[proofplan] We construct $u$ by duality. The a priori estimate makes the assignment $T^* g \mapsto (f, g)_{\mathcal{H}_2}$ a well-defined bounded linear functional $\ell$ on the subspace $T^*\bigl(\operatorname{Dom}(T^*) \cap \ker S\bigr) \subseteq \mathcal{H}_1$, with operator norm at most $C\|f\|$. By the [Norm-Preserving Extension of Linear Functionals](/theorems/880), $\ell$ extends to a bounded functional $\tilde\ell$ on all of $\mathcal{H}_1$ with the same norm; the [Riesz Representation Theorem](/theorems/221) then yields $u \in \mathcal{H}_1$ representing $\tilde\ell$, with $\|u\| \le C\|f\|$. To verify $Tu = f$ we test against arbitrary $g \in \operatorname{Dom}(T^*)$: the closed complex condition $\operatorname{Range}(T) \subseteq \ker S$ forces $T^*$ to vanish on $(\ker S)^\perp \cap \operatorname{Dom}(T^*)$, which lets us reduce to $g \in \ker S$ and identify $(T^* g, u) = (f, g)$. [/proofplan]

[step:Define a linear functional on $T^*\bigl(\operatorname{Dom}(T^*) \cap \ker S\bigr)$]Set \begin{align*} V := T^*\bigl(\operatorname{Dom}(T^*) \cap \ker S\bigr) \subseteq \mathcal{H}_1 \end{align*} and define \begin{align*} \ell : V &\to \mathbb{C}, \\ T^* g &\mapsto (f, g)_{\mathcal{H}_2}, \qquad g \in \operatorname{Dom}(T^*) \cap \ker S. \end{align*} We verify that $\ell$ is well defined, i.e., that $\ell(T^* g)$ depends only on $T^* g$ and not on the chosen $g$. Suppose $g, g' \in \operatorname{Dom}(T^*) \cap \ker S$ satisfy $T^* g = T^* g'$. Then $h := g - g'$ lies in $\operatorname{Dom}(T^*) \cap \ker S$ (since both intersected sets are linear subspaces) and $T^* h = 0$. Applying the hypothesised a priori estimate to $h$: \begin{align*} \|h\|_{\mathcal{H}_2} \le C\, \|T^* h\|_{\mathcal{H}_1} = 0, \end{align*} hence $h = 0$, i.e., $g = g'$. In particular $(f, g)_{\mathcal{H}_2} = (f, g')_{\mathcal{H}_2}$, so $\ell$ is unambiguously defined on $V$. Linearity of $\ell$ on $V$ follows immediately from linearity of $T^*$ on its domain and of the inner product in its first argument.[/step]

[guided]The key technical issue at this stage is well-definedness: the rule $T^* g \mapsto (f, g)$ assigns a value to an element of $V$ by choosing a *representative* $g$, and we must check that different representatives give the same value. Suppose $g, g' \in \operatorname{Dom}(T^*) \cap \ker S$ produce the same image, $T^* g = T^* g'$. Their difference $h := g - g'$ lies in $\operatorname{Dom}(T^*) \cap \ker S$ (an intersection of two linear subspaces, hence a linear subspace) and satisfies $T^* h = T^* g - T^* g' = 0$. This is precisely the situation in which the a priori estimate applies: the estimate \begin{align*} \|h\|_{\mathcal{H}_2} \le C\, \|T^* h\|_{\mathcal{H}_1} \end{align*} forces $\|h\| \le 0$, so $h = 0$ and $g = g'$. Note that the estimate is doing real work here: it is the *only* mechanism we have for converting information about $T^* g$ (a vector in $\mathcal{H}_1$) back into information about $g$ (a vector in $\mathcal{H}_2$). Without it, the map $g \mapsto T^* g$ could collapse genuinely distinct elements of $\operatorname{Dom}(T^*) \cap \ker S$ to the same vector, and the value $(f, g)$ would depend on the representative. In fact this argument shows something stronger: $T^*\big|_{\operatorname{Dom}(T^*) \cap \ker S}$ is *injective*, which is why we can pull the inverse rule $T^* g \mapsto (f, g)$ apart at all. Linearity of $\ell$ then follows: if $T^* g_1, T^* g_2 \in V$ and $\alpha, \beta \in \mathbb{C}$, then $\alpha g_1 + \beta g_2 \in \operatorname{Dom}(T^*) \cap \ker S$ (linear subspace) with $T^*(\alpha g_1 + \beta g_2) = \alpha T^* g_1 + \beta T^* g_2$, and \begin{align*} \ell(\alpha T^* g_1 + \beta T^* g_2) = (f, \alpha g_1 + \beta g_2)_{\mathcal{H}_2} = \bar\alpha\, \ell(T^* g_1) + \bar\beta\, \ell(T^* g_2) \end{align*} (or with $\alpha, \beta$ in place of $\bar\alpha, \bar\beta$ depending on the linearity convention for the inner product — in either case $\ell$ is a linear or conjugate-linear functional, which suffices for the subsequent Hahn–Banach argument).[/guided]

