[proofplan] We first enlarge $K$ to a compact set $K_0$ whose polynomial hull remains inside $\Omega$, and we cut off $f$ by a smooth function equal to $1$ near $K_0$. The resulting $\bar\partial$-error is supported away from $K$, and polynomial convexity supplies a smooth subharmonic weight that is negative near $K$ and positive on that support. Hörmander's weighted $L^2$ estimate for $\bar\partial$ on $\mathbb C$ then produces entire functions converging to $f$ in $L^2(K)$. Finally, the Taylor polynomials of one of these entire functions give polynomial approximants. [/proofplan] [step:Enlarge $K$ and localize $f$ by a compactly supported cutoff] Here $d\mathcal L^2$ denotes two-dimensional Lebesgue measure on $\mathbb C\cong\mathbb R^2$, and $\widehat K_{0,\mathrm{poly}}$ denotes the polynomial hull of $K_0$. Since $\Omega$ is Runge, choose a compact set $K_0 \subset \Omega$ such that \begin{align*} K \subset K_0^\circ \subset K_0 \Subset \Omega, \qquad \widehat K_{0,\mathrm{poly}} \Subset \Omega. \end{align*} Define the polynomial hull of $K_0$ by \begin{align*} \widehat K_{0,\mathrm{poly}} := \left\{z \in \mathbb C : |q(z)| \le \sup_{w \in K_0}|q(w)| \text{ for every } q \in \mathbb C[z]\right\}. \end{align*} Choose an [open set](/page/Open%20Set) $V \subset \mathbb C$ with \begin{align*} \widehat K_{0,\mathrm{poly}} \Subset V \Subset \Omega. \end{align*} By smooth cutoff construction, choose \begin{align*} \chi: \mathbb C \to [0,1] \end{align*} with $\chi \in C_c^\infty(\Omega)$ and $\chi = 1$ on $V$. Define the smooth compactly supported function \begin{align*} g: \mathbb C &\to \mathbb C \\ z &\mapsto \begin{cases} \chi(z)f(z), & z \in \Omega,\\ 0, & z \in \mathbb C \setminus \operatorname{supp}\chi, \end{cases} \end{align*} where the definition is smooth because $\operatorname{supp}\chi \Subset \Omega$. Since $f$ is holomorphic on $\Omega$, the compactly supported $(0,1)$-form \begin{align*} \alpha = a\,d\bar z \end{align*} defined by \begin{align*} a: \mathbb C &\to \mathbb C \\ z &\mapsto \frac{\partial g}{\partial \bar z}(z) \end{align*} satisfies \begin{align*} a(z)= f(z)\frac{\partial \chi}{\partial \bar z}(z) \end{align*} for $z \in \Omega$. Because $\chi=1$ on $V$ and $\widehat K_{0,\mathrm{poly}} \Subset V$, we have \begin{align*} \operatorname{supp} a \cap \widehat K_{0,\mathrm{poly}} = \varnothing. \end{align*} In particular, $\operatorname{supp} a$ is a compact subset of $\Omega \setminus \widehat K_{0,\mathrm{poly}}$. [guided] The purpose of the cutoff is to turn the local [holomorphic function](/page/Holomorphic%20Function) $f$ into a globally defined smooth function on $\mathbb C$, at the cost of introducing a controlled $\bar\partial$-error. The measure $d\mathcal L^2$ is two-dimensional Lebesgue measure on $\mathbb C\cong\mathbb R^2$, and the notation $\widehat K_{0,\mathrm{poly}}$ means the polynomial hull of $K_0$. Since $\Omega$ is Runge, we may enlarge $K$ to a compact set $K_0$ with \begin{align*} K \subset K_0^\circ \subset K_0 \Subset \Omega, \qquad \widehat K_{0,\mathrm{poly}} \Subset \Omega. \end{align*} Define the polynomial hull of $K_0$ by \begin{align*} \widehat K_{0,\mathrm{poly}} := \left\{z \in \mathbb C : |q(z)| \le \sup_{w \in K_0}|q(w)| \text{ for every } q \in \mathbb C[z]\right\}. \end{align*} We then choose an [open set](/page/Open%20Set) $V \subset \mathbb C$ satisfying \begin{align*} \widehat K_{0,\mathrm{poly}} \Subset V \Subset \Omega, \end{align*} and a smooth cutoff \begin{align*} \chi: \mathbb C \to [0,1] \end{align*} with $\chi \in C_c^\infty(\Omega)$ and $\chi=1$ on $V$. Define \begin{align*} g: \mathbb C &\to \mathbb C \\ z &\mapsto \begin{cases} \chi(z)f(z), & z \in \Omega,\\ 0, & z \in \mathbb C \setminus \operatorname{supp}\chi. \end{cases} \end{align*} This is a globally smooth compactly supported function because $\operatorname{supp}\chi$ is compactly contained in $\Omega$. Its $\bar\partial$-derivative is the compactly supported $(0,1)$-form $\alpha=a\,d\bar z$, where \begin{align*} a: \mathbb C &\to \mathbb C \\ z &\mapsto \frac{\partial g}{\partial \bar z}(z). \end{align*} On $\Omega$, the product rule gives \begin{align*} a(z)=\frac{\partial}{\partial \bar z}(\chi f)(z) = f(z)\frac{\partial \chi}{\partial \bar z}(z) +\chi(z)\frac{\partial f}{\partial \bar z}(z). \end{align*} Since $f$ is holomorphic, $\partial f/\partial \bar z=0$, hence \begin{align*} a(z)=f(z)\frac{\partial\chi}{\partial \bar z}(z). \end{align*} Because $\chi=1$ on $V$, its $\bar\partial$-derivative vanishes on $V$. Since $\widehat K_{0,\mathrm{poly}} \Subset V$, this gives \begin{align*} \operatorname{supp}a \cap \widehat K_{0,\mathrm{poly}}=\varnothing. \end{align*} This separation is the geometric input needed for the weighted estimate: the error lies away from the entire polynomial hull of the set on which approximation is required. [/guided] [/step] [step:Construct a subharmonic weight separating $K$ from the $\bar\partial$-error] Let \begin{align*} L := \operatorname{supp} a. \end{align*} If $L=\varnothing$, define \begin{align*} \psi: \mathbb C &\to \mathbb R \\ z &\mapsto -1, \end{align*} and set $\eta:=1$. Then $\psi$ is smooth and subharmonic, $\psi\le -\eta$ on $K$, and $\psi\ge \eta$ on $L$ holds vacuously. Hence assume for the rest of this step that $L\ne\varnothing$. Since $L$ is compact and $L \cap \widehat K_{0,\mathrm{poly}}=\varnothing$, polynomial convexity gives, for each $\zeta \in L$, a polynomial \begin{align*} h_\zeta: \mathbb C \to \mathbb C \end{align*} such that \begin{align*} T_\zeta := |h_\zeta(\zeta)| > S_\zeta := \sup_{z \in K_0}|h_\zeta(z)|. \end{align*} Choose an integer $m_\zeta\ge1$ such that \begin{align*} T_\zeta^{m_\zeta}>4S_\zeta^{m_\zeta}, \end{align*} where this is automatic with $m_\zeta=1$ if $S_\zeta=0$, and otherwise follows from $(T_\zeta/S_\zeta)^{m_\zeta}\to\infty$. Choose a constant $\lambda_\zeta>0$ satisfying \begin{align*} \frac{2}{T_\zeta^{m_\zeta}}<\lambda_\zeta\le \frac{1}{2S_\zeta^{m_\zeta}}, \end{align*} with the upper bound interpreted as $+\infty$ when $S_\zeta=0$. This interval is non-empty precisely because $T_\zeta^{m_\zeta}>4S_\zeta^{m_\zeta}$. Define