Series convergence in $L_2$
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Is it true that if $sum_{n=1}^{infty} s_n^2$ converges and $u_n in L_2[a;b]$ are bounded then the series $sum_{n=1}^{infty} s_n u_n$ converges to $L_2$ function?
If so, how can I prove it?
convergence lp-spaces
asked Dec 12 '18 at 19:16
Anton ZagrivinAnton Zagrivin
1548
$endgroup$
add a comment |
$begingroup$
Is it true that if $sum_{n=1}^{infty} s_n^2$ converges and $u_n in L_2[a;b]$ are bounded then the series $sum_{n=1}^{infty} s_n u_n$ converges to $L_2$ function?
If so, how can I prove it?
convergence lp-spaces
asked Dec 12 '18 at 19:16
Anton ZagrivinAnton Zagrivin
1548
$endgroup$
add a comment |
$begingroup$
Is it true that if $sum_{n=1}^{infty} s_n^2$ converges and $u_n in L_2[a;b]$ are bounded then the series $sum_{n=1}^{infty} s_n u_n$ converges to $L_2$ function?
If so, how can I prove it?
convergence lp-spaces
asked Dec 12 '18 at 19:16
Anton ZagrivinAnton Zagrivin
1548
$endgroup$
Is it true that if $sum_{n=1}^{infty} s_n^2$ converges and $u_n in L_2[a;b]$ are bounded then the series $sum_{n=1}^{infty} s_n u_n$ converges to $L_2$ function?
If so, how can I prove it?
convergence lp-spaces
convergence lp-spaces
asked Dec 12 '18 at 19:16
Anton ZagrivinAnton Zagrivin
1548
asked Dec 12 '18 at 19:16
Anton ZagrivinAnton Zagrivin
1548
asked Dec 12 '18 at 19:16
Anton ZagrivinAnton Zagrivin
1548
asked Dec 12 '18 at 19:16
Anton ZagrivinAnton Zagrivin
1548
asked Dec 12 '18 at 19:16
Anton ZagrivinAnton Zagrivin
1548
1548
add a comment |
add a comment |
1 Answer
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It does hold, using Pythagores’ theorem and completeness, if the $u_n$ are pairwise orthogonal.
It is false in general though, take for instance $s_n=n^{-1}$ and $u_n=1$.
answered Dec 12 '18 at 19:36
MindlackMindlack
4,760210
$endgroup$
$begingroup$
Oh, my system ${u_n}$ is really pairwise orthogonal and I just forgot about this condition. Thanks a lot
$endgroup$
– Anton Zagrivin
Dec 13 '18 at 20:06
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
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oldest
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active
oldest
votes
$begingroup$
It does hold, using Pythagores’ theorem and completeness, if the $u_n$ are pairwise orthogonal.
It is false in general though, take for instance $s_n=n^{-1}$ and $u_n=1$.
answered Dec 12 '18 at 19:36
MindlackMindlack
4,760210
$endgroup$
$begingroup$
Oh, my system ${u_n}$ is really pairwise orthogonal and I just forgot about this condition. Thanks a lot
$endgroup$
– Anton Zagrivin
Dec 13 '18 at 20:06
add a comment |
$begingroup$
It does hold, using Pythagores’ theorem and completeness, if the $u_n$ are pairwise orthogonal.
It is false in general though, take for instance $s_n=n^{-1}$ and $u_n=1$.
answered Dec 12 '18 at 19:36
MindlackMindlack
4,760210
$endgroup$
$begingroup$
Oh, my system ${u_n}$ is really pairwise orthogonal and I just forgot about this condition. Thanks a lot
$endgroup$
– Anton Zagrivin
Dec 13 '18 at 20:06
add a comment |
$begingroup$
It does hold, using Pythagores’ theorem and completeness, if the $u_n$ are pairwise orthogonal.
It is false in general though, take for instance $s_n=n^{-1}$ and $u_n=1$.
answered Dec 12 '18 at 19:36
MindlackMindlack
4,760210
$endgroup$
It does hold, using Pythagores’ theorem and completeness, if the $u_n$ are pairwise orthogonal.
It is false in general though, take for instance $s_n=n^{-1}$ and $u_n=1$.
answered Dec 12 '18 at 19:36
MindlackMindlack
4,760210
answered Dec 12 '18 at 19:36
MindlackMindlack
4,760210
answered Dec 12 '18 at 19:36
MindlackMindlack
4,760210
answered Dec 12 '18 at 19:36
MindlackMindlack
4,760210
4,760210
$begingroup$
Oh, my system ${u_n}$ is really pairwise orthogonal and I just forgot about this condition. Thanks a lot
$endgroup$
– Anton Zagrivin
Dec 13 '18 at 20:06
add a comment |
$begingroup$
Oh, my system ${u_n}$ is really pairwise orthogonal and I just forgot about this condition. Thanks a lot
$endgroup$
– Anton Zagrivin
Dec 13 '18 at 20:06
$begingroup$
Oh, my system ${u_n}$ is really pairwise orthogonal and I just forgot about this condition. Thanks a lot
$endgroup$
– Anton Zagrivin
Dec 13 '18 at 20:06
$begingroup$
Oh, my system ${u_n}$ is really pairwise orthogonal and I just forgot about this condition. Thanks a lot
$endgroup$
– Anton Zagrivin
Dec 13 '18 at 20:06
add a comment |
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