$L_1$ inequality












1












$begingroup$


$tilde{f} =lim_{ntoinfty} frac{1}{n} sum_{k=0}^{n-1} f(x+k)$, where $f$ is an $L_1$ function with Lebesgue measure $mu$. Show that $||tilde{f} ||_{L_1} leq ||f ||_{L_1}$






real-analysis functional-analysis measure-theory







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  • $begingroup$
    Please add your attempts.
    $endgroup$
    – Jack D'Aurizio
    Mar 9 '17 at 13:48










  • $begingroup$
    This is not much more than a straightforward consequence of the triangle inequality and the fact that $int_{mathbb{R}}|f(x+k)|,dx = int_{mathbb{R}}|f(x)|,dx$ if $fin L^1(mathbb{R})$.
    $endgroup$
    – Jack D'Aurizio
    Mar 9 '17 at 13:50
















1












$begingroup$


$tilde{f} =lim_{ntoinfty} frac{1}{n} sum_{k=0}^{n-1} f(x+k)$, where $f$ is an $L_1$ function with Lebesgue measure $mu$. Show that $||tilde{f} ||_{L_1} leq ||f ||_{L_1}$






real-analysis functional-analysis measure-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Please add your attempts.
    $endgroup$
    – Jack D'Aurizio
    Mar 9 '17 at 13:48










  • $begingroup$
    This is not much more than a straightforward consequence of the triangle inequality and the fact that $int_{mathbb{R}}|f(x+k)|,dx = int_{mathbb{R}}|f(x)|,dx$ if $fin L^1(mathbb{R})$.
    $endgroup$
    – Jack D'Aurizio
    Mar 9 '17 at 13:50














1












1








1


1



$begingroup$


$tilde{f} =lim_{ntoinfty} frac{1}{n} sum_{k=0}^{n-1} f(x+k)$, where $f$ is an $L_1$ function with Lebesgue measure $mu$. Show that $||tilde{f} ||_{L_1} leq ||f ||_{L_1}$






real-analysis functional-analysis measure-theory







share|cite|improve this question









$endgroup$




$tilde{f} =lim_{ntoinfty} frac{1}{n} sum_{k=0}^{n-1} f(x+k)$, where $f$ is an $L_1$ function with Lebesgue measure $mu$. Show that $||tilde{f} ||_{L_1} leq ||f ||_{L_1}$






real-analysis functional-analysis measure-theory




real-analysis functional-analysis measure-theory






share|cite|improve this question













share|cite|improve this question











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asked Mar 9 '17 at 13:43









MAN-MADEMAN-MADE

3,63311031




3,63311031












  • $begingroup$
    Please add your attempts.
    $endgroup$
    – Jack D'Aurizio
    Mar 9 '17 at 13:48










  • $begingroup$
    This is not much more than a straightforward consequence of the triangle inequality and the fact that $int_{mathbb{R}}|f(x+k)|,dx = int_{mathbb{R}}|f(x)|,dx$ if $fin L^1(mathbb{R})$.
    $endgroup$
    – Jack D'Aurizio
    Mar 9 '17 at 13:50


















  • $begingroup$
    Please add your attempts.
    $endgroup$
    – Jack D'Aurizio
    Mar 9 '17 at 13:48










  • $begingroup$
    This is not much more than a straightforward consequence of the triangle inequality and the fact that $int_{mathbb{R}}|f(x+k)|,dx = int_{mathbb{R}}|f(x)|,dx$ if $fin L^1(mathbb{R})$.
    $endgroup$
    – Jack D'Aurizio
    Mar 9 '17 at 13:50
















$begingroup$
Please add your attempts.
$endgroup$
– Jack D'Aurizio
Mar 9 '17 at 13:48




$begingroup$
Please add your attempts.
$endgroup$
– Jack D'Aurizio
Mar 9 '17 at 13:48












