[proofplan] Local integrability of plurisubharmonic functions makes the convolution finite on $\Omega_\varepsilon$, and smoothness follows from the Leibniz rule for differentiation under the integral. Plurisubharmonicity of $\phi_\varepsilon$ is proved by restricting to an affine complex line and verifying the circular sub-mean inequality: the convolution of the sub-mean inequality for $\phi$ against $\eta_\varepsilon$ yields, via Fubini, the sub-mean inequality for $\phi_\varepsilon$. The monotone decrease $\phi_\varepsilon \downarrow \phi$ is obtained by rewriting $\phi_\varepsilon(z)$ in polar coordinates as a weighted integral of the real spherical means $S(\phi, z, r) := \fint_{S^{2n-1}} \phi(z + r\omega)\, d\sigma(\omega)$, then invoking that $r \mapsto S(\phi, z, r)$ is non-decreasing (because plurisubharmonic functions are subharmonic on $\mathbb{R}^{2n}$) and tends to $\phi(z)$ as $r \downarrow 0$ by upper semicontinuity. [/proofplan]

[step:Establish that $\phi_\varepsilon$ is well-defined and finite on $\Omega_\varepsilon$]Since $\phi$ is plurisubharmonic on $\Omega$ and not identically $-\infty$ on any connected component, $\phi$ is locally integrable: $\phi \in L^1_{\mathrm{loc}}(\Omega; \mathcal{L}^{2n})$ (a standard consequence of the sub-mean inequality applied on small Euclidean balls, viewing $\phi$ as subharmonic on $\mathbb{R}^{2n}$; see the [Stability Properties of PSH Functions](/theorems/3404)). Fix $z \in \Omega_\varepsilon$. Because $\operatorname{supp}\, \eta_\varepsilon \subseteq \overline{B}(0,\varepsilon)$, the integrand $w \mapsto \phi(z-w)\, \eta_\varepsilon(w)$ is supported in $\overline{B}(0,\varepsilon)$, and the substitution $u = z - w$ identifies this with $\int_{\overline{B}(z,\varepsilon)} \phi(u)\, \eta_\varepsilon(z-u)\, d\mathcal{L}^{2n}(u)$. Since $\operatorname{dist}(z, \partial\Omega) > \varepsilon$, the closed ball $\overline{B}(z, \varepsilon)$ is a compact subset of $\Omega$, on which $\phi$ is integrable and $\eta_\varepsilon$ is bounded. Hence \begin{align*} |\phi_\varepsilon(z)| \leq \|\eta_\varepsilon\|_{L^\infty} \int_{\overline{B}(z, \varepsilon)} |\phi(u)|\, d\mathcal{L}^{2n}(u) < \infty, \end{align*} so $\phi_\varepsilon(z) \in \mathbb{R}$.[/step]

[guided]The first thing to check is that the integral defining $\phi_\varepsilon(z)$ even makes sense — psh functions can take the value $-\infty$ (e.g. $\log|f|$ for $f$ holomorphic), so we need to know that we are not integrating $-\infty$ on a set of positive measure. The key input is that a psh function which is not identically $-\infty$ on a connected component lies in $L^1_{\mathrm{loc}}$. We invoke this from the [Stability Properties of PSH Functions](/theorems/3404). The point is that psh implies subharmonic on $\mathbb{R}^{2n}$ (we will use this again in Step 4), and an upper-semicontinuous subharmonic function not identically $-\infty$ is in $L^1_{\mathrm{loc}}$. Now we