Richberg Regularisation Theorem (Theorem # 3681)
Theorem
Let $n \geq 1$ be an integer and let $\Omega \subseteq \mathbb{C}^n$ be open. Identify $\mathbb{C}^n$ with $\mathbb{R}^{2n}$, write $\mathcal{L}^{2n}$ for Lebesgue measure, and set
\begin{align*}
\omega_0 := i\partial\bar\partial |z|^2 = i\sum_{k=1}^n dz_k \wedge d\bar z_k .
\end{align*}
For $u \in C^2$ and $\xi \in \mathbb{C}^n$ let $\mathcal{L}_u(z)(\xi,\xi) := \sum_{j,k=1}^n \partial_{z_j}\partial_{\bar z_k} u(z)\,\xi_j\overline{\xi_k}$ denote the Levi form, and write $i\partial\bar\partial u \geq \lambda\,\omega_0$ to mean $\mathcal{L}_u(z)(\xi,\xi) \geq \lambda|\xi|^2$ for all $z,\xi$.
Call a continuous function $\phi:\Omega \to \mathbb{R}$ **strictly plurisubharmonic** if every point of $\Omega$ has an open neighbourhood $V$ and a constant $\lambda = \lambda(V) > 0$ such that $\phi - \lambda|z|^2$ is plurisubharmonic on $V$ (equivalently, $i\partial\bar\partial\phi \geq \lambda\,\omega_0$ in the sense of currents on $V$).
Let $\phi:\Omega \to \mathbb{R}$ be continuous and strictly plurisubharmonic, and let $\varepsilon:\Omega \to (0,\infty)$ be continuous. Then there exists $\psi \in C^\infty(\Omega)$ that is strictly plurisubharmonic — indeed $i\partial\bar\partial\psi$ is locally bounded below by a positive multiple of $\omega_0$ — and satisfies
\begin{align*}
\phi \leq \psi \leq \phi + \varepsilon \qquad \text{on } \Omega.
\end{align*}
Discussion
Every continuous strictly plurisubharmonic function can be approximated from above by a smooth strictly plurisubharmonic function with arbitrarily small error. This is useful because it lets one replace rough strictly plurisubharmonic data by smooth data without losing positivity.
Proof
[proofplan]
We approximate $\phi$ locally and patch the pieces with a smooth regularised maximum. First we choose a locally finite cover of $\Omega$ by balls on which $\phi - 2\beta_j|z|^2$ is plurisubharmonic, so that strict plurisubharmonicity survives mollification: convolving $\phi$ produces smooth functions $h_j$ obeying $i\partial\bar\partial h_j \geq 2\beta_j\,\omega_0$ and $\|h_j-\phi\|_{C^0}<\tau_j$. We then bend each $h_j$ by a small radial quadratic tent $\chi_j$, obtaining smooth strictly psh functions $g_j$ that exceed $\phi$ on the centre of their ball but plunge below $\phi$ near its edge; the descent localises them. Demailly's regularised maximum $M_\eta$ — a smooth, convex, non-decreasing approximation of $\max$ that ignores any argument lying definitely below the others — is applied to the family $(g_j)$. A locality lemma shows that near each point only finitely many $g_j$ are active and all of them are smooth there, so the glued function $\psi$ is smooth and strictly plurisubharmonic; the calibration of the errors keeps $\phi \leq \psi \leq \phi+\varepsilon$.
[/proofplan]
[step:Record the working notion of strict plurisubharmonicity and the regularised maximum]
Throughout, $\rho \in C_c^\infty(\mathbb{R}^{2n})$ is a fixed [standard mollifier](/page/Standard%20Mollifier): non-negative, radial, $\operatorname{supp}\rho \subseteq \overline{B}(0,1)$, $\int_{\mathbb{R}^{2n}}\rho\,d\mathcal{L}^{2n}=1$, and $\rho_\delta(w):=\delta^{-2n}\rho(w/\delta)$ for $\delta>0$. We record the gluing device.
[claim:The regularised maximum $M_\eta$ exists with the five properties below]
Fix once and for all an even function $\theta \in C_c^\infty(\mathbb{R})$ with $\theta \geq 0$, $\operatorname{supp}\theta \subseteq [-1,1]$, and $\int_{\mathbb{R}}\theta\,d\mathcal{L}^1 = 1$. For an integer $p\geq 1$ and $\eta = (\eta_1,\dots,\eta_p)\in(0,\infty)^p$ define
\begin{align*}
M_\eta : \mathbb{R}^p &\to \mathbb{R}\\
t = (t_1,\dots,t_p) &\mapsto \int_{\mathbb{R}^p} \max\{t_1+\sigma_1,\dots,t_p+\sigma_p\}\,\prod_{k=1}^p \tfrac{1}{\eta_k}\theta\!\left(\tfrac{\sigma_k}{\eta_k}\right) d\mathcal{L}^p(\sigma).
\end{align*}
Then:
**(a)** $M_\eta \in C^\infty(\mathbb{R}^p)$, is convex, and is non-decreasing in each variable.
**(b)** $\max_k t_k \leq M_\eta(t) \leq \max_k (t_k+\eta_k)$.
**(c)** $M_\eta(t_1+a,\dots,t_p+a) = M_\eta(t)+a$ for every $a\in\mathbb{R}$; consequently $\sum_{k=1}^p \partial_{t_k}M_\eta \equiv 1$.
**(d)** (Locality) If $j_0$ satisfies $t_{j_0}+\eta_{j_0} \leq \max_{k\neq j_0}(t_k-\eta_k)$, then $M_\eta(t)$ does not depend on $t_{j_0}$ and equals $M_{\eta'}(t')$, where $t',\eta'$ omit the $j_0$-th entry.
