A Cauchy-Schwarz-type inequality for $intprod_n|f_n|$












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If $X_1,X_2$ have finite second moments then Cauchy-Schwarz gives $langle |X_1||X_2|rangle^2 leq langle |X_1|^2rangle langle |X_2|^2rangle $



If $(X_n)_{n=1}^N$ have their $N$th moments is it so that $langleprod_n|X_n|rangle^N leq prod_nlangle |X_n|^Nrangle $?






real-analysis probability-theory reference-request cauchy-schwarz-inequality holder-inequality







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  • 2




    $begingroup$
    I think this is probably what you are looking for: en.wikipedia.org/wiki/…
    $endgroup$
    – b00n heT
    Jan 2 at 8:33










  • $begingroup$
    Thanks, it does appear much closer to Holder's inequality. I will think about it in the morning.
    $endgroup$
    – enthdegree
    Jan 2 at 8:36






  • 1




    $begingroup$
    Cauchy-Schwarz is indeed a special case of Hölder's inequality. Good luck :)
    $endgroup$
    – b00n heT
    Jan 2 at 8:37


















1












$begingroup$


If $X_1,X_2$ have finite second moments then Cauchy-Schwarz gives $langle |X_1||X_2|rangle^2 leq langle |X_1|^2rangle langle |X_2|^2rangle $



If $(X_n)_{n=1}^N$ have their $N$th moments is it so that $langleprod_n|X_n|rangle^N leq prod_nlangle |X_n|^Nrangle $?






real-analysis probability-theory reference-request cauchy-schwarz-inequality holder-inequality







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I think this is probably what you are looking for: en.wikipedia.org/wiki/…
    $endgroup$
    – b00n heT
    Jan 2 at 8:33










  • $begingroup$
    Thanks, it does appear much closer to Holder's inequality. I will think about it in the morning.
    $endgroup$
    – enthdegree
    Jan 2 at 8:36






  • 1




    $begingroup$
    Cauchy-Schwarz is indeed a special case of Hölder's inequality. Good luck :)
    $endgroup$
    – b00n heT
    Jan 2 at 8:37
















1












1








1





$begingroup$


If $X_1,X_2$ have finite second moments then Cauchy-Schwarz gives $langle |X_1||X_2|rangle^2 leq langle |X_1|^2rangle langle |X_2|^2rangle $



If $(X_n)_{n=1}^N$ have their $N$th moments is it so that $langleprod_n|X_n|rangle^N leq prod_nlangle |X_n|^Nrangle $?






real-analysis probability-theory reference-request cauchy-schwarz-inequality holder-inequality







share|cite|improve this question











$endgroup$




If $X_1,X_2$ have finite second moments then Cauchy-Schwarz gives $langle |X_1||X_2|rangle^2 leq langle |X_1|^2rangle langle |X_2|^2rangle $



If $(X_n)_{n=1}^N$ have their $N$th moments is it so that $langleprod_n|X_n|rangle^N leq prod_nlangle |X_n|^Nrangle $?






real-analysis probability-theory reference-request cauchy-schwarz-inequality holder-inequality




real-analysis probability-theory reference-request cauchy-schwarz-inequality holder-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 8:36







enthdegree

















asked Jan 2 at 8:13









enthdegreeenthdegree

2,61821335




2,61821335








  • 2




    $begingroup$
    I think this is probably what you are looking for: en.wikipedia.org/wiki/…
    $endgroup$
    – b00n heT
    Jan 2 at 8:33










  • $begingroup$
    Thanks, it does appear much closer to Holder's inequality. I will think about it in the morning.
    $endgroup$
    – enthdegree
    Jan 2 at 8:36






  • 1




    $begingroup$
    Cauchy-Schwarz is indeed a special case of Hölder's inequality. Good luck :)
    $endgroup$
    – b00n heT
    Jan 2 at 8:37
















  • 2




    $begingroup$
    I think this is probably what you are looking for: en.wikipedia.org/wiki/…
    $endgroup$
    – b00n heT
    Jan 2 at 8:33










  • $begingroup$
    Thanks, it does appear much closer to Holder's inequality. I will think about it in the morning.
    $endgroup$
    – enthdegree
    Jan 2 at 8:36






  • 1




    $begingroup$
    Cauchy-Schwarz is indeed a special case of Hölder's inequality. Good luck :)
    $endgroup$
    – b00n heT
    Jan 2 at 8:37










2




2




$begingroup$
I think this is probably what you are looking for: en.wikipedia.org/wiki/…
$endgroup$
– b00n heT
Jan 2 at 8:33




$begingroup$
I think this is probably what you are looking for: en.wikipedia.org/wiki/…
$endgroup$
– b00n heT
Jan 2 at 8:33












$begingroup$
Thanks, it does appear much closer to Holder's inequality. I will think about it in the morning.
$endgroup$
– enthdegree
Jan 2 at 8:36




