Pressure Schur equivalence from inf-sup, continuity, and stiffness scaling

Fix $q\in Q$ and put $w=A^{-1}{B}^{T}q$. By [F1] and symmetry of $A$, $b(v,q)=q^{T}Bv=v^T{B}^{T}q=v^{T}Aw=a(v,w)$ for every $v\in V$. By [F3], $\mathrm{\mid }a(v,w)\mathrm{\mid }\le \mathrm{\parallel }v{\mathrm{\parallel }}_A\mathrm{\parallel }w{\mathrm{\parallel }}_A$. If $w\mathrm{\ne }0$, equality is attained at $v=w$; if $w=0$, both sides are zero. Hence ${\sup }_{v\mathrm{\ne }0}\frac{\mathrm{\mid }b(v,q)\mathrm{\mid }}{\sqrt{a(v,v)}}=\sqrt{a(w,w)}=\sqrt{q^{T}B{A}^{-1}{B}^{T}q}\mathrm{.}$

The upper energy inequality in the statement implies $\sqrt{a(v,v)}\le \sqrt{\mathrm{\Lambda }}\mathrm{\parallel }v{\mathrm{\parallel }}_V$, so the supremum in step 1.1 is at least $\beta \mathrm{\parallel }q{\mathrm{\parallel }}_Q\mathrm{/}\sqrt{\mathrm{\Lambda }}$. The lower energy inequality implies $\sqrt{a(v,v)}\ge \sqrt{\alpha }\mathrm{\parallel }v{\mathrm{\parallel }}_V$; continuity of $b$ therefore makes that supremum at most $C_b\mathrm{\parallel }q{\mathrm{\parallel }}_Q\mathrm{/}\sqrt{\alpha }$. Squaring these inequalities and using step 1.1 proves both displayed Schur bounds.

For nonzero $q$, the lower bound in step 2.1 is positive because $\beta >0$ and $M_Q$ is positive definite. Thus $S$ is positive definite, and the two quadratic-form inequalities are the stated pressure-mass spectral equivalence.