text:
\section{Introduction}\label{S:Intro}

The Bergman space, $A^2(\Omega)$, of an open set $\Omega\subset\mathbb{C}^n$ is the
vector space of $L^2(\Omega)$-integrable holomorphic functions on $\Omega$.
Through work of Carleson \cite{Ca67} and Wiegerinck \cite{Wi84} the dimension of the Bergman space of an open set in the complex plane is completely characterized by the polarity of the complement, see the equivalences $(a)$--$(c)$ in Theorem \ref{T:WC}. Wiegerinck \cite{Wi84} also constructs domains in  $\mathbb{C}^2$ whose Bergman spaces are nontrivial but finite dimensional; see \cite{PZ05,Dinew07,Jucha12,GaHaHe17,PZ17,Boud21} for further partial results on  the dimension of Bergman spaces of open sets in higher dimensions.

The results on the dimension of the Bergman space of open sets in $\mathbb{C}$ may be consolidated as the following theorem.

\begin{theorem}[\cite{Ca67},\cite{Wi84},\cite{GaHaHe17},\cite{GaLeRa21}]\label{T:WC} Let $K\subsetneq\C$ be a closed subset. Then the following are equivalent.
\begin{itemize}
\item[$(a)$] $K$ is polar.
\item [$(b)$] $A^2(\C\setminus K)= \{0\}$.
\item [$(c)$] $\dim A^2(\C\setminus K)<\infty$.
\item [$(d)$] There exists no bounded $\psi\in\mathcal{C}^\infty(\mathbb{C}\setminus K)$ such that $\triangle\psi>0$ on $\mathbb{C}\setminus K$.
\end{itemize}
\end{theorem}
 That the Bergman space is either trivial or infinite dimensional, i.e., the equivalence of $(b)$ and $(c)$, is due to Wiegerinck, see \cite{Wi84}. Carleson shows  in \cite[Theorem 1, \S VI]{Ca67} that the Bergman space is nontrivial if and only if the logarithmic capacity of $K$ is positive, that is, the equivalence of $(a)$ and $(c)$. Property $(d)$ is an additional, potential-theoretic characterization of  open sets in $\mathbb{C}$ with finite dimensional Bergman spaces. A proof of implication $(c)\Rightarrow(d)$ may be found in \cite{GaHaHe17}, while implication $(d)\Rightarrow(c)$ is due to \cite[Prop. 5.1]{GaLeRa21}.

 Recently, Sz\H{o}ke \cite{Sz20} considered the dimension problem
 in the setting of compact Riemann surfaces. He conjectures that, given a  holomorphic line bundle $L$ on a compact Riemann surface $M$,   the Bergman space, $A^2(D;L)$, of holomorphic sections of the restriction of  $L$  to an open set $D\subseteq M$ either coincides with the space of global holomorphic sections of $L$, or is infinite dimensional; see Subsection~\ref{SS:BSforSections} for the precise definition of $A^2(D;L)$.  Furthermore, he proves this conjecture in the case when $M$ is $\mathbb{P}^1$,  the Riemann sphere.

 \begin{theorem}\cite{Sz20}\label{T:Sz}
   Let $L$ be a holomorphic line bundle  on $\mathbb{P}^1$, and $D\subset\mathbb{P}^1$ be an open set. Then $A^2(D;L)$ is either equal to $\Gamma(\mathbb{P}^1;L)$, i.e., the space of global holomorphic sections on $\mathbb{P}^1$, or infinite dimensional.
 \end{theorem}

Sz\H{o}ke's proof is a modification of Wiegerinck's proof in \cite{Wi84}, and relies on the relatively simple algebraic structure of holomorphic line bundles on $\mathbb{P}^1$. While this proof is hard to generalize to other compact Riemann surfaces, Sz\H{o}ke gives a partial result for general compact Riemann surfaces. However, he does not provide any potential-theoretic characterizations, akin to Theorem~\ref{T:WC}, for the general case; see the remarks above Proposition 1.3 in \cite{Sz20}.  An objective of this paper is to fill this gap for $\mathbb{P}^1$. In fact, our main result is a complete analogue of Theorem~\ref{T:WC} for Bergman spaces of holomorphic sections on open subsets of $\mathbb{P}^1$.

\begin{theorem}\label{T:main} Let $\Pi:L\rightarrow \mathbb{P}^1$ be a holomorphic line bundle.
Suppose that $K\subsetneq\mathbb{P}^1$ is a compact subset. Then the following are equivalent.
	\begin{itemize}
\item[$(a)$] $K$ is polar.
\item[$(b)$] $A^2(\mathbb{P}^1\setminus K;L)=\Gamma(\mathbb{P}^1;L)$.
\item[$(c)$] $\dim A^2(\mathbb{P}^1\setminus K;L)<\infty$.
\item[$(d)$] There exists no bounded function $\psi\in\mathcal{C}^\infty(\mathbb{P}^1\setminus K)$ such that $i\partial\bar{\partial}\psi\geq \omega$ on $\mathbb{P}^1\setminus K$ for some volume form $\omega$ on $\mathbb{P}^1$.
\end{itemize}
\end{theorem}

The implications $(a)\Rightarrow(b)\Rightarrow(c)\Rightarrow(a)$ are proved using some basic results in potential theory.
In particular, our proof of $(c)\Rightarrow(a)$ in Subsection~\ref{SS:CT} is inspired by Wiegerinck's proof in \cite{Wi84}, but uses Carleson's construction of a nontrivial $L^2$-integrable holomorphic function as seen in \cite[Theorem 9.5, Ch. 21]{Ca67}. We include this proof of the implication $(c)\Rightarrow(a)$, because it does not rely on partial differential equations techniques, unlike our proof of $(c)\Rightarrow(d)\Rightarrow(a)$.

As in the proof of Theorem~\ref{T:WC}, the proof of the implication  $(c)\Rightarrow(d)$ in
 Theorem~\ref{T:main} is done by using H\"ormander's weighted $L^2$-method for solving the Cauchy--Riemann equations.
 We note that the equivalence of $(a)$ and $(d)$ can be paraphrased as: $\mathbb{P}^1\setminus K$ is nonpolar if and only if, for every Hermitian metric $h$ on $L$ and  volume form $\omega$ on $\mathbb{P}^1$, there is a bounded function  $\psi\in\mathcal{C}^\infty(\mathbb{P}^1\setminus K)$ satisfying
  \begin{align*}
     i\Theta_{e^{-\psi}h}\geq \omega\;\;\text{ on }\mathbb{P}^1\setminus K.
  \end{align*}
That is, whenever $K$ is nonpolar, the given Hermitian metric $h$ on $L$ may be twisted such that the newly obtained Hermitian metric on $L|_{D}$ has positive curvature, see \cite[Theorem 3.11.2]{NaRa11} for a related result on general open Riemann surfaces.
This positivity lets one use
H\"ormander's method successfully; the boundedness of $\psi$ ensures that the resultant estimates are for the Bergman spaces associated to the given Hermitian metric $h$ on $L$.

The proof of $(d)\Rightarrow(a)$ in Theorem~\ref{T:main} is also motivated by the proof of the same implication in Theorem~\ref{T:main}. In the latter, one uses the potential function associated to the equilibrium measure of a compact, nonpolar set to construct the strictly subharmonic, bounded weight function $\psi$. This construction may be done in a local chart to prove the implication
 $(d)\Rightarrow(a)$ in Theorem~\ref{T:main}.
In fact, the proofs of the equivalences in Theorem~\ref{T:main} are all done in a local chart. However, these proofs, except for the direct proof of $(c)\Rightarrow(a)$  in Subsection~\ref{SS:CT},  do not depend on the particular structure of $L$ and $h$ so that an extension of Theorem~\ref{T:main} to compact Riemann surfaces seems feasible. In fact, we will consider this question for compact Riemann surfaces in a forthcoming paper.

As a byproduct of the proof of the implication $(c)\Rightarrow(d)$ of Theorem~\ref{T:main}, one obtains a complete description of the dimension of the weighted Bergman space $A^2_{e^{-\psi}}(\Omega)$, where $\Omega\subset\mathbb{C}$ is an open set, and $\psi$ is a subharmonic function on $\Omega$. This is a combination of Corollary~\ref{C:LeThesisminusK}, which is a
 version of Theorem~\ref{T:WC} for $A^2_{e^{-\psi}}(\Omega)$, and the complete characterization of the dimension of $A_{e^{-\psi}}(\mathbb{C})$ due to Borichev, Le, and Youssfi; see \cite[Theorem 2.6]{Le21}.

In \cite{Wi84}, Wiegerinck gives examples of domains in $\C^2$ that have nontrivial, but finite dimensional Bergman spaces. These examples immediately show that the dichotomy conjectured for compact Riemann surfaces does not hold for the Bergman spaces of holomorphic sections of a specific holomorphic line bundle, $L=\mathcal{O}(-3)$, on the complex projective space $\mathbb{P}^2$. Motivated by Wiegerinck's construction, we produce examples in $\mathbb{P}^2$ to show that this dichotomy is absent irrespective of the choice of holomorphic line bundle on $\mathbb{P}^2$.

The paper is structured as follows. In Section~\ref{S:prelim}, we detail the required background material and notation on Bergman spaces, potential theory, and holomorphic line bundles on $\mathbb{P}^1$. The first four subsections of Section~\ref{S:proofs} contain the proofs of the equivalences of Theorem~\ref{T:main}. This is followed by Subsection~\ref{SS:Le}, where the results on the dimension of weighted Bergman spaces for open sets in $\mathbb{C}$ are given.  In Section \ref{S:counter}, we show by example that for each holomorphic line bundle on $\mathbb{P}^2$, there is a domain in $\mathbb{P}^2$ such that the Bergman space of the corresponding holomorphic sections does not equal the space of global holomorphic sections but is finite dimensional.

\noindent {\bf Acknowledgements.}  The authors are grateful to Peter Pflug for his helpful comments on Lemma~\ref{L:removable}.

\section{Background and preliminaries}\label{S:prelim}

\subsection{Bergman spaces in $\mathbb{C}^n$ and potential theory in $\mathbb{C}$} Let $\Omega\subseteq \mathbb{C}^n$ be an open subset, and $\lambda$ be the standard volume form on $\C$. For a positive function $\phi$ on $\Omega$, the {\em weighted Bergman space} of $\Omega$ with weight $\phi$ is the Banach space
	\bes
		A^2_\phi(\Omega):=\left\{f\in\mathcal{O}(\Omega):||f||:=\left(\int_D|f(z)|^2\phi(z)\lambda(z)
	\right)^{1/2}<\infty\right\}.
	\ees
For $\phi\equiv 1$, the space $A^2_\phi(\Omega)$ is denoted by $A^2(\Omega)$, and referred to as the Bergman space of $\Omega$.

A set $K\subset\C$ is said to be {\em polar} if there is a nonconstant subharmonic function $s$ on $\C$ such that $K\subseteq\{z\in\C:s(z)=-\infty\}$. Polar sets admit an alternate characterization via logarithmic potential theory. For a finite Borel measure $\nu$ with compact support in $\C$, its {\em potential} is the function $p_\nu:\C\rightarrow [-\infty,\infty)$ given by
	\bes
		p_\nu(z)=\int_\C\ln|z-w|\,d\nu(w),\qquad z\in\C.
	\ees
The {\em energy} of $\nu$ is the quantity
	\bes
		I(\nu)=\int_\C p_{\nu}(z)\ d\nu(z).
	\ees
The {\em (logarithmic) capacity} of $K\subset \C$ is defined as
	\bes
		\operatorname{cap}(K):=\sup\{e^{I(\nu)}:\nu\ \text{is a Borel probability measure with compact support in}\ K\}.
	\ees
For a compact nonpolar set $K\subset\C$, it is known that there is a unique Borel probability measure $\nu_K$ supported on $K$ such that
$$I(\nu_K)=\sup\{\nu:\nu\ \text{is a Borel probability measure on}\ K\}.$$
This measure is called the {\em equilibrium measure} of $K$.
The following result relates the notion of polarity, local polarity, and capacity.

\begin{theorem}\label{T:polar} Let $K\subset\C$ be a Borel subset. Then the following are equivalent.
\begin{itemize}
\item [$(i)$] $K$ is polar.
\item [$(ii)$] There is an open set $G$ containing $K$, and an $s\in\operatorname{sh}(G)$, nonconstant on any component of $G$, such that $K\subset \{x\in G:s(x)=-\infty\}$.
\item [$(iii)$] For every $\zeta\in \C$, there is a connected open neighborhood $V_\zeta\subseteq\C$ of $\zeta$ and a nonconstant function $s_\zeta\in\operatorname{sh}(V_\zeta)$ such that $K\cap V_\zeta\subseteq\{x\in V_\zeta:s_\zeta(x)=-\infty\}$.
\item [$(iv)$] $\operatorname{cap}(K)=0$.
\end{itemize}
\end{theorem}
For the equivalences of $(i)$ and $(ii)$, $(i)$ and $(iii)$, and $(i)$ and $(iv)$, see for instance Proposition~$5.5$, Lemma~$5.6$ and Theorem~$7.5$, respectively, in \cite[Ch.~21]{Co95}.

