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 9 hours ago









      user64494

      1,660517




      1,660517










      asked 10 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
            10 hours ago











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






          • 1




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










          Your Answer





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          1 Answer
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          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
            10 hours ago











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






          • 1




            $begingroup$
            Thanks, that is pretty elegant. Even though I am suprised that the statement is not true :)
            $endgroup$
            – Severin Schraven
            10 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
            10 hours ago











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






          • 1




            $begingroup$
            Thanks, that is pretty elegant. Even though I am suprised that the statement is not true :)
            $endgroup$
            – Severin Schraven
            10 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 10 hours ago

























          answered 10 hours ago









          Piotr HajlaszPiotr Hajlasz

          9,90343974




          9,90343974











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











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






          • 1




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
















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











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






          • 1




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















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





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













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




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




          1




          1




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




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

















          draft saved

          draft discarded
















































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