# basis functions do not lie in the space they form

For example, any continuous function in $\mathbb{L}^2(-\infty,\infty)$ space can be expanded by delta functions $\delta(x-a)$ or Fourier basis $e^{ikx}$. However, the basis functions, both $\delta(x-a)$ and $e^{ikx}$, are not square-integrable, thus not in $\mathbb{L}^2(-\infty,\infty)$ space. It's hard to understand there is a space whose has basis that are not in the space. Is there an intuitive explanation?

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In case your question was inspired by quantum mechanics you may find this wikipedia page useful. – user10001 Jul 12 '12 at 23:00
To add, in the complex exponential case the subspace in the rigged Hilbert space "sandwich" is the Schwarz space, so one has $S \subset L^2 \subset S'$. By Plancharel's theorem the Fourier transform is an automorphism of $S'$, the space where the functionals $\phi \rightarrow \int e^{ikx} \phi$ live. – Nick Alger Jul 13 '12 at 4:04

I think there is a confusion between two similar-sounding statements:

• every element of space $X$ can be written in terms of functions in the set $S$
• the set $S$ spans $X$

The second statement has a precise meaning, and it does require $S$ to be a subset of $X$. The meaning of the first is determined from context: "written in terms of" means whatever the writer wants it to mean. Here is a simple example without exotic things like delta functions:

Fact. Every function $f\in C[0,1]$ is the sum of a uniformly convergent series $f=\sum_{k=1}^\infty c_k g_k$, where $g_1,g_2,\dots$ are characteristic functions of the dyadic intervals $[(j-1)/2^m,j/2^m]$ enumerated in some way.

Note that the topology of $C[0,1]$ is indeed the topology of uniform convergence, so the infinite sum is almost like an expansion of $f$ with respect to a spanning set of $C[0,1]$. But it's not, precisely because $g_k$ do not belong to $C[0,1]$. Not being contained in $C[0,1]$, the set $\{g_k\}$ does not satisfy the definition of a spanning set or of a basis.

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Is there a name/theory for this more general notion of expanding a function in a basis from a larger space? Dushya has already mentioned the idea of a "rigged hilbert space", but your example makes it clear the idea applys to more general things than hilbert spaces. – Nick Alger Jul 15 '12 at 13:34
@NickAlger There is a body of literature on integral representation of functions, as a subset of literature on function spaces. I don't know if the subject is unified enough to be called a theory. – user31373 Jul 15 '12 at 16:35

When talking about $L^2$-spaces, the domain of definition of the member functions matters. $L^2(\mathbb{R})$ is different from $L^2(a,b)$ for example. What the theory of Fourier sequences asserts is that the functions $e^{ikx}$ are a basis for $L^2(-\pi, \pi)$ and they definitely do lie in this space since

$$\int_{-\pi}^\pi \! |e^{ikx}|^2\, dx = 2\pi$$

and thus $\|e^{ikx}\|_2 = \sqrt{2\pi}$. It is not asserted that they are a basis for $L^2(\mathbb{R})$, and this is as you have pointed out obviously not true.

As for the $\delta$s, these are not actually functions. (It should be an easy exercise to show that every measurable $f: \mathbb{R} \to \mathbb{C}$ whose support is a single point has integral equal to $0$.) A rigorous treatment of these "functions" comes from the theory of distributions, a thorough discussion of which is too long for this post. As an accessible reference, I recommend Appendix A and Chapter 2 of the book Partial Differential Equations by J. Rauch.

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See here. It seems that $e^{ikx}$ are basis of $\mathbb{L}^2(-\infty,\infty)$ – chaohuang Jul 12 '12 at 22:55
@chaohuang, no, they are not. They are not even in that space. – Mariano Suárez-Alvarez Jul 13 '12 at 0:40
@MarianoSuárez-Alvarez I mean continuous functions in $\mathbb{L}^2(-\infty,\infty)$ space can be expanded by $e^{ikx}$. – chaohuang Jul 13 '12 at 0:59
@chaohuang the author of that page puts quotation marks around the words "basis functions" and "dirac delta functions" indicating that this is not rigorous mathematics, but some motivation for why some things should be true. The word basis functions arises in this context because one can view the Fourier transform as a "limit" of the Fourier sequences of periodic functions on an interval as the interval gets large. – user12014 Jul 13 '12 at 21:52

I am not sure if this analogy is right:

Every real number $x$ can be written as $x=\frac{(z+z^*)}{2}$, where $z=x+yi$ is a complex number, $y$ is a non-zero real number, $*$ denotes complex conjugate. So every real number can be written in terms of a pair of conjugate complex numbers. Could this explain that every element of a space $X$ could be written in terms of elements outside $X$?

EDIT: Elements in a vector space $V_1$ could be expanded by basis of $V_2$ if $V_1\subset V_2$; on the other hand, Elements that are not in a vector space $V_1$ cannot be expanded by basis of $V_1$ since $V_1$ is spanned by its basis.

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