verify that the integrand has compact support inside $\Omega$. Since $\eta$ is supported in $\overline{B}(0,1)$, the scaled mollifier $\eta_\varepsilon$ is supported in $\overline{B}(0,\varepsilon)$. So in the integral \begin{align*} \phi_\varepsilon(z) = \int_{\mathbb{C}^n} \phi(z-w)\, \eta_\varepsilon(w) \, d\mathcal{L}^{2n}(w), \end{align*} only $w \in \overline{B}(0,\varepsilon)$ contribute, i.e. only $u = z-w \in \overline{B}(z,\varepsilon)$. The defining condition $z \in \Omega_\varepsilon$ — i.e. $\operatorname{dist}(z, \partial\Omega) > \varepsilon$ — is precisely what guarantees $\overline{B}(z, \varepsilon) \subset \Omega$, so $\phi$ is integrable on this ball. Combined with the obvious bound $\eta_\varepsilon \leq \|\eta\|_{L^\infty}\varepsilon^{-2n}$, we get \begin{align*} |\phi_\varepsilon(z)| \leq \|\eta\|_{L^\infty}\, \varepsilon^{-2n} \int_{\overline{B}(z,\varepsilon)} |\phi(u)|\, d\mathcal{L}^{2n}(u) < \infty. \end{align*} This is exactly the role of restricting to $\Omega_\varepsilon$: any closer to $\partial\Omega$ and the support of the mollification would spill outside $\Omega$, leaving $\phi$ undefined.[/guided]

[step:Differentiate under the integral to obtain smoothness] We claim $\phi_\varepsilon \in C^\infty(\Omega_\varepsilon)$. Fix $z_0 \in \Omega_\varepsilon$ and choose $\delta > 0$ with $\overline{B}(z_0, \delta) \subset \Omega_\varepsilon$. Then for every $z \in B(z_0, \delta)$ and every multi-index $\alpha \in \mathbb{N}_0^{2n}$ (in real coordinates on $\mathbb{C}^n \cong \mathbb{R}^{2n}$), \begin{align*} \partial_z^\alpha \big[ \eta_\varepsilon(z - u) \big] \quad \text{is continuous in } (z, u) \text{ and supported in } u \in \overline{B}(z_0, \delta+\varepsilon). \end{align*} Since $\phi \in L^1(\overline{B}(z_0, \delta+\varepsilon))$ and $\partial_z^\alpha \eta_\varepsilon(z-u)$ is uniformly bounded on $\overline{B}(z_0, \delta) \times \overline{B}(z_0, \delta+\varepsilon)$, the hypotheses of the [Leibniz Integral Rule](/theorems/831) are satisfied (the dominating function being $\sup_{z, u} |\partial_z^\alpha \eta_\varepsilon(z-u)| \cdot |\phi(u)| \mathbb{1}_{\overline{B}(z_0,\delta+\varepsilon)}(u) \in L^1$). Hence \begin{align*} \partial_z^\alpha \phi_\varepsilon(z) = \int_{\mathbb{C}^n} \phi(u)\, \partial_z^\alpha \eta_\varepsilon(z-u) \, d\mathcal{L}^{2n}(u) \end{align*} exists and is continuous on $B(z_0, \delta)$. As $z_0$ was arbitrary, $\phi_\varepsilon \in C^\infty(\Omega_\varepsilon)$. This is the standard [Smoothness of Mollification (Local)](/theorems/35) applied to $\phi \in L^1_{\mathrm{loc}}(\Omega)$. [/step]

