Mixing
Converges in inner product space
$begingroup$
Assume $(f_i)_{iin I}$ is an orthonormal/orthogonal system in an (complex) inner product space. Does $$sum_{iin I}langle f_i,frangle f_i$$ always converges for any $f$ (may not to $f$)? Especially, when $I$ is uncountable.
And we define convergency of a sum $sum_I x_i$ on an arbitrary set by: if there exists $g$ and for any $epsilon>0$, there exists a finite set $F$ and for all other finite set $Hsupseteq F,|g-sum_H x_i|<epsilon$ as an analogue to the limit of series in analysis.
functional-analysis hilbert-spaces
asked Jan 17 at 23:01
CO2CO2
1629
$endgroup$
add a comment |
$begingroup$
Assume $(f_i)_{iin I}$ is an orthonormal/orthogonal system in an (complex) inner product space. Does $$sum_{iin I}langle f_i,frangle f_i$$ always converges for any $f$ (may not to $f$)? Especially, when $I$ is uncountable.
And we define convergency of a sum $sum_I x_i$ on an arbitrary set by: if there exists $g$ and for any $epsilon>0$, there exists a finite set $F$ and for all other finite set $Hsupseteq F,|g-sum_H x_i|<epsilon$ as an analogue to the limit of series in analysis.
functional-analysis hilbert-spaces
asked Jan 17 at 23:01
CO2CO2
1629
$endgroup$
$begingroup$
I believe it only converges to $f$ if the orthonormal system is "complete". That is, the span of $f_i$ is dense in your vector space.
$endgroup$
– Jeffery Opoku-Mensah
Jan 17 at 23:07
$begingroup$
@JefferyOpoku-Mensah Completeness does not imply anything. When the space is Hilbert then completeness is enough and sufficient. But I just want to know for any arbitrary system whether this sum converges to something.
$endgroup$
– CO2
Jan 17 at 23:11
$begingroup$
Ah yeah. I implicitly read your problem as in a hilbert space, given the tag.
$endgroup$
– Jeffery Opoku-Mensah
Jan 17 at 23:13
$begingroup$
@JefferyOpoku-Mensah Well, yes it is on the way there, as preparation knowledge. Maybe I should remove the tag.
$endgroup$
– CO2
Jan 17 at 23:23
$begingroup$
Bessel's inequality.
$endgroup$
– copper.hat
Jan 18 at 0:15
add a comment |
$begingroup$
Assume $(f_i)_{iin I}$ is an orthonormal/orthogonal system in an (complex) inner product space. Does $$sum_{iin I}langle f_i,frangle f_i$$ always converges for any $f$ (may not to $f$)? Especially, when $I$ is uncountable.
And we define convergency of a sum $sum_I x_i$ on an arbitrary set by: if there exists $g$ and for any $epsilon>0$, there exists a finite set $F$ and for all other finite set $Hsupseteq F,|g-sum_H x_i|<epsilon$ as an analogue to the limit of series in analysis.
functional-analysis hilbert-spaces
asked Jan 17 at 23:01
CO2CO2
1629
$endgroup$
Assume $(f_i)_{iin I}$ is an orthonormal/orthogonal system in an (complex) inner product space. Does $$sum_{iin I}langle f_i,frangle f_i$$ always converges for any $f$ (may not to $f$)? Especially, when $I$ is uncountable.
And we define convergency of a sum $sum_I x_i$ on an arbitrary set by: if there exists $g$ and for any $epsilon>0$, there exists a finite set $F$ and for all other finite set $Hsupseteq F,|g-sum_H x_i|<epsilon$ as an analogue to the limit of series in analysis.
functional-analysis hilbert-spaces
functional-analysis hilbert-spaces
asked Jan 17 at 23:01
CO2CO2
1629
asked Jan 17 at 23:01
CO2CO2
1629
edited Jan 17 at 23:27
CO2
asked Jan 17 at 23:01
CO2CO2
1629
asked Jan 17 at 23:01
CO2CO2
1629
asked Jan 17 at 23:01
CO2CO2
1629
1629
$begingroup$
I believe it only converges to $f$ if the orthonormal system is "complete". That is, the span of $f_i$ is dense in your vector space.
