Why weak convergence and a.e. convergence imply the convergence of this integral?












1












$begingroup$


In the proof of the Brezis-Nirenberg theorem on the book "Nonlinear Analysis and Semilinear Elliptic Problems" by Ambrosetti and Malchiodi, it is used lemma 11.11 at pag. 183, in whose proof we have a sequence $(u_n)_{ninmathbb{N}}$ that converges to $bar u$ in the following senses:




  • weakly in $H^1_0(Omega)$;

  • almost everywhere;


where $Omega$ is a non-empty bounded open subset of $mathbb{R}^N$ with $Nge3$ and so, by Sobolev embedding theorem, $(u_n)_{ninmathbb{N}}$ converges weakly in $L^{2^*}(Omega)$ to $bar u$, where $2^*=frac{2N}{N-2}$. Now, it is clamed that:
$$forall vin H^1_0(Omega), int_Omega |u_n(x)|^{2^*-2}u_n(x) v(x)operatorname{d}xto int_Omega |bar u(x)|^{2^*-2}bar u(x) v(x)operatorname{d}x, nto infty.$$
I tried to figure out why it is so, using Hölder inequality or trying to find a dominating function like in the proof of Riesz-Fischer theorem, but failed in both cases... it seems to me that what I actually need to prove such a claim using that instruments it is that $|u_n-u|_{2^*}rightarrow0,nrightarrowinfty$, a thing that we don't have here.
Obviously I'm missing something... can anyone help me to figure out what I'm missing?










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    In the proof of the Brezis-Nirenberg theorem on the book "Nonlinear Analysis and Semilinear Elliptic Problems" by Ambrosetti and Malchiodi, it is used lemma 11.11 at pag. 183, in whose proof we have a sequence $(u_n)_{ninmathbb{N}}$ that converges to $bar u$ in the following senses:




    • weakly in $H^1_0(Omega)$;

    • almost everywhere;


    where $Omega$ is a non-empty bounded open subset of $mathbb{R}^N$ with $Nge3$ and so, by Sobolev embedding theorem, $(u_n)_{ninmathbb{N}}$ converges weakly in $L^{2^*}(Omega)$ to $bar u$, where $2^*=frac{2N}{N-2}$. Now, it is clamed that:
    $$forall vin H^1_0(Omega), int_Omega |u_n(x)|^{2^*-2}u_n(x) v(x)operatorname{d}xto int_Omega |bar u(x)|^{2^*-2}bar u(x) v(x)operatorname{d}x, nto infty.$$
    I tried to figure out why it is so, using Hölder inequality or trying to find a dominating function like in the proof of Riesz-Fischer theorem, but failed in both cases... it seems to me that what I actually need to prove such a claim using that instruments it is that $|u_n-u|_{2^*}rightarrow0,nrightarrowinfty$, a thing that we don't have here.
    Obviously I'm missing something... can anyone help me to figure out what I'm missing?










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      In the proof of the Brezis-Nirenberg theorem on the book "Nonlinear Analysis and Semilinear Elliptic Problems" by Ambrosetti and Malchiodi, it is used lemma 11.11 at pag. 183, in whose proof we have a sequence $(u_n)_{ninmathbb{N}}$ that converges to $bar u$ in the following senses:




      • weakly in $H^1_0(Omega)$;

      • almost everywhere;


      where $Omega$ is a non-empty bounded open subset of $mathbb{R}^N$ with $Nge3$ and so, by Sobolev embedding theorem, $(u_n)_{ninmathbb{N}}$ converges weakly in $L^{2^*}(Omega)$ to $bar u$, where $2^*=frac{2N}{N-2}$. Now, it is clamed that:
      $$forall vin H^1_0(Omega), int_Omega |u_n(x)|^{2^*-2}u_n(x) v(x)operatorname{d}xto int_Omega |bar u(x)|^{2^*-2}bar u(x) v(x)operatorname{d}x, nto infty.$$
      I tried to figure out why it is so, using Hölder inequality or trying to find a dominating function like in the proof of Riesz-Fischer theorem, but failed in both cases... it seems to me that what I actually need to prove such a claim using that instruments it is that $|u_n-u|_{2^*}rightarrow0,nrightarrowinfty$, a thing that we don't have here.
      Obviously I'm missing something... can anyone help me to figure out what I'm missing?










      share|cite|improve this question











      $endgroup$




      In the proof of the Brezis-Nirenberg theorem on the book "Nonlinear Analysis and Semilinear Elliptic Problems" by Ambrosetti and Malchiodi, it is used lemma 11.11 at pag. 183, in whose proof we have a sequence $(u_n)_{ninmathbb{N}}$ that converges to $bar u$ in the following senses:




      • weakly in $H^1_0(Omega)$;

      • almost everywhere;


      where $Omega$ is a non-empty bounded open subset of $mathbb{R}^N$ with $Nge3$ and so, by Sobolev embedding theorem, $(u_n)_{ninmathbb{N}}$ converges weakly in $L^{2^*}(Omega)$ to $bar u$, where $2^*=frac{2N}{N-2}$. Now, it is clamed that:
      $$forall vin H^1_0(Omega), int_Omega |u_n(x)|^{2^*-2}u_n(x) v(x)operatorname{d}xto int_Omega |bar u(x)|^{2^*-2}bar u(x) v(x)operatorname{d}x, nto infty.$$
      I tried to figure out why it is so, using Hölder inequality or trying to find a dominating function like in the proof of Riesz-Fischer theorem, but failed in both cases... it seems to me that what I actually need to prove such a claim using that instruments it is that $|u_n-u|_{2^*}rightarrow0,nrightarrowinfty$, a thing that we don't have here.
      Obviously I'm missing something... can anyone help me to figure out what I'm missing?







      real-analysis functional-analysis sobolev-spaces weak-convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 2 '18 at 10:23







      Bob

















      asked Dec 1 '18 at 16:05









      BobBob

      1,4641624




      1,4641624






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Edit The proof originally incorrectly applied a theorem concerning linear operators to a non-linear one, which has been fixed in the edit.