$begingroup$
Thanks, it does appear much closer to Holder's inequality. I will think about it in the morning.
$endgroup$
– enthdegree
Jan 2 at 8:36




1




1




$begingroup$
Cauchy-Schwarz is indeed a special case of Hölder's inequality. Good luck :)
$endgroup$
– b00n heT
Jan 2 at 8:37






$begingroup$
Cauchy-Schwarz is indeed a special case of Hölder's inequality. Good luck :)
$endgroup$
– b00n heT
Jan 2 at 8:37












1 Answer
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This can be proven using Holder's inequality. Let's do it by induction, obviously this is true for $N=1,2$ so I focus on the induction step. Suppose we are given $(X_n)_{n=1}^{N+1}$ and have finite $N+1$'th moments, then
begin{align*}
leftlangle prod_{n=1}^{N+1} |X_n|rightrangle&leq leftlangle prod_{n=1}^N |X_n|^{frac{N+1}{N}} rightrangle^{frac{N}{N+1}}cdotlangle |X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
&leq prod_{n=1}^{N+1} langle |X_n|^{N+1} rangle^{frac{1}{N+1}} cdot langle|X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
end{align*}
where the first inequality is Holder's inequality with $p=frac{N+1}{N}$ and $q=N+1$, the second one is induction hypothesis.



Taking all this to the power $N+1$ you get your desired result.






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    $begingroup$

    This can be proven using Holder's inequality. Let's do it by induction, obviously this is true for $N=1,2$ so I focus on the induction step. Suppose we are given $(X_n)_{n=1}^{N+1}$ and have finite $N+1$'th moments, then
    begin{align*}
    leftlangle prod_{n=1}^{N+1} |X_n|rightrangle&leq leftlangle prod_{n=1}^N |X_n|^{frac{N+1}{N}} rightrangle^{frac{N}{N+1}}cdotlangle |X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
    &leq prod_{n=1}^{N+1} langle |X_n|^{N+1} rangle^{frac{1}{N+1}} cdot langle|X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
    end{align*}
    where the first inequality is Holder's inequality with $p=frac{N+1}{N}$ and $q=N+1$, the second one is induction hypothesis.



    Taking all this to the power $N+1$ you get your desired result.






    share|cite|improve this answer









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      0












      $begingroup$

      This can be proven using Holder's inequality. Let's do it by induction, obviously this is true for $N=1,2$ so I focus on the induction step. Suppose we are given $(X_n)_{n=1}^{N+1}$ and have finite $N+1$'th moments, then
      begin{align*}
      leftlangle prod_{n=1}^{N+1} |X_n|rightrangle&leq leftlangle prod_{n=1}^N |X_n|^{frac{N+1}{N}} rightrangle^{frac{N}{N+1}}cdotlangle |X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
      &leq prod_{n=1}^{N+1} langle |X_n|^{N+1} rangle^{frac{1}{N+1}} cdot langle|X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
      end{align*}
      where the first inequality is Holder's inequality with $p=frac{N+1}{N}$ and $q=N+1$, the second one is induction hypothesis.



      Taking all this to the power $N+1$ you get your desired result.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        This can be proven using Holder's inequality. Let's do it by induction, obviously this is true for $N=1,2$ so I focus on the induction step. Suppose we are given $(X_n)_{n=1}^{N+1}$ and have finite $N+1$'th moments, then
        begin{align*}
        leftlangle prod_{n=1}^{N+1} |X_n|rightrangle&leq leftlangle prod_{n=1}^N |X_n|^{frac{N+1}{N}} rightrangle^{frac{N}{N+1}}cdotlangle |X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
        &leq prod_{n=1}^{N+1} langle |X_n|^{N+1} rangle^{frac{1}{N+1}} cdot langle|X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
        end{align*}
        where the first inequality is Holder's inequality with $p=frac{N+1}{N}$ and $q=N+1$, the second one is induction hypothesis.



        Taking all this to the power $N+1$ you get your desired result.






        share|cite|improve this answer









        $endgroup$



        This can be proven using Holder's inequality. Let's do it by induction, obviously this is true for $N=1,2$ so I focus on the induction step. Suppose we are given $(X_n)_{n=1}^{N+1}$ and have finite $N+1$'th moments, then
        begin{align*}
        leftlangle prod_{n=1}^{N+1} |X_n|rightrangle&leq leftlangle prod_{n=1}^N |X_n|^{frac{N+1}{N}} rightrangle^{frac{N}{N+1}}cdotlangle |X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
        &leq prod_{n=1}^{N+1} langle |X_n|^{N+1} rangle^{frac{1}{N+1}} cdot langle|X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
        end{align*}
        where the first inequality is Holder's inequality with $p=frac{N+1}{N}$ and $q=N+1$, the second one is induction hypothesis.



        Taking all this to the power $N+1$ you get your desired result.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 8:40









        P. QuintonP. Quinton

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