Bergman kernel and oscillation theory of plurisubharmonic functions

Abstract

Based on Harnack’s inequality and convex analysis we show that each plurisubharmonic function is locally BUO (bounded upper oscillation) with respect to polydiscs of finite type but not for arbitrary polydiscs. We also show that each function in the Lelong class is globally BUO with respect to all polydiscs. A dimension-free BUO estimate is obtained for the logarithm of the modulus of a complex polynomial. As an application we obtain an approximation formula for the Bergman kernel that preserves all directional Lelong numbers. For smooth plurisubharmonic functions we derive a new asymptotic identity for the Bergman kernel from Berndtsson’s complex Brunn–Minkowski theory, which also yields a slightly better version of the sharp Ohsawa–Takegoshi extension theorem in some special cases.

1 Introduction

Let $\Omega$ be a domain in ${\mathbb {C}}^n$ and $PSH(\Omega )$ the set of plurisubharmonic (psh) functions on $\Omega$. Recall that each $\phi \in PSH(\Omega )$ satisfies the following mean-value inequality:

$$\begin{aligned} \phi (z)\le \frac{1}{|S|} \int _S \phi =:\phi _S \end{aligned}$$

whenever S is a ball or a polydisc, with center z. Here |S| denotes the Lebesgue measure of S and $\int _S$ means the Lebesgue integral. The above inequality implies $\phi \in L^1_\mathrm{loc}(\Omega )$ and suggests to estimate the difference $|\phi -\phi _S|$. The concept of BMO functions then enters naturally. Let ${{\mathcal {S}}}={{\mathcal {S}}}(\Omega )$ be a family of relatively compact open subsets in $\Omega$. We say that $\phi \in L^1_\mathrm{loc}(\Omega )$ has bounded mean oscillation (BMO) with respect to ${{\mathcal {S}}}$ if

$$\begin{aligned} {\sup }_{S\in {{\mathcal {S}}}}\, MO_{S}(\phi ) <\infty ,\quad MO_{S}(\phi ):=\frac{1}{|S|} \int _S |\phi -\phi _S|. \end{aligned}$$

Let $BMO(\Omega ,{{\mathcal {S}}})$ denote the set of functions which are BMO with respect to ${{\mathcal {S}}}$. When ${{\mathcal {S}}}$ is the set of balls in $\Omega$, this is the original definition of BMO functions due to John–Nirenberg [13]. A classical example of BMO functions is $\log |z|$. It is also convenient to introduce local BMO functions as follows. For an open set $\Omega _0\subset \subset \Omega$ we define ${{\mathcal {S}}}|_{\Omega _0}$ to be the sets of all $S\in {{\mathcal {S}}}$ which are relatively compact in $\Omega _0$. Let $BMO_\mathrm{loc}(\Omega ,{{\mathcal {S}}})$ be the set of functions on $\Omega$ which belong to $BMO(\Omega _0,{{\mathcal {S}}}|_{\Omega _0})$ for every open set $\Omega _0\subset \subset \Omega$.

By using pluripotential theory, Brudnyi [6] was able to show that each psh function is locally BMO with respect to balls (see also [7] for stronger results concerning subharmonic functions in the plane). Recently, the first author found another approach to local BMO properties of psh functions by using the Riesz decomposition theorem and some basic facts of psh functions (cf. [9]). Benelkourchi et al. [1] showed that every function in the Lelong class ${{\mathcal {L}}}$ is globally BMO with respect to balls. Recall that

$$\begin{aligned} {{\mathcal {L}}}=\left\{ u\in PSH({\mathbb {C}}^n): {\limsup }_{|z|\rightarrow \infty }\, (u(z)-\log |z|)<\infty \right\} . \end{aligned}$$

In this paper we propose a new and simpler approach based on the following basic observation:

It is easier to look at the upper oscillation instead of the mean oscillation for psh functions.

