If $f(x)$ is continuous on $[a,b]$ and $M=max ; |f(x)|$, is $M=lim limits_{ntoinfty}...












15












$begingroup$


Let $f(x)$ be a continuous real-valued function on $[a,b]$ and $M=max{|f(x)| ; :; x in [a,b]}$. Is it true that:
$$
M= lim_{ntoinfty}left(int_a^b|f(x)|^n,mathrm dxright)^{1/n} ?
$$



Thanks!






real-analysis







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Did you forget $limlimits_{ntoinfty}$ somewhere?
    $endgroup$
    – Ilya
    Dec 4 '11 at 17:46






  • 6




    $begingroup$
    In other words, you are asking if $| f |_n to | f |_infty$ for a continuous $f$.
    $endgroup$
    – Srivatsan
    Dec 4 '11 at 19:09


















15












$begingroup$


Let $f(x)$ be a continuous real-valued function on $[a,b]$ and $M=max{|f(x)| ; :; x in [a,b]}$. Is it true that:
$$
M= lim_{ntoinfty}left(int_a^b|f(x)|^n,mathrm dxright)^{1/n} ?
$$



Thanks!






real-analysis







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Did you forget $limlimits_{ntoinfty}$ somewhere?
    $endgroup$
    – Ilya
    Dec 4 '11 at 17:46






  • 6




    $begingroup$
    In other words, you are asking if $| f |_n to | f |_infty$ for a continuous $f$.
    $endgroup$
    – Srivatsan
    Dec 4 '11 at 19:09
















15












15








15


10



$begingroup$


Let $f(x)$ be a continuous real-valued function on $[a,b]$ and $M=max{|f(x)| ; :; x in [a,b]}$. Is it true that:
$$
M= lim_{ntoinfty}left(int_a^b|f(x)|^n,mathrm dxright)^{1/n} ?
$$



Thanks!






real-analysis







share|cite|improve this question











$endgroup$




Let $f(x)$ be a continuous real-valued function on $[a,b]$ and $M=max{|f(x)| ; :; x in [a,b]}$. Is it true that:
$$
M= lim_{ntoinfty}left(int_a^b|f(x)|^n,mathrm dxright)^{1/n} ?
$$



Thanks!






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '11 at 19:04









Srivatsan

20.9k371125




20.9k371125










asked Dec 4 '11 at 17:45









tomergtomerg

738820




738820








  • 3




    $begingroup$
    Did you forget $limlimits_{ntoinfty}$ somewhere?
    $endgroup$
    – Ilya
    Dec 4 '11 at 17:46






  • 6




    $begingroup$
    In other words, you are asking if $| f |_n to | f |_infty$ for a continuous $f$.
    $endgroup$
    – Srivatsan
    Dec 4 '11 at 19:09
















  • 3




    $begingroup$
    Did you forget $limlimits_{ntoinfty}$ somewhere?
    $endgroup$
    – Ilya
    Dec 4 '11 at 17:46






  • 6




    $begingroup$
    In other words, you are asking if $| f |_n to | f |_infty$ for a continuous $f$.
    $endgroup$
    – Srivatsan
    Dec 4 '11 at 19:09










3




3




$begingroup$
Did you forget $limlimits_{ntoinfty}$ somewhere?
$endgroup$
– Ilya
Dec 4 '11 at 17:46




$begingroup$
Did you forget $limlimits_{ntoinfty}$ somewhere?
$endgroup$
– Ilya
Dec 4 '11 at 17:46




6




6




$begingroup$
In other words, you are asking if $| f |_n to | f |_infty$ for a continuous $f$.
$endgroup$
– Srivatsan
Dec 4 '11 at 19:09






$begingroup$
In other words, you are asking if $| f |_n to | f |_infty$ for a continuous $f$.
$endgroup$
– Srivatsan
Dec 4 '11 at 19:09












2 Answers
2






active

oldest

votes


















18












$begingroup$

Put $S_{delta}:={xinleft[a,bright], |f(x)|geq M-delta}$ for any $delta>0$. Then we have for all $n$
$$Mcdot (b-a)^{frac 1n}geqleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq left(int_{S_{delta}}|f(x)|^ndxright)^{frac 1n}geq (M-delta)left(lambda(S_{delta})right)^{frac 1n}.$$
Since $f$ is continuous, the measure of $S_{delta}$ is positive and taking the $liminf$ and $limsup$, we can see that
$$Mgeq limsup_n left(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-deltaquad mbox{and }quad Mgeq liminf_nleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-delta$$
for all $delta>0$, so $lim_{ntoinfty}left(int_a^b|f(x)|^ndxright)^{frac 1n}=M$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why do you absolutely need to take the lim inf and lim sup? All the limits exists.
    $endgroup$
    – Patrick Da Silva
    Jan 5 '12 at 10:42






