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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?

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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.

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No. Consider $\mathbb RP^1= S^1$ as the 1-point 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.

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    $\begingroup$ 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

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