$begingroup$
This is not much more than a straightforward consequence of the triangle inequality and the fact that $int_{mathbb{R}}|f(x+k)|,dx = int_{mathbb{R}}|f(x)|,dx$ if $fin L^1(mathbb{R})$.
$endgroup$
– Jack D'Aurizio
Mar 9 '17 at 13:50




$begingroup$
This is not much more than a straightforward consequence of the triangle inequality and the fact that $int_{mathbb{R}}|f(x+k)|,dx = int_{mathbb{R}}|f(x)|,dx$ if $fin L^1(mathbb{R})$.
$endgroup$
– Jack D'Aurizio
Mar 9 '17 at 13:50










2 Answers
2






active

oldest

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0












$begingroup$

since I can't post comments I write this as an answer.



I am sure that this won't work, because you need additionally that $tilde{f} in L_1$.
Then you can follow the hint of Jack D'Aurizio.



But if you can't assume that, then $f = 1_{[0,1)}$ is a counterexample.
Because
$$
tilde{f}_{n} := frac{1}{n} sum_{k=0}^{n-1} 1_{[0,1)}(x+k)
= frac{1}{n} 1_{[-n+1,1)}(x).
$$
Now it is easy to check that $|tilde{f}_n| = 1$ for all $ninmathbb{N}$. So if $tilde{f}_n$ converts in $L_1$ then the norm must be $1$. But if you regard the series $tilde{f}_n$ pointwise then the limit is $0$ which contradicts with the previous statement.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But OP didn't say that $lim_n sum_{0le k< n} f(x+k)$ converges in $L^1$. The convergence is only pointwise. So $lim_n tilde{f_n}=0$ does not contradict the statement: $|lim_n tilde{f}_n|le |f|$.
    $endgroup$
    – Song
    Jan 13 at 14:17












  • $begingroup$
    That is true, but I wanted to point out that the given sequence does not automatically converge if you start with an $f in L^1(mathbb{R})$, since the suggested proof of Jack D'Aurizio uses that property.
    $endgroup$
    – Nathanael Skrepek
    Jan 13 at 15:59












  • $begingroup$
    Oh, I just realized that this answer was written almost 2 years ago ... Thank you for your response.
    $endgroup$
    – Song
    Jan 13 at 16:47



















0












$begingroup$

Assuming $g(x)=lim_{ntoinfty} frac{1}{n}sum_{k=0}^{n-1}f(x+k)$ converges almost everywhere pointwise, we can show that $|g|_{L^1}le |f|_{L^1}$ as follows. Let $g_n(x)=frac{1}{n}sum_{k=0}^{n-1}f(x+k)$. Note that
$$
|g_n(x)|le frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|
$$ and hence
$$
|g(x)|=lim_{ntoinfty}|g_n(x)|le liminf_{ntoinfty}frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|.
$$ By Fatou's lemma it follows
$$
|g|_{L^1}le int_mathbb{R} left[liminf_{ntoinfty}frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|right];dmuleliminf_{ntoinfty}int_mathbb{R}frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|dmu=|f|_{L^1}.
$$






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






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    0












    $begingroup$

    since I can't post comments I write this as an answer.



    I am sure that this won't work, because you need additionally that $tilde{f} in L_1$.
    Then you can follow the hint of Jack D'Aurizio.



    But if you can't assume that, then $f = 1_{[0,1)}$ is a counterexample.
    Because
    $$
    tilde{f}_{n} := frac{1}{n} sum_{k=0}^{n-1} 1_{[0,1)}(x+k)
    = frac{1}{n} 1_{[-n+1,1)}(x).
    $$
    Now it is easy to check that $|tilde{f}_n| = 1$ for all $ninmathbb{N}$. So if $tilde{f}_n$ converts in $L_1$ then the norm must be $1$. But if you regard the series $tilde{f}_n$ pointwise then the limit is $0$ which contradicts with the previous statement.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But OP didn't say that $lim_n sum_{0le k< n} f(x+k)$ converges in $L^1$. The convergence is only pointwise. So $lim_n tilde{f_n}=0$ does not contradict the statement: $|lim_n tilde{f}_n|le |f|$.
      $endgroup$
      – Song
      Jan 13 at 14:17












