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Does there exist a continuous map $\gamma:[0,1] \to \mathbb{R}^n$ s.t. $$|\gamma(x)-\gamma(y)| \ge |x-y|^{\frac 1 2} $$ for all $x,y \in [0,1]$?

user64494
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2 Answers2

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Yes, there is such a function. You should be able to construct one explicitly by an appropriate modification of the Koch snowflake construction, although I don't want to work through the details.

In fact, much more is true. Take any metric space $(X,d)$ which satisfies a certain finite-dimensionality condition called "doubling". (This means that, for some constant $N$, every ball of radius $r$ can be covered by at most $N$ balls of radius $r/2$.)

Now take any $0<\alpha <1$. Then there is a constant $C\geq 1$, a positive integer $n$, and a map $f:X\rightarrow \mathbb{R}^n$ which satisfies $$ C^{-1} d(x,y)^\alpha \leq |f(x) - f(y)| \leq Cd(x,y)^\alpha.$$

This is called Assouad's embedding theorem.

What does this have to do with your question? Well, let $(X,d)$ be $[0,1]$ with the usual metric given by $|\cdot|$. This is a doubling metric space, so by Assouad's theorem (with $\alpha=1/2$) we get a map $f:[0,1]\rightarrow\mathbb{R}^n$ satisfying $$ C^{-1} |x-y|^{1/2} \leq |f(x) - f(y)| \leq C|x-y|^{1/2}.$$

If you just rescale $f$ by a factor of $C$, you get $$ |x-y|^{1/2} \leq |f(x) - f(y)| \leq C^2|x-y|^{1/2},$$ which in particular is what you wanted.

GCD
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  • Could you give an accessible reference to the Assouad's theorem? – user64494 Jun 20 '13 at 14:06
  • I would like to note that the Assouad's theorem claims the existence of the dimension $n$. This means that its application does not produce the answer to the question (where $n$ is arbitrary). I am not able to construct the required function, using the Koch flake. – user64494 Jun 20 '13 at 14:13
  • A proof of Assouad's theorem can be found in Heinonen's book Lectures on Analysis on Metric Spaces. – GCD Jun 20 '13 at 14:51
  • It was not really clear from your question whether $n$ was meant to be arbitrary or whether any $n$ sufficed. Assouad's theorem gives such curves for all $n$ sufficiently large. Maybe my remark about the snowflake was a bit ambitious. If you would like such a curve for $n=2$, it must have the property that the image of every sub-arc has positive Lebesgue measure; such curves exist (see http://mathoverflow.net/questions/24264/a-question-about-the-osgood-curve) but I'm not sure whether the construction referenced there satisfies your condition. – GCD Jun 20 '13 at 14:54
  • @ GCD: Could you give a reference to the so called Assouad theorem in the book "Lectures on Analysis on Metric Spaces" by Juha Heinonen , indicating pages and number of this Theorem: Juha Heinonen uses notation like Preposition 3.19. Because in the question under consideration n is not restricted by the existsence quantor, it should be understood as an arbitrary positive integer number. – user64494 Jun 20 '13 at 17:16
  • It is Theorem 12.2 in Heinonen's book. This was merely the reference I know; I'm sure that it can be found in other places, besides Assouad's original paper of course. – GCD Jun 20 '13 at 17:46
  • Unfortunately, Theorem 12.2 does not coincide with the statement in your answer. – user64494 Jun 20 '13 at 18:03
  • Yes, Theorem 12.2 in your link to Heinonen's book is exactly what I have said, just in different words. Try reading the text for a discussion--I don't plan on writing a full exposition here. I hope you find a satisfactory answer to your question, whether using my thoughts above or otherwise. – GCD Jun 20 '13 at 21:59
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While Heinonen's exposition of Assouad's theorem is excellent, one should not forget the original 1983 paper by Assouad, Plongements lipschitziens dans $\mathbb R^n$. Proposition 4.4 answers your question in a more general form. I restate it below, changing terminology to make it self-contained:

Let $k\ge 1$ be an integer, and $p\in (0,1)$. Suppose $n>1/p$ is an integer. Then there exists a map $f:[0,1]^k\to \mathbb R^n$ such that $$c|x-y|^p \le |f(x)-f(y)|\le C|x-y|^p \tag1$$ for all $x,y\in [0,1]^k$, with some positive constants $c$ and $C$.

You can take $k=1$, $p=1/2$, and $n=3$. Then multiply $f$ by a constant to get $c=1$.

As GCD hinted, the construction by Assouad is an $n$-dimensional form of von Koch snowflake.

It is known that $f$ as in (1) does not exist for $n=1/p$ (in particular, when $n=2$ and $p=1/2$). Indeed, (1) implies that the range of $f$ must be a porous set and therefore its $n$-dimensional measure must be zero; when $n=1/p$ the latter is incompatible with the lower bound in (1). See Porous sets and quasisymmetric maps by Väisälä.

The above paragraph leaves open the question of whether you can take $n=2$ if you only want the lower bound in (1) (plus continuity of $f$). I do not know the answer, but suspect it is negative.

One more reference: Geometric embeddings of metric spaces by Heinonen, specifically Chapter 3.

40 votes
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  • For $n = 2$, the answer is indeed negative (https://projecteuclid.org/download/pdf_1/euclid.acta/1485889215) – charmd Jun 21 '18 at 08:01