Hilbert inf-sup gives a bounded constraint right inverse

If $Q=\{0\}$, the unique map $R:Q\to V$ is a bounded right inverse and the claim is complete. Otherwise, since the scalar field is real, replacing $v$ by $-v$ changes the sign of $b(v,q)$; hence the displayed supremum equals the supremum of $\mathrm{\mid }b(v,q)\mathrm{\mid }\mathrm{/}(\mathrm{\parallel }v\mathrm{\parallel }\mathrm{\parallel }q\mathrm{\parallel })$. By [F3], $\mathrm{\parallel }{B}^{\ast }q{\mathrm{\parallel }}_V\ge \beta \mathrm{\parallel }q{\mathrm{\parallel }}_Q$.

On $Q$, define $c(q,r):=(B^{\ast }q,B^{\ast }r{)}_V$. It is bounded by $\mathrm{\parallel }B{\mathrm{\parallel }}^2\mathrm{\parallel }q\mathrm{\parallel }\mathrm{\parallel }r\mathrm{\parallel }$ and coercive with constant ${\beta }^2$ by step 1.1. For arbitrary $g\in Q$, the functional $r\mapsto (g,r{)}_Q$ is bounded, so [F2] gives a unique $q\in Q$ with $(B^{\ast }q,B^{\ast }r{)}_V=(g,r{)}_Q$ for all $r$. The adjoint identity implies $B{B}^{\ast }q=g$.

Set $Rg:=B^{\ast }q$, with $q$ from step 2.1. Uniqueness in Lax–Milgram makes this assignment linear, and the adjoint identity in [F3] gives $BRg=g$. Also $\mathrm{\parallel }Rg{\mathrm{\parallel }}^2=(g,q)\le \mathrm{\parallel }g\mathrm{\parallel }\mathrm{\parallel }q\mathrm{\parallel }$ while step 2.1 gives $\beta \mathrm{\parallel }q\mathrm{\parallel }\le \mathrm{\parallel }Rg\mathrm{\parallel }$. If $Rg=0$ the desired bound is immediate; otherwise combine and divide to obtain $\mathrm{\parallel }Rg\mathrm{\parallel }\le {\beta }^{-1}\mathrm{\parallel }g\mathrm{\parallel }$. Thus $R$ is the claimed bounded right inverse.