\subsection{Bergman spaces of holomorphic sections}\label{SS:BSforSections}
 Let $M$ be a complex manifold. Given a holomorphic line bundle $\Pi:L\rightarrow M$ and an open set $D\subset M$,
the space of holomorphic sections of $L|_D$ is denoted by $\Gamma(D;L)$.

Let $\omega$ be a volume form on $M$. Given a holomorphic line bundle $\Pi:L\rightarrow M$, and a smooth Hermitian metric $h$ on $L$,
we define the Bergman space of sections of $(L|_D,h)$ as
	\bes
		A^2_{h,\omega}(D;L)=\left\{s\in\Gamma(D;L):||s||:=\left(\int_D h(s,s)\,\omega\right)^{1/2}<\infty\right\}.
	\ees
$A^2_{h,\omega}(D;L)$ is a reproducing kernel Hilbert space.
If $M=\mathbb{C}^n$, then $\Gamma(D;L)=\mathcal{O}(D)$ and $h(s,s)\omega=|s|^2\phi\lambda$, for some positive function $\phi$ on $\mathbb{C}^n$. In this case, $A^2_{h,\omega}(D;L)$ is the weighted Bergman space $A^2_\phi(D)$.

\begin{lemma}\label{L:choice} Let $M, D, L, h,\omega$ be as above. Suppose that $M$ is compact. Then, as a vector space, $A^2_{h,\omega}(D;L)$ is independent of the choices of $h$ and $\omega$. In particular, $A^2_{h,\omega}(M;L)=\Gamma(M;L)$.
\end{lemma}
\begin{proof} Given any two smooth volume forms, $\omega$ and $\omega'$, on $M$, and any two smooth Hermitian metrics, $h$ and $h'$, on $L$, there exist smooth positive functions, $f$ and $g$, on $M$, such that $\omega=f\omega'$ and $h=gh'$. Thus, the Bergman spaces of the sections of $L|_D$ are isomorphic for different choices of the volume form on $M$ and Hermitian metric on~$L$.

The final claim holds since $h(s,s)$ is a continuous positive function on $M$ for any choice of $h$ and $s$.
\end{proof}

\begin{remark} In this paper, $M$ is either $\mathbb P^1$ or $\mathbb P^2$. Thus, in view of Lemma~\ref{L:choice}, $A^2_{h,\omega}(D;L)$ is simply denoted by $A^2(D;L)$.
\end{remark}

\subsection{Line bundles on $\mathbb{P}^1$} We recall certain standard facts about $\mathbb{P}^1$. Given any $q\in\mathbb{P}^1$, one may choose coordinates $\zeta=[\zeta_0:\zeta_1]$ on $\mathbb{P}^1$ so that $q=[0:1]$. In this case, we often identify $U^z=\mathbb{P}^1\setminus\{q\}$ with the complex plane $\C$ via
	\bes
		\varphi_z:\zeta=[\zeta_0:\zeta_1]\mapsto z=
							\frac{\zeta_0}{\zeta_1},
	\ees
and refer to $z$ as the local coordinate on the affine chart $\mathbb{P}^1\setminus \{q\}$. On occasion, the chart $(U^{1/z},\varphi_{1/z})$ given by $\varphi_{1/z}:\zeta\mapsto \zeta_1/\zeta_0$ on $U^{1/z}=\mathbb{P}^1\setminus\{[1:0]\}$ will also be used. In this case, we will use the coordinate $w$ on $\varphi_{1/z}(U^{1/z})=\C$.

A smooth volume form $\omega$ on $\mathbb{P}^1$ is of the form
		$f\,\omega_{\text{FS}}$,
where $\omega_{\text{FS}}$ is the Fubini--Study volume form, and $f$ is a smooth positive function on $\mathbb{P}^1$. Thus, in the local coordinate $z$,
	\bes
		\omega(z)=f(z)\,\dfrac{idz\wedge d\overline z}{2(1+|z|^2)^2},\quad z\in\C,
	\ees
where $f:\C\rightarrow(0,\infty)$ is smooth, and $f\circ\varphi_z$ admits a smooth positive extension to $\mathbb{P}^1$. Given a smooth function $\psi$ on an open set in $\mathbb P^1$, $\partial\overline\partial \psi$ is the unique smooth $(1,1)$-form on $\mathbb{P}^1$ satisfying $(\varphi_z)_*\left(\partial\overline\partial \psi\right)(z)=\frac{\partial^2(\psi\circ\varphi_z)}{\partial z\partial\overline z}(z)\, d\overline z\wedge dz$ for all $z\in U^z$.

Next, any holomorphic line bundle $\Pi:L\rightarrow\mathbb{P}^1$ is of the form $\mathcal{O}(k)$, for some $k\in\Z$, where $\mathcal{O}(k)$ is the line bundle associated to the divisor $k\{p\}$ for any fixed $p\in\mathbb{P}^1$.
In the local coordinate $z$, a global section of $\mathcal{O}(k)$ is given by a pair, $s=(s_1,s_2)$, of holomorphic functions on $\C$ such that $s_1(z)=z^{k}s_2(1/z)$ if $z\in\C^*$. Similarly, a Hermitian metric on $L=\mathcal{O}(k)$ is given by a pair, $h=(h_1,h_2)$, of smooth positive functions on $\C$, such that $h_1(z)=|z|^{-2k}h_2(1/z)$ if $z\in\C^*$. Owing to the relationship between $h_1$ and $h_2$, $\partial\overline{\partial} h_1(z)=\partial\overline{\partial} h_2(1/z)$, $z\in\C^*$, where $\partial\overline\partial f(z)=\frac{\partial^2f}{\partial z\partial\overline z}(z)\, d\overline z\wedge dz$ for any $\mathcal{C}^2$-smooth function $f$ on $\C$. Thus, there is a smooth $(1,1)$-form $\Theta_h$ on $\mathbb{P}^1$ such that $(\varphi_z)_*\Theta_h=\partial\overline{\partial} h_1$ and $\left(\varphi_{1/z}\right)_*\Theta_h=\partial\overline{\partial} h_2$. The form $\Theta_h$ is referred to as the curvature of the Hermitian metric $h$. Note that it suffices to specify $s_1$, $h_1$ and $\partial\overline\partial h_1$, which is the convention we will employ. Lastly, recall the following description of the space of global sections of $\mathcal{O}(k)$, $k\in\Z$.

\begin{lemma}\label{L:Bergman} Let $k\in\Z$. In the local coordinate $z$,
	\bes
		A^2(\mathbb{P}^1;\mathcal{O}(k))=\Gamma(\mathbb{P}^1;\mathcal{O}(k))=\begin{cases}
				\{\text{polynomials of degree at most}\ k\},& \text{if}\ k\geq 0,\\
				\{f\equiv 0\},& \text{if}\ k< 0.
				\end{cases}
	\ees
In particular, $\dim A^2(\mathbb{P}^1;\mathcal{O}(k))=\dim\Gamma(\mathbb{P}^1;\mathcal{O}(k))=\max\{0,k+1\}$.
\end{lemma}

While Lemma~\ref{L:Bergman} is a standard result, see \cite[p. 81]{Varolin11}, we direct the reader to the remarks before Corollary~\ref{C:LeThesisminusK} for a proof in the case $k\geq -2$.

\subsection{Potential theory on $\mathbb{P}^1$} Given an open set $U\subseteq\mathbb{P}^1$, and an upper semicontinuous function $s:U\rightarrow [-\infty, \infty)$ with $s\nequiv -\infty$, $s$ is said to be {\em subharmonic on $U$} if, for every coordinate chart $(V,\varphi)$ that intersects $U$, $s\circ\varphi^{-1}$ is subharmonic on $\varphi(U\cap V)\subset \C$. The set of all subharmonic functions on $U$ is denoted by $\operatorname{sh}(U)$.

\begin{definition}\label{D:polarP} Given a set $K\subset\mathbb{P}^1$, $K$ is said to be {\em polar} if for each $\zeta\in\mathbb{P}^1$, there is an open neighborhood $V_\zeta\subset\mathbb{P}^1$ of $\zeta$ and a function $s_\zeta\in\operatorname{sh}(V_\zeta)$ such that $K\cap V_\zeta\subset\{x\in V_\zeta:s_\zeta(x)=-\infty\}$.
\end{definition}

\begin{lemma}\label{L:polar1} Let $K\subset\mathbb{P}^1$ be compact. For any fixed $q=[0:1]\in\mathbb{P}^1$, $K$ is polar in $\mathbb{P}^1$ if and only if $K_*:=\varphi_z(K)$ is polar in $\C$.
\end{lemma}
\begin{proof} First, suppose that $K$ is polar. Let $\zeta_*\in\C$ and $\zeta=\varphi_z^{-1}(\zeta_*)$. Then it is clear that
	\bes
		K_*\cap U_{\zeta_*}\subset\{x\in U_{\zeta_*}:s_{\zeta_*\,}(x)=-\infty\},
	\ees
where $U_{\zeta_*}=\varphi_z(V_\zeta\setminus\{q\})$, and $s_{\zeta_*}=s_\zeta\circ\varphi_z^{-1}$. Thus, $K_*$ is polar in $\C$.

Next, assume that $K_*$ is polar. By a similar reasoning as above, we obtain that for every $\zeta\in\mathbb{P}^1\setminus\{q\}$, there exist $V_\zeta\subset\mathbb{P}^1$ and $s_\zeta\in\operatorname{sh}(V_\zeta)$ that satisfy the condition in Definition~\ref{D:polarP}. It remains to produce such a pair $(V_\zeta,s_\zeta)$ for $\zeta=q$. Since subharmonicity is preserved by biholomorphic maps, it follows that the set $K_{**}:=\varphi_{1/z}\circ \varphi_z^{-1}(K_*)$ is polar in $\C^*$. In fact, by Theorem~\ref{T:polar}, $K_{**}$ is polar in $\C$. On the other hand, $\{0\}=\varphi_{1/z}(\{q\})$ is also polar in $\C$. Now, using the fact that the union of two polar sets in $\C$ is polar in $\C$, see \cite[Lemma~5.6, Ch.~21]{Co95}, $\varphi_{1/z}(K)=K_{**}\cup\{0\}$ is polar in $\C$, i.e., there is an $s\in\operatorname{sh}(\C)$ such that $\varphi_{1/z}(K)\subset\{x\in\C:s(x)=-\infty\}$. To complete the proof, we simply observe that $V_q=\varphi_{1/z}^{-1}(\C)$ and $s_q=s\circ\varphi_{1/z}$ satisfy the condition required at $q\in K$ in Definition~\ref{D:polarP}.
\end{proof}

\section{Proof(s) of the main result}\label{S:proofs}
 By Lemma~\ref{L:Bergman}, $(b)\Rightarrow(c)$. In Subsections~\ref{SS:a=>b}--\ref{SS:(d)=>(a)}, we prove the implications $(a)\Rightarrow(b)$, $(c)\Rightarrow(a)$, $(c)\Rightarrow(d)$, and $(d)\Rightarrow(a)$, respectively. This section concludes with the characterization of weighted Bergman spaces, $A^2_{e^{-\psi}}(\Omega)$, for subharmonic $\psi$ and $\Omega\subset\mathbb{C}$ open.

\subsection{Proof of $\mathbf{(a)}$ implies $\mathbf{(b)}$}\label{SS:a=>b}
In order to prove that $A^2(\mathbb{P}^1\setminus K;L)=\Gamma(\mathbb{P}^1;L)$ whenever $K\subset\mathbb{P}^1$ is a compact polar set, we first prove a version of this result for sets in $\mathbb{C}$ in Lemma~\ref{L:removable} below.
A readily available result in the literature is that  $A^{2}(\Omega)=A^2(\Omega\setminus X)$ for any open set $\Omega\subset\C$ and polar, compact set $X\subset \Omega$, see for instance Theorem~9.5 in \cite[Ch.~21]{Co95}.
We present a generalization of this result to the case when $X$ is a polar and relatively closed subset of $\Omega$.\footnote{After the first draft of this manuscript appeared on the arXiv, Pflug brought to our attention that Lemma~\ref{L:removable} has appeared in \cite{Si82}, where Siciak attributes it to Sakai and Skwarczy{\' n}ski. Since neither \cite{Si82} nor the source cited therein are easily accessible, we give a proof of Lemma~\ref{L:removable} in this work. This proof is simpler than our original one, and follows an argument suggested by Pflug. This argument resembles that of Siciak's.}

\begin{lemma}[\cite{Si82}]\label{L:removable}
Let $\Omega\subseteq\C$ be an open subset, and $X\subset \C$ be a closed polar subset. Then, $A^2(\Omega\setminus X)=A^2(\Omega)$, i.e., for any $f\in A^2(\Omega\setminus X)$, there is an $F\in A^2(\Omega)$ such that $F|_{\Omega\setminus X}=f$.
\end{lemma}
\begin{proof} The proof uses the fact that for any $a\in \Omega\cap X$, there exists an $r_a>0$ such that $\overline{\mathbb{D}(a,r_a)}\subset\Omega$, and $b\mathbb{D}(a,r_a)\cap X=\emptyset$, see Theorem 7.3.9 in \cite{ArGa12}.

