Let $S:=(0,1)^2$ be the unit square in $\mathbb{R}^2$, and let $M:=\{u\in L^2(S)\mid \int_S u=0\}$ be the space of (real-valued) $L^2$-functions with mean value zero. On $M$ we can consider the $L^2(S)$-norm, and the norm $u\mapsto\|Du\|_{W^{-1,2}(S)}$.

**Question:** Are these two norms equivalent on $M$?

I would like to think of $W^{-1,2}(S)$ as the dual space to $W_0^{1,2}(S)$. Therefore, $$\|Dv\|_{W^{-1,2}(S)}:=\sup \{\left < Dv,\phi\right >\mid \phi\in W_0^{1,2}(S,\mathbb{R}^2), \|\phi\|_{W^{1,2}(S,\mathbb{R}^2)}\le 1\}$$ and $$\left < Dv,\phi\right > = -\int_S v \operatorname{div} \phi$$

and the question is whether $\|Dv\|_{W^{-1,2}(S)}\ge C\|v\|_{L^2(S)}$ (the reverse inequality is easy).

Note that in one dimension (taking $S$ to be an interval $[0,1]$), the statement can be proven by choosing $\phi(t):=\int_0 ^t v(s) ds$, and since $v$ has mean zero, $\phi$ is zero at the end points. But this does not seem to generalize.