To define the upper oscillation one simply uses $\sup _S \phi$ instead of $\phi _S$:

$$\begin{aligned} UO_{S}(\phi ):=\frac{1}{|S|} \int _S \left| \phi -{\sup }_S\, \phi \right| ={\sup }_S\, \phi -\phi _S. \end{aligned}$$

(1.1)

Note that $-UO_S(-\phi )$ is exactly the lower oscillation introduced by Coiffman–Rochberg (cf. [10], see also [16] for further properties). Since

$$\begin{aligned} MO_{S}(\phi )=\frac{1}{|S|} \int _S |\phi -\phi _S|=\frac{2}{|S|} \int _{\phi <\phi _S} (\phi _S-\phi ) \le 2\, UO_S(\phi ), \end{aligned}$$

(1.2)

we see that bounded upper oscillation (BUO) implies BMO. One may define $BUO(\Omega ,{{\mathcal {S}}})$ and $BUO_\mathrm{loc}(\Omega ,{{\mathcal {S}}})$ analogously as the case of BMO.

Let ${{\mathcal {P}}}={{\mathcal {P}}}(\Omega )$ denote the set of relatively compact polydiscs in $\Omega$ and ${{\mathcal {P}}}_N$ the set of polydiscs $P\subset \subset \Omega$ of finite type N, i.e.,

$$\begin{aligned} \max \{r_j\} \le \min \{r_j^{1/N}\}, \end{aligned}$$

where $N>0$ and $\{r_j\}_{1\le j\le n}$ is the polyradius of P.

Based on Harnack’s inequality and convex analysis, we are able to show the following

Theorem 1.1

• (1) $PSH(\Omega )\subset BUO_\mathrm{loc}(\Omega ,{{\mathcal {P}}}_N) \subset BMO_\mathrm{loc}(\Omega ,{{\mathcal {P}}}_N)$;

• (2) $PSH({\mathbb {D}}^n)\nsubseteq BMO_\mathrm{loc}({\mathbb {D}}^n,{{\mathcal {P}}})$ for $n\ge 2$, where ${\mathbb {D}}^n$ is the unit polydisc;

• (3) ${{\mathcal {L}}}\subset BUO({\mathbb {C}}^n, {\mathcal {P}})$; more precisely, for every $\phi \in PSH({\mathbb {C}}^n)$ with

  $$\begin{aligned} \phi (z_1,\ldots , z_n) \le c+\max _{1\le j\le n} \log (1+|z_j|), \ \ \forall \ (z_1, \ldots , z_n)\in {\mathbb {C}}^n, \end{aligned}$$

  where c is a constant, we have $UO_P(\phi ) < 3^n$ for all polydiscs P in ${\mathbb {C}}^n$.

For $(\mathrm{deg\,}p)^{-1}\log |p|\in {\mathcal {L}}$ where p is a complex polynomial, we even obtain a dimension-free BUO estimate with respect to all compact convex sets.

Theorem 1.2

For every non-empty compact convex set A in ${\mathbb {C}}^n$, we have

$$\begin{aligned} UO_{A} (\log |p|) \le \gamma \cdot \deg p, \end{aligned}$$

for all $p\in {\mathbb {C}}[z_1, \ldots , z_n]$. Here the constant $\gamma \in (1,2)$ is determined by

$$\begin{aligned} \gamma +\log (\gamma -1)=0. \end{aligned}$$

Remark

• (i) The above estimate is sharp, in fact, there exists a line segment A in ${\mathbb {C}}$ such that

  $$\begin{aligned} UO_A(\log |z|)=\gamma . \end{aligned}$$

• (ii) In particular, if A is a compact convex set in ${\mathbb {R}}^n \subset {\mathbb {C}}^n$ and all coefficients of p are real, then we have

  $$\begin{aligned} UO_A (\log |p|) \le \gamma \cdot \deg p <2\deg p, \end{aligned}$$

  which is closely related the classical Remez inequality for real polynomials. Theorem 1.2 also suggests to study the Remez inequality for complex polynomials (see [1] and [8] for related results).

