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}(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.


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.


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

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