the rescaled polynomial \begin{align*} q_\zeta: \mathbb C &\to \mathbb C \\ z &\mapsto \lambda_\zeta h_\zeta(z)^{m_\zeta}. \end{align*} Then \begin{align*} \sup_{z \in K_0}|q_\zeta(z)| \le \lambda_\zeta S_\zeta^{m_\zeta} \le \frac{1}{2}, \qquad |q_\zeta(\zeta)| = \lambda_\zeta T_\zeta^{m_\zeta} > 2. \end{align*} By continuity and compactness of $L$, choose $\zeta_1,\dots,\zeta_N \in L$ such that, for every $z \in L$, at least one index $j \in \{1,\dots,N\}$ satisfies \begin{align*} |q_{\zeta_j}(z)| > 1. \end{align*} Choose an integer $M \ge 1$ such that \begin{align*} N4^{-M}<\frac14, \end{align*} and define polynomial maps \begin{align*} r_j: \mathbb C &\to \mathbb C \\ z &\mapsto q_{\zeta_j}(z)^M \end{align*} for $j\in\{1,\dots,N\}$. Then \begin{align*} \sum_{j=1}^N |r_j(z)|^2 < \frac14 \quad \text{for all } z \in K_0, \end{align*} while for every $z\in L$ at least one $j$ satisfies $|r_j(z)|>1$. Define \begin{align*} \Psi: \mathbb C &\to \mathbb R \\ z &\mapsto \log\left(\sum_{j=1}^N |r_j(z)|^2 + \frac{1}{4}\right). \end{align*} Then $\Psi$ is smooth and subharmonic. Moreover, \begin{align*} \Psi(z) < \log\left(\frac{1}{2}\right) \end{align*} on $K_0$, while \begin{align*} \Psi(z) > \log\left(\frac{5}{4}\right) \end{align*} on $L$. After replacing $\Psi$ by an affine rescaling \begin{align*} \psi: \mathbb C &\to \mathbb R \\ z &\mapsto A\Psi(z)-B \end{align*} with constants $A>0$ and $B \in \mathbb R$ chosen large enough, there exists $\eta>0$ such that \begin{align*} \psi(z) \le -\eta \quad \text{for all } z \in K, \qquad \psi(z) \ge \eta \quad \text{for all } z \in L. \end{align*} The affine rescaling preserves smoothness and subharmonicity. [guided] We need a weight that rewards smallness on $K$ while penalising the support of the $\bar\partial$-error. Set \begin{align*} L := \operatorname{supp}a. \end{align*} If $L=\varnothing$, there is no error support to separate. In that case define \begin{align*} \psi: \mathbb C &\to \mathbb R \\ z &\mapsto -1, \end{align*} and set $\eta:=1$. Then $\psi$ is smooth and subharmonic, $\psi\le -\eta$ on $K$, and the condition $\psi\ge\eta$ on $L$ is vacuous. We therefore assume from now on in this construction that $L\ne\varnothing$. The set $L$ is compact, and from the previous step it is disjoint from $\widehat K_{0,\mathrm{poly}}$. Since each point of $L$ lies outside the polynomial hull of $K_0$, polynomial convexity first gives only strict separation: for every $\zeta \in L$, there is a polynomial map \begin{align*} h_\zeta: \mathbb C \to \mathbb C \end{align*} such that \begin{align*} T_\zeta:=|h_\zeta(\zeta)|>S_\zeta:=\sup_{z\in K_0}|h_\zeta(z)|. \end{align*} To get the stronger numerical constants needed later, we first amplify the strict inequality by taking powers. Choose $m_\zeta\ge1$ so that \begin{align*} T_\zeta^{m_\zeta}>4S_\zeta^{m_\zeta}; \end{align*} if $S_\zeta=0$, this holds with $m_\zeta=1$, while if $S_\zeta>0$ it follows from $(T_\zeta/S_\zeta)^{m_\zeta}\to\infty$. Now choose $\lambda_\zeta>0$ with \begin{align*} \frac{2}{T_\zeta^{m_\zeta}}<\lambda_\zeta\le \frac{1}{2S_\zeta^{m_\zeta}}, \end{align*} interpreting the upper bound as $+\infty$ when $S_\zeta=0$. The interval is non-empty because $T_\zeta^{m_\zeta}>4S_\zeta^{m_\zeta}$. Define \begin{align*} q_\zeta: \mathbb C &\to \mathbb C \\ z &\mapsto \lambda_\zeta h_\zeta(z)^{m_\zeta}. \end{align*} Then the upper bound on $\lambda_\zeta$ gives \begin{align*} \sup_{z \in K_0}|q_\zeta(z)| \le \frac{1}{2}, \end{align*} and the lower bound gives \begin{align*} |q_\zeta(\zeta)| > 2. \end{align*} By continuity, the inequality $|q_\zeta|>1$ holds on an open neighbourhood of $\zeta$. Compactness of $L$ gives finitely many points $\zeta_1,\dots,\zeta_N \in L$ such that for every $z \in L$ there is some $j$ with \begin{align*} |q_{\zeta_j}(z)|>1. \end{align*} The finite number $N$ may be large, so the estimates must be normalised before summing. Choose an integer $M\ge1$ such that \begin{align*} N4^{-M}<\frac14, \end{align*} and define polynomial maps \begin{align*} r_j: \mathbb C &\to \mathbb C \\ z &\mapsto q_{\zeta_j}(z)^M \end{align*} for $j\in\{1,\dots,N\}$. On $K_0$, each $|q_{\zeta_j}|<1/2$, hence \begin{align*} \sum_{j=1}^N |r_j(z)|^2 = \sum_{j=1}^N |q_{\zeta_j}(z)|^{2M} < N4^{-M} < \frac14. \end{align*} On $L$, at least one $|q_{\zeta_j}(z)|$ is greater than $1$, so the corresponding $|r_j(z)|$ is also greater than $1$. Now define \begin{align*} \Psi: \mathbb C &\to \mathbb R \\ z &\mapsto \log\left(\sum_{j=1}^N |r_j(z)|^2+\frac{1}{4}\right). \end{align*} The function $\Psi$ is smooth because the quantity inside the logarithm is everywhere at least $1/4$. It is subharmonic because logarithms of sums of squared moduli of holomorphic functions, with a positive constant added, are subharmonic. On $K_0$ we have \begin{align*} \sum_{j=1}^N |r_j(z)|^2+\frac14 < \frac12, \end{align*} and hence \begin{align*} \Psi(z)<\log\left(\frac12\right) \end{align*} for $z \in K_0$. On the other hand, if $z \in L$, then one of the polynomials $r_j$ has modulus greater than $1$, so \begin{align*} \sum_{j=1}^N |r_j(z)|^2+\frac14 > \frac54, \end{align*} and therefore \begin{align*} \Psi(z)>\log\left(\frac54\right). \end{align*} Finally choose constants $A>0$ and $B \in \mathbb R$ so that \begin{align*} \psi(z):=A\Psi(z)-B \end{align*} satisfies, for some $\eta>0$, \begin{align*} \psi(z)\le -\eta \quad \text{on } K, \qquad \psi(z)\ge \eta \quad \text{on } L. \end{align*} Affine rescaling by a positive factor preserves subharmonicity, so $\psi$ is still a smooth subharmonic function. [/guided] [/step] [step:Solve the $\bar\partial$-equation with weights that force smallness on $K$] For each integer $m \ge 1$, define the smooth subharmonic weight \begin{align*} \varphi_m: \mathbb C &\to \mathbb R \\ z &\mapsto m\psi(z)+|z|^2. \end{align*} We use the following complete-plane form of Hörmander's weighted $L^2$ estimate for $\bar\partial$ on $\mathbb C$ (citing a result not yet in the wiki: Hörmander $L^2$ estimate for the $\bar\partial$-equation). This is the one-dimensional complete Kähler case with the standard Euclidean metric