Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Coming from a completely different world, I am trying to learn some very basic Fourier analysis, and have been scratching my head around this (it may be a very stupid question with an obvious answer):

Let $(f_n(t))_{n\in\mathbb{N}}$ be defined as $f_n(t)=\exp(it\cdot x_n)$ on some compact $T\subset\mathbb R^d$ for a given sequence $(x_n)_{n\in\mathbb N}$ in $\mathbb R^d$. Is it possible to choose $T$ and construct a measure $\lambda$ on it in such a way to make $(f_n(t))_{n\in\mathbb{N}}$ an orthonormal basis in $L_2(T,\lambda)$? This would be in analogy with the usual Fourier basis on the unit circle where $d=1$ and $x_n$ are the integers. So, we need that $\int\exp(it\cdot(x_n-x_m))d\lambda(t)=0$, for $n\neq m$, which seems impossible for generic $(x_n)_{n\in\mathbb N}$, but it is not obvious that it is since we can choose T and the measure on it freely?

Many thanks, P.S.

share|improve this question
1  
The integral you are setting to $0$ is just the $L_2$ dot product of $e^{it (x_n-x_m)$ and $\lambda(t)$ (if $\lambda$ can be represented by an $L_2$ function). Thus, there is a solution if (but not necessarily only if) the functions $e^{it(x_n-x_m)}$ do not span a dense subspace of $L^2$. Note, for instance, that the functions $e^{it(n-m)}$, $N\neq m$, span the space of functions with integral $0$. –  Will Sawin Nov 11 '11 at 19:00

1 Answer 1

Such T and measure do exist and in certain cases the measure is the restriction of usual Lebeasgue measure in $\mathbb{R}^n$. This kinds of sets are called spectral sets and spectral measures respectively, i.e., sets/ measures for which an complete orthogonal system of exponentials exist. The study of such sets began with the work of Fuglede (Fuglede, B. "Commuting Self-Adjoint Partial Differential Operators and a Group Theoretic Problem." J. Func. Anal. 16, 101-121, 1974.) who conjectured that a set is spectral if and only if it tiles and proved it in the special case when the spectrum or the tiling set is a lattice. The conjecture however has been disproved in dimension bigger than 3 following the work of Tao Tao, T. "Fuglede's Conjecture Is False in 5 and Higher Dimensions." ( http://arxiv.org/abs/math.CO/0306134.) but remain open in dimension 1,2. Further under additional hypothesis on the set T many positive results are known. The study of spectral measure began with the work of Jorgensen and Pedersen and Strichartz.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.