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?



Density of Laplace image
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}(y-x)}{\sqrt{\xi}} f(y) \, dy , \\
 G^+\_\xi[f](x) &= \int\_{\mathbb{R}} \Theta(x-y) \frac{\sin\sqrt{\xi}(y-x)}{\sqrt{\xi}} f(y) \, dy , \\
 G^-\_\xi[f](x) &= \int\_{\mathbb{R}} \Theta(y-x) \frac{\sin\sqrt{\xi}(y-x)}{\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 2-dimensional, 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.