on $\mathbb C$ and the auxiliary exhaustion factor $(1+|z|^2)^{-2}$ included in the estimate. If $\varphi:\mathbb C\to\mathbb R$ is smooth and subharmonic, and if $a:\mathbb C\to\mathbb C$ is measurable with \begin{align*} \int_{\mathbb C}|a(z)|^2e^{-\varphi(z)}\,d\mathcal L^2(z)<\infty, \end{align*} then there exists $u\in L^2_{\mathrm{loc}}(\mathbb C)$ with $\partial u/\partial\bar z=a$ in the sense of distributions and \begin{align*} \int_{\mathbb C}\frac{|u(z)|^2 e^{-\varphi(z)}}{(1+|z|^2)^2}\,d\mathcal L^2(z) \le \int_{\mathbb C}|a(z)|^2 e^{-\varphi(z)}\,d\mathcal L^2(z). \end{align*} For $\varphi=\varphi_m$, the smooth subharmonicity has just been verified. Since $a$ is continuous with compact support $L$, the weighted integral is finite. Therefore there exists \begin{align*} u_m: \mathbb C \to \mathbb C \end{align*} with $u_m \in L^2_{\mathrm{loc}}(\mathbb C)$ and \begin{align*} \frac{\partial u_m}{\partial \bar z}=a \end{align*} in the sense of distributions, such that \begin{align*} \int_{\mathbb C}\frac{|u_m(z)|^2 e^{-\varphi_m(z)}}{(1+|z|^2)^2}\,d\mathcal L^2(z) \le \int_{\mathbb C}|a(z)|^2 e^{-\varphi_m(z)}\,d\mathcal L^2(z). \end{align*} Let \begin{align*} R_K := \sup_{z \in K}|z|, \qquad M_L := \begin{cases} \sup_{z \in L} e^{-|z|^2}|a(z)|^2, & L\ne\varnothing,\\ 0, & L=\varnothing. \end{cases} \end{align*} Since $\psi \ge \eta$ on $L$, the right-hand side satisfies \begin{align*} \int_{\mathbb C}|a(z)|^2e^{-\varphi_m(z)}\,d\mathcal L^2(z) = \int_L |a(z)|^2e^{-m\psi(z)-|z|^2}\,d\mathcal L^2(z) \le M_L\,\mathcal L^2(L)e^{-m\eta}. \end{align*} Since $\psi \le -\eta$ on $K$, for $z \in K$ we have \begin{align*} \frac{e^{-\varphi_m(z)}}{(1+|z|^2)^2} = \frac{e^{-m\psi(z)-|z|^2}}{(1+|z|^2)^2} \ge \frac{e^{m\eta-R_K^2}}{(1+R_K^2)^2}. \end{align*} Therefore \begin{align*} \int_K |u_m(z)|^2\,d\mathcal L^2(z) \le (1+R_K^2)^2 e^{R_K^2} M_L\mathcal L^2(L)e^{-2m\eta}. \end{align*} Thus \begin{align*} \lim_{m\to\infty}\int_K |u_m(z)|^2\,d\mathcal L^2(z)=0. \end{align*} [guided] For each integer $m\ge1$, define \begin{align*} \varphi_m: \mathbb C &\to \mathbb R \\ z &\mapsto m\psi(z)+|z|^2. \end{align*} The function $\varphi_m$ is smooth and subharmonic because $\psi$ is smooth and subharmonic, $|z|^2$ is smooth and subharmonic, and positive linear combinations preserve subharmonicity. We now apply the complete-plane form of Hörmander's weighted $L^2$ estimate for $\bar\partial$ on $\mathbb C$ (citing a result not yet in the wiki: Hörmander $L^2$ estimate for the $\bar\partial$-equation). This is the one-dimensional complete Kähler case for the standard Euclidean metric on $\mathbb C$; the auxiliary exhaustion in that formulation is responsible for the factor $(1+|z|^2)^{-2}$ on the left-hand side. The version used says that a smooth subharmonic weight $\varphi:\mathbb C\to\mathbb R$ and a measurable coefficient $a:\mathbb C\to\mathbb C$ with \begin{align*} \int_{\mathbb C}|a(z)|^2e^{-\varphi(z)}\,d\mathcal L^2(z)<\infty \end{align*} produce a distributional solution $u$ of $\partial u/\partial\bar z=a$ satisfying the