$endgroup$
– Jeffery Opoku-Mensah
Jan 17 at 23:07
$begingroup$
@JefferyOpoku-Mensah Completeness does not imply anything. When the space is Hilbert then completeness is enough and sufficient. But I just want to know for any arbitrary system whether this sum converges to something.
$endgroup$
– CO2
Jan 17 at 23:11
$begingroup$
Ah yeah. I implicitly read your problem as in a hilbert space, given the tag.
$endgroup$
– Jeffery Opoku-Mensah
Jan 17 at 23:13
$begingroup$
@JefferyOpoku-Mensah Well, yes it is on the way there, as preparation knowledge. Maybe I should remove the tag.
$endgroup$
– CO2
Jan 17 at 23:23
$begingroup$
Bessel's inequality.
$endgroup$
– copper.hat
Jan 18 at 0:15
add a comment |
$begingroup$
I believe it only converges to $f$ if the orthonormal system is "complete". That is, the span of $f_i$ is dense in your vector space.
$endgroup$
– Jeffery Opoku-Mensah
Jan 17 at 23:07
$begingroup$
@JefferyOpoku-Mensah Completeness does not imply anything. When the space is Hilbert then completeness is enough and sufficient. But I just want to know for any arbitrary system whether this sum converges to something.
$endgroup$
– CO2
Jan 17 at 23:11
$begingroup$
Ah yeah. I implicitly read your problem as in a hilbert space, given the tag.
$endgroup$
– Jeffery Opoku-Mensah
Jan 17 at 23:13
$begingroup$
@JefferyOpoku-Mensah Well, yes it is on the way there, as preparation knowledge. Maybe I should remove the tag.
$endgroup$
– CO2
Jan 17 at 23:23
$begingroup$
Bessel's inequality.
$endgroup$
– copper.hat
Jan 18 at 0:15
$begingroup$
I believe it only converges to $f$ if the orthonormal system is "complete". That is, the span of $f_i$ is dense in your vector space.
$endgroup$
– Jeffery Opoku-Mensah
Jan 17 at 23:07
$begingroup$
I believe it only converges to $f$ if the orthonormal system is "complete". That is, the span of $f_i$ is dense in your vector space.
$endgroup$
– Jeffery Opoku-Mensah
Jan 17 at 23:07
$begingroup$
@JefferyOpoku-Mensah Completeness does not imply anything. When the space is Hilbert then completeness is enough and sufficient. But I just want to know for any arbitrary system whether this sum converges to something.
$endgroup$
– CO2
Jan 17 at 23:11
$begingroup$
@JefferyOpoku-Mensah Completeness does not imply anything. When the space is Hilbert then completeness is enough and sufficient. But I just want to know for any arbitrary system whether this sum converges to something.
$endgroup$
– CO2
Jan 17 at 23:11
$begingroup$
Ah yeah. I implicitly read your problem as in a hilbert space, given the tag.
$endgroup$
– Jeffery Opoku-Mensah
Jan 17 at 23:13
$begingroup$
Ah yeah. I implicitly read your problem as in a hilbert space, given the tag.
$endgroup$
– Jeffery Opoku-Mensah
Jan 17 at 23:13
$begingroup$
@JefferyOpoku-Mensah Well, yes it is on the way there, as preparation knowledge. Maybe I should remove the tag.
$endgroup$
– CO2
Jan 17 at 23:23
$begingroup$
@JefferyOpoku-Mensah Well, yes it is on the way there, as preparation knowledge. Maybe I should remove the tag.
$endgroup$
– CO2
Jan 17 at 23:23
$begingroup$
Bessel's inequality.
$endgroup$
– copper.hat
Jan 18 at 0:15
$begingroup$
Bessel's inequality.
