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







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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







    share|cite|improve this question
























      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







      share|cite|improve this question














      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






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Nov 19 at 23:49









      sound wave

      1548




      1548






















          1 Answer
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          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.






          share|cite|improve this answer





















          • 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













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          1 Answer
          1






          active

          oldest

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          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.






          share|cite|improve this answer





















          • 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

















          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.






          share|cite|improve this answer





















          • 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















          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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




















          • 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




















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