Prove that a pointwise converging sequence, bounded in $L^p$, converge uniformly in $L^q$, using Egorov's...
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Let $p > 1$, $E$ be a measurable subset of $mathbb{R}^n$ with finite
Lebesgue measure $mu$, ${f_n}_{nin N}$ be a sequence of measurable
functions $f_n : E → mathbb{R}$, bounded in $L^p(E)$, and pointwise converging to $f : E → mathbb{R}$.
Prove that ${f_n}_{n∈N}$ converges to $f$ in
$L^q(E)$ for all $q ∈(1, p)$.
All the hypothesis of Egorov's theorem are satisfied, then $forallε > 0$ $exists N_ε ⊆ Ω$ measurable such that $mu(N_ε) ≤ ε$ and such that the sequence $f_n$ converges uniformly to $f$ in $Ω setminus N_ε$.
Any hint on how to start the exercise?
functional-analysis convergence
asked Nov 19 at 23:49
sound wave
1548
add a comment |
up vote
0
down vote
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Let $p > 1$, $E$ be a measurable subset of $mathbb{R}^n$ with finite
Lebesgue measure $mu$, ${f_n}_{nin N}$ be a sequence of measurable
functions $f_n : E → mathbb{R}$, bounded in $L^p(E)$, and pointwise converging to $f : E → mathbb{R}$.
Prove that ${f_n}_{n∈N}$ converges to $f$ in
$L^q(E)$ for all $q ∈(1, p)$.
All the hypothesis of Egorov's theorem are satisfied, then $forallε > 0$ $exists N_ε ⊆ Ω$ measurable such that $mu(N_ε) ≤ ε$ and such that the sequence $f_n$ converges uniformly to $f$ in $Ω setminus N_ε$.
Any hint on how to start the exercise?
functional-analysis convergence
asked Nov 19 at 23:49
sound wave
1548
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $p > 1$, $E$ be a measurable subset of $mathbb{R}^n$ with finite
Lebesgue measure $mu$, ${f_n}_{nin N}$ be a sequence of measurable
functions $f_n : E → mathbb{R}$, bounded in $L^p(E)$, and pointwise converging to $f : E → mathbb{R}$.
Prove that ${f_n}_{n∈N}$ converges to $f$ in
$L^q(E)$ for all $q ∈(1, p)$.
All the hypothesis of Egorov's theorem are satisfied, then $forallε > 0$ $exists N_ε ⊆ Ω$ measurable such that $mu(N_ε) ≤ ε$ and such that the sequence $f_n$ converges uniformly to $f$ in $Ω setminus N_ε$.
Any hint on how to start the exercise?
functional-analysis convergence
asked Nov 19 at 23:49
sound wave
1548
Let $p > 1$, $E$ be a measurable subset of $mathbb{R}^n$ with finite
Lebesgue measure $mu$, ${f_n}_{nin N}$ be a sequence of measurable
functions $f_n : E → mathbb{R}$, bounded in $L^p(E)$, and pointwise converging to $f : E → mathbb{R}$.
Prove that ${f_n}_{n∈N}$ converges to $f$ in
$L^q(E)$ for all $q ∈(1, p)$.
All the hypothesis of Egorov's theorem are satisfied, then $forallε > 0$ $exists N_ε ⊆ Ω$ measurable such that $mu(N_ε) ≤ ε$ and such that the sequence $f_n$ converges uniformly to $f$ in $Ω setminus N_ε$.
Any hint on how to start the exercise?
functional-analysis convergence
functional-analysis convergence
asked Nov 19 at 23:49
sound wave
1548
asked Nov 19 at 23:49
sound wave
1548
asked Nov 19 at 23:49
sound wave
1548
asked Nov 19 at 23:49
sound wave
1548
asked Nov 19 at 23:49
sound wave
1548
1548
add a comment |
add a comment |
1 Answer
1
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votes
up vote
1
down vote
accepted
Here's a hint on how to start. Use Egorov's theorem to split the domain into two parts, one with $epsilon$ measure on which convergence is bad, and one with the majority of the measure on which convergence is good. Then bound the total $q$-norm of the difference of the sequence from the limit by bounding the $q$-norm of the difference on each of these sets.
answered Nov 19 at 23:59
jgon
11.6k21839
Thanks. I would do this: since the sequence converge almost everywhere, take the set $N_epsilon$ of measure zero where the seq does not converge, and $Esetminus N_epsilon$ where the seq converge. Then from what you wrote I got that we have to bound $||f_n-f||_{L^q(E)}$ with $||f_n-f||_{L^q(N_epsilon)}+||f_n-f||_{L^q(Esetminus N_epsilon)}$. But the first norm is infinite because on $N_epsilon$ the sequence does not converge. Do we need to change the choice of the two sets?
– sound wave
Nov 20 at 12:28
Could you say what it the $L^q$ norm of $f_n-f$ on the set $Esetminus N_varepsilon$ ?
– sound wave
Nov 21 at 14:20
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Here's a hint on how to start. Use Egorov's theorem to split the domain into two parts, one with $epsilon$ measure on which convergence is bad, and one with the majority of the measure on which convergence is good. Then bound the total $q$-norm of the difference of the sequence from the limit by bounding the $q$-norm of the difference on each of these sets.
answered Nov 19 at 23:59
jgon
11.6k21839
Thanks. I would do this: since the sequence converge almost everywhere, take the set $N_epsilon$ of measure zero where the seq does not converge, and $Esetminus N_epsilon$ where the seq converge. Then from what you wrote I got that we have to bound $||f_n-f||_{L^q(E)}$ with $||f_n-f||_{L^q(N_epsilon)}+||f_n-f||_{L^q(Esetminus N_epsilon)}$. But the first norm is infinite because on $N_epsilon$ the sequence does not converge. Do we need to change the choice of the two sets?