[step:Bound $\ell$ by $C\|f\|$ on $V$] By the [Cauchy-Schwarz Inequality](/theorems/432) in $\mathcal{H}_2$ and the a priori estimate applied to $g \in \operatorname{Dom}(T^*) \cap \ker S$: \begin{align*} |\ell(T^* g)| = |(f, g)_{\mathcal{H}_2}| \le \|f\|_{\mathcal{H}_2}\, \|g\|_{\mathcal{H}_2} \le C\, \|f\|_{\mathcal{H}_2}\, \|T^* g\|_{\mathcal{H}_1}. \end{align*} Thus $\ell$ is a bounded linear functional on $V$ (with respect to the $\mathcal{H}_1$-norm inherited as a subspace) with operator norm \begin{align*} \|\ell\|_{V^*} := \sup_{0 \ne v \in V} \frac{|\ell(v)|}{\|v\|_{\mathcal{H}_1}} \le C\, \|f\|_{\mathcal{H}_2}. \end{align*} [/step]

[step:Extend $\ell$ to a bounded functional on $\mathcal{H}_1$ via Hahn–Banach]We apply the [Norm-Preserving Extension of Linear Functionals](/theorems/880) (the normed-space [Hahn–Banach theorem](/theorems/2751)) to the bounded linear functional $\ell \in V^*$ on the linear subspace $V \subseteq \mathcal{H}_1$. The hypotheses of the theorem are: $\mathcal{H}_1$ is a normed space (indeed a [Hilbert space](/page/Hilbert%20Space)), $V$ is a linear subspace, and $\ell$ is a bounded linear functional on $V$. All three hold by Steps 1 and 2. The conclusion yields a bounded linear functional \begin{align*} \tilde\ell : \mathcal{H}_1 \to \mathbb{C} \end{align*} with $\tilde\ell\big|_V = \ell$ and \begin{align*} \|\tilde\ell\|_{\mathcal{H}_1^*} = \|\ell\|_{V^*} \le C\, \|f\|_{\mathcal{H}_2}. \end{align*}[/step]

[guided]Hahn–Banach is needed because [Riesz representation](/theorems/67) in the next step requires a functional defined on the *whole* [Hilbert space](/page/Hilbert%20Space), not merely on a subspace. The subspace $V = T^*(\operatorname{Dom}(T^*) \cap \ker S)$ has no reason to be closed, much less all of $\mathcal{H}_1$, so we cannot apply Riesz directly to $\ell$. The [Hahn–Banach theorem](/theorems/2751)'s defining feature — and what makes it indispensable here — is that the extension $\tilde\ell$ has *the same norm* as the original $\ell$. This is what propagates the bound $C\|f\|$ from Step 2 all the way to the final estimate $\|u\| \le C\|f\|$. A generic extension (say, by setting $\tilde\ell = 0$ on the complement of $\overline{V}$) would lose this control and break the proof. The complex case of Hahn–Banach (which we invoke since $\ell$ is potentially complex-valued) reduces to the real case applied to $\operatorname{Re}\,\ell$, and rebuilds the complex functional via $\tilde\ell(v) = \widetilde{\operatorname{Re}\,\ell}(v) - i\,\widetilde{\operatorname{Re}\,\ell}(iv)$; the bound on the norm is preserved by a factor we absorb into $\|\ell\|$. Theorem 880 handles this packaging.[/guided]

[step:Represent $\tilde\ell$ via Riesz to obtain a candidate $u$] The [Hilbert space](/page/Hilbert%20Space) $\mathcal{H}_1$ satisfies the hypothesis of the [Riesz Representation Theorem](/theorems/221) (every [Hilbert space](/page/Hilbert%20Space) has the property that bounded linear functionals are represented by inner products with a unique vector). Applying it to $\tilde\ell$ produces a unique $u \in \mathcal{H}_1$ such that \begin{align*} \tilde\ell(\phi) = (\phi, u)_{\mathcal{H}_1} \qquad \text{for all } \phi \in \mathcal{H}_1, \end{align*} and \begin{align*} \|u\|_{\mathcal{H}_1} = \|\tilde\ell\|_{\mathcal{H}_1^*} \le C\, \|f\|_{\mathcal{H}_2}. \end{align*} This is the norm bound claimed in the theorem; it remains to verify $u \in \operatorname{Dom}(T)$ and $Tu = f$. [/step]