• (iii) Notice that $1.278<\gamma <1.279.$ By (1.2) we have

  $$\begin{aligned} MO_A (\log |p|) \le 2\gamma \cdot \deg p< 2.558\cdot \deg p. \end{aligned}$$

  Such dimension-free estimate (with a slightly better constant $2+\log 2\approx 2.301$) was first obtained by Nazarov et al. [15]. Our proof of Theorem 1.2 is elementary, however.

For $\phi \in PSH(\Omega )$ we define the (weighted) Bergman kernel by

$$\begin{aligned} K_{\phi ,\Omega }(z)=\sup \left\{ |f(z)|^2: f\in {{\mathcal {O}}}(\Omega ),\ \int _\Omega |f|^2 e^{-\phi }\le 1\right\} . \end{aligned}$$

For a vector $a=(a_1,\ldots ,a_n)$ with all $a_j>0$ we set

$$\begin{aligned} P_{r^a}:=\{z\in {\mathbb {C}}^n: |z_j|\le r^{a_j},\,1\le j\le n\}. \end{aligned}$$

It was shown in [9] that if $\phi$ is psh on the closure of the unit ball ${\mathbb {B}}^n$ and $a_0=(1,1/2,\ldots ,1/2)$ then

$$\begin{aligned} \lim _{r\rightarrow 0+} \frac{\log K_{\varepsilon \phi ,\mathbb B^n}(1-r,0,\ldots ,0)}{\log 1/r}=n+1-\varepsilon \cdot \lim _{r\rightarrow 0+} \frac{\sup _{z\in P_{r^{a_0}}} \phi (1+z)}{\log r} \end{aligned}$$

provided $\varepsilon \ll 1$, where $1+z=(1+z_1,z_2,\ldots ,z_n)$. The limit in RHS of the above inequality is called the $a_0-$directional Lelong number of $\phi$ at $(1,0,\ldots ,0)$ (see [14]).

Here we will present an analogous but independent result, as an application of Theorem 1.1. For $\phi \in PSH({\mathbb D}^n)$ and $t\in {\mathbb {D}}^n$ we define

$$\begin{aligned} \phi ^t(z):=\phi (tz), \quad tz:=(t_1z_1, \ldots , t_n z_n). \end{aligned}$$

A fundamental result of Berndtsson [2] implies that

$$\begin{aligned} F(\phi ): \, (t,z) \mapsto \log K_{\phi ^t, \, {\mathbb {D}}^n}(z) \end{aligned}$$

is psh on ${\mathbb {D}}^n \times {\mathbb {D}}^n$.

Theorem 1.3

For each $a=(a_1,\ldots ,a_n)$ with all $a_j>0$, there exists a number $\varepsilon _0=\varepsilon _0(a,\phi ,\Omega )$ such that

$$\begin{aligned} \lim _{r\rightarrow 0+} \frac{\sup _{t\in P_{r^a}} F(\varepsilon \phi )(t,0)/\varepsilon }{\log r}=\lim _{r\rightarrow 0+} \frac{\sup _{z\in P_{r^a}} \phi (z) }{\log r} \end{aligned}$$

(1.3)

holds for all $\varepsilon \le \varepsilon _0$.

Although Theorem 1.3 makes sense only when $\phi$ is singular at the origin, it is of independent interest to study the relation between $F(\phi )$ and $\phi$ for smooth $\phi$.

Theorem 1.4

Let $\phi$ be a smooth psh function on ${\mathbb {D}}^n$. Then

$$\begin{aligned} \lim _{t\rightarrow 0} \frac{\partial ^2 F(\phi )}{\partial t_j \partial \bar{t}_k} (t,0)= {\left\{ \begin{array}{ll} \frac{1}{2} \cdot \frac{\partial ^2 \phi }{\partial z_j \partial {\bar{z}}_j} (0), \quad \text {if}\ \, j=k; \\ 0, \quad \text {if}\ \, j\ne k. \end{array}\right. } \end{aligned}$$

(1.4)

In particular $F(\phi )(t,0)$ is strictly psh at $t=0$ if $\phi$ is strictly psh at $z=0$.

