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5.15.1 Plurifine topology on domainsC9F4423A543A

Let $\Omega \subseteq \mathbb {C}^ n$ ($n\in \mathbb {N}$) be a domain.

Definition 5.47.14.09A00D4B8096 The plurifine topology on $\Omega $ is the weakest topology making all $\mathbb {R}$-valued plurisubharmonic functions on $\Omega $ continuous.

We want to distinguish the Euclidean topology from the plurifine topology. here, topological notions without adjectives refer to those with respect to the Euclidean topology. We include the symbol $\mathcal{F}$ in order to denote those with respect to the plurifine topology. For example, we will say $\mathcal{F}$-open subset, $\mathcal{F}$-neighborhood, $\mathcal{F}$-closure, etc. The $\mathcal{F}$-closure of a set $E\subseteq \Omega $ will be denoted by $\overline{E}^{\mathcal{F}}$.

A neighborhood need not be open. Similarly, an $\mathcal F$-neighborhood need not be $\mathcal F$-open.

A priori, we should include $\Omega $ into the notations as well, but as we will see shortly in corollary 5.15.14, this is usually unnecessary.

When $n=0$, the domain $\Omega $ is a singleton and all assertions in this subsection are trivial. In the proofs below we therefore assume $n>0$.

Proposition 5.15.2.0E4FD8AF1256 The plurifine topology on $\Omega $ is finer than the Euclidean topology.

Proof. It suffices to show that the unit ball $\{ z\in \mathbb {C}^ n:|z|<1\} $ is $\mathcal{F}$-open. This follows from the observation that this set can be written as

\[ \{ \psi <0\} \quad \textup{with}\quad \psi (z)\coloneqq (\log |z|) \lor (-1). \]

This completes the proof. $\square$

Example 5.15.3.28F2383E9245 Let $\varphi \in \mathrm{PSH}(\Omega )$ and $C\in \mathbb {R}$. Then the sets $\{ \varphi >C\} $ and $\{ \varphi <C\} $ are both $\mathcal{F}$-open.

In fact, the latter case follows from proposition 5.15.2, while the former follows from the observation

\[ \{ \varphi >C\} =\{ \varphi \lor (C-1)>C\} . \]

Definition 5.15.4.EC0662753F9C A subset $E\subseteq \Omega $ is thin1 at $x\in \Omega $ if one of the following conditions holds:

  1. 2F25DFCC541C

    $x\not\in \overline{E}$;

  2. 9A680A93372A

    $x\in \overline{E}$ and there is an open neighborhood $U\subseteq \Omega $ of $x$ and $\varphi \in \mathrm{PSH}(U)$ such that

    \[ \varlimsup _{(U\cap E)\setminus \{ x\} \ni y\to x} \varphi (y)<\varphi (x). \]

We say $E$ is thin if it is thin at all $x\in \Omega $.

Remark 5.15.5.3833CFAB6397 In the second case, we can always arrange that

\[ \varphi |_{(E\setminus \{ x\} )\cap U}=0. \]

In fact, we may assume that $\varphi \leq 0$ and $C<0$ is such that

\[ \varlimsup _{(U\cap E)\setminus \{ x\} \ni y\to x} \varphi (y)<C<\varphi (x). \]

We let

\[ \psi =(-C)^{-1}(\varphi \lor C)+1. \]

Then $\psi $ satisfies our requirements for a smaller $U$.

In the second case, the function $\varphi $ can be very much improved.

Proposition 5.15.6 (Bedford–Taylor).80A580DBD009 Consider a set $E\subseteq \Omega $ and $x\in \overline{E}$. Assume that $E$ is thin at $x$. Then there is $\varphi \in \mathrm{PSH}(\mathbb {C}^ n)$ with the following properties:

  1. 855F0342BA11

    $\varphi $ is locally bounded outside $x$;

  2. B158426A08F5

    $\varphi (x)>-\infty $;

  3. 268FE0CF955A
    \[ \lim _{E\setminus \{ x\} \ni y\to x} \varphi (y)=-\infty . \]

Proof. By 5.15.5, there is an open neighborhood $U\subseteq \Omega $ of $x$ and $\psi \in \mathrm{PSH}(U)$ such that

\[ \psi |_{U\cap (E\setminus \{ x\} )}=-1 \quad \textup{and}\quad \psi (x)=0. \]

Without loss of generality we may assume $x=0$ and that $U=B_ n(0,1)$ is the open unit ball.

