Nice orthonormal basis for L^2(Cantor set) - MathOverflow

Nice orthonormal basis for L^2(Cantor set)

2010-08-03T15:22:43Z

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?

Answer by Willie Wong for Nice orthonormal basis for L^2(Cantor set)

2010-08-03T15:39:32Z

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 $L^2(X,\mu)$.

Answer by BS for Nice orthonormal basis for L^2(Cantor set)

2010-08-03T16:20:25Z

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.

Answer by Vagabond for Nice orthonormal basis for L^2(Cantor set)

2010-08-03T16:43:18Z

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

P. E. T. Jorgensen and S. Pedersen , Dense analytic subspaces in fractal L2-spaces. J. Anal. Math. 75 (1998), pp. 185–228. http://www.springerlink.com/content/211651p66833m7j7/

R. Strichartz , Remarks on Dense analytic subspaces in fractal L2-spaces. J. Anal. Math. 75 (1998), pp. 229–231 http://www.springerlink.com/content/a53685h15g17x509/

Izabella Laba and Yang Wang On Spectral Cantor Measures Journal of Functional Analysis Volume 193, Issue 2, 20 August 2002, Pages 409-420 http://www.math.ubc.ca/~ilaba/preprints/meas1.dvi