– sound wave
Nov 20 at 12:28
Could you say what it the $L^q$ norm of $f_n-f$ on the set $Esetminus N_varepsilon$ ?
– sound wave
Nov 21 at 14:20
add a comment |
up vote
1
down vote
accepted
Here's a hint on how to start. Use Egorov's theorem to split the domain into two parts, one with $epsilon$ measure on which convergence is bad, and one with the majority of the measure on which convergence is good. Then bound the total $q$-norm of the difference of the sequence from the limit by bounding the $q$-norm of the difference on each of these sets.
answered Nov 19 at 23:59
jgon
11.6k21839
Thanks. I would do this: since the sequence converge almost everywhere, take the set $N_epsilon$ of measure zero where the seq does not converge, and $Esetminus N_epsilon$ where the seq converge. Then from what you wrote I got that we have to bound $||f_n-f||_{L^q(E)}$ with $||f_n-f||_{L^q(N_epsilon)}+||f_n-f||_{L^q(Esetminus N_epsilon)}$. But the first norm is infinite because on $N_epsilon$ the sequence does not converge. Do we need to change the choice of the two sets?
– sound wave
Nov 20 at 12:28
Could you say what it the $L^q$ norm of $f_n-f$ on the set $Esetminus N_varepsilon$ ?
– sound wave
Nov 21 at 14:20
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Here's a hint on how to start. Use Egorov's theorem to split the domain into two parts, one with $epsilon$ measure on which convergence is bad, and one with the majority of the measure on which convergence is good. Then bound the total $q$-norm of the difference of the sequence from the limit by bounding the $q$-norm of the difference on each of these sets.
answered Nov 19 at 23:59
jgon
11.6k21839
Here's a hint on how to start. Use Egorov's theorem to split the domain into two parts, one with $epsilon$ measure on which convergence is bad, and one with the majority of the measure on which convergence is good. Then bound the total $q$-norm of the difference of the sequence from the limit by bounding the $q$-norm of the difference on each of these sets.
answered Nov 19 at 23:59
jgon
11.6k21839
answered Nov 19 at 23:59
jgon
11.6k21839
answered Nov 19 at 23:59
jgon
11.6k21839
answered Nov 19 at 23:59
jgon
11.6k21839
11.6k21839
Thanks. I would do this: since the sequence converge almost everywhere, take the set $N_epsilon$ of measure zero where the seq does not converge, and $Esetminus N_epsilon$ where the seq converge. Then from what you wrote I got that we have to bound $||f_n-f||_{L^q(E)}$ with $||f_n-f||_{L^q(N_epsilon)}+||f_n-f||_{L^q(Esetminus N_epsilon)}$. But the first norm is infinite because on $N_epsilon$ the sequence does not converge. Do we need to change the choice of the two sets?
– sound wave
Nov 20 at 12:28
Could you say what it the $L^q$ norm of $f_n-f$ on the set $Esetminus N_varepsilon$ ?
– sound wave
Nov 21 at 14:20
add a comment |
Thanks. I would do this: since the sequence converge almost everywhere, take the set $N_epsilon$ of measure zero where the seq does not converge, and $Esetminus N_epsilon$ where the seq converge. Then from what you wrote I got that we have to bound $||f_n-f||_{L^q(E)}$ with $||f_n-f||_{L^q(N_epsilon)}+||f_n-f||_{L^q(Esetminus N_epsilon)}$. But the first norm is infinite because on $N_epsilon$ the sequence does not converge. Do we need to change the choice of the two sets?
– sound wave
Nov 20 at 12:28
Could you say what it the $L^q$ norm of $f_n-f$ on the set $Esetminus N_varepsilon$ ?
– sound wave
Nov 21 at 14:20
Thanks. I would do this: since the sequence converge almost everywhere, take the set $N_epsilon$ of measure zero where the seq does not converge, and $Esetminus N_epsilon$ where the seq converge. Then from what you wrote I got that we have to bound $||f_n-f||_{L^q(E)}$ with $||f_n-f||_{L^q(N_epsilon)}+||f_n-f||_{L^q(Esetminus N_epsilon)}$. But the first norm is infinite because on $N_epsilon$ the sequence does not converge. Do we need to change the choice of the two sets?
– sound wave
Nov 20 at 12:28
Thanks. I would do this: since the sequence converge almost everywhere, take the set $N_epsilon$ of measure zero where the seq does not converge, and $Esetminus N_epsilon$ where the seq converge. Then from what you wrote I got that we have to bound $||f_n-f||_{L^q(E)}$ with $||f_n-f||_{L^q(N_epsilon)}+||f_n-f||_{L^q(Esetminus N_epsilon)}$. But the first norm is infinite because on $N_epsilon$ the sequence does not converge. Do we need to change the choice of the two sets?
– sound wave
Nov 20 at 12:28
Could you say what it the $L^q$ norm of $f_n-f$ on the set $Esetminus N_varepsilon$ ?
– sound wave
Nov 21 at 14:20
Could you say what it the $L^q$ norm of $f_n-f$ on the set $Esetminus N_varepsilon$ ?
– sound wave
Nov 21 at 14:20
add a comment |
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