Since $\psi $ is upper semicontinuous at $0$ with $\psi (0)=0$, we may choose a strictly decreasing sequence $(\delta _ j)_{j\geq 1}\subseteq (0,\mathrm{e}^{-1})$ such that

921249630360 5.15.7
\begin{equation} \psi (y)<2^{-j-2}\quad \textup{whenever } |y|<\delta _ j. \end{equation}

In particular $|\log \delta _ j|>1$ for every $j\geq 1$, so the series of constants

1BCCC754A81F 5.15.8
\begin{equation} \sum _{j=1}^\infty \frac{2^{-j-1}}{|\log \delta _ j|} \leq \sum _{j=1}^\infty 2^{-j-1}=\frac{1}{2} \end{equation}

converges. We further set

\[ \gamma _ j\coloneqq \exp \bigl (2^{j+1}\log \delta _ j\bigr )=\delta _ j^{2^{j+1}}\in (0,1), \]

which is also decreasing in $j$.

For each $j\geq 1$ define

\[ \varphi _ j(z)\coloneqq \begin{cases} \frac{2^{-j-1}}{|\log \delta _ j|}\log |z|\lor \bigl (\psi (z)-2^{-j}\bigr ), & \textup{if } |z|<\delta _ j,\\ \frac{2^{-j-1}}{|\log \delta _ j|}\log |z|,& \textup{if } |z|\geq \delta _ j. \end{cases} \]

On the boundary $|z|=\delta _ j$, the logarithmic term tends to $-2^{-j-1}$. By (5.15.7),

\[ \varlimsup _{|z|<\delta _ j,\ z\to z_0}(\psi (z)-2^{-j})\leq 2^{-j-2}-2^{-j}<-2^{-j-1} \]

for every $z_0$ with $|z_0|=\delta _ j$. Hence the boundary condition in 5.14.16 is satisfied. We obtain that $\varphi _ j\in \mathrm{PSH}(\mathbb {C}^ n)$.

By construction, $\varphi _ j\leq 0$ on $U$, $\varphi _ j(0)=-2^{-j}$, and for $z\in E\setminus \{ 0\} $ with $|z|<\gamma _ j$ we have $\varphi _ j(z)\leq -1$, since

\[ \frac{2^{-j-1}}{|\log \delta _ j|}\log |z|<\frac{2^{-j-1}}{|\log \delta _ j|}\, (2^{j+1}\log \delta _ j)=-1 \]

and, on $E\setminus \{ 0\} $, $\psi (z)-2^{-j}=-1-2^{-j}<-1$.

We then define

\[ \varphi \coloneqq \sum _{j=1}^{\infty }\varphi _ j. \]

On the unit ball $U$ each $\varphi _ j\leq 0$, so the partial sums $S_ N\coloneqq \sum _{j=1}^ N\varphi _ j$ form a decreasing sequence of plurisubharmonic functions with

\[ S_ N(0)=-\sum _{j=1}^ N 2^{-j}\to -1>-\infty ,\quad \textup{as } N\to \infty \]

so $\varphi |_ U\in \mathrm{PSH}(U)$ by proposition 5.14.2. On $\mathbb {C}^ n\setminus B_ n(0,\delta _1)$ each $\varphi _ j$ equals $\frac{2^{-j-1}}{|\log \delta _ j|}\log |z|$, and (5.15.8) shows $S_ N$ converges uniformly on every compact subset to $\bigl (\sum _{j\geq 1}\tfrac {2^{-j-1}}{|\log \delta _ j|}\bigr )\log |z|$, a plurisubharmonic limit. Both formulas agree on $U\setminus B_ n(0,\delta _1)$, so altogether $\varphi \in \mathrm{PSH}(\mathbb {C}^ n)$.

We now verify the three claims.

Property 2. This is the equality $\varphi (0)=-\sum _{j\geq 1}2^{-j}=-1>-\infty $ obtained above.

