Expectation of $XY$ bounded for all bounded $Y$ implies $X$ is $L^p$












1












$begingroup$


I'm trying to prove:




Let $X$ be a real random variable, $p, q in (1,infty)$, $frac 1 p + frac 1 q = 1$. If there is $C < infty$ such that $|mathbb E[XY]| leq C ||Y||_q$ for any bounded random variable $Y$, then $X$ is in $mathcal L^p$.




My idea is to use the fact that $left(L^q(mathbb P)right)' cong L^p(mathbb P)$, and to show $F : L^q(mathbb P) to mathbb R$ defined by $F(Y) = mathbb E[XY]$ is continuous. For then, the isomorphism in particular is the isometry $kappa(f) = left( Y mapsto mathbb E[fY]right)$, so we must have $f = X$. But I'm not sure if that conclusion is correct, nor am I sure how to prove $F$ is continuous over $L^q(mathbb P)$; only over bounded functions in $L^q(mathbb P)$. Any suggestions?






real-analysis functional-analysis probability-theory lebesgue-integral lp-spaces







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I'm trying to prove:




    Let $X$ be a real random variable, $p, q in (1,infty)$, $frac 1 p + frac 1 q = 1$. If there is $C < infty$ such that $|mathbb E[XY]| leq C ||Y||_q$ for any bounded random variable $Y$, then $X$ is in $mathcal L^p$.




    My idea is to use the fact that $left(L^q(mathbb P)right)' cong L^p(mathbb P)$, and to show $F : L^q(mathbb P) to mathbb R$ defined by $F(Y) = mathbb E[XY]$ is continuous. For then, the isomorphism in particular is the isometry $kappa(f) = left( Y mapsto mathbb E[fY]right)$, so we must have $f = X$. But I'm not sure if that conclusion is correct, nor am I sure how to prove $F$ is continuous over $L^q(mathbb P)$; only over bounded functions in $L^q(mathbb P)$. Any suggestions?






    real-analysis functional-analysis probability-theory lebesgue-integral lp-spaces







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I'm trying to prove:




      Let $X$ be a real random variable, $p, q in (1,infty)$, $frac 1 p + frac 1 q = 1$. If there is $C < infty$ such that $|mathbb E[XY]| leq C ||Y||_q$ for any bounded random variable $Y$, then $X$ is in $mathcal L^p$.




      My idea is to use the fact that $left(L^q(mathbb P)right)' cong L^p(mathbb P)$, and to show $F : L^q(mathbb P) to mathbb R$ defined by $F(Y) = mathbb E[XY]$ is continuous. For then, the isomorphism in particular is the isometry $kappa(f) = left( Y mapsto mathbb E[fY]right)$, so we must have $f = X$. But I'm not sure if that conclusion is correct, nor am I sure how to prove $F$ is continuous over $L^q(mathbb P)$; only over bounded functions in $L^q(mathbb P)$. Any suggestions?






      real-analysis functional-analysis probability-theory lebesgue-integral lp-spaces







      share|cite|improve this question









      $endgroup$




      I'm trying to prove:




      Let $X$ be a real random variable, $p, q in (1,infty)$, $frac 1 p + frac 1 q = 1$. If there is $C < infty$ such that $|mathbb E[XY]| leq C ||Y||_q$ for any bounded random variable $Y$, then $X$ is in $mathcal L^p$.




      My idea is to use the fact that $left(L^q(mathbb P)right)' cong L^p(mathbb P)$, and to show $F : L^q(mathbb P) to mathbb R$ defined by $F(Y) = mathbb E[XY]$ is continuous. For then, the isomorphism in particular is the isometry $kappa(f) = left( Y mapsto mathbb E[fY]right)$, so we must have $f = X$. But I'm not sure if that conclusion is correct, nor am I sure how to prove $F$ is continuous over $L^q(mathbb P)$; only over bounded functions in $L^q(mathbb P)$. Any suggestions?






      real-analysis functional-analysis probability-theory lebesgue-integral lp-spaces




      real-analysis functional-analysis probability-theory lebesgue-integral lp-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 10 at 19:54









