If grad f is in (H^(-1)(U))^3 is f in L^2(U)?

Question

I know that if $f \in L^2(U)$ then $\nabla f \in H^{-1}(U)$. However, is the converse true? I was thinking that maybe I could use the characterization of $H^{-1}(U)$ that says that if $f \in H^{-1}(U)$ then there exists $f_0 \in L^2(U)$, $f_1=(L^2(U))^3$ such that $$\langle f, v \rangle=(f_0, v)_{L^2(U)}+(f_1, \nabla v)_{L^2(U)}.$$ However, I'm not sure exactly how. Does anyone know of any books that would have details related to this?

Answer 1

This is a highly nontrivial result. It is stated in Temam, Navier-Stokes equations, Proposition I.1.2. For the proof it refers to Necas, Equations aux derivees partielles, which I was not able to verify. Basically, every book that treats Stokes or Navier-Stokes problems should deal with this result. But most of them do not provide a proof.