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Let $A\subset \mathbb{R}^n$ a set with Hausdorff–Besicovitch fractal dimension $d \le n$, then be a continuous function $f:A\to \mathbb{R}^m$ with $m\ge n$.

  1. Under what conditions the HB fractal dimension is preserved? It seems that if $f$ is Lipschitz continuous with inverse $f^{-1}$also being Lipschitz continuos, then both $A\subset \mathbb{R}^n$ and $f(A)\subset \mathbb{R}^m$ have the same HB fractal dimension (equivalently $f$ is a bi-Lipshchtiz continuous or Lipeomorphism).

  2. Suppose that $f:\mathbb{R}^n\to \mathbb{R}^n$ is a homeomorphism such that for each subset $A\subset\mathbb{R}^n$, $\dim(A)=\dim(f(A))$. Does it follow that $f$ is a Lipeomorphism? [edited in response to a comment]

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    $\begingroup$ Such maps are usually called "bilipschitz" (or "lipeomorphisms") and it is an easy exercise to see that they preserve Hausdorff dimension. Your question 2 is very unclear, I truly do not understand what are you asking in this item. As for 3, then no, it would not make sense. $\endgroup$ Jul 9, 2022 at 1:27
  • $\begingroup$ @MoisheKohan with question 2, I meant: "can we weaken the requirement to be a lipeomorphism and still having the same HD fractal dimension?" $\endgroup$
    – Davius
    Jul 9, 2022 at 12:26
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    $\begingroup$ I think, a meaningful form of (2) can be: Suppose that $f: R^n\to R^n$ is a homeomorphism such that for each subset $A\subset R^n$, $dim(A)=dim(f(A))$. Does it follow that $f$ is a Lipeomorphism? I do not know an answer even if $n=1$. $\endgroup$ Jul 9, 2022 at 13:19
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    $\begingroup$ $f : \mathbb R \to \mathbb R$ defined by $f(x) = x^3$ and its inverse both preserve H-B dimension, but neither $f$ nor its inverse $x^{1/3}$ are Lipschitz. $\endgroup$
    – GEdgar
    Jul 9, 2022 at 15:39

1 Answer 1

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$f : \mathbb R \to \mathbb R$ defined by $f(x) = x^3$ and its inverse both preserve H-B dimension, but neither $f$ nor its inverse $x^{1/3}$ are Lipschitz.

A known condition on a map $f$ so that $\dim f(A) \le \dim A$ for all subsets $A \subseteq \mathbb R^n$ is: There exist sets $T_n$ so that $\mathbb R^n = \bigcup_{n=1}^\infty T_n$ and $f$ is Lipschitz on $T_n$ for all $n$. Both $f$ and $f^{-1}$ have this property in the example above.

This can also be weakened. For example, things like $$ |f(x)-f(y)| \le C|x-y|(-\log|x-y|)\quad\text{when }|x-y|<\delta $$ replacing Lipschitz.

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