It is well known that the set of functions $\left\{ e^{^{inx}}\right\}$, for integer $n$, is an othonormal basis for the space of square integrable real functions in the interval $[-\pi,\pi]$.

Now let $\left\{ k_{n}\right\}$ be a sequence of real numbers and consider the set of functions $\left\{ e^{ik_{n}x}\right\}$. For what sequences $\left\{ k_{n}\right\}$ do the functions $\left\{ e^{ik_{n}x}\right\}$ form a basis (not necessarily orthonormal) for the space of square integrable real functions in some interval (not necessarily $[-\pi,\pi]$)?


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    $\begingroup$ I know that ${k}_{n} = 2\pi n/T$ gives an orthonormal basis for functions in $[-T,T]$. I'm looking for other sequences besides these. $\endgroup$ – becko Oct 29 '11 at 4:10
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    $\begingroup$ By the way, for the infinite interval thing, your domain is not compact anymore, so you lose maaaaaaaaaaaaaaany properties which makes things more complicated. Don't expect things to become as nice as they used to over there. $\endgroup$ – Patrick Da Silva Oct 29 '11 at 4:16
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    $\begingroup$ @PatrickDaSilva: I said that the basis I was looking for did not need to be orthogonal. My question is: Are there any other non-orthogonal bases? I think your argument does not apply. $\endgroup$ – becko Oct 29 '11 at 4:20
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    $\begingroup$ @becko: Your remark that the basis need not be orthonormal requires a bit of clarification. There are two distinct notions of basis in an inner product space, a Hilbert basis (such as $\left\{\mathrm e^{\mathrm inx}\right\}$) and a Hamel basis (which $\left\{\mathrm e^{\mathrm inx}\right\}$ isn't). A Hilbert basis is usually taken to be orthonormal by definition. Do I understand correctly that what you have in mind is a sequence of functions which, when orthonormalized, would form a Hilbert basis? $\endgroup$ – joriki Oct 29 '11 at 16:50
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    $\begingroup$ @becka, I presume that you are asking about Schauder bases? en.wikipedia.org/wiki/Schauder_basis $\endgroup$ – Johan Nov 2 '11 at 9:28

The question originally asked about infinite intervals, but the functions $e^{ikx}$ aren't square integrable over infinite intervals, so they can't form a basis for $\mathcal L ^2(\mathbb R)$. The analogous object for infinite intervals is the Fourier transform.

For finite intervals, the key phrase to search for is nonharmonic fourier series. There are entire books dedicated solely to this subject, but one can start with the 1952 paper by Duffin & Schaefer. There they introduce the notion of a frame, which can be easier to work with than a basis.

Another topic to search for is nonuniform sampling. By inverting the roles of the time domain and frequency domain, one can show that your question is equivalent to "What sets of samples are sufficient to reconstruct a bandlimited function?". From an article on nonuniform sampling by Aldroubi and Gröchenig:

Kadec’s theorem states that if $X = \{x_k ∈ \mathbb R : |x_k−k| \le L < 1/4\}$ for all $k \in \mathbb Z$, then the set $\{e^{ i 2 \pi x_k \xi}: k \in \mathbb Z\}$ is a Riesz basis of $\mathcal L^2(−1/2, 1/2)$.

(Kadec's original paper is in Russian.) Converting to your notation: the collection $\{e^{ i k_n x}: n \in \mathbb Z\}$ is a Riesz basis of $\mathcal L^2(−\pi, \pi)$ if for all $n \in \mathbb Z$, $|k_n-n|\le L <1/4$. In other words, if you perturb the frequencies slightly from the integers, you still get a basis. This isn't a necessary condition, but see the literature if you're interested in more general conditions.

  • $\begingroup$ Yes. Sorry I let that slip by. Forget about infinite intervals. I'm editing that on my question, but it is still unanswered. $\endgroup$ – becko Nov 2 '11 at 6:31
  • $\begingroup$ Thanks. I'm checking out what I can find about nonharmonic fourier series. (BTW, why did you remove your previous answer using Sturm-Liouville theory? I thought it was useful. Would you consider adding it again, perhaps as a separate answer?) $\endgroup$ – becko Nov 5 '11 at 17:00
  • $\begingroup$ It seemed like a bit of a long digression, since it's sort of just a very special case of Kadec's theorem. But I added it as a second answer and expanded on it. $\endgroup$ – p.s. Nov 5 '11 at 19:01

For every $k\in\mathbb R$, let $e_k:x\mapsto \mathrm e^{\mathrm ikx}$. Let $C=\{e_k;k\in\mathbb R\}$. Fix a positive $T$ and consider the space $E$ of square integrable $T$-periodic functions. Let $K=2\pi\mathbb Z/T$ and $B=\{e_k;k\in K\}$.