$endgroup$
– copper.hat
Jan 18 at 0:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose you let $X$ be the set of sequences $x_n in ell^2$ such that for some constant $C$, $x_n = C/n$ for all but finitely many $n$; and the orthonormal system is ${ e_{2n} mid n in mathbb{N}^+ }$. Then for $f = (frac{1}{n})_{n=1}^infty in X$, the sum of $sum_{n=1}^infty langle f, e_{2n} rangle e_{2n}$ would have to be $g$ where $g_n = frac{1}{n}$ if $n$ is even or $g_n = 0$ if $n$ is odd; but that is not an element of $X$.
answered Jan 18 at 1:20
Daniel ScheplerDaniel Schepler
8,9591620
$endgroup$
$begingroup$
What is your $X$? Is it a space only finitely many terms are $C/n$ all others are zero? Something like $(2/1,2/2,2/3,0,2/5,0,0,2/8,0,0,...)$?
$endgroup$
– CO2
Jan 18 at 1:47
$begingroup$
No, it's the one where all but finitely many terms are $C/n$. Another way to say it would be that $n x_n$ is eventually constant. An example of an element would be $(0, 3, 2, pi, frac{3}{5}, frac{3}{6}, frac{3}{7}, frac{3}{8}, ldots)$.
$endgroup$
– Daniel Schepler
Jan 18 at 1:52
add a comment |
$begingroup$
Completeness is necessary. Otherwise the series converges in the completion, not necessarily in $X$. [For a counterexample you can take any element in the completion of an inner product space that does not belong to the original space and look at its expansion with respect to an orthonormal basis].
Now assume that $X$ is a Hilbert space.
Hints for proof of convergence: for any finite subcollection ${g_1,g_2,..,g_n}$ of ${f_i}$ we have $|f-sumlimits_{k=1}^{n} langle f, g_krangle g_k|^{2} geq 0 $. Expanding this we get $sumlimits_{k=1}^{n} |langle f, g_krangle |^{2} leq |f|^{2}$. This implies that ${i: langle f, f_irangle neq 0}$ is at most countable. If this set is $i_1,i_2,...$ then $sumlimits_{k=1}^{infty} langle f, f_{i_k}rangle f_{i_k}$ converges because its partial sums are Cauchy. Now verify that the required uncountable sum exists and equals $sumlimits_{k=1}^{infty} langle f, f_{i_k}rangle f_{i_k}$.
answered Jan 17 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.8k42262
$endgroup$
$begingroup$
The OP requires a series involving $f in X$. So I don't think your counter-example of the first paragraph works because you require $f notin X$
$endgroup$
– DisintegratingByParts
Jan 20 at 6:37
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
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oldest
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$begingroup$
Suppose you let $X$ be the set of sequences $x_n in ell^2$ such that for some constant $C$, $x_n = C/n$ for all but finitely many $n$; and the orthonormal system is ${ e_{2n} mid n in mathbb{N}^+ }$. Then for $f = (frac{1}{n})_{n=1}^infty in X$, the sum of $sum_{n=1}^infty langle f, e_{2n} rangle e_{2n}$ would have to be $g$ where $g_n = frac{1}{n}$ if $n$ is even or $g_n = 0$ if $n$ is odd; but that is not an element of $X$.
answered Jan 18 at 1:20
Daniel ScheplerDaniel Schepler
8,9591620
$endgroup$
$begingroup$
What is your $X$? Is it a space only finitely many terms are $C/n$ all others are zero? Something like $(2/1,2/2,2/3,0,2/5,0,0,2/8,0,0,...)$?
$endgroup$
– CO2
Jan 18 at 1:47
$begingroup$
No, it's the one where all but finitely many terms are $C/n$. Another way to say it would be that $n x_n$ is eventually constant. An example of an element would be $(0, 3, 2, pi, frac{3}{5}, frac{3}{6}, frac{3}{7}, frac{3}{8}, ldots)$.