  • 3




    $begingroup$
    @PatrickDaSilva We don't know a priori that the limit $lim_nleft(int_a^b|f(x)|^nright)^{frac 1n}$ exists.
    $endgroup$
    – Davide Giraudo
    Jan 5 '12 at 12:36










  • $begingroup$
    Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out
    $endgroup$
    – Patrick Da Silva
    Jan 5 '12 at 23:13



















0












$begingroup$

It can be seen as a particular case of Laplace's principle. For simplicity assume that $f(x) ge 0$.



$$lim_{nrightarrowinfty} log frac{1}{n} int_a^b exp(n log f(x))mathrm d x = max_{xin[a,b]} log f(x)$$



Taking $exp$ of both sides gives the result.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    18












    $begingroup$

    Put $S_{delta}:={xinleft[a,bright], |f(x)|geq M-delta}$ for any $delta>0$. Then we have for all $n$
    $$Mcdot (b-a)^{frac 1n}geqleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq left(int_{S_{delta}}|f(x)|^ndxright)^{frac 1n}geq (M-delta)left(lambda(S_{delta})right)^{frac 1n}.$$
    Since $f$ is continuous, the measure of $S_{delta}$ is positive and taking the $liminf$ and $limsup$, we can see that
    $$Mgeq limsup_n left(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-deltaquad mbox{and }quad Mgeq liminf_nleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-delta$$
    for all $delta>0$, so $lim_{ntoinfty}left(int_a^b|f(x)|^ndxright)^{frac 1n}=M$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Why do you absolutely need to take the lim inf and lim sup? All the limits exists.
      $endgroup$
      – Patrick Da Silva
      Jan 5 '12 at 10:42






    • 3




      $begingroup$
      @PatrickDaSilva We don't know a priori that the limit $lim_nleft(int_a^b|f(x)|^nright)^{frac 1n}$ exists.
      $endgroup$
      – Davide Giraudo
      Jan 5 '12 at 12:36










    • $begingroup$
      Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out
      $endgroup$
      – Patrick Da Silva
      Jan 5 '12 at 23:13
















    18












    $begingroup$

    Put $S_{delta}:={xinleft[a,bright], |f(x)|geq M-delta}$ for any $delta>0$. Then we have for all $n$
    $$Mcdot (b-a)^{frac 1n}geqleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq left(int_{S_{delta}}|f(x)|^ndxright)^{frac 1n}geq (M-delta)left(lambda(S_{delta})right)^{frac 1n}.$$
    Since $f$ is continuous, the measure of $S_{delta}$ is positive and taking the $liminf$ and $limsup$, we can see that
    $$Mgeq limsup_n left(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-deltaquad mbox{and }quad Mgeq liminf_nleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-delta$$
    for all $delta>0$, so $lim_{ntoinfty}left(int_a^b|f(x)|^ndxright)^{frac 1n}=M$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Why do you absolutely need to take the lim inf and lim sup? All the limits exists.
      $endgroup$
      – Patrick Da Silva
      Jan 5 '12 at 10:42






    • 3




      $begingroup$
      @PatrickDaSilva We don't know a priori that the limit $lim_nleft(int_a^b|f(x)|^nright)^{frac 1n}$ exists.
      $endgroup$
      – Davide Giraudo
      Jan 5 '12 at 12:36










    • $begingroup$
      Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out
      $endgroup$
      – Patrick Da Silva
      Jan 5 '12 at 23:13














    18












    18








    18





    $begingroup$

    Put $S_{delta}:={xinleft[a,bright], |f(x)|geq M-delta}$ for any $delta>0$. Then we have for all $n$
    $$Mcdot (b-a)^{frac 1n}geqleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq left(int_{S_{delta}}|f(x)|^ndxright)^{frac 1n}geq (M-delta)left(lambda(S_{delta})right)^{frac 1n}.$$
    Since $f$ is continuous, the measure of $S_{delta}$ is positive and taking the $liminf$ and $limsup$, we can see that
    $$Mgeq limsup_n left(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-deltaquad mbox{and }quad Mgeq liminf_nleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-delta$$
    for all $delta>0$, so $lim_{ntoinfty}left(int_a^b|f(x)|^ndxright)^{frac 1n}=M$.






