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.
Interpreting the Space of Square-Integrable Functions
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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
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 $L^2$.
$L^2[-a,a]$ 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.