Prove $int_Rfg,dmleq|f|_p^{1-p/r}|g|_p^{1-q/r}(int_Rf^pg^q,dm)^{1/r}$, where $1leq pleqinfty$ and...












-2












$begingroup$


Let $f$, $g$ be positive real functions. And $f in L^p(R)$, $g in L^q(R)$, and $1 leqslant p,q <infty$. Then $fg in L^1(R)$ and
$$ int_R fg ,dm;leqslant; |f|_p^{1-p/r}|g|_p^{1-q/r}left(int_R f^pg^q ,dmright)^{1/r}$$
Where
$$1leqslant p leqslant +infty quadtext{and}quadfrac{1}{r}=frac{1}{p}+frac{1}{q}-1$$






integration measure-theory lebesgue-integral holder-inequality







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$endgroup$












  • $begingroup$
    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    $endgroup$
    – robjohn♦
    Dec 27 '18 at 11:03










  • $begingroup$
    Shouldn't the norm for $g$ be $q$? While you're correcting that, why not add some context? I have an answer, but unless there is more context, it would not be right to post it.
    $endgroup$
    – robjohn♦
    Dec 28 '18 at 8:11


















-2












$begingroup$


Let $f$, $g$ be positive real functions. And $f in L^p(R)$, $g in L^q(R)$, and $1 leqslant p,q <infty$. Then $fg in L^1(R)$ and
$$ int_R fg ,dm;leqslant; |f|_p^{1-p/r}|g|_p^{1-q/r}left(int_R f^pg^q ,dmright)^{1/r}$$
Where
$$1leqslant p leqslant +infty quadtext{and}quadfrac{1}{r}=frac{1}{p}+frac{1}{q}-1$$






integration measure-theory lebesgue-integral holder-inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    $endgroup$
    – robjohn♦
    Dec 27 '18 at 11:03










  • $begingroup$
    Shouldn't the norm for $g$ be $q$? While you're correcting that, why not add some context? I have an answer, but unless there is more context, it would not be right to post it.
    $endgroup$
    – robjohn♦
    Dec 28 '18 at 8:11
















-2












-2








-2





$begingroup$


Let $f$, $g$ be positive real functions. And $f in L^p(R)$, $g in L^q(R)$, and $1 leqslant p,q <infty$. Then $fg in L^1(R)$ and
$$ int_R fg ,dm;leqslant; |f|_p^{1-p/r}|g|_p^{1-q/r}left(int_R f^pg^q ,dmright)^{1/r}$$
Where
$$1leqslant p leqslant +infty quadtext{and}quadfrac{1}{r}=frac{1}{p}+frac{1}{q}-1$$






integration measure-theory lebesgue-integral holder-inequality







share|cite|improve this question











$endgroup$




Let $f$, $g$ be positive real functions. And $f in L^p(R)$, $g in L^q(R)$, and $1 leqslant p,q <infty$. Then $fg in L^1(R)$ and
$$ int_R fg ,dm;leqslant; |f|_p^{1-p/r}|g|_p^{1-q/r}left(int_R f^pg^q ,dmright)^{1/r}$$
Where
$$1leqslant p leqslant +infty quadtext{and}quadfrac{1}{r}=frac{1}{p}+frac{1}{q}-1$$






integration measure-theory lebesgue-integral holder-inequality




integration measure-theory lebesgue-integral holder-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 10:56









Blue

48.6k870154




48.6k870154










asked Dec 27 '18 at 10:24









nnMannnMan

464




464












  • $begingroup$
    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    $endgroup$
    – robjohn♦
    Dec 27 '18 at 11:03










  • $begingroup$
    Shouldn't the norm for $g$ be $q$? While you're correcting that, why not add some context? I have an answer, but unless there is more context, it would not be right to post it.
    $endgroup$
    – robjohn♦
    Dec 28 '18 at 8:11




















  • $begingroup$
    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    $endgroup$
    – robjohn♦
    Dec 27 '18 at 11:03