$endgroup$
– Daniel Schepler
Jan 18 at 1:52
add a comment |
$begingroup$
Suppose you let $X$ be the set of sequences $x_n in ell^2$ such that for some constant $C$, $x_n = C/n$ for all but finitely many $n$; and the orthonormal system is ${ e_{2n} mid n in mathbb{N}^+ }$. Then for $f = (frac{1}{n})_{n=1}^infty in X$, the sum of $sum_{n=1}^infty langle f, e_{2n} rangle e_{2n}$ would have to be $g$ where $g_n = frac{1}{n}$ if $n$ is even or $g_n = 0$ if $n$ is odd; but that is not an element of $X$.
answered Jan 18 at 1:20
Daniel ScheplerDaniel Schepler
8,9591620
$endgroup$
$begingroup$
What is your $X$? Is it a space only finitely many terms are $C/n$ all others are zero? Something like $(2/1,2/2,2/3,0,2/5,0,0,2/8,0,0,...)$?
$endgroup$
– CO2
Jan 18 at 1:47
$begingroup$
No, it's the one where all but finitely many terms are $C/n$. Another way to say it would be that $n x_n$ is eventually constant. An example of an element would be $(0, 3, 2, pi, frac{3}{5}, frac{3}{6}, frac{3}{7}, frac{3}{8}, ldots)$.
$endgroup$
– Daniel Schepler
Jan 18 at 1:52
add a comment |
$begingroup$
Suppose you let $X$ be the set of sequences $x_n in ell^2$ such that for some constant $C$, $x_n = C/n$ for all but finitely many $n$; and the orthonormal system is ${ e_{2n} mid n in mathbb{N}^+ }$. Then for $f = (frac{1}{n})_{n=1}^infty in X$, the sum of $sum_{n=1}^infty langle f, e_{2n} rangle e_{2n}$ would have to be $g$ where $g_n = frac{1}{n}$ if $n$ is even or $g_n = 0$ if $n$ is odd; but that is not an element of $X$.
answered Jan 18 at 1:20
Daniel ScheplerDaniel Schepler
8,9591620
$endgroup$
Suppose you let $X$ be the set of sequences $x_n in ell^2$ such that for some constant $C$, $x_n = C/n$ for all but finitely many $n$; and the orthonormal system is ${ e_{2n} mid n in mathbb{N}^+ }$. Then for $f = (frac{1}{n})_{n=1}^infty in X$, the sum of $sum_{n=1}^infty langle f, e_{2n} rangle e_{2n}$ would have to be $g$ where $g_n = frac{1}{n}$ if $n$ is even or $g_n = 0$ if $n$ is odd; but that is not an element of $X$.
answered Jan 18 at 1:20
Daniel ScheplerDaniel Schepler
8,9591620
answered Jan 18 at 1:20
Daniel ScheplerDaniel Schepler
8,9591620
answered Jan 18 at 1:20
Daniel ScheplerDaniel Schepler
8,9591620
answered Jan 18 at 1:20
Daniel ScheplerDaniel Schepler
8,9591620
8,9591620
$begingroup$
What is your $X$? Is it a space only finitely many terms are $C/n$ all others are zero? Something like $(2/1,2/2,2/3,0,2/5,0,0,2/8,0,0,...)$?
$endgroup$
– CO2
Jan 18 at 1:47
$begingroup$
No, it's the one where all but finitely many terms are $C/n$. Another way to say it would be that $n x_n$ is eventually constant. An example of an element would be $(0, 3, 2, pi, frac{3}{5}, frac{3}{6}, frac{3}{7}, frac{3}{8}, ldots)$.
$endgroup$
– Daniel Schepler
Jan 18 at 1:52
add a comment |
$begingroup$
What is your $X$? Is it a space only finitely many terms are $C/n$ all others are zero? Something like $(2/1,2/2,2/3,0,2/5,0,0,2/8,0,0,...)$?
$endgroup$
– CO2
Jan 18 at 1:47
$begingroup$
No, it's the one where all but finitely many terms are $C/n$. Another way to say it would be that $n x_n$ is eventually constant. An example of an element would be $(0, 3, 2, pi, frac{3}{5}, frac{3}{6}, frac{3}{7}, frac{3}{8}, ldots)$.