[step:Reduce arbitrary $g \in \operatorname{Dom}(T^*)$ to its $\ker S$-component]To identify $Tu$ we will pair $u$ against $T^* g$ for $g$ ranging over $\operatorname{Dom}(T^*)$ — but $\ell$ was only defined for $g \in \ker S$. We bridge this gap using the [orthogonal decomposition](/theorems/436) by $\ker S$. Since $S$ is closed, $\ker S$ is a closed linear subspace of $\mathcal{H}_2$. By the [Orthogonal Decomposition Theorem](/theorems/241), \begin{align*} \mathcal{H}_2 = \ker S \oplus (\ker S)^\perp, \end{align*} so every $g \in \mathcal{H}_2$ admits a unique decomposition \begin{align*} g = g_1 + g_2, \qquad g_1 \in \ker S,\ g_2 \in (\ker S)^\perp. \end{align*} We claim that if $g \in \operatorname{Dom}(T^*)$, then both $g_1, g_2 \in \operatorname{Dom}(T^*)$, with \begin{align*} T^* g_2 = 0 \quad\text{and}\quad T^* g_1 = T^* g. \end{align*} Indeed, for any $v \in \operatorname{Dom}(T)$ we have $Tv \in \operatorname{Range}(T) \subseteq \ker S$ by the complex condition, hence $(g_2, Tv)_{\mathcal{H}_2} = 0$ since $g_2 \perp \ker S$. By definition of the adjoint, the map $v \mapsto (g_2, Tv)_{\mathcal{H}_2}$ is bounded (it is identically zero), so $g_2 \in \operatorname{Dom}(T^*)$ with $T^* g_2 = 0$. Then $g_1 = g - g_2 \in \operatorname{Dom}(T^*)$ as a difference of elements of the linear subspace $\operatorname{Dom}(T^*)$, and $T^* g_1 = T^* g - T^* g_2 = T^* g$. Moreover, since $f \in \ker S$ and $g_2 \in (\ker S)^\perp$: \begin{align*} (f, g_2)_{\mathcal{H}_2} = 0, \qquad (f, g)_{\mathcal{H}_2} = (f, g_1)_{\mathcal{H}_2}. \end{align*}[/step]

[guided]The mismatch is structural: our functional $\ell$ is built to "see" only the part of $\mathcal{H}_2$ inside $\ker S$, but identifying $Tu$ requires testing against *all* $g \in \operatorname{Dom}(T^*)$. The complex condition $\operatorname{Range}(T) \subseteq \ker S$ is precisely what allows us to discard the $(\ker S)^\perp$ component of any test element. To make this rigorous, we need $\ker S$ to be closed (so the [orthogonal decomposition](/theorems/436) exists), which holds because $S$ is a closed operator: $\ker S = S^{-1}(\{0\})$ is the preimage of a [closed set](/page/Closed%20Set) under a closed map, equivalently the intersection of the graph of $S$ with $\mathcal{H}_2 \times \{0\}$, projected to $\mathcal{H}_2$, and one verifies this is closed in $\mathcal{H}_2$. The decisive computation is showing $T^* g_2 = 0$ for $g_2 \in (\ker S)^\perp \cap \operatorname{Dom}(T^*)$. The argument is: for any $v \in \operatorname{Dom}(T)$, \begin{align*} (g_2, Tv)_{\mathcal{H}_2} = 0 \end{align*} because $Tv \in \operatorname{Range}(T) \subseteq \ker S$ and $g_2 \perp \ker S$. By definition of the adjoint, $g_2 \in \operatorname{Dom}(T^*)$ iff the linear functional $v \mapsto (g_2, Tv)$ is bounded on $\operatorname{Dom}(T)$ in the norm of $\mathcal{H}_1$; here the functional is *identically zero*, hence bounded with bound $0$, so $g_2 \in \operatorname{Dom}(T^*)$ and the Riesz vector representing it (which is $T^* g_2$) is $0$. (Note this is automatic and does not require any further hypothesis on $g_2$ beyond lying in $(\ker S)^\perp$ — but here we additionally know $g_2 \in \operatorname{Dom}(T^*)$ by subtraction, and we have just re-derived $T^* g_2 = 0$.) Symmetrically, since $f \in \ker S$, $f \perp g_2$, so $(f, g_2) = 0$ and $(f, g) = (f, g_1)$. This means the right-hand side $(f, g)$ of the equation we are about to test depends only on $g_1$, exactly matching the dependence of $T^* g = T^* g_1$ on the left. The reduction is therefore consistent.[/guided]

