[proofplan] We avoid any closed-range theorem by restricting $T$ to the closed subspace $K=\ker S$. On $K$, the assumed estimate becomes the lower bound $\|v\|_{H_2}\le C^{1/2}\|T^*v\|_{H_1}$ for vectors in the adjoint domain of the restricted operator. This lower bound makes the functional $T^*v\mapsto (f,v)_{H_2}$ well-defined and bounded with norm at most $C^{1/2}\|f\|_{H_2}$. Hahn-Banach, [Riesz representation](/theorems/67), and the double-adjoint theorem then produce a vector $u\in\operatorname{Dom}(T)$ with $Tu=f$ and the same constant. [/proofplan] [step:Restrict $T$ to the closed space of $S$-closed vectors] Define \begin{align*} K=\ker S=\{v\in\operatorname{Dom}(S)\mid Sv=0\}\subset H_2. \end{align*} Since $S$ is closed, $K$ is a closed subspace of $H_2$ with the inherited Hilbert norm and inner product. The hypotheses $\operatorname{Range}(T)\subset\operatorname{Dom}(S)$ and $ST=0$ on $\operatorname{Dom}(T)$ imply that \begin{align*} Tu\in K \end{align*} for every $u\in\operatorname{Dom}(T)$. Hence $T$ defines an operator \begin{align*} T_K:\operatorname{Dom}(T_K)=\operatorname{Dom}(T)\subset H_1&\to K\\ u&\mapsto Tu. \end{align*} The domain of $T_K$ is dense in $H_1$ because $T$ is densely defined. The operator $T_K$ is closed: if $u_m\in\operatorname{Dom}(T)$, $u_m\to u$ in $H_1$, and $T_Ku_m\to g$ in $K$, then $T u_m\to g$ in $H_2$, so the closedness of $T$ gives $u\in\operatorname{Dom}(T)$ and $Tu=g$. [/step] [step:Identify the adjoint of the restricted operator] The adjoint $T_K^*$ is \begin{align*} \operatorname{Dom}(T_K^*)=\operatorname{Dom}(T^*)\cap K, \qquad T_K^*v=T^*v. \end{align*} Indeed, if $v\in\operatorname{Dom}(T^*)\cap K$, then for every $u\in\operatorname{Dom}(T)$, \begin{align*} (T_Ku,v)_K &=(Tu,v)_{H_2}\\ &=(u,T^*v)_{H_1}, \end{align*} so $v\in\operatorname{Dom}(T_K^*)$ and $T_K^*v=T^*v$. Conversely, if $v\in\operatorname{Dom}(T_K^*)$, then $v\in K\subset H_2$ and there exists $w\in H_1$ such that \begin{align*} (Tu,v)_{H_2}=(T_Ku,v)_K=(u,w)_{H_1} \end{align*} for every $u\in\operatorname{Dom}(T)$. This is the defining condition for $v\in\operatorname{Dom}(T^*)$ with $T^*v=w$. For $v\in\operatorname{Dom}(T_K^*)$, we have $v\in K$, hence $Sv=0$. The assumed estimate therefore gives \begin{align*} \|v\|_{H_2}^2 &\le C\bigl(\|T^*v\|_{H_1}^2+\|Sv\|_{H_3}^2\bigr)\\ &=C\|T_K^*v\|_{H_1}^2. \end{align*} Taking square roots gives the explicit bound \begin{align*} \|v\|_{H_2}\le C^{1/2}\|T_K^*v\|_{H_1}. \end{align*} [/step] [step:Represent the bounded conjugate-linear functional] Fix $f\in\ker S=K$. Define the subspace \begin{align*} \mathcal R=T_K^*(\operatorname{Dom}(T_K^*))\subset H_1. \end{align*} Define a conjugate-linear functional $\Lambda:\mathcal R\to\mathbb C$ by \begin{align*} \Lambda(T_K^*v)=(f,v)_{H_2}, \qquad v\in\operatorname{Dom}(T_K^*). \end{align*} This definition is independent of the representative. If $T_K^*v=0$, then the estimate from the previous step gives \begin{align*} \|v\|_{H_2}\le C^{1/2}\|T_K^*v\|_{H_1}=0, \end{align*} so $v=0$ and $(f,v)_{H_2}=0$. The same estimate tracks the norm of $\Lambda$: \begin{align*} |\Lambda(T_K^*v)| &=|(f,v)_{H_2}|\\ &\le \|f\|_{H_2}\|v\|_{H_2}\\ &\le C^{1/2}\|f\|_{H_2}\|T_K^*v\|_{H_1}. \end{align*} The complex [Hahn-Banach theorem](/page/Hahn-Banach%20Theorem) for conjugate-linear functionals extends $\Lambda$ to a bounded conjugate-linear functional on all of $H_1$ with norm at most $C^{1/2}\|f\|_{H_2}$. The [Riesz representation theorem](/theorems/221) with the convention that inner products are linear in the first argument says that there is $u\in H_1$ such that \begin{align*} \Lambda(g)=(u,g)_{H_1} \end{align*} for every $g\in H_1$, and \begin{align*} \|u\|_{H_1}=\|\Lambda\|\le C^{1/2}\|f\|_{H_2}. \end{align*} Consequently, \begin{align*} (u,T_K^*v)_{H_1}=(f,v)_{H_2} \end{align*} for every $v\in\operatorname{Dom}(T_K^*)$. [/step] [step:Use the double-adjoint theorem to recover $Tu=f$] The last identity gives, after conjugate symmetry of the Hilbert inner products, \begin{align*} (T_K^*v,u)_{H_1}=(v,f)_{H_2} \end{align*} for every $v\in\operatorname{Dom}(T_K^*)$. This is the definition of membership in the domain of the double adjoint: \begin{align*} u\in\operatorname{Dom}\bigl((T_K^*)^*\bigr), \qquad (T_K^*)^*u=f. \end{align*} Here $f$ is viewed as an element of the [Hilbert space](/page/Hilbert%20Space) $K$. Since $T_K$ is densely defined and closed, the double-adjoint theorem for closed densely defined operators gives \begin{align*} (T_K^*)^*=T_K. \end{align*} Therefore \begin{align*} u\in\operatorname{Dom}(T_K)=\operatorname{Dom}(T), \qquad Tu=T_Ku=f. \end{align*} The norm estimate obtained from [Riesz representation](/theorems/67) is \begin{align*} \|u\|_{H_1}\le C^{1/2}\|f\|_{H_2}. \end{align*} Thus $f\in\operatorname{Range}(T)$, and the primitive lies in $\operatorname{Dom}(T)$ with the stated constant. [/step]