I'am trying to construct a space-filling curve in the unit square $I^{2}:=[0,1]\times [01,]$ following the results shown in the Barnsley's book "Fractals Everywhere". Thus, let $\Delta:=\big\{ (0,0),(0,0.5),(0.5,0.5),(1,0.5),(1,0) \big\}$ and define the functions (contractions)
$f _{1}(x,y):=(0.5y,0.5x), \quad f _{2}(x,y):=(0.5x,0.5y+0.5)$,
$f _{3}(x,y):=(0.5x+0.5,0.5y+0.5),\quad f _{4}(x,y):=(-0.5y+1,-0.5x+0.5)$,
for each $x,y\in I$. These functions define a set function $F(K):=\cup_{i=1}^{4}f_{i}(K)$ for each compact (non-void) set $K\subset I^{2}$. Let $g_{0}:I\longrightarrow I^{2}$ be a simple curve such that $g(0)=(0,0),g_{0}(1)=(1,0)$ and $g_{0}$ meet the boundary of $I^{2}$ only in these points. One can to show that each $F^{n}(g_{0}(I))$ (the exponent mean for the composition) is a simple curve that connects the points $(0,0)$ and $(1,0)$, and the "attractor" (i.e. the fixed point) set of the sequence $(F^{n}(g_{0}(I)))_{n\geq 1}$. Then, the attractor set of the above sequence is the graph of a continuous function (and therefore such function is a space-filling curve). Such curve interpolate the above data set $\Delta$.
Then, I think that the above space-filling curve can be constructed, actually, as follows: let, for each $n\geq 1$, $G_{n}$ the set of all permutations with repetitions of $\{1,2,3,4\}$ of length $n$, so $G_{1}=\{[1],[2],[3],[4]\}$, $G_{2}=\{[1,1],[1,2],\ldots, [4,3],[4,4]\}$ and so on. Divide the interval $I$ into $4^{n}$ equal subintervals, and let $\tau_{j}:[\frac{j-1}{4^{n}},\frac{j}{4^{n}}]\longrightarrow I$, for $j=1,\ldots, 4^{n}$, be the mapping $\tau_{j}(t):=4^{n}t-j+1$. Then, define
$g_{n}(t):=(f_{i_{1}}\circ \ldots f_{i_{n}})(g_{0}(\tau_{j}(t))),$
for $t\in[\frac{j-1}{4^{n}},\frac{j}{4^{n}}]$, where $[i_{1},\ldots,i_{n}]\in G_{n}$ is the element in $j$-th position. Each $g_{n}$ is a simple curve that connect the points $(0,0)$ and $(1,0)$. In fact, $g_{n}(I)=F^{n}(I)$ , and because as for each $t\in[\frac{j-1}{4^{n}},\frac{j}{4^{n}}]$, $g_{n}(t)$ lies in a sub-square of $I^{2}$ of side-length $2^{-n}$, $g_{n}$ converges uniformly to the above space-filling curve $g$. More specifically,
$(1) \quad \quad |g(t)-g_{n}(t)|\leq 2^{-n}\sqrt{2}$
for each $t\in I$, this fact follow from the above fact that $g_{n}(t)$ lies in a sub-square of $I^{2}$ of side-length $2^{-n}$ (and so $|g(t)-g_{n}(t)|$ is less than its diameter).
What do you think about this "informal proof"? Is the sequence of functions $g_{n}$ uniformly convergent to the space-filling curve $g$ induce by the iterated function system $f_{1},\ldots f_{4}$, in the sense given in (1)?
Thank you for your time.