Does Young's inequality hold only for conjugate exponents?












3












$begingroup$


Suppose that $ab leq frac{1}{p}a^p+frac{1}{q}b^q$ holds for every real numbers $a,bge 0$. (where $p,q>0$ are some fixed numbers).



Is it true that $ frac{1}{p}+frac{1}{q}=1$?



I guess so, and I would like to find an easy proof of that fact. Plugging in $a=b=1$, we get $ frac{1}{p}+frac{1}{q}ge1$. Is there an easy way to see that the converse inequality must hold?



To be clear, I am not looking for proofs of Young's inequality; only for a way to see why the relation between the conjugate exponents is the only possible option.



Comment: I know that the relation $ frac{1}{p}+frac{1}{q}=1$ is necessary for Holder's inequality in general; this can be seen by scaling the measure. However, I don't think this approach is applicable here.






real-analysis inequality symmetry young-inequality







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    3












    $begingroup$


    Suppose that $ab leq frac{1}{p}a^p+frac{1}{q}b^q$ holds for every real numbers $a,bge 0$. (where $p,q>0$ are some fixed numbers).



    Is it true that $ frac{1}{p}+frac{1}{q}=1$?



    I guess so, and I would like to find an easy proof of that fact. Plugging in $a=b=1$, we get $ frac{1}{p}+frac{1}{q}ge1$. Is there an easy way to see that the converse inequality must hold?



    To be clear, I am not looking for proofs of Young's inequality; only for a way to see why the relation between the conjugate exponents is the only possible option.



    Comment: I know that the relation $ frac{1}{p}+frac{1}{q}=1$ is necessary for Holder's inequality in general; this can be seen by scaling the measure. However, I don't think this approach is applicable here.






    real-analysis inequality symmetry young-inequality







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      0



      $begingroup$


      Suppose that $ab leq frac{1}{p}a^p+frac{1}{q}b^q$ holds for every real numbers $a,bge 0$. (where $p,q>0$ are some fixed numbers).



      Is it true that $ frac{1}{p}+frac{1}{q}=1$?



      I guess so, and I would like to find an easy proof of that fact. Plugging in $a=b=1$, we get $ frac{1}{p}+frac{1}{q}ge1$. Is there an easy way to see that the converse inequality must hold?



      To be clear, I am not looking for proofs of Young's inequality; only for a way to see why the relation between the conjugate exponents is the only possible option.



      Comment: I know that the relation $ frac{1}{p}+frac{1}{q}=1$ is necessary for Holder's inequality in general; this can be seen by scaling the measure. However, I don't think this approach is applicable here.






      real-analysis inequality symmetry young-inequality







      share|cite|improve this question









      $endgroup$




      Suppose that $ab leq frac{1}{p}a^p+frac{1}{q}b^q$ holds for every real numbers $a,bge 0$. (where $p,q>0$ are some fixed numbers).



      Is it true that $ frac{1}{p}+frac{1}{q}=1$?



      I guess so, and I would like to find an easy proof of that fact. Plugging in $a=b=1$, we get $ frac{1}{p}+frac{1}{q}ge1$. Is there an easy way to see that the converse inequality must hold?



      To be clear, I am not looking for proofs of Young's inequality; only for a way to see why the relation between the conjugate exponents is the only possible option.



      Comment: I know that the relation $ frac{1}{p}+frac{1}{q}=1$ is necessary for Holder's inequality in general; this can be seen by scaling the measure. However, I don't think this approach is applicable here.






      real-analysis inequality symmetry young-inequality




      real-analysis inequality symmetry young-inequality






      share|cite|improve this question













      share|cite|improve this question











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      asked 1 hour ago









      Asaf ShacharAsaf Shachar

      5,2323941




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          3 Answers
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          $begingroup$

          If you apply the inequality for $a=lambda^{frac{1}{p}}, b=lambda^{frac{1}{q}}$, you get something like
          $$ lambda^{frac{1}{p}+frac{1}{q}}leq left( frac{1}{p}+frac{1}{q} right)lambda, qquad forall lambda>0. $$
          If you take $lambda to infty$, it's clear that $frac{1}{p}+frac{1}{q}leq 1$. Otherwise, after dividing by $lambda$ you get
          $$ lambda^{frac{1}{p}+frac{1}{q}-1} leq frac{1}{p}+frac{1}{q}, $$
          which is false for $lambda$ large enough, as the right side is constant. Taking $lambda to 0$ gets the other inequality.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            I think the following can help.



