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Maybe this question is not appropriate here.

Let R be real numbers, and L^2(R) the square integrable functions, now what's the space of smooth functions in L^2(R)?

Edit:Sorry for the ambiguity. Let's consider the following question. V be the smooth functions in L^2, and let V' be the closure of the subspace generated by f(x)-f(x-c) where f belongs to V and c is any real number. Now the question I want to know is what's the quotient of V modulo by V'?

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  • $\begingroup$ If you're looking for a notational name, the only thing that comes to mind is $L^2(R)\cap C^\infty(R)$. $\endgroup$
    – Alex R.
    Commented Feb 13, 2010 at 3:34
  • $\begingroup$ Some motivation might make your question easier to answer. $\endgroup$
    – j.c.
    Commented Feb 13, 2010 at 3:47
  • $\begingroup$ Smooth square-integrable functions is the only thing that comes to mind. $\endgroup$
    – Harry Gindi
    Commented Feb 13, 2010 at 4:14
  • $\begingroup$ I don't see anything wrong with this as an MO question per se: presumably it's asking if said space of functions is a more familiar one under a different guise. Unfortunately I don't know if there's anything better than Tracer Tong's answer $\endgroup$
    – Yemon Choi
    Commented Feb 13, 2010 at 4:19
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    $\begingroup$ What was wrong with the question was that what he intended to ask, which is presumably what he asked in his edit, was quite impossible to guess from what he'd originally written! :) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 13, 2010 at 4:40

2 Answers 2

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$V'$ is indeed dense in $L^2$. Taking Fourier transforms, note that any bounded measurable function with compact support is the Fourier transform of a function in $V$. And the Fourier transform of $f(x)-f(x-c)$ is $(1-e^{c\xi})\hat f(\xi)$. For the clincher, note that the space of bounded measurable functions with compact support contained in the complement of $(2\pi/n)\mathbb{Z}$ is dense in $L^2$. (Adjust signs and factors $2\pi$ accroding to your taste in Fourier transform conventions.)

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  • $\begingroup$ Or more general, let $\Omega\subseteq\mathbb{R}^d$ and $1\leq p\leq\infty$. Then $C_c^{\infty}(\Omega)$ is dense in $L^p{(\Omega,{\lambda}_d)}$, where ${\lambda}_d$ denotes the d-dimensional Borel-Lebesgue measure. $\endgroup$
    – M.G.
    Commented Feb 13, 2010 at 16:03
  • $\begingroup$ Indeed. But my version is sufficient and has the advantage of being utterly trivial. $\endgroup$
    – Harald Hanche-Olsen
    Commented Feb 13, 2010 at 16:45
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N.B. this answer was in response to an earlier version of the question, which only had the first two paragraphs -- hence it doesn't address what appears to have been the original poster's actual question. For that, see the answers of Leonid or Harald.

I'm not sure if this answers your question, but it might be worth noting that a measurable function $f$ on the real line is in $L^2({\mathbb R})$ if and only if its Fourier transform $\widehat{f}$ is (Plancherel theorem), while it is in $C^\infty({\mathbb R})$ if and only if we have

$$ \int_{-\infty}^\infty | \widehat{f}(x) |^2 (1+ |x|^2)^{k} < \infty \quad{\rm for }\ k=1,2,\dots $$

(this is a form of Sobolev embedding, albeit in a very special case). In particular, if I've correctly understood the notation from the wikipedia page for Sobolev spaces, the space you're after seems to be the intersection $\bigcap_{k=0}^\infty H^k({\mathbb R})$. I don't know if this goes by a particular name.

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    $\begingroup$ I've seen this called the space of smooth vectors for the operator $(Af)(x) = (1+x^2)f(x)$. In this case, it's actually the space of Schwarz test functions of sufficiently rapid descent. $\endgroup$
    – user1504
    Commented Feb 13, 2010 at 18:00
  • $\begingroup$ What's the background (or reference) of the operator A? How about the function 1/(x^2+1)? $\endgroup$
    – user1832
    Commented Mar 9, 2010 at 4:36

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