      D FordD Ford

      568213




      568213






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Hint: This is about how you can apply Riesz representation theorem saying that $[L^q]^*=L^p$. Let $Omega$ denote the underlying space. First note that by choosing an appropriate $theta(omega)$ for each $omega$ (in a measurable way), we can make
          $$
          |X(omega)||Y(omega)|= X(omega)Y(omega)e^{itheta(omega)}.
          $$           By letting $Y'(omega)=Y(omega)e^{itheta(omega)}$, we can improve the inequality to
          $$
          E[|X||Y|]le C|Y|_{L^q}
          $$           for all bounded $Y$. Then we can use monotone convergence theorem to conclude
          $$
          E[|X||Y|]le C|Y|_{L^q}
          $$           for all $Yin L^q$. Now deduce the conclusion that $Xin L^p$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
            $endgroup$
            – D Ford
            Jan 11 at 21:30






          • 1




            $begingroup$
            In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
            $endgroup$
            – Song
            Jan 11 at 21:37



















          2












          $begingroup$

          Hint: for each fixed $n$, apply the assumption to the random variable
          $$
          Y=Y_n= operatorname{sgn}left(Xright)leftlvert Xrightrvert^{p-1}mathbf 1{leftlvert Xrightrvertleqslant n},
          $$
          where $operatorname{sgn}left(Xright)=1$ if $X$ is positive, $-1$ if $X$ is negative and $0$ for $X=0$.
          This will give a bounded on $mathbb Eleft[leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}right]$ which does not depend on $n$.



          Indeed, let $X_n:=leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}$ and $x_n:=mathbb Eleft[X_nright]$, Then we got that $x_nleqslant cx_n^{1/q}$.






          share|cite|improve this answer











          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3069102%2fexpectation-of-xy-bounded-for-all-bounded-y-implies-x-is-lp%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Hint: This is about how you can apply Riesz representation theorem saying that $[L^q]^*=L^p$. Let $Omega$ denote the underlying space. First note that by choosing an appropriate $theta(omega)$ for each $omega$ (in a measurable way), we can make
            $$
            |X(omega)||Y(omega)|= X(omega)Y(omega)e^{itheta(omega)}.
            $$             By letting $Y'(omega)=Y(omega)e^{itheta(omega)}$, we can improve the inequality to
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all bounded $Y$. Then we can use monotone convergence theorem to conclude
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all $Yin L^q$. Now deduce the conclusion that $Xin L^p$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
              $endgroup$
              – D Ford
              Jan 11 at 21:30






            • 1




              $begingroup$
              In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
              $endgroup$
              – Song
              Jan 11 at 21:37
















            1












            $begingroup$

            Hint: This is about how you can apply Riesz representation theorem saying that $[L^q]^*=L^p$. Let $Omega$ denote the underlying space. First note that by choosing an appropriate $theta(omega)$ for each $omega$ (in a measurable way), we can make
            $$
            |X(omega)||Y(omega)|= X(omega)Y(omega)e^{itheta(omega)}.
            $$             By letting $Y'(omega)=Y(omega)e^{itheta(omega)}$, we can improve the inequality to
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all bounded $Y$. Then we can use monotone convergence theorem to conclude
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all $Yin L^q$. Now deduce the conclusion that $Xin L^p$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
              $endgroup$
              – D Ford
              Jan 11 at 21:30






            • 1




              $begingroup$
              In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
              $endgroup$
              – Song
              Jan 11 at 21:37














            1












            1








            1





            $begingroup$

            Hint: This is about how you can apply Riesz representation theorem saying that $[L^q]^*=L^p$. Let $Omega$ denote the underlying space. First note that by choosing an appropriate $theta(omega)$ for each $omega$ (in a measurable way), we can make
            $$
            |X(omega)||Y(omega)|= X(omega)Y(omega)e^{itheta(omega)}.
            $$             By letting $Y'(omega)=Y(omega)e^{itheta(omega)}$, we can improve the inequality to
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all bounded $Y$. Then we can use monotone convergence theorem to conclude
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all $Yin L^q$. Now deduce the conclusion that $Xin L^p$.






            share|cite|improve this answer











            $endgroup$



            Hint: This is about how you can apply Riesz representation theorem saying that $[L^q]^*=L^p$. Let $Omega$ denote the underlying space. First note that by choosing an appropriate $theta(omega)$ for each $omega$ (in a measurable way), we can make
            $$
            |X(omega)||Y(omega)|= X(omega)Y(omega)e^{itheta(omega)}.
            $$             By letting $Y'(omega)=Y(omega)e^{itheta(omega)}$, we can improve the inequality to
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all bounded $Y$. Then we can use monotone convergence theorem to conclude
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all $Yin L^q$. Now deduce the conclusion that $Xin L^p$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 10 at 20:30

























            answered Jan 10 at 20:22









            SongSong

            9,644627




            9,644627












            • $begingroup$
              Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
              $endgroup$
              – D Ford
              Jan 11 at 21:30