**(e)** If $u_1,\dots,u_p \in C^\infty(W)$ on an open $W\subseteq\mathbb{C}^n$, then $v:=M_\eta(u_1,\dots,u_p)\in C^\infty(W)$, and if $i\partial\bar\partial u_a \geq \beta\,\omega_0$ for every $a$, then $i\partial\bar\partial v \geq \beta\,\omega_0$.
[proof]
**(a)** Substituting $\mu_k = t_k+\sigma_k$ gives $M_\eta(t)=\int_{\mathbb{R}^p}\max_k(\mu_k)\prod_k \tfrac1{\eta_k}\theta(\tfrac{\mu_k-t_k}{\eta_k})\,d\mathcal{L}^p(\mu)$; the integrand has compact $\mu$-support and is $C^\infty$ in $t$ with all $t$-derivatives dominated by integrable functions, so differentiation under the integral sign gives $M_\eta\in C^\infty$. For each fixed $\sigma$, the map $t\mapsto\max_k(t_k+\sigma_k)$ is convex and non-decreasing in each $t_k$; integrating against the non-negative density preserves both properties.
**(b)** On $\operatorname{supp}\prod_k\theta(\sigma_k/\eta_k)$ we have $|\sigma_k|\leq\eta_k$, hence $\max_k(t_k+\sigma_k)\leq\max_k(t_k+\eta_k)$, giving the upper bound after integration. For the lower bound, the product density is a probability measure on $\mathbb{R}^p$ under which each coordinate has mean $\int_{\mathbb{R}}\sigma\,\tfrac1{\eta_k}\theta(\sigma/\eta_k)\,d\mathcal{L}^1(\sigma)=0$ ($\theta$ even); since $t\mapsto\max_k(t_k+\sigma_k)$ is convex, [Jensen's inequality](/theorems/9) yields $M_\eta(t)\geq\max_k(t_k+0)=\max_k t_k$.
**(c)** $\max_k(t_k+a+\sigma_k)=\max_k(t_k+\sigma_k)+a$; integrate. Differentiating $M_\eta(t+a\mathbf{1})=M_\eta(t)+a$ at $a=0$ gives $\sum_k\partial_{t_k}M_\eta=1$.
**(d)** On the support of the density, $|\sigma_k|\leq\eta_k$, so
\begin{align*}
t_{j_0}+\sigma_{j_0}\leq t_{j_0}+\eta_{j_0}\leq \max_{k\neq j_0}(t_k-\eta_k)\leq \max_{k\neq j_0}(t_k+\sigma_k).
\end{align*}
Thus $\max_k(t_k+\sigma_k)=\max_{k\neq j_0}(t_k+\sigma_k)$ is independent of $\sigma_{j_0}$; integrating out $\sigma_{j_0}$ (whose density integrates to $1$) leaves exactly $M_{\eta'}(t')$.
**(e)** Smoothness of $v$ is composition of smooth maps. For the Levi form, write $\partial_a M$, $\partial_a\partial_b M$ for the derivatives of $M_\eta$ evaluated at $(u_1,\dots,u_p)$. The chain rule gives
\begin{align*}
\partial_{z_j}\partial_{\bar z_k} v = \sum_{a}(\partial_a M)\,\partial_{z_j}\partial_{\bar z_k}u_a + \sum_{a,b}(\partial_a\partial_b M)\,\partial_{z_j}u_a\,\partial_{\bar z_k}u_b.
\end{align*}
Contract with $\xi_j\overline{\xi_k}$ and set $\zeta_a := \sum_{j}\partial_{z_j}u_a\,\xi_j \in \mathbb{C}$:
\begin{align*}
\mathcal{L}_v(\xi,\xi) = \sum_a (\partial_a M)\,\mathcal{L}_{u_a}(\xi,\xi) + \sum_{a,b}(\partial_a\partial_b M)\,\zeta_a\overline{\zeta_b}.
\end{align*}
The Hessian $H=(\partial_a\partial_b M)$ is real, symmetric, and positive semidefinite by convexity (a); for any $\zeta\in\mathbb{C}^p$ the imaginary part of $\sum_{a,b}H_{ab}\zeta_a\overline{\zeta_b}$ vanishes (antisymmetry against symmetric $H$) and the real part is $\sum_{a,b}H_{ab}(\operatorname{Re}\zeta_a\operatorname{Re}\zeta_b+\operatorname{Im}\zeta_a\operatorname{Im}\zeta_b)\geq 0$; hence the second sum is $\geq 0$. In the first sum $\partial_a M\geq 0$ by (a) and $\sum_a\partial_a M=1$ by (c), so using $\mathcal{L}_{u_a}(\xi,\xi)\geq\beta|\xi|^2$,
\begin{align*}
\mathcal{L}_v(\xi,\xi)\ \geq\ \sum_a(\partial_a M)\,\beta|\xi|^2 = \beta|\xi|^2 .
\end{align*}
Thus $i\partial\bar\partial v\geq\beta\,\omega_0$.
[/proof]
[/claim]
[guided]
We will glue countably many local smooth approximations $g_j$ of $\phi$. The naive choice $\max_j g_j$ is continuous and plurisubharmonic but not $C^1$ along the seams $\{g_j=g_k\}$. Demailly's **regularised maximum** repairs exactly this: it is a smooth average of $\max$ over a box of size $\eta$, so it agrees with $\max$ away from the seams and rounds the corners inside a band of width $\eta$. Let us build it and verify everything we will need.
