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Let X be the Cantor set, which we view as the space $2^\mathbb{N}$ (the set of all infinite binary sequences), equipped with the product topology. We can construct a Borel probability measure $\mu$ on this space by defining $\mu(C_{a_i})=1/2$, where the $C_{a_i}=\{x\in X | x_i=a_i\}$ are the open subbase cylinders of the product topology, and extending to a $\sigma$-algebra in the standard fashion.

Now, consider the Hilbert space $L^2(X,\mu)$. We can obtain orthonormal bases for it using the measure-space isomorphism between $(X,\mu)$ and $[0,1]$ (with Lesbesgue measure) via the binary decimal representations of real numbers. However, the ordinary bases (e.g., the trigonometric basis) on $L^2([0,1])$ are quite ugly when viewed on the Cantor set.

Is there an orthonormal basis for $L^2(X,\mu)$ with nice properties (continuity? simply expressible functions?) relative to the structure of the Cantor set?

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Since the Cantor set with your measure is also the compact group $(\mathbb{Z}/2)^\mathbb{N}$ with Haar measure, a natural orthonormal basis is the (continuous) characters $\alpha:X\to S^1$, namely the finite products of coordinates $c_n(x)$, $n\in\mathbb{N}$ if you view $\mathbb{Z}/2$ as {-1,1}. These form the discrete group $(\mathbb{Z}/2)^{(\mathbb{N})}$.

If you view $X$ as the Cantor middle third, $c_n(x)$ corresponds to $a_n(x)-1$, the $n$-th base $3$ digit of $x$ minus 1 (all digits are 0 or 2 by definition). These correspond to Walsh functions mentioned by Willie Wong when you use the measure isomorphism $X\to I$, which maps $x$ to ${1\over2} \sum_n a_n(x) 2^{-n}$.

Another possible model is $\mathbb{Z}_2$, the compact group of 2-adic integers, and the characters are then identified to power-of-two roots of unity, forming a group isomorphic to $\mathbb{Z}[1/2]/\mathbb{Z}$. This seems to lead to the same basis, although indexed differently. EDIT: as remarked by Greg Kuperberg in a comment, this can't be true.

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    $\begingroup$ There are a variety of interesting compact abelian groups that are homeomorphic to a Cantor set, and for each such you get an orthonormal basis of characters. Two of them are $\mathbb{Z}/2$ and $\mathbb{Z}_2$. You do not get the same basis, though, since as you say characters themselves form a discrete group and they are not the same group. $\endgroup$
    – Greg Kuperberg
    Aug 3, 2010 at 17:15
  • $\begingroup$ Of course, Greg, you are right. The values of the functions of the first basis are $\pm 1$, whereas they are all power-of-two roots of unity for the second, hence they cannot be the same under any measure isomorphism of two Cantor sets (which, as you point out, would entail isomorphism of character groups). I thought boldly that carry problems wouldn't harm... Since Pontryagin duality is involutive, any two non-isomorphic compact abelian groups homeomorphic to the Cantor set will give "different" basis in this sense. $\endgroup$
    – BS.
    Aug 3, 2010 at 19:03
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Perhaps the Walsh functions? http://en.wikipedia.org/wiki/Walsh_function

They are defined by dyadic intervals on $L^2([0,1])$, so is relatively well behaved under binary decimal representations. And thus should give a fairly nice description of you $L^2(X,\mu)$.

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Though not always, for certain Cantor measures $\mu$ there exists orthonormal basis for $L^2(\mu)$ consisting of complex exponentials $\{e^{2 \pi i \lambda_n t}: \lambda \in \Lambda \}$ where $\Lambda \subset \mathbb R$. These are called spectral Cantor measures.

See the following papers for details:

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