A coercive Hilbert-complex estimate solves the closed equation

Since $S$ is closed, its kernel is closed: if $x_k\in \ker S$ and $x_k\to x$ in $H_1$ then $(x_k,S{x}_k)=(x_k,0)\to (x,0)$ lies in the closed graph of $S$, so $x\in D(S)$ and $Sx=0$; the same argument shows that $\ker T$ is a closed subspace of $H_0$, and $T(D(T))\subseteq \ker S$ says exactly that the composition $S\circ T$ is zero on $D(T)$.

Semidefinite Cauchy–Schwarz. For all $x,y\in H_1$ one has $\mathrm{\mid }⟨Ax,y⟩\mathrm{\mid }\le ⟨Ax,x{⟩}^{1\mathrm{/}2}⟨Ay,y{⟩}^{1\mathrm{/}2}$: fix $x,y$ and note that for every real $t$, using [F10] and $⟨Ax,x⟩\ge 0$, $0\le ⟨A(x+ty),x+ty⟩=⟨Ax,x⟩+2t\mathrm{Re}⟨Ax,y⟩+t^2⟨Ay,y⟩$; if $⟨Ay,y⟩>0$ the discriminant of this nonnegative quadratic is $\le 0$, so $(\mathrm{Re}⟨Ax,y⟩{)}^2\le ⟨Ax,x⟩⟨Ay,y⟩$, while if $⟨Ay,y⟩=0$ the same quadratic forces $\mathrm{Re}⟨Ax,y⟩=0$ and the inequality holds trivially; choosing $\lambda$ with $\mathrm{\mid }\lambda \mathrm{\mid }=1$ and $\lambda ⟨Ax,y⟩=\mathrm{\mid }⟨Ax,y⟩\mathrm{\mid }$ and applying the real-part inequality to the pair $(\lambda x,y)$, whose quadratic values are unchanged because $A$ is linear and $⟨A(\lambda x),\lambda x⟩=⟨Ax,x⟩$, gives the stated modulus bound.

The adjoint defined in [F1] has a linear domain. Directly from its defining identity, $(\mathrm{ran}T{)}^{\perp }=\ker T^{\ast }$: vanishing of $⟨Tx,y⟩$ for every $x\in D(T)$ is equivalent to $T^{\ast }y=0$. More precisely, in $H_0\oplus H_1$ one has $\mathrm{\Gamma }(T{)}^{\perp }=\{(-T^{\ast }y,y):y\in D(T^{\ast })\},$ because $(a,y)$ is orthogonal to every $(x,Tx)$ exactly when $⟨Tx,y⟩=⟨x,-a⟩$ for every $x\in D(T)$. Since $T$ is closed, [F2] gives $\mathrm{\Gamma }(T)=\mathrm{\Gamma }(T{)}^{\perp \perp }$.

For $x\in D(T^{\ast })$ decompose $x=x^{\mathrm{\prime }}+x^{\mathrm{\prime }\mathrm{\prime }}$ with $x^{\mathrm{\prime }}\in \ker S$ and $x^{\mathrm{\prime }\mathrm{\prime }}\in (\ker S{)}^{\perp }$ by [F3] applied to the closed subspace $\ker S$ of step 1.1; since $T(D(T))\subseteq \ker S$, taking orthogonal complements gives $(\ker S{)}^{\perp }\subseteq (\mathrm{ran}T{)}^{\perp }=\ker T^{\ast }$ by step 1.3, so $x^{\mathrm{\prime }\mathrm{\prime }}\in \ker T^{\ast }\subseteq D(T^{\ast })$, hence $x^{\mathrm{\prime }}=x-x^{\mathrm{\prime }\mathrm{\prime }}\in D(T^{\ast })$ as well, with $T^{\ast }{x}^{\mathrm{\prime }}=T^{\ast }x-T^{\ast }{x}^{\mathrm{\prime }\mathrm{\prime }}=T^{\ast }x$; moreover $⟨x,f⟩=⟨x^{\mathrm{\prime }},f⟩+⟨x^{\mathrm{\prime }\mathrm{\prime }},f⟩=⟨x^{\mathrm{\prime }},f⟩$ because $x^{\mathrm{\prime }\mathrm{\prime }}\perp \ker S\ni f$.

For every $x\in D(T^{\ast })$ the vector $x^{\mathrm{\prime }}$ of step 2.1 lies in $D(T^{\ast })\cap D(S)$ and $S{x}^{\mathrm{\prime }}=0$, so the hypothesis and step 2.1 give $\mathrm{\parallel }{x}^{\mathrm{\prime }}{\mathrm{\parallel }}^2\le C^{-1}(\mathrm{\parallel }{T}^{\ast }{x}^{\mathrm{\prime }}{\mathrm{\parallel }}^2+\mathrm{\parallel }S{x}^{\mathrm{\prime }}{\mathrm{\parallel }}^2)=C^{-1}\mathrm{\parallel }{T}^{\ast }x{\mathrm{\parallel }}^2$, and [F7] together with step 2.1 gives $\mathrm{\mid }⟨x,f⟩{\mathrm{\mid }}^2=\mathrm{\mid }⟨x^{\mathrm{\prime }},f⟩{\mathrm{\mid }}^2\le \mathrm{\parallel }{x}^{\mathrm{\prime }}{\mathrm{\parallel }}^2\mathrm{\parallel }f{\mathrm{\parallel }}^2\le C^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}^2\mathrm{\parallel }{T}^{\ast }x{\mathrm{\parallel }}^2$.