            • 1




              $begingroup$
              In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
              $endgroup$
              – Song
              Jan 11 at 21:37


















            • $begingroup$
              Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
              $endgroup$
              – D Ford
              Jan 11 at 21:30






            • 1




              $begingroup$
              In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
              $endgroup$
              – Song
              Jan 11 at 21:37
















            $begingroup$
            Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
            $endgroup$
            – D Ford
            Jan 11 at 21:30




            $begingroup$
            Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
            $endgroup$
            – D Ford
            Jan 11 at 21:30




            1




            1




            $begingroup$
            In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
            $endgroup$
            – Song
            Jan 11 at 21:37




            $begingroup$
            In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
            $endgroup$
            – Song
            Jan 11 at 21:37











            2












            $begingroup$

            Hint: for each fixed $n$, apply the assumption to the random variable
            $$
            Y=Y_n= operatorname{sgn}left(Xright)leftlvert Xrightrvert^{p-1}mathbf 1{leftlvert Xrightrvertleqslant n},
            $$
            where $operatorname{sgn}left(Xright)=1$ if $X$ is positive, $-1$ if $X$ is negative and $0$ for $X=0$.
            This will give a bounded on $mathbb Eleft[leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}right]$ which does not depend on $n$.



            Indeed, let $X_n:=leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}$ and $x_n:=mathbb Eleft[X_nright]$, Then we got that $x_nleqslant cx_n^{1/q}$.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Hint: for each fixed $n$, apply the assumption to the random variable
              $$
              Y=Y_n= operatorname{sgn}left(Xright)leftlvert Xrightrvert^{p-1}mathbf 1{leftlvert Xrightrvertleqslant n},
              $$
              where $operatorname{sgn}left(Xright)=1$ if $X$ is positive, $-1$ if $X$ is negative and $0$ for $X=0$.
              This will give a bounded on $mathbb Eleft[leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}right]$ which does not depend on $n$.



              Indeed, let $X_n:=leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}$ and $x_n:=mathbb Eleft[X_nright]$, Then we got that $x_nleqslant cx_n^{1/q}$.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Hint: for each fixed $n$, apply the assumption to the random variable
                $$
                Y=Y_n= operatorname{sgn}left(Xright)leftlvert Xrightrvert^{p-1}mathbf 1{leftlvert Xrightrvertleqslant n},
                $$
                where $operatorname{sgn}left(Xright)=1$ if $X$ is positive, $-1$ if $X$ is negative and $0$ for $X=0$.
                This will give a bounded on $mathbb Eleft[leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}right]$ which does not depend on $n$.



                Indeed, let $X_n:=leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}$ and $x_n:=mathbb Eleft[X_nright]$, Then we got that $x_nleqslant cx_n^{1/q}$.






                share|cite|improve this answer











                $endgroup$



                Hint: for each fixed $n$, apply the assumption to the random variable
                $$
                Y=Y_n= operatorname{sgn}left(Xright)leftlvert Xrightrvert^{p-1}mathbf 1{leftlvert Xrightrvertleqslant n},
                $$
                where $operatorname{sgn}left(Xright)=1$ if $X$ is positive, $-1$ if $X$ is negative and $0$ for $X=0$.
                This will give a bounded on $mathbb Eleft[leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}right]$ which does not depend on $n$.



                Indeed, let $X_n:=leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}$ and $x_n:=mathbb Eleft[X_nright]$, Then we got that $x_nleqslant cx_n^{1/q}$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 12 at 10:34

























                answered Jan 10 at 20:25









                Davide GiraudoDavide Giraudo

                126k16150261




                126k16150261






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3069102%2fexpectation-of-xy-bounded-for-all-bounded-y-implies-x-is-lp%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Mario Kart Wii

                    Image
                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
                    Read more

                    The Binding of Isaac: Rebirth/Afterbirth

                    Image
                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
                    Read more

                    What does “Dominus providebit” mean?

                    Image
                    5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
                    Read more