Fix an even $\theta\in C_c^\infty(\mathbb{R})$ with $\theta\ge0$, $\operatorname{supp}\theta\subseteq[-1,1]$, and $\int_{\mathbb{R}}\theta\,d\mathcal{L}^1=1$. For an integer $p\ge1$ and $\eta=(\eta_1,\dots,\eta_p)\in(0,\infty)^p$ set
\begin{align*}
M_\eta(t)=\int_{\mathbb{R}^p}\max\{t_1+\sigma_1,\dots,t_p+\sigma_p\}\prod_{k=1}^p\tfrac1{\eta_k}\theta\!\left(\tfrac{\sigma_k}{\eta_k}\right)d\mathcal{L}^p(\sigma),
\end{align*}
i.e. we smear the kinked function $\max$ against a smooth product bump of width $\eta_k$ in the $k$-th slot. Why does this help? Each factor integrates to $1$, so the density is a probability measure and $M_\eta$ is a weighted average of nearby values of $\max$; averaging a kinked convex function against a smooth density produces a smooth convex function.
*Smoothness, convexity, monotonicity (a).* Substitute $\mu_k=t_k+\sigma_k$ to move the $t$-dependence into the density:
\begin{align*}
M_\eta(t)=\int_{\mathbb{R}^p}\max_k(\mu_k)\prod_k\tfrac1{\eta_k}\theta\!\left(\tfrac{\mu_k-t_k}{\eta_k}\right)d\mathcal{L}^p(\mu).
\end{align*}
The integrand has compact $\mu$-support and is $C^\infty$ in $t$ with all $t$-derivatives dominated by integrable functions, so we may differentiate under the integral sign: $M_\eta\in C^\infty(\mathbb{R}^p)$. For each fixed $\sigma$ the map $t\mapsto\max_k(t_k+\sigma_k)$ is convex and non-decreasing in each $t_k$; integrating against the non-negative density preserves both properties.
*The sandwich (b).* On the support of the density $|\sigma_k|\le\eta_k$, so $\max_k(t_k+\sigma_k)\le\max_k(t_k+\eta_k)$, giving $M_\eta(t)\le\max_k(t_k+\eta_k)$ after integration. For the lower bound, $\theta$ even forces each coordinate mean to vanish, $\int_{\mathbb{R}}\sigma\,\tfrac1{\eta_k}\theta(\sigma/\eta_k)\,d\mathcal{L}^1(\sigma)=0$; since $t\mapsto\max_k(t_k+\sigma_k)$ is convex, [Jensen's inequality](/theorems/1977) against this mean-zero probability density gives $M_\eta(t)\ge\max_k(t_k+0)=\max_k t_k$. The sandwich $\max_k t_k\le M_\eta(t)\le\max_k(t_k+\eta_k)$ is exactly what later produces $\phi\le\psi\le\phi+\varepsilon$.
*Translation and the derivative sum (c).* Since $\max_k(t_k+a+\sigma_k)=\max_k(t_k+\sigma_k)+a$, integrating gives $M_\eta(t+a\mathbf{1})=M_\eta(t)+a$; differentiating at $a=0$ yields $\sum_k\partial_{t_k}M_\eta\equiv1$. This identity is the algebraic reason $M_\eta$ preserves — rather than degrades — a uniform lower bound on the Levi form.
*Locality (d).* Suppose $t_{j_0}+\eta_{j_0}\le\max_{k\neq j_0}(t_k-\eta_k)$. On the support $|\sigma_k|\le\eta_k$, so
\begin{align*}
t_{j_0}+\sigma_{j_0}\le t_{j_0}+\eta_{j_0}\le\max_{k\neq j_0}(t_k-\eta_k)\le\max_{k\neq j_0}(t_k+\sigma_k),
\end{align*}
hence $\max_k(t_k+\sigma_k)=\max_{k\neq j_0}(t_k+\sigma_k)$ is independent of $\sigma_{j_0}$; integrating out $\sigma_{j_0}$ (its density integrates to $1$) leaves $M_{\eta'}(t')$, with $t',\eta'$ omitting the $j_0$-th entry. So an argument sitting more than $\eta$ below the leader is simply forgotten — this is what lets $\psi$ be, near any point, a regularised maximum of only finitely many functions, even though infinitely many $g_j$ are globally in play and each $g_j$ lives only on its own ball.
*The plurisubharmonic payoff (e).* Let $u_1,\dots,u_p\in C^\infty(W)$ on an open $W\subseteq\mathbb{C}^n$ with $i\partial\bar\partial u_a\ge\beta\,\omega_0$ for every $a$, and put $v:=M_\eta(u_1,\dots,u_p)$. Smoothness of $v$ is composition of smooth maps. Writing $\partial_aM,\partial_a\partial_bM$ for derivatives of $M_\eta$ evaluated at $(u_1,\dots,u_p)$, the chain rule gives
\begin{align*}
\partial_{z_j}\partial_{\bar z_k}v=\sum_a(\partial_aM)\,\partial_{z_j}\partial_{\bar z_k}u_a+\sum_{a,b}(\partial_a\partial_bM)\,\partial_{z_j}u_a\,\partial_{\bar z_k}u_b.
\end{align*}
Contracting with $\xi_j\overline{\xi_k}$ and setting $\zeta_a:=\sum_j\partial_{z_j}u_a\,\xi_j\in\mathbb{C}$,
\begin{align*}
\mathcal{L}_v(\xi,\xi)=\sum_a(\partial_aM)\,\mathcal{L}_{u_a}(\xi,\xi)+\sum_{a,b}(\partial_a\partial_bM)\,\zeta_a\overline{\zeta_b}.
\end{align*}
The Hessian $H=(\partial_a\partial_bM)$ is real, symmetric, and positive semidefinite by convexity (a); for any $\zeta\in\mathbb{C}^p$ the imaginary part of $\sum_{a,b}H_{ab}\zeta_a\overline{\zeta_b}$ vanishes against symmetric $H$, and the real part equals $\sum_{a,b}H_{ab}(\operatorname{Re}\zeta_a\operatorname{Re}\zeta_b+\operatorname{Im}\zeta_a\operatorname{Im}\zeta_b)\ge0$, so the second sum is $\ge0$. In the first sum the weights $\partial_aM\ge0$ by (a) and sum to $1$ by (c), so it is a convex combination of the input Levi forms; using $\mathcal{L}_{u_a}(\xi,\xi)\ge\beta|\xi|^2$,
\begin{align*}
\mathcal{L}_v(\xi,\xi)\ge\sum_a(\partial_aM)\,\beta|\xi|^2=\beta|\xi|^2,
\end{align*}
that is $i\partial\bar\partial v\ge\beta\,\omega_0$. Feeding smooth strictly psh inputs with a common Levi bound into $M_\eta$ returns a smooth strictly psh output with the *same* bound — the property that makes the whole gluing work.