    • $begingroup$
      That is true, but I wanted to point out that the given sequence does not automatically converge if you start with an $f in L^1(mathbb{R})$, since the suggested proof of Jack D'Aurizio uses that property.
      $endgroup$
      – Nathanael Skrepek
      Jan 13 at 15:59












    • $begingroup$
      Oh, I just realized that this answer was written almost 2 years ago ... Thank you for your response.
      $endgroup$
      – Song
      Jan 13 at 16:47
















    0












    $begingroup$

    since I can't post comments I write this as an answer.



    I am sure that this won't work, because you need additionally that $tilde{f} in L_1$.
    Then you can follow the hint of Jack D'Aurizio.



    But if you can't assume that, then $f = 1_{[0,1)}$ is a counterexample.
    Because
    $$
    tilde{f}_{n} := frac{1}{n} sum_{k=0}^{n-1} 1_{[0,1)}(x+k)
    = frac{1}{n} 1_{[-n+1,1)}(x).
    $$
    Now it is easy to check that $|tilde{f}_n| = 1$ for all $ninmathbb{N}$. So if $tilde{f}_n$ converts in $L_1$ then the norm must be $1$. But if you regard the series $tilde{f}_n$ pointwise then the limit is $0$ which contradicts with the previous statement.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But OP didn't say that $lim_n sum_{0le k< n} f(x+k)$ converges in $L^1$. The convergence is only pointwise. So $lim_n tilde{f_n}=0$ does not contradict the statement: $|lim_n tilde{f}_n|le |f|$.
      $endgroup$
      – Song
      Jan 13 at 14:17












    • $begingroup$
      That is true, but I wanted to point out that the given sequence does not automatically converge if you start with an $f in L^1(mathbb{R})$, since the suggested proof of Jack D'Aurizio uses that property.
      $endgroup$
      – Nathanael Skrepek
      Jan 13 at 15:59












    • $begingroup$
      Oh, I just realized that this answer was written almost 2 years ago ... Thank you for your response.
      $endgroup$
      – Song
      Jan 13 at 16:47














    0












    0








    0





    $begingroup$

    since I can't post comments I write this as an answer.



    I am sure that this won't work, because you need additionally that $tilde{f} in L_1$.
    Then you can follow the hint of Jack D'Aurizio.



    But if you can't assume that, then $f = 1_{[0,1)}$ is a counterexample.
    Because
    $$
    tilde{f}_{n} := frac{1}{n} sum_{k=0}^{n-1} 1_{[0,1)}(x+k)
    = frac{1}{n} 1_{[-n+1,1)}(x).
    $$
    Now it is easy to check that $|tilde{f}_n| = 1$ for all $ninmathbb{N}$. So if $tilde{f}_n$ converts in $L_1$ then the norm must be $1$. But if you regard the series $tilde{f}_n$ pointwise then the limit is $0$ which contradicts with the previous statement.






    share|cite|improve this answer











    $endgroup$



    since I can't post comments I write this as an answer.



    I am sure that this won't work, because you need additionally that $tilde{f} in L_1$.
    Then you can follow the hint of Jack D'Aurizio.



    But if you can't assume that, then $f = 1_{[0,1)}$ is a counterexample.
    Because
    $$
    tilde{f}_{n} := frac{1}{n} sum_{k=0}^{n-1} 1_{[0,1)}(x+k)
    = frac{1}{n} 1_{[-n+1,1)}(x).
    $$
    Now it is easy to check that $|tilde{f}_n| = 1$ for all $ninmathbb{N}$. So if $tilde{f}_n$ converts in $L_1$ then the norm must be $1$. But if you regard the series $tilde{f}_n$ pointwise then the limit is $0$ which contradicts with the previous statement.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 12 at 22:59

























    answered Mar 9 '17 at 16:58









    Nathanael SkrepekNathanael Skrepek

    1,8021615




    1,8021615












    • $begingroup$
      But OP didn't say that $lim_n sum_{0le k< n} f(x+k)$ converges in $L^1$. The convergence is only pointwise. So $lim_n tilde{f_n}=0$ does not contradict the statement: $|lim_n tilde{f}_n|le |f|$.
      $endgroup$
      – Song
      Jan 13 at 14:17