          You need to exploit the weak convergence of $u_n.$ Observe that since $|(|u_n|^{2^*-2}u_n)| = |u_n|^{2^*-1},$ for each $n$ we get,
          $$|u_n|^{2^*-2}u_n in L^{frac{2^*}{2^*-1}}(Omega) = L^{frac{2N}{N+2}}.$$
          Moreover the sequence $|u_n|^{2^*-2}u_n$ is uniformly bounded in $L^p$ where $p = frac{2N}{N+2} > 1,$ so by reflexivity we get a weakly convergent subsequence
          $$|u_{n_k}|^{2^*-2}u_{n_k} rightharpoonup v in L^p(Omega).$$
          Since it also convergences a.e., we get $v = |overline u|^{2^*-2}overline u.$ As the above holds for every subsequence, we conclude that the entire sequence $|u_n|^{2^*-2}u_n$ converges weakly to $|overline u|^{2^*-2}overline u$ in $L^p(Omega).$



          As the conjugate dual of $p=frac{2N}{N+2}$ is $2^*,$ for all $v in L^{2^*}(Omega)$ we have,
          $$ int_{Omega} |u_n|^{2^*-2}u_n v ,mathrm{d} x rightarrow int_{Omega} |overline u|^{2^*-2}overline u v ,mathrm{d}x. $$
          As $H^1_0(Omega) hookrightarrow L^{2^*}(Omega)$ by Sobolev embedding, the result follows.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for the answer. Just a question: being the map $umapsto |u|^{2^*-2}u$ not linear, don't we need the compactness of the map $F:H^1_0to L^{frac{2N}{N+2}}$ to get that $umapsto|u|^{2^*-2}u$ is weakly continuous?
            $endgroup$
            – Bob
            Dec 1 '18 at 22:41












          • $begingroup$
            I don't think that your answer actually works. In general it is not true that if $V,W$ are reflexive Banach spaces and $F:Vto W$ is continuous with respect to strong-strong topologies, then $F$ is also continuous with respect to weak-weak topologies. If I correctly understand your answer, this seems the result you used
            $endgroup$
            – Bob
            Dec 2 '18 at 8:57












          • $begingroup$
            @Bob You're right, my previous proof was incorrect. I take it you noticed how to fix the argument already, but I've also edited my answer accordingly.
            $endgroup$
            – ktoi
            Dec 2 '18 at 17:41





















          1












          $begingroup$

          Let's use $Vert . Vert_{alpha}$ to denote the usual norm on $L^{alpha}(Omega)$ for every $1leq alpha leq infty $. It is sufficient to show that:
          $$
          Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})vVert_1to 0 text{ as } n to infty
          $$

          Now, for any $phiin C^{infty}_0(Omega)$:
          $$
          Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})vVert_1\
          leq
          Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})phiVert_1
          +
          Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}}) (v - phi)Vert_1\
          $$



          The function $tto t|t|^{{2^*-2}}$ is Lipschitz on every subset of the form $[-M,M]$ where its Lipschitz constant is $(2^*-1)M^{{2^*-2}}$. This gives us the following estimate:
          $$|a|a|^{{2^*-2}} -b|b|^{{2^*-2}}|
          leq (2^*-1)(|a|+|b|)^{{2^*-2}}|a-b| qquad forall a,b in mathbb{R}
          $$

          Therefore, thanks to this relation and Hölder inequality we are able to prove:
          begin{align}
          Vert (u_n|u_n|^{{2^*-2}} &-bar{u}|bar{u}|^{{2^*-2}})phiVert_1
          leq Vert phiVert_{infty} Vert u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}}Vert_1 \
          &leq Vert phiVert_{infty} Vert (2^*-1)(|u_n| + |bar{u}|)^{2^*-2} (u_n-bar{u}) Vert_1 \
          &leq (2^*-1) Vert phiVert_{infty} Vert (|u_n| + |bar{u}|)^{2^*-2} Vert_{frac{2^*-1}{2^*-2}} Vert u_n-bar{u} Vert_{2^*-1}\
          &= (2^*-1) Vert phiVert_{infty} Vert|u_n| + |bar{u}|Vert^{2^*-2}_{2^*-1} Vert u_n-bar{u} Vert_{2^*-1} \
          &leq (2^*-1) Vert phiVert_{infty} (Vert u_nVert_{2^*-1} + Vert bar{u}Vert_{2^*-1})^{2^*-2} Vert u_n-bar{u} Vert_{2^*-1} \
          &leq (2^*-1) Vert phiVert_{infty} (2C_1)^{2^*-2} Vert u_n-bar{u} Vert_{2^*-1}
          end{align}