$endgroup$
– Daniel Schepler
Jan 18 at 1:52
$begingroup$
What is your $X$? Is it a space only finitely many terms are $C/n$ all others are zero? Something like $(2/1,2/2,2/3,0,2/5,0,0,2/8,0,0,...)$?
$endgroup$
– CO2
Jan 18 at 1:47
$begingroup$
What is your $X$? Is it a space only finitely many terms are $C/n$ all others are zero? Something like $(2/1,2/2,2/3,0,2/5,0,0,2/8,0,0,...)$?
$endgroup$
– CO2
Jan 18 at 1:47
$begingroup$
No, it's the one where all but finitely many terms are $C/n$. Another way to say it would be that $n x_n$ is eventually constant. An example of an element would be $(0, 3, 2, pi, frac{3}{5}, frac{3}{6}, frac{3}{7}, frac{3}{8}, ldots)$.
$endgroup$
– Daniel Schepler
Jan 18 at 1:52
$begingroup$
No, it's the one where all but finitely many terms are $C/n$. Another way to say it would be that $n x_n$ is eventually constant. An example of an element would be $(0, 3, 2, pi, frac{3}{5}, frac{3}{6}, frac{3}{7}, frac{3}{8}, ldots)$.
$endgroup$
– Daniel Schepler
Jan 18 at 1:52
add a comment |
$begingroup$
Completeness is necessary. Otherwise the series converges in the completion, not necessarily in $X$. [For a counterexample you can take any element in the completion of an inner product space that does not belong to the original space and look at its expansion with respect to an orthonormal basis].
Now assume that $X$ is a Hilbert space.
Hints for proof of convergence: for any finite subcollection ${g_1,g_2,..,g_n}$ of ${f_i}$ we have $|f-sumlimits_{k=1}^{n} langle f, g_krangle g_k|^{2} geq 0 $. Expanding this we get $sumlimits_{k=1}^{n} |langle f, g_krangle |^{2} leq |f|^{2}$. This implies that ${i: langle f, f_irangle neq 0}$ is at most countable. If this set is $i_1,i_2,...$ then $sumlimits_{k=1}^{infty} langle f, f_{i_k}rangle f_{i_k}$ converges because its partial sums are Cauchy. Now verify that the required uncountable sum exists and equals $sumlimits_{k=1}^{infty} langle f, f_{i_k}rangle f_{i_k}$.
answered Jan 17 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.8k42262
$endgroup$
$begingroup$
The OP requires a series involving $f in X$. So I don't think your counter-example of the first paragraph works because you require $f notin X$
$endgroup$
– DisintegratingByParts
Jan 20 at 6:37
add a comment |
$begingroup$
Completeness is necessary. Otherwise the series converges in the completion, not necessarily in $X$. [For a counterexample you can take any element in the completion of an inner product space that does not belong to the original space and look at its expansion with respect to an orthonormal basis].
Now assume that $X$ is a Hilbert space.
Hints for proof of convergence: for any finite subcollection ${g_1,g_2,..,g_n}$ of ${f_i}$ we have $|f-sumlimits_{k=1}^{n} langle f, g_krangle g_k|^{2} geq 0 $. Expanding this we get $sumlimits_{k=1}^{n} |langle f, g_krangle |^{2} leq |f|^{2}$. This implies that ${i: langle f, f_irangle neq 0}$ is at most countable. If this set is $i_1,i_2,...$ then $sumlimits_{k=1}^{infty} langle f, f_{i_k}rangle f_{i_k}$ converges because its partial sums are Cauchy. Now verify that the required uncountable sum exists and equals $sumlimits_{k=1}^{infty} langle f, f_{i_k}rangle f_{i_k}$.
answered Jan 17 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.8k42262
$endgroup$
$begingroup$
The OP requires a series involving $f in X$. So I don't think your counter-example of the first paragraph works because you require $f notin X$
$endgroup$
– DisintegratingByParts
Jan 20 at 6:37
add a comment |
$begingroup$
Completeness is necessary. Otherwise the series converges in the completion, not necessarily in $X$. [For a counterexample you can take any element in the completion of an inner product space that does not belong to the original space and look at its expansion with respect to an orthonormal basis].
