Equality case in Hölder's inequality












2












$begingroup$


Let $p$ and $q$ be dual exponents and let $f in L^p(0,1)$ and $g in L^q(0,1)$ be non-zero functions. Further assume that $||fg||_1=||f||_p||g||_q$. Show that $f=cg^{q-1}$ where $c$ is constant.



By Hölder we know that $||fg||_1 leq||f||_p||g||_q$, but I'm not sure how this helps us. By setting $f=cg^{q-1}$ I was able to verify the equality, but I don't know how to prove this is the only solution.






measure-theory inequality lp-spaces







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$endgroup$












  • $begingroup$
    This result is standard. See the discussion after Theorem 3.5 in Rudin's RCA.
    $endgroup$
    – Kavi Rama Murthy
    Dec 9 '18 at 23:35










  • $begingroup$
    I looked at Rudin and I don't understand.
    $endgroup$
    – Trobjon
    Dec 9 '18 at 23:48
















2












$begingroup$


Let $p$ and $q$ be dual exponents and let $f in L^p(0,1)$ and $g in L^q(0,1)$ be non-zero functions. Further assume that $||fg||_1=||f||_p||g||_q$. Show that $f=cg^{q-1}$ where $c$ is constant.



By Hölder we know that $||fg||_1 leq||f||_p||g||_q$, but I'm not sure how this helps us. By setting $f=cg^{q-1}$ I was able to verify the equality, but I don't know how to prove this is the only solution.






measure-theory inequality lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    This result is standard. See the discussion after Theorem 3.5 in Rudin's RCA.
    $endgroup$
    – Kavi Rama Murthy
    Dec 9 '18 at 23:35










  • $begingroup$
    I looked at Rudin and I don't understand.
    $endgroup$
    – Trobjon
    Dec 9 '18 at 23:48














2












2








2





$begingroup$


Let $p$ and $q$ be dual exponents and let $f in L^p(0,1)$ and $g in L^q(0,1)$ be non-zero functions. Further assume that $||fg||_1=||f||_p||g||_q$. Show that $f=cg^{q-1}$ where $c$ is constant.



By Hölder we know that $||fg||_1 leq||f||_p||g||_q$, but I'm not sure how this helps us. By setting $f=cg^{q-1}$ I was able to verify the equality, but I don't know how to prove this is the only solution.






measure-theory inequality lp-spaces







share|cite|improve this question











$endgroup$




Let $p$ and $q$ be dual exponents and let $f in L^p(0,1)$ and $g in L^q(0,1)$ be non-zero functions. Further assume that $||fg||_1=||f||_p||g||_q$. Show that $f=cg^{q-1}$ where $c$ is constant.



By Hölder we know that $||fg||_1 leq||f||_p||g||_q$, but I'm not sure how this helps us. By setting $f=cg^{q-1}$ I was able to verify the equality, but I don't know how to prove this is the only solution.






measure-theory inequality lp-spaces




measure-theory inequality lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 12:44









Davide Giraudo

126k16150261




126k16150261










asked Dec 9 '18 at 22:06









TrobjonTrobjon

234




234












  • $begingroup$
    This result is standard. See the discussion after Theorem 3.5 in Rudin's RCA.
    $endgroup$
    – Kavi Rama Murthy
    Dec 9 '18 at 23:35










  • $begingroup$
    I looked at Rudin and I don't understand.
    $endgroup$
    – Trobjon
    Dec 9 '18 at 23:48


















  • $begingroup$
    This result is standard. See the discussion after Theorem 3.5 in Rudin's RCA.
    $endgroup$
    – Kavi Rama Murthy
    Dec 9 '18 at 23:35










  • $begingroup$
    I looked at Rudin and I don't understand.
    $endgroup$
    – Trobjon
    Dec 9 '18 at 23:48
















$begingroup$
This result is standard. See the discussion after Theorem 3.5 in Rudin's RCA.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:35




$begingroup$
This result is standard. See the discussion after Theorem 3.5 in Rudin's RCA.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:35












$begingroup$
I looked at Rudin and I don't understand.
$endgroup$
– Trobjon
Dec 9 '18 at 23:48




$begingroup$
I looked at Rudin and I don't understand.
$endgroup$
– Trobjon
Dec 9 '18 at 23:48










1 Answer
1






active

oldest

votes


















1












$begingroup$

I will assume that the function are non-negative. Otherwise, the result may not hold, as we can have when $q$ is an even integer $f=1$ and $g=1$ on $(0,1/2)$ and $-1$ on $(1/2,1)$ or even worse $g^{q-1}$ may not be well-defined if $q$ is not rational and $g$ take negative values.



Since $f$ and $g$ are not zero, by scaling, it suffices to consider the case where
$leftlVert frightrVert_p=leftlVert qrightrVert_q=1$.



Observe that for all positive numbers $a$ and $b$, by convexity of the exponential function,
$$
ab=expleft(frac 1pln left(a^pright)+frac 1qln left(b^qright)right)
leqslant frac 1pexpleft(ln left(a^pright) right)+
frac 1qexpleft(ln left(b^qright) right)
$$
hence
$$
tag{*}
ableqslant frac{a^p}p+frac{b^q}q, quad a,bgeqslant 0
$$
and by strict convexity, equality holds if and only if $a^p=b^q$.
Let
$$
h:=frac{f^p}p+frac{g^q}q-fg.
$$
By $(*)$, $h$ is non-negative and by assumption, $int h=0$ hence $h=0$ almost everywhere. By the equality case in $(*)$, we derive that $f^p=g^q$ almost everywhere.






share|cite|improve this answer









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    1 Answer
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    active

    oldest

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    1












    $begingroup$

    I will assume that the function are non-negative. Otherwise, the result may not hold, as we can have when $q$ is an even integer $f=1$ and $g=1$ on $(0,1/2)$ and $-1$ on $(1/2,1)$ or even worse $g^{q-1}$ may not be well-defined if $q$ is not rational and $g$ take negative values.



