Hausdorff dimension of the boundary of fibres of Lipschitz mapsHausdorff dimension vs. cardinalityHausdorff measure of the zero setLipschitz boundary vs rectifiable curve boundaryCan Hausdorff dimension make sets into a Tropical Semiring?Hausdorff dimension of R x XThe relation between Hausdorff dimension of an $n$-manifold and $n$Hausdorff dimension of a Cantor-like setControlling the size of the balls in Hausdorff dimension/measureCompact sets of Hausdorff dimension zeroabout the Hausdorff dimension of Removable singularities of PDE

Hausdorff dimension of the boundary of fibres of Lipschitz maps


Hausdorff dimension vs. cardinalityHausdorff measure of the zero setLipschitz boundary vs rectifiable curve boundaryCan Hausdorff dimension make sets into a Tropical Semiring?Hausdorff dimension of R x XThe relation between Hausdorff dimension of an $n$-manifold and $n$Hausdorff dimension of a Cantor-like setControlling the size of the balls in Hausdorff dimension/measureCompact sets of Hausdorff dimension zeroabout the Hausdorff dimension of Removable singularities of PDE













6












$begingroup$


Let $f: mathbbR^mrightarrow mathbbR^m-k$ be a Lipschitz map.




Can we get a uniform estimate on the Hausdorff dimension of the boundaries of fibres of $f$? I.e. do we have an upper bound for
$$ sup_yin mathbbR^n-k dim_H(partial f^-1(y)) ?$$




Theorem 2.5 in [1] tells us, that for almost every $yin mathbbR^n-k$ we have that $dim_H(f^-1(y))leq k$. This tells us
$$ textessup_yin mathbbR^n-k dim_H(partial f^-1(y)) leq k.$$
Can we pass to the supremum? And are there even better bounds? I mean, I used $partial f^-1(y)subseteq f^-1(y)$ as $f$ is continuous and the monotonicity of the Hausdorff dimension, but I guess that one can do better than this.



[1] G. Alberti, S. Bianchini, G. Crippa, Structure of level sets and Sard-type properties of Lipschitz maps: results and counterexamples.
Ann. Sc. Norm. Super. Pisa Cl. Sci. (5) 12 (2013), no. 4, 863–902.










share|cite|improve this question











$endgroup$
















    6












    $begingroup$


    Let $f: mathbbR^mrightarrow mathbbR^m-k$ be a Lipschitz map.




    Can we get a uniform estimate on the Hausdorff dimension of the boundaries of fibres of $f$? I.e. do we have an upper bound for
    $$ sup_yin mathbbR^n-k dim_H(partial f^-1(y)) ?$$




    Theorem 2.5 in [1] tells us, that for almost every $yin mathbbR^n-k$ we have that $dim_H(f^-1(y))leq k$. This tells us
    $$ textessup_yin mathbbR^n-k dim_H(partial f^-1(y)) leq k.$$
    Can we pass to the supremum? And are there even better bounds? I mean, I used $partial f^-1(y)subseteq f^-1(y)$ as $f$ is continuous and the monotonicity of the Hausdorff dimension, but I guess that one can do better than this.



    [1] G. Alberti, S. Bianchini, G. Crippa, Structure of level sets and Sard-type properties of Lipschitz maps: results and counterexamples.
    Ann. Sc. Norm. Super. Pisa Cl. Sci. (5) 12 (2013), no. 4, 863–902.










    share|cite|improve this question











    $endgroup$














      6












      6








      6


      1



      $begingroup$


      Let $f: mathbbR^mrightarrow mathbbR^m-k$ be a Lipschitz map.




      Can we get a uniform estimate on the Hausdorff dimension of the boundaries of fibres of $f$? I.e. do we have an upper bound for
      $$ sup_yin mathbbR^n-k dim_H(partial f^-1(y)) ?$$




      Theorem 2.5 in [1] tells us, that for almost every $yin mathbbR^n-k$ we have that $dim_H(f^-1(y))leq k$. This tells us
      $$ textessup_yin mathbbR^n-k dim_H(partial f^-1(y)) leq k.$$
      Can we pass to the supremum? And are there even better bounds? I mean, I used $partial f^-1(y)subseteq f^-1(y)$ as $f$ is continuous and the monotonicity of the Hausdorff dimension, but I guess that one can do better than this.



