L^2 convergence and pointwise convergence
$begingroup$
I have seen an informal note which says if $f_n to f$ under $L^2$ norm it is pointwise convergent almost everywhere.
I know that pointwise convergence implies $L^2$ convergence under LDCT conditions but I haven’t heard anything about its converse. Is it true or not?
If it is true I think there can be some relations with Littlewood’s principles. Could someone please illuminate me about this situation? If it is true could you please prove it in the easiest way?
Thanks a lot
functional-analysis fourier-analysis lebesgue-integral lebesgue-measure
asked Dec 19 '18 at 7:57
esrabasaresrabasar
420110
$endgroup$
add a comment |
$begingroup$
I have seen an informal note which says if $f_n to f$ under $L^2$ norm it is pointwise convergent almost everywhere.
I know that pointwise convergence implies $L^2$ convergence under LDCT conditions but I haven’t heard anything about its converse. Is it true or not?
If it is true I think there can be some relations with Littlewood’s principles. Could someone please illuminate me about this situation? If it is true could you please prove it in the easiest way?
Thanks a lot
functional-analysis fourier-analysis lebesgue-integral lebesgue-measure
asked Dec 19 '18 at 7:57
esrabasaresrabasar
420110
$endgroup$
add a comment |
$begingroup$
I have seen an informal note which says if $f_n to f$ under $L^2$ norm it is pointwise convergent almost everywhere.
I know that pointwise convergence implies $L^2$ convergence under LDCT conditions but I haven’t heard anything about its converse. Is it true or not?
If it is true I think there can be some relations with Littlewood’s principles. Could someone please illuminate me about this situation? If it is true could you please prove it in the easiest way?
Thanks a lot
functional-analysis fourier-analysis lebesgue-integral lebesgue-measure
asked Dec 19 '18 at 7:57
esrabasaresrabasar
420110
$endgroup$
I have seen an informal note which says if $f_n to f$ under $L^2$ norm it is pointwise convergent almost everywhere.
I know that pointwise convergence implies $L^2$ convergence under LDCT conditions but I haven’t heard anything about its converse. Is it true or not?
If it is true I think there can be some relations with Littlewood’s principles. Could someone please illuminate me about this situation? If it is true could you please prove it in the easiest way?
Thanks a lot
functional-analysis fourier-analysis lebesgue-integral lebesgue-measure
functional-analysis fourier-analysis lebesgue-integral lebesgue-measure
asked Dec 19 '18 at 7:57
esrabasaresrabasar
420110
asked Dec 19 '18 at 7:57
esrabasaresrabasar
420110
asked Dec 19 '18 at 7:57
esrabasaresrabasar
420110
asked Dec 19 '18 at 7:57
esrabasaresrabasar
420110
asked Dec 19 '18 at 7:57
esrabasaresrabasar
420110
420110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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False. Let $f_n$ be listing of the characteristic functions of the intervals $[frac {i-1} n, frac i n)$ ($1 leq i leq n, ngeq 1$) in a sequence . Then $f_n to 0$ in $L^{2}$ (w.r.t. Lebesgue measure on $(0,1)$) but the sequence does not converge at any point.
answered Dec 19 '18 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$begingroup$
Thanks a lot...
$endgroup$
– esrabasar
Dec 19 '18 at 18:50
add a comment |
$begingroup$
False. Kavi Rama gave you a counterexample.
If $f_n to f$ in $L^2$, then there is a subsequence $(f_{n_k})$, which is pointwise convergent almost everywhere.
answered Dec 19 '18 at 8:08
FredFred
45.2k1847
$endgroup$
$begingroup$
Thanks a lot...
$endgroup$
– esrabasar
Dec 19 '18 at 18:49
add a comment |
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2 Answers
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active
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2 Answers
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active
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$begingroup$
False. Let $f_n$ be listing of the characteristic functions of the intervals $[frac {i-1} n, frac i n)$ ($1 leq i leq n, ngeq 1$) in a sequence . Then $f_n to 0$ in $L^{2}$ (w.r.t. Lebesgue measure on $(0,1)$) but the sequence does not converge at any point.
answered Dec 19 '18 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$begingroup$
Thanks a lot...
