Uncountable sum of vectors in a Hilbert Space
I am currently reading Hilbert Spaces and confused about a thing. Say, $C={e_alpha : alphainmathcal{A}}$ be a complete orthonormal set of a Hilbert Space $H$, possibly uncountable. Is $sum_{alphainmathcal{A}}e_alpha$ well defined ? I think it should be, is there some kind of convergence needed for these sums?
functional-analysis hilbert-spaces
add a comment |
I am currently reading Hilbert Spaces and confused about a thing. Say, $C={e_alpha : alphainmathcal{A}}$ be a complete orthonormal set of a Hilbert Space $H$, possibly uncountable. Is $sum_{alphainmathcal{A}}e_alpha$ well defined ? I think it should be, is there some kind of convergence needed for these sums?
functional-analysis hilbert-spaces
2
Even when $mathcal A$ is countably infinite the sum $sum e_n$ does not exist either in the norm or weakly.
– Kavi Rama Murthy
45 mins ago
Oh, I see if it was a well defined vector, then the norm would be in $mathbf{R}$, which is not the case here. Is that ok? @KaviRamaMurthy
– PSG
42 mins ago
1
If $sum e_n=x$, say in weak topology, then we have $sum langle y,e_nrangle =langle y,xrangle$ for all $y$. You get a contradiction by taking $y=sum frac 1 j e_j$.
– Kavi Rama Murthy
39 mins ago
add a comment |
I am currently reading Hilbert Spaces and confused about a thing. Say, $C={e_alpha : alphainmathcal{A}}$ be a complete orthonormal set of a Hilbert Space $H$, possibly uncountable. Is $sum_{alphainmathcal{A}}e_alpha$ well defined ? I think it should be, is there some kind of convergence needed for these sums?
functional-analysis hilbert-spaces
I am currently reading Hilbert Spaces and confused about a thing. Say, $C={e_alpha : alphainmathcal{A}}$ be a complete orthonormal set of a Hilbert Space $H$, possibly uncountable. Is $sum_{alphainmathcal{A}}e_alpha$ well defined ? I think it should be, is there some kind of convergence needed for these sums?
functional-analysis hilbert-spaces
functional-analysis hilbert-spaces
asked 51 mins ago
PSG
3609
3609
2
Even when $mathcal A$ is countably infinite the sum $sum e_n$ does not exist either in the norm or weakly.
– Kavi Rama Murthy
45 mins ago
Oh, I see if it was a well defined vector, then the norm would be in $mathbf{R}$, which is not the case here. Is that ok? @KaviRamaMurthy
– PSG
42 mins ago
1
If $sum e_n=x$, say in weak topology, then we have $sum langle y,e_nrangle =langle y,xrangle$ for all $y$. You get a contradiction by taking $y=sum frac 1 j e_j$.
– Kavi Rama Murthy
39 mins ago
add a comment |
2
Even when $mathcal A$ is countably infinite the sum $sum e_n$ does not exist either in the norm or weakly.
– Kavi Rama Murthy
45 mins ago
Oh, I see if it was a well defined vector, then the norm would be in $mathbf{R}$, which is not the case here. Is that ok? @KaviRamaMurthy
– PSG
42 mins ago
1
If $sum e_n=x$, say in weak topology, then we have $sum langle y,e_nrangle =langle y,xrangle$ for all $y$. You get a contradiction by taking $y=sum frac 1 j e_j$.
– Kavi Rama Murthy
39 mins ago
2
2
Even when $mathcal A$ is countably infinite the sum $sum e_n$ does not exist either in the norm or weakly.
– Kavi Rama Murthy
45 mins ago
Even when $mathcal A$ is countably infinite the sum $sum e_n$ does not exist either in the norm or weakly.
– Kavi Rama Murthy
45 mins ago
Oh, I see if it was a well defined vector, then the norm would be in $mathbf{R}$, which is not the case here. Is that ok? @KaviRamaMurthy
– PSG
42 mins ago
Oh, I see if it was a well defined vector, then the norm would be in $mathbf{R}$, which is not the case here. Is that ok? @KaviRamaMurthy
– PSG
42 mins ago
1
1
If $sum e_n=x$, say in weak topology, then we have $sum langle y,e_nrangle =langle y,xrangle$ for all $y$. You get a contradiction by taking $y=sum frac 1 j e_j$.
– Kavi Rama Murthy
39 mins ago
If $sum e_n=x$, say in weak topology, then we have $sum langle y,e_nrangle =langle y,xrangle$ for all $y$. You get a contradiction by taking $y=sum frac 1 j e_j$.
– Kavi Rama Murthy
39 mins ago
add a comment |
2 Answers
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In an infinite dimensional Hilbert space the term
$$
sum_{ain mathcal A} e_a
$$
is not well defined.
Even for countable $mathcal A$ this sum does not converge.
For defining uncountable sums it is usually required that at most countable many summands are nonzero
and that the countable sum over the nonzero entries converges absolutely.
add a comment |
The background of your question is how do you define $displaystyle sum_{i in I} c_i$ in a Banach space for any set $I$?
The usual way is to say that the sum $displaystyle sum_{i in I} c_alpha$ of a set $mathcal C = {c_i ; i in I}$ of vectors exists when following Cauchy criteria is met:
$$(forall epsilon > 0) , (exists J_0 in mathcal F(I)) , (forall K in mathcal F(I setminus J_0)) , leftVert displaystyle sum_{k in K} c_k rightVert< epsilon $$
Where $mathcal F(A)$ is defined as the sets of finite subsets of $A$.
This has interesting consequences.
- Such a sum $displaystyle sum_{i in I} c_alpha$ does make sense only if the set of non zero elements of $mathcal C$ is at most countable.