Property 3. Fix $j_0\geq 1$. For $z\in E\setminus \{ 0\} $ with $|z|<\gamma _{j_0}$, the bound $\varphi _ j(z)\leq -1$ holds for every $j\leq j_0$ (since the $\gamma _ j$ are decreasing), while $\varphi _ j(z)\leq 0$ for $j>j_0$. Hence $\varphi (z)\leq -j_0$, and letting $j_0\to \infty $ yields

\[ \varlimsup _{E\setminus \{ 0\} \ni y\to 0}\varphi (y)=-\infty . \]

Property 1. Fix any $z_0\in \mathbb {C}^ n$ with $z_0\neq 0$ and choose $J\geq 1$ such that $\delta _ J<|z_0|/2$. On the open set $W=\{ w\in \mathbb {C}^ n:|w|>|z_0|/2\} $, all $\varphi _ j$ with $j\geq J$ lie in the outer regime, and (5.15.8) shows $\sum _{j\geq J}\varphi _ j(w) =\bigl (\sum _{j\geq J}\tfrac {2^{-j-1}}{|\log \delta _ j|}\bigr )\log |w|$ is locally bounded on $W$. The remaining finitely many terms $\varphi _1,\ldots ,\varphi _{J-1}$ are plurisubharmonic on $\mathbb {C}^ n$, hence locally bounded above on $W$, while $\varphi _ j(w)\geq \tfrac {2^{-j-1}}{|\log \delta _ j|}\log |w|$ keeps them locally bounded below as well. Therefore $\varphi $ is locally bounded on $W$. Since $z_0\neq 0$ was arbitrary, $\varphi $ is locally bounded on $\mathbb {C}^ n\setminus \{ 0\} $, which is property 1. $\square$

Lemma 5.15.9.D3F2BF3D82BD Let $E_1,E_2\subseteq \Omega $. Assume that $E_1$, $E_2$ are both thin at $x\in \Omega $. Then so is $E_1\cup E_2$.

Proof. We may assume that $x\in \overline{E_1}\cap \overline{E_2}$. Take an open neighborhood $U\subseteq \Omega $ of $x$ and $\varphi _1,\varphi _2\in \mathrm{PSH}(U)$ such that

\[ \varlimsup _{y\to x,y\in E_ i\setminus \{ x\} } \varphi _ i(y)<\varphi _ i(x),\quad i=1,2. \]

Then $\varphi _1+\varphi _2\in \mathrm{PSH}(U)$ and

\[ \varlimsup _{y\to x,y\in (E_1\cup E_2)\setminus \{ x\} } (\varphi _1+\varphi _2)(y)<\varphi _1(x)+\varphi _2(x). \]

In particular, $E_1\cup E_2$ is thin at $x$. $\square$

Theorem 5.15.10 (H. Cartan).458C93F5F6CD Consider a set $E\subseteq \Omega $ and $x\in E$. Then the following are equivalent:

  1. 15B95CB674C7

    $E$ is an $\mathcal{F}$-neighborhood of $x$;

  2. 53048930270F

    $\Omega \setminus E$ is thin at $x$.

Proof. 2 $\implies $ 1. Assume 2. We may assume that $x\in \overline{\Omega \setminus E}$. Otherwise, our assertion follows from proposition 5.15.2.

By proposition 5.15.6, there is an open neighborhood $U$ of $x$ in $\Omega $ and $\varphi \in \mathrm{PSH}(\mathbb {C}^ n)$ such that

\[ \varphi (x)>\sup _{y\in U\cap (\Omega \setminus E)} \varphi (y)\eqqcolon \lambda . \]

Let $F=\{ y\in \Omega : \varphi (y)>\lambda \} $. Then $x\in F$ and $F$ is $\mathcal{F}$-open by example 5.15.3. Moreover, $U\cap F\subseteq E$. By proposition 5.15.2, we conclude 1.

1 $\implies $ 2. Assume 1. We may always replace $E$ by smaller $\mathcal{F}$-neighborhoods of $x$. In particular, we may assume that $E$ has the following form

\[ \{ y\in U:\varphi _1(y)>\lambda _1,\ldots ,\varphi _ m(y)>\lambda _ m\} , \]

where $U\subseteq \Omega $ is an open neighborhood of $x$, and $\varphi _1,\ldots ,\varphi _ m$ are $\mathbb {R}$-valued psh functions on $\Omega $, and $\lambda _1,\ldots ,\lambda _ m\in \mathbb {R}$. Since $x\in U$, the set $\Omega \setminus U$ is thin at $x$. Moreover,

\[ \Omega \setminus E=(\Omega \setminus U)\cup \bigcup _{i=1}^ m\{ y\in U:\varphi _ i(y)\leq \lambda _ i\} . \]

By 5.15.9, it is enough to prove that each set $\{ y\in U:\varphi _ i(y)\leq \lambda _ i\} $ is thin at $x$. Thus we may assume $m=1$. In this case, $\Omega \setminus E$ is clearly thin at $x$. $\square$

Theorem 5.15.11.4388B85DDD67 A base of the plurifine topology on $\Omega $ is given by sets of the following form:

E3CEA68870B6 5.15.12
\begin{equation} \left\{ x\in U: \varphi (x)>0\right\} , \end{equation}

where $U\subseteq \Omega $ is an open subset and $\varphi \in \mathrm{PSH}(U)$.