    • $begingroup$
      That is true, but I wanted to point out that the given sequence does not automatically converge if you start with an $f in L^1(mathbb{R})$, since the suggested proof of Jack D'Aurizio uses that property.
      $endgroup$
      – Nathanael Skrepek
      Jan 13 at 15:59












    • $begingroup$
      Oh, I just realized that this answer was written almost 2 years ago ... Thank you for your response.
      $endgroup$
      – Song
      Jan 13 at 16:47


















    • $begingroup$
      But OP didn't say that $lim_n sum_{0le k< n} f(x+k)$ converges in $L^1$. The convergence is only pointwise. So $lim_n tilde{f_n}=0$ does not contradict the statement: $|lim_n tilde{f}_n|le |f|$.
      $endgroup$
      – Song
      Jan 13 at 14:17












    • $begingroup$
      That is true, but I wanted to point out that the given sequence does not automatically converge if you start with an $f in L^1(mathbb{R})$, since the suggested proof of Jack D'Aurizio uses that property.
      $endgroup$
      – Nathanael Skrepek
      Jan 13 at 15:59












    • $begingroup$
      Oh, I just realized that this answer was written almost 2 years ago ... Thank you for your response.
      $endgroup$
      – Song
      Jan 13 at 16:47
















    $begingroup$
    But OP didn't say that $lim_n sum_{0le k< n} f(x+k)$ converges in $L^1$. The convergence is only pointwise. So $lim_n tilde{f_n}=0$ does not contradict the statement: $|lim_n tilde{f}_n|le |f|$.
    $endgroup$
    – Song
    Jan 13 at 14:17






    $begingroup$
    But OP didn't say that $lim_n sum_{0le k< n} f(x+k)$ converges in $L^1$. The convergence is only pointwise. So $lim_n tilde{f_n}=0$ does not contradict the statement: $|lim_n tilde{f}_n|le |f|$.
    $endgroup$
    – Song
    Jan 13 at 14:17














    $begingroup$
    That is true, but I wanted to point out that the given sequence does not automatically converge if you start with an $f in L^1(mathbb{R})$, since the suggested proof of Jack D'Aurizio uses that property.
    $endgroup$
    – Nathanael Skrepek
    Jan 13 at 15:59






    $begingroup$
    That is true, but I wanted to point out that the given sequence does not automatically converge if you start with an $f in L^1(mathbb{R})$, since the suggested proof of Jack D'Aurizio uses that property.
    $endgroup$
    – Nathanael Skrepek
    Jan 13 at 15:59














    $begingroup$
    Oh, I just realized that this answer was written almost 2 years ago ... Thank you for your response.
    $endgroup$
    – Song
    Jan 13 at 16:47




    $begingroup$
    Oh, I just realized that this answer was written almost 2 years ago ... Thank you for your response.
    $endgroup$
    – Song
    Jan 13 at 16:47