Define $\mathrm{\ell }$ on the subspace $V_0:=T^{\ast }(D(T^{\ast }))\subseteq H_0$ by $\mathrm{\ell }(T^{\ast }x):=⟨x,f⟩$; if $T^{\ast }x=0$ then step 3.1 gives $\mathrm{\mid }⟨x,f⟩\mathrm{\mid }\le C^{-1\mathrm{/}2}\mathrm{\parallel }f\mathrm{\parallel }\cdot 0=0$, so $\mathrm{\ell }$ is well defined, and step 3.1 says $\mathrm{\mid }\mathrm{\ell }(w)\mathrm{\mid }\le C^{-1\mathrm{/}2}\mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }w\mathrm{\parallel }$ for every $w\in V_0$, so $\mathrm{\ell }$ is linear and bounded on $V_0$; let $V:=\overline{V_0}\subseteq H_0$ and define $\tilde{\mathrm{\ell }}(y):={\lim }_k\mathrm{\ell }(w_k)$ for $y\in V$ and any sequence $w_k\in V_0$ with $w_k\to y$: the sequence $(\mathrm{\ell }(w_k))$ is Cauchy because $\mathrm{\mid }\mathrm{\ell }(w_k)-\mathrm{\ell }(w_l)\mathrm{\mid }\le C^{-1\mathrm{/}2}\mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }{w}_k-w_l\mathrm{\parallel }$, and the limit exists by [F8]; the countable instance of [F9] is the choice principle consumed by [F1], [F3], [F4] and [F8] here, and the construction selects nothing further; it is independent of the sequence because two sequences for the same $y$ have difference tending to $0$, and passing to limits of the defining inequalities preserves linearity (limits of sums and scalar multiples) and gives $\mathrm{\mid }\tilde{\mathrm{\ell }}(y)\mathrm{\mid }\le C^{-1\mathrm{/}2}\mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$.

$A$-weighted estimate. For every $x\in D(T^{\ast })$, with $x=x^{\mathrm{\prime }}+x^{\mathrm{\prime }\mathrm{\prime }}$ as in step 2.1, the vector $x^{\mathrm{\prime }}$ lies in $D(T^{\ast })\cap D(S)$ with $S{x}^{\mathrm{\prime }}=0$ and $T^{\ast }{x}^{\mathrm{\prime }}=T^{\ast }x$ (step 3.1 for the first two, step 2.1 for the third), so by step 1.2 applied to the pair $(x^{\mathrm{\prime }},g)$, by [F10], by $f=Ag$ and by the domination hypothesis for (iv) on $x^{\mathrm{\prime }}$: $\mathrm{\mid }⟨x,f⟩\mathrm{\mid }=\mathrm{\mid }⟨x^{\mathrm{\prime }},f⟩\mathrm{\mid }=\mathrm{\mid }⟨x^{\mathrm{\prime }},Ag⟩\mathrm{\mid }=\mathrm{\mid }⟨A{x}^{\mathrm{\prime }},g⟩\mathrm{\mid }\le ⟨A{x}^{\mathrm{\prime }},x^{\mathrm{\prime }}{⟩}^{1\mathrm{/}2}⟨Ag,g{⟩}^{1\mathrm{/}2}\le (\mathrm{\parallel }{T}^{\ast }{x}^{\mathrm{\prime }}{\mathrm{\parallel }}^2+\mathrm{\parallel }S{x}^{\mathrm{\prime }}{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}⟨f,g{⟩}^{1\mathrm{/}2}=\mathrm{\parallel }{T}^{\ast }x\mathrm{\parallel }⟨f,g{⟩}^{1\mathrm{/}2}$, where $⟨Ag,g⟩=⟨f,g⟩$ and $⟨f,g⟩\ge 0$ by positivity of $A$.

Let $P:H_0\to V$ be the orthogonal projection onto the closed subspace $V$, defined for $y\in H_0$ by the unique decomposition $y=Py+(y-Py)$ with $y-Py\in V^{\perp }$ of [F3], and put $\mathrm{\Lambda }:=\tilde{\mathrm{\ell }}\circ P$; then $\mathrm{\Lambda }$ is linear and $\mathrm{\mid }\mathrm{\Lambda }(y)\mathrm{\mid }\le C^{-1\mathrm{/}2}\mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$ for all $y\in H_0$, so by [F4] there is a unique $u\in H_0$ with $\mathrm{\Lambda }(y)=⟨y,u⟩$ for all $y\in H_0$ and $\mathrm{\parallel }u\mathrm{\parallel }=\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }\le C^{-1\mathrm{/}2}\mathrm{\parallel }f\mathrm{\parallel }$.

