[proofplan] The proof is the standard correction argument for solving extension problems by the $\bar\partial$ equation. We subtract from the preliminary smooth extension $\widetilde F$ a solution $u$ of the same $\bar\partial$ equation, so the difference has vanishing distributional $\bar\partial$. Local Weyl regularity for the $\bar\partial$ operator then upgrades this weak holomorphicity to genuine holomorphicity. Finally, the imposed vanishing of $u$ along $H$ preserves the prescribed boundary value $f$. [/proofplan] [step:Define the corrected extension as a locally square-integrable function] Define the pointwise corrected function \begin{align*} F_0: \Omega &\to \mathbb{C} \\ z &\mapsto \widetilde F(z) - u(z). \end{align*} Since $\widetilde F: \Omega \to \mathbb{C}$ is smooth and $u: \Omega \to \mathbb{C}$ is continuous, $F_0$ is continuous on $\Omega$. Since $\widetilde F \in C^\infty(\Omega) \subset L^2_{\mathrm{loc}}(\Omega)$ and $u \in L^2_{\mathrm{loc}}(\Omega)$, the same function also defines an element of $L^2_{\mathrm{loc}}(\Omega)$. [guided] We first identify the object whose holomorphicity must be proved. The preliminary extension \begin{align*} \widetilde F: \Omega &\to \mathbb{C} \end{align*} is smooth, hence locally square-integrable on $\Omega$. The correction term is a continuous map \begin{align*} u: \Omega &\to \mathbb{C} \end{align*} which is also assumed to belong to $L^2_{\mathrm{loc}}(\Omega)$. Therefore their pointwise difference defines a continuous function \begin{align*} F_0: \Omega &\to \mathbb{C}, \\ z &\mapsto \widetilde F(z)-u(z), \end{align*} and the same function determines an element of $L^2_{\mathrm{loc}}(\Omega)$. At this point $F_0$ is continuous and locally square-integrable, not yet known to be holomorphic. The purpose of the next step is to show that its distributional Cauchy-Riemann equations vanish. [/guided] [/step] [step:Cancel the distributional $\bar\partial$ derivatives] The distributional $\bar\partial$ operator is linear on $L^2_{\mathrm{loc}}(\Omega)$. Hence, for the locally square-integrable function $F_0=\widetilde F-u$, \begin{align*} \bar\partial F_0 &= \bar\partial(\widetilde F - u) \\ &= \bar\partial \widetilde F - \bar\partial u \\ &= \bar\partial \widetilde F - \bar\partial \widetilde F \\ &= 0 \end{align*} in the sense of distributions on $\Omega$. [guided] The correction $u$ was chosen to have exactly the same distributional $\bar\partial$ derivative as the preliminary extension $\widetilde F$. Since the distributional $\bar\partial$ operator is linear, we compute in the space of distributions on $\Omega$: \begin{align*} \bar\partial F &= \bar\partial(\widetilde F - u) \\ &= \bar\partial \widetilde F - \bar\partial u. \end{align*} The hypothesis gives \begin{align*} \bar\partial u = \bar\partial \widetilde F, \end{align*} so substitution gives \begin{align*} \bar\partial F &= \bar\partial \widetilde F - \bar\partial \widetilde F \\ &= 0. \end{align*} Thus $F$ satisfies the Cauchy-Riemann equations weakly, that is, in the distributional sense. [/guided] [/step] [step:Upgrade weak holomorphicity to holomorphicity] Since $F_0 \in L^2_{\mathrm{loc}}(\Omega) \subset L^1_{\mathrm{loc}}(\Omega)$ and $\bar\partial F_0 = 0$ distributionally on $\Omega$, Weyl regularity for the $\bar\partial$ operator applies in the following precise form: if $G \in L^1_{\mathrm{loc}}(\Omega)$ and $\bar\partial G=0$ distributionally, then $G$ agrees almost everywhere on $\Omega$ with a [holomorphic function](/page/Holomorphic%20Function). Therefore $F_0$ agrees almost everywhere on $\Omega$ with a [holomorphic function](/page/Holomorphic%20Function) \begin{align*} F_{\mathrm{hol}}: \Omega &\to \mathbb{C}. \end{align*} Here $\mathcal{O}(\Omega)$ denotes the set of holomorphic maps from $\Omega$ to $\mathbb{C}$. [guided] The previous step proves only weak holomorphicity: \begin{align*} \bar\partial F = 0 \end{align*} as a distribution. To convert this into ordinary holomorphicity, we use Weyl regularity for the $\bar\partial$ operator (citing a result not yet in the wiki: Weyl regularity for the $\bar\partial$ operator). The hypotheses required for this regularity statement are precisely that $F$ is locally integrable, or in particular locally square-integrable, and that $\bar\partial F=0$ distributionally. We have already verified both conditions: \begin{align*} F \in L^2_{\mathrm{loc}}(\Omega) \end{align*} and \begin{align*} \bar\partial F = 0 \end{align*} in the sense of distributions. Therefore there exists a [holomorphic function](/page/Holomorphic%20Function) \begin{align*} F_{\mathrm{hol}}: \Omega &\to \mathbb{C} \end{align*} such that $F_{\mathrm{hol}} = F$ almost everywhere on $\Omega$. Since functions in $L^2_{\mathrm{loc}}$ are understood up to almost-everywhere equality, we replace $F$ by this holomorphic representative. [/guided] [/step] [step:Verify that the corrected extension has the prescribed restriction to $H$] First compare the holomorphic representative with the continuous pointwise correction. Define \begin{align*} E: \Omega &\to \mathbb{C} \\ z &\mapsto F_{\mathrm{hol}}(z)-F_0(z). \end{align*} The function $E$ is continuous because $F_{\mathrm{hol}}$ is holomorphic, hence continuous, and $F_0$ is continuous. Weyl regularity gave $E=0$ almost everywhere with respect to Lebesgue measure $\mathcal{L}^{2n}$ on $\Omega$. If there were a point $z_0\in\Omega$ with $E(z_0)\ne 0$, continuity would give an open ball $B(z_0,r)\subset\Omega$ on which $|E|>|E(z_0)|/2$, and this ball has positive $\mathcal{L}^{2n}$-measure, contradicting $E=0$ $\mathcal{L}^{2n}$-a.e. Hence $F_{\mathrm{hol}}=F_0$ pointwise on $\Omega$. Using the hypotheses $\widetilde F|_H=f$ and $u|_H=0$, and using the compatibility of restriction with subtraction for the pointwise continuous functions $\widetilde F$ and $u$, we obtain \begin{align*} F_{\mathrm{hol}}|_H &= F_0|_H \\ &= (\widetilde F-u)|_H \\ &= \widetilde F|_H-u|_H \\ &= f-0 \\ &= f. \end{align*} Thus the [holomorphic function](/page/Holomorphic%20Function) $F_{\mathrm{hol}}$ extends $f$ from $H$ to $\Omega$. [/step]