Mean-zero pressure recovery from a solenoidal residual

For almost every $t$, define a bounded linear functional ${\mathrm{\ell }}_t$ on $Q$ by ${\mathrm{\ell }}_t(q)=R(t)(B_{D}q)$. From [F1], $\mathrm{\parallel }{\mathrm{\ell }}_t{\mathrm{\parallel }}_{Q^{\ast }}\le C_{\mathrm{d}\mathrm{i}\mathrm{v}}(D)\mathrm{\parallel }R(t){\mathrm{\parallel }}_{V^{\ast }}$. By [F4], there is a unique $\widehat{p}(t)\in Q$ such that $(\widehat{p}(t),q{)}_{L^2}={\mathrm{\ell }}_t(q)$ for all $q\in Q$. The map $R(t)\mapsto \widehat{p}(t)$ is bounded linear, so $\widehat{p}\in L^s(0,T;Q)$ with the corresponding norm bound.

For $v\in V$, put $q=\mathrm{div}v\in Q$ by [F2]. Then $v-B_{D}q\in V_{\sigma }$, so the annihilator assumption gives $R(t)(v)=R(t)(B_{D}q)=(\widehat{p}(t),q{)}_{L^2}=-b(v,\widehat{p}(t))\mathrm{.}$ Set $p=-\widehat{p}$ to obtain $R(t)(v)=b(v,p(t))$.

If $p_1,p_2\in Q$ both represent the same residual, then $b(B_{D}q,p_1-p_2)=0$ for every $q\in Q$. By [F1], $\mathrm{div}{B}_{D}q=q$, so $(p_1-p_2,q{)}_{L^2}=0$ for all $q\in Q$; choosing $q=p_1-p_2$ proves equality. This also confirms uniqueness in the zero-mean gauge.