- For a family of uncountable vectors of norm equal to $1$, which is your initial question, the sum cannot exist: take $epsilon = 1/2$ in the definition above.
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
In an infinite dimensional Hilbert space the term
$$
sum_{ain mathcal A} e_a
$$
is not well defined.
Even for countable $mathcal A$ this sum does not converge.
For defining uncountable sums it is usually required that at most countable many summands are nonzero
and that the countable sum over the nonzero entries converges absolutely.
add a comment |
In an infinite dimensional Hilbert space the term
$$
sum_{ain mathcal A} e_a
$$
is not well defined.
Even for countable $mathcal A$ this sum does not converge.
For defining uncountable sums it is usually required that at most countable many summands are nonzero
and that the countable sum over the nonzero entries converges absolutely.
add a comment |
In an infinite dimensional Hilbert space the term
$$
sum_{ain mathcal A} e_a
$$
is not well defined.
Even for countable $mathcal A$ this sum does not converge.
For defining uncountable sums it is usually required that at most countable many summands are nonzero
and that the countable sum over the nonzero entries converges absolutely.
In an infinite dimensional Hilbert space the term
$$
sum_{ain mathcal A} e_a
$$
is not well defined.
Even for countable $mathcal A$ this sum does not converge.
For defining uncountable sums it is usually required that at most countable many summands are nonzero
and that the countable sum over the nonzero entries converges absolutely.
answered 43 mins ago
supinf
5,9991027
5,9991027
add a comment |
add a comment |
The background of your question is how do you define $displaystyle sum_{i in I} c_i$ in a Banach space for any set $I$?
The usual way is to say that the sum $displaystyle sum_{i in I} c_alpha$ of a set $mathcal C = {c_i ; i in I}$ of vectors exists when following Cauchy criteria is met:
$$(forall epsilon > 0) , (exists J_0 in mathcal F(I)) , (forall K in mathcal F(I setminus J_0)) , leftVert displaystyle sum_{k in K} c_k rightVert< epsilon $$
Where $mathcal F(A)$ is defined as the sets of finite subsets of $A$.
This has interesting consequences.
- Such a sum $displaystyle sum_{i in I} c_alpha$ does make sense only if the set of non zero elements of $mathcal C$ is at most countable.
- For a family of uncountable vectors of norm equal to $1$, which is your initial question, the sum cannot exist: take $epsilon = 1/2$ in the definition above.
add a comment |
The background of your question is how do you define $displaystyle sum_{i in I} c_i$ in a Banach space for any set $I$?
The usual way is to say that the sum $displaystyle sum_{i in I} c_alpha$ of a set $mathcal C = {c_i ; i in I}$ of vectors exists when following Cauchy criteria is met:
$$(forall epsilon > 0) , (exists J_0 in mathcal F(I)) , (forall K in mathcal F(I setminus J_0)) , leftVert displaystyle sum_{k in K} c_k rightVert< epsilon $$
Where $mathcal F(A)$ is defined as the sets of finite subsets of $A$.
This has interesting consequences.
- Such a sum $displaystyle sum_{i in I} c_alpha$ does make sense only if the set of non zero elements of $mathcal C$ is at most countable.
- For a family of uncountable vectors of norm equal to $1$, which is your initial question, the sum cannot exist: take $epsilon = 1/2$ in the definition above.
add a comment |
The background of your question is how do you define $displaystyle sum_{i in I} c_i$ in a Banach space for any set $I$?
The usual way is to say that the sum $displaystyle sum_{i in I} c_alpha$ of a set $mathcal C = {c_i ; i in I}$ of vectors exists when following Cauchy criteria is met:
$$(forall epsilon > 0) , (exists J_0 in mathcal F(I)) , (forall K in mathcal F(I setminus J_0)) , leftVert displaystyle sum_{k in K} c_k rightVert< epsilon $$
Where $mathcal F(A)$ is defined as the sets of finite subsets of $A$.
This has interesting consequences.
- Such a sum $displaystyle sum_{i in I} c_alpha$ does make sense only if the set of non zero elements of $mathcal C$ is at most countable.
- For a family of uncountable vectors of norm equal to $1$, which is your initial question, the sum cannot exist: take $epsilon = 1/2$ in the definition above.
The background of your question is how do you define $displaystyle sum_{i in I} c_i$ in a Banach space for any set $I$?
The usual way is to say that the sum $displaystyle sum_{i in I} c_alpha$ of a set $mathcal C = {c_i ; i in I}$ of vectors exists when following Cauchy criteria is met:
$$(forall epsilon > 0) , (exists J_0 in mathcal F(I)) , (forall K in mathcal F(I setminus J_0)) , leftVert displaystyle sum_{k in K} c_k rightVert< epsilon $$
Where $mathcal F(A)$ is defined as the sets of finite subsets of $A$.
This has interesting consequences.
- Such a sum $displaystyle sum_{i in I} c_alpha$ does make sense only if the set of non zero elements of $mathcal C$ is at most countable.
- For a family of uncountable vectors of norm equal to $1$, which is your initial question, the sum cannot exist: take $epsilon = 1/2$ in the definition above.
answered 15 mins ago
mathcounterexamples.net
24.8k21753
24.8k21753
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2
Even when $mathcal A$ is countably infinite the sum $sum e_n$ does not exist either in the norm or weakly.
– Kavi Rama Murthy
45 mins ago
Oh, I see if it was a well defined vector, then the norm would be in $mathbf{R}$, which is not the case here. Is that ok? @KaviRamaMurthy
– PSG
42 mins ago
1
If $sum e_n=x$, say in weak topology, then we have $sum langle y,e_nrangle =langle y,xrangle$ for all $y$. You get a contradiction by taking $y=sum frac 1 j e_j$.
– Kavi Rama Murthy
39 mins ago