[/guided]
[/step]
[step:Choose a locally finite cover of $\Omega$ adapted to the strict plurisubharmonicity of $\phi$]
By strict plurisubharmonicity, each $p\in\Omega$ has an open ball $B(p,R_p)$ with $\overline{B(p,R_p)}\subset\Omega$ and a constant $\lambda_p>0$ such that $\phi-\lambda_p|z|^2$ is plurisubharmonic on a neighbourhood of $\overline{B(p,R_p)}$. The balls $\{B(p,R_p/8)\}_{p\in\Omega}$ cover $\Omega$. Since $\Omega$ is an open subset of $\mathbb{C}^n$, it is second countable and locally compact, hence paracompact; choose a countable, locally finite refinement. This produces:
- centres $a_j$ and radii $0<r_j'<s_j<\rho_j<R_j$ ($j\in\mathbb{N}$) with $r_j'=R_j/4$, $s_j=R_j/2$, $\rho_j=3R_j/4$, such that $\overline{B(a_j,R_j)}\subset\Omega$ and $\phi-2\beta_j|z|^2$ is plurisubharmonic on a neighbourhood of $\overline{W_j}$, where $W_j:=B(a_j,R_j)$ and $2\beta_j:=\lambda_{p_j}>0$;
- the small balls $B_j':=B(a_j,r_j')$ cover $\Omega$;
- the family $(W_j)_{j}$ is locally finite.
Because $\overline{W_j}\Subset\Omega$ is compact and $\varepsilon$ is continuous and positive, $m_j := \inf_{\overline{W_j}}\varepsilon > 0$.
We now fix all constants. Put
\begin{align*}
c_j := \min\Big\{\beta_j,\ \frac{m_j}{4\,s_j^2}\Big\} > 0,\qquad \kappa_j := c_j\,(s_j^2 - r_j'^2) > 0,\qquad d_j := c_j\,(\rho_j^2 - s_j^2) > 0 .
\end{align*}
Then choose $\tau_j,\eta_j>0$ so small that
\begin{align*}
\tau_j < \min\{\tfrac14 m_j,\ \kappa_j,\ d_j\},\qquad \eta_j < \min\{\tfrac14 m_j,\ \kappa_j-\tau_j,\ d_j-\tau_j\}.
\end{align*}
These choices are possible because every right-hand side is a fixed positive number. We record three consequences for later: $c_j\leq\beta_j$; $\;\tau_j+\eta_j<d_j$; and $\tau_j + c_j s_j^2 + \eta_j < \tfrac14 m_j+\tfrac14 m_j+\tfrac14 m_j < m_j$ (using $c_j s_j^2\leq \tfrac14 m_j$).
[/step]
[step:Construct smooth strictly plurisubharmonic local approximations by mollification]
Fix $j$. Choose $\delta_j>0$ smaller than the distance from $\overline{W_j}$ to the boundary of the neighbourhood on which $\phi-2\beta_j|z|^2$ is plurisubharmonic, and define
\begin{align*}
h_j : \overline{W_j} &\to \mathbb{R}\\
z &\mapsto (\phi * \rho_{\delta_j})(z) = \int_{\mathbb{R}^{2n}} \phi(z-w)\,\rho_{\delta_j}(w)\,d\mathcal{L}^{2n}(w).
\end{align*}
By [Properties of Mollification](/theorems/461)(1), $h_j\in C^\infty$ on a neighbourhood of $\overline{W_j}$. Since $\phi$ is continuous and $\overline{W_j}$ is compact, [Uniform Convergence of Mollification on Compact Sets](/theorems/47) lets us shrink $\delta_j$ so that
\begin{align*}
\phi - \tau_j \ \leq\ h_j \ \leq\ \phi + \tau_j \qquad \text{on } \overline{W_j}.
\end{align*}
[claim:$i\partial\bar\partial h_j \geq 2\beta_j\,\omega_0$ on $W_j$]
[proof]
The function $\phi - 2\beta_j|z|^2$ is plurisubharmonic on a neighbourhood of $\overline{W_j}$ containing $\overline{B}(z,\delta_j)$ for every $z\in\overline{W_j}$. Mollification preserves plurisubharmonicity: for any complex line $z=\alpha+\zeta\,b$ ($\zeta\in\mathbb{C}$), the function $\zeta\mapsto (\phi-2\beta_j|z|^2)(\alpha+\zeta b - w)$ is subharmonic for each $w$, and
\begin{align*}
\big((\phi-2\beta_j|z|^2)*\rho_{\delta_j}\big)(\alpha+\zeta b) = \int_{\mathbb{R}^{2n}} (\phi-2\beta_j|z|^2)(\alpha+\zeta b - w)\,\rho_{\delta_j}(w)\,d\mathcal{L}^{2n}(w)
\end{align*}
is a superposition with non-negative weights of subharmonic functions of $\zeta$, hence (by Tonelli applied to the sub-mean-value inequality) subharmonic in $\zeta$; as this holds on every complex line, $(\phi-2\beta_j|z|^2)*\rho_{\delta_j}$ is plurisubharmonic. Now, because $\rho_{\delta_j}$ is radial,
\begin{align*}
(|z|^2*\rho_{\delta_j})(w) = \int_{\mathbb{R}^{2n}}|w-y|^2\rho_{\delta_j}(y)\,d\mathcal{L}^{2n}(y) = |w|^2 + \int_{\mathbb{R}^{2n}}|y|^2\rho_{\delta_j}(y)\,d\mathcal{L}^{2n}(y) = |w|^2 + C_j,
\end{align*}
since the linear term $-2\operatorname{Re}\big(w\cdot\overline{\int y\,\rho_{\delta_j}\,d\mathcal{L}^{2n}}\big)$ vanishes ($\int y\,\rho_{\delta_j}\,d\mathcal{L}^{2n}=0$ by radial symmetry). Therefore $(\phi-2\beta_j|z|^2)*\rho_{\delta_j} = h_j - 2\beta_j|z|^2 - 2\beta_j C_j$, so the smooth function $h_j - 2\beta_j|z|^2$ is plurisubharmonic. By the [Levi Form Criterion for Smooth PSH Functions](/theorems/3403), its Levi form is positive semidefinite, i.e. $\mathcal{L}_{h_j}(\xi,\xi) - 2\beta_j|\xi|^2 \geq 0$ for all $\xi$, which is $i\partial\bar\partial h_j \geq 2\beta_j\,\omega_0$.