[step:Reduce plurisubharmonicity to the sub-mean inequality on every complex line] Since $\phi_\varepsilon \in C^\infty(\Omega_\varepsilon)$ in particular it is upper semicontinuous. To prove $\phi_\varepsilon$ is plurisubharmonic on $\Omega_\varepsilon$ it suffices to verify the circular sub-mean inequality on every affine complex line: for every $z \in \Omega_\varepsilon$, every $v \in \mathbb{C}^n \setminus \{0\}$, and every $r > 0$ such that the closed disc \begin{align*} \overline{D}_{z,v,r} := \{ z + \zeta v : \zeta \in \mathbb{C}, |\zeta| \leq r \} \end{align*} is contained in $\Omega_\varepsilon$, \begin{align*} \phi_\varepsilon(z) \;\leq\; \frac{1}{2\pi} \int_0^{2\pi} \phi_\varepsilon(z + r e^{i\theta} v) \, d\theta. \tag{$\ast$} \end{align*} We prove $(\ast)$ in the next step. [/step]

[step:Verify the sub-mean inequality on a complex line by Fubini]Fix $z \in \Omega_\varepsilon$, $v \in \mathbb{C}^n \setminus \{0\}$, and $r > 0$ with $\overline{D}_{z,v,r} \subset \Omega_\varepsilon$. For each fixed $w \in \overline{B}(0,\varepsilon)$, the translate $\zeta \mapsto \phi(z + \zeta v - w)$ is defined on the [open set](/page/Open%20Set) $\{\zeta \in \mathbb{C} : z + \zeta v - w \in \Omega\}$, which contains the closed disc $\{|\zeta| \leq r\}$ because $\overline{D}_{z,v,r} \subset \Omega_\varepsilon$ implies $\overline{D}_{z,v,r} - w \subset \Omega$ (using $\operatorname{dist}(\overline{D}_{z,v,r}, \partial\Omega) > \varepsilon \geq |w|$). Since $\phi$ is plurisubharmonic on $\Omega$, the function $\zeta \mapsto \phi(z + \zeta v - w)$ is subharmonic on this disc; the sub-mean inequality at $\zeta = 0$ on the circle of radius $r$ gives \begin{align*} \phi(z - w) \;\leq\; \frac{1}{2\pi} \int_0^{2\pi} \phi(z + r e^{i\theta} v - w) \, d\theta. \tag{$\ast\ast$} \end{align*} Multiplying $(\ast\ast)$ by $\eta_\varepsilon(w) \geq 0$ and integrating over $w \in \mathbb{C}^n$: \begin{align*} \phi_\varepsilon(z) = \int_{\mathbb{C}^n} \phi(z-w)\, \eta_\varepsilon(w)\, d\mathcal{L}^{2n}(w) \;\leq\; \int_{\mathbb{C}^n} \eta_\varepsilon(w) \cdot \frac{1}{2\pi}\int_0^{2\pi} \phi(z + re^{i\theta}v - w)\, d\theta\, d\mathcal{L}^{2n}(w). \end{align*} We exchange the order of integration via the [Fubini Theorem](/theorems/513). To justify Fubini, we apply it to $|\phi(z + re^{i\theta}v - w)|\, \eta_\varepsilon(w)$: the integration is over the compact set $w \in \overline{B}(0,\varepsilon)$ and $\theta \in [0, 2\pi]$, the integrand is $\mathcal{L}^{2n} \otimes d\theta$-measurable, and as $\theta$ varies the point $z + re^{i\theta}v - w$ ranges over the compact set $K := \overline{D}_{z,v,r} - \overline{B}(0,\varepsilon) \subset \Omega$; since $\phi \in L^1(K)$ and $\eta_\varepsilon \in L^\infty$, the absolute integral is finite. Hence \begin{align*} \phi_\varepsilon(z) \;\leq\; \frac{1}{2\pi}\int_0^{2\pi} \int_{\mathbb{C}^n} \phi(z + re^{i\theta}v - w)\, \eta_\varepsilon(w)\, d\mathcal{L}^{2n}(w)\, d\theta = \frac{1}{2\pi}\int_0^{2\pi} \phi_\varepsilon(z + re^{i\theta}v)\, d\theta, \end{align*} which is $(\ast)$. Combined with the smoothness and upper semicontinuity of $\phi_\varepsilon$ from Step 2, this establishes that $\phi_\varepsilon$ is plurisubharmonic on $\Omega_\varepsilon$, proving claim (ii).[/step]

