Space-filling curves and multiple integrals

Question

Dears,

It is a well known fact that a multiple integral can be “replaced” by a single variable one using the so called “space-filling curves”. Indeed, if $g$ is a measure preserving space-filling curve, such as the Hilbert or Peano curve, and $f:[0,1]^{2} \longrightarrow \mathbb{R}$ is a continuous function, then

$\int_{[0,1]^{2}}f(u,v)dudv=\int_{0}^{1} f(g(t))dt$

A skecth of the proof (see, for instance, http://www.math.bas.bg/smb/2004_2007_PK/2007/pdf/321-325.pdf) is the following: as

$\int_{[0,1]^{2}}f(u,v)dudv=\lim_{n}\frac{1}{4^{n}}\sum_{i=1}^{2^{n}-1}\sum_{j=1}^{2^{n}-1}f(\frac{i}{2^{n}},\frac{j}{2^{n}})$

and, if $g$ is the Hilbert space-filling curve, then for each $p$ there is a pair $i,j$ such that

$g(\frac{p}{2^{n}})=(\frac{i}{2^{n}},\frac{j}{2^{n}})$,

we find

$\int_{[0,1]^{2}}f(u,v)dudv=\lim_{n}\frac{1}{4^{n}}\sum_{i=1}^{4^{n}-1}f(g(\frac{p}{2^{n}}))$,

and the result holds. However, I have not found a similar result for the integral

$I(x,y):=\int_{0}^{y}\int_{0}^{x}f(u,v)dudv$,

for each $x,y\in[0,1]$. Of course, the map $\tilde{g}(t):=(xg(t),yg(t))$ is a space-filling curve in $[0,x]\times[0,y]$, but I have not found a similar formula as the above for the integral $I(x,y)$.

What do you think?

Thanks in advance for your help!

Answer 1

There seems to be quite a bit of trickery behind this result (which I would not call well known, or which at least I never heard of). Nevertheless the change does not look to hard.

In general this result will only work because your space filling curve is somewhat uniform. If you think about it, for any interval $I$, the area of the image $g(I)$ is roughly the same as the length of $I$. If you think of it as a kind of transformation formula, you will note that the volume element is missing. This is because this ratio plays the roll of the volume element and actually is 1 in this case.

If you use your scaled version $\tilde{g}$, the area of $\tilde{g}(I)$ will scale accordingly to $xy$ times the length of $I$. So you should expect something like $$I(x,y) = \int_0^1 f(g(t)) xy dt.$$

Indeed looking at your sketch of the proof, it will work the same with little changes.

Answer 2

First at all, many thanks for your answer. I agree with you in the "factor" $xy$. In fact, when I approached the problem I thought that an approximation of the type

$I(x,y)\approx xy\int_{0}^{t}f(g(t))dt$,

It'd be enough. However, taking $f(t,s):=(t^{2}-s)^{4}$, for $x:=0.1$ and $y:=0.6$, I have found with the Maple software the approximations

$I(x,y)\approx 0.001512853781 \quad xy\int_{0}^{t}f(g(t))dt \approx 0.005402234428$,

taking the 4-th iteration of the Hilbert space-filling curve and the Gaussian quadrature

$\int_{0}^{1}h(t)dt \approx \frac{1}{n+1}\sum_{k=0}^{n}h(k/n)$

for integrals of a single variable. For other values of $x,y,f(t,s)$ and the results are even worse

Again, thanks!