Proof. Observe that sets of the form (5.15.12) are $\mathcal{F}$-open.2 By theorem 5.15.10, it suffices to show its complement in $\Omega $ is thin at each point of (5.15.12), which is obvious.

Now consider $x\in \Omega $ and an $\mathcal{F}$-open neighborhood $V\subseteq \Omega $ of $x$. We want to find a set of the form (5.15.12) contained in $V$ and containing $x$.

Write $E=\Omega \setminus V$. In case $x\in \operatorname{Int}V$, there is nothing to prove. So we may assume that $x\in \overline{E}$. By theorem 5.15.10, $E$ is thin at $x$. By definition, there is an open neighborhood $U\subseteq \Omega $ of $x$ and $\varphi \in \mathrm{PSH}(U)$ such that

\[ \varlimsup _{y\to x,y\in U\cap (E\setminus \{ x\} )}\varphi (y)<\varphi (x). \]

We may assume that $\varphi |_{E\cap U}\leq 0<\varphi (x)$. Then the set $\{ y\in U:\varphi (y)>0\} $ suffices for our purpose. $\square$

Remark 5.15.13.2EDAD1EF1694Recall that in general, an $\mathcal{F}$-open set is not a countable union of sets of the form (5.15.12). In fact, an $\mathcal{F}$-open set is not a Borel set in general. See [EK23] for a concrete example.

Corollary 5.15.14.49909C9A34E2 Let $\Omega _1\subseteq \Omega _2\subseteq \mathbb {C}^ n$ be two non-empty open subsets. Then the plurifine topology on $\Omega _1$ is the same as the subspace topology induced from the plurifine topology on $\Omega _2$.

In particular, when we talk about an $\mathcal{F}$-open set $U$ in $\mathbb {C}^ n$, we no longer have to specify the domain $\Omega \supseteq U$.

Corollary 5.15.15.C94ADCA6E19D Let $\Omega \subseteq \mathbb {C}^ n$ be an $\mathcal{F}$-open subset and $x\in \Omega $. Then $x$ has a compact $\mathcal{F}$-neighborhood contained in $\Omega $.

Proof. By theorem 5.15.11, it suffices to prove the assertion after replacing $\Omega $ by a smaller $\mathcal F$-open neighborhood of $x$. Thus we may assume that

\[ \Omega =\{ y\in U:\varphi (y)>0\} \]

for some open set $U\subseteq \mathbb C^ n$ and some $\varphi \in \mathrm{PSH}(U)$.

Take a compact neighborhood $K$ of $x$ in $U$. Now $\{ y\in K: \varphi (y)\geq \varphi (x)/2\} $ is a compact $\mathcal{F}$-neighborhood of $x$ contained in $\Omega $. $\square$

Corollary 5.15.16.DF37542BAB03 Let $\Omega \subseteq \mathbb {C}^ n$, $\Omega '\subseteq \mathbb {C}^{n'}$ be two domains and $F\colon \Omega '\rightarrow \Omega $ be a holomorphic map. Then $F$ is $\mathcal{F}$-continuous.

Proof. It suffices to show that the inverse image $F^{-1}(U)$ of each $\mathcal{F}$-open subset $U\subseteq \Omega $ is $\mathcal{F}$-open. By theorem 5.15.11, after possibly shrinking $\Omega $ and $\Omega '$, we may assume that $U$ has the form $\{ x\in \Omega :\psi (x)>0\} $, where $\psi \in \mathrm{PSH}(\Omega )$. Clearly, we may assume that $F^{-1}(U)$ is non-empty. Then $F^*\psi \in \mathrm{PSH}(\Omega ')$ by proposition 5.13.17, we find that

\[ F^{-1}(U)=\{ y\in \Omega ': F^*\psi (y)>0\} \]

is $\mathcal{F}$-open. $\square$

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