Are the singular points of harmonic function on the disk always of measure zero?












1












$begingroup$


Let $f : mathbb D^2 to mathbb R$ be a smooth function with no singular points, i.e. $df neq 0$ on $mathbb D^2$. (Here $mathbb D^2$ is the closed unit disk in $mathbb R^2$).
Let $omega:mathbb D^2 to mathbb R$ be the harmonic function corresponding to the Dirichlet problem imposed by $f$, i.e $omega|_{partial mathbb D^2}=f|_{partial mathbb D^2}$.



Is it true that $domega neq 0$ on a set of full measure in $M$?



*By harmonic I mean $Delta omega=omega_{xx}+omega_{yy}=0$.






differential-topology riemannian-geometry harmonic-functions singularity







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $f : mathbb D^2 to mathbb R$ be a smooth function with no singular points, i.e. $df neq 0$ on $mathbb D^2$. (Here $mathbb D^2$ is the closed unit disk in $mathbb R^2$).
    Let $omega:mathbb D^2 to mathbb R$ be the harmonic function corresponding to the Dirichlet problem imposed by $f$, i.e $omega|_{partial mathbb D^2}=f|_{partial mathbb D^2}$.



    Is it true that $domega neq 0$ on a set of full measure in $M$?



    *By harmonic I mean $Delta omega=omega_{xx}+omega_{yy}=0$.






    differential-topology riemannian-geometry harmonic-functions singularity







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $f : mathbb D^2 to mathbb R$ be a smooth function with no singular points, i.e. $df neq 0$ on $mathbb D^2$. (Here $mathbb D^2$ is the closed unit disk in $mathbb R^2$).
      Let $omega:mathbb D^2 to mathbb R$ be the harmonic function corresponding to the Dirichlet problem imposed by $f$, i.e $omega|_{partial mathbb D^2}=f|_{partial mathbb D^2}$.



      Is it true that $domega neq 0$ on a set of full measure in $M$?



      *By harmonic I mean $Delta omega=omega_{xx}+omega_{yy}=0$.






      differential-topology riemannian-geometry harmonic-functions singularity







      share|cite|improve this question









      $endgroup$




      Let $f : mathbb D^2 to mathbb R$ be a smooth function with no singular points, i.e. $df neq 0$ on $mathbb D^2$. (Here $mathbb D^2$ is the closed unit disk in $mathbb R^2$).
      Let $omega:mathbb D^2 to mathbb R$ be the harmonic function corresponding to the Dirichlet problem imposed by $f$, i.e $omega|_{partial mathbb D^2}=f|_{partial mathbb D^2}$.



      Is it true that $domega neq 0$ on a set of full measure in $M$?



      *By harmonic I mean $Delta omega=omega_{xx}+omega_{yy}=0$.






      differential-topology riemannian-geometry harmonic-functions singularity




      differential-topology riemannian-geometry harmonic-functions singularity






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 7 at 13:17









      Asaf ShacharAsaf Shachar

      5,79931145




      5,79931145






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          After speaking with a colleague, here is a possible solution:



          Sine $omega$ is harmonic, it is real-analytic. In particular its partial derivatives $omega_x,omega_y:mathbb D^2 to mathbb R$ are real-analytic. Define $A={ pin mathbb D^2 , | , omega_x(p)=omega_y(p)=0}$. We shall prove $A$ has measure zero. Indeed, $A$ is closed. So, if it had positive measure, then it would contain an accumulation point in $operatorname{Int} mathbb D^2$ (see the lemma below). By the identity theorem, this would imply $omega_x=omega_y=0$ on $operatorname{Int} mathbb D^2$, which would imply $omega$ is constant on $mathbb D^2$. However, $f$ cannot be constant* on $partial mathbb D^2$ which is a contradiction to $f|_{partial mathbb D^2}=omega|_{partial mathbb D^2}$.






          • $f$ is not constant on $partial mathbb D^2$: Indeed, we assumed that $df neq 0$ everywhere, so $min f,max f$ are both obtained on $partial mathbb D^2$, and are distinct, since $f$ is not constant on the whole disk.



          Lemma: A closed subset $A subseteq mathbb D^2$ of positive measure contains an accumulation point in $operatorname{Int} mathbb D^2$.



          Proof:



          Assume by contradiction that $A$ does not contain an accumulation point in $operatorname{Int} mathbb D^2$. This implies (by compactness), that for any natural $n$, $A cap B_{1-1/n}$ is finite, where $B_r$ is the closed disk of radius $r$. Thus, $A cap operatorname{Int} mathbb D^2=cup_n A cap B_{1-1/n}$ is countable, hence of measure zero. Thus $A$ has measure zero, contradiction.






