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
yesterday
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
yesterday
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
yesterday
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 yesterday
PSG
3749
3749
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
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
Even when $mathcal A$ is countably infinite the sum $sum e_n$ does not exist either in the norm or weakly.
– Kavi Rama Murthy
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
add a comment |
3 Answers
3
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The sum $sum_{alphain A} v_alpha$ does make sense. It is defined to converge to $L in H$ if
$$forall varepsilon > 0 ,exists F_0 subseteq A text{ finite such that }forall F subseteq A text{ finite}, F supseteq F_0 text{ we have} left|sum_{alphain F}v_alpha - Lright| < varepsilon$$
However, the sum $sum_{alphain A} e_alpha$ of an orthonormal set only converges if $A$ is finite.
Indeed, assume that $sum_{alphain A} e_alpha = L$. For $varepsilon = frac12$ there exists $F_0 subseteq A$ finite such that for all $F subseteq A$ finite, $F supseteq F_0$ implies $left|sum_{alphain F} e_alpha -Lright| < frac12$.
For any $F_0 subseteq F subseteq A$ finite we have
$$sqrt{|F setminus F_0|}= left|sum_{alphain Fsetminus F_0}e_alpharight| = left|sum_{alphain F}e_alpha - sum_{alphain F_0}e_alpharight| le
left|sum_{alphain F}e_alpha - Lright| +left|L- sum_{alphain F_0}e_alpharight| < 1$$
so $F = F_0$. Therefore it necessarily holds $F_0 = A$ so $A$ is finite.
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 |
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; you'll get a contradiction.
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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The sum $sum_{alphain A} v_alpha$ does make sense. It is defined to converge to $L in H$ if
$$forall varepsilon > 0 ,exists F_0 subseteq A text{ finite such that }forall F subseteq A text{ finite}, F supseteq F_0 text{ we have} left|sum_{alphain F}v_alpha - Lright| < varepsilon$$
However, the sum $sum_{alphain A} e_alpha$ of an orthonormal set only converges if $A$ is finite.
Indeed, assume that $sum_{alphain A} e_alpha = L$. For $varepsilon = frac12$ there exists $F_0 subseteq A$ finite such that for all $F subseteq A$ finite, $F supseteq F_0$ implies $left|sum_{alphain F} e_alpha -Lright| < frac12$.
For any $F_0 subseteq F subseteq A$ finite we have
$$sqrt{|F setminus F_0|}= left|sum_{alphain Fsetminus F_0}e_alpharight| = left|sum_{alphain F}e_alpha - sum_{alphain F_0}e_alpharight| le
left|sum_{alphain F}e_alpha - Lright| +left|L- sum_{alphain F_0}e_alpharight| < 1$$
so $F = F_0$. Therefore it necessarily holds $F_0 = A$ so $A$ is finite.
add a comment |
The sum $sum_{alphain A} v_alpha$ does make sense. It is defined to converge to $L in H$ if
$$forall varepsilon > 0 ,exists F_0 subseteq A text{ finite such that }forall F subseteq A text{ finite}, F supseteq F_0 text{ we have} left|sum_{alphain F}v_alpha - Lright| < varepsilon$$
However, the sum $sum_{alphain A} e_alpha$ of an orthonormal set only converges if $A$ is finite.
Indeed, assume that $sum_{alphain A} e_alpha = L$. For $varepsilon = frac12$ there exists $F_0 subseteq A$ finite such that for all $F subseteq A$ finite, $F supseteq F_0$ implies $left|sum_{alphain F} e_alpha -Lright| < frac12$.
For any $F_0 subseteq F subseteq A$ finite we have
$$sqrt{|F setminus F_0|}= left|sum_{alphain Fsetminus F_0}e_alpharight| = left|sum_{alphain F}e_alpha - sum_{alphain F_0}e_alpharight| le
left|sum_{alphain F}e_alpha - Lright| +left|L- sum_{alphain F_0}e_alpharight| < 1$$
so $F = F_0$. Therefore it necessarily holds $F_0 = A$ so $A$ is finite.
add a comment |
The sum $sum_{alphain A} v_alpha$ does make sense. It is defined to converge to $L in H$ if
$$forall varepsilon > 0 ,exists F_0 subseteq A text{ finite such that }forall F subseteq A text{ finite}, F supseteq F_0 text{ we have} left|sum_{alphain F}v_alpha - Lright| < varepsilon$$
However, the sum $sum_{alphain A} e_alpha$ of an orthonormal set only converges if $A$ is finite.
Indeed, assume that $sum_{alphain A} e_alpha = L$. For $varepsilon = frac12$ there exists $F_0 subseteq A$ finite such that for all $F subseteq A$ finite, $F supseteq F_0$ implies $left|sum_{alphain F} e_alpha -Lright| < frac12$.
For any $F_0 subseteq F subseteq A$ finite we have
$$sqrt{|F setminus F_0|}= left|sum_{alphain Fsetminus F_0}e_alpharight| = left|sum_{alphain F}e_alpha - sum_{alphain F_0}e_alpharight| le
left|sum_{alphain F}e_alpha - Lright| +left|L- sum_{alphain F_0}e_alpharight| < 1$$
so $F = F_0$. Therefore it necessarily holds $F_0 = A$ so $A$ is finite.
The sum $sum_{alphain A} v_alpha$ does make sense. It is defined to converge to $L in H$ if
$$forall varepsilon > 0 ,exists F_0 subseteq A text{ finite such that }forall F subseteq A text{ finite}, F supseteq F_0 text{ we have} left|sum_{alphain F}v_alpha - Lright| < varepsilon$$
However, the sum $sum_{alphain A} e_alpha$ of an orthonormal set only converges if $A$ is finite.
Indeed, assume that $sum_{alphain A} e_alpha = L$. For $varepsilon = frac12$ there exists $F_0 subseteq A$ finite such that for all $F subseteq A$ finite, $F supseteq F_0$ implies $left|sum_{alphain F} e_alpha -Lright| < frac12$.
For any $F_0 subseteq F subseteq A$ finite we have
$$sqrt{|F setminus F_0|}= left|sum_{alphain Fsetminus F_0}e_alpharight| = left|sum_{alphain F}e_alpha - sum_{alphain F_0}e_alpharight| le
left|sum_{alphain F}e_alpha - Lright| +left|L- sum_{alphain F_0}e_alpharight| < 1$$
so $F = F_0$. Therefore it necessarily holds $F_0 = A$ so $A$ is finite.
answered yesterday
mechanodroid
26.9k62447
26.9k62447
add a comment |
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.
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 yesterday
supinf
6,0791028
6,0791028
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; you'll get a contradiction.
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; you'll get a contradiction.
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; you'll get a contradiction.
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; you'll get a contradiction.
edited yesterday
answered yesterday
mathcounterexamples.net
25.1k21953
25.1k21953
add a comment |
add a comment |
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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
yesterday
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
yesterday
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
yesterday