displayed weighted estimate below. Here $\varphi_m$ is smooth and subharmonic, and $a$ is continuous with compact support $L$, so the weighted integral is finite for this fixed $m$. The theorem gives a function \begin{align*} u_m: \mathbb C \to \mathbb C \end{align*} with $u_m \in L^2_{\mathrm{loc}}(\mathbb C)$ and \begin{align*} \frac{\partial u_m}{\partial \bar z}=a \end{align*} in the distributional sense, satisfying \begin{align*} \int_{\mathbb C}\frac{|u_m(z)|^2 e^{-\varphi_m(z)}}{(1+|z|^2)^2}\,d\mathcal L^2(z) \le \int_{\mathbb C}|a(z)|^2 e^{-\varphi_m(z)}\,d\mathcal L^2(z). \end{align*} The point of the sign choice is now visible. On the support $L$ of $a$, the weight has $\psi \ge \eta$, so the right-hand side is exponentially small. Define \begin{align*} M_L := \begin{cases} \sup_{z \in L} e^{-|z|^2}|a(z)|^2, & L\ne\varnothing,\\ 0, & L=\varnothing. \end{cases} \end{align*} Since $L$ is compact and $a$ is continuous, $M_L<\infty$. Because $a=0$ outside $L$, \begin{align*} \int_{\mathbb C}|a(z)|^2e^{-\varphi_m(z)}\,d\mathcal L^2(z) = \int_L |a(z)|^2e^{-m\psi(z)-|z|^2}\,d\mathcal L^2(z) \le M_L\,\mathcal L^2(L)e^{-m\eta}. \end{align*} On $K$, the weight has the opposite sign: $\psi \le -\eta$. Define \begin{align*} R_K := \sup_{z \in K}|z|. \end{align*} Then for $z \in K$, \begin{align*} \frac{e^{-\varphi_m(z)}}{(1+|z|^2)^2} = \frac{e^{-m\psi(z)-|z|^2}}{(1+|z|^2)^2} \ge \frac{e^{m\eta-R_K^2}}{(1+R_K^2)^2}. \end{align*} Combining this lower bound on $K$ with Hörmander's estimate gives \begin{align*} \frac{e^{m\eta-R_K^2}}{(1+R_K^2)^2} \int_K |u_m(z)|^2\,d\mathcal L^2(z) \le M_L\,\mathcal L^2(L)e^{-m\eta}. \end{align*} Hence \begin{align*} \int_K |u_m(z)|^2\,d\mathcal L^2(z) \le (1+R_K^2)^2 e^{R_K^2}M_L\mathcal L^2(L)e^{-2m\eta}. \end{align*} The right-hand side tends to $0$ as $m\to\infty$, so $u_m \to 0$ in $L^2(K)$. [/guided] [/step] [step:Subtract the solution to obtain entire approximants] Define \begin{align*} F_m: \mathbb C &\to \mathbb C \\ z &\mapsto g(z)-u_m(z). \end{align*} In the sense of distributions, \begin{align*} \frac{\partial F_m}{\partial \bar z} = \frac{\partial g}{\partial \bar z} - \frac{\partial u_m}{\partial \bar z} = a-a = 0. \end{align*} By the Weyl lemma for the $\bar\partial$-operator (citing a result not yet in the wiki: Weyl lemma for $\bar\partial$), $F_m$ agrees almost everywhere with an entire function on $\mathbb C$; we replace $F_m$ by that entire representative. Since $\chi=1$ on a neighbourhood of $K$, we have $g=f$ on $K$. Therefore \begin{align*} \int_K |F_m(z)-f(z)|^2\,d\mathcal L^2(z) = \int_K |u_m(z)|^2\,d\mathcal L^2(z) \to 0 \end{align*} as $m\to\infty$. [guided] We now cancel the $\bar\partial$-error. Define \begin{align*} F_m: \mathbb C &\to \mathbb C \\ z &\mapsto g(z)-u_m(z). \end{align*} Both terms are locally square-integrable, so the distributional $\bar\partial$ derivative is defined. Using the equation $\partial u_m/\partial\bar z=a$ and the definition $a=\partial g/\partial\bar z$, we get \begin{align*} \frac{\partial F_m}{\partial \bar z} = \frac{\partial