Remark

Since $F(\phi )(t,0)$ depends only on $(|t_1|,\ldots , |t_n|)$, it follows from the psh property of $F(\phi )$ that

$$\begin{aligned} \log \frac{e^{\phi (0)}}{\pi ^n}=F(\phi )(0,0)\le F(\phi )(t,0)=\log K_{\phi ^t, \,{\mathbb {D}}^n}(0). \end{aligned}$$

(1.5)

Letting t tend to $(1,\ldots , 1)$, we obtain the sharp Ohsawa–Takegoshi estimate (cf. [5]; see also [4, 12]):

$$\begin{aligned} K_{\phi ,\, {\mathbb {D}}^n}(0) \ge \frac{e^{\phi (0)}}{\pi ^n}. \end{aligned}$$

(1.6)

Theorem 1.4 suggests that one should have a better lower bound for $K_{\phi , \,{\mathbb {D}}^n}$ in case $\phi$ is strictly psh.

2 An enlightening example

To explain why BUO is easier than BMO, we will show that the upper oscillation of $\log |z|$ with respect to discs is computable. Recall that

$$\begin{aligned} UO_{B} (\log |z|):={\sup }_B \log |z|-(\log |z|)_B \end{aligned}$$

for every disc B in ${\mathbb {C}}$.

Lemma 2.0.1

Fix $\hat{z}\in {\mathbb {C}}$ and set

$$\begin{aligned} I(c):=\frac{1}{2\pi } \int _{0}^{2\pi } \log |{\hat{z}}+c e^{i\theta }| \, d\theta , \quad c>0. \end{aligned}$$

Then we have

$$\begin{aligned} I(c)= {\left\{ \begin{array}{ll} \begin{array}{ll} \log |{\hat{z}}| &{}\quad \text {if}\ \ c\le |{\hat{z}}| \\ \log c &{}\quad \text {if}\ \ c> |{\hat{z}}|. \end{array} \end{array}\right. } \end{aligned}$$

Proof

If $c\le |{\hat{z}}|$ then $\log |z|$ is harmonic in the disc $\{z:|z-{\hat{z}}| < c\}$, so that $I(c)=\log |{\hat{z}}|$, in view of the mean-value equality. For $c>|{\hat{z}}|$ we may write

$$\begin{aligned} I(c)=\frac{1}{2\pi }\int _{0}^{2\pi } \log |{\hat{z}} e^{i\theta }+c|\, d\theta . \end{aligned}$$

As $\log |z|$ is harmonic in $\{z:|z-c|< |{\hat{z}}|\}$, we get $I(c)=\log c$. $\square$

Proposition 2.0.1

For any disc B we have

$$\begin{aligned} UO_{B} (\log |z|) \le \log \frac{\sqrt{5}+1}{2} +\frac{\sqrt{5}-1}{4}. \end{aligned}$$

Moreover, the bound is sharp.

Proof

Suppose $B=\{z:|z-{\hat{z}}|<b\}$. By Lemma 2.0.1 we have

$$\begin{aligned} (\log |z|)_{B}= \log |{\hat{z}}|, \quad \text {if}\ \ b\le |{\hat{z}}|, \end{aligned}$$

and if $b>|{\hat{z}}|$ then

$$\begin{aligned} (\log |z|)_{B}= & {} \frac{1}{\pi b^2}\int _{0}^b 2\pi c \cdot I(c)\, dc\\= & {} \log b-\frac{1}{2} \left( 1-\frac{|{\hat{z}}|^2}{b^2}\right) . \end{aligned}$$

It follows that

$$\begin{aligned} UO_{B} (\log |z|) = {\left\{ \begin{array}{ll} \begin{array}{ll} \log (b+|{\hat{z}}|)-\log |{\hat{z}}| &{}\quad \text {if} \ \ b\le |{\hat{z}}|\\ \log (b+|{\hat{z}}|)-\log b+\frac{1}{2} \left( 1-\frac{|{\hat{z}}|^2}{b^2}\right) &{}\quad \text {if}\ \ b>|{\hat{z}}| . \end{array} \end{array}\right. } \end{aligned}$$