          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%2f3064993%2fare-the-singular-points-of-harmonic-function-on-the-disk-always-of-measure-zero%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            After speaking with a colleague, here is a possible solution:



            Sine $omega$ is harmonic, it is real-analytic. In particular its partial derivatives $omega_x,omega_y:mathbb D^2 to mathbb R$ are real-analytic. Define $A={ pin mathbb D^2 , | , omega_x(p)=omega_y(p)=0}$. We shall prove $A$ has measure zero. Indeed, $A$ is closed. So, if it had positive measure, then it would contain an accumulation point in $operatorname{Int} mathbb D^2$ (see the lemma below). By the identity theorem, this would imply $omega_x=omega_y=0$ on $operatorname{Int} mathbb D^2$, which would imply $omega$ is constant on $mathbb D^2$. However, $f$ cannot be constant* on $partial mathbb D^2$ which is a contradiction to $f|_{partial mathbb D^2}=omega|_{partial mathbb D^2}$.






            • $f$ is not constant on $partial mathbb D^2$: Indeed, we assumed that $df neq 0$ everywhere, so $min f,max f$ are both obtained on $partial mathbb D^2$, and are distinct, since $f$ is not constant on the whole disk.



            Lemma: A closed subset $A subseteq mathbb D^2$ of positive measure contains an accumulation point in $operatorname{Int} mathbb D^2$.



            Proof:



            Assume by contradiction that $A$ does not contain an accumulation point in $operatorname{Int} mathbb D^2$. This implies (by compactness), that for any natural $n$, $A cap B_{1-1/n}$ is finite, where $B_r$ is the closed disk of radius $r$. Thus, $A cap operatorname{Int} mathbb D^2=cup_n A cap B_{1-1/n}$ is countable, hence of measure zero. Thus $A$ has measure zero, contradiction.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              After speaking with a colleague, here is a possible solution:



              Sine $omega$ is harmonic, it is real-analytic. In particular its partial derivatives $omega_x,omega_y:mathbb D^2 to mathbb R$ are real-analytic. Define $A={ pin mathbb D^2 , | , omega_x(p)=omega_y(p)=0}$. We shall prove $A$ has measure zero. Indeed, $A$ is closed. So, if it had positive measure, then it would contain an accumulation point in $operatorname{Int} mathbb D^2$ (see the lemma below). By the identity theorem, this would imply $omega_x=omega_y=0$ on $operatorname{Int} mathbb D^2$, which would imply $omega$ is constant on $mathbb D^2$. However, $f$ cannot be constant* on $partial mathbb D^2$ which is a contradiction to $f|_{partial mathbb D^2}=omega|_{partial mathbb D^2}$.






              • $f$ is not constant on $partial mathbb D^2$: Indeed, we assumed that $df neq 0$ everywhere, so $min f,max f$ are both obtained on $partial mathbb D^2$, and are distinct, since $f$ is not constant on the whole disk.



              Lemma: A closed subset $A subseteq mathbb D^2$ of positive measure contains an accumulation point in $operatorname{Int} mathbb D^2$.



              Proof:



              Assume by contradiction that $A$ does not contain an accumulation point in $operatorname{Int} mathbb D^2$. This implies (by compactness), that for any natural $n$, $A cap B_{1-1/n}$ is finite, where $B_r$ is the closed disk of radius $r$. Thus, $A cap operatorname{Int} mathbb D^2=cup_n A cap B_{1-1/n}$ is countable, hence of measure zero. Thus $A$ has measure zero, contradiction.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                After speaking with a colleague, here is a possible solution:



                Sine $omega$ is harmonic, it is real-analytic. In particular its partial derivatives $omega_x,omega_y:mathbb D^2 to mathbb R$ are real-analytic. Define $A={ pin mathbb D^2 , | , omega_x(p)=omega_y(p)=0}$. We shall prove $A$ has measure zero. Indeed, $A$ is closed. So, if it had positive measure, then it would contain an accumulation point in $operatorname{Int} mathbb D^2$ (see the lemma below). By the identity theorem, this would imply $omega_x=omega_y=0$ on $operatorname{Int} mathbb D^2$, which would imply $omega$ is constant on $mathbb D^2$. However, $f$ cannot be constant* on $partial mathbb D^2$ which is a contradiction to $f|_{partial mathbb D^2}=omega|_{partial mathbb D^2}$.






                • $f$ is not constant on $partial mathbb D^2$: Indeed, we assumed that $df neq 0$ everywhere, so $min f,max f$ are both obtained on $partial mathbb D^2$, and are distinct, since $f$ is not constant on the whole disk.



                Lemma: A closed subset $A subseteq mathbb D^2$ of positive measure contains an accumulation point in $operatorname{Int} mathbb D^2$.



                Proof:



                Assume by contradiction that $A$ does not contain an accumulation point in $operatorname{Int} mathbb D^2$. This implies (by compactness), that for any natural $n$, $A cap B_{1-1/n}$ is finite, where $B_r$ is the closed disk of radius $r$. Thus, $A cap operatorname{Int} mathbb D^2=cup_n A cap B_{1-1/n}$ is countable, hence of measure zero. Thus $A$ has measure zero, contradiction.






                share|cite|improve this answer









                $endgroup$



                After speaking with a colleague, here is a possible solution:



                Sine $omega$ is harmonic, it is real-analytic. In particular its partial derivatives $omega_x,omega_y:mathbb D^2 to mathbb R$ are real-analytic. Define $A={ pin mathbb D^2 , | , omega_x(p)=omega_y(p)=0}$. We shall prove $A$ has measure zero. Indeed, $A$ is closed. So, if it had positive measure, then it would contain an accumulation point in $operatorname{Int} mathbb D^2$ (see the lemma below). By the identity theorem, this would imply $omega_x=omega_y=0$ on $operatorname{Int} mathbb D^2$, which would imply $omega$ is constant on $mathbb D^2$. However, $f$ cannot be constant* on $partial mathbb D^2$ which is a contradiction to $f|_{partial mathbb D^2}=omega|_{partial mathbb D^2}$.






                • $f$ is not constant on $partial mathbb D^2$: Indeed, we assumed that $df neq 0$ everywhere, so $min f,max f$ are both obtained on $partial mathbb D^2$, and are distinct, since $f$ is not constant on the whole disk.



                Lemma: A closed subset $A subseteq mathbb D^2$ of positive measure contains an accumulation point in $operatorname{Int} mathbb D^2$.



                Proof:



                Assume by contradiction that $A$ does not contain an accumulation point in $operatorname{Int} mathbb D^2$. This implies (by compactness), that for any natural $n$, $A cap B_{1-1/n}$ is finite, where $B_r$ is the closed disk of radius $r$. Thus, $A cap operatorname{Int} mathbb D^2=cup_n A cap B_{1-1/n}$ is countable, hence of measure zero. Thus $A$ has measure zero, contradiction.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 7 at 14:36









                Asaf ShacharAsaf Shachar

                5,79931145




                5,79931145






























                    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%2f3064993%2fare-the-singular-points-of-harmonic-function-on-the-disk-always-of-measure-zero%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

                    Index of /

                    This page is only for reference, If you need detailed information, please check here
                    Read more

                    Tribalistas

                    Image
                      Nota: Para o álbum homônimo, veja Tribalistas (álbum de 2002). Para o segundo álbum homônimo, veja Tribalistas (álbum de 2017). Para o tipo de organização tribal, veja Tribalismo. Tribalistas Informação geral Origem Salvador e Rio de Janeiro País   Brasil Gênero(s) MPB Pop Samba Período em atividade 2002–2004 2017–atualmente Gravadora(s) EMI Phonomotor Records Universal Music Afiliação(ões) Dadi Pretinho da Serrinha Margareth Menezes Integrantes Arnaldo Antunes Carlinhos Brown Marisa Monte Página oficial www2.uol.com.br/tribalistas/ ...
                    Read more

                    Filisteus

                    Image
                    Esta página ou secção cita fontes confiáveis e independentes, mas que não cobrem todo o conteúdo, o que compromete a verificabilidade (desde Outubro de 2011) . Por favor, insira mais referências no texto. Material sem fontes poderá ser removido. — Encontre fontes: Google (notícias, livros e acadêmico) Estados Filisteus ca. 830 a.C. Pentápolis filisteia (em vermelho) Os filisteus (do פְּלְשְׁתִּים; transl.: plishtim ) foram um povo que ocupou a costa sudoeste de Canaã, com seu território nomeado como Filístia em contextos posteriores. Teoriza-se que eles se originaram entre os "povos do mar" [ 1 ] . A arqueologia moderna também sugeriu os primeiros laços culturais com o mundo micênico na Grécia. Apesar de eles terem adotado a língua e a cultura local cananeia antes de deixar quaisquer textos escritos (e posteriormente adotarem a língua Aramaica), uma origem indo-europeia tem sido sugerida para um grande número de palavras filisteias conhecidas que so...
                    Read more