simple application of the fact that we can always extend a linearly independent set to a basis (assuming AC) is the existence of a discontinuous linear functional on any infinite dimensional normed space.
– Rhys Steele
Aug 1 at 20:20
I think "none" is at least an exaggeration, if not just false. For example, a simple application of the fact that we can always extend a linearly independent set to a basis (assuming AC) is the existence of a discontinuous linear functional on any infinite dimensional normed space.
– Rhys Steele
Aug 1 at 20:20
add a comment |Â
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1
When the dimension is continuum, I think there is almost no hope for writing such explicit basis. Choice of one that includes a specific countable linearly independent set is possible though.
– i707107
Jul 29 at 5:32
Possible duplicate of A Hamel basis for $ell^p$?
– adfriedman
Jul 29 at 6:02
Thank you for the link! I changed the question to ask what (if any) 'benefits' there might be to being able to choose a Hamel basis including a certain linearly independent set of vectors. This is not a duplicate as far as I can tell.
– ertl
Jul 29 at 6:26
Yes, an arbitrary linearly independent subset $S$ can be extended to a Hamel basis. Just apply Zorn's lemma to the family of all linearly independent sets which contain $S$, ordered by inclusion. The maximal element will be a Hamel basis.
– mechanodroid
Jul 29 at 12:38