# The radius of an interval's image through a space-filling curve

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?

$$$$\max_{|a-b|<\varepsilon} \|f(a)-f(b)\|.$$$$

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 $$\varepsilon$$?

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.
• 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. Commented Jul 9, 2019 at 18:52