$endgroup$
– esrabasar
Dec 19 '18 at 18:50
add a comment |
$begingroup$
False. Let $f_n$ be listing of the characteristic functions of the intervals $[frac {i-1} n, frac i n)$ ($1 leq i leq n, ngeq 1$) in a sequence . Then $f_n to 0$ in $L^{2}$ (w.r.t. Lebesgue measure on $(0,1)$) but the sequence does not converge at any point.
answered Dec 19 '18 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$begingroup$
Thanks a lot...
$endgroup$
– esrabasar
Dec 19 '18 at 18:50
add a comment |
$begingroup$
False. Let $f_n$ be listing of the characteristic functions of the intervals $[frac {i-1} n, frac i n)$ ($1 leq i leq n, ngeq 1$) in a sequence . Then $f_n to 0$ in $L^{2}$ (w.r.t. Lebesgue measure on $(0,1)$) but the sequence does not converge at any point.
answered Dec 19 '18 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
False. Let $f_n$ be listing of the characteristic functions of the intervals $[frac {i-1} n, frac i n)$ ($1 leq i leq n, ngeq 1$) in a sequence . Then $f_n to 0$ in $L^{2}$ (w.r.t. Lebesgue measure on $(0,1)$) but the sequence does not converge at any point.
answered Dec 19 '18 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Dec 19 '18 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Dec 19 '18 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Dec 19 '18 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56k42158
56k42158
$begingroup$
Thanks a lot...
$endgroup$
– esrabasar
Dec 19 '18 at 18:50
add a comment |
$begingroup$
Thanks a lot...
$endgroup$
– esrabasar
Dec 19 '18 at 18:50
$begingroup$
Thanks a lot...
$endgroup$
– esrabasar
Dec 19 '18 at 18:50
$begingroup$
Thanks a lot...
$endgroup$
– esrabasar
Dec 19 '18 at 18:50
add a comment |
$begingroup$
False. Kavi Rama gave you a counterexample.
If $f_n to f$ in $L^2$, then there is a subsequence $(f_{n_k})$, which is pointwise convergent almost everywhere.
answered Dec 19 '18 at 8:08
FredFred
45.2k1847
$endgroup$
$begingroup$
Thanks a lot...
$endgroup$
– esrabasar
Dec 19 '18 at 18:49
add a comment |
$begingroup$
False. Kavi Rama gave you a counterexample.
If $f_n to f$ in $L^2$, then there is a subsequence $(f_{n_k})$, which is pointwise convergent almost everywhere.
answered Dec 19 '18 at 8:08
FredFred
45.2k1847
$endgroup$
$begingroup$
Thanks a lot...
$endgroup$
– esrabasar
Dec 19 '18 at 18:49
add a comment |
$begingroup$
False. Kavi Rama gave you a counterexample.
If $f_n to f$ in $L^2$, then there is a subsequence $(f_{n_k})$, which is pointwise convergent almost everywhere.
answered Dec 19 '18 at 8:08
FredFred
45.2k1847
$endgroup$
False. Kavi Rama gave you a counterexample.
If $f_n to f$ in $L^2$, then there is a subsequence $(f_{n_k})$, which is pointwise convergent almost everywhere.
answered Dec 19 '18 at 8:08
FredFred
45.2k1847
answered Dec 19 '18 at 8:08
FredFred
45.2k1847
answered Dec 19 '18 at 8:08
FredFred
45.2k1847
answered Dec 19 '18 at 8:08
FredFred
45.2k1847
45.2k1847
$begingroup$
Thanks a lot...
$endgroup$
– esrabasar
Dec 19 '18 at 18:49
add a comment |
$begingroup$
Thanks a lot...
$endgroup$
– esrabasar
Dec 19 '18 at 18:49
$begingroup$
Thanks a lot...
$endgroup$
– esrabasar
Dec 19 '18 at 18:49
$begingroup$
Thanks a lot...
$endgroup$
– esrabasar
Dec 19 '18 at 18:49
add a comment |
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