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Take $f:[0,1]\to [0,1]^n$ a continuous tour around $[0,1]^n,$ say, some iteration of a Hilbert curve. For $\varepsilon \in (0,1)$ what is the following thing called and are there any nontrivial upper bounds?

\begin{equation} \max_{|a-b|<\varepsilon} \|f(a)-f(b)\|. \end{equation}

Or if not a maximum, then the typical value for such $a,b$.

It seems that most research focuses on the opposite, more impressive direction. That is, for $\varepsilon>0$ characterizing how often $p,q\in [0,1]^n$ have their closest tour points $f(a)\approx p,f(b)\approx q$ such that $|a-b|<\varepsilon.$

For a k-th approximation to a Hilbert curve over $[0,1]^n$ is it true that for any $\varepsilon$-length interval roughly traverses not much more than a cube of ($\mathbb{R}^n$-)volume $\varepsilon$?

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The Peano curve $f:[0,1]\to [0,1]^2$ is Holder continuous with exponent $1/2$ and one can have an $n$-dimensional analogue $f:[0,1]\to [0,1]^n$ which is Holder continuous with exponent $1/n$. That is $|f(a)-f(b)|\leq C|a-b|^{1/n}$ so you get the estimate $$ \max_{|a-b|<\varepsilon} |f(a)-f(b)|\leq C\varepsilon^{1/n}. $$ This `thing' is known as the modulus of continuity.

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  • $\begingroup$ Thanks. Do you know anywhere these things are constructed and proven? $\endgroup$ – enthdegree Apr 20 '18 at 3:33
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    $\begingroup$ I do not know the reference on top of my head. Just Google: Peano curve Holder exponent. $\endgroup$ – Piotr Hajlasz Apr 20 '18 at 3:43
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    $\begingroup$ And let's add that $1/n$ is the best exponent: any $\alpha$-Hölder curve $[0,1]\to[0,1]^n$ with $\alpha>1/n$ can't be surjective. $\endgroup$ – Pietro Majer Jul 9 at 18:52

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