          where $C_1=max { sup_{n} { Vert u_n Vert_{2^*-1} },Vert bar{u} Vert_{2^*-1} }$ is finite since $H^1_0(Omega)$ embeds continuously in $L^{2^*-1}(Omega)$ and $u_n$ weakly converges to $u$ in $H^1_0(Omega)$.
          We have to estimate the other term now:
          begin{align}
          Vert (u_n|u_n|^{{2^*-2}} &-bar{u}|bar{u}|^{{2^*-2}})(v - phi)Vert_1
          leq Vert u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}}Vert_{frac{2^*}{2^*-1}} Vert v - phiVert_{2^*}\
          &leq (Vert u_n|u_n|^{{2^*-2}}Vert_{frac{2^*}{2^*-1}} +Vert bar{u}|bar{u}|^{{2^*-2}}Vert_{frac{2^*}{2^*-1}}) Vert v - phiVert_{2^*}\
          &= (Vert u_nVert^{2^*-1}_{2^*} +Vert bar{u}Vert^{2^*-1}_{2^*}) Vert v - phiVert_{2^*}\
          &leq (2C_2)^{2^*-1} Vert v - phiVert_{2^*}
          end{align}

          Where $C_2:=max{ sup_{n}{Vert u_nVert_{2^*}},Vert bar{u}Vert_{2^*} }$ that is finite thanks to Sobolev embedding of $H^1_0(Omega)$ in $L^{2^*}(Omega)$ combined, once again, with the weak convergence of $u_n$ in $H^1_0(Omega)$.
          $$
          0leqlimsup_{n to infty} Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})vVert_1\
          leq
          limsup_{n to infty} (2^*-1) Vert phiVert_{infty} (2C_1)^{2^*-2} Vert u_n-bar{u} Vert_{2^*-1}
          +\
          + limsup_{n to infty} (2C_2)^{2^*-1} Vert v - phiVert_{2^*}\
          =
          (2C_2)^{2^*-1} Vert v - phiVert_{2^*}
          $$

          using that $u_n$ converges to $bar{u}$ stongly in $L^{2^*-1}(Omega)$ due to Rellich-Kondrachov compact embeddings.
          Since $Vert v - phiVert_{2^*}$ can be chosen arbitrarily small as $C^{infty}_0(Omega)$ is dense in $L^{2^*}(Omega)$, the thesis follows.






          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%2f3021506%2fwhy-weak-convergence-and-a-e-convergence-imply-the-convergence-of-this-integral%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









            2












            $begingroup$

            Edit The proof originally incorrectly applied a theorem concerning linear operators to a non-linear one, which has been fixed in the edit.



            You need to exploit the weak convergence of $u_n.$ Observe that since $|(|u_n|^{2^*-2}u_n)| = |u_n|^{2^*-1},$ for each $n$ we get,
            $$|u_n|^{2^*-2}u_n in L^{frac{2^*}{2^*-1}}(Omega) = L^{frac{2N}{N+2}}.$$
            Moreover the sequence $|u_n|^{2^*-2}u_n$ is uniformly bounded in $L^p$ where $p = frac{2N}{N+2} > 1,$ so by reflexivity we get a weakly convergent subsequence
            $$|u_{n_k}|^{2^*-2}u_{n_k} rightharpoonup v in L^p(Omega).$$
            Since it also convergences a.e., we get $v = |overline u|^{2^*-2}overline u.$ As the above holds for every subsequence, we conclude that the entire sequence $|u_n|^{2^*-2}u_n$ converges weakly to $|overline u|^{2^*-2}overline u$ in $L^p(Omega).$



            As the conjugate dual of $p=frac{2N}{N+2}$ is $2^*,$ for all $v in L^{2^*}(Omega)$ we have,
            $$ int_{Omega} |u_n|^{2^*-2}u_n v ,mathrm{d} x rightarrow int_{Omega} |overline u|^{2^*-2}overline u v ,mathrm{d}x. $$
            As $H^1_0(Omega) hookrightarrow L^{2^*}(Omega)$ by Sobolev embedding, the result follows.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks for the answer. Just a question: being the map $umapsto |u|^{2^*-2}u$ not linear, don't we need the compactness of the map $F:H^1_0to L^{frac{2N}{N+2}}$ to get that $umapsto|u|^{2^*-2}u$ is weakly continuous?
              $endgroup$
              – Bob
              Dec 1 '18 at 22:41












            • $begingroup$
              I don't think that your answer actually works. In general it is not true that if $V,W$ are reflexive Banach spaces and $F:Vto W$ is continuous with respect to strong-strong topologies, then $F$ is also continuous with respect to weak-weak topologies. If I correctly understand your answer, this seems the result you used
              $endgroup$
              – Bob
              Dec 2 '18 at 8:57












            • $begingroup$
              @Bob You're right, my previous proof was incorrect. I take it you noticed how to fix the argument already, but I've also edited my answer accordingly.
              $endgroup$
              – ktoi
              Dec 2 '18 at 17:41


















            2












            $begingroup$

            Edit The proof originally incorrectly applied a theorem concerning linear operators to a non-linear one, which has been fixed in the edit.



