# "Morphisms" for fractal dimension

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]

• 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. Jul 9, 2022 at 1:27
• @MoisheKohan with question 2, I meant: "can we weaken the requirement to be a lipeomorphism and still having the same HD fractal dimension?" Jul 9, 2022 at 12:26
• 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 $n=1$. Jul 9, 2022 at 13:19
• $f : \mathbb R \to \mathbb R$ 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. Jul 9, 2022 at 15:39

$$f : \mathbb R \to \mathbb R$$ 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.