Equivalence of negative Sobolev norm of derivative to $L^2$-norm

Question

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.

Answer 1

Hint.

If $f\in M$, then $f=\sum_{m,n\in\mathbb Z}\hat f_{m,n}\mathrm{e}^{2\pi i(mx+ny)}$, with $\hat f_{0,0}=0$, and $\|f\|_{L^2}^2=\sum_{m,n\in\mathbb Z}\hat |f_{m,n}|^2$.

Meanwhile $\nabla f=2\pi i\sum_{m,n\in\mathbb Z}\hat f_{m,n}\mathrm{e}^{2\pi i(mx+ny)}(m,n)$, and $$ \|f_x\|_{H^{-1}}=\sup_{\|g\|_{H^1}=1}\langle f_x,g\rangle_{L^2}= \sup_{\sum |g_{m,n}|^2=1}\sum_{m,n\in\mathbb Z}m\,\hat f_{m,n}\,\overline{\hat g_{m,n}}. $$