What's the space of smooth functions in L^2(R)? 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'? 
 A: 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.
A: $V'$ 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.)
