[proofplan] We localise the prescribed polynomial near the point $a$ by multiplying it by a smooth cutoff that is identically $1$ near $a$. The resulting function is square-integrable but not holomorphic; its $\bar{\partial}$-error is supported away from $a$. We solve this $\bar{\partial}$-equation using Hörmander’s $L^2$ estimate with a plurisubharmonic logarithmic pole at $a$. The singular weight forces the solution to vanish to order at least $m+1$ at $a$, so subtracting the solution removes the $\bar{\partial}$-error without changing the prescribed $m$-jet. [/proofplan]

[step:Choose a cutoff that preserves the polynomial near $a$] Since $\Omega$ is open and $a\in\Omega$, choose $r>0$ such that $\overline{B}(a,r)\subset\Omega$, where \begin{align*} B(a,r):=\{z\in\mathbb{C}^n: |z-a|<r\}. \end{align*} Choose a function \begin{align*} \chi:\mathbb{C}^n\to[0,1] \end{align*} with $\chi\in C_c^\infty(B(a,r))$ and $\chi=1$ on $B(a,r/2)$. Define \begin{align*} F:\Omega\to\mathbb{C},\qquad F(z):=\chi(z)P(z-a). \end{align*} The function $F$ is smooth and compactly supported in $B(a,r)$, hence \begin{align*} F\in L^2(\Omega,\mathcal{B}(\Omega),\mathcal{L}^{2n}). \end{align*} Moreover, on $B(a,r/2)$ we have $F(z)=P(z-a)$. [/step]

[step:Express the failure of holomorphicity as a compactly supported $\bar{\partial}$-closed form] Define the $(0,1)$-form \begin{align*} g:=\bar{\partial}F. \end{align*} Since $P(z-a)$ is holomorphic in $z$, we have \begin{align*} g=P(z-a)\bar{\partial}\chi. \end{align*} Thus $g$ is a smooth $(0,1)$-form compactly supported in the annulus \begin{align*} K:=\operatorname{supp}\bar{\partial}\chi\subset B(a,r)\setminus \overline{B}(a,r/2). \end{align*} In particular, there is no singular behaviour of $g$ at $a$. Also, \begin{align*} \bar{\partial}g=\bar{\partial}^2F=0. \end{align*} [/step]

[step:Solve the $\bar{\partial}$ equation with a logarithmic pole at $a$] Let \begin{align*} N:=m+n. \end{align*} Define the plurisubharmonic weight \begin{align*} \varphi:\Omega\to[-\infty,\infty),\qquad \varphi(z):=N\log |z-a|^2. \end{align*} The function $\varphi$ is plurisubharmonic because $z\mapsto \log |z-a|^2$ is plurisubharmonic on $\mathbb{C}^n$. Since $g$ is supported in $K$ and $|z-a|\ge r/2$ on $K$, the weighted norm of $g$ is finite: \begin{align*} \int_\Omega |g(z)|^2 e^{-\varphi(z)}\,d\mathcal{L}^{2n}(z) = \int_K |g(z)|^2 |z-a|^{-2N}\,d\mathcal{L}^{2n}(z) <\infty. \end{align*} By Hörmander’s $L^2$ existence theorem for the $\bar{\partial}$-equation on bounded pseudoconvex domains with possibly singular plurisubharmonic weights (citing a result not yet in the wiki: Hörmander $L^2$ estimate for $\bar{\partial}$), applied to the $\bar{\partial}$-closed $(0,1)$-form $g$ and the plurisubharmonic weight $\varphi$, there exists \begin{align*} u\in L^2_{\mathrm{loc}}(\Omega) \end{align*} such that \begin{align*} \bar{\partial}u=g \end{align*} in the sense of distributions and \begin{align*} \int_\Omega |u(z)|^2 e^{-\varphi(z)}\,d\mathcal{L}^{2n}(z) = \int_\Omega |u(z)|^2 |z-a|^{-2N}\,d\mathcal{L}^{2n}(z) <\infty. \end{align*} Because $\Omega$ is bounded, there is a constant \begin{align*} R:=\sup_{z\in\Omega}|z-a|<\infty. \end{align*} Then $|z-a|^{-2N}\ge R^{-2N}$ on $\Omega$, so \begin{align*} \int_\Omega |u(z)|^2\,d\mathcal{L}^{2n}(z) \le R^{2N}\int_\Omega |u(z)|^2 |z-a|^{-2N}\,d\mathcal{L}^{2n}(z) <\infty. \end{align*} Hence $u\in L^2(\Omega,\mathcal{B}(\Omega),\mathcal{L}^{2n})$. [/step]

