An isoperimetric-type inequality inside a cube Planned maintenance scheduled April 23, 2019 at...



An isoperimetric-type inequality inside a cube



Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Name for an inequality of isoperimetric typeLevy's isoperimetric inequality for sphereStronger version of the isoperimetric inequalityIsoperimetric-like inequality for non-connected setsHypercube isoperimetric inequality for non-increasing eventsPeculiar vertex-isoperimetric inequality on the discrete torus (and generalization)Isoperimetric inequality via Crofton's formulaAn isoperimetric type of inequality in terms of Wasserstein distance/Optimal transportA cube is placed inside another cubeA question of Ahlswede and Katona: known lower bounds on $beta(d,n)$?












6












$begingroup$


I am looking for a reference for the following inequality: if $Omega subset [0,1]^d$ satisfies $mbox{vol}(Omega) leq 1/2$, then
$$ mathcal{H}^{d-1}left( partialOmega cap (0,1)^dright) geq c_d mbox{vol}(Omega)^{frac{d-1}{d}},$$
where $mathcal{H}^{d-1}$ is the $(d-1)-$dimensional Hausdorff measure and $c_d > 0$ is a universal constant depending only on $d$.



This is a variation of the classical isoperimetric inequality with the interesting addition that surface 'on the boundary of the cube' does not count. This seems like it should be known. A discrete version of this inequality (for subsets of the grid graph) was proven by Bollobas and Leader (Edge-isoperimetric inequalities in the grid, Combinatorica 1991) and it seems there is a wealth of information for the discrete case. Has anybody seen the continuous case stated somewhere?










share|cite|improve this question









New contributor




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







$endgroup$

















    6












    $begingroup$


    I am looking for a reference for the following inequality: if $Omega subset [0,1]^d$ satisfies $mbox{vol}(Omega) leq 1/2$, then
    $$ mathcal{H}^{d-1}left( partialOmega cap (0,1)^dright) geq c_d mbox{vol}(Omega)^{frac{d-1}{d}},$$
    where $mathcal{H}^{d-1}$ is the $(d-1)-$dimensional Hausdorff measure and $c_d > 0$ is a universal constant depending only on $d$.



    This is a variation of the classical isoperimetric inequality with the interesting addition that surface 'on the boundary of the cube' does not count. This seems like it should be known. A discrete version of this inequality (for subsets of the grid graph) was proven by Bollobas and Leader (Edge-isoperimetric inequalities in the grid, Combinatorica 1991) and it seems there is a wealth of information for the discrete case. Has anybody seen the continuous case stated somewhere?










    share|cite|improve this question









    New contributor




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







    $endgroup$















      6












      6








      6


      1



      $begingroup$


      I am looking for a reference for the following inequality: if $Omega subset [0,1]^d$ satisfies $mbox{vol}(Omega) leq 1/2$, then
      $$ mathcal{H}^{d-1}left( partialOmega cap (0,1)^dright) geq c_d mbox{vol}(Omega)^{frac{d-1}{d}},$$
      where $mathcal{H}^{d-1}$ is the $(d-1)-$dimensional Hausdorff measure and $c_d > 0$ is a universal constant depending only on $d$.



      This is a variation of the classical isoperimetric inequality with the interesting addition that surface 'on the boundary of the cube' does not count. This seems like it should be known. A discrete version of this inequality (for subsets of the grid graph) was proven by Bollobas and Leader (Edge-isoperimetric inequalities in the grid, Combinatorica 1991) and it seems there is a wealth of information for the discrete case. Has anybody seen the continuous case stated somewhere?










      share|cite|improve this question









      New contributor




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







      $endgroup$




      I am looking for a reference for the following inequality: if $Omega subset [0,1]^d$ satisfies $mbox{vol}(Omega) leq 1/2$, then
      $$ mathcal{H}^{d-1}left( partialOmega cap (0,1)^dright) geq c_d mbox{vol}(Omega)^{frac{d-1}{d}},$$
      where $mathcal{H}^{d-1}$ is the $(d-1)-$dimensional Hausdorff measure and $c_d > 0$ is a universal constant depending only on $d$.