            You need to exploit the weak convergence of $u_n.$ Observe that since $|(|u_n|^{2^*-2}u_n)| = |u_n|^{2^*-1},$ for each $n$ we get,
            $$|u_n|^{2^*-2}u_n in L^{frac{2^*}{2^*-1}}(Omega) = L^{frac{2N}{N+2}}.$$
            Moreover the sequence $|u_n|^{2^*-2}u_n$ is uniformly bounded in $L^p$ where $p = frac{2N}{N+2} > 1,$ so by reflexivity we get a weakly convergent subsequence
            $$|u_{n_k}|^{2^*-2}u_{n_k} rightharpoonup v in L^p(Omega).$$
            Since it also convergences a.e., we get $v = |overline u|^{2^*-2}overline u.$ As the above holds for every subsequence, we conclude that the entire sequence $|u_n|^{2^*-2}u_n$ converges weakly to $|overline u|^{2^*-2}overline u$ in $L^p(Omega).$



            As the conjugate dual of $p=frac{2N}{N+2}$ is $2^*,$ for all $v in L^{2^*}(Omega)$ we have,
            $$ int_{Omega} |u_n|^{2^*-2}u_n v ,mathrm{d} x rightarrow int_{Omega} |overline u|^{2^*-2}overline u v ,mathrm{d}x. $$
            As $H^1_0(Omega) hookrightarrow L^{2^*}(Omega)$ by Sobolev embedding, the result follows.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks for the answer. Just a question: being the map $umapsto |u|^{2^*-2}u$ not linear, don't we need the compactness of the map $F:H^1_0to L^{frac{2N}{N+2}}$ to get that $umapsto|u|^{2^*-2}u$ is weakly continuous?
              $endgroup$
              – Bob
              Dec 1 '18 at 22:41












            • $begingroup$
              I don't think that your answer actually works. In general it is not true that if $V,W$ are reflexive Banach spaces and $F:Vto W$ is continuous with respect to strong-strong topologies, then $F$ is also continuous with respect to weak-weak topologies. If I correctly understand your answer, this seems the result you used
              $endgroup$
              – Bob
              Dec 2 '18 at 8:57












            • $begingroup$
              @Bob You're right, my previous proof was incorrect. I take it you noticed how to fix the argument already, but I've also edited my answer accordingly.
              $endgroup$
              – ktoi
              Dec 2 '18 at 17:41
















            2












            2








            2





            $begingroup$

            Edit The proof originally incorrectly applied a theorem concerning linear operators to a non-linear one, which has been fixed in the edit.



            You need to exploit the weak convergence of $u_n.$ Observe that since $|(|u_n|^{2^*-2}u_n)| = |u_n|^{2^*-1},$ for each $n$ we get,
            $$|u_n|^{2^*-2}u_n in L^{frac{2^*}{2^*-1}}(Omega) = L^{frac{2N}{N+2}}.$$
            Moreover the sequence $|u_n|^{2^*-2}u_n$ is uniformly bounded in $L^p$ where $p = frac{2N}{N+2} > 1,$ so by reflexivity we get a weakly convergent subsequence
            $$|u_{n_k}|^{2^*-2}u_{n_k} rightharpoonup v in L^p(Omega).$$
            Since it also convergences a.e., we get $v = |overline u|^{2^*-2}overline u.$ As the above holds for every subsequence, we conclude that the entire sequence $|u_n|^{2^*-2}u_n$ converges weakly to $|overline u|^{2^*-2}overline u$ in $L^p(Omega).$



            As the conjugate dual of $p=frac{2N}{N+2}$ is $2^*,$ for all $v in L^{2^*}(Omega)$ we have,
            $$ int_{Omega} |u_n|^{2^*-2}u_n v ,mathrm{d} x rightarrow int_{Omega} |overline u|^{2^*-2}overline u v ,mathrm{d}x. $$
            As $H^1_0(Omega) hookrightarrow L^{2^*}(Omega)$ by Sobolev embedding, the result follows.






            share|cite|improve this answer











            $endgroup$



            Edit The proof originally incorrectly applied a theorem concerning linear operators to a non-linear one, which has been fixed in the edit.



            You need to exploit the weak convergence of $u_n.$ Observe that since $|(|u_n|^{2^*-2}u_n)| = |u_n|^{2^*-1},$ for each $n$ we get,
            $$|u_n|^{2^*-2}u_n in L^{frac{2^*}{2^*-1}}(Omega) = L^{frac{2N}{N+2}}.$$
            Moreover the sequence $|u_n|^{2^*-2}u_n$ is uniformly bounded in $L^p$ where $p = frac{2N}{N+2} > 1,$ so by reflexivity we get a weakly convergent subsequence
            $$|u_{n_k}|^{2^*-2}u_{n_k} rightharpoonup v in L^p(Omega).$$
            Since it also convergences a.e., we get $v = |overline u|^{2^*-2}overline u.$ As the above holds for every subsequence, we conclude that the entire sequence $|u_n|^{2^*-2}u_n$ converges weakly to $|overline u|^{2^*-2}overline u$ in $L^p(Omega).$



            As the conjugate dual of $p=frac{2N}{N+2}$ is $2^*,$ for all $v in L^{2^*}(Omega)$ we have,
            $$ int_{Omega} |u_n|^{2^*-2}u_n v ,mathrm{d} x rightarrow int_{Omega} |overline u|^{2^*-2}overline u v ,mathrm{d}x. $$
            As $H^1_0(Omega) hookrightarrow L^{2^*}(Omega)$ by Sobolev embedding, the result follows.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 2 '18 at 17:40

























            answered Dec 1 '18 at 21:53









            ktoiktoi

            2,3861616




            2,3861616












            • $begingroup$
              Thanks for the answer. Just a question: being the map $umapsto |u|^{2^*-2}u$ not linear, don't we need the compactness of the map $F:H^1_0to L^{frac{2N}{N+2}}$ to get that $umapsto|u|^{2^*-2}u$ is weakly continuous?
              $endgroup$
              – Bob
              Dec 1 '18 at 22:41












