Showing $f:=f_{1},…,f_{n}in L^{p}(mu)$ for $f_{i} in L^{p_{i}}$












0












$begingroup$


Let $f_{1},...,f_{n}:X to bar{mathbb R}$ measurable, while $f:=f_{1}times...times f_{n}$, and $p_{1},...,p_{n} in [1,infty]$ where $f_{i}in L^{p_{i}}(mu), forall iin {1,...,n}$.



Note $frac{1}{p}=sum_{i=1}^{n}frac{1}{p_{i}}$



Show:



i) $f_{1}times...times f_{n}=f in L^{p}$



ii) $||f||_{p} leq prod_{i}^{n}||f_{i}||_{p_{i}}$



My ideas:



ii) In class we have proven the Hölder inequality for two functions, i.e. $||fg||_{r}leq||f||_{p}||g||_{q}$, whereby $frac{1}{r}=frac{1}{p}+frac{1}{q}$. I think that the extension of this would be to merely use induction to show that Hölder is valid for any $n in mathbb N$, but this seems too easy.



i) I am stuck on $f_{1}times...times f_{n}in L^p$.



$(int_{mathbb R}|f|^p)^frac{1}{p}=(int_{mathbb R}|prod_{i=1}^{n}f_{i}|^p)^{sum_{i=1}^{n}frac{1}{p_{i}}}=prod_{j=1}^{n}(int_{mathbb R}prod_{i=1}^{n}|f_{i}|^p)^{frac{1}{p_{j}}}$



and from $frac{1}{p}=sum_{i=1}^{n}frac{1}{p_{i}}$, we glean that $p leq p_{i}$ , $i=1,...,n$



Therefore, can I say $prod_{j=1}^{n}(int_{mathbb R}prod_{i=1}^{n}|f_{i}|^p)^{frac{1}{p_{j}}}leq prod_{j=1}^{n}(int_{mathbb R}prod_{i=1}^{n}|f_{i}|^{p_{i}})^{frac{1}{p_{j}}}$ and that is $< infty$



Therefore $f in L^{p}$. I am not sure on this proof.






real-analysis measure-theory norm holder-inequality







share|cite|improve this question









$endgroup$












  • $begingroup$
    "[...], but this seems too easy". It is easy.
    $endgroup$
    – Federico
    Dec 6 '18 at 17:47


















0












$begingroup$


Let $f_{1},...,f_{n}:X to bar{mathbb R}$ measurable, while $f:=f_{1}times...times f_{n}$, and $p_{1},...,p_{n} in [1,infty]$ where $f_{i}in L^{p_{i}}(mu), forall iin {1,...,n}$.



Note $frac{1}{p}=sum_{i=1}^{n}frac{1}{p_{i}}$



Show:



i) $f_{1}times...times f_{n}=f in L^{p}$



ii) $||f||_{p} leq prod_{i}^{n}||f_{i}||_{p_{i}}$



My ideas:



ii) In class we have proven the Hölder inequality for two functions, i.e. $||fg||_{r}leq||f||_{p}||g||_{q}$, whereby $frac{1}{r}=frac{1}{p}+frac{1}{q}$. I think that the extension of this would be to merely use induction to show that Hölder is valid for any $n in mathbb N$, but this seems too easy.



i) I am stuck on $f_{1}times...times f_{n}in L^p$.



$(int_{mathbb R}|f|^p)^frac{1}{p}=(int_{mathbb R}|prod_{i=1}^{n}f_{i}|^p)^{sum_{i=1}^{n}frac{1}{p_{i}}}=prod_{j=1}^{n}(int_{mathbb R}prod_{i=1}^{n}|f_{i}|^p)^{frac{1}{p_{j}}}$



and from $frac{1}{p}=sum_{i=1}^{n}frac{1}{p_{i}}$, we glean that $p leq p_{i}$ , $i=1,...,n$



Therefore, can I say $prod_{j=1}^{n}(int_{mathbb R}prod_{i=1}^{n}|f_{i}|^p)^{frac{1}{p_{j}}}leq prod_{j=1}^{n}(int_{mathbb R}prod_{i=1}^{n}|f_{i}|^{p_{i}})^{frac{1}{p_{j}}}$ and that is $< infty$



Therefore $f in L^{p}$. I am not sure on this proof.






real-analysis measure-theory norm holder-inequality







share|cite|improve this question









$endgroup$












  • $begingroup$
    "[...], but this seems too easy". It is easy.
    $endgroup$
    – Federico
    Dec 6 '18 at 17:47
















0












0








0





$begingroup$


Let $f_{1},...,f_{n}:X to bar{mathbb R}$ measurable, while $f:=f_{1}times...times f_{n}$, and $p_{1},...,p_{n} in [1,infty]$ where $f_{i}in L^{p_{i}}(mu), forall iin {1,...,n}$.



