2
$\begingroup$

I know that to construct the space $L^2( [-a,a) ) $, and to appreciate its richness, we need the machinery of lebesgue integration. However, I would like to work and talk about this space without ever having to invoke results from the lebesgue theory. What then is the best way to interpret $L^2( [-a,a) )$? Is it correct to say that $L^2( [-a,a) ) $ is obtained by completing the space of continuous functions on $[-a,a)$ (w.r.t. to the $L^2$ norm)? Can I say the same about $L^2(\mathbb{R})$? One result that I would really like to use is that any $L^2$ function can be approximated in norm by a step function, and this can be easily proved for a continuous function. My point is, I would like to work with $L^2$ without being hand-wavy with Lebesgue theory, whose machinery I don't really need to develop for my purposes.

  • 0
    Continuous functions with compact support are dense in $L^2$ (and hence step functions are).2012-03-10
  • 0
    Well, but how can you define the norm without Lebesgue integrals?2012-03-10
  • 0
    continuous functions with compact support and step functions are Riemann integrable. The Riemann integral and Lebesgue integral agree when they are both defined.2012-03-10
  • 0
    @ShawnD Do you mean that every $L^2$ function is the pointwise limit of compactly supported continuous functions? Or is the limit in the mean-square sense?2012-03-10
  • 1
    in the mean-square sense. Functions in $L^2$ can be approximated pointwise by continuous functions, but a sequence of continuous $L^2$ functions that converges pointwise does not necessarily converge in the mean-square sense (so pointwise convergence is in some sense not the "right" concept when talking about $L^2$)2012-03-10
  • 3
    Peter, if you would like to work on $L^2$ without ever appealing to the Lebesgue theory, then just treat it as the separable Hilbert Space (which basically characterizes its interesting properties). If you happen to need to work with functions and integration though, there is really no getting around the Lebesgue theory. I cannot imagine trying to work on integration without access to the dominated and monotone convergence theorems, both of which depend on using the Lebesgue integral (as opposed to the Riemann integral).2012-03-10
  • 0
    By the way most (in some sense) $L^2$ functions cannot be constructed as the pointwise limit of continuous functions since the pointwise limit of continuous functions must be continuous on a dense $G_\delta$ set. This is not terribly hard to show as a consequence of Baire's theorem. A simple counterexample to the claim on $L^2[0,1]$ is the indicator function of the rationals which is nowhere continuous. It is also possible to construct less trivial counterexamples (ones where you cannot modify the function on a set of measure zero to make it continuous).2012-03-10
  • 0
    @Chris If $f_n \rightarrow f$ in $L^2$ (where $f_n$ is continuous), then $f_n$ converges to $f$ in measure and (assuming $\sigma$-finite) there is a sub-sequence converging almost everywhere. Of course maybe we need a precise definition of what it means for an element of $L^2$ to be a pointwise limit of continuous functions.2012-03-10
  • 0
    @ShawnD Does the proof showing that the space of compactly supported continuous funcs is dense in $L^2$ require lebesgue theory? Can I get away by defining $L^2$ to be the completion in mean-square of the first space?2012-03-10
  • 0
    You could (see Chris's comment) define $L^2$ in this way (depending on what you want to do).2012-03-10
  • 0
    You have $L^2\equiv l^2$ since it is a Separable Hilbert space.2012-03-10
  • 0
    @ShawnD, the difference is indeed in our definitions of pointwise convergence. I mean pointwise convergence classically while your definition (a.e.) probably makes more sense in the $L^2$ context. You do not need the measure space to be sigma finite for that subsequence by the way. The sigma finiteness is needed for that subsequence condition to characterize convergence in measure. Peter yes it requires the Lebesgue theory. The proof generally runs through Lusin's theorem, though you can do it without it. In any case it doesn't make sense to ask the question without the Lebesgue theory.2012-03-10
  • 0
    @ShawnD I am interested in using Fourier analysis as a tool for studying some theoretical applications of image and signal processing. In my work, I would like to freely talk about $L^2$ functions, because they are the natural function space of choice. In particular, I want to say that every $L^2$ function can be well approximated in $L^2$-norm by a trigonometric function. Since it is within my sensibility to show this is true for continuous functions, I would like to use the density property and say it is true on $L^2$ as well. Is this admissible?2012-03-11

2 Answers 2

1

For any (good enough) measure space that is also locally compact Hausdorff space $X$, the set $C_c(X)$ (compactly supported continuous functions) is dense in $L^2(X)$. The space $\mathbb{R}^n$ and any of its open or closed subsets are like that.

Also, any bounded function is a uniform limit of step functions. Unbounded functions are $L^2$ limits of step functions (i.e. step functions are dense).

Whether you can avoid Lebesgue theory or not, depends on what you want to do with your $L^2$.

0

$L^2[-a,a]$ is the smallest possible space whose elements (ie. functions) can be treated as geometric objects.

  • $L^1[-a,a]$ (integrable functions) is only a Banach space, with no basis. It is so packed with so many kinds of functions that it is not a very useful space to work on.

  • In contrast, $L^2[-a,a]$ is a Hilbert space. Because $L^2$ supports an inner product, $L^2$ is automatically endowed with the five basic axioms of the euclidian geometry. Hence, $L^2$ and its functions behave similarly to the euclidian space we are all familiar with : functions of $L^2$ can be "orthogonal" to one another, $L^2$ can be endowed with a basis ($L^1$ cannot), in $L^2$ function operations such as Fourier transforms behave similar to rotations or reflections (isometries), Pythagoras' theorem can be applied and much more. This richness makes $L^2$ the ideal framework for functional analysis.