            • $begingroup$
              I don't think that your answer actually works. In general it is not true that if $V,W$ are reflexive Banach spaces and $F:Vto W$ is continuous with respect to strong-strong topologies, then $F$ is also continuous with respect to weak-weak topologies. If I correctly understand your answer, this seems the result you used
              $endgroup$
              – Bob
              Dec 2 '18 at 8:57












            • $begingroup$
              @Bob You're right, my previous proof was incorrect. I take it you noticed how to fix the argument already, but I've also edited my answer accordingly.
              $endgroup$
              – ktoi
              Dec 2 '18 at 17:41




















            • $begingroup$
              Thanks for the answer. Just a question: being the map $umapsto |u|^{2^*-2}u$ not linear, don't we need the compactness of the map $F:H^1_0to L^{frac{2N}{N+2}}$ to get that $umapsto|u|^{2^*-2}u$ is weakly continuous?
              $endgroup$
              – Bob
              Dec 1 '18 at 22:41












            • $begingroup$
              I don't think that your answer actually works. In general it is not true that if $V,W$ are reflexive Banach spaces and $F:Vto W$ is continuous with respect to strong-strong topologies, then $F$ is also continuous with respect to weak-weak topologies. If I correctly understand your answer, this seems the result you used
              $endgroup$
              – Bob
              Dec 2 '18 at 8:57












            • $begingroup$
              @Bob You're right, my previous proof was incorrect. I take it you noticed how to fix the argument already, but I've also edited my answer accordingly.
              $endgroup$
              – ktoi
              Dec 2 '18 at 17:41


















            $begingroup$
            Thanks for the answer. Just a question: being the map $umapsto |u|^{2^*-2}u$ not linear, don't we need the compactness of the map $F:H^1_0to L^{frac{2N}{N+2}}$ to get that $umapsto|u|^{2^*-2}u$ is weakly continuous?
            $endgroup$
            – Bob
            Dec 1 '18 at 22:41






            $begingroup$
            Thanks for the answer. Just a question: being the map $umapsto |u|^{2^*-2}u$ not linear, don't we need the compactness of the map $F:H^1_0to L^{frac{2N}{N+2}}$ to get that $umapsto|u|^{2^*-2}u$ is weakly continuous?
            $endgroup$
            – Bob
            Dec 1 '18 at 22:41














            $begingroup$
            I don't think that your answer actually works. In general it is not true that if $V,W$ are reflexive Banach spaces and $F:Vto W$ is continuous with respect to strong-strong topologies, then $F$ is also continuous with respect to weak-weak topologies. If I correctly understand your answer, this seems the result you used
            $endgroup$
            – Bob
            Dec 2 '18 at 8:57






            $begingroup$
            I don't think that your answer actually works. In general it is not true that if $V,W$ are reflexive Banach spaces and $F:Vto W$ is continuous with respect to strong-strong topologies, then $F$ is also continuous with respect to weak-weak topologies. If I correctly understand your answer, this seems the result you used
            $endgroup$
            – Bob
            Dec 2 '18 at 8:57














            $begingroup$
            @Bob You're right, my previous proof was incorrect. I take it you noticed how to fix the argument already, but I've also edited my answer accordingly.
            $endgroup$
            – ktoi
            Dec 2 '18 at 17:41






            $begingroup$
            @Bob You're right, my previous proof was incorrect. I take it you noticed how to fix the argument already, but I've also edited my answer accordingly.
            $endgroup$
            – ktoi
            Dec 2 '18 at 17:41













            1












            $begingroup$

            Let's use $Vert . Vert_{alpha}$ to denote the usual norm on $L^{alpha}(Omega)$ for every $1leq alpha leq infty $. It is sufficient to show that:
            $$
            Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})vVert_1to 0 text{ as } n to infty
            $$

            Now, for any $phiin C^{infty}_0(Omega)$:
            $$
            Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})vVert_1\
            leq
            Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})phiVert_1
            +
            Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}}) (v - phi)Vert_1\
            $$



            The function $tto t|t|^{{2^*-2}}$ is Lipschitz on every subset of the form $[-M,M]$ where its Lipschitz constant is $(2^*-1)M^{{2^*-2}}$. This gives us the following estimate:
            $$|a|a|^{{2^*-2}} -b|b|^{{2^*-2}}|
            leq (2^*-1)(|a|+|b|)^{{2^*-2}}|a-b| qquad forall a,b in mathbb{R}
            $$

            Therefore, thanks to this relation and Hölder inequality we are able to prove:
            begin{align}
            Vert (u_n|u_n|^{{2^*-2}} &-bar{u}|bar{u}|^{{2^*-2}})phiVert_1
            leq Vert phiVert_{infty} Vert u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}}Vert_1 \
            &leq Vert phiVert_{infty} Vert (2^*-1)(|u_n| + |bar{u}|)^{2^*-2} (u_n-bar{u}) Vert_1 \
            &leq (2^*-1) Vert phiVert_{infty} Vert (|u_n| + |bar{u}|)^{2^*-2} Vert_{frac{2^*-1}{2^*-2}} Vert u_n-bar{u} Vert_{2^*-1}\
            &= (2^*-1) Vert phiVert_{infty} Vert|u_n| + |bar{u}|Vert^{2^*-2}_{2^*-1} Vert u_n-bar{u} Vert_{2^*-1} \
            &leq (2^*-1) Vert phiVert_{infty} (Vert u_nVert_{2^*-1} + Vert bar{u}Vert_{2^*-1})^{2^*-2} Vert u_n-bar{u} Vert_{2^*-1} \
            &leq (2^*-1) Vert phiVert_{infty} (2C_1)^{2^*-2} Vert u_n-bar{u} Vert_{2^*-1}
            end{align}