Now, let $D_a:=\mathbb{D}(a,r_a)$ and $X_a:=\mathbb{D}(a,r_a)\cap X$, where $r_a$ is as above. Then, $X_a$ is a compact subset of $D_a$. Using the result for compact, polar sets mentioned before the statement of this lemma, for any $f\in A^2(\Omega)$, there is an $F_a\in A^2(D_a)$ such that $F_a|_{D_a\setminus X_a}=f|_{D_a\setminus X_a}$. For any $a,b\in X$ such that $D_a\cap D_b\neq \emptyset$, we have that $(D_a\cap D_b)\setminus X$ is a nonempty open set because $X$ is a closed and polar set. Thus, it follows from the identity theorem that $F_a=F_b$ on $D_a\cap D_b$. Hence, the following function is well-defined and holomorphic on $\Omega$:
	\bes
		F(z)=\begin{cases}
				F_a(z),&\ z\in D_a,\\
					f(z),&\ z\in\Omega\setminus X.
			\end{cases}
	\ees
 Moreover, since $f$ and $F$ differ only on the zero measure set $X$, it follows that $||F||_{A^2(\Omega)}=||f||_{A^2(\Omega\setminus X)}$.
\end{proof}

We are now set to prove that $(a)$ implies $(b)$ in Theorem~\ref{T:main}.

\begin{proof}[Proof of $(a)\Rightarrow(b)$ in Theorem~\ref{T:main}]
Suppose that $K\subsetneq \mathbb{P}^1$ is polar. If $K$ is empty, then $(b)$ follows from $(a)$, see Lemma~\ref{L:Bergman}. Thus, we assume that $K$ is a nonempty polar set, and fix a $q\in K$. With $q=[0:1]$, we fix the Hermitian metric on $L=\mathcal{O}(k)$ as $h_1(z)={1}/(1+|z|^2)^k$ in the local coordinate $z$. To account for the volume form in the local coordinate $z$, we set
\be\label{E:metric}
		\phi_k(z):=\frac{1}{(1+|z|^2)^{k+2}},\qquad z\in\C.
\ee
Setting $D_*:=\C\setminus K_*$, it then follows that  $A^2(D;\mathcal{O}(k))$ is isomorphic to $A^2_{\phi_k}(D_*)$.

 Next, we show that $A^2_{\phi_k}(D_*)=A^2_{\phi_k}(\C)$. For this, let $f\in A^2_{\phi_k}(D_*)$. For any $R>0$, set $D_*(R)=D_*\cap \mathbb{D}(0;R)$ and $f_R=f\cdot\chi_{_{D_*(R)}}$. Then, since
	\bes
		\phi_k(z)\geq \begin{cases}
			 1,& \text{if}\ k<-2,\\
			(1+R^2)^{-k-2},& \text{if}\ k\geq -2,
		\end{cases}\qquad z\in \mathbb{D}(0;R),
	\ees
it follows that $f_R\in A^2(D_*(R))$. However, $K_*$ is a closed polar subset of $\C$.
Thus, by Lemma~\ref{L:removable}, there exists an $F_R\in A^2(\mathbb{D}(0;R))$ such that $F_R|_{D_*(R)}=f_R$. We abuse notation, and assume that $F_R$ is defined on $\C$ by setting it to be $0$ on
$\C\setminus \mathbb{D}(0;R)$.  Since $$F_S|_{D_*(R)}=F_{R}|_{D_*(R)}=f|_{D_*(R)}\;\;\text{ whenever }\;\;S\geq R>0,$$
it follows from the identity theorem that $F_S|_{\mathbb{D}(0;R)}=F_R|_{\mathbb{D}(0;R)}$ for all $S\geq R>0$.
Thus, the sequence $\{F_N\}_{n\in\N}$ admits a pointwise limit, say $F$, on $\C$. Moreover, since $F|_{\mathbb{D}(0;N)}=F_N$ for all $N\in\N$, $F\in\mathcal{O}(\C)$ and $F|_{D_*}=f$. In particular, $F$ is measurable so that
\begin{align*}
  \|F\|_{L^2_{\phi_k}(\mathbb{C})}=\|F\|_{L^2_{\phi_k}(D_{*})}=\|f\|_{L^2_{\phi_k}(D_{*})}<\infty
\end{align*}
follows. Therefore, $A^2_{\phi_k}(D_*)$ is isomorphic to $A^2_{\phi_k}(\mathbb{C})$.  Finally, note that
$A^2_{\phi_k}(\C)$ is isomorphic to $A^2(\mathbb{P}^1;\mathcal{O}(k))$, which is $\Gamma(\mathbb{P}^1;\mathcal{O}(k))$, by Lemma~\ref{L:Bergman}. Thus, $(a)$ implies $(b)$ in Theorem~\ref{T:main}.
\end{proof}

\subsection{Proof of $\mathbf{(c)}$ implies $\mathbf{(a)}$ via the Cauchy transform approach}\label{SS:CT}

\begin{proof}
We prove the implication by contraposition, i.e., we prove that if $K\subsetneq\mathbb{P}^1$ is nonpolar, then $A^2(\mathbb{P}^1\setminus K;\mathcal{O}(k))$ is infinite dimensional. As in the proof of $(a)\Rightarrow(b)$ in Subsection~\ref{SS:a=>b}, we assume that $q=[0:1]\in K$ and fix the Hermitian metric $h_1(z)=1/(1+|z|^2)^k$ on $\mathcal{O}(k)$. Then, it suffices to show that $A^2_{\phi_k}(D_*)$ is infinite dimensional, where $\phi_k$ is as in \eqref{E:metric}, and $D_*=\C\setminus K_*$.

\noindent {\bf Case 1.} Suppose that $k\geq -2$. Then, since $\phi_k(z)\leq 1$ for all $z\in \C$, $A^2(D_*)\subseteq A^2_{\phi_k}(D_*)$. Since, by Lemma~\ref{L:polar1}, $\C\setminus D_*=K_*$ is nonpolar, Theorem~\ref{T:WC} yields that $A^2(D_*)$, and therefore, $A^2_{\phi_k}(D_*)$ is infinite dimensional.

\noindent {\bf Case 2.} Suppose that $k<-2$. In this case,  $A^2_{\phi_k}(D_*)\subseteq A^2(D_*)$, so Theorem~\ref{T:WC},  do not directly yield our claim. However, we use the techniques of Carleson and Wiegerinck to produce infinitely many independent functions in $A^2_{\phi_k}(D_*)$.

First, we recall Carleson's construction of a nontrivial function in $A^2(D_*)$; see \cite[Theorem~9.5, Ch.~21]{Co95} for a detailed exposition. Set $K_*:=\varphi_z(K)=\C\setminus D_*$. Then, $K_*$ is a Borel nonpolar set in $\C$. Thus, it contains a compact set, say $E$, with positive logarithmic capacity, see for instance \cite[Theorem 7.5, Ch.~21]{Co95}. Let $E_1,E_2$ be disjoint compact subsets of $E$, each of which has positive logarithmic capacity. For $j\in\{1,2\}$, let $\mu_j$ be the equilibrium measure of $E_j$, and set $\mu=\mu_1-\mu_2$. Set $f$ to be the Cauchy transform of $\mu$, i.e.,
	\begin{align}\label{E:Cauchytransform}
		f(z)=\int_E\frac{d\mu(\xi)}{\xi-z}.
	\end{align}
Then, $f$ is analytic on $\C_\infty\setminus E$, with $f(\infty)=f'(\infty)=0$. Carleson further shows that  $f\in A^2(\C\setminus E)$ which is contained in $A^2(D_*)$.

We next use Wiegerinck's technique to produce a sequence of linearly independent functions $\{g_j\}_{j\in\mathbb{N}}\subset A^2(\C\setminus E)$ such that $g_j$ vanishes at $\infty$, and its order of vanishing at $\infty$ is at least $j$. Assuming the existence of this sequence for the moment, we claim that $g_j\in A^2_{\phi_k	}(D_*)$ for all $j\geq -k$. To see this, fix a $j\geq -k$, and note that there exist $\{c_{j,\ell}\}_{\ell\geq j}\subset\C$ and $R>0$ so that
	\bes
		g^j(z)=\sum_{\ell=j}^\infty c_{\ell,j}\, z^{-\ell}
			\qquad \text{for}\ |z|>R.
	\ees
 Thus, since $j\geq -k$ and $k<-2$, it follows that
	\begin{align*}
		\int_{D_*}|g_j(z)|^2&\frac{idz\wedge d\overline z}{2(1+|z|^2)^{k+2}}\\
		&\leq
		(1+R^2)^{-2-k}\int_{D(0;R)\setminus E}|g_j(z)|^2\,\lambda(z)
			+\int_{|z|>R}|g_j(z)|^2\frac{idz\wedge d\overline z}{2(1+|z|^2)^{k+2}}\\
		&\leq	||g_j||^2_{A^2(\C\setminus E)}+2\pi \int_{r>R}\sum_{\ell=j}|c_{\ell,j}|^2r^{-2\ell} r(1+r^2)^{-2-k} dr<\infty.
	\end{align*}
This gives the infinite dimensionality of $A^2_{\phi_k}(D_*)$. It remains to construct the sequence $\{g_j\}_{j\in\N}$.

\noindent{\em Subcase 1. Suppose that $f$ in \eqref{E:Cauchytransform} is rational.} Since $f$ is $L^2$-integrable on $\C\setminus E$, but not on $\C$, $E$ must have positive Lebesgue measure. In this case, the function
	\bes
		g(z)=\int_E\frac{\lambda(\xi)}{\xi-z}
	\ees
is bounded and analytic on $\C\setminus E$, with $g(\infty)=0$ and $g'(\infty)=-\lambda(E)$, see \cite[Page 2]{Ga72}. Thus, $g^j\in A^2(\C\setminus E)$ for all $j\in\N$, and each $g^j$ has order of vanishing $j$ at $\infty$. In this case, we set $g_j=g^j$.

\noindent{\em Subcase 2. Suppose that $f$ in \eqref{E:Cauchytransform} is not rational.} Expanding $f$ in a Laurent series around $\infty$, one gets
	\bes
		f(z)=\sum_{\ell=p}^\infty c_\ell\,z^{-\ell},\qquad \text{ for some }p\geq 2, c_p\neq 0.
	\ees
Now, we produce a nontrivial function $g\in A^2(\C\setminus E)$ whose Laurent expansion at $\infty$ does not contain any terms in $z^{-1}$,\ldots,$z^{-p}$. Let $z_1,\ldots,z_{p+1}$ be distinct points in $\C\setminus E$, and
	\bes
		g(z)=\sum_{\ell=1}^{p+1}b_\ell\:\frac{(f(z)-f(z_\ell))}{z-z_\ell}.
	\ees
Then, expanding $g$ as a Laurent series around $\infty$, one obtains $g(z)=\sum_{\ell=1}^\infty a_\ell\,z^\ell$, where
	\be\label{E:coeff}
		a_m=\sum_{\ell=1}^{p+1}-b_\ell f(z_\ell)\,z_\ell^{m-1},\qquad
			m\in\{1,\ldots,p\}.
	\ee
Here, the constants $b_1,\ldots,b_{p+1}$ are chosen to solve the homogeneous system of $p$ linear equations obtained by setting $a_m=0$, $m\in\{1,\ldots,p\}$. Moreover, $g$ cannot be trivial, else $f$ will be rational. Thus, $g$ has the desired properties.