If $b\le |{\hat{z}}|$ then

$$\begin{aligned} UO_{B} (\log |z|) =\log \left( \frac{b}{|{\hat{z}}|}+1\right) \le \log 2. \end{aligned}$$

For $b> |{\hat{z}}|$ we set $x=|{\hat{z}}|/b$ and write $UO_{B} (\log |z|)$ as

$$\begin{aligned} f(x)=\log (1+x)+\frac{1}{2} \cdot (1-x^2), \quad 0<x<1. \end{aligned}$$

Since

$$\begin{aligned} f'(x)=\frac{1}{1+x}-x, \end{aligned}$$

we see that f is increasing on $[0, {\hat{x}}]$ and decreasing on $[{\hat{x}}, 1]$, where ${\hat{x}}=\frac{\sqrt{5}-1}{2}.$ Notice that

$$\begin{aligned} f({\hat{x}})=\log \frac{\sqrt{5}+1}{2} +\frac{\sqrt{5}-1}{4}. \end{aligned}$$

Thus

$$\begin{aligned} UO_{B} (\log |z|) \le \log \frac{\sqrt{5}+1}{2} +\frac{\sqrt{5}-1}{4} \end{aligned}$$

and the equality holds if and only if

$$\begin{aligned} \frac{|{\hat{z}}|}{b}=\frac{\sqrt{5}-1}{2}. \end{aligned}$$

This finishes the proof. $\square$

3 Proof of Theorem 1.1

3.1 One dimensional case

Let $\Omega$ be a domain in ${\mathbb {C}}$ and $\phi$ a subharmonic function on $\Omega$. Recall that

$$\begin{aligned} UO_B(\phi )={\sup }_B\, \phi -\phi _B \end{aligned}$$

where $B=\{z:|z-{\hat{z}}| <r\}\subset \Omega .$ The idea is to use Harnack’s inequality and a convexity lemma. Let us write

$$\begin{aligned} UO_B(\phi )=I_1+I_2, \end{aligned}$$

where

$$\begin{aligned} I_1={\sup }_{B} \,\phi -\phi _{\partial B}, \quad I_2:=\phi _{\partial B}-\phi _{B}, \end{aligned}$$

with $\phi _{\partial B}$ being the mean-value of $\phi$ over the boundary $\partial B$. For each $\tau >0$ we set

$$\begin{aligned} \tau B=\{z:|z-{\hat{z}}|<\tau r\}. \end{aligned}$$

Applying Harnack’s inequality to the nonpositive subharmonic function $\psi :=\phi -\sup _{B} \phi$, we get

$$\begin{aligned} {\sup }_{\frac{1}{2} B}\, \psi = {\sup }_{\partial (\frac{1}{2} B)}\, \psi \le \frac{1}{3} \cdot \psi _{\partial \, B}, \end{aligned}$$

i.e.,

$$\begin{aligned} I_1 \le 3\left( {\sup }_{B}\, \phi -{\sup }_{\frac{1}{2} B}\, \phi \right) . \end{aligned}$$

Here the constant 1/3 comes from the Poisson kernel of the unit disc since

$$\begin{aligned} \inf _{|z|=1/2} \frac{1-|z|^2}{|1-z|^2}=\frac{1}{3}. \end{aligned}$$

The following fact explains why we need such an estimate.

Fact 1 $J_1:=\sup _{B} \phi - \sup _{\frac{1}{2} B} \phi$ is continuous in ${\hat{z}}$ and r respectively; moreover, it is increasing with respect to r.