            where $C_1=max { sup_{n} { Vert u_n Vert_{2^*-1} },Vert bar{u} Vert_{2^*-1} }$ is finite since $H^1_0(Omega)$ embeds continuously in $L^{2^*-1}(Omega)$ and $u_n$ weakly converges to $u$ in $H^1_0(Omega)$.
            We have to estimate the other term now:
            begin{align}
            Vert (u_n|u_n|^{{2^*-2}} &-bar{u}|bar{u}|^{{2^*-2}})(v - phi)Vert_1
            leq Vert u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}}Vert_{frac{2^*}{2^*-1}} Vert v - phiVert_{2^*}\
            &leq (Vert u_n|u_n|^{{2^*-2}}Vert_{frac{2^*}{2^*-1}} +Vert bar{u}|bar{u}|^{{2^*-2}}Vert_{frac{2^*}{2^*-1}}) Vert v - phiVert_{2^*}\
            &= (Vert u_nVert^{2^*-1}_{2^*} +Vert bar{u}Vert^{2^*-1}_{2^*}) Vert v - phiVert_{2^*}\
            &leq (2C_2)^{2^*-1} Vert v - phiVert_{2^*}
            end{align}

            Where $C_2:=max{ sup_{n}{Vert u_nVert_{2^*}},Vert bar{u}Vert_{2^*} }$ that is finite thanks to Sobolev embedding of $H^1_0(Omega)$ in $L^{2^*}(Omega)$ combined, once again, with the weak convergence of $u_n$ in $H^1_0(Omega)$.
            $$
            0leqlimsup_{n to infty} Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})vVert_1\
            leq
            limsup_{n to infty} (2^*-1) Vert phiVert_{infty} (2C_1)^{2^*-2} Vert u_n-bar{u} Vert_{2^*-1}
            +\
            + limsup_{n to infty} (2C_2)^{2^*-1} Vert v - phiVert_{2^*}\
            =
            (2C_2)^{2^*-1} Vert v - phiVert_{2^*}
            $$

            using that $u_n$ converges to $bar{u}$ stongly in $L^{2^*-1}(Omega)$ due to Rellich-Kondrachov compact embeddings.
            Since $Vert v - phiVert_{2^*}$ can be chosen arbitrarily small as $C^{infty}_0(Omega)$ is dense in $L^{2^*}(Omega)$, the thesis follows.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Let's use $Vert . Vert_{alpha}$ to denote the usual norm on $L^{alpha}(Omega)$ for every $1leq alpha leq infty $. It is sufficient to show that:
              $$
              Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})vVert_1to 0 text{ as } n to infty
              $$

              Now, for any $phiin C^{infty}_0(Omega)$:
              $$
              Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})vVert_1\
              leq
              Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})phiVert_1
              +
              Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}}) (v - phi)Vert_1\
              $$



              The function $tto t|t|^{{2^*-2}}$ is Lipschitz on every subset of the form $[-M,M]$ where its Lipschitz constant is $(2^*-1)M^{{2^*-2}}$. This gives us the following estimate:
              $$|a|a|^{{2^*-2}} -b|b|^{{2^*-2}}|
              leq (2^*-1)(|a|+|b|)^{{2^*-2}}|a-b| qquad forall a,b in mathbb{R}
              $$

              Therefore, thanks to this relation and Hölder inequality we are able to prove:
              begin{align}
              Vert (u_n|u_n|^{{2^*-2}} &-bar{u}|bar{u}|^{{2^*-2}})phiVert_1
              leq Vert phiVert_{infty} Vert u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}}Vert_1 \
              &leq Vert phiVert_{infty} Vert (2^*-1)(|u_n| + |bar{u}|)^{2^*-2} (u_n-bar{u}) Vert_1 \
              &leq (2^*-1) Vert phiVert_{infty} Vert (|u_n| + |bar{u}|)^{2^*-2} Vert_{frac{2^*-1}{2^*-2}} Vert u_n-bar{u} Vert_{2^*-1}\
              &= (2^*-1) Vert phiVert_{infty} Vert|u_n| + |bar{u}|Vert^{2^*-2}_{2^*-1} Vert u_n-bar{u} Vert_{2^*-1} \
              &leq (2^*-1) Vert phiVert_{infty} (Vert u_nVert_{2^*-1} + Vert bar{u}Vert_{2^*-1})^{2^*-2} Vert u_n-bar{u} Vert_{2^*-1} \
              &leq (2^*-1) Vert phiVert_{infty} (2C_1)^{2^*-2} Vert u_n-bar{u} Vert_{2^*-1}
              end{align}

              where $C_1=max { sup_{n} { Vert u_n Vert_{2^*-1} },Vert bar{u} Vert_{2^*-1} }$ is finite since $H^1_0(Omega)$ embeds continuously in $L^{2^*-1}(Omega)$ and $u_n$ weakly converges to $u$ in $H^1_0(Omega)$.
              We have to estimate the other term now:
              begin{align}
              Vert (u_n|u_n|^{{2^*-2}} &-bar{u}|bar{u}|^{{2^*-2}})(v - phi)Vert_1
              leq Vert u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}}Vert_{frac{2^*}{2^*-1}} Vert v - phiVert_{2^*}\
              &leq (Vert u_n|u_n|^{{2^*-2}}Vert_{frac{2^*}{2^*-1}} +Vert bar{u}|bar{u}|^{{2^*-2}}Vert_{frac{2^*}{2^*-1}}) Vert v - phiVert_{2^*}\
              &= (Vert u_nVert^{2^*-1}_{2^*} +Vert bar{u}Vert^{2^*-1}_{2^*}) Vert v - phiVert_{2^*}\
              &leq (2C_2)^{2^*-1} Vert v - phiVert_{2^*}
              end{align}