Setting $g_1=\cdots =g_p=f$ and $g_{p+1}=g$, we construct $g_j$ inductively for $j\geq p+1$ by repeating the above procedure for $g_{j-1}$ in place of $f$. This completes the construction of the sequence in all cases, and hence, the proof of Theorem~\ref{T:main}.
\end{proof}

\subsection{Proof of $\mathbf{(c)}$ implies $\mathbf{(d)}$}\label{SS:dbar}
The proof of this implication follows from a similar result for sets in $\mathbb{C}$.

\begin{lemma}\label{L:GaHaHemodification}
  Let $\Omega\subset\mathbb{C}$ be an open set and $A_{e^{-\psi_1}}^2(\Omega)$ be the weighted Bergman space for some $\psi_1\in\mathcal{C}^\infty(\Omega,\mathbb{R})$. Suppose that
  there exists a subharmonic function $\psi_2\in\mathcal{C}^\infty(\Omega,\mathbb{R})$  such that it is bounded above, $\psi_1+\psi_2$ is subharmonic on $\Omega$, and
  $$\triangle\psi_2 >0\;\;\text{on }\overline{U}$$
for an open set $U\Subset\Omega$.  Then $A_{e^{-\psi_1}}^{2}(\Omega)$ is infinite dimensionial.
\end{lemma}

\begin{proof}
  This follows from the  proof of Theorem 1 and a minor modification in the proof of Lemma 7 in \cite{GaHaHe17}. This modification consists of replacing $K\varphi(z)$ by
  $\psi_1(z)+K\psi_2(z)$, and using the boundedness from above of $\psi_2$ to show that the constructed function $u$ belongs to $L^2_{e^{-\psi_1}}(\Omega)$.
\end{proof}

\begin{proof}[Proof of $(c)\Rightarrow(d)$ in Theorem~\ref{T:main}]
  The proof is done by contraposition.
  Thus, we assume that there exists a bounded function $\psi\in\mathcal{C}^\infty(\mathbb{P}^1\setminus K)$ such that $i\partial\bar{\partial}\psi\geq \omega$ holds on $\mathbb{P}^1\setminus K$ for some volume form $\omega$ on $\mathbb{P}^1$. Note that, if $K$ was empty, then it would follow that there is a bounded, nonconstant subharmonic function on $\mathbb{C}$. Hence, we may assume that $K$ is nonempty. As before, we choose a  $q\in K$ and coordinates such that $q=[0:1]$

  Next, we note that, as in Subsection~\ref{SS:CT}, it suffices to show that $A^2_{\phi_k}(D_*)$, for $D_*:=\mathbb{P}^1\setminus K$, is infinite dimensional for each $k\in\Z$.

 For that, we use Lemma~\ref{L:GaHaHemodification}. In particular, we set $\widetilde\psi_2=\psi\circ\varphi_z$. Then, clearly, $\widetilde\psi_2\in\mathcal{C}^\infty(D_*)$ is a bounded, strictly subharmonic function.
 Moreover, it follows that there is a constant $c_1>0$ such that
 \begin{align*}
   \triangle\widetilde\psi_2(z)\geq c_1(1+|z|^2)^{-2}\;\;\text{ for }\;\;z\in D_*.
 \end{align*}
 Next, set  $\psi_1:=-\ln(\phi_k)$.
Then $\psi_1\in\mathcal{C}^\infty(D_*)$ and
 \begin{align}\label{E:Laplacemetric}
   \triangle\psi_1(z)=4(k+2)(1+|z|^2)^{-2}
 \end{align}
 on $D_*$.
 Hence, there exists a constant $c>0$ such that
 \begin{align*}
   \triangle(\psi_1+c \widetilde\psi_2)>0\;\;\text{on }D_*.
 \end{align*}
Setting $\psi_2=c\widetilde\psi_2$,
 it follows that the hypotheses of Lemma~\ref{L:GaHaHemodification} are satisfied for any open set $U\Subset D_*$. Thus, $A^2_{\phi_k}(D_*)$ is infinite dimensional.
\end{proof}

\subsection{Proof of $\mathbf{(d)}$ implies $\mathbf{(a)}$}\label{SS:(d)=>(a)}
\begin{proof}
This proof is also done by contraposition. That is, we assume that $K\subset\mathbb{P}^1$ is nonpolar. Since $K$ is nonempty, we may construct $\psi$ in a chart first. As before, we let $q\in K$ and choose coordinates
such that $q=[0:1]$. We set $K_*=\varphi_z(K)$ and $D_*=\mathbb{C}\setminus K_*$.
We shall use a function which was initially constructed in the proof of Proposition 5.1 in \cite{GaLeRa21}.
For that, let $G\subset K_*$ be a nonpolar, compact set and  $\nu=\nu_{G}$ be the equilibrium measure of $G$. The associated potential function
 $p=p_{G}:\mathbb{C}\longrightarrow[-\infty,\infty)$ is defined as
 \begin{align*}
   p(z)=\int_{\mathbb{C}}\ln|z-w|\;d\nu(w)\;\;\text{for }z\in\mathbb{C}.
 \end{align*}
 Since $p$ is harmonic on $G^c$, see for instance \cite[Theorem 3.1.2]{Ra95}, it follows that
 $\triangle e^{-p}=e^{-p}|\nabla p|^2$ on $G^c$.
 Thus, $e^{-p}$ is subharmonic on $G^c$, and strictly subharmonic at all points in $G^c$ at which the gradient of $p$  is nonvanishing.
 It follows from the proof of Proposition~5.1 in \cite{GaLeRa21} that $|\nabla p|$ is strictly positive outside a sufficiently large disc containing $G$. In particular, there exist constants $\tau_1, R>0$ such that
 $G\subset\mathbb{D}(0,R)$
 and
$$|\nabla p(z)|>\frac{\tau_1}{|z|}\;\;\text{for } z\in\mathbb{D}(0,2R)^c.$$
Furthermore, since $\nu$ is an equilibrium measure, one obtains for $z\in\mathbb{D}(0,2R)^c$ that
\begin{align*}
  p(z)=\int_{\mathbb{C}}\ln|z-w|\;d\nu(w)\leq\ln(3|z|/2).
\end{align*}
Therefore,
$$\triangle e^{-p(z)}\geq\frac{2}{3}\frac{\tau_1^2}{|z|^3} \;\;\text{for }z\in\mathbb{D}(0,2R)^c.$$
Strict subharmonicity on all of $G^c$ may now be achieved by adding to $e^{-p}$ a particular compactly supported function which is strictly subharmonic on $\mathbb{D}(0,2R)$. For instance, let $\chi\in\mathcal{C}^\infty_0(\mathbb{D}(0,R'))$ for some $R'>2R$ such that $\chi(z)=|z|^2$ on $\mathbb{D}(0,2R)$. Then, for $\epsilon>0$ sufficiently small, there exist $\tau_2,\tau_3>0$ such that the function
$\psi_*(z):=e^{-p(z)}+\epsilon\chi(z)$ satisfies
\begin{align}\label{E:Laplacepsi2}
  \triangle\psi_*(z)\geq
  \left\{
     \begin{array}{lr}
       \frac{\tau_2}{|z|^3} & \text{for } z\in\mathbb{D}(0,2R)^c\\
       \tau_3 & \text{for } z\in\mathbb{D}(0,2R)
     \end{array}
   \right. .
\end{align}
It follows that $\psi:=\varphi_z^{-1}(\psi_*)$ satisfies $i\partial\bar{\partial}\psi\geq \omega$ on $\mathbb{P}^1\setminus K$ for some volume form $\omega$ on $\mathbb{P}^1$.
Moreover, Frostman's theorem implies that $e^{-p}$ is bounded on $G^c$, and hence, $\psi_*$ is bounded on $G^c$. That is, $\psi$ is bounded on $\mathbb{P}^1\setminus K$, and hence satisfies  $(d)$ in
Theorem~\ref{T:main}.
\end{proof}

\subsection{A corollary to Lemma~\ref{L:GaHaHemodification} on weighted Bergman spaces}\label{SS:Le}

A variation of Lemma~\ref{L:GaHaHemodification} yields an analogue of Theorem~\ref{T:WC} for the Bergman space $A^2_{e^{-\psi}}(\Omega)$ for $\psi$ subharmonic on $\Omega$, see
Corollary~\ref{C:LeThesisminusK} below.

The dimension of $A^2_{e^{-\psi}}(\C)$ for $\psi$ subharmonic on $\C$ is completely described by Borichev, Le, and Youssfi in \cite[Theorem 2.6]{Le21} in terms of the continuous
part $(\mu_\psi)^c$  of the Riesz measure $\mu_\psi$ of $\psi$:
\begin{itemize}
  \item[$(a)$] $\dim A^2_{e^{-\psi}}(\C)<\infty$ iff $(\mu_\psi)^c(\C)<\infty$,
  \item[$(b)$] if $(\mu_\psi)^c(\C)<\infty$, then
    \begin{align}\label{E:Le}
      \dim A^2_{e^{-\psi}}(\C)=\left\lceil\frac{(\mu_\psi)^c(\C)}{4\pi}\right\rceil,
    \end{align}
  where $\lceil x\rceil$ is the largest integer less than $x$ for $x>0$ and $\lceil 0 \rceil=0$
\end{itemize}

Recall that $ A^2(\mathbb{P}^1;\mathcal{O}(k))$ is isomorphic to $A^2_{\phi_k}(\C)$, and that Lemma~\ref{L:Bergman} states that $\dim A^2(\mathbb{P}^1, \mathcal{O}(k))=\max\{0,k+1\}$. Using \cite[Theorem 2.6]{Le21}, we can now
rediscover Lemma~\ref{L:Bergman} for $k\geq-2$. In particular, for $k\geq -2$, the function $\psi:=-\ln(\phi_k)$ is subharmonic since
$$(\triangle\psi)(z)=4(k+2)(1+|z|^2)^{-2}.$$
It follows that $(\mu_\psi)^c=4(k+2)(1+|z|^2)^{-2}\lambda$, so that $(\mu_\psi)^c(\C)=4\pi(k+2)$. By \eqref{E:Le}, it follows that
$\dim A^2_{\phi_k}(\C)=\max\{0,k+1\}$.

As in the case of unweighted Bergman spaces, the dimension of a weighted Bergman space for an open set in $\mathbb{C}$ may be determined through the polarity property of its complement as follows.

\begin{cor}\label{C:LeThesisminusK}
  Let $K\subset\mathbb{C}$ be a closed set, $\psi$ a subharmonic function on $\mathbb{C}$. Then the following hold.
  \begin{itemize}
    \item[(1)] If $K$ is nonpolar, then $A^2_{e^{-\psi}}(\C\setminus K)$ is infinite dimensional.
    \item[(2)] If $K$ is polar, then $A^2_{e^{-\psi}}(\C\setminus K)$ is isomorphic to $A^2_{e^{-\psi}}(\C)$.
  \end{itemize}
\end{cor}

\begin{proof}
  To prove (1), we note first that Lemma~\ref{L:GaHaHemodification} still holds if $\psi_1$ is subharmonic, not necessarily smooth, on $\Omega$ as long as the open set $U\Subset\Omega$ may be chosen such
  that $e^{-\psi_1}$ is
  integrable on $U$.

  Next, we note that Propositions 2.1 and 2.2 in \cite{Jucha12} imply that $e^{-\psi}$ is integrable except near finitely many points. Thus, we may choose an open set $U$ in $\C\setminus K$ such that
  $e^{-\psi}$ is integrable on $U$. Now, choose $\psi_2$ equal to $\psi_{*}$ as defined above \eqref{E:Laplacepsi2}. Then $\psi_2$ is a bounded above, strictly subharmonic, smooth function on $\C\setminus K$.
  Thus, with $\psi_1:=\psi$, the claim follows from the above mentioned variation of Lemma~\ref{L:GaHaHemodification}

  For the proof of (2) we first note that $\psi$ is upper semicontinuous, which implies that $\psi$ is bounded from above on any compact subset of $\mathbb{C}$.  Write $\Omega$ for $\C\setminus K$, and let $f\in A^2_{e^{-\psi}}(\Omega)$. It then follows that
  $f\in A^2(\Omega\cap\mathbb{D}(0,R))$ for any $R>0$. We now may proceed as in the paragraph following \eqref{E:metric} and construct a function $F\in\mathcal{O}(\C)$ such that
  $F|_{\Omega}=f$. Since $K$ is of Lebesgue measure $0$, it then follows that $\|F\|_{L^2_{e^{-\psi}}(\C)}=\|f\|_{L^2_{e^{-\psi}}(\Omega)}$. Thus,
  $A^2_{e^{-\psi}}(\C\setminus K)$ is isomorphic to $A^2_{e^{-\psi}}(\C)$.
\end{proof}

\section{Finite dimensional Bergman spaces in $\mathbb P^2$}\label{S:counter}

In this section, we show that the dichotomy displayed by Bergman spaces on $\mathbb P^1$ does not hold in higher dimensional projective spaces. In particular, for every holomorphic line bundle $L$ on $\mathbb P^2$, there exists a domain $D\subset \mathbb P^2$ such that the dimension of the Bergman space $A^2(D;L)$ is finite, but strictly larger than that of the space of global holomorphic sections of $L$. The examples in this section are entirely motivated by Wiegerinck's examples in \cite{Wi84}.