    share|cite|improve this answer











    $endgroup$



    Put $S_{delta}:={xinleft[a,bright], |f(x)|geq M-delta}$ for any $delta>0$. Then we have for all $n$
    $$Mcdot (b-a)^{frac 1n}geqleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq left(int_{S_{delta}}|f(x)|^ndxright)^{frac 1n}geq (M-delta)left(lambda(S_{delta})right)^{frac 1n}.$$
    Since $f$ is continuous, the measure of $S_{delta}$ is positive and taking the $liminf$ and $limsup$, we can see that
    $$Mgeq limsup_n left(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-deltaquad mbox{and }quad Mgeq liminf_nleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-delta$$
    for all $delta>0$, so $lim_{ntoinfty}left(int_a^b|f(x)|^ndxright)^{frac 1n}=M$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 5 '11 at 15:27

























    answered Dec 4 '11 at 18:06









    Davide GiraudoDavide Giraudo

    125k16150261




    125k16150261












    • $begingroup$
      Why do you absolutely need to take the lim inf and lim sup? All the limits exists.
      $endgroup$
      – Patrick Da Silva
      Jan 5 '12 at 10:42






    • 3




      $begingroup$
      @PatrickDaSilva We don't know a priori that the limit $lim_nleft(int_a^b|f(x)|^nright)^{frac 1n}$ exists.
      $endgroup$
      – Davide Giraudo
      Jan 5 '12 at 12:36










    • $begingroup$
      Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out
      $endgroup$
      – Patrick Da Silva
      Jan 5 '12 at 23:13


















    • $begingroup$
      Why do you absolutely need to take the lim inf and lim sup? All the limits exists.
      $endgroup$
      – Patrick Da Silva
      Jan 5 '12 at 10:42






    • 3




      $begingroup$
      @PatrickDaSilva We don't know a priori that the limit $lim_nleft(int_a^b|f(x)|^nright)^{frac 1n}$ exists.
      $endgroup$
      – Davide Giraudo
      Jan 5 '12 at 12:36










    • $begingroup$
      Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out
      $endgroup$
      – Patrick Da Silva
      Jan 5 '12 at 23:13
















    $begingroup$
    Why do you absolutely need to take the lim inf and lim sup? All the limits exists.
    $endgroup$
    – Patrick Da Silva
    Jan 5 '12 at 10:42




    $begingroup$
    Why do you absolutely need to take the lim inf and lim sup? All the limits exists.
    $endgroup$
    – Patrick Da Silva
    Jan 5 '12 at 10:42




    3




    3




    $begingroup$
    @PatrickDaSilva We don't know a priori that the limit $lim_nleft(int_a^b|f(x)|^nright)^{frac 1n}$ exists.
    $endgroup$
    – Davide Giraudo
    Jan 5 '12 at 12:36




    $begingroup$
    @PatrickDaSilva We don't know a priori that the limit $lim_nleft(int_a^b|f(x)|^nright)^{frac 1n}$ exists.
    $endgroup$
    – Davide Giraudo
    Jan 5 '12 at 12:36












    $begingroup$
    Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out
    $endgroup$
    – Patrick Da Silva
    Jan 5 '12 at 23:13




    $begingroup$
    Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out
    $endgroup$
    – Patrick Da Silva
    Jan 5 '12 at 23:13











    0












    $begingroup$

    It can be seen as a particular case of Laplace's principle. For simplicity assume that $f(x) ge 0$.



    $$lim_{nrightarrowinfty} log frac{1}{n} int_a^b exp(n log f(x))mathrm d x = max_{xin[a,b]} log f(x)$$



    Taking $exp$ of both sides gives the result.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      It can be seen as a particular case of Laplace's principle. For simplicity assume that $f(x) ge 0$.



      $$lim_{nrightarrowinfty} log frac{1}{n} int_a^b exp(n log f(x))mathrm d x = max_{xin[a,b]} log f(x)$$



      Taking $exp$ of both sides gives the result.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        It can be seen as a particular case of Laplace's principle. For simplicity assume that $f(x) ge 0$.



        $$lim_{nrightarrowinfty} log frac{1}{n} int_a^b exp(n log f(x))mathrm d x = max_{xin[a,b]} log f(x)$$



        Taking $exp$ of both sides gives the result.






        share|cite|improve this answer









        $endgroup$



        It can be seen as a particular case of Laplace's principle. For simplicity assume that $f(x) ge 0$.



        $$lim_{nrightarrowinfty} log frac{1}{n} int_a^b exp(n log f(x))mathrm d x = max_{xin[a,b]} log f(x)$$



        Taking $exp$ of both sides gives the result.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 21:38









        beckobecko

        2,34931942




        2,34931942






























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