    Since $f$ and $g$ are not zero, by scaling, it suffices to consider the case where
    $leftlVert frightrVert_p=leftlVert qrightrVert_q=1$.



    Observe that for all positive numbers $a$ and $b$, by convexity of the exponential function,
    $$
    ab=expleft(frac 1pln left(a^pright)+frac 1qln left(b^qright)right)
    leqslant frac 1pexpleft(ln left(a^pright) right)+
    frac 1qexpleft(ln left(b^qright) right)
    $$
    hence
    $$
    tag{*}
    ableqslant frac{a^p}p+frac{b^q}q, quad a,bgeqslant 0
    $$
    and by strict convexity, equality holds if and only if $a^p=b^q$.
    Let
    $$
    h:=frac{f^p}p+frac{g^q}q-fg.
    $$
    By $(*)$, $h$ is non-negative and by assumption, $int h=0$ hence $h=0$ almost everywhere. By the equality case in $(*)$, we derive that $f^p=g^q$ almost everywhere.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      I will assume that the function are non-negative. Otherwise, the result may not hold, as we can have when $q$ is an even integer $f=1$ and $g=1$ on $(0,1/2)$ and $-1$ on $(1/2,1)$ or even worse $g^{q-1}$ may not be well-defined if $q$ is not rational and $g$ take negative values.



      Since $f$ and $g$ are not zero, by scaling, it suffices to consider the case where
      $leftlVert frightrVert_p=leftlVert qrightrVert_q=1$.



      Observe that for all positive numbers $a$ and $b$, by convexity of the exponential function,
      $$
      ab=expleft(frac 1pln left(a^pright)+frac 1qln left(b^qright)right)
      leqslant frac 1pexpleft(ln left(a^pright) right)+
      frac 1qexpleft(ln left(b^qright) right)
      $$
      hence
      $$
      tag{*}
      ableqslant frac{a^p}p+frac{b^q}q, quad a,bgeqslant 0
      $$
      and by strict convexity, equality holds if and only if $a^p=b^q$.
      Let
      $$
      h:=frac{f^p}p+frac{g^q}q-fg.
      $$
      By $(*)$, $h$ is non-negative and by assumption, $int h=0$ hence $h=0$ almost everywhere. By the equality case in $(*)$, we derive that $f^p=g^q$ almost everywhere.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        I will assume that the function are non-negative. Otherwise, the result may not hold, as we can have when $q$ is an even integer $f=1$ and $g=1$ on $(0,1/2)$ and $-1$ on $(1/2,1)$ or even worse $g^{q-1}$ may not be well-defined if $q$ is not rational and $g$ take negative values.



        Since $f$ and $g$ are not zero, by scaling, it suffices to consider the case where
        $leftlVert frightrVert_p=leftlVert qrightrVert_q=1$.



        Observe that for all positive numbers $a$ and $b$, by convexity of the exponential function,
        $$
        ab=expleft(frac 1pln left(a^pright)+frac 1qln left(b^qright)right)
        leqslant frac 1pexpleft(ln left(a^pright) right)+
        frac 1qexpleft(ln left(b^qright) right)
        $$
        hence
        $$
        tag{*}
        ableqslant frac{a^p}p+frac{b^q}q, quad a,bgeqslant 0
        $$
        and by strict convexity, equality holds if and only if $a^p=b^q$.
        Let
        $$
        h:=frac{f^p}p+frac{g^q}q-fg.
        $$
        By $(*)$, $h$ is non-negative and by assumption, $int h=0$ hence $h=0$ almost everywhere. By the equality case in $(*)$, we derive that $f^p=g^q$ almost everywhere.






        share|cite|improve this answer









        $endgroup$



        I will assume that the function are non-negative. Otherwise, the result may not hold, as we can have when $q$ is an even integer $f=1$ and $g=1$ on $(0,1/2)$ and $-1$ on $(1/2,1)$ or even worse $g^{q-1}$ may not be well-defined if $q$ is not rational and $g$ take negative values.



        Since $f$ and $g$ are not zero, by scaling, it suffices to consider the case where
        $leftlVert frightrVert_p=leftlVert qrightrVert_q=1$.



        Observe that for all positive numbers $a$ and $b$, by convexity of the exponential function,
        $$
        ab=expleft(frac 1pln left(a^pright)+frac 1qln left(b^qright)right)
        leqslant frac 1pexpleft(ln left(a^pright) right)+
        frac 1qexpleft(ln left(b^qright) right)
        $$
        hence
        $$
        tag{*}
        ableqslant frac{a^p}p+frac{b^q}q, quad a,bgeqslant 0
        $$
        and by strict convexity, equality holds if and only if $a^p=b^q$.
        Let
        $$
        h:=frac{f^p}p+frac{g^q}q-fg.
        $$
        By $(*)$, $h$ is non-negative and by assumption, $int h=0$ hence $h=0$ almost everywhere. By the equality case in $(*)$, we derive that $f^p=g^q$ almost everywhere.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 10:33









        Davide GiraudoDavide Giraudo

        126k16150261




        126k16150261






























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