In analogy with $\mathbb{P}^1$, we use the coordinates $\zeta=[\zeta_0:\zeta_1:\zeta_2]$ on $\mathbb P^2$. The open set $U^{(z,w)}=\mathbb P^2\setminus\{\zeta:\zeta_2=0\}$ is identified with $\C^2$ via
	\bes
		\varphi_{(z,w)}:\zeta=[\zeta_0:\zeta_1:\zeta_2]\mapsto (z,w)=
							\left(\frac{\zeta_0}{\zeta_2},\frac{\zeta_1}{\zeta_2}\right),
	\ees
and $(z,w)$ are referred to as the local coordinates on the affine chart $U^{(z,w)}$. Any holomorphic line bundle $\Pi:L\rightarrow\mathbb P^2$ is of the form $\mathcal{O}(k)$, for some $k\in\Z$, where $\mathcal{O}(k)$ is the line bundle associated to the divisor $k\{\ell\}$ for any fixed hyperplane $\ell\subset\mathbb P^1$. In the local coordinates $(z,w)$, a global section of $\mathcal{O}(k)$ is given by a triplet, $s=(s_1,s_2,s_3)$, of holomorphic functions on $\C^2$ such that
	\be\label{E:sectionsP2}
		s_1(z,w)=
		\begin{cases}
			z^{k}s_2(1/z,w/z),& \text{if}\ (z,w)\in\C^2\setminus\{z=0\},\\
						w^{k}s_3(z/w,1/w),& \text{if}\ (z,w)\in\C^2\setminus\{w=0\}.
		\end{cases}
			\ee
Similarly, a Hermitian metric on $\mathcal{O}(k)$ is given by a triplet, $h=(h_1,h_2,h_3)$, of smooth positive functions on $\C^2$ that satisfy compatibility conditions analogous to \eqref{E:sectionsP2}. In view of Lemma~\ref{L:choice}, we fix the following smooth volume form on $\mathbb P^2$, and Hermitian metric on $\mathcal{O}(k)$, $k\in\Z$, respectively, in the local coordinates $(z,w)$ on $U^{(z,w)}$:
\begin{eqnarray*}
	\omega_{\text{FS}}(z,w)&=&(1+|z|^2+|w|^2)^{-3}\,dz\wedge d\overline z\wedge dw\wedge d\overline w\\
h_1(z,w)&=&(1+|z|^2+|w|^2)^{-k}.
\end{eqnarray*}
Finally, recall the following description of the space of global sections of $\mathcal{O}(k)$, see \cite[p. 81]{Varolin11}.

\begin{lemma}\label{L:Bergman2} Let $k\in\Z$. In the local coordinates $(z,w)$ on the affine chart $U^{(z,w)}$,
	\bes
		A^2(\mathbb P^2;\mathcal{O}(k))=\Gamma(\mathbb P^2;\mathcal{O}(k))=\begin{cases}
				\text{span}\{z^pw^q:(p,q)\in\N^2,\ p+q\leq k\},& \text{if}\ k\geq 0,\\
				\{f\equiv 0\},& \text{if}\ k< 0.
				\end{cases}
	\ees
In particular, $\dim A^2(\mathbb P^2;\mathcal{O}(k))=\dim\Gamma(\mathbb P^2;\mathcal{O}(k))=\max\left\{0,\frac{(k+1)(k+2)}{2}\right\}$.
\end{lemma}

\begin{theorem}\label{T:WiegP2} Given $k\in\Z$, there exists a domain $\Omega_{k}\subset\mathbb P^2$ such that
	\bes
		\dim A^2(\mathbb P^2;\mathcal{O}(k))<\dim A^2(\Omega_{k};\mathcal{O}(k))<\infty.
	\ees
In particular, $\dim A^2(\Omega_{k};\mathcal{O}(k))=1$ for $k<-2$, and
$\dim A^2(\Omega_{k};\mathcal{O}(k))=(k+3)(k+4)/2$ for $k\geq -2$.
\end{theorem}
\begin{proof} It suffices to produce a domain in the chart $U^{(z,w)}$, i.e., a domain $\Omega_{k}\subset\C^2$ such that $\dim A^2_{\phi_k}(\C^2)<\dim A^2_{\phi_k}(\Omega_{k})<\infty$, where
	\bes
		\phi_k(z,w)=(1+|z|^2+|w|^2)^{k+3},\qquad (z,w)\in\C^2.
	\ees
 The desired domain will be of the form $B\cup X_\ell\cup Y\cup Z_m$, for appropriately chosen $\ell,m\in\N$, where
	\begin{eqnarray*}
		B&=&\left\{(z,w)\in\C^2:\max\{|z|,|w|\}<2\right\},\\
		X_\ell&=&\left\{(z,w)\in\C^2:|z|>\sqrt{2},\ |w|<1/|z|^\ell\right\},\quad \ell\in\N,\\
		Y&=&\left\{(z,w)\in\C^2:|w|>\sqrt{2},\ |z|<1/|w|\right\},\\
		Z_m&=&\left\{(z,w)\in\C^2:|z|^2+|w|^2>2,\ \big||z|-|w|\big|<\frac{1}{(|z|+|w|)^m}\right\},\quad m\in\N\setminus\{0,1\}.
	\end{eqnarray*}

In the rest of the proof, $A^2_{k,\text{mon}}(\Omega)$ denotes the set of monomials in $A^2_{\phi_k}(\Omega)$, for any $\Omega\subset\C^2$. We claim that
	\bea
		A^2_{k,\text{mon}}(B)&=&\{z^pw^q:(p,q)\in\N^2\},\label{E:B}\\
		A^2_{k,\text{mon}}(X_\ell)&=&\{z^pw^q:(p,q)\in\N^2,\ p-\ell q\leq \ell+k +1\}, \label{E:X}\\
		A^2_{k,\text{mon}}(Y)&=&\{z^pw^q:(p,q)\in\N^2,\ q-p\leq k+2\}, \label{E:Y}\\
		A^2_{k,\text{mon}}(Z_m)&=&\{z^pw^q:(p,q)\in\N^2,\ 2(p+q)\leq m+2k+2\}. \label{E:Z}
	\eea
Assuming \eqref{E:B}-\eqref{E:Z} for the moment, set
	\be\label{E:dom}
		\Omega_{k}=
		\begin{cases}
			B\cup X_1\cup Y\cup Z_2,\ & \text{if}\ k\geq -2,\\
			B\cup X_{1-2(k+2)}\cup Y\cup Z_{-2(2k+3)},\ & \text{if}\ k<-2.
		\end{cases}
	\ee
Since $\Omega_k$ is Reinhardt for any $k\in\Z$, $A^2_{k,\text{mon}}(\Omega_k)$ is a basis for $A^2_{\phi_k}(\Omega_k)$. Thus, $\dim A^2_{\phi_k}(\Omega_k)$ is the cardinality of  $A^2_{k,\text{mon}}(\Omega_k)$. From \eqref{E:dom}, if $k\geq -2$,
	\bes
		A^2_{k,\text{mon}}(\Omega_k)=\{z^pw^q:(p,q)\in\N^2,\ |p-q|\leq k+2,\ p+q\leq k+2\},
	\ees
which is of cardinality  $\frac{(k+3)(k+4)}{2}$. On the other hand, if $k<-2$, i.e., $k+2=-t$ for some $t\geq 1$, then
\bes
A^2_{k,\text{mon}}(\Omega_k)=\left\{z^pw^q:(p,q)\in\N^2,\ \frac{p-t}{1+2t}\leq q\leq p-t,\ p+q\leq t\right\}=\{z^t\}.
\ees
Thus, recalling Lemma~\ref{L:Bergman2}, $\dim A^2_{\phi_k}(\C^2)<\dim A^2_{\phi_k}(\Omega_{k})<\infty$, in either case.

It now remains to prove \eqref{E:B}-\eqref{E:Z}. We use the notation $L\approx M$ for $L,M\in\mathbb{R}$ to mean that there exist constants $c,d>0$ such that
$cM\leq L\leq dM$.
 Since \eqref{E:B} is clear, and \eqref{E:Y} follows from \eqref{E:X}, we only need to prove \eqref{E:X} and \eqref{E:Z}.
To find the monomials which are contained in $A^2_{\phi_k}(X_\ell)$, consider
	\begin{eqnarray*}
		\widetilde X_\ell&=&\left\{(r,s): r>\sqrt{2}, 0< s <1/r^\ell\right\}.
\end{eqnarray*}
For $p,q,\ell\in\N$, we get
$$||z^pw^q||^2_{A^2_{\phi_k}(X_\ell)}=(2\pi)^2\int_{\widetilde X_\ell}
		\frac{r^{2p+1}s^{2q+1}}{(1+r^2+s^2)^{3+k}}\;ds\;dr=:(2\pi)^2\cdot \mathcal{J}.$$
To determine the range of $p$ and $q$ for which $\mathcal{J}$ is finite for a given $\ell$, we consider $\widetilde X_\ell$ in polar coordinates, i.e.,
$$\widetilde X_\ell=\{(R\cos\theta, R\sin\theta): R\cos\theta>\sqrt{2},\; R^{\ell+1}\sin\theta(\cos\theta)^\ell<1, \theta\in(0,\pi/2)\},$$
and introduce
$$\mathcal{J}(a,g):=\int_{a}^{\infty}\int_{0}^{g(R)}\frac{R^{2p+2q+2}}{(1+R^2)^{3+k}}(\cos\theta)^{2p+1}(\sin\theta)^{2q+1} R\;d\theta\;dR,$$
where $a>0$ is a constant and $g$ is a positive, continuous function. We notice that $\theta\in(0,\pi/4)$ on $\widetilde X_\ell$. It then follows that
$$ \mathcal{J}(2,\arcsin(R^{-\ell-1}))\leq \mathcal{J}\leq\mathcal{J}(\sqrt{2}, \arcsin(\sqrt{2}^\ell R^{-\ell-1})),$$
 because, firstly, any point in $\widetilde X_\ell$ satisfies $R>\sqrt{2}$ and $\theta\in(0,\arcsin(\sqrt{2}^\ell R^{-\ell-1}))$, and, secondly,
 any point with $R>2$ and $\theta\in(0,\arcsin(R^{-\ell-1}))$ is a point in $\widetilde X_\ell$. Hence, it remains to estimate $\mathcal{J}(a,\arcsin(cR^{-\ell-1})$ for $c\in\{1,\sqrt{2}^\ell\}$. Again,
 since $\theta\in(0,\pi/4)$ on $\widetilde X_\ell$, it follows that
 \begin{align*}
   \mathcal{J}(a,\arcsin(c R^{-\ell-1}))
   &=\int_{a}^\infty\frac{R^{2p+2q+2}}{(1+R^2)^{3+k}}\left(\int_{0}^{\arcsin (cR^{-\ell-1})} (\cos\theta)^{2p+1}(\sin\theta)^{2q+1}\;d\theta\right)R\;dR\\
&\approx \int_{a}^\infty\frac{R^{2p+2q+2}}{(1+R^2)^{3+k}}\left(\int_{0}^{\arcsin (cR^{-\ell-1})} \cos\theta (\sin\theta)^{2q+1} \;d\theta\right)R\;dR\\
&\approx \int_{a}^\infty R^{2p+2q+3-(\ell+1)(2q+2)-2(3+k)}\;dR,
 \end{align*}
which converges if and only if $2p+2q+3-(\ell+1)(2q+2)-2(3+k)<-1$. It follows that $z^pw^q\in A^2_{\phi_k}(X_\ell)$ if and only if
	\be\label{E:cond1}
		p-\ell q\leq k+\ell+1.
	\ee
In a similar fashion, we  find the monomials in $A^2_{\phi_k}(Z_m)$. For that, we write
	\begin{eqnarray*}
	\widetilde Z_m=\left\{(r,s):r^2+s^2>2,\ |r-s|(r+s)^m<1\right\}
	\end{eqnarray*}
and note that
for $p,q,m\in\N$
	\begin{align*}
		||z^pw^q||^2_{A^2_{\phi_k}(Z_m)}
	=(2\pi)^2\int_{\widetilde Z_m}\frac{r^{2p+1}s^{2q+1}}{(1+r^2+s^2)^{3+k}}\;dr\;ds:=(2\pi)^2\cdot\mathcal{I}.
	\end{align*}
As before, we introduce polar coordinates on $\widetilde Z_m$. It is straightforward to check that then
\begin{align*}
  \widetilde Z_m
  &= \left\{(R\cos\theta,R\sin\theta):R>\sqrt{2},\, \left|\sin\left(\textstyle\frac{\pi}{4}-\theta\right)\right|
  \cos^m\left(\textstyle\frac{\pi}{4}-\theta\right)<(\sqrt{2}R)^{-m-1},\theta\in(0,\textstyle\frac{\pi}{2})\right\}\\
  &=\left\{\left(R\cos\left(\textstyle\frac{\pi}{4}-\psi\right), R\sin\left(\textstyle\frac{\pi}{4}-\psi\right)\right):
  R>\sqrt{2},\,\left|\sin\psi\right|\cos^m\psi<(\sqrt{2} R)^{-m-1}, \psi\in(-\textstyle\frac{\pi}{4},\textstyle\frac{\pi}{4})
  \right\}.
\end{align*}
We work in the coordinates $(R,\psi)$ to estimate $\mathcal{I}$, utilizing integrals of the form
\begin{align*}
  \mathcal{I}(g):=\int_{\sqrt{2}}^\infty\int_{-g(R)}^{g(R)}
  \frac{R^{2p+2q+3}}{(1+R^2)^{3+k}}
	 \cos^{2p+1}\left(\textstyle\frac{\pi}{4}-\psi\right) \sin^{2q+1}\left(\textstyle\frac{\pi}{4}-\psi\right)  \;d\psi \;dR
\end{align*}
for some positive, continuous function $g$. Since $\psi\in[-\pi/4,\pi/4)$, it follows that
\begin{align*}
  \mathcal{I}\Bigl(\arcsin\bigl((\sqrt{2}R)^{-m-1}\bigr)\Bigr)\leq \mathcal{I}\leq\mathcal{I}\left(\arcsin\left(\sqrt{2}R^{-m-1}\right)\right).
\end{align*}
Since $m\geq 2$, both trigonometric functions in the definition of $\mathcal{I}(g)$ are strictly positive for $|\psi|\leq|\arcsin(\sqrt{2}R^{-m-1})|$.
Therefore,
\begin{align*}
  \mathcal{I}\left(\arcsin(cR^{-m-1})\right)\approx
  \int_{\sqrt{2}}^\infty\int_{-\arcsin(cR^{-m-1})}^{\arcsin(cR^{-m-1})}
  \frac{R^{2p+2q+3}}{(1+R^2)^{3+k}}
	  \;d\psi \;dR
\end{align*}
for $c\in\{\sqrt{2},\sqrt{2}^{-m-1}\}$.
Moreover, $|t/2|\leq|\sin t|\leq |t|$ holds for $|t|\leq\pi/4$. Thus
\begin{align*}
  \mathcal{I}\left(\arcsin(cR^{-m-1}\right)&\approx\int_{\sqrt{2}}^\infty R^{2p+2q+3-(m+1)-2(3+k)}dR,
\end{align*}
which converges if and only if
	\be\label{E:cond2}
		2(p+q)\leq m+2k+2.
	\ee
This completes the proof of \eqref{E:Z}, and thus, of Theorem~\ref{T:WiegP2}.
\end{proof}