Note $frac{1}{p}=sum_{i=1}^{n}frac{1}{p_{i}}$



Show:



i) $f_{1}times...times f_{n}=f in L^{p}$



ii) $||f||_{p} leq prod_{i}^{n}||f_{i}||_{p_{i}}$



My ideas:



ii) In class we have proven the Hölder inequality for two functions, i.e. $||fg||_{r}leq||f||_{p}||g||_{q}$, whereby $frac{1}{r}=frac{1}{p}+frac{1}{q}$. I think that the extension of this would be to merely use induction to show that Hölder is valid for any $n in mathbb N$, but this seems too easy.



i) I am stuck on $f_{1}times...times f_{n}in L^p$.



$(int_{mathbb R}|f|^p)^frac{1}{p}=(int_{mathbb R}|prod_{i=1}^{n}f_{i}|^p)^{sum_{i=1}^{n}frac{1}{p_{i}}}=prod_{j=1}^{n}(int_{mathbb R}prod_{i=1}^{n}|f_{i}|^p)^{frac{1}{p_{j}}}$



and from $frac{1}{p}=sum_{i=1}^{n}frac{1}{p_{i}}$, we glean that $p leq p_{i}$ , $i=1,...,n$



Therefore, can I say $prod_{j=1}^{n}(int_{mathbb R}prod_{i=1}^{n}|f_{i}|^p)^{frac{1}{p_{j}}}leq prod_{j=1}^{n}(int_{mathbb R}prod_{i=1}^{n}|f_{i}|^{p_{i}})^{frac{1}{p_{j}}}$ and that is $< infty$



Therefore $f in L^{p}$. I am not sure on this proof.






real-analysis measure-theory norm holder-inequality







share|cite|improve this question









$endgroup$




Let $f_{1},...,f_{n}:X to bar{mathbb R}$ measurable, while $f:=f_{1}times...times f_{n}$, and $p_{1},...,p_{n} in [1,infty]$ where $f_{i}in L^{p_{i}}(mu), forall iin {1,...,n}$.



Note $frac{1}{p}=sum_{i=1}^{n}frac{1}{p_{i}}$



Show:



i) $f_{1}times...times f_{n}=f in L^{p}$



ii) $||f||_{p} leq prod_{i}^{n}||f_{i}||_{p_{i}}$



My ideas:



ii) In class we have proven the Hölder inequality for two functions, i.e. $||fg||_{r}leq||f||_{p}||g||_{q}$, whereby $frac{1}{r}=frac{1}{p}+frac{1}{q}$. I think that the extension of this would be to merely use induction to show that Hölder is valid for any $n in mathbb N$, but this seems too easy.



i) I am stuck on $f_{1}times...times f_{n}in L^p$.



$(int_{mathbb R}|f|^p)^frac{1}{p}=(int_{mathbb R}|prod_{i=1}^{n}f_{i}|^p)^{sum_{i=1}^{n}frac{1}{p_{i}}}=prod_{j=1}^{n}(int_{mathbb R}prod_{i=1}^{n}|f_{i}|^p)^{frac{1}{p_{j}}}$



and from $frac{1}{p}=sum_{i=1}^{n}frac{1}{p_{i}}$, we glean that $p leq p_{i}$ , $i=1,...,n$



Therefore, can I say $prod_{j=1}^{n}(int_{mathbb R}prod_{i=1}^{n}|f_{i}|^p)^{frac{1}{p_{j}}}leq prod_{j=1}^{n}(int_{mathbb R}prod_{i=1}^{n}|f_{i}|^{p_{i}})^{frac{1}{p_{j}}}$ and that is $< infty$



Therefore $f in L^{p}$. I am not sure on this proof.






