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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 $[-\pi,\pi]$)?

Thanks.

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    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.2011-10-29
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    Well if you stay on $[-\pi, \pi]$ and want to find all families of functions of the form $e^{ik_n x}$ then you'll realize that since two functions need to be orthogonal with respect to your scalar product (the integral of $f \, \bar g$ over $[-\pi,\pi]$) you need that for all integers $n,m$, $k_n = k_m + \alpha(n,m)$ for some integer $\alpha \in \mathbb Z$. In other words that means this is the only basis you'll get for a set of functions of the form $e^{ik_n x}$ (up to trivial isomorphisms, i.e. multiplying by constants)2011-10-29
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    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.2011-10-29
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    @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.2011-10-29
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    @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?2011-10-29
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    @joriki: You understand correctly. What I have in mind is "a sequence of functions which, when orthonormalized, would form a Hilbert basis;" exactly that.2011-10-29
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    @becko : You said not need to be orthonormal. I assumed you just meant that the basis vectors needed not have norm 1, which I just didn't really took care noticing. If you want to forget about orthogonality of your basis then your question becomes a lot more... say, unclear. If you tell us why you want an answer to this question perhaps we could go in one direction more than another and find something that suits your purposes, but if not, then this is just a wandering question amongst others... which can be fun to study, but can waste our time very easily without giving satisfying answers. =)2011-11-01
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    @PatrickDaSilva: I think the question is interesting until we find the answer, and that's enough motivation. Why do you say that the question is unclear? Tell me and I will do my best to fix it.2011-11-02
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    @becka, I presume that you are asking about Schauder bases? http://en.wikipedia.org/wiki/Schauder_basis2011-11-02
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    @Johan: I'm not sure if a Schauder basis requires orthogonality. From the definition at wikipedia I think it does not. So yes, I'm asking about Shauder bases.2011-11-02

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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.

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    Yes. Sorry I let that slip by. Forget about infinite intervals. I'm editing that on my question, but it is still unanswered.2011-11-02
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    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?)2011-11-05
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    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.2011-11-05
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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 $B$.

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    What does "$B$ is free" mean?2011-11-02
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    You mean that the vectors in B are independent? Then I get it...2011-11-02
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    Yes I do. $ $ $ $2011-11-02
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    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.2011-11-04
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    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.2011-11-04
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    See also the other answer by p.s., which gives a method to construct other bases, which would be counterexamples to your argument.2011-11-04
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    becko, please read what I wrote. That $e_k$ is not in $E$ if $k$ is not in $K$ is a fact. Note that I took care to define precisely the objects my answer applies to (something we should all do, answerers **and askers**, don't you think?), then the results it reaches. Obviously the latter do not apply to anything else than the former, so I see no "counterexamples" there.2011-11-04
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    I'm not saying your answer is wrong. I'm just saying it doesn't answer my question. You say that $e_k$ is not in $E$ if $k$ is not in $K$, and of course I agree. What I'm saying is that I don't want a basis for $E$ (the space of square integrable $T$-periodic functions). I want a basis for the space of square integrable functions defined on some interval.2011-11-04
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    Aren't "square-integrable T-periodic functions" and "square integrable functions on $[0,T)$" basically the same things anyway? That is, restriction to $[0,T)$ and periodic extension are vector space isomorphisms, right?2011-11-04
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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 $\mathcal{L}^2(a,b)$.

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    + 1. Even though Kadek's theorem seems more informative, I don't see how this is just a special case.2011-11-07
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    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.)2011-11-07