[/proof]
[/claim]
[/step]
[step:Bend each approximation by a radial quadratic tent to localise it]
Define the tents and the bent approximations: for each $j$,
\begin{align*}
\chi_j : \mathbb{C}^n &\to \mathbb{R}, & z &\mapsto c_j\big(s_j^2 - |z-a_j|^2\big),\\
g_j : W_j &\to \mathbb{R}, & z &\mapsto h_j(z) + \chi_j(z).
\end{align*}
Since $i\partial\bar\partial\chi_j = -c_j\,\omega_0$ and $i\partial\bar\partial h_j\geq 2\beta_j\,\omega_0$, while $c_j\leq\beta_j$,
\begin{align*}
i\partial\bar\partial g_j = i\partial\bar\partial h_j - c_j\,\omega_0 \ \geq\ (2\beta_j - c_j)\,\omega_0 \ \geq\ \beta_j\,\omega_0 \qquad \text{on } W_j,
\end{align*}
so each $g_j\in C^\infty(W_j)$ is strictly plurisubharmonic with Levi bound $\beta_j$. Using $\phi-\tau_j\le h_j\le\phi+\tau_j$ on $\overline{W_j}$, we record three estimates:
**Centre (on $\overline{B_j'}$, where $|z-a_j|\leq r_j'$):** $\chi_j \geq c_j(s_j^2 - r_j'^2) = \kappa_j$, hence
\begin{align*}
g_j \ \geq\ (\phi - \tau_j) + \kappa_j \ =\ \phi + (\kappa_j-\tau_j),\qquad \text{and so}\quad g_j - \eta_j \geq \phi + (\kappa_j-\tau_j-\eta_j) > \phi .
\end{align*}
**Collar (where $\rho_j\leq |z-a_j| < R_j$):** $\chi_j \leq c_j(s_j^2 - \rho_j^2) = -d_j$, hence
\begin{align*}
g_j + \eta_j \ \leq\ (\phi+\tau_j) - d_j + \eta_j \ <\ \phi \qquad (\text{using } \tau_j+\eta_j<d_j).
\end{align*}
**Cap (on all of $\overline{W_j}$):** $\chi_j \leq c_j s_j^2$, hence
\begin{align*}
g_j + \eta_j \ \leq\ \phi + \big(\tau_j + c_j s_j^2 + \eta_j\big) \ <\ \phi + m_j \ \leq\ \phi + \varepsilon(z)\quad\text{for } z\in\overline{W_j}.
\end{align*}
[guided]
Why a tent, and why these three estimates? After mollifying we have functions $h_j\approx\phi$, but they overlap on different balls and we cannot feed *infinitely many* of them into one regularised maximum unless all but finitely many drop out near each point. Property (d) drops an argument only when it lies a definite amount *below* the leader. The tent $\chi_j$ manufactures exactly this gap: it is positive (height up to $c_j s_j^2$) at the centre of $B_j$ and decreases quadratically, becoming as negative as $-d_j$ in the collar near $\partial B_j$.
The decisive point is that we may bend $h_j$ *downward* by $c_j|z-a_j|^2$ and still stay strictly psh, because strictness gave us the surplus $i\partial\bar\partial h_j\ge 2\beta_j\omega_0$ and we only spend $c_j\le\beta_j$ of it. (A psh function cannot be bent down to a localised bump while remaining psh — but here we are bending a function that has Levi-form room to spare, leaving $\ge\beta_j\omega_0$.)
Read the three estimates as the three jobs of the tent. The **centre** estimate says $g_j$ sticks up above $\phi$ on $\overline{B_j'}$, so wherever $B_j'$ covers a point, $g_j$ is a genuine contender for the maximum — this will deliver the lower bound $\psi\ge\phi$. The **collar** estimate says $g_j$ has already dropped more than $\eta_j$ below $\phi$ by the time we approach $\partial B_j$ — this is what lets property (d) forget $g_j$ before it would have to be evaluated outside its domain $W_j$. The **cap** estimate says $g_j$ never rises more than the local error budget $m_j\le\varepsilon$ above $\phi$ — this will deliver the upper bound $\psi\le\phi+\varepsilon$. The calibration of Step 2 was reverse-engineered precisely so all three hold simultaneously.
