[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]