real-analysis measure-theory norm holder-inequality




real-analysis measure-theory norm holder-inequality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 6 '18 at 17:29









SABOYSABOY

541311




541311












  • $begingroup$
    "[...], but this seems too easy". It is easy.
    $endgroup$
    – Federico
    Dec 6 '18 at 17:47




















  • $begingroup$
    "[...], but this seems too easy". It is easy.
    $endgroup$
    – Federico
    Dec 6 '18 at 17:47


















$begingroup$
"[...], but this seems too easy". It is easy.
$endgroup$
– Federico
Dec 6 '18 at 17:47






$begingroup$
"[...], but this seems too easy". It is easy.
$endgroup$
– Federico
Dec 6 '18 at 17:47












1 Answer
1






active

oldest

votes


















3












$begingroup$

To show how trivial this is, let me switch the notation, something that people should have already done probably.



I use $q=p^{-1}in[0,1]$ and $q_i=p_i^{-1}in[0,1]$ and introduce
$$
[![f]!]_q := |f|_{q^{-1}} = |f|_p.
$$



Then Holder inequality says that $[![f_1f_2]!]_{q_1+q_2}leq[![f_1]!]_{q_1}[![f_2]!]_{q_2}$ if $q_iin[0,1]$ and $q_1+q_2leq 1$.



Now, in our case, $q=q_1+dots+q_n = (q_1+dots+q_{n-1})+q_n$, so
$$
[![f_1dotsm f_n]!]_q leq [![f_1dotsm f_{n-1}]!]_{q_1+dots+q_{n-1}}[![f_n]!]_{q_n}
leq dots
leq [![f_1]!]_{q_1} dotsm [![f_n]!]_{q_n}
$$
by bringing outside one function at a time with Holder.




No more ugly $frac1{left(frac1{p_1}+dots+frac1{p_{n-1}}right)^{-1}}+frac1{p_n}=1$ expressions for the World.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Since you have not commented on i) can I assume it is correct?
    $endgroup$
    – SABOY
    Dec 6 '18 at 18:36






  • 1




    $begingroup$
    i) is a consequence of the inequality
    $endgroup$
    – Federico
    Dec 6 '18 at 18:38











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

To show how trivial this is, let me switch the notation, something that people should have already done probably.



I use $q=p^{-1}in[0,1]$ and $q_i=p_i^{-1}in[0,1]$ and introduce
$$
[![f]!]_q := |f|_{q^{-1}} = |f|_p.
$$



Then Holder inequality says that $[![f_1f_2]!]_{q_1+q_2}leq[![f_1]!]_{q_1}[![f_2]!]_{q_2}$ if $q_iin[0,1]$ and $q_1+q_2leq 1$.



Now, in our case, $q=q_1+dots+q_n = (q_1+dots+q_{n-1})+q_n$, so
$$
[![f_1dotsm f_n]!]_q leq [![f_1dotsm f_{n-1}]!]_{q_1+dots+q_{n-1}}[![f_n]!]_{q_n}
leq dots
leq [![f_1]!]_{q_1} dotsm [![f_n]!]_{q_n}
$$
by bringing outside one function at a time with Holder.




No more ugly $frac1{left(frac1{p_1}+dots+frac1{p_{n-1}}right)^{-1}}+frac1{p_n}=1$ expressions for the World.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Since you have not commented on i) can I assume it is correct?
    $endgroup$
    – SABOY
    Dec 6 '18 at 18:36






  • 1




    $begingroup$
    i) is a consequence of the inequality
    $endgroup$
    – Federico
    Dec 6 '18 at 18:38
















3












$begingroup$

To show how trivial this is, let me switch the notation, something that people should have already done probably.



I use $q=p^{-1}in[0,1]$ and $q_i=p_i^{-1}in[0,1]$ and introduce
$$
[![f]!]_q := |f|_{q^{-1}} = |f|_p.
$$



Then Holder inequality says that $[![f_1f_2]!]_{q_1+q_2}leq[![f_1]!]_{q_1}[![f_2]!]_{q_2}$ if $q_iin[0,1]$ and $q_1+q_2leq 1$.