Now assume that $X$ is a Hilbert space.
Hints for proof of convergence: for any finite subcollection ${g_1,g_2,..,g_n}$ of ${f_i}$ we have $|f-sumlimits_{k=1}^{n} langle f, g_krangle g_k|^{2} geq 0 $. Expanding this we get $sumlimits_{k=1}^{n} |langle f, g_krangle |^{2} leq |f|^{2}$. This implies that ${i: langle f, f_irangle neq 0}$ is at most countable. If this set is $i_1,i_2,...$ then $sumlimits_{k=1}^{infty} langle f, f_{i_k}rangle f_{i_k}$ converges because its partial sums are Cauchy. Now verify that the required uncountable sum exists and equals $sumlimits_{k=1}^{infty} langle f, f_{i_k}rangle f_{i_k}$.
answered Jan 17 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.8k42262
$endgroup$
Completeness is necessary. Otherwise the series converges in the completion, not necessarily in $X$. [For a counterexample you can take any element in the completion of an inner product space that does not belong to the original space and look at its expansion with respect to an orthonormal basis].
Now assume that $X$ is a Hilbert space.
Hints for proof of convergence: for any finite subcollection ${g_1,g_2,..,g_n}$ of ${f_i}$ we have $|f-sumlimits_{k=1}^{n} langle f, g_krangle g_k|^{2} geq 0 $. Expanding this we get $sumlimits_{k=1}^{n} |langle f, g_krangle |^{2} leq |f|^{2}$. This implies that ${i: langle f, f_irangle neq 0}$ is at most countable. If this set is $i_1,i_2,...$ then $sumlimits_{k=1}^{infty} langle f, f_{i_k}rangle f_{i_k}$ converges because its partial sums are Cauchy. Now verify that the required uncountable sum exists and equals $sumlimits_{k=1}^{infty} langle f, f_{i_k}rangle f_{i_k}$.
answered Jan 17 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.8k42262
edited Jan 18 at 23:57
answered Jan 17 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.8k42262
answered Jan 17 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.8k42262
answered Jan 17 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.8k42262
62.8k42262
$begingroup$
The OP requires a series involving $f in X$. So I don't think your counter-example of the first paragraph works because you require $f notin X$
$endgroup$
– DisintegratingByParts
Jan 20 at 6:37
add a comment |
$begingroup$
The OP requires a series involving $f in X$. So I don't think your counter-example of the first paragraph works because you require $f notin X$
$endgroup$
– DisintegratingByParts
Jan 20 at 6:37
$begingroup$
The OP requires a series involving $f in X$. So I don't think your counter-example of the first paragraph works because you require $f notin X$
$endgroup$
– DisintegratingByParts
Jan 20 at 6:37
$begingroup$
The OP requires a series involving $f in X$. So I don't think your counter-example of the first paragraph works because you require $f notin X$
$endgroup$
– DisintegratingByParts
Jan 20 at 6:37
add a comment |
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$begingroup$
I believe it only converges to $f$ if the orthonormal system is "complete". That is, the span of $f_i$ is dense in your vector space.
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– Jeffery Opoku-Mensah
Jan 17 at 23:07
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@JefferyOpoku-Mensah Completeness does not imply anything. When the space is Hilbert then completeness is enough and sufficient. But I just want to know for any arbitrary system whether this sum converges to something.
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– CO2
Jan 17 at 23:11
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Ah yeah. I implicitly read your problem as in a hilbert space, given the tag.
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– Jeffery Opoku-Mensah
Jan 17 at 23:13
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@JefferyOpoku-Mensah Well, yes it is on the way there, as preparation knowledge. Maybe I should remove the tag.
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– CO2
Jan 17 at 23:23
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Bessel's inequality.
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– copper.hat
Jan 18 at 0:15