            By AM-GM $$frac{1}{p}a^p+frac{1}{q}b^qgeqleft(frac{1}{p}+frac{1}{q}right)left(left(a^pright)^{frac{1}{p}}left(b^qright)^{frac{1}{q}}right)^{frac{1}{frac{1}{p}+frac{1}{q}}}=left(frac{1}{p}+frac{1}{q}right)(ab)^{frac{1}{frac{1}{p}+frac{1}{q}}}.$$
            The equality occurs, of course.






            share|cite|improve this answer











            $endgroup$





















              0












              $begingroup$

              I found an argument that works for $p,qge 1$, since I need convexity of $xto x^p,x^q$.
              Assume by contradiction that $1/p+1/q>1$. Then, if $p'$ is the conjugate of $q$ we know that $p'>p$.



              Now, for any $a>0$ we have
              $$
              a^{p'}/p'=maxlimits_{b>0}{ab-b^q/q}
              $$
              (just because $ato a^{p'}/p'$ is the conjugate (Legendre transform) of $bto b^q/q$). Then we can pick $a$ large enough such that
              $$
              frac{a^p}{p}<frac{a^{p'}}{p'}
              $$
              and for such $a$ we would have
              $$
              a^{p}/p<maxlimits_{b>0}{ab-b^q/q}
              $$
              which implies that for some $b>0$ (and the given $a$) Young's inequality is violated.



              Hope this helps.






              share|cite|improve this answer








              New contributor




              GReyes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






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                3 Answers
                3






                active

                oldest

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                3 Answers
                3






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                active

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                4












                $begingroup$

                If you apply the inequality for $a=lambda^{frac{1}{p}}, b=lambda^{frac{1}{q}}$, you get something like
                $$ lambda^{frac{1}{p}+frac{1}{q}}leq left( frac{1}{p}+frac{1}{q} right)lambda, qquad forall lambda>0. $$
                If you take $lambda to infty$, it's clear that $frac{1}{p}+frac{1}{q}leq 1$. Otherwise, after dividing by $lambda$ you get
                $$ lambda^{frac{1}{p}+frac{1}{q}-1} leq frac{1}{p}+frac{1}{q}, $$
                which is false for $lambda$ large enough, as the right side is constant. Taking $lambda to 0$ gets the other inequality.






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  If you apply the inequality for $a=lambda^{frac{1}{p}}, b=lambda^{frac{1}{q}}$, you get something like
                  $$ lambda^{frac{1}{p}+frac{1}{q}}leq left( frac{1}{p}+frac{1}{q} right)lambda, qquad forall lambda>0. $$
                  If you take $lambda to infty$, it's clear that $frac{1}{p}+frac{1}{q}leq 1$. Otherwise, after dividing by $lambda$ you get
                  $$ lambda^{frac{1}{p}+frac{1}{q}-1} leq frac{1}{p}+frac{1}{q}, $$
                  which is false for $lambda$ large enough, as the right side is constant. Taking $lambda to 0$ gets the other inequality.