      [1] G. Alberti, S. Bianchini, G. Crippa, Structure of level sets and Sard-type properties of Lipschitz maps: results and counterexamples.
      Ann. Sc. Norm. Super. Pisa Cl. Sci. (5) 12 (2013), no. 4, 863–902.










      share|cite|improve this question











      $endgroup$




      Let $f: mathbbR^mrightarrow mathbbR^m-k$ be a Lipschitz map.




      Can we get a uniform estimate on the Hausdorff dimension of the boundaries of fibres of $f$? I.e. do we have an upper bound for
      $$ sup_yin mathbbR^n-k dim_H(partial f^-1(y)) ?$$




      Theorem 2.5 in [1] tells us, that for almost every $yin mathbbR^n-k$ we have that $dim_H(f^-1(y))leq k$. This tells us
      $$ textessup_yin mathbbR^n-k dim_H(partial f^-1(y)) leq k.$$
      Can we pass to the supremum? And are there even better bounds? I mean, I used $partial f^-1(y)subseteq f^-1(y)$ as $f$ is continuous and the monotonicity of the Hausdorff dimension, but I guess that one can do better than this.



      [1] G. Alberti, S. Bianchini, G. Crippa, Structure of level sets and Sard-type properties of Lipschitz maps: results and counterexamples.
      Ann. Sc. Norm. Super. Pisa Cl. Sci. (5) 12 (2013), no. 4, 863–902.







      geometric-measure-theory hausdorff-dimension hausdorff-measure






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 13 hours ago









      user64494

      1,660517




      1,660517










      asked 14 hours ago









      Severin SchravenSeverin Schraven

      29619




      29619




















          1 Answer
          1






          active

          oldest

          votes


















          7












          $begingroup$

          Unfortunately, you can always find a Lipschitz map
          $$
          f:mathbbR^mtomathbbR^m-k
          quad
          textand
          quad
          yinmathbbR^m-k
          $$

          such that $partial f^-1(y)$ has positive $m$-dimensional measure so
          $dim_H partial f^-1(y)=m$.



          Here is an example. Let $KsubsetmathbbR^m$ be a Cantor set (i.e. a set homeomorphic to the ternary Cantor set) of positive $m$-dimensional measure. Existence of such a set $K$ is standard. Let $f(x)=operatornamedist(x,K)$. Then $f:mathbbR^mtomathbbR$ is $1$-Lipschitz and it vanishes precisely on $K$. That is $f^-1(0)=K=partial K$ (the boundary of a Cantor set is the Cantor set itself) has positive $m$-dimensional measure. Now, assuming that $mathbbRsubsetmathbbR^m-k$ we can regard $f$ as a mapping $f:mathbbR^mtomathbbR^m-k$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Is it difficult to show the existence of such a map?
            $endgroup$
            – Severin Schraven
            14 hours ago











          • $begingroup$
            @SeverinSchraven I added details for the construction.
            $endgroup$
            – Piotr Hajlasz
            14 hours ago






          • 1




            $begingroup$
            Thanks, that is pretty elegant. Even though I am suprised that the statement is not true :)
            $endgroup$
            – Severin Schraven
            14 hours ago










          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: "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
          );



          );













          draft saved

          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f325624%2fhausdorff-dimension-of-the-boundary-of-fibres-of-lipschitz-maps%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









          7












          $begingroup$

          Unfortunately, you can always find a Lipschitz map
          $$
          f:mathbbR^mtomathbbR^m-k
          quad
          textand
          quad
          yinmathbbR^m-k
          $$

          such that $partial f^-1(y)$ has positive $m$-dimensional measure so
          $dim_H partial f^-1(y)=m$.