[/guided]
[/step]
[step:Define the glued function as a regularised maximum of the active local pieces]
For $z\in\Omega$ set $J(z):=\{\,j\in\mathbb{N} : z\in W_j\,\}$. Since $(W_j)$ is locally finite, $J(z)$ is finite; since the $B_j'$ cover $\Omega$, $J(z)\neq\varnothing$. Define
\begin{align*}
\psi : \Omega &\to \mathbb{R}\\
z &\mapsto M_{\,\eta|_{J(z)}}\big( (g_j(z))_{j\in J(z)} \big),
\end{align*}
where $\eta|_{J(z)} := (\eta_j)_{j\in J(z)}$ and $M$ is the regularised maximum of the Claim in Step 1. This is well defined because every $g_j$, $j\in J(z)$, is defined at $z$ (as $z\in W_j$). The next step proves $\psi$ is smooth; the present definition is purely pointwise.
[/step]
[step:Localise the regularised maximum to finitely many active smooth functions]
[claim:Localisation Lemma]
For every $p\in\Omega$ there are an [open set](/page/Open%20Set) $U$ with $p\in U\Subset\Omega$ and a finite set $A\subseteq\mathbb{N}$ such that $U\subseteq W_a$ (so $g_a\in C^\infty(U)$) for every $a\in A$, and
\begin{align*}
\psi(z) = M_{\,\eta|_A}\big( (g_a(z))_{a\in A} \big)\qquad\text{for all } z\in U .
\end{align*}
Consequently $\psi\in C^\infty(U)$ and $i\partial\bar\partial\psi \geq \big(\min_{a\in A}\beta_a\big)\,\omega_0$ on $U$.
[proof]
By local finiteness of $(W_j)$ the family $(\overline{W_j})$ is also locally finite, so $F:=\{j: p\in\overline{W_j}\}$ is finite. For each $j\in F$, $h_j\in C^\infty$ on a neighbourhood of $\overline{W_j}$ and $\chi_j\in C^\infty(\mathbb{C}^n)$, so $g_j=h_j+\chi_j$ extends to a smooth function on an [open set](/page/Open%20Set) $N_j\supseteq\overline{W_j}$. Since $\phi-\tau_j\le h_j\le\phi+\tau_j$ holds on all of $\overline{W_j}$, the centre estimate holds on $\overline{B_j'}$ and the collar estimate $g_j+\eta_j<\phi$ holds on the closed annulus $\{\rho_j\leq|z-a_j|\leq R_j\}$, in particular on the boundary sphere $\{|z-a_j|=R_j\}=\partial W_j$. Since $(\overline{W_j})$ is locally finite and $p\notin\overline{W_j}$ for $j\notin F$, we may choose an open $U_0$ with $p\in U_0\Subset\Omega$, $U_0\subseteq\bigcap_{j\in F}N_j$, and $U_0\cap\overline{W_j}=\varnothing$ for every $j\notin F$. Then every $g_j$ ($j\in F$) is smooth on $U_0$, and for $z\in U_0$ the membership $z\in W_j$ forces $j\in F$, so $J(z)\subseteq F$. Fix an index $k_0$ with $p\in B_{k_0}'$ (the $B_j'$ cover); then $k_0\in F$.
*Active indices lie in the core of their ball.* Call $j\in J(p)$ **active** if $g_j(p)+\eta_j \geq \max_{k\in J(p),\,k\neq j}\big(g_k(p)-\eta_k\big)$, and set $A:=\{j\in J(p): j \text{ active}\}\cup\{k_0\}$. We claim every $a\in A$ satisfies $|p-a_a|<\rho_a$. For $a=k_0$ this holds since $p\in B_{k_0}'$ gives $|p-a_{k_0}|<r_{k_0}'<\rho_{k_0}$. For an active $a\neq k_0$: the central estimate gives $g_{k_0}(p)-\eta_{k_0}>\phi(p)$, so
\begin{align*}
g_a(p)+\eta_a \ \geq\ \max_{k\in J(p),\,k\neq a}\big(g_k(p)-\eta_k\big) \ \geq\ g_{k_0}(p)-\eta_{k_0} \ >\ \phi(p),
\end{align*}
hence $g_a(p)>\phi(p)-\eta_a$. But the collar estimate says $g_a+\eta_a<\phi$, i.e. $g_a<\phi-\eta_a$, wherever $\rho_a\le|z-a_a|<R_a$; thus $|p-a_a|<\rho_a$. In particular $p\in B(a_a,\rho_a)$ and $\overline{B(a_a,\rho_a)}\subset W_a$ for every $a\in A$.
*The leader is active.* If $k^\ast\in J(p)$ attains $\max_{k\in J(p)}(g_k(p)-\eta_k)$, then $g_{k^\ast}(p)+\eta_{k^\ast}\geq g_{k^\ast}(p)-\eta_{k^\ast}=\max_{k\in J(p)}(g_k(p)-\eta_k)\geq\max_{k\neq k^\ast}(g_k(p)-\eta_k)$, so $k^\ast$ is active, i.e. $k^\ast\in A$. Hence
\begin{align*}
\max_{k\in J(p)}\big(g_k(p)-\eta_k\big) = \max_{a\in A}\big(g_a(p)-\eta_a\big).
\end{align*}
We now show the **strict** inequality $g_j(p)+\eta_j < \max_{a\in A}(g_a(p)-\eta_a)$ for every $j\in F\setminus A$, in two cases. If $j\in J(p)\setminus A$, then $j$ is non-active and the leader $k^\ast\in A$ differs from $j$, so the non-activeness condition gives
\begin{align*}
g_j(p)+\eta_j < \max_{k\in J(p),\,k\neq j}\big(g_k(p)-\eta_k\big) = \max_{a\in A}\big(g_a(p)-\eta_a\big).
\end{align*}
If instead $j\in F\setminus J(p)$, then $p\in\overline{W_j}\setminus W_j$, i.e. $|p-a_j|=R_j\geq\rho_j$, so the collar estimate (valid on $\overline{W_j}$) gives
\begin{align*}
g_j(p)+\eta_j < \phi(p) < g_{k_0}(p)-\eta_{k_0} \leq \max_{a\in A}\big(g_a(p)-\eta_a\big),
\end{align*}
the middle inequality being the central estimate at $p\in B_{k_0}'$. In all cases the strict inequality holds.
