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
$begingroup$
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.
geometric-measure-theory hausdorff-dimension hausdorff-measure
$endgroup$
add a comment |
$begingroup$
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.
geometric-measure-theory hausdorff-dimension hausdorff-measure
$endgroup$
add a comment |
$begingroup$
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.
geometric-measure-theory hausdorff-dimension hausdorff-measure
$endgroup$
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
geometric-measure-theory hausdorff-dimension hausdorff-measure
edited 13 hours ago
user64494
1,660517
1,660517
asked 14 hours ago
Severin SchravenSeverin Schraven
29619
29619
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $f:mathbbR^mtomathbbR^m-k$.
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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 $f:mathbbR^mtomathbbR^m-k$.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $f:mathbbR^mtomathbbR^m-k$.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $f:mathbbR^mtomathbbR^m-k$.
$endgroup$
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 $f:mathbbR^mtomathbbR^m-k$.
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
add a comment |
$begingroup$
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
add a comment |
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