Proof

Since $\sup _{B} \phi$ is a convex function of $\log r$ (see [11, Corollary 5.14]), it follows that $J_1$ is a continuous increasing function of r. The continuity of $J_1$ in $\hat{z}$ is obvious. $\square$

Let $\Omega _0$ be a relatively compact open subset in $\Omega$. Let $\delta _0$ denote the distance between $\overline{\Omega _0}$ and $\partial \Omega$. By the above fact we see that if the radius r of $B\subset \Omega _0$ is less than $\delta _0/2$ then

$$\begin{aligned} I_1\le 3\, {\sup }_{\hat{z}\in \Omega _0}\, J_1(\hat{z},\delta _0/2)<\infty , \end{aligned}$$

and if $r\ge \delta _0/2$ then

$$\begin{aligned} I_1 \le 3\,{\sup }_{\Omega _0} \phi -3\, {\inf }_{\hat{z}\in \Omega _0}\, {\sup }_{\{|z-\hat{z}|<\delta _0/4\}}\,\phi <\infty . \end{aligned}$$

To estimate $I_2$, we need the following convexity lemma which was communicated to the second author by Bo Berndtsson:

Lemma 3.0.2

Let $d\mu$ be a probability measure on a Borel measurable subset S in ${\mathbb {R}}^n$ with barycenter ${\hat{t}} \in {\mathbb {R}}^n$. Let f be a convex function on ${\mathbb {R}}^n$. Then

$$\begin{aligned} \int _S f \, d\mu \ge f({\hat{t}}). \end{aligned}$$

Proof

Since f is convex, there exists an affine function l such that $f({\hat{t}})=l({\hat{t}})$ and $f\ge l$ on ${\mathbb {R}}^n$, which implies

$$\begin{aligned} \int _S f\, d\mu \ge \int _S l\, d\mu = l({\hat{t}})=f({\hat{t}}), \end{aligned}$$

where the first equality follows from the definition of barycenter. $\square$

With $f(t):=\phi _{\{z:|z-{\hat{z}}|=e^t r\}}$ we have

$$\begin{aligned} I_2= & {} f(0)-\frac{1}{\pi r^2}\int _{-\infty }^0 2\pi e^{t} r\cdot f(t) \, d(e^t r)\\= & {} f(0)-\int _{-\infty }^0 f(t) \, d (e^{2t}). \end{aligned}$$

Since f(t) is convex and $d (e^{2t})$ is a probability measure on $(-\infty , 0)$ with barycenter at $t=-1/2$, it follows from Lemma 3.0.2 that

$$\begin{aligned} \int _{-\infty }^0 f(t) \, d (e^{2t}) \ge f(-1/2), \end{aligned}$$

(3.1)

which implies

$$\begin{aligned} I_2 \le J_2:= f(0)-f(-1/2). \end{aligned}$$

Since f is convex, we get an analogous conclusion as Fact 1:

Fact 2 $J_2$ is continuous in ${\hat{z}}$ and r respectively; moreover, it is increasing with respect to r.

By a similar argument as above, we may verify that

$$\begin{aligned} {\sup }_{B\subset \Omega _0}\, I_2 <\infty . \end{aligned}$$

3.2 High dimensional case

The following result plays the role of Fact 1, 2.

Lemma 3.0.3

Let $g(t)=g(t_1, \ldots , t_n)$ be a convex function on $(-\infty , 2)^n$ which is increasing in each variable. Then

$$\begin{aligned} {\sup }_{t\in A_N} \left[ g(t)-g(t-1)\right] \le nN \left[ g(1,\ldots , 1)-g(0)\right] , \end{aligned}$$

where $t-1:=(t_1-1, \ldots , t_n-1)$, $N\ge 1$ and

$$\begin{aligned} A_N:=\left\{ t\in (-\infty , 0]^n: \max \{-t_j\} \le N\min \{-t_j\}\right\} . \end{aligned}$$