      This is a variation of the classical isoperimetric inequality with the interesting addition that surface 'on the boundary of the cube' does not count. This seems like it should be known. A discrete version of this inequality (for subsets of the grid graph) was proven by Bollobas and Leader (Edge-isoperimetric inequalities in the grid, Combinatorica 1991) and it seems there is a wealth of information for the discrete case. Has anybody seen the continuous case stated somewhere?







      reference-request mg.metric-geometry geometric-measure-theory isoperimetric-problems






      share|cite|improve this question









      New contributor




      Stefan Steinerberger 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 question









      New contributor




      Stefan Steinerberger 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 question




      share|cite|improve this question








      edited 5 hours ago







      Stefan Steinerberger













      New contributor




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









      asked 7 hours ago









      Stefan SteinerbergerStefan Steinerberger

      333




      333




      New contributor




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





      New contributor





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






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






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          This result is known as the relative isoperimetric inequality, see e.g. Functions of Bounded Variation by L. Ambrosio, N. Fusco and D. Pallara (2000), Eq. (3.43).



          It follows from Poincare inequality (see e.g. Eq. (3.41) in the cited book) applied to $chi_Omega$ (the indicator of the set $Omega$). Indeed, by Poincare inequality it holds
          $|chi_Omega - mbox{vol}(Omega)|_{L^p((0,1)^d)} le C |Dchi_Omega|((0,1)^d)$, where $p=frac{d}{d-1}$. Here $|Dchi_Omega|((0,1)^d)=mathcal{H}^{d-1}(partial Omega cap (0,1^d))$ if $partial Omega$ is sufficiently smooth.
          And
          $$
          |chi_Omega - mbox{vol}(Omega)|_p = bigl((1 - mbox{vol}(Omega))^p mbox{vol}(Omega) + mbox{vol}(Omega)^p (1 - mbox{vol}(Omega))bigr)^{1/p} ge frac{1}{2} mbox{vol}(Omega)^{1/p}
          $$

          since $mbox{vol}(Omega) le frac{1}{2}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the reference!
            $endgroup$
            – Stefan Steinerberger
            3 hours ago












          Your Answer








          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "504"
          };
          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
          });


          }
          });






          Stefan Steinerberger is a new contributor. Be nice, and check out our Code of Conduct.










          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f328607%2fan-isoperimetric-type-inequality-inside-a-cube%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









          4












          $begingroup$

          This result is known as the relative isoperimetric inequality, see e.g. Functions of Bounded Variation by L. Ambrosio, N. Fusco and D. Pallara (2000), Eq. (3.43).



          It follows from Poincare inequality (see e.g. Eq. (3.41) in the cited book) applied to $chi_Omega$ (the indicator of the set $Omega$). Indeed, by Poincare inequality it holds
          $|chi_Omega - mbox{vol}(Omega)|_{L^p((0,1)^d)} le C |Dchi_Omega|((0,1)^d)$, where $p=frac{d}{d-1}$. Here $|Dchi_Omega|((0,1)^d)=mathcal{H}^{d-1}(partial Omega cap (0,1^d))$ if $partial Omega$ is sufficiently smooth.
          And
          $$
          |chi_Omega - mbox{vol}(Omega)|_p = bigl((1 - mbox{vol}(Omega))^p mbox{vol}(Omega) + mbox{vol}(Omega)^p (1 - mbox{vol}(Omega))bigr)^{1/p} ge frac{1}{2} mbox{vol}(Omega)^{1/p}
          $$

          since $mbox{vol}(Omega) le frac{1}{2}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the reference!
            $endgroup$
            – Stefan Steinerberger
            3 hours ago
















          4












          $begingroup$

          This result is known as the relative isoperimetric inequality, see e.g. Functions of Bounded Variation by L. Ambrosio, N. Fusco and D. Pallara (2000), Eq. (3.43).



          It follows from Poincare inequality (see e.g. Eq. (3.41) in the cited book) applied to $chi_Omega$ (the indicator of the set $Omega$). Indeed, by Poincare inequality it holds
          $|chi_Omega - mbox{vol}(Omega)|_{L^p((0,1)^d)} le C |Dchi_Omega|((0,1)^d)$, where $p=frac{d}{d-1}$. Here $|Dchi_Omega|((0,1)^d)=mathcal{H}^{d-1}(partial Omega cap (0,1^d))$ if $partial Omega$ is sufficiently smooth.
          And
          $$
          |chi_Omega - mbox{vol}(Omega)|_p = bigl((1 - mbox{vol}(Omega))^p mbox{vol}(Omega) + mbox{vol}(Omega)^p (1 - mbox{vol}(Omega))bigr)^{1/p} ge frac{1}{2} mbox{vol}(Omega)^{1/p}
          $$

          since $mbox{vol}(Omega) le frac{1}{2}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the reference!
            $endgroup$
            – Stefan Steinerberger
            3 hours ago














          4












          4








          4





          $begingroup$

          This result is known as the relative isoperimetric inequality, see e.g. Functions of Bounded Variation by L. Ambrosio, N. Fusco and D. Pallara (2000), Eq. (3.43).