Now, in our case, $q=q_1+dots+q_n = (q_1+dots+q_{n-1})+q_n$, so
$$
[![f_1dotsm f_n]!]_q leq [![f_1dotsm f_{n-1}]!]_{q_1+dots+q_{n-1}}[![f_n]!]_{q_n}
leq dots
leq [![f_1]!]_{q_1} dotsm [![f_n]!]_{q_n}
$$
by bringing outside one function at a time with Holder.




No more ugly $frac1{left(frac1{p_1}+dots+frac1{p_{n-1}}right)^{-1}}+frac1{p_n}=1$ expressions for the World.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Since you have not commented on i) can I assume it is correct?
    $endgroup$
    – SABOY
    Dec 6 '18 at 18:36






  • 1




    $begingroup$
    i) is a consequence of the inequality
    $endgroup$
    – Federico
    Dec 6 '18 at 18:38














3












3








3





$begingroup$

To show how trivial this is, let me switch the notation, something that people should have already done probably.



I use $q=p^{-1}in[0,1]$ and $q_i=p_i^{-1}in[0,1]$ and introduce
$$
[![f]!]_q := |f|_{q^{-1}} = |f|_p.
$$



Then Holder inequality says that $[![f_1f_2]!]_{q_1+q_2}leq[![f_1]!]_{q_1}[![f_2]!]_{q_2}$ if $q_iin[0,1]$ and $q_1+q_2leq 1$.



Now, in our case, $q=q_1+dots+q_n = (q_1+dots+q_{n-1})+q_n$, so
$$
[![f_1dotsm f_n]!]_q leq [![f_1dotsm f_{n-1}]!]_{q_1+dots+q_{n-1}}[![f_n]!]_{q_n}
leq dots
leq [![f_1]!]_{q_1} dotsm [![f_n]!]_{q_n}
$$
by bringing outside one function at a time with Holder.




No more ugly $frac1{left(frac1{p_1}+dots+frac1{p_{n-1}}right)^{-1}}+frac1{p_n}=1$ expressions for the World.






share|cite|improve this answer











$endgroup$



To show how trivial this is, let me switch the notation, something that people should have already done probably.



I use $q=p^{-1}in[0,1]$ and $q_i=p_i^{-1}in[0,1]$ and introduce
$$
[![f]!]_q := |f|_{q^{-1}} = |f|_p.
$$



Then Holder inequality says that $[![f_1f_2]!]_{q_1+q_2}leq[![f_1]!]_{q_1}[![f_2]!]_{q_2}$ if $q_iin[0,1]$ and $q_1+q_2leq 1$.



Now, in our case, $q=q_1+dots+q_n = (q_1+dots+q_{n-1})+q_n$, so
$$
[![f_1dotsm f_n]!]_q leq [![f_1dotsm f_{n-1}]!]_{q_1+dots+q_{n-1}}[![f_n]!]_{q_n}
leq dots
leq [![f_1]!]_{q_1} dotsm [![f_n]!]_{q_n}
$$
by bringing outside one function at a time with Holder.




No more ugly $frac1{left(frac1{p_1}+dots+frac1{p_{n-1}}right)^{-1}}+frac1{p_n}=1$ expressions for the World.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 17:50

























answered Dec 6 '18 at 17:45









FedericoFederico

4,889514




4,889514












  • $begingroup$
    Since you have not commented on i) can I assume it is correct?
    $endgroup$
    – SABOY
    Dec 6 '18 at 18:36






  • 1




    $begingroup$
    i) is a consequence of the inequality
    $endgroup$
    – Federico
    Dec 6 '18 at 18:38


















  • $begingroup$
    Since you have not commented on i) can I assume it is correct?
    $endgroup$
    – SABOY
    Dec 6 '18 at 18:36






  • 1




    $begingroup$
    i) is a consequence of the inequality
    $endgroup$
    – Federico
    Dec 6 '18 at 18:38
















$begingroup$
Since you have not commented on i) can I assume it is correct?
$endgroup$
– SABOY
Dec 6 '18 at 18:36




$begingroup$
Since you have not commented on i) can I assume it is correct?
$endgroup$
– SABOY
Dec 6 '18 at 18:36




1




1




$begingroup$
i) is a consequence of the inequality
$endgroup$
– Federico
Dec 6 '18 at 18:38




$begingroup$
i) is a consequence of the inequality
$endgroup$
– Federico
Dec 6 '18 at 18:38


















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