[guided]We want to show that the restriction of $\phi_\varepsilon$ to any affine complex line is subharmonic, and the simplest route is the circular sub-mean inequality. The idea is: we already know it for $\phi$ along every line; averaging that inequality against $\eta_\varepsilon$ — i.e. mollifying — should preserve it, provided we can swap integrals. **Setting up the line.** For $z, v, r$ as fixed, and any $w$ in the support of $\eta_\varepsilon$ (so $|w| \leq \varepsilon$), the *translated* disc $\overline{D}_{z,v,r} - w$ also lies in $\Omega$ — this is precisely why we restricted to $\Omega_\varepsilon$: the buffer $\varepsilon$ around $\overline{D}_{z,v,r}$ stays inside $\Omega$. So the function $\zeta \mapsto \phi(z + \zeta v - w)$ is subharmonic on a neighbourhood of $\{|\zeta| \leq r\}$, and we have the sub-mean inequality $(\ast\ast)$ at $\zeta = 0$. **Averaging.** Multiply $(\ast\ast)$ by $\eta_\varepsilon(w)$ — this preserves the inequality because $\eta_\varepsilon \geq 0$ — and integrate $dw$. The LHS becomes $\phi_\varepsilon(z)$. The RHS is a double integral (over $w$ and over the circle $\theta \in [0, 2\pi]$). **Why Fubini applies.** We need to swap $\int dw$ and $\int d\theta$. Fubini–Tonelli requires absolute integrability of the integrand on the product space. Both integrations are over compact sets: $w \in \overline{B}(0,\varepsilon)$, $\theta \in [0, 2\pi]$. The integrand $\phi(z + re^{i\theta}v - w)\eta_\varepsilon(w)$ is a product of a locally integrable function of one variable and a bounded function of the other; uniformly in $\theta$, the argument $z + re^{i\theta}v - w$ ranges over the compact set $K = \overline{D}_{z,v,r} - \overline{B}(0,\varepsilon)$, which is contained in $\Omega$, so $\phi \in L^1(K)$. Thus the absolute integral is bounded by $2\pi \|\eta_\varepsilon\|_{L^\infty} \|\phi\|_{L^1(K)} < \infty$ and Fubini applies. **Swap and re-recognise.** After the swap, the inner integral over $w$ is exactly $\phi_\varepsilon(z + re^{i\theta}v)$, yielding $(\ast)$. **Why this gives plurisubharmonicity.** A smooth function on a complex domain whose restriction to every affine complex disc satisfies the circular sub-mean inequality is plurisubharmonic (this is one of several equivalent definitions). The upper semicontinuity required in the definition is automatic from $\phi_\varepsilon \in C^\infty$.[/guided]

[step:Express $\phi_\varepsilon$ as a weighted integral of spherical means] For $\psi \in L^1_{\mathrm{loc}}(\Omega)$, $z \in \Omega$, and $0 < r < \operatorname{dist}(z, \partial\Omega)$, define the **real spherical mean** on the Euclidean sphere of $\mathbb{R}^{2n} \cong \mathbb{C}^n$: \begin{align*} S: \{(z, r) : z \in \Omega,\, 0 < r < \operatorname{dist}(z, \partial\Omega)\} &\to [-\infty, \infty), \\ (z, r) &\mapsto S(\psi, z, r) := \frac{1}{\sigma_{2n-1}(S^{2n-1})} \int_{S^{2n-1}} \psi(z + r\omega)\, d\sigma_{2n-1}(\omega), \end{align*} where $\sigma_{2n-1}$ is the surface measure on $S^{2n-1} \subset \mathbb{R}^{2n}$ with total mass $\sigma_{2n-1}(S^{2n-1}) =: \omega_{2n-1}$. Fix $z \in \Omega_\varepsilon$. Writing $w = s\omega$ with $s \in (0, \varepsilon)$ and $\omega \in S^{2n-1}$ in polar coordinates on $\mathbb{R}^{2n}$, the Lebesgue measure decomposes as $d\mathcal{L}^{2n}(w) = s^{2n-1}\, d\sigma_{2n-1}(\omega)\, d\mathcal{L}^1(s)$, and $\eta_\varepsilon(w) = \tilde{\eta}_\varepsilon(s)$ is independent of $\omega$ by radiality, where $\tilde{\eta}_\varepsilon(s) := \varepsilon^{-2n} \tilde{\eta}(s/\varepsilon)$. Using that $\omega \mapsto -\omega$ preserves $d\sigma_{2n-1}$, we compute: \begin{align*} \phi_\varepsilon(z) &= \int_{\mathbb{C}^n} \phi(z-w)\, \eta_\varepsilon(w)\, d\mathcal{L}^{2n}(w) \\ &= \int_0^\varepsilon \tilde{\eta}_\varepsilon(s)\, s^{2n-1} \int_{S^{2n-1}} \phi(z - s\omega)\, d\sigma_{2n-1}(\omega) \, d\mathcal{L}^1(s) \\ &= \int_0^\varepsilon \tilde{\eta}_\varepsilon(s)\, s^{2n-1}\, \omega_{2n-1}\, S(\phi, z, s)\, d\mathcal{L}^1(s). \end{align*} Substituting $s = \varepsilon t$, so $d\mathcal{L}^1(s) = \varepsilon\, d\mathcal{L}^1(t)$ and $\tilde{\eta}_\varepsilon(\varepsilon t)\, (\varepsilon t)^{2n-1} \varepsilon = \tilde{\eta}(t) t^{2n-1}$: \begin{align*} \phi_\varepsilon(z) \;=\; \omega_{2n-1} \int_0^1 \tilde{\eta}(t)\, t^{2n-1}\, S(\phi, z, \varepsilon t)\, d\mathcal{L}^1(t). \tag{$\dagger$} \end{align*} [/step]

[step:Apply monotonicity of spherical means of subharmonic functions] Plurisubharmonic functions on $\Omega \subseteq \mathbb{C}^n$ are subharmonic on $\Omega$ viewed as an open subset of $\mathbb{R}^{2n}$ — see the [Stability Properties of PSH Functions](/theorems/3404). A standard consequence of subharmonicity is that for fixed $z \in \Omega$, the spherical-mean function \begin{align*} r \longmapsto S(\phi, z, r), \qquad 0 < r < \operatorname{dist}(z, \partial\Omega), \end{align*} is non-decreasing in $r$, and $\lim_{r \downarrow 0} S(\phi, z, r) = \phi(z) \in [-\infty, \infty)$. (Proof sketch of the monotonicity, included for completeness: for $0 < r_1 < r_2$, apply the sub-mean inequality on the sphere of radius $r_2$ centred at $z$ to the average of $\phi$ on the sphere of radius $r_1$ — equivalently, integrate the sub-mean inequality of $\phi$ over the inner sphere; this gives $S(\phi, z, r_1) \leq S(\phi, z, r_2)$. The limit $r \downarrow 0$ uses upper semicontinuity of $\phi$ to obtain $\limsup_{r \downarrow 0} S(\phi, z, r) \leq \phi(z)$, while the sub-mean inequality gives the reverse inequality $\phi(z) \leq S(\phi, z, r)$, forcing equality in the limit.) Hence for fixed $z \in \Omega$ and $t \in (0,1]$, the function $\varepsilon \mapsto S(\phi, z, \varepsilon t)$ is non-decreasing in $\varepsilon$ (on the range where $z \in \Omega_{\varepsilon t}$, in particular $z \in \Omega_\varepsilon$) and decreases to $\phi(z)$ as $\varepsilon \downarrow 0$. [/step]

