# Is the Laplacian surjective on $C_0^{\infty}$?

Let $M := C_0^{\infty}(\mathbb{R}^n)$ denote the smooth maps with compact support. Then we have a map

$\Delta:M\rightarrow M,\,\, f\mapsto \Delta f$,

where $\Delta f = \sum_{i=1}^{n} \frac{\partial^2}{\partial x_i^2}f$ is the Laplacian. I am wondering if $\Delta$ is surjective, i.e. if for any $f\in M$ there exists an $F\in M$ with $\Delta F = f$. Is that true?

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Laplacian is the divergence of gradient. By the divergence theorem the integral of Laplacian must be zero. –  user31373 Sep 13 '12 at 11:14
@LKV: I don't understand how this helps me. Could you please elaborate? Thanks. –  Sh4pe Sep 13 '12 at 13:02
re LVK's comment , start with $\mathbb R^1$ –  mike Sep 13 '12 at 14:30

It is very far from being surjective. Note that if $f\in C^\infty_0$ and $u$ is any harmonic function in the entire space, then $\int (\Delta f)u=\int f(\Delta u)=0$ (integration by parts or Green). This imposes infinitely many independent restrictions on the functions that can be represented as Laplacians of smooth compactly supported functions in every dimension above $1$ (in dimension $1$ the only harmonic functions are linear).

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Could you please specify what exactly you mean by "infinitely many independent restrictions"? How can I construct an $f$ from you comment that cannot be written as $\Delta F$? –  Sh4pe Sep 13 '12 at 12:56
Ah - yeah. Got it now. Thank you! –  Sh4pe Sep 13 '12 at 14:30