$1$-Lipschitz $f:[0,1]^k\to[0,1]^n$ With Maximally Dense Image

Question

Is there anything known about which $1$-Lipschitz functions $$f:(X:=[0,1]^m)\to(Y:=[0,1]^n)$$ for $m$ < $n$ fill the codomain maximally dense, i.e. I want to minimize $\sup_{y\in Y}d(y, f(X))$ where $d(a, B)$ is defined as $\inf_{b\in B}d(a, b)$ and we use Euclidean distance for $d$. What about $K$-Lipschitz functions for $K\neq 1$ (or equivalently $X:=[0,K]^m$)?

If this is too hard to answer, what about $f$ being of the form $$y_i=\sum_j\sin^*(a_j x_j+b_j)$$ where $\sin^*(x):=(\sin(x)+1)/2$?

Answer 1

Is there anything known about which $1$-Lipschitz functions $$f:(X:=[0,1]^m)\to(Y:=[0,1]^n)$$ for $m$ < $n$ fill the codomain maximally dense, i.e. I want to minimize $\sup_{y\in Y}d(y, f(X))$ where $d(a, B)$ is defined as $\inf_{b\in B}d(a, b)$ and we use Euclidean distance for $d$. What about $K$-Lipschitz functions for $K\neq 1$ (or equivalently $X:=[0,K]^m$)?

I know a good lower bound $n=2$ and $m=1$, which is as follows:

• For $K\leq\frac12$, consider the function $f\colon [0,K]\to[0,1]^2$ given by $f(t)=\left(\frac{1-K}{2}+t,\frac12\right)$. It has a maximal distance $\frac12\sqrt{2-2K+K^2}$, which I believe is maximal for this range of $K$.
• For $\frac12\le K\le\frac32$, consider the function given by $f(t)=\begin{cases}\left(\frac14,\frac{5-2K}8+t\right)&t\le\frac{2K-1}4,\\\left(\frac{1-K}2+t,\frac{2K-1}{8}\right)&\frac{2K-1}4\le t\le\frac{2K+3}4,\\\left(\frac34,t-\frac{2K+7}8\right)&\frac{2K+3}4\le t\le K.\end{cases}$
• For $\frac32\le K\le\frac{15}4$, set $a=\frac{6-K}9$ and $b=\frac{2K-3}{18}$ (such that $a+b=\frac12$ and $K=12a+3b$) and consider the function given by $$f(t)=\begin{cases}\left(\frac14-a+t,\frac14-a\right)&0\le t\le a\\\left(\frac14+a,\frac14-a+(t-a)\right)&a\le t\le 2a\\\left(\frac14+a-(t-2a),\frac14+a\right)&2a\le t\le 3a\\\left(\frac14-a,\frac14+a+(t-3a)\right)&3a\le t\le4a+b\\\left(\frac14-a+(t-4a-b),\frac34+a\right)&4a+b\le t\le5a+b\\\left(\frac14+a,\frac34+a-(t-5a-b)\right)&5a+b\le t\le 6a+b\\\left(\frac14+a+(t-6a-b),\frac34-a\right)&6a+b\le t\le 6a+2b\\\left(\frac34-a,\frac34-a+(t-6a-2b)\right)&6a+2b\le t\le 7a+2b\\\left(\frac34-a+(t-7a-2b),\frac34+a\right)&7a+2b\le t\le8a+2b\\\left(\frac34+a,\frac34+a-(t-8a-2b)\right)&8a+2b\le t\le 9a+3b\\\left(\frac34+a-(t-9a-3b),\frac14+a\right)&9a+3b\le t\le 10a+3b\\\left(\frac34-a,\frac14+a-(t-10a-3b)\right)&10a+3b\le t\le 11a+3b\\\left(\frac34-a+(t-11a-3b),\frac14-a\right)&11a+3b\le t\le K.\end{cases}$$ Note that for $K=\frac32$ this is the first iteration of the Hilbert curve, and for $K=\frac{15}4$ this is the second iteration. It has minimal distance $(\frac14-\frac a2)\sqrt2$. You can see in a drawing that this is not the precise minimum, because e.g. the distance to $(\frac12,\frac12)$ or $(0,\frac12)$ is strictly smaller than the minimal distance, so some length can be taken away to make distances smaller.
• For larger $K$, interpolate the Hilbert curve in a similar fashion (see also this image).

For different values of $m$, while keeping $n=1$ you should end up with space filling curves of $m$-dimensional space when $K\to\infty$. For $K=1$ I believe the first-iteration Hilbert curve is still optimal though.

When changing $n$ and keeping $K=1$, I believe your best bet is taking a solution for a 1-dimensional domain and just extending it in the other directions. When $K$ becomes larger, you could change the domain by finding a map $[0,K]^m\hookrightarrow[0,1]^{m-1}\times[0,K']$ with $K'\ge K$ and then still doing the same.

If this is too hard to answer, what about $f$ being of the form $$y_i=\sum_j\sin^*(a_j x_j+b_j)$$ where $\sin^*(x):=(\sin(x)+1)/2$?

I don't think this makes things simpler. Proving something for the current case is hard enough.