[step:Deduce monotone decrease and pointwise convergence to $\phi$]Fix $z \in \Omega$. By the formula $(\dagger)$ from the previous step, for any $\varepsilon$ with $z \in \Omega_\varepsilon$, \begin{align*} \phi_\varepsilon(z) = \omega_{2n-1} \int_0^1 \tilde{\eta}(t)\, t^{2n-1}\, S(\phi, z, \varepsilon t)\, d\mathcal{L}^1(t). \end{align*} The integrand is non-negative-weighted by $\tilde{\eta}(t)\, t^{2n-1} \geq 0$, and for each $t \in (0,1]$ the function $\varepsilon \mapsto S(\phi, z, \varepsilon t)$ is non-decreasing in $\varepsilon$ and converges to $\phi(z)$ as $\varepsilon \downarrow 0$. Therefore $\varepsilon \mapsto \phi_\varepsilon(z)$ is non-decreasing in $\varepsilon$, proving claim (iii). For the pointwise limit, we apply the [Monotone Convergence Theorem](/theorems/509) to the family $f_\varepsilon(t) := \omega_{2n-1}\, \tilde{\eta}(t)\, t^{2n-1}\,[S(\phi, z, \varepsilon_0 t) - S(\phi, z, \varepsilon t)]$, indexed by $\varepsilon \in (0, \varepsilon_0]$ for a fixed $\varepsilon_0$ small enough that $z \in \Omega_{\varepsilon_0}$. We verify the hypotheses of MCT: 1. Non-negativity: $S(\phi, z, \varepsilon t) \leq S(\phi, z, \varepsilon_0 t)$ for $\varepsilon \leq \varepsilon_0$ by Step 6, so $f_\varepsilon \geq 0$. 2. Monotone increase as $\varepsilon \downarrow 0$: again by Step 6. 3. Pointwise limit: $f_\varepsilon(t) \uparrow \omega_{2n-1}\, \tilde{\eta}(t)\, t^{2n-1}\,[S(\phi, z, \varepsilon_0 t) - \phi(z)]$ as $\varepsilon \downarrow 0$. The MCT yields \begin{align*} \int_0^1 f_\varepsilon \, d\mathcal{L}^1 \;\xrightarrow{\varepsilon \downarrow 0}\; \omega_{2n-1} \int_0^1 \tilde{\eta}(t)\, t^{2n-1}\, [S(\phi, z, \varepsilon_0 t) - \phi(z)]\, d\mathcal{L}^1(t). \end{align*} Rewriting in terms of $\phi_\varepsilon(z)$ via $(\dagger)$, this says \begin{align*} \phi_{\varepsilon_0}(z) - \phi_\varepsilon(z) \;\xrightarrow{\varepsilon \downarrow 0}\; \phi_{\varepsilon_0}(z) - \phi(z) \cdot \omega_{2n-1}\int_0^1 \tilde{\eta}(t)\, t^{2n-1}\, d\mathcal{L}^1(t), \end{align*} provided the integral $\omega_{2n-1}\int_0^1 \tilde{\eta}(t)\, t^{2n-1}\, d\mathcal{L}^1(t)$ is finite — which is immediate since the integrand is continuous on $[0,1]$. We now compute this constant: applying the same polar-coordinate decomposition as in Step 5 to the unit-mass integral $\int_{\mathbb{C}^n} \eta\, d\mathcal{L}^{2n} = 1$, \begin{align*} 1 = \int_{\mathbb{C}^n} \eta(w)\, d\mathcal{L}^{2n}(w) = \omega_{2n-1} \int_0^1 \tilde{\eta}(t)\, t^{2n-1}\, d\mathcal{L}^1(t). \end{align*} Substituting back, $\phi_\varepsilon(z) \to \phi(z)$ as $\varepsilon \downarrow 0$. (The argument allows $\phi(z) = -\infty$: in that case, $S(\phi, z, \varepsilon t) \downarrow -\infty$ as $\varepsilon \downarrow 0$ for every $t \in (0,1]$, and $\phi_\varepsilon(z) \downarrow -\infty$ accordingly.) This proves claim (iv) and completes the proof.[/step]