For every $x\in D(T^{\ast })$, steps 4.1 and 5.1 give $⟨T^{\ast }x,u⟩=⟨x,f⟩$. Therefore $(u,f)$ is orthogonal to every $(-T^{\ast }x,x)\in \mathrm{\Gamma }(T{)}^{\perp }$ by step 1.3 (take complex conjugates of the displayed identity). Since $T$ is closed, $(u,f)\in \mathrm{\Gamma }(T{)}^{\perp \perp }=\mathrm{\Gamma }(T)$; hence $u\in D(T)$ and $Tu=f$, proving (i).

Define $\mathrm{\ell }$ on $V_0:=T^{\ast }(D(T^{\ast }))$ by $\mathrm{\ell }(T^{\ast }x):=⟨x,f⟩$ for $x\in D(T^{\ast })$ and put $\beta :=⟨f,g{⟩}^{1\mathrm{/}2}$: step 4.2 with $T^{\ast }x=0$ shows $\mathrm{\ell }$ is well defined, and it shows $\mathrm{\mid }\mathrm{\ell }(w)\mathrm{\mid }\le \beta \mathrm{\parallel }w\mathrm{\parallel }$ for every $w\in V_0$, so $\mathrm{\ell }$ is linear and bounded; the extension $\tilde{\mathrm{\ell }}$ of $\mathrm{\ell }$ to $V:=\overline{V_0}$ by limits along sequences in $V_0$, its independence of the chosen sequence, its linearity and its bound $\mathrm{\mid }\tilde{\mathrm{\ell }}(y)\mathrm{\mid }\le \beta \mathrm{\parallel }y\mathrm{\parallel }$ are verified by the same computation as in step 4.1 with $\beta$ in place of $C^{-1\mathrm{/}2}\mathrm{\parallel }f\mathrm{\parallel }$, with the Cauchy property supplied by [F8] and the same countable instance of [F9]; letting $P$ be the orthogonal projection of [F3] onto $V$ as in step 5.1, the functional $\mathrm{\Lambda }:=\tilde{\mathrm{\ell }}\circ P$ is linear with $\mathrm{\mid }\mathrm{\Lambda }(y)\mathrm{\mid }\le \beta \mathrm{\parallel }y\mathrm{\parallel }$ on $H_0$, so [F4] gives a unique $u\in H_0$ with $\mathrm{\Lambda }(y)=⟨y,u⟩$ for all $y\in H_0$ and $\mathrm{\parallel }u\mathrm{\parallel }=\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }\le \beta =⟨f,g{⟩}^{1\mathrm{/}2}$.

By step 1.1 the subspace $\ker T$ is closed, so [F3] decomposes the solution $u$ of step 6.1 as $u=u_1+u_2$ with $u_1\in \ker T$ and $u_2\in (\ker T{)}^{\perp }$; then $T{u}_2=Tu-T{u}_1=f$, so $u_2$ is a solution lying in $(\ker T{)}^{\perp }$, every solution has the form $u_2+k$ with $k\in \ker T$, and since $k\perp u_2$ the Pythagorean identity gives $\mathrm{\parallel }{u}_2+k{\mathrm{\parallel }}^2=\mathrm{\parallel }{u}_2{\mathrm{\parallel }}^2+\mathrm{\parallel }k{\mathrm{\parallel }}^2\ge \mathrm{\parallel }{u}_2{\mathrm{\parallel }}^2$ with equality only for $k=0$; thus $u_2$ is the unique least-norm solution and $\mathrm{\parallel }{u}_2\mathrm{\parallel }\le \mathrm{\parallel }u\mathrm{\parallel }\le C^{-1\mathrm{/}2}\mathrm{\parallel }f\mathrm{\parallel }$ by step 5.1, proving (ii) and (iii).

For every $x\in D(T^{\ast })$, step 6.2 gives $⟨T^{\ast }x,u⟩=⟨x,f⟩$. As in step 6.1, the graph identity of step 1.3 and closedness of $T$ yield $(u,f)\in \mathrm{\Gamma }(T)$, so $u\in D(T)$ and $Tu=f$. This proves existence in (iv).

By step 1.1 the subspace $\ker T$ is closed, so [F3] decomposes the solution $u$ of step 7.2 as $u=u_1+u_0$ with $u_1\in \ker T$ and $u_0\in (\ker T{)}^{\perp }$; then $T{u}_0=Tu-T{u}_1=f$ and the argument of step 7.1 shows that $u_0$ is the unique least-norm solution, with $\mathrm{\parallel }{u}_0\mathrm{\parallel }\le \mathrm{\parallel }u\mathrm{\parallel }\le ⟨f,g{⟩}^{1\mathrm{/}2}$ by step 6.2, hence $\mathrm{\parallel }{u}_0{\mathrm{\parallel }}^2\le ⟨f,g⟩$, which proves (iv); moreover with $A:=C\mathrm{i}\mathrm{d}$ and $g:=f\mathrm{/}C$ the hypotheses of (iv) hold by the coercivity assumption, and its conclusion specializes to $\mathrm{\parallel }{u}_0{\mathrm{\parallel }}^2\le ⟨f,f\mathrm{/}C⟩=C^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}^2$, in agreement with (iii).