Proof

A standard regularization process reduces to the case when g is smooth. Set

$$\begin{aligned} f(a)=g(t_1+a,\ldots , t_n+a):=g(t+a). \end{aligned}$$

We have

$$\begin{aligned} f(0)-f(-1)=\int _{-1}^0 f'(a)\, da=\int _{-1}^0 \sum g_j(t+a)\, da \end{aligned}$$

where $g_j:=\frac{\partial g}{\partial t_j}$. Notice that

$$\begin{aligned} \sum g_j(t+a) \le \frac{1}{\min \{-t_j-a\}} \sum (-t_j-a)g_j(t+a) \end{aligned}$$

and

$$\begin{aligned} \sum (-t_j-a)g_j(-s(t+a)) =\frac{dg(-s(t+a))}{d s} \end{aligned}$$

is an increasing function of $s\in (-\infty , 0)$ by convexity of g. Thus we have

$$\begin{aligned} \sum (-t_j-a)g_j(t+a) \le \sum (-t_j-a)g_j(0) \le \max \{-t_j-a\} \sum g_j(0), \end{aligned}$$

which implies

$$\begin{aligned} \sum g_j(t+a) \le \frac{\max \{-t_j-a\}}{\min \{-t_j-a\}} \sum g_j(0). \end{aligned}$$

For any $t\in A_N$, we have $t+a \in A_N$ (since $a\le 0$), so that

$$\begin{aligned} \frac{\max \{-t_j-a\}}{\min \{-t_j-a\}} \le N. \end{aligned}$$

Thus

$$\begin{aligned} g(t)-g(t-1) \le N \sum g_j(0). \end{aligned}$$

Since g is convex and increasing, we have

$$\begin{aligned} g_j(0) \le g(1,\ldots , 1)-g(0), \end{aligned}$$

which finishes the proof. $\square$

Let

$$\begin{aligned} P:=\left\{ z\in {\mathbb {C}}^n: |z_j-{\hat{z}}_j|<r_j, \ 1\le j\le n\right\} \subset \Omega \end{aligned}$$

be a polydisc of type N, i.e.,

$$\begin{aligned} \max \{r_j\} \le \min \left\{ r_j^{1/N}\right\} . \end{aligned}$$

Similar as above, we write

$$\begin{aligned} UO_{P} (\phi )={\sup }_P\, \phi -\phi _P =I_1+I_2, \end{aligned}$$

where

$$\begin{aligned} I_1:={\sup }_P\, \phi -\phi _{\partial P}, \quad I_2:=\phi _{\partial P}-\phi _{P}, \end{aligned}$$

and

$$\begin{aligned} \partial P:=\{z\in {\mathbb {C}}^n: |z_j-{\hat{z}}_j|=r_j, \ 1\le j\le n\} \end{aligned}$$

is the Shilov boundary of P. Applying Harnack’s inequality (see [14, p. 186]) n-times, we get the following

Lemma 3.0.4

$I_1\le 3^n J_1$, where $J_1:=\sup _{P} \phi -\sup _{\frac{1}{2} P} \phi$.

Using (3.1) repeatedly we get

Lemma 3.0.5

$I_2\le J_2$, where $J_2:=f(0)-f(-1/2, \ldots , -1/2)$ with

$$\begin{aligned} f(t):=\phi _{\{z\, :\, |z_j-{\hat{z}}_j|=e^{t_j} r_j,\ 1\le j\le n\}}. \end{aligned}$$

Since both $\sup _{P} \phi$ and $\phi _{\partial P}$ are continuous in ${\hat{z}}_j$ and convex increasing with respect to $\log r_j$ for all j, it follows from Lemma 3.0.3 (through a similar argument as the one-dimensional case) that

$$\begin{aligned} {\sup }_{P\in {{\mathcal {P}}}_N|_{\Omega _0} } \left( J_1+J_2\right) <\infty , \end{aligned}$$

for every open set $\Omega _0\subset \subset \Omega$, which finishes the proof of the first part of Theorem 1.1.

3.3 A counterexample

For the second part of Theorem 1.1, we need to construct a counterexample. For the sake of simplicity, we only consider the case $n=2$. It suffices to verify the following

Theorem 3.1

Set $\phi (z, w):=-\sqrt{(\log |z|+\log |w|)\log |w|},$ $z,w\in {\mathbb {D}}$. Then we have $\phi \in PSH({\mathbb {D}