[guided]Step 5 reduced $\phi_\varepsilon(z)$ to an integral over $t \in (0,1]$ of the spherical means $S(\phi, z, \varepsilon t)$, weighted by the radial profile $\tilde\eta(t) t^{2n-1}$. Step 6 told us each spherical mean is monotone non-decreasing in its radius and approaches $\phi(z)$ in the limit. Now we just have to pass the limit through the $t$-integral. **Monotonicity in $\varepsilon$ is immediate.** Since $\tilde\eta(t) t^{2n-1} \geq 0$ and $\varepsilon \mapsto S(\phi, z, \varepsilon t)$ is non-decreasing for each $t$, the same is true after integrating in $t$. So $\phi_{\varepsilon_1}(z) \leq \phi_{\varepsilon_2}(z)$ for $\varepsilon_1 < \varepsilon_2$, which is claim (iii). **Passing to the limit.** We want to compute $\lim_{\varepsilon \downarrow 0} \int_0^1 \tilde\eta(t)t^{2n-1} S(\phi, z, \varepsilon t)\, d\mathcal{L}^1(t)$. The integrand is non-decreasing in $\varepsilon$ but might go to $-\infty$ (if $\phi(z) = -\infty$), so MCT cannot be applied directly to $S(\phi, z, \varepsilon t)$ itself. The standard trick is to subtract from a fixed upper bound: let $\varepsilon_0$ be small enough that $z \in \Omega_{\varepsilon_0}$, and look at the **gap** \begin{align*} g_\varepsilon(t) := S(\phi, z, \varepsilon_0 t) - S(\phi, z, \varepsilon t) \geq 0. \end{align*} As $\varepsilon \downarrow 0$, $g_\varepsilon(t)$ increases pointwise to $S(\phi, z, \varepsilon_0 t) - \phi(z)$ (well-defined in $[0, +\infty]$). Multiply by the non-negative weight $\omega_{2n-1}\tilde\eta(t) t^{2n-1}$ — the MCT now applies cleanly. **Computing the limit.** The MCT gives \begin{align*} \lim_{\varepsilon \downarrow 0} \left[\phi_{\varepsilon_0}(z) - \phi_\varepsilon(z)\right] = \omega_{2n-1}\int_0^1 \tilde\eta(t)t^{2n-1}[S(\phi, z, \varepsilon_0 t) - \phi(z)]\, d\mathcal{L}^1(t). \end{align*} The RHS is $\phi_{\varepsilon_0}(z) - \phi(z) \cdot C$ where $C = \omega_{2n-1}\int_0^1 \tilde\eta(t)t^{2n-1}\, d\mathcal{L}^1(t)$. **The constant $C$ equals $1$.** This is the radial polar-coordinate computation of the unit mass of $\eta$: writing $\int_{\mathbb{C}^n}\eta\, d\mathcal{L}^{2n} = 1$ in polar coordinates and using radiality, \begin{align*} 1 = \int_{S^{2n-1}}\int_0^1 \tilde\eta(s)\, s^{2n-1}\, d\mathcal{L}^1(s)\, d\sigma_{2n-1}(\omega) = \omega_{2n-1}\int_0^1 \tilde\eta(s) s^{2n-1}\, d\mathcal{L}^1(s), \end{align*} so $C = 1$. **Conclusion.** Substituting back, $\phi_{\varepsilon_0}(z) - \phi_\varepsilon(z) \to \phi_{\varepsilon_0}(z) - \phi(z)$, i.e. $\phi_\varepsilon(z) \to \phi(z)$. Combined with monotonicity, this is $\phi_\varepsilon(z) \downarrow \phi(z)$, which is claim (iv). **The $\phi(z) = -\infty$ case.** Nothing fails: the limit identity still holds in the extended real sense. Concretely, $\phi_\varepsilon(z)$ is a finite real number for each $\varepsilon > 0$ (Step 1), and by monotonicity is non-increasing as $\varepsilon \downarrow 0$, so it converges in $[-\infty, +\infty)$. The argument above forces the limit to be $\phi(z) = -\infty$.[/guided]