We know that $B$ is a basis of $E$. If $k\notin K$, $e_k$ is not $T$-periodic hence $e_k\notin E$. Thus a basis of $E$ cannot contain functions in $C\setminus B$. If $k\in K$, $e_k$ is not in the vector space spanned by $B\setminus\{e_k\}$ because $B$ is free. Finally the only basis of $E$ included in $C$ is $B$.

  • $\begingroup$ What does "$B$ is free" mean? $\endgroup$ – becko Nov 2 '11 at 18:51
  • $\begingroup$ You mean that the vectors in B are independent? Then I get it... $\endgroup$ – becko Nov 2 '11 at 18:54
  • $\begingroup$ Yes I do. $ $ $ $ $\endgroup$ – Did Nov 2 '11 at 19:04
  • $\begingroup$ However this does not answer my question, since I want a basis of the space of square integrable functions in some interval, say $[0,T]$, not of the space of $T$-periodic functions. $\endgroup$ – becko Nov 4 '11 at 5:35
  • $\begingroup$ You say $e_k$ is not $T$-periodic, and that allows you to take it out of $E$. But $e_k$ does belong to the space of square integrable functions in the interval considered. You can't take it out. $\endgroup$ – becko Nov 4 '11 at 5:40

One generalization of Fourier series comes from Sturm Liouville theory. This arises naturally in the study of PDEs. For example, consider a 1-D wave equation on a finite interval: $$ \begin{eqnarray} u_{tt}(x,t) &=& u_{xx}(x,t) \\ \forall t, \ \alpha_1 u(a,t) + \alpha_2 u_x(a,t)=0 & \quad \quad & \forall t, \ \beta_1 u(b,t) + \beta_2 u_x(b,t) = 0\\ \forall x \in [a,b], \ u(x,0) = f(x) & \quad \quad & \forall x \in [a,b], \ u_t(x,0) = g(x) \end{eqnarray} $$ One way to solve this is to look for a solution of the form $u(x,t) = \sum_{n=1}^\infty w_n(t) \phi_n(x)$ where $\phi_n(x)$ are the eigenmodes of the system. If we had periodic boundary conditions, then these would be the standard Fourier basis functions. It's important that the eigenmodes form a basis so that the initial conditions can be satisfied for all $f,g \in \mathcal L^2 (a,b)$.

To find the eigenmodes, you solve for nontrivial solutions to $y''=-\lambda y$ with the boundary conditions $\alpha_1 y(a)+ \alpha_2 y'(a)=0$, and $\beta_1 y(b)+ \beta_2 y'(b)=0$. These solutions are: $\mathcal B = \{c_n e^{i \sqrt{\lambda_n} x} + d_ne^{-i \sqrt{\lambda_n} x}\}_{n=1}^\infty$, where where $\lambda_n$ are the positive solutions to

$$ \tan((b-a)\sqrt{\lambda_n})=\frac{\sqrt{\lambda_n}(\alpha_1 \beta_2+\alpha_2 \beta_1)}{\lambda_n \alpha_2 \beta_2-\alpha_1 \beta_1}. $$

For general boundary conditions there's no closed-form solution for $\lambda_n$ other than the above implicit definition. (There are formulas for $c_n$ and $d_n$, but I'm omitting them.) Then by the Sturm Liouville theorem, the set $\mathcal B$ is an orthonormal basis for $\mathcal{L}^2(a,b)$.

In the basic case, $\alpha_1=\beta_1=1,$ $\ \alpha_2=\beta_2=0$, we get $\lambda_n = (n\pi/(b-a))^2$, and the orthonormal basis $\left\{ \sin\left(n \pi\frac{x-a}{b-a}\right)\right\}_{n=1}^\infty$. For the wave equation, this corresponds to the eigenmodes when the ends are held at $0$.

Admittedly, this isn't the same thing as the collection $\mathcal{\tilde{B}}=\{e^{\pm i \sqrt{\lambda_n}x}\}_{n=1}^\infty$ being a basis, but it does show that $\mathcal{\tilde{B}}$ spans $\mathcal{L}^2(a,b)$.

  • $\begingroup$ + 1. Even though Kadek's theorem seems more informative, I don't see how this is just a special case. $\endgroup$ – becko Nov 7 '11 at 21:43
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    $\begingroup$ Well, it's not quite the same due the difference between $\mathcal B$ and $\mathcal{\tilde{B}}$. What I was getting at is the solutions to $\tan((b-a)k) = ck/(d k^2 - e)$ will be almost regularly spaced. Roughly $k_{n+1}-k_{n} \approx \pi/(b-a)$. (Think about the plot of the left hand side versus the right hand side.) $\endgroup$ – p.s. Nov 7 '11 at 22:27

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