              Where $C_2:=max{ sup_{n}{Vert u_nVert_{2^*}},Vert bar{u}Vert_{2^*} }$ that is finite thanks to Sobolev embedding of $H^1_0(Omega)$ in $L^{2^*}(Omega)$ combined, once again, with the weak convergence of $u_n$ in $H^1_0(Omega)$.
              $$
              0leqlimsup_{n to infty} Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})vVert_1\
              leq
              limsup_{n to infty} (2^*-1) Vert phiVert_{infty} (2C_1)^{2^*-2} Vert u_n-bar{u} Vert_{2^*-1}
              +\
              + limsup_{n to infty} (2C_2)^{2^*-1} Vert v - phiVert_{2^*}\
              =
              (2C_2)^{2^*-1} Vert v - phiVert_{2^*}
              $$

              using that $u_n$ converges to $bar{u}$ stongly in $L^{2^*-1}(Omega)$ due to Rellich-Kondrachov compact embeddings.
              Since $Vert v - phiVert_{2^*}$ can be chosen arbitrarily small as $C^{infty}_0(Omega)$ is dense in $L^{2^*}(Omega)$, the thesis follows.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Let's use $Vert . Vert_{alpha}$ to denote the usual norm on $L^{alpha}(Omega)$ for every $1leq alpha leq infty $. It is sufficient to show that:
                $$
                Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})vVert_1to 0 text{ as } n to infty
                $$

                Now, for any $phiin C^{infty}_0(Omega)$:
                $$
                Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})vVert_1\
                leq
                Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})phiVert_1
                +
                Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}}) (v - phi)Vert_1\
                $$



                The function $tto t|t|^{{2^*-2}}$ is Lipschitz on every subset of the form $[-M,M]$ where its Lipschitz constant is $(2^*-1)M^{{2^*-2}}$. This gives us the following estimate:
                $$|a|a|^{{2^*-2}} -b|b|^{{2^*-2}}|
                leq (2^*-1)(|a|+|b|)^{{2^*-2}}|a-b| qquad forall a,b in mathbb{R}
                $$

                Therefore, thanks to this relation and Hölder inequality we are able to prove:
                begin{align}
                Vert (u_n|u_n|^{{2^*-2}} &-bar{u}|bar{u}|^{{2^*-2}})phiVert_1
                leq Vert phiVert_{infty} Vert u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}}Vert_1 \
                &leq Vert phiVert_{infty} Vert (2^*-1)(|u_n| + |bar{u}|)^{2^*-2} (u_n-bar{u}) Vert_1 \
                &leq (2^*-1) Vert phiVert_{infty} Vert (|u_n| + |bar{u}|)^{2^*-2} Vert_{frac{2^*-1}{2^*-2}} Vert u_n-bar{u} Vert_{2^*-1}\
                &= (2^*-1) Vert phiVert_{infty} Vert|u_n| + |bar{u}|Vert^{2^*-2}_{2^*-1} Vert u_n-bar{u} Vert_{2^*-1} \
                &leq (2^*-1) Vert phiVert_{infty} (Vert u_nVert_{2^*-1} + Vert bar{u}Vert_{2^*-1})^{2^*-2} Vert u_n-bar{u} Vert_{2^*-1} \
                &leq (2^*-1) Vert phiVert_{infty} (2C_1)^{2^*-2} Vert u_n-bar{u} Vert_{2^*-1}
                end{align}

                where $C_1=max { sup_{n} { Vert u_n Vert_{2^*-1} },Vert bar{u} Vert_{2^*-1} }$ is finite since $H^1_0(Omega)$ embeds continuously in $L^{2^*-1}(Omega)$ and $u_n$ weakly converges to $u$ in $H^1_0(Omega)$.
                We have to estimate the other term now:
                begin{align}
                Vert (u_n|u_n|^{{2^*-2}} &-bar{u}|bar{u}|^{{2^*-2}})(v - phi)Vert_1
                leq Vert u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}}Vert_{frac{2^*}{2^*-1}} Vert v - phiVert_{2^*}\
                &leq (Vert u_n|u_n|^{{2^*-2}}Vert_{frac{2^*}{2^*-1}} +Vert bar{u}|bar{u}|^{{2^*-2}}Vert_{frac{2^*}{2^*-1}}) Vert v - phiVert_{2^*}\
                &= (Vert u_nVert^{2^*-1}_{2^*} +Vert bar{u}Vert^{2^*-1}_{2^*}) Vert v - phiVert_{2^*}\
                &leq (2C_2)^{2^*-1} Vert v - phiVert_{2^*}
                end{align}

                Where $C_2:=max{ sup_{n}{Vert u_nVert_{2^*}},Vert bar{u}Vert_{2^*} }$ that is finite thanks to Sobolev embedding of $H^1_0(Omega)$ in $L^{2^*}(Omega)$ combined, once again, with the weak convergence of $u_n$ in $H^1_0(Omega)$.
                $$
                0leqlimsup_{n to infty} Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})vVert_1\
                leq
                limsup_{n to infty} (2^*-1) Vert phiVert_{infty} (2C_1)^{2^*-2} Vert u_n-bar{u} Vert_{2^*-1}
                +\
                + limsup_{n to infty} (2C_2)^{2^*-1} Vert v - phiVert_{2^*}\
                =
                (2C_2)^{2^*-1} Vert v - phiVert_{2^*}
                $$

                using that $u_n$ converges to $bar{u}$ stongly in $L^{2^*-1}(Omega)$ due to Rellich-Kondrachov compact embeddings.
                Since $Vert v - phiVert_{2^*}$ can be chosen arbitrarily small as $C^{infty}_0(Omega)$ is dense in $L^{2^*}(Omega)$, the thesis follows.