          Here is an example. Let $KsubsetmathbbR^m$ be a Cantor set (i.e. a set homeomorphic to the ternary Cantor set) of positive $m$-dimensional measure. Existence of such a set $K$ is standard. Let $f(x)=operatornamedist(x,K)$. Then $f:mathbbR^mtomathbbR$ is $1$-Lipschitz and it vanishes precisely on $K$. That is $f^-1(0)=K=partial K$ (the boundary of a Cantor set is the Cantor set itself) has positive $m$-dimensional measure. Now, assuming that $mathbbRsubsetmathbbR^m-k$ we can regard $f$ as a mapping $f:mathbbR^mtomathbbR^m-k$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Is it difficult to show the existence of such a map?
            $endgroup$
            – Severin Schraven
            14 hours ago











          • $begingroup$
            @SeverinSchraven I added details for the construction.
            $endgroup$
            – Piotr Hajlasz
            14 hours ago






          • 1




            $begingroup$
            Thanks, that is pretty elegant. Even though I am suprised that the statement is not true :)
            $endgroup$
            – Severin Schraven
            14 hours ago















          7












          $begingroup$

          Unfortunately, you can always find a Lipschitz map
          $$
          f:mathbbR^mtomathbbR^m-k
          quad
          textand
          quad
          yinmathbbR^m-k
          $$

          such that $partial f^-1(y)$ has positive $m$-dimensional measure so
          $dim_H partial f^-1(y)=m$.



          Here is an example. Let $KsubsetmathbbR^m$ be a Cantor set (i.e. a set homeomorphic to the ternary Cantor set) of positive $m$-dimensional measure. Existence of such a set $K$ is standard. Let $f(x)=operatornamedist(x,K)$. Then $f:mathbbR^mtomathbbR$ is $1$-Lipschitz and it vanishes precisely on $K$. That is $f^-1(0)=K=partial K$ (the boundary of a Cantor set is the Cantor set itself) has positive $m$-dimensional measure. Now, assuming that $mathbbRsubsetmathbbR^m-k$ we can regard $f$ as a mapping $f:mathbbR^mtomathbbR^m-k$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Is it difficult to show the existence of such a map?
            $endgroup$
            – Severin Schraven
            14 hours ago











          • $begingroup$
            @SeverinSchraven I added details for the construction.
            $endgroup$
            – Piotr Hajlasz
            14 hours ago






          • 1




            $begingroup$
            Thanks, that is pretty elegant. Even though I am suprised that the statement is not true :)
            $endgroup$
            – Severin Schraven
            14 hours ago













          7












          7








          7





          $begingroup$

          Unfortunately, you can always find a Lipschitz map
          $$
          f:mathbbR^mtomathbbR^m-k
          quad
          textand
          quad
          yinmathbbR^m-k
          $$

          such that $partial f^-1(y)$ has positive $m$-dimensional measure so
          $dim_H partial f^-1(y)=m$.



          Here is an example. Let $KsubsetmathbbR^m$ be a Cantor set (i.e. a set homeomorphic to the ternary Cantor set) of positive $m$-dimensional measure. Existence of such a set $K$ is standard. Let $f(x)=operatornamedist(x,K)$. Then $f:mathbbR^mtomathbbR$ is $1$-Lipschitz and it vanishes precisely on $K$. That is $f^-1(0)=K=partial K$ (the boundary of a Cantor set is the Cantor set itself) has positive $m$-dimensional measure. Now, assuming that $mathbbRsubsetmathbbR^m-k$ we can regard $f$ as a mapping $f:mathbbR^mtomathbbR^m-k$.






          share|cite|improve this answer











          $endgroup$



          Unfortunately, you can always find a Lipschitz map
          $$
          f:mathbbR^mtomathbbR^m-k
          quad
          textand
          quad
          yinmathbbR^m-k
          $$

          such that $partial f^-1(y)$ has positive $m$-dimensional measure so
          $dim_H partial f^-1(y)=m$.