redpajama_set_name: RedPajamaArXiv


Currently, the Academic Senate on the Urbana-Champaign campus is a legislative body comprised of 200 faculty and 50 undergraduate and graduate students and performs and important role in shaping educational policy for our institution that has a strong tradition of shared governance.
For more information please visit the Senate's website at: senate.illinois.edu.
Senate meetings are also rebroadcast on UI-7.
Are Academic Professionals new to the Academic Senate?
Recognizing the substantive role that Academic Professionals play in advancing the academic mission of our institution, the Academic Senate included an Academic Professional in each of its committees to assist them in their work and represent the interests of the Council of Academic Professionals (CAP) and the AP community. The Council (CAP) solicits self-nominations from the community of Academic Professionals and determines the AP representative that is then considered for approval by the Senate Committee on Committees and the Senate as a whole. So even though up until now, Academic Professionals have not been part of the Senate, they participate in the key functions performed by its Committees.
Who will conduct the elections?
In accordance with the above-mentioned SP.10.09–B , the Senate Committee on University Statutes and Senate Procedures has delegated to the Council of Academic Professionals the authority to administer the elections for the seven (7) AP senators.
How many Academic Professionals will get elected to the Academic Senate and from which CAP districts?
Seven (7) Academic Professionals will get elected to the Senate; the eligible electorate in each of these CAP districts will elect one AP Senator: 2, 3, 5, 6, 8, 9 and 11. Anyone eligible to vote is also eligible to run for a Senate seat.
Why doesn't every district get a seat on the Senate?
Unfortunately, to get any representation on the Senate, some compromises had to be made. The Senate's guiding principle was that any academic professional on the Senate must be engaged in and responsible for the educational function of the Urbana-Champaign campus. University Administration (UA) employees, such as those in CAP District 1, the University of Illinois Press, and the Institute for Government and Public Affairs, are not UIUC employees and thus are ineligible to serve on the Senate. Ordinarily, the educational function of the University will involve teaching and research. The Senate felt that districts dominated by high-ranking administrative units, e.g., District 4 (e.g., Vice Chancellor for Research), District 7 (e.g., Vice Chancellor for Student Affairs), and District 10 (e.g., Offices of the Provost and Chancellor) were inherently administrative and should not be included. This is not to diminish the work of any AP; all of us contribute in a meaningful and essential way to the mission of the university. But, at least right now the Senate seeks to include in the electorate only those APs whose work can be directly tied to the academic mission. It should be noted though that staff in these districts/units may still seek to serve in Senate committees (as noted above) and/or as a district representative on CAP.
Of these criteria, the educational function criterion is the most difficult to define. The University Statutes and Senate Procedures Committee has advocated regular contact with students, which involves teaching or furthering the student's research. With respect to PRI (District 11), the person must be actively involved in research. With respect to Extension (District 6), the person must be actively involved in teaching.
Ultimately the Deans of each College (or the executive officer of Extension and PRI) determine who meets these criteria. However, it is anticipated that many Deans will delegate this responsibility to their respective department heads. Colleges are allowed latitude in interpreting these guidelines because it is impossible to have a single set of criteria that is applicable to every academic unit on campus. For example, activities involved in the teaching and research mission of the University might be very different in the College of Fine and Applied Arts (Music, Dance, Art and Design, etc.) than in the College of Engineering.
Are Deans and unit heads eligible to vote?
Deans are already ex-officio members of the Senate. Thus they are not part of the AP electorate. Many department heads are eligible to vote and run for faculty Senate seats; thus, those individuals cannot simultaneously vote and run for AP Senate seats. In other words, you can't be a member of both the faculty and the academic professional electoral groups.
As an Academic Professional, what do I need to do in order to be prepared for these elections?
You need to be aware of which CAP district you belong to. You can determine that by visiting this page on the CAP website. If it is determined that you are part of the electorate, you will receive more information about the election including directions on submitting a self-nomination. Consider attending our February town hall meeting about the elections. If you do not plan to be a candidate, consider encouraging someone who you think would be a strong candidate. Ideally, CAP would like to see two strong candidates from each district. We would like to see AP senators prepared to make a real contribution to the Senate, represent the concerns of their fellow APs, add their unique expertise and perspective to the Senate, and demonstrate above all that an academic professional presence on the Senate does not detract from faculty and student representation, but in fact makes the institution stronger.
What are the future prospects for APs in the Senate?
As a future objective, we hope that someday the campus' academic professionals, regardless of district/unit/function, will be included in the Senate electorate. We believe that the composition of this initial Senate electorate is but a first step in securing a voice and representation for APs in the Senate. Efforts will continue to extend this privilege to those APs not currently eligible to participate. For now, we should be encouraged by this significant first step that serves to strengthen the voice and impact of APs in the shared governance process. And, we should strive to make the best use of this opportunity so that it might lead to even greater participation for APs in the future.


Each module in the CPA Professional Education Program (CPA PEP) includes a mandatory, in-person workshop, as indicated in CPA Canada's Harmonized Education Policies. All candidates are also required to attend the Orientation Workshop on the first two days of Core 1.
Each workshop occurs over a two-day period which runs from 8:30 am to 4:30 pm each day.
Candidates who miss the workshop are not permitted to write the module-end examination. This counts as a module attempt (fail) and the candidate is required to repeat the module.
All workshop sessions follow the schedule below. The workshop session leader will inform candidates of times for lunch and any breaks.
Sunday 8:30 a.m. to 4:30 p.m.
Failure to attend any portion of the workshop counts as an automatic failure of the module.
In special circumstances, candidates can request an accommodation to miss part or an entire workshop. Accommodations are only granted for medical reasons, religious requirements, or extenuating circumstances. Candidates must provide proof of the circumstances for CPAWSB to approve an accommodation. Additional information about accommodations is available here.
Candidates are expected to maintain a level of professional behaviour while attending workshops. Incidents of academic misconduct and/or inappropriate behaviour may be recorded on candidate records.
Attending another location is not permitted unless you have received prior permission from CPAWSB.
Candidates are required to complete some work before attending the Core 1 and Capstone 1 workshops and will receive access to the module materials one-to-three business days before the module starts. Candidates enrolled in all other modules will receive access to module content one business day before the module starts.
An electronic copy of the workshop files, downloaded from D2L-Brightspace as an internet connection may not be available.
When registering for a module during the module registration period, candidates select their workshop city. Candidates who register after the close of a registration period are required to pay a waitlist fee and select their top three locations and will be placed in a location based on availability.
Location availability depends on candidate registrations and locations without enough candidates may be cancelled. Candidates in cancelled locations will be emailed following the close of the registration period and given alternative options.
Generally, CPA PEP workshop venues are post-secondary institutions or hotel conference centres. The specific locations are confirmed two weeks before the workshop when the workshop locations are posted on the CPAWSB website. Candidates require permission from CPAWSB to attend a workshop in another location.
Candidates seeking to participate in workshops in other regions (Ontario, Quebec, Atlantic Canada) should contact cpamodule@cpawsb.ca at least three weeks before the scheduled workshop date.


