Let $\Omega \subset \mathbb{R}^{n}$ be open and bounded with Lipschitz boundary. Is the gradient operator $\nabla :H^{1} ( \Omega ) \rightarrow L^{2} ( \Omega )$ surjective? Here $H^{1} ( \Omega ) =W^{1,2} ( \Omega )$ the Sobolev space of real valued functions in $L^{2}$ with weak derivative in $L^{2}$.
For unbounded sets in $\mathbb{R}$ the answer is clearly no, since any compactly supported continuous function is in $L^{2} ( \mathbb{R} )$ but has non integrable primitive.
So, if I try to disprove this for bounded sets, my usual simple test for these things, the set $( 0,1 ) \subset \mathbb{R}$ and the function $x^{\alpha}$, which is in $L^{p}$ iff $\alpha >-n/p$ doesn't really help much as a candidate. Integrating is always possible and yields a function in $H^{1} ( \Omega )$.
Ideas? Solution anyone? What about vector valued functions?
Let $\mathcal{D}(\Omega)^n$ denote the set of $C^\infty$, compact supported vector fields defined in $\Omega$. Define $$\mathcal{V}=\{\varphi\in \mathcal{D}(\Omega)^n:\ \operatorname{div}\varphi=0 \}$$
Denote by $H$ the closure of $\mathcal{V}$ in $L^2(\Omega)^n$. It can be showed that $$H=\{u\in L^2(\Omega)^n:\ \operatorname{div} u=0,\ \gamma(u)=0\}$$
$$H^{\perp}=\{u\in L^2(\Omega)^n:\ \exists\ p\in H^1(\Omega),\ u=\nabla p\}$$
where $\gamma$ is defined here and $H^{\perp}$ is the orthogonal of $H$ in $L^2(\Omega)^n$. Therefore $$L^2(\Omega)^n=H\oplus H^\perp$$
For references take a look in this book of Boyer and Fabrie, in section 3.3.