*Choice of $U$.* Let
\begin{align*}
U := U_0 \cap \bigcap_{a\in A} B(a_a,\rho_a) \cap \Big\{ z : g_j(z)+\eta_j < \max_{a\in A}\big(g_a(z)-\eta_a\big)\ \text{for all } j\in F\setminus A \Big\}.
\end{align*}
The last set is open and contains $p$ by the strict inequalities just established and continuity of the finitely many $g_j$ ($j\in F$) on $U_0$; the middle intersection is open and contains $p$. Thus $U$ is an open neighbourhood of $p$, and $U\subseteq B(a_a,\rho_a)\subset W_a$ for every $a\in A$, so each $g_a\in C^\infty(U)$.
*The formula on $U$.* Fix $z\in U$. Then $J(z)\subseteq F$, and $A\subseteq J(z)$ (since for $a\in A$, $z\in B(a_a,\rho_a)\subseteq W_a$). For every $j\in J(z)\setminus A\subseteq F\setminus A$ the defining condition of $U$ gives
\begin{align*}
g_j(z)+\eta_j \ <\ \max_{a\in A}\big(g_a(z)-\eta_a\big) \ \leq\ \max_{k\in J(z),\,k\neq j}\big(g_k(z)-\eta_k\big),
\end{align*}
so by the locality property (d) we may delete index $j$ from $M_{\eta|_{J(z)}}$ without changing its value; deletion leaves the values $(g_a(z))_{a\in A}$ and the quantity $\max_{a\in A}(g_a(z)-\eta_a)$ untouched, so the remaining non-active indices stay deletable. Removing all of $J(z)\setminus A$ one at a time yields
\begin{align*}
\psi(z) = M_{\eta|_{J(z)}}\big((g_k(z))_{k\in J(z)}\big) = M_{\eta|_A}\big((g_a(z))_{a\in A}\big).
\end{align*}
Finally, $A$ and $\eta|_A$ are fixed and each $g_a\in C^\infty(U)$ with $i\partial\bar\partial g_a\geq\beta_a\,\omega_0\geq(\min_{b\in A}\beta_b)\,\omega_0$; by property (e), $\psi\in C^\infty(U)$ and $i\partial\bar\partial\psi\geq(\min_{a\in A}\beta_a)\,\omega_0$ on $U$.
[/proof]
[/claim]
Since smoothness and the Levi bound are local properties and $p\in\Omega$ was arbitrary, $\psi\in C^\infty(\Omega)$ and $\psi$ is strictly plurisubharmonic, with $i\partial\bar\partial\psi$ bounded below near each point by a positive multiple of $\omega_0$.
[guided]
This is the heart of the patching. The function $\psi$ is defined by a formula whose very shape — the index set $J(z)=\{j:z\in W_j\}$ — changes from point to point, so smoothness is not automatic. The lemma shows that near any fixed $p$ the formula is *constant in shape*: it equals one fixed finite regularised maximum $M_{\eta|_A}$ of smooth strictly psh functions.
First we pin down which indices can possibly appear near $p$. By local finiteness of $(W_j)$ the family $(\overline{W_j})$ is locally finite, so $F:=\{j:p\in\overline{W_j}\}$ is finite. Each $g_j=h_j+\chi_j$ extends smoothly past $\overline{W_j}$ (since $h_j\in C^\infty$ near $\overline{W_j}$ and $\chi_j\in C^\infty(\mathbb{C}^n)$), to a smooth function on an open $N_j\supseteq\overline{W_j}$; because $\phi-\tau_j\le h_j\le\phi+\tau_j$ on $\overline{W_j}$, the centre, collar and cap estimates hold on $\overline{W_j}$, the collar estimate $g_j+\eta_j<\phi$ holding on the closed annulus $\{\rho_j\le|z-a_j|\le R_j\}$. Shrinking, choose $p\in U_0\Subset\Omega$ with $U_0\subseteq\bigcap_{j\in F}N_j$ and $U_0\cap\overline{W_j}=\varnothing$ for $j\notin F$; then $J(z)\subseteq F$ for $z\in U_0$ and every $g_j$ ($j\in F$) is smooth on $U_0$. Fix $k_0$ with $p\in B_{k_0}'$, so $k_0\in F$.
Now we decide which indices in $J(p)$ matter. Call $j\in J(p)$ **active** if $g_j(p)+\eta_j\ge\max_{k\in J(p),\,k\neq j}(g_k(p)-\eta_k)$, and let $A:=\{j\in J(p):j\text{ active}\}\cup\{k_0\}$. Every $a\in A$ in fact sits in the *core* $|p-a_a|<\rho_a$: for $a=k_0$ this is immediate from $p\in B_{k_0}'$; for an active $a\neq k_0$, the centre estimate $g_{k_0}(p)-\eta_{k_0}>\phi(p)$ together with activeness gives
\begin{align*}
g_a(p)+\eta_a\ge\max_{k\in J(p),\,k\neq a}(g_k(p)-\eta_k)\ge g_{k_0}(p)-\eta_{k_0}>\phi(p),
\end{align*}
so $g_a(p)>\phi(p)-\eta_a$, contradicting the collar estimate $g_a<\phi-\eta_a$ on $\{\rho_a\le|z-a_a|<R_a\}$; hence $|p-a_a|<\rho_a$ and $\overline{B(a_a,\rho_a)}\subset W_a$. We insist the retained indices lie in the core precisely so that $U\subseteq B(a_a,\rho_a)\subset W_a$ and no $g_a$ is ever evaluated off its domain.