← Lordship Salvation: Why Am I Here?
Is Lordship Salvation a False Gospel of Works? No Question!
Romans 3:26: To declare, I say, at this time his righteousness: that he might be just, and the justifier of him which believeth in Jesus.
I happened back across a website last night called Got Questions? I had seen it before, and knew it to be unreliable, with a definitive Lordship "salvation" (LS) bent. However, I was startled by the starkness of its false messages, and attempts to reconcile LS (which is a man-centered false gospel of works) with the gospel of grace.
Just as salvation cannot be earned by good works, neither does it need good works to be maintained or sustained. Good works and changed lives are the inevitable results of salvation.
My comment: This is a harbinger of the false messages and internal contradictions that can be found throughout their website.
If good words and changed lives are the inevitable results of salvation, then no one would have assurance of salvation until the good works manifested themselves in sufficient quantity and nature to meet the arbitrary, differential, and fickle standards of the person who is making the determination of either himself or others.
The Bible makes no such claim. In fact, the Bible says that assurance of eternal life comes from God's promises alone.
John 5:24: Verily, verily, I say unto you, He that heareth my word, and believeth on him that sent me, hath everlasting life, and shall not come into condemnation; but is passed from death unto life.
Free Grace theologians consider their position more biblical than Lordship Salvation, which they consider to be a works-based theology. According to Free Grace theologians, Lordship Salvation holds that saving faith includes inherently the "act" of accomplishing radical internal change leading to good works.
This leads to the Free Grace emphasis on assurance of salvation, again based on the basic promises in John's Gospel, that belief is all that is necessary for salvation. To the Free Grace theologian, this is a simple, cut-and-dried issue—if you believe, you are saved. For the Lordship Salvation camp, assurance of salvation comes through the observation of change in the professing believer, i.e., that he is accomplishing good works. Each camp views the other as possibly leading to heresy.
Although Free Grace Theology and Lordship Salvation are terms that have developed only recently, they represent concerns that have been around since the beginning of the church. At the end of the day, there is no question about the basic salvation of those who hold either view. Both views are within the limits of orthodoxy.
Titus 3:8: This is a faithful saying, and these things I will that thou affirm constantly, that they which have believed in God might be careful to maintain good works. These things are good and profitable unto men.
Romans 6:12: Let not sin therefore reign in your mortal body, that ye should obey it in the lusts thereof.
Romans 12:2: And be not conformed to this world: but be ye transformed by the renewing of your mind, that ye may prove what is that good, and acceptable, and perfect, will of God.
Galatians 5:16: This I say then, Walk in the Spirit, and ye shall not fulfill the lust of the flesh.
Second, the errant Got Questions? website has admitted that LS forces its sad adherents to search their performance for assurance of eternal life. This is no basis for assurance.
Third, there IS a question about the salvation of those who hold the LS view.
My comment: Contrary to the claims of Got Questions?, Lordship "salvation" IS a salvation-by-works doctrine. Their claims are both non-biblical, and internally inconsistent. Faith itself is NOT a gift of God, faith need not involve a commitment to Christ in order to result in eternal life, and faith is NOT more than being convinced of the truth of the gospel. None of the biblical references that they have provided, nor any other passages from the Bible, support their erroneous claims.
The Got Questions? site is managed by a group that is either ignorant of grace or enemies of grace. Either way, the site, and those involved with it, should be marked and avoided.
So, is Lordship "salvation" a false gospel of works? Despite what the false teachers from Got Questions? have to say, the Biblical answer is ABSOLUTELY!
If you would like the truth regarding how to have eternal life, click here: THE GOSPEL.
This entry was posted in Assurance, Assurance of salvation, Lordship Salvation, Salvation. Bookmark the permalink.
Kakash, doubts often come by listening to the teaching of men. It also comes from difficult passages that people misconstrue regarding losing eternal life, or proof of eternal life.
Please openly ask any questions. There is a difficult passages section on this blog, which will help answer some of those questions. Always when you're reading a passage, ask WHO it is being spoken to (i.e. eternally secure believers). Ask whether the 'salvation' spoken of speaks of being saved from eternal hell or a deliverance from something else.
Eternal life is free, and is God's promise. He will never take it away. I'm reformatting an old article on Eternal Security on my blog Redeeming moments (linked on the right) and hope to have it done today Lord willing.
Kakashi, some believers have doubts, some more than others. Some have very few, or none. It is not the quality of our faith that saves us, it is the object – Jesus Christ. We do not have to come to the point that we know we will never have any doubts again in order to be saved. If a believer has doubts, the last thing that will help is trying to believe harder.
When someone is experiencing doubts about whether or not he has actually believed, asking himself if he is sure may only cause more doubt.
I would recommend focusing on Christ, and not on your doubts.
I believe that Jesus died on the cross to save me, that he was buried and that he rose again on the third day. It just happens that I constantly doubt whether or not I'm saved. I read the Bible and my doubts go away but they come back. Does this mean I'm not saved? But I believe and trust Jesus and what he did to save me. That the only thing I cling on to. I only cling on his promises. But I am afraid because I think I don't believe him enough because of this doubts.
I just saw this. Many of us have battled the confusion and some of still battle the confusion. I am praying for as well.
Kakashi, praying you will be able to rest in Him, as you get more time in His Word and prayer. Trust in His grace and mercy, and ask Him to show you over time how to rightly divide His Word. Stick with us here, we'll do our best to answer questions.
helpedbygrace, I agree with Tom that Ephesians 2:8-9 is referring to eternal life, rather than faith, as the gift of God.
If faith itself were the gift, and faith is also the means by which we appropriate eternal life given to us by Christ Jesus, then eternal life would not really be a gift received freely – it would be an imposition.
If faith itself were the gift, and not everyone receives the gift, it would mean that God withholds eternal life from some people, even though the Bible says that eternal life is available to everyone (Revelation 22:17).
If faith itself were the gift, that would mean that God has not given us free will. If God hasn't given us free will, then any choice we make is really Him making it for us. That would necessarily include all sin.
If faith itself were the gift, and I didn't believe in Jesus, that would mean that God doesn't want all people to be saved. Yet, the Bible clearly states otherwise (2 Peter 3:9, 1 Timothy 2:4).
I saw one of Tom's sermons where he quoted a Bible scholar saying that salvation is the gift in Ephesians 2:8-9, not faith. If you have any experience with latin languages like french or spanish, you know that some nouns are feminine and some nouns are masculine or neuter. In these languages, you would be able to tell which noun an adverb is referring to based on whether the adverb is feminine or masculine/neuter. According to this Bible scholar, the construction of Ephesians 2:8-9 in greek shows that the gift is salvation, not faith. This would make sense since it also agrees with other Bible passages like Romans 6:23.
Kakashi, welcome. I have prayed for you.
A blessed day. Can anyone please recommend some articles that I can read and verses on the Bible that supports the faith alone in Christ alone? I have been battling with all of these confusion and I would want nothing more but to get these off me. I would also appreciate it if anyone can pray for me. Thank you very much! God bless.
Holly, "Got Questions" gives lots of non-biblical answers that corrupt the gospel. It is LS through and through.
Johninnc – if gotquestions was right in that regard (they are not), then I wonder what the guy in 1 Cor 5 was like before his sin 'decreased'?
The hallmark of a true Christian is the decreasing presence of sin in his life.
My comment: If that statement were unequivocally true, than the basis for assurance of eternal life would be the decreasing presence of sin in one's life.
The Bible says that we can have assurance of eternal life based on God's promise alone, without reference to the extent, if any, to which our lives begin to manifest change.
Larry – you are right to be troubled. It does affect their gospel, not only how they give it out (rarely consistently the same) but also how they troublt others who may have believed). One site I saw recently was called 'reason for hope'. On their statement of faith they had what seemed to be a fairly normal gospel (but no real detail or defense of certain truths). Yet on another page, they start off o.k., then continue with their laundry list of works.
We can't earn our salvation. We are saved by God's grace alone, through faith in Jesus. If you don't know Jesus, 1) repent of your sins today. 2) Confess your sins and 3)ask Jesus for forgiveness. 4) Turn from your sins and turn to Jesus. Trust that He loves you, that He paid the penalty for your sins, that He desires you to come to Him in faith, and that He will forgive you and give you eternal life.
Larry, I think someone who believes in Lordship "salvation"(LS)would believe in LS whether or not those parables were even part of scripture. I think they would simply find other proof texts to support their false doctrine.
Many LSers believe that the prodigal son was saved before he left his father's house, and that his return proves that a "true Christian" may stray for a while, but that he will always return.
The parable of the sower clearly says that people represented by soil 2 believe for a while. Since we know that anyone who believes has everlasting life, it seems clear to me that those represented by soils 2-4 have everlasting life.
I believe "Lordship/Reform doctrine has greatly influenced theological viewpoint even though they may say they are not Reform or Lordship.
My son is attending a Baptist church and the pastor takes the position that the prodigal son was lost (on the way to hell) before he returned to the father. And in the parable of the sower the only "genuine" believers were the seeds that fell on good ground. This troubles me as I believe it will greatly impact how they give out the gospel and how they determine whether one is saved or lost.
The account of the demon possessed man in Mark 5:18-20 who was healed by Jesus debunks Lordship Salvation's false teaching that one cannot be saved unless one makes Jesus as "Lord".
Mark 5:18-20 – 18 And when he was come into the ship, he that had been possessed with the devil PRAYED HIM THAT HE MIGHT BE WITH HIM.
The man who was demon possessed and healed by Jesus obviously wanted to follow Jesus as Lord as the context of verse 18 clearly indicates, BUT Jesus told him "NO", but that he should go home and tell his family and friends how God has been gracious to him.
They don't have the Five Solas. They have redefined "solus -a -um" as MEtheism.
"Got Questions" is no place to get answers! I often hear Calvinism being praised as "logical". I could never see it as anything but an illogical system that contradicts scripture when read in context. Calvinism is not logical but it is highly systematized. It can and will always circle back on itself and there are many "boogie men" that threaten the system that they must warn of (like easy believism) with no shortage of catchy nonsensical phrases to use like "salvation is certainly free, but, at the same time, it costs us everything" or "we are saved by faith alone, but saving faith is never alone". "Free" redefined. "Faith" redefined. "Alone" redefined.
The system is arranged around the Five Solas one of them being Scripture alone (where they say they get the "doctrines of grace"). What they really mean (in their conversations) is Scripture as (insert name of famous Calvinist) reads it (and adds to it) and they do this even after you have sorted out the "you" "we" "us" "them" of their favorite proof text (Rom 8, Eph 2, Jam 2, 1John etc). It's a hard circle to break.
Oh, and how about this gem?—"If there is no evidence of growth and good works, we have reason to doubt that salvation ever truly took place." This makes "no works" the opposite of faith when in reality "doubt" is the opposite of faith. "Got Questions" is pushing doubt, boasting in works and not preaching faith alone in Christ alone because "alone" simply does not mean alone. Duh!
Following are excerpts from an article at "Got Questions" entitled "What is Easy Believism?" I bolded some sections for emphasis.
So we see that faith, given as a gift by God, is what saves us. But the next verse tells of the results of that salvation: "For we are his workmanship, created in Christ Jesus unto good works, which God hath before ordained that we should walk in them." Rather than being saved by some easy act of our own wills, we are saved by the hand of God Almighty, by His will and for His use. We are His servants, and from the moment of salvation by faith, we embark on a journey of pre-ordained good works that are the evidence of that salvation. If there is no evidence of growth and good works, we have reason to doubt that salvation ever truly took place. "Faith without works is dead" (James 2:20), and a dead faith is not a saving faith.
My comment: Eternal life, not faith, is the gift of God.
Romans 6:23: For the wages of sin is death; but the gift of God is eternal life through Jesus Christ our Lord.
No "evidence of growth and good works" is necessary to confirm that one has received the free gift of eternal life.
Salvation is certainly free, but, at the same time, it costs us everything. We are to die to ourselves as we change into the likeness of Christ. Where easy believism fails is its lack of recognition that a person with faith in Jesus will lead a progressively changed life. Salvation is a free gift from God to those who believe, but discipleship and obedience are the response that will no doubt occur when one truly comes to Christ in faith.
My comment: NONSENSE! Something cannot be free to someone, yet cost them ANYTHING or EVERYTHING. Easy believism does not fail in its lack of recognition that a person with faith in Jesus will lead a progressively changed life. There is simply no guarantee in the Bible that a Christian will lead a progressively changed life.
Further, if faith is the gift of God, as "Got Questions" so falsely asserts, then why would they say that anyone "comes to Christ in faith"? According to their false doctrine, Christ comes to them with faith.
"Got Questions" is scratching for answers that simply aren't there. It is a heretical Lordship "salvation" site that should be avoided at all costs.
Yes, I can relate!! I struggle reading Scripture now because of the LSer's & the lies Satan uses them to teach! Satan has a fine time in my mind. But, I hope you will be encouraged that no matter his tricks & lies you KNOW in whom you have put your faith & you will not be disappointed.
I encourage you to take John's advice….by the way, I have had difficulties in doing the things John listed still, so don't get discouraged.
Yeah, I looked over the site, at their SOF. After I'd posted there I saw that confusing bit you quoted. I agree, it does have a "repent of sins" sound to it. My posted questions haven't drawn any further response so far. Doesn't look like a site I'd waste anymore time with.
Thanks for the prayerful back-up, everyone.
When I was on Black Voices, where I didn't belong, I met some Pentecostals who believed in eternal security. One of them was even Word of Faith. Unfortunately, the fine nuances between grace and Zellerism were never brought up. In fact, I probably confused them. They made me feel better because of the extreme LS views that had me down. One of them had a doctrine of reprobation that he applied to me when I said that I would not repent of an issue where he was weak. Needless to say, they were an unstable bunch. Hebrew Israelites, Muslims, goddess worshipers, and Egyptians were participants in the general brawl. I heard of Obama five years before he ran for president, and they were already hopeful. What a mess.
I found a SOF on their 'grace' site, and it appears they do believe in a separate baptism of the Spirit. They also say this confusing bit.
Since they separate repenting to the end, it suggests they believe repentance means turning from sin. And I still didn't see any 'gospel' (whether true or not) in the SOF.
Chas, just a thought. Whether Pentecostals don't hold with the hyper WOF Pentecostals, they still are all basically Arminian. Some will tell you that eternal life is eternal, but you can 'walk away'. I haven't ever met a Pentecostal pastor who has that one right, which means they don't have the gospel right. I'd find his gospel, because likely it has a turn from sin requirement, and possibly a (if-you-don't-speak-in-tongues-you-haven't-been-baptized-by-the-Spirit caveat). I don't know much about him, just quickly perused his blog. Whenever I can't find an easy gospel, I just don't spend time listening to anything they have.
Chas, I'm praying for you.
Just to be clear though, there are Christians who are licentious. And, others are licentious sometimes. As you said, sin in the lives of believers has consequences – both natural and eternal. But, even licentiousness does not annul grace.
Oh well, looks like I've stepped in it again.
That, of course, is a popular distorted caricature of the doctrine of salvation by grace alone, and as such it struck me as just what a LS-er would say. The commenter also quoted Titus 2:11-15 and "Jude 1:4" (sic).
"Thank you for taking the time to reply, [name withheld]. You've cited what some falsely attribute to the 'Free Grace' view of justification, and stated what is essentially John MacArthur's 'Lordship Salvation' view. It's not clear to me that Pastor Randles holds MacArthur's view. I certainly hope not.
Re: Titus 2:11-15: The passage is not a definition of grace, but a statement of God's intent in giving His grace–His "UN-merited favor"–which cannot be earned in any way; not by works, not by "turning from sin" and not by any promises to do so.
Those of the Free Grace view do not turn the grace of God into licentiousness. We acknowledge that a believer's continuing in sin will incur God's discipline, and a lack of Spirit-led good works will result in a lack of rewards in heaven. But neither good works, turning from sin, nor any promises to do either play a part in justification. Eternal life is a gift; received by grace alone, through faith alone in the sacrifice of Christ alone. A gift requires no payment, reimbursement or pledge of future compensation.
Naturally, my response didn't cover all points of contention. It's probable that the commenter will fire away with the complete arsenal of LS "proofs". So, what I'm saying is, pray for me. I think I've opened a can of worms. And I still don't know what Randles' views are because he has not responded as of this post.
Luke, I've prayed for the people with whom you shared the gospel.
I am also thankful for Holly's comment. I hated to post my comments but I had to think seriously about recommending an entire series for those recovering from false doctrine. If it wasn't helping me, could it help my neighbor? Exposure to persecution gospels in the past has made me hypersensitive. I have been helped by both Tom and Ralph, and being negative is far from me.
When Jesus says that the world will hate us, He is not saying anything along the lines of "a Christian will" but simply being realistic, much like we ourselves would be if we said that life is not a bowl of cherries. He cannot contradict Himself and surely would not teach, "Judge not or ye shall be judged" (as in Matthew 7:1), and then encourage us to judge each other over a fruit.
Holly, thank you for a wise and thoughtful comment.
One thing I know is I'm constantly cleaning up my past stuff, as we quote people not knowing their full doctrine. My parents went to J. Vernon McGee's church and did the Bible studies in our home when I was a little girl. I remember sneaking out to the hallway to listen to his drawl, and remember about one thing, and that is 'this is where the rubber meets the road' (something to that effect). The preacher before I 'think' was Talbot when my dad was a little boy. The issue is a lot of these men learned from other men's commentaries without proving all things by the Word. We all need to do that, and one thing I know about Tom is that he is open always to hearing and is not quoting people now if he can help it. I am also trying to be careful and also have to spend some time to clean up my recommended list, although we all have the responsibility to be Bereans, to test the spirits, to prove all things.
We also need to be cautious in bringing accusations against elders, and realize sometimes someone else see's things from a different way than we view it. For example the first thing I thought with 'you can't be a Christian and be popular' is that we'll always be hated (by the enemy) whether it seems that way in public or not. Carnal believers of course can be popular. But Luke 6:26 came to mind as well as others, but I see what Jason is saying, so he saw it another way, and really, he is correct there.
A believer can be popular. Who knows how these men mean things when they say them and since we're human, we're apt to make mistakes anytime we start using our own statements (or other's words) vs. teaching His Word. I've done it, and about the time I think I'm being careful, I do it again, but that's what the body is for to hold each other accountable and help each other be careful in all these things.
For what it's worth Jason, I found this sermon by Yankee very encouraging recently: The Necessary Blood of Christ https://www.youtube.com/watch?v=kFKDvcT8z1I We can all use some encouragement in these trying times I'd say.
Indeed. I listened to that series on John partly to recover from listening last week to the cult that I brought up on this thread. What I am hearing in these parts is not helping.
Note: Please remember to consider all interpretations of scripture in light of consistency with the clear biblical tenets of grace, eternal security, and assurance based solely on God's promises – no matter who the source is.
See number 4 under my previous comment in this thread.
Part 46 pos 15:55 is shocking, especially coming from a careful preacher. It is a thought I myself had during a careless moment and later regretted.
A more general annoyance is that the later parts of the series are sounding too much like the epistles. How is John 12-14 part of that which was written that we may know we have eternal life? I am unenlightened. Part 45 even sounds like the Sermon on the Mount!
Jason, being a believer who waffles on grace also improves one's appeal to the religious community.
It is difficult to be popular without being carnal, but being popular and a carnal believer is easy.
Jason, Tom is usually careful with references, but that one isn't good, nor accurate.
I'm not a big fan of McGee and consider him to have been inconsistent on grace, at best, and more likely LS.
It is difficult to be out as a grace believer and still be poplular, particularly with the churchianity set. But, it is not impossible to be a believer and be popular.
I am listening to Tom's 58 part series on John. The early parts were helpful. But starting in part 45, I'm feeling condemnation. Too many McGee quotes, some of which are plain awful. Example: "it is impossible to believe and be popular". Really? If I have learned anything here, I have learned that it is possible to be a believer and do or be pretty much anything, not that one should.
Jason, thanks for your encouraging words for Lise!
A year ago, I was where Lise is now. Spending time with Biblical teaching has helped. It amased me how some Bible passages read differently, later.
lise, I have a few suggestions.
1. Pray for God to guide you where you are confused.
2. Remember that God neither lies, nor contradicts himself. Since the Bible is the word of God, there are no lies or contradictions.
3. Use clear passages to explain major tenets of the faith (grace, eternal security, assurance), rather than using less clear passages.
– Is it consistent with eternal life by Grace alone through Faith alone in Christ alone?
– Is it consistent with eternal security?
– Is it consistent with assurance of eternal life, based on God's promises alone (i.e., it is not internally-focused on changes in attitudes, behavior, etc.)?
If an interpretations meets ALL THREE of these criteria, it COULD be correct. If it does not meet ALL THREE, it CANNOT be correct. Realize that any previous LS interpretation that you may have heard is not true.
5. Read "THE GOSPEL" booklet by Ron Shea once a day for a whole month. Then, read the book of Galatians through multiple times.
6. Listen to some sermons on "Sermon Audio" from a reliable pastor, such as Tom Cucuzza. Start with his verse-by-verse series on the book of John.
7. Ask questions here, but remember not to rely on what people tell you, unless it lines up with scripture and clearly meets the criteria of grace, eternal security, and assurance.
8.You may not understand a verse or passage completely, and that's OK. More understanding will come with time, as you are better grounded.
Is there a Bible teacher anywhere who can lead me through scriptures and help me to find hope when all I see when I look is condemnation?
I don't find encouragement in the Bible. Where is the good news?
I have over 30 years of LS teaching that's been drilled into me. I feel like I need every passage to be re-taught.
I can believe that God loves me and all is forgiven– past, present, and future, and that all I have to do is trust him.
But then I open my Bible and I'm instantly deflated.
Can anyone else relate to this?
Jon, no. Neither good works nor spiritual growth are inevitable in the life of a Christian.
Is there a difference between someone suggesting works are inevitable vs spiritual growth is inevitable for a believer?
Ha, ha. Good one, johninnc. Very true. 🙂 The false gospel message is ages older for sure.
Jason, like Got Questions, all of the others you mentioned are bad news.
Got Questions articles were items I used to stumble upon a lot in my searches. Just months before I came here, they were one of the winds of doctrine that were blowing me, along with Anderson, JesusIsSavior, and Zeller.
Jason, Satan is very subtle. He likes to have people dwell on false teachings to either keep people lost or to trip up believers.
Me too. The sublety of these cults is not that they teach LS in its naked deformity, but that they challenge our fundamental assumptions about what God really said. Did God really say, leave your life and follow me? Did he really say that to Peter the fisherman? Is Peter the model for every believer? It is very disarming. They are wise as serpents and not harmless as doves.
Fryingpan, it is always good to hear from you!
Yeah, their parody of "Got Milk" is really old. Not as old as their works-for-salvation doctrine, but really old.
Fantastic article, johninnc! You once again knock it out of the park. Not only is "Got Questions" errant, but THEY'RE LS through and through!
Praise be to God that the nearly 5 years since discovering this blog have given me the ability to not be taken in by these folks!
I'd actually mark and avoid them just for using the Got Milk? parody. But that's my personal bias. 🙂 It was cute 25 years ago but it's been done approximate 17 quadrillion times now. Show some initiative and find some original ideas, guys.
Jason, if the guys you referenced believe they need to do these things to have eternal life, then they are not following Jesus.
I try to stay away from cults.
Brady, the huge problem with the view that "faith" is the gift of God is that it is used to support the Calvinist tenet of unconditional election.
The Bible says that eternal life, not faith,is the gift of God (Romans 6:23). One receives the Spirit upon believing in Jesus as Savior (Galatians 3:2).
And yes, Christians are instructed to walk in the Spirit, and to not grieve the Spirit.
Hebrews 12:2 refers to Jesus as the author and finisher of our faith, but that does not mean that God chooses who will believe in Jesus as Savior.
Is Lordship Salvation a teaching of Jesus? The cult, Jesus Christians, sincerely thinks so. And they think they are the only followers of Jesus. No other group takes the teachings of Jesus as literally as they do: quit your job, sell all your possessions, give all your money to the poor, hate your family, live like the sparrows, do not keep a spare kidney, and if your right hand causes you to worship mammon during the great tribulation, then cut it off literally. Talk about radical. I was unfortunate to stumble upon one of their youtube channels, A Voice in the Desert. They teach that America will be nuked right before the tribulation, with no rapture.
It is impossible to follow their teachings without living like bums, and when I dig deeper, that is exactly what they do. They teach dogmatically that having a job and getting paid is worshiping mammon. They also engage in illegal activities, kidnapping, and self-flagellation. Such a lifestyle would cause me to worry about food, and they tell me it won't? They get thrills from offending others. John 5:24 is a teaching of Jesus, and they do not follow it.
Thank you for this excellent response to Got Questions. I also noticed the completely contradictory language when I came across their definitions pertaining to grace and Lordship Theology several months ago. Biblical grace is completely absent from their attempt to answer these questions. However, I don't think that receiving the gift of faith as a problem for Free Grace proponents anymore than receiving the Holy Spirit. But we should walk in the spirit so that we don't grieve the Holy Spirit. And doesn't the Bible say that God is the author and perfector of our faith?
Sometimes, I find it frightening how subtle the LS "gospel" can be. It is so seductive to substitute "works are inevitable" for "works are necessary" and think one is still believing the gospel of grace.
Whether it's "necessary" works or "inevitable" works, the question is the same: "How much works?" When that unanswerable question is raised, the problem with either view is shown to be the same, and both views are shown to be false.
Lordship Salvation: This Easter, Are You a Basket Case?


