Let $D$ define the differential operator $\frac{\partial^2}{\partial x^2}$ on $\mathbb R$. Let $\xi\notin\mathbb R$ be a complex number. Is it true that $$ (D\xi)C_c^\infty({\mathbb R}) $$ is dense in $C_c({\mathbb R})$, where the latter space has the usual locally convex inductive limit topology?
I introduce the following notation: let $D_\xi = D\xi$; let $\chi\colon \mathbb{R} \to \mathbb{R}$ be any smooth function such that $\chi(x) \equiv 0$, for $x<1$, and $\chi(x) \equiv 1$, for $x>1$; also let $G_\xi$ and $G^\pm_\xi$ be the following integral operators \begin{align} G_\xi[f](x) &= \int_{\mathbb{R}} \frac{\sin\sqrt{\xi}(yx)}{\sqrt{\xi}} f(y) \, dy , \\ G^+_\xi[f](x) &= \int_{\mathbb{R}} \Theta(xy) \frac{\sin\sqrt{\xi}(yx)}{\sqrt{\xi}} f(y) \, dy , \\ G^_\xi[f](x) &= \int_{\mathbb{R}} \Theta(yx) \frac{\sin\sqrt{\xi}(yx)}{\sqrt{\xi}} f(y) \, dy , \end{align} where $\Theta(x)$ is the Heaviside step function equal to $1$ for $x>0$ and to $0$ for $x<0$.
Then the following is an exact sequence of vector spaces (not necessarily taking any topology into account), as evidenced by the contracting homotopy indicated with the dashed arrows:
$$\require{AMScd} \begin{CD} 0 \to C^\infty_c(\mathbb{R}) @>{D_\xi}>{\stackrel{\dashleftarrow}{{}^\chi G_\xi}}> C^\infty_c(\mathbb{R}) @>{G_\xi}>{\stackrel{\dashleftarrow}{D_\xi^\chi}}> C^\infty(\mathbb{R}) @>{D_\xi}>{\stackrel{\dashleftarrow}{G_\xi^\chi}}> C^\infty(\mathbb{R}) \to 0 , \end{CD} $$
where $$ \begin{aligned} {}^\chi G_\xi[f] &= \chi G^_\xi[f] + (1\chi)G^+_\xi, \\ D_\xi^\chi[g] &= D_\xi[\chi g]  \chi D_\xi[g], \\ G_\xi^\chi[f] &= G^+_\xi[\chi f] + G^_\xi[(1\chi)f], \end{aligned} \quad \text{and} \quad \begin{aligned} {}^\chi G_\xi \circ D_\xi &= \mathrm{id} , \\ D_\xi \circ {}^\chi G_\xi + D_\xi^\chi \circ G_\xi &= \mathrm{id} , \\ G_\xi \circ D_\xi^\chi + G_\xi^\chi \circ D_\xi &= \mathrm{id} , \\ D_\xi \circ G_\xi^\chi &= \mathrm{id} . \end{aligned} $$
Once one chooses topologies, it remains to check that $G_\xi$ is a continuous map, which is not so hard with the standard topologies, so that its kernel is a closed subspace. By the exactness of the above sequence, this kernel coincides with the image of $D_\xi$. Again, following from the exactness of the above sequence, $G_\xi$ descends to an isomorphism between $\operatorname{coker} D_\xi_{C^\infty_c}$ and $\ker D_\xi_{C^\infty}$. Of course, everybody knows that the latter space is 2dimensional, which shows that the image of $D_\xi$ is not dense in $C^\infty_c(\mathbb{R})$.
The above identities may appear mysterious at first, but they are easily checked if one starts with the observation that $G^\pm_\xi$ are Green functions for $D_\xi$, with particular support properties.
No. Consider $\mathbb RP^1= S^1$ as the 1point compactification of $\mathbb R$. Then we have the short exact sequence (It would be nice to be able to write a diagram) $$ C^\infty_c(\mathbb R) \to C^\infty(S^1) \to C^\infty_\infty(\mathbb RP^1) $$ where the last space is the space of all germs at the point $\infty$.
$D\xi$ acts on each of these spaces:
On the middle space it is elliptic of index 1, since we may move $\xi$ around without changing the index, and for $\xi=0$ the operator $D$ is elliptic, has exactly the constants in its kernel, and the image has codimension 2 (integral over $S^1$ has to vanish, so that the antiderivative is again periodic, and once more the antiderivative has to have vanishing integral). By elliptic regularity all this holds in $C^\infty(S^1) = \bigcap_k H^k(S^1)$.
On the space of germs $D\xi$ is surjective, since we just have to find local solutions.
If $D\xi$ would be surjective on the left space, then a diagram chase would show that also on the middle space it should be surjective which is not true.
EDIT:
The image is also not dense, because on the middle space there is a nonzero continuous linear functional vanishing on the image of $D\xi$, because Fredholm operators have closed range. Restricted to $C^\infty_c$, it is still nonzero (since it cannot factor to the space of germs) and still vanishes on the image of $D\xi$.
In fact, I believe that $D\xi$ has closed range on the left hand side.

1$\begingroup$ I didn't ask for surjectivity. I asked if the image be dense in $C_c({\mathbb R})$. $\endgroup$ – user1688 Jun 11 '15 at 17:45

$\begingroup$ Ok its not dense in $C_c^\infty({\mathbb R})$ but could the image still be dense in $C_c({\mathbb R})$? $\endgroup$ – user1688 Jun 11 '15 at 18:43

$\begingroup$ I cannot see how the operator is elliptic in the middle as the principal symbol vanishes at $\infty$? $\endgroup$ – user1688 Jun 11 '15 at 18:57