          Here is an example. Let $KsubsetmathbbR^m$ be a Cantor set (i.e. a set homeomorphic to the ternary Cantor set) of positive $m$-dimensional measure. Existence of such a set $K$ is standard. Let $f(x)=operatornamedist(x,K)$. Then $f:mathbbR^mtomathbbR$ is $1$-Lipschitz and it vanishes precisely on $K$. That is $f^-1(0)=K=partial K$ (the boundary of a Cantor set is the Cantor set itself) has positive $m$-dimensional measure. Now, assuming that $mathbbRsubsetmathbbR^m-k$ we can regard $f$ as a mapping $f:mathbbR^mtomathbbR^m-k$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 14 hours ago

























          answered 14 hours ago









          Piotr HajlaszPiotr Hajlasz

          9,90343974




          9,90343974











          • $begingroup$
            Is it difficult to show the existence of such a map?
            $endgroup$
            – Severin Schraven
            14 hours ago











          • $begingroup$
            @SeverinSchraven I added details for the construction.
            $endgroup$
            – Piotr Hajlasz
            14 hours ago






          • 1




            $begingroup$
            Thanks, that is pretty elegant. Even though I am suprised that the statement is not true :)
            $endgroup$
            – Severin Schraven
            14 hours ago
















          • $begingroup$
            Is it difficult to show the existence of such a map?
            $endgroup$
            – Severin Schraven
            14 hours ago











          • $begingroup$
            @SeverinSchraven I added details for the construction.
            $endgroup$
            – Piotr Hajlasz
            14 hours ago






          • 1




            $begingroup$
            Thanks, that is pretty elegant. Even though I am suprised that the statement is not true :)
            $endgroup$
            – Severin Schraven
            14 hours ago















          $begingroup$
          Is it difficult to show the existence of such a map?
          $endgroup$
          – Severin Schraven
          14 hours ago





          $begingroup$
          Is it difficult to show the existence of such a map?
          $endgroup$
          – Severin Schraven
          14 hours ago













          $begingroup$
          @SeverinSchraven I added details for the construction.
          $endgroup$
          – Piotr Hajlasz
          14 hours ago




          $begingroup$
          @SeverinSchraven I added details for the construction.
          $endgroup$
          – Piotr Hajlasz
          14 hours ago




          1




          1




          $begingroup$
          Thanks, that is pretty elegant. Even though I am suprised that the statement is not true :)
          $endgroup$
          – Severin Schraven
          14 hours ago




          $begingroup$
          Thanks, that is pretty elegant. Even though I am suprised that the statement is not true :)
          $endgroup$
          – Severin Schraven
          14 hours ago

















          draft saved

          draft discarded
















































          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%2f325624%2fhausdorff-dimension-of-the-boundary-of-fibres-of-lipschitz-maps%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

          The Calvary Singular or Plural The 2019 Stack Overflow Developer Survey Results Are InAre collective nouns always plural, or are certain ones singular?Is “audience” singular or plural?“Wasn't” vs. “weren't” in a vernacular sentence“My last couple of years” — singular or plural?Is 'rest' singular or plural?Is “all but one” singular or plural?Whether to use the singular or plural form of basis?Singular and Plural for numbersIs there a plural form of teeth?performance: plural vs singular?singular or plural nouns?Singular and Plural

          How does one intimidate enemies without having the capacity for violence?Ideas for how aliens would approach this fight?How to convey the scale of my humanoid without science or units?How does the “space drive” conserve momentum?Planet Vanishes - How does this affect the orbiting starships?How does a community of a Universe Simulator have the same language as its creator?How would US Presidential elections be affected if voters could choose the state their vote for President was counted in?How Does One Ensures the Immortality of Their ConsciousnessHow do I retain national independence while also having a one world government?How can Ganymede have an Earth-like gravity without us having realized it?How would one make a lion mount for a fantasy world?

          Output visual diagram of pictureASCII-art logic gate diagramBooks on a ShelfDetermine the Dimensions of a Rotated RectangleDraw a Houndstooth PatternDraw and label an ASCII hexagonal gridGolf me an ASCII AlphabetASCII Jigsaw PuzzleOutput a pretty boxASCII-Art Venn DiagramASCII Exact Cover with Rectangles