    0












    $begingroup$

    Assuming $g(x)=lim_{ntoinfty} frac{1}{n}sum_{k=0}^{n-1}f(x+k)$ converges almost everywhere pointwise, we can show that $|g|_{L^1}le |f|_{L^1}$ as follows. Let $g_n(x)=frac{1}{n}sum_{k=0}^{n-1}f(x+k)$. Note that
    $$
    |g_n(x)|le frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|
    $$     and hence
    $$
    |g(x)|=lim_{ntoinfty}|g_n(x)|le liminf_{ntoinfty}frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|.
    $$     By Fatou's lemma it follows
    $$
    |g|_{L^1}le int_mathbb{R} left[liminf_{ntoinfty}frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|right];dmuleliminf_{ntoinfty}int_mathbb{R}frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|dmu=|f|_{L^1}.
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Assuming $g(x)=lim_{ntoinfty} frac{1}{n}sum_{k=0}^{n-1}f(x+k)$ converges almost everywhere pointwise, we can show that $|g|_{L^1}le |f|_{L^1}$ as follows. Let $g_n(x)=frac{1}{n}sum_{k=0}^{n-1}f(x+k)$. Note that
      $$
      |g_n(x)|le frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|
      $$       and hence
      $$
      |g(x)|=lim_{ntoinfty}|g_n(x)|le liminf_{ntoinfty}frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|.
      $$       By Fatou's lemma it follows
      $$
      |g|_{L^1}le int_mathbb{R} left[liminf_{ntoinfty}frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|right];dmuleliminf_{ntoinfty}int_mathbb{R}frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|dmu=|f|_{L^1}.
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Assuming $g(x)=lim_{ntoinfty} frac{1}{n}sum_{k=0}^{n-1}f(x+k)$ converges almost everywhere pointwise, we can show that $|g|_{L^1}le |f|_{L^1}$ as follows. Let $g_n(x)=frac{1}{n}sum_{k=0}^{n-1}f(x+k)$. Note that
        $$
        |g_n(x)|le frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|
        $$         and hence
        $$
        |g(x)|=lim_{ntoinfty}|g_n(x)|le liminf_{ntoinfty}frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|.
        $$         By Fatou's lemma it follows
        $$
        |g|_{L^1}le int_mathbb{R} left[liminf_{ntoinfty}frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|right];dmuleliminf_{ntoinfty}int_mathbb{R}frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|dmu=|f|_{L^1}.
        $$






        share|cite|improve this answer









        $endgroup$



        Assuming $g(x)=lim_{ntoinfty} frac{1}{n}sum_{k=0}^{n-1}f(x+k)$ converges almost everywhere pointwise, we can show that $|g|_{L^1}le |f|_{L^1}$ as follows. Let $g_n(x)=frac{1}{n}sum_{k=0}^{n-1}f(x+k)$. Note that
        $$
        |g_n(x)|le frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|
        $$         and hence
        $$
        |g(x)|=lim_{ntoinfty}|g_n(x)|le liminf_{ntoinfty}frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|.
        $$         By Fatou's lemma it follows
        $$
        |g|_{L^1}le int_mathbb{R} left[liminf_{ntoinfty}frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|right];dmuleliminf_{ntoinfty}int_mathbb{R}frac{1}{n}sum_{k=0}^{n-1}|f(x+k)|dmu=|f|_{L^1}.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 13 at 16:08









        SongSong

        18.6k21651




        18.6k21651






























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            VLT Carioca

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            VLT Carioca Informações Proprietário Prefeitura do Rio de Janeiro Local [[Rio de Janeiro, RJ]] País {{country data Brasil| country flag2| name = Brasil| variant = | size = }} Tipo de transporte Bonde (Veículo leve sobre trilhos) Número de linhas 2 (+1 em construção) Número de estações 26 (+3 em construção) Tráfego 80 mil passageiros/dia [ 1 ] (dezembro/2018) Website www.vltrio.com.br Funcionamento Início de funcionamento 5 de junho de 2016 (2 anos) Operadora(s) Concessionária do VLT Carioca S.A. Número de veículos 32 Alstom Citadis Comprimento dos veículos 44 m (144 ft) Dados técnicos Extensão do sistema 28 km (17,4 mi) [ 2 ] Frequência 5–10 minutos Bitola Bitola padrão/internacional 1 435 mm (4,71 ft) Eletrificação APS, 750V DC Velocidade média 15 km/h (9,32 mph) [ 2 ] Velocidade máxima 50 km/h (31,1 mph) ...
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