g}{\partial \bar z} - \frac{\partial u_m}{\partial \bar z} = a-a = 0 \end{align*} as distributions. A function with vanishing distributional $\bar\partial$ derivative is holomorphic after modification on a null set, by the Weyl lemma for the $\bar\partial$-operator (citing a result not yet in the wiki: Weyl lemma for $\bar\partial$). Thus we regard $F_m$ as an entire function on $\mathbb C$. On $K$, the cutoff has introduced no error because $\chi=1$ on a neighbourhood of $K_0$ and $K \subset K_0$. Therefore $g=f$ on $K$, and \begin{align*} F_m-f = g-u_m-f = -u_m \end{align*} on $K$. Hence \begin{align*} \int_K |F_m(z)-f(z)|^2\,d\mathcal L^2(z) = \int_K |u_m(z)|^2\,d\mathcal L^2(z). \end{align*} The previous step proves that the right-hand side tends to $0$, so the entire functions $F_m$ approximate $f$ in $L^2(K)$. [/guided] [/step] [step:Approximate one entire approximant by a polynomial and upgrade on a larger compact set] Choose $m$ large enough that \begin{align*} \int_K |F_m(z)-f(z)|^2\,d\mathcal L^2(z) < \frac{\varepsilon}{4}. \end{align*} Let \begin{align*} R := 1+\sup_{z \in K}|z|. \end{align*} Since $F_m$ is entire, its Taylor series at $0$ converges uniformly on the compact closed disk $\overline{B}(0,R)$. Hence there exists a Taylor polynomial \begin{align*} p: \mathbb C &\to \mathbb C \end{align*} such that \begin{align*} \sup_{z \in K}|F_m(z)-p(z)| < \left(\frac{\varepsilon}{4\max\{1,\mathcal L^2(K)\}}\right)^{1/2}. \end{align*} Therefore \begin{align*} \int_K |F_m(z)-p(z)|^2\,d\mathcal L^2(z) < \frac{\varepsilon}{4}. \end{align*} Using the inequality $|A+B|^2 \le 2|A|^2+2|B|^2$ with \begin{align*} A=f-F_m, \qquad B=F_m-p, \end{align*} we obtain \begin{align*} \int_K |f(z)-p(z)|^2\,d\mathcal L^2(z) &\le 2\int_K |f(z)-F_m(z)|^2\,d\mathcal L^2(z) + 2\int_K |F_m(z)-p(z)|^2\,d\mathcal L^2(z)\\ &< \varepsilon. \end{align*} This proves the $L^2(K)$ polynomial approximation. For the uniform assertion, replace the original compact set by a compact set $K_1$ with \begin{align*} K \subset K_1^\circ \subset K_1 \Subset \Omega, \qquad \widehat K_{1,\mathrm{poly}}\Subset\Omega. \end{align*} The preceding construction applied to $K_1$ gives entire functions $F_m$ such that \begin{align*} \int_{K_1}|F_m(z)-f(z)|^2\,d\mathcal L^2(z)\to0. \end{align*} Choose $r>0$ such that $\overline{B}(z,r)\subset K_1^\circ$ for every $z\in K$. This is possible because $K$ is compact and $K\subset K_1^\circ$. Since $F_m-f$ is holomorphic on $K_1^\circ$, the submean-value inequality for the non-negative subharmonic function $|F_m-f|^2$ gives, for every $z\in K$, \begin{align*} |F_m(z)-f(z)|^2 \le \frac{1}{\pi r^2}\int_{B(z,r)}|F_m(w)-f(w)|^2\,d\mathcal L^2(w) \le \frac{1}{\pi r^2}\int_{K_1}|F_m(w)-f(w)|^2\,d\mathcal L^2(w). \end{align*} Taking the supremum over $z\in K$ shows $F_m\to f$ uniformly on $K$. Choose $m$ with \begin{align*} \sup_{z\in K}|F_m(z)-f(z)|<\frac{\varepsilon}{2}. \end{align*} Since $F_m$ is entire, its Taylor polynomials converge uniformly on a closed disk containing $K$, so choose a Taylor polynomial $p$ with \begin{align*} \sup_{z\in K}|F_m(z)-p(z)|<\frac{\varepsilon}{2}. \end{align*} Then \begin{align*} \sup_{z\in K}|f(z)-p(z)| \le \sup_{z\in K}|f(z)-F_m(z)|+ \sup_{z\in K}|F_m(z)-p(z)| <\varepsilon. \end{align*} This proves the stated uniform approximation after enlarging the compact set used in the construction. [guided] First prove the requested $L^2$ approximation. Choose $m$ so large that \begin{align*} \int_K |F_m(z)-f(z)|^2\,d\mathcal L^2(z) < \frac{\varepsilon}{4}. \end{align*} Let \begin{align*} R := 1+\sup_{z \in K}|z|. \end{align*} The compact set $K$ is contained in the closed disk $\overline{B}(0,R)$. Since $F_m$ is entire, its Taylor series at $0$ converges uniformly on $\overline{B}(0,R)$. Therefore some Taylor polynomial \begin{align*} p: \mathbb C &\to \mathbb C \end{align*} satisfies \begin{align*} \sup_{z \in K}|F_m(z)-p(z)| < \left(\frac{\varepsilon}{4\max\{1,\mathcal L^2(K)\}}\right)^{1/2}. \end{align*} It follows that \begin{align*} \int_K |F_m(z)-p(z)|^2\,d\mathcal L^2(z) \le \mathcal L^2(K)\sup_{z\in K}|F_m(z)-p(z)|^2 < \frac{\varepsilon}{4}. \end{align*} Now decompose the error as \begin{align*} f-p=(f-F_m)+(F_m-p). \end{align*} The elementary inequality $|A+B|^2\le2|A|^2+2|B|^2$ gives \begin{align*} \int_K |f(z)-p(z)|^2\,d\mathcal L^2(z) &\le 2\int_K |f(z)-F_m(z)|^2\,d\mathcal L^2(z) + 2\int_K |F_m(z)-p(z)|^2\,d\mathcal L^2(z)\\ &<\varepsilon. \end{align*} This proves the $L^2(K)$ assertion. Now prove the uniform assertion rather than merely citing the same argument. Choose a larger compact set $K_1$ with \begin{align*} K \subset K_1^\circ \subset K_1 \Subset \Omega, \qquad \widehat K_{1,\mathrm{poly}}\Subset\Omega. \end{align*} Applying the already proved weighted construction with $K_1$ in place of $K$ gives entire functions $F_m$ satisfying \begin{align*} \int_{K_1}|F_m(z)-f(z)|^2\,d\mathcal L^2(z)\to0. \end{align*} Why does this imply [uniform convergence](/page/Uniform%20Convergence) on the smaller set $K$? The difference $F_m-f$ is holomorphic on $K_1^\circ$, so $|F_m-f|^2$ is subharmonic there. Since $K$ is compactly contained in $K_1^\circ$, choose $r>0$ such that every closed disk $\overline{B}(z,r)$ with $z\in K$ lies in $K_1^\circ$. The submean-value inequality gives \begin{align*} |F_m(z)-f(z)|^2 \le \frac{1}{\pi r^2}\int_{B(z,r)}|F_m(w)-f(w)|^2\,d\mathcal L^2(w) \le \frac{1}{\pi r^2}\int_{K_1}|F_m(w)-f(w)|^2\,d\mathcal L^2(w) \end{align*} for every $z\in K$. Taking the supremum over $K$ proves $F_m\to f$ uniformly on $K$. Choose $m$ so large that \begin{align*} \sup_{z\in K}|F_m(z)-f(z)|<\frac{\varepsilon}{2}. \end{align*} Since $F_m$ is entire, choose a Taylor polynomial $p$ approximating $F_m$ uniformly on a closed disk containing $K$ with \begin{align*} \sup_{z\in K}|F_m(z)-p(z)|<\frac{\varepsilon}{2}. \end{align*} Then the triangle inequality gives \begin{align*} \sup_{z\in K}|f(z)-p(z)| \le \sup_{z\in K}|f(z)-F_m(z)|+ \sup_{z\in K}|F_m(z)-p(z)| <\varepsilon. \end{align*} Thus the uniform approximation follows after enlarging the compact set used in the $L^2$ construction. [/guided] [/step]