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    If you apply the inequality for $a=lambda^{frac{1}{p}}, b=lambda^{frac{1}{q}}$, you get something like
                    $$ lambda^{frac{1}{p}+frac{1}{q}}leq left( frac{1}{p}+frac{1}{q} right)lambda, qquad forall lambda>0. $$
                    If you take $lambda to infty$, it's clear that $frac{1}{p}+frac{1}{q}leq 1$. Otherwise, after dividing by $lambda$ you get
                    $$ lambda^{frac{1}{p}+frac{1}{q}-1} leq frac{1}{p}+frac{1}{q}, $$
                    which is false for $lambda$ large enough, as the right side is constant. Taking $lambda to 0$ gets the other inequality.






                    share|cite|improve this answer









                    $endgroup$



                    If you apply the inequality for $a=lambda^{frac{1}{p}}, b=lambda^{frac{1}{q}}$, you get something like
                    $$ lambda^{frac{1}{p}+frac{1}{q}}leq left( frac{1}{p}+frac{1}{q} right)lambda, qquad forall lambda>0. $$
                    If you take $lambda to infty$, it's clear that $frac{1}{p}+frac{1}{q}leq 1$. Otherwise, after dividing by $lambda$ you get
                    $$ lambda^{frac{1}{p}+frac{1}{q}-1} leq frac{1}{p}+frac{1}{q}, $$
                    which is false for $lambda$ large enough, as the right side is constant. Taking $lambda to 0$ gets the other inequality.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    Nicolás VilchesNicolás Vilches

                    43117




                    43117























                        1












                        $begingroup$

                        I think the following can help.



                        By AM-GM $$frac{1}{p}a^p+frac{1}{q}b^qgeqleft(frac{1}{p}+frac{1}{q}right)left(left(a^pright)^{frac{1}{p}}left(b^qright)^{frac{1}{q}}right)^{frac{1}{frac{1}{p}+frac{1}{q}}}=left(frac{1}{p}+frac{1}{q}right)(ab)^{frac{1}{frac{1}{p}+frac{1}{q}}}.$$
                        The equality occurs, of course.






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          I think the following can help.



                          By AM-GM $$frac{1}{p}a^p+frac{1}{q}b^qgeqleft(frac{1}{p}+frac{1}{q}right)left(left(a^pright)^{frac{1}{p}}left(b^qright)^{frac{1}{q}}right)^{frac{1}{frac{1}{p}+frac{1}{q}}}=left(frac{1}{p}+frac{1}{q}right)(ab)^{frac{1}{frac{1}{p}+frac{1}{q}}}.$$
                          The equality occurs, of course.






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            I think the following can help.



                            By AM-GM $$frac{1}{p}a^p+frac{1}{q}b^qgeqleft(frac{1}{p}+frac{1}{q}right)left(left(a^pright)^{frac{1}{p}}left(b^qright)^{frac{1}{q}}right)^{frac{1}{frac{1}{p}+frac{1}{q}}}=left(frac{1}{p}+frac{1}{q}right)(ab)^{frac{1}{frac{1}{p}+frac{1}{q}}}.$$
                            The equality occurs, of course.






                            share|cite|improve this answer











                            $endgroup$



                            I think the following can help.



                            By AM-GM $$frac{1}{p}a^p+frac{1}{q}b^qgeqleft(frac{1}{p}+frac{1}{q}right)left(left(a^pright)^{frac{1}{p}}left(b^qright)^{frac{1}{q}}right)^{frac{1}{frac{1}{p}+frac{1}{q}}}=left(frac{1}{p}+frac{1}{q}right)(ab)^{frac{1}{frac{1}{p}+frac{1}{q}}}.$$
                            The equality occurs, of course.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 38 mins ago

























                            answered 59 mins ago









                            Michael RozenbergMichael Rozenberg

                            98.1k1590188




                            98.1k1590188























                                0












                                $begingroup$

                                I found an argument that works for $p,qge 1$, since I need convexity of $xto x^p,x^q$.
                                Assume by contradiction that $1/p+1/q>1$. Then, if $p'$ is the conjugate of $q$ we know that $p'>p$.