                share|cite|improve this answer









                $endgroup$



                Let's use $Vert . Vert_{alpha}$ to denote the usual norm on $L^{alpha}(Omega)$ for every $1leq alpha leq infty $. It is sufficient to show that:
                $$
                Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})vVert_1to 0 text{ as } n to infty
                $$

                Now, for any $phiin C^{infty}_0(Omega)$:
                $$
                Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})vVert_1\
                leq
                Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})phiVert_1
                +
                Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}}) (v - phi)Vert_1\
                $$



                The function $tto t|t|^{{2^*-2}}$ is Lipschitz on every subset of the form $[-M,M]$ where its Lipschitz constant is $(2^*-1)M^{{2^*-2}}$. This gives us the following estimate:
                $$|a|a|^{{2^*-2}} -b|b|^{{2^*-2}}|
                leq (2^*-1)(|a|+|b|)^{{2^*-2}}|a-b| qquad forall a,b in mathbb{R}
                $$

                Therefore, thanks to this relation and Hölder inequality we are able to prove:
                begin{align}
                Vert (u_n|u_n|^{{2^*-2}} &-bar{u}|bar{u}|^{{2^*-2}})phiVert_1
                leq Vert phiVert_{infty} Vert u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}}Vert_1 \
                &leq Vert phiVert_{infty} Vert (2^*-1)(|u_n| + |bar{u}|)^{2^*-2} (u_n-bar{u}) Vert_1 \
                &leq (2^*-1) Vert phiVert_{infty} Vert (|u_n| + |bar{u}|)^{2^*-2} Vert_{frac{2^*-1}{2^*-2}} Vert u_n-bar{u} Vert_{2^*-1}\
                &= (2^*-1) Vert phiVert_{infty} Vert|u_n| + |bar{u}|Vert^{2^*-2}_{2^*-1} Vert u_n-bar{u} Vert_{2^*-1} \
                &leq (2^*-1) Vert phiVert_{infty} (Vert u_nVert_{2^*-1} + Vert bar{u}Vert_{2^*-1})^{2^*-2} Vert u_n-bar{u} Vert_{2^*-1} \
                &leq (2^*-1) Vert phiVert_{infty} (2C_1)^{2^*-2} Vert u_n-bar{u} Vert_{2^*-1}
                end{align}

                where $C_1=max { sup_{n} { Vert u_n Vert_{2^*-1} },Vert bar{u} Vert_{2^*-1} }$ is finite since $H^1_0(Omega)$ embeds continuously in $L^{2^*-1}(Omega)$ and $u_n$ weakly converges to $u$ in $H^1_0(Omega)$.
                We have to estimate the other term now:
                begin{align}
                Vert (u_n|u_n|^{{2^*-2}} &-bar{u}|bar{u}|^{{2^*-2}})(v - phi)Vert_1
                leq Vert u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}}Vert_{frac{2^*}{2^*-1}} Vert v - phiVert_{2^*}\
                &leq (Vert u_n|u_n|^{{2^*-2}}Vert_{frac{2^*}{2^*-1}} +Vert bar{u}|bar{u}|^{{2^*-2}}Vert_{frac{2^*}{2^*-1}}) Vert v - phiVert_{2^*}\
                &= (Vert u_nVert^{2^*-1}_{2^*} +Vert bar{u}Vert^{2^*-1}_{2^*}) Vert v - phiVert_{2^*}\
                &leq (2C_2)^{2^*-1} Vert v - phiVert_{2^*}
                end{align}

                Where $C_2:=max{ sup_{n}{Vert u_nVert_{2^*}},Vert bar{u}Vert_{2^*} }$ that is finite thanks to Sobolev embedding of $H^1_0(Omega)$ in $L^{2^*}(Omega)$ combined, once again, with the weak convergence of $u_n$ in $H^1_0(Omega)$.
                $$
                0leqlimsup_{n to infty} Vert (u_n|u_n|^{{2^*-2}} -bar{u}|bar{u}|^{{2^*-2}})vVert_1\
                leq
                limsup_{n to infty} (2^*-1) Vert phiVert_{infty} (2C_1)^{2^*-2} Vert u_n-bar{u} Vert_{2^*-1}
                +\
                + limsup_{n to infty} (2C_2)^{2^*-1} Vert v - phiVert_{2^*}\
                =
                (2C_2)^{2^*-1} Vert v - phiVert_{2^*}
                $$

                using that $u_n$ converges to $bar{u}$ stongly in $L^{2^*-1}(Omega)$ due to Rellich-Kondrachov compact embeddings.
                Since $Vert v - phiVert_{2^*}$ can be chosen arbitrarily small as $C^{infty}_0(Omega)$ is dense in $L^{2^*}(Omega)$, the thesis follows.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 6 '18 at 23:48









                FormulaWriterFormulaWriter

                412




                412






























                    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%2f3021506%2fwhy-weak-convergence-and-a-e-convergence-imply-the-convergence-of-this-integral%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

                    Bundesstraße 106

                    Verónica Boquete

                    Ida-Boy-Ed-Garten