text:
Tanya Singh
Original Articles July 14, 2020 No comments
The Ice Age in relations is here to stay: how US-China tensions are expanding from trade to everything.
Written By: Tanya Singh
The US-China trade relationship has expanded immensely after China's reformation of its economy and liberalization in 1979.The world's two largest economies have persevered bitter rivalry locked by a trade battle for a year now. The onset of the pandemic revealed new dynamics in the US-China trade wars, overshadowing the agreement made in January that was meant to draw President Trump's policy still endeavours to encourage domestic production and purchases leading it to impose more than $360bn worth of tariffs on Chinese goods. According to the "phase one" deal signed in January, China vowed to boost US imports by $200bn above 2017 levels and extend intellectual property rules, however, Trump seems to have delayed the upliftment of tariffs and with the recent indictments regarding Covid-19, it seems that both the countries are back to square one. Here's how we can break this situation down:
What do the accusations look like?
The US denounces China of unfair trade practices and for a breach of its intellectual property and demands some major changes in its economic policies to call a truce. It also holds allegations against China pertaining to mixed records on implementation of WTO obligations, establishment of procedures for impacting the value of its currency and restrictions on FDI. US demands that China must increase its US-based imports to clear its $321.2bn trade deficit with China. Tariffs on $200bn worth of Chinese goods was due to rise to 25% form 10% this year but this hike halted amidst the breakout of a pandemic. China on the other hand has targeted most US goods ranging from chemicals and vegetables to whiskey, while also targeting alternative imports for goods like soybeans.
What does the policy failure look like for both countries?
The United States had been pursuing a containment policy marked by a three-tier strategy of the trade war, technological blockade, and ideological attacks, which failed to create a significant impact against China. This was mostly because, the principal cause of a dip in the Chinese economy is caused by the differentiation in the domestic supply and demand, the financial bubble caused by land financing. Hence, both times China's GDP growth rate surpassed its target of 6 percent, majorly unaffected by the US's policy against it.
With the omission of these policies amidst the pandemic outbreak, the US has been a victim to severe economic losses, American stock markets saw the successive triggering of their circuit-breaker mechanisms in 10 consecutive days in March. The Dow Jones, S&P 500, and the Nasdaq indexes leaked to the levels they were at before Trump's term began. This delivers it almost impossible for the US to launch new large-scale suppressive policies.
What does the fate of this conflict look like, who is the beneficiary?
The International Monetary Fund declared that this war has led to a "significantly weakened global expansion" as it cut its 2019 global growth forecast. This trade war has led to a chain reaction of trade wars amongst other nations as well. Rising tensions in the market have led to a mutual downfall of both the nation's demand in the market. The yuan fell over 5% versus the US dollar last year, before stabilizing in 2019, reported by Reuters. A huge shift in the outsourcing culture is anticipated to occur, for example, last year Enphase (solar energy firm) based in California stated that it would shift some of its production to Mexico, to dodge tariffs. Furthermore, Apple plans on moving its assembly from Taiwanese manufacturing company Foxconn's facilities in China to its facilities in India. Exports out of Vietnam have risen to $290.4 billion last year, according to the International Trade Centre, and Mexico grew to replace China as the largest trading partner for the US. Amidst these rising tensions south-east-Asian nations such as Thailand, Philippines, India, Vietnam, etc., are bound to benefit, with large firms such as Nintendo, sharp, Apple, Kyocera announcing their shift of production from China to other nations.
It is expected that, Donald Trump, who is seeking re-election, will definitely adjust his China policy to supress the Chinese high-tech enterprises and educational institutes, as the technological competition between the two nations would continue to grow. This would spread the economic plans for giant firms to other nations, however, these giants would also need to account for the time and money it takes to establish new production sites in other nations and the opportunity cost involved in the same. While the struggle to curb a pandemic continues, the questions still remains if these nations are doing more harm than good for their economies by engaging in a trade war.
Jain, Manjula & Saraswat, Saloni. (2019). US–China Trade War: Chinese Perspective. Management and Economics Research Journal. 5. 1-8. 10.18639/MERJ.2019.895478.
(2020, July 08). The US-China rivalry is broadening from trade to everything. Retrieved July 13, 2020, from https://economictimes.indiatimes.com/news/international/business/the-us-china-rivalryis-broadening-from-trade-to-everything/articleshow/76852973.cms
Chen, D. (2020, June 09). Why a US-China Détente Is Coming in 2021: The Failure of Trump's Policy. Retrieved July 13, 2020, from https://thediplomat.com/2020/06/why-aus-china-detente-is-coming-in-2021-the-failure-of-trumps-policy/
Vaswani, K. (2019, July 30). Can fresh US-China talks end the trade war? Retrieved July13, 2020, from https://www.bbc.com/news/business-49148685
Finley, K. (n.d.). Trump's Trade War Isn't Just a US–China Problem. Retrieved July 13, 2020, from https://www.wired.com/story/us-china-trade-war-spills-over/
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