  • $begingroup$
    Shouldn't the norm for $g$ be $q$? While you're correcting that, why not add some context? I have an answer, but unless there is more context, it would not be right to post it.
    $endgroup$
    – robjohn♦
    Dec 28 '18 at 8:11


















$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Dec 27 '18 at 11:03




$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Dec 27 '18 at 11:03












$begingroup$
Shouldn't the norm for $g$ be $q$? While you're correcting that, why not add some context? I have an answer, but unless there is more context, it would not be right to post it.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:11






$begingroup$
Shouldn't the norm for $g$ be $q$? While you're correcting that, why not add some context? I have an answer, but unless there is more context, it would not be right to post it.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:11












1 Answer
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$begingroup$

Hints: I suppose $1+frac 1 r =frac 1 p +frac 1 q$. Write $fg$ as $f^{p/r}g^{q/r} f^{1-p/r}g^{1-q/r}$ and apply Holder with conjugate indices $r$ and $frac r {r-1}$. You will get $(int f^{p}g^{q})^{1/r}$ as one of the factors. Now applying another Holder to complete the proof is fairly straightfoward.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Actually, the condition is $frac1r=frac1p+frac1q-1$.
    $endgroup$
    – robjohn♦
    Dec 28 '18 at 8:09











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1 Answer
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1 Answer
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$begingroup$

Hints: I suppose $1+frac 1 r =frac 1 p +frac 1 q$. Write $fg$ as $f^{p/r}g^{q/r} f^{1-p/r}g^{1-q/r}$ and apply Holder with conjugate indices $r$ and $frac r {r-1}$. You will get $(int f^{p}g^{q})^{1/r}$ as one of the factors. Now applying another Holder to complete the proof is fairly straightfoward.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Actually, the condition is $frac1r=frac1p+frac1q-1$.
    $endgroup$
    – robjohn♦
    Dec 28 '18 at 8:09
















0












$begingroup$

Hints: I suppose $1+frac 1 r =frac 1 p +frac 1 q$. Write $fg$ as $f^{p/r}g^{q/r} f^{1-p/r}g^{1-q/r}$ and apply Holder with conjugate indices $r$ and $frac r {r-1}$. You will get $(int f^{p}g^{q})^{1/r}$ as one of the factors. Now applying another Holder to complete the proof is fairly straightfoward.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Actually, the condition is $frac1r=frac1p+frac1q-1$.
    $endgroup$
    – robjohn♦
    Dec 28 '18 at 8:09














0












0








0





$begingroup$

Hints: I suppose $1+frac 1 r =frac 1 p +frac 1 q$. Write $fg$ as $f^{p/r}g^{q/r} f^{1-p/r}g^{1-q/r}$ and apply Holder with conjugate indices $r$ and $frac r {r-1}$. You will get $(int f^{p}g^{q})^{1/r}$ as one of the factors. Now applying another Holder to complete the proof is fairly straightfoward.






share|cite|improve this answer











$endgroup$



Hints: I suppose $1+frac 1 r =frac 1 p +frac 1 q$. Write $fg$ as $f^{p/r}g^{q/r} f^{1-p/r}g^{1-q/r}$ and apply Holder with conjugate indices $r$ and $frac r {r-1}$. You will get $(int f^{p}g^{q})^{1/r}$ as one of the factors. Now applying another Holder to complete the proof is fairly straightfoward.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 28 '18 at 8:20

























answered Dec 27 '18 at 10:34









Kavi Rama MurthyKavi Rama Murthy

62.6k42262




62.6k42262












  • $begingroup$
    Actually, the condition is $frac1r=frac1p+frac1q-1$.
    $endgroup$
    – robjohn♦
    Dec 28 '18 at 8:09


















  • $begingroup$
    Actually, the condition is $frac1r=frac1p+frac1q-1$.
    $endgroup$
    – robjohn♦
    Dec 28 '18 at 8:09
















$begingroup$
Actually, the condition is $frac1r=frac1p+frac1q-1$.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:09




$begingroup$
Actually, the condition is $frac1r=frac1p+frac1q-1$.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:09


















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