[step:Show that the weighted estimate forces $u$ to vanish to order at least $m+1$ at $a$] On $B(a,r/2)$, the form $g$ vanishes because $\chi=1$ there. Thus \begin{align*} \bar{\partial}u=0 \end{align*} in the sense of distributions on $B(a,r/2)$. By the local Weyl lemma for $\bar{\partial}$ (citing a result not yet in the wiki: distributional $\bar{\partial}$-closed $L^2_{\mathrm{loc}}$ functions are holomorphic), $u$ has a holomorphic representative on $B(a,r/2)$. We claim that every Taylor coefficient of $u$ at $a$ of total degree at most $m$ is zero. Suppose not. Let $k\le m$ be the smallest total degree for which the homogeneous degree-$k$ part of the Taylor expansion of $u$ at $a$ is nonzero. Then there is a nonzero homogeneous holomorphic polynomial $Q_k$ of degree $k$ and a holomorphic remainder function $R_{k+1}:B(0,r/2)\to\mathbb{C}$ such that \begin{align*} u(a+\zeta)=Q_k(\zeta)+R_{k+1}(\zeta), \qquad \lim_{\zeta\to0}\frac{|R_{k+1}(\zeta)|}{|\zeta|^k}=0. \end{align*} Equivalently, after writing $\zeta=\rho\omega$ with $\rho>0$ and $|\omega|=1$, the convergence \begin{align*} \rho^{-k}u(a+\rho\omega)\to Q_k(\omega) \end{align*} is uniform for $\omega$ on the unit sphere. Let \begin{align*} S^{2n-1}:=\{\omega\in\mathbb{C}^n:|\omega|=1\}. \end{align*} Since $Q_k$ is not identically zero, the continuous function $|Q_k|^2$ is not identically zero on $S^{2n-1}$, and therefore \begin{align*} A_k:=\int_{S^{2n-1}} |Q_k(\omega)|^2\,d\mathcal{H}^{2n-1}(\omega)>0. \end{align*} Using polar coordinates $\zeta=\rho\omega$, where $\rho\in(0,r/2)$ and $\omega\in S^{2n-1}$, the Lebesgue measure decomposes as \begin{align*} d\mathcal{L}^{2n}(\zeta)=\rho^{2n-1}\,d\mathcal{H}^{2n-1}(\omega)\,d\mathcal{L}^1(\rho). \end{align*} The [uniform convergence](/page/Uniform%20Convergence) on $S^{2n-1}$ implies convergence in $L^2(S^{2n-1},\mathcal{H}^{2n-1})$, so for sufficiently small $\rho>0$, \begin{align*} \int_{S^{2n-1}} |u(a+\rho\omega)|^2\,d\mathcal{H}^{2n-1}(\omega) \ge \frac{A_k}{2}\rho^{2k}. \end{align*} Choose $\varepsilon\in(0,r/2)$ so that this lower bound holds for every $\rho\in(0,\varepsilon)$. Since the integrand is nonnegative, Tonelli's theorem justifies the iterated integral after applying the polar-coordinate change of variables. Thus \begin{align*} \int_{B(a,\varepsilon)} |u(z)|^2|z-a|^{-2N}\,d\mathcal{L}^{2n}(z) &= \int_0^\varepsilon \int_{S^{2n-1}} |u(a+\rho\omega)|^2 \rho^{-2N} \rho^{2n-1} \,d\mathcal{H}^{2n-1}(\omega)\,d\mathcal{L}^1(\rho)\\ &\ge \frac{A_k}{2} \int_0^\varepsilon \rho^{2k-2N+2n-1}\,d\mathcal{L}^1(\rho). \end{align*} Since $N=m+n$ and $k\le m$, \begin{align*} 2k-2N+2n-1 = 2k-2m-1 \le -1. \end{align*} The last one-dimensional integral diverges. This contradicts the weighted integrability of $u$. Hence $u$ vanishes at $a$ to order at least $m+1$. [/step]

[step:Subtract the correction and recover the prescribed jet] Define \begin{align*} f:\Omega\to\mathbb{C},\qquad f(z):=F(z)-u(z). \end{align*} Since $F,u\in L^2(\Omega,\mathcal{B}(\Omega),\mathcal{L}^{2n})$, we have \begin{align*} f\in L^2(\Omega,\mathcal{B}(\Omega),\mathcal{L}^{2n}). \end{align*} Moreover, \begin{align*} \bar{\partial}f=\bar{\partial}F-\bar{\partial}u=g-g=0 \end{align*} in the sense of distributions on $\Omega$. Applying again the local Weyl lemma for $\bar{\partial}$, $f$ is holomorphic on $\Omega$. Thus \begin{align*} f\in A^2(\Omega). \end{align*} Finally, on $B(a,r/2)$ we have $F(z)=P(z-a)$, while the previous step shows that $u$ vanishes to order at least $m+1$ at $a$. Therefore \begin{align*} f(z)-P(z-a)=-u(z) \end{align*} vanishes to order at least $m+1$ at $a$. Equivalently, \begin{align*} \partial_z^\alpha f(a)=\partial_z^\alpha P(0) \end{align*} for every multi-index $\alpha\in\mathbb{N}_0^n$ with $|\alpha|\le m$. This is exactly the required local jet realisation. [/step]