Next, the leader is active: if $k^\ast$ attains $\max_{k\in J(p)}(g_k(p)-\eta_k)$ then $g_{k^\ast}(p)+\eta_{k^\ast}\ge g_{k^\ast}(p)-\eta_{k^\ast}=\max_{k\in J(p)}(g_k(p)-\eta_k)$, so $k^\ast\in A$ and
\begin{align*}
\max_{k\in J(p)}(g_k(p)-\eta_k)=\max_{a\in A}(g_a(p)-\eta_a).
\end{align*}
We now check that every *other* index of $F$ sits strictly below the $A$-leader at $p$, in two cases. If $j\in J(p)\setminus A$, then $j$ is non-active and $k^\ast\neq j$, so
\begin{align*}
g_j(p)+\eta_j<\max_{k\in J(p),\,k\neq j}(g_k(p)-\eta_k)=\max_{a\in A}(g_a(p)-\eta_a).
\end{align*}
If $j\in F\setminus J(p)$, then $p\in\overline{W_j}\setminus W_j$, i.e. $|p-a_j|=R_j\ge\rho_j$, so the collar estimate gives
\begin{align*}
g_j(p)+\eta_j<\phi(p)<g_{k_0}(p)-\eta_{k_0}\le\max_{a\in A}(g_a(p)-\eta_a).
\end{align*}
Thus $g_j(p)+\eta_j<\max_{a\in A}(g_a(p)-\eta_a)$ for every $j\in F\setminus A$. This handles the subtle danger that a new index could *enter* $J(z)$ as $z$ moves: entering $W_j$ happens across its boundary sphere $|z-a_j|=R_j$, which lies in the collar, where $g_j$ has already dropped more than $\eta_j$ below $\phi$ — so any newly-appearing index is born already-deletable.
These strict inequalities and the continuity of the finitely many $g_j$ ($j\in F$) on $U_0$ let us define
\begin{align*}
U:=U_0\cap\bigcap_{a\in A}B(a_a,\rho_a)\cap\Big\{z:g_j(z)+\eta_j<\max_{a\in A}(g_a(z)-\eta_a)\ \text{for all }j\in F\setminus A\Big\},
\end{align*}
an open neighbourhood of $p$ with $U\subseteq B(a_a,\rho_a)\subset W_a$, so each $g_a\in C^\infty(U)$.
Finally, fix $z\in U$. Then $J(z)\subseteq F$ and $A\subseteq J(z)$ (as $z\in B(a_a,\rho_a)\subseteq W_a$). For each $j\in J(z)\setminus A\subseteq F\setminus A$,
\begin{align*}
g_j(z)+\eta_j<\max_{a\in A}(g_a(z)-\eta_a)\le\max_{k\in J(z),\,k\neq j}(g_k(z)-\eta_k),
\end{align*}
so locality (d) lets us delete $j$ from $M_{\eta|_{J(z)}}$ without changing its value; deletion leaves the $A$-values and $\max_{a\in A}(g_a(z)-\eta_a)$ untouched, so the remaining non-active indices stay deletable. Removing all of $J(z)\setminus A$ one at a time gives
\begin{align*}
\psi(z)=M_{\eta|_{J(z)}}\big((g_k(z))_{k\in J(z)}\big)=M_{\eta|_A}\big((g_a(z))_{a\in A}\big).
\end{align*}
Since $A$ and $\eta|_A$ are now fixed and each $g_a\in C^\infty(U)$ with $i\partial\bar\partial g_a\ge\beta_a\,\omega_0\ge(\min_{b\in A}\beta_b)\,\omega_0$, property (e) gives $\psi\in C^\infty(U)$ and $i\partial\bar\partial\psi\ge(\min_{a\in A}\beta_a)\,\omega_0$ on $U$, with $\min_{a\in A}\beta_a>0$. The step "active $\Rightarrow$ core" guaranteed every retained $a$ has $U\subseteq W_a$, so no $g_a$ is evaluated outside its domain.
[/guided]
[/step]
[step:Verify the two-sided estimate $\phi \leq \psi \leq \phi + \varepsilon$]
Fix $z\in\Omega$.
**Lower bound.** Choose $k_0$ with $z\in B_{k_0}'$; then $k_0\in J(z)$. By property (b) and the centre estimate,
\begin{align*}
\psi(z) = M_{\eta|_{J(z)}}\big((g_j(z))_{j\in J(z)}\big) \ \geq\ \max_{j\in J(z)} g_j(z) \ \geq\ g_{k_0}(z) \ \geq\ \phi(z) + (\kappa_{k_0}-\tau_{k_0}) \ >\ \phi(z).
\end{align*}
**Upper bound.** By property (b) and the cap estimate (valid for each $j\in J(z)$, since $z\in W_j$),
\begin{align*}
\psi(z) \ \leq\ \max_{j\in J(z)}\big(g_j(z)+\eta_j\big) \ \leq\ \max_{j\in J(z)}\Big(\phi(z) + \big(\tau_j + c_j s_j^2 + \eta_j\big)\Big) \ <\ \phi(z) + m_j \ \leq\ \phi(z) + \varepsilon(z),
\end{align*}
the strict inequality using $\tau_j+c_js_j^2+\eta_j<m_j=\inf_{\overline{W_j}}\varepsilon\leq\varepsilon(z)$ for every $j\in J(z)$ (recall $z\in\overline{W_j}$).
Hence $\phi \leq \psi \leq \phi + \varepsilon$ on all of $\Omega$.
[/step]
[step:Conclude]
By the Localisation Lemma of Step 5, $\psi\in C^\infty(\Omega)$ and $i\partial\bar\partial\psi$ is locally bounded below by a positive multiple of $\omega_0$, so $\psi$ is a smooth strictly plurisubharmonic function on $\Omega$. By Step 6 it satisfies $\phi\leq\psi\leq\phi+\varepsilon$. This is exactly the assertion of the theorem. $\blacksquare$
[/step]
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