[step:Test against $T^* g$ for $g \in \operatorname{Dom}(T^*)$ to conclude $u \in \operatorname{Dom}(T)$ and $Tu = f$]Fix $g \in \operatorname{Dom}(T^*)$ and let $g_1 \in \operatorname{Dom}(T^*) \cap \ker S$ be its component as in Step 5, so that $T^* g_1 = T^* g$ and $(f, g_1)_{\mathcal{H}_2} = (f, g)_{\mathcal{H}_2}$. Then $T^* g_1 \in V$, and using $\tilde\ell\big|_V = \ell$: \begin{align*} (T^* g, u)_{\mathcal{H}_1} = (T^* g_1, u)_{\mathcal{H}_1} = \tilde\ell(T^* g_1) = \ell(T^* g_1) = (f, g_1)_{\mathcal{H}_2} = (f, g)_{\mathcal{H}_2}. \end{align*} Thus, for every $g \in \operatorname{Dom}(T^*)$, \begin{align*} (u, T^* g)_{\mathcal{H}_1} = \overline{(T^* g, u)_{\mathcal{H}_1}} = \overline{(f, g)_{\mathcal{H}_2}} = (g, f)_{\mathcal{H}_2}. \end{align*} In particular, \begin{align*} |(u, T^* g)_{\mathcal{H}_1}| = |(f, g)_{\mathcal{H}_2}| \le \|f\|_{\mathcal{H}_2}\, \|g\|_{\mathcal{H}_2}, \end{align*} so the map $g \mapsto (u, T^* g)_{\mathcal{H}_1}$ is bounded on $\operatorname{Dom}(T^*)$ in the $\mathcal{H}_2$-norm. By definition of the adjoint of $T^*$, this is exactly the condition $u \in \operatorname{Dom}((T^*)^*) = \operatorname{Dom}(T^{**})$, with \begin{align*} (T^{**} u, g)_{\mathcal{H}_2} = (u, T^* g)_{\mathcal{H}_1} = (g, f)_{\mathcal{H}_2}^{*} \end{align*} for all $g \in \operatorname{Dom}(T^*)$, equivalently $(T^{**} u - f, g)_{\mathcal{H}_2} = 0$ for all $g$ in the dense subspace $\operatorname{Dom}(T^*) \subseteq \mathcal{H}_2$ (densely defined because $T$ is densely defined and closed). Hence $T^{**} u = f$. Since $T$ is densely defined and closed, $T^{**} = T$, so $u \in \operatorname{Dom}(T)$ and $Tu = f$. Combined with the bound $\|u\|_{\mathcal{H}_1} \le C\|f\|_{\mathcal{H}_2}$ from Step 4, this is the conclusion of the theorem. $\blacksquare$[/step]

[guided]The final identification rests on three standard facts about closed densely defined operators on Hilbert spaces: (i) **The adjoint $T^*$ is densely defined.** Because $T$ is densely defined and closed, $T^*$ is closed and densely defined; this is the symmetric statement that closed densely defined operators have closed densely defined adjoints. (ii) **The double adjoint equals the original: $T^{**} = T$.** This holds for any densely defined closed operator on a [Hilbert space](/page/Hilbert%20Space). (iii) **Characterisation of $\operatorname{Dom}(T^{**})$.** A vector $u \in \mathcal{H}_1$ lies in $\operatorname{Dom}((T^*)^*)$ precisely when the linear functional $g \mapsto (u, T^* g)_{\mathcal{H}_1}$ is bounded on $\operatorname{Dom}(T^*)$ in the $\mathcal{H}_2$-norm; if so, the unique vector $w$ satisfying $(u, T^* g) = (w, g)$ for all $g \in \operatorname{Dom}(T^*)$ is $w = T^{**} u$. We have established in Step 6 that $(u, T^* g) = (g, f)^*$ (in inner-product conventions where the inner product is linear in the first slot) — equivalently $(T^* g, u) = (f, g)$. Reading off the boundedness from $|(f, g)| \le \|f\| \|g\|$ gives $u \in \operatorname{Dom}(T^{**})$. Reading off the representing vector $f$ gives $T^{**} u = f$. Combining with $T^{**} = T$ yields $Tu = f$. The norm bound $\|u\| \le C\|f\|$ was already established in Step 4 by the [Riesz representation](/theorems/67), with the constant $C$ tracing all the way back to the a priori estimate via Hahn–Banach norm preservation. No further estimate is needed. The proof is non-constructive in two essential ways: Hahn–Banach (which in general requires the [axiom of choice](/page/Axiom%20of%20Choice) for the extension to exist) and [Riesz representation](/theorems/67) (which is constructive, but only after the functional has been extended). Both opacities are inherent to the duality strategy; they are the cost of converting an a priori estimate (a one-way bound) into actual solvability.[/guided]