                                Now, for any $a>0$ we have
                                $$
                                a^{p'}/p'=maxlimits_{b>0}{ab-b^q/q}
                                $$
                                (just because $ato a^{p'}/p'$ is the conjugate (Legendre transform) of $bto b^q/q$). Then we can pick $a$ large enough such that
                                $$
                                frac{a^p}{p}<frac{a^{p'}}{p'}
                                $$
                                and for such $a$ we would have
                                $$
                                a^{p}/p<maxlimits_{b>0}{ab-b^q/q}
                                $$
                                which implies that for some $b>0$ (and the given $a$) Young's inequality is violated.



                                Hope this helps.






                                share|cite|improve this answer








                                New contributor




                                GReyes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.






                                $endgroup$


















                                  0












                                  $begingroup$

                                  I found an argument that works for $p,qge 1$, since I need convexity of $xto x^p,x^q$.
                                  Assume by contradiction that $1/p+1/q>1$. Then, if $p'$ is the conjugate of $q$ we know that $p'>p$.



                                  Now, for any $a>0$ we have
                                  $$
                                  a^{p'}/p'=maxlimits_{b>0}{ab-b^q/q}
                                  $$
                                  (just because $ato a^{p'}/p'$ is the conjugate (Legendre transform) of $bto b^q/q$). Then we can pick $a$ large enough such that
                                  $$
                                  frac{a^p}{p}<frac{a^{p'}}{p'}
                                  $$
                                  and for such $a$ we would have
                                  $$
                                  a^{p}/p<maxlimits_{b>0}{ab-b^q/q}
                                  $$
                                  which implies that for some $b>0$ (and the given $a$) Young's inequality is violated.



                                  Hope this helps.






                                  share|cite|improve this answer








                                  New contributor




                                  GReyes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.






                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    I found an argument that works for $p,qge 1$, since I need convexity of $xto x^p,x^q$.
                                    Assume by contradiction that $1/p+1/q>1$. Then, if $p'$ is the conjugate of $q$ we know that $p'>p$.



                                    Now, for any $a>0$ we have
                                    $$
                                    a^{p'}/p'=maxlimits_{b>0}{ab-b^q/q}
                                    $$
                                    (just because $ato a^{p'}/p'$ is the conjugate (Legendre transform) of $bto b^q/q$). Then we can pick $a$ large enough such that
                                    $$
                                    frac{a^p}{p}<frac{a^{p'}}{p'}
                                    $$
                                    and for such $a$ we would have
                                    $$
                                    a^{p}/p<maxlimits_{b>0}{ab-b^q/q}
                                    $$
                                    which implies that for some $b>0$ (and the given $a$) Young's inequality is violated.



                                    Hope this helps.






                                    share|cite|improve this answer








                                    New contributor




                                    GReyes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.






                                    $endgroup$



                                    I found an argument that works for $p,qge 1$, since I need convexity of $xto x^p,x^q$.
                                    Assume by contradiction that $1/p+1/q>1$. Then, if $p'$ is the conjugate of $q$ we know that $p'>p$.



                                    Now, for any $a>0$ we have
                                    $$
                                    a^{p'}/p'=maxlimits_{b>0}{ab-b^q/q}
                                    $$
                                    (just because $ato a^{p'}/p'$ is the conjugate (Legendre transform) of $bto b^q/q$). Then we can pick $a$ large enough such that
                                    $$
                                    frac{a^p}{p}<frac{a^{p'}}{p'}
                                    $$
                                    and for such $a$ we would have
                                    $$
                                    a^{p}/p<maxlimits_{b>0}{ab-b^q/q}
                                    $$
                                    which implies that for some $b>0$ (and the given $a$) Young's inequality is violated.



                                    Hope this helps.







                                    share|cite|improve this answer








                                    New contributor




                                    GReyes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.









                                    share|cite|improve this answer



                                    share|cite|improve this answer






                                    New contributor




                                    GReyes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.









                                    answered 1 hour ago









                                    GReyesGReyes

                                    2463




                                    2463




                                    New contributor




                                    GReyes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.





                                    New contributor





                                    GReyes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.






                                    GReyes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.






























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