          It follows from Poincare inequality (see e.g. Eq. (3.41) in the cited book) applied to $chi_Omega$ (the indicator of the set $Omega$). Indeed, by Poincare inequality it holds
          $|chi_Omega - mbox{vol}(Omega)|_{L^p((0,1)^d)} le C |Dchi_Omega|((0,1)^d)$, where $p=frac{d}{d-1}$. Here $|Dchi_Omega|((0,1)^d)=mathcal{H}^{d-1}(partial Omega cap (0,1^d))$ if $partial Omega$ is sufficiently smooth.
          And
          $$
          |chi_Omega - mbox{vol}(Omega)|_p = bigl((1 - mbox{vol}(Omega))^p mbox{vol}(Omega) + mbox{vol}(Omega)^p (1 - mbox{vol}(Omega))bigr)^{1/p} ge frac{1}{2} mbox{vol}(Omega)^{1/p}
          $$

          since $mbox{vol}(Omega) le frac{1}{2}$.






          share|cite|improve this answer









          $endgroup$



          This result is known as the relative isoperimetric inequality, see e.g. Functions of Bounded Variation by L. Ambrosio, N. Fusco and D. Pallara (2000), Eq. (3.43).



          It follows from Poincare inequality (see e.g. Eq. (3.41) in the cited book) applied to $chi_Omega$ (the indicator of the set $Omega$). Indeed, by Poincare inequality it holds
          $|chi_Omega - mbox{vol}(Omega)|_{L^p((0,1)^d)} le C |Dchi_Omega|((0,1)^d)$, where $p=frac{d}{d-1}$. Here $|Dchi_Omega|((0,1)^d)=mathcal{H}^{d-1}(partial Omega cap (0,1^d))$ if $partial Omega$ is sufficiently smooth.
          And
          $$
          |chi_Omega - mbox{vol}(Omega)|_p = bigl((1 - mbox{vol}(Omega))^p mbox{vol}(Omega) + mbox{vol}(Omega)^p (1 - mbox{vol}(Omega))bigr)^{1/p} ge frac{1}{2} mbox{vol}(Omega)^{1/p}
          $$

          since $mbox{vol}(Omega) le frac{1}{2}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 4 hours ago









          SkeeveSkeeve

          985514




          985514












          • $begingroup$
            Thanks for the reference!
            $endgroup$
            – Stefan Steinerberger
            3 hours ago


















          • $begingroup$
            Thanks for the reference!
            $endgroup$
            – Stefan Steinerberger
            3 hours ago
















          $begingroup$
          Thanks for the reference!
          $endgroup$
          – Stefan Steinerberger
          3 hours ago




          $begingroup$
          Thanks for the reference!
          $endgroup$
          – Stefan Steinerberger
          3 hours ago










          Stefan Steinerberger is a new contributor. Be nice, and check out our Code of Conduct.










          draft saved

          draft discarded


















          Stefan Steinerberger is a new contributor. Be nice, and check out our Code of Conduct.













          Stefan Steinerberger is a new contributor. Be nice, and check out our Code of Conduct.












          Stefan Steinerberger is a new contributor. Be nice, and check out our Code of Conduct.
















          Thanks for contributing an answer to MathOverflow!


          • 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%2fmathoverflow.net%2fquestions%2f328607%2fan-isoperimetric-type-inequality-inside-a-cube%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

          VNC viewer RFB protocol error: bad desktop size 0x0I Cannot Type the Key 'd' (lowercase) in VNC Viewer...

          Couldn't open a raw socket. Error: Permission denied (13) (nmap)Is it possible to run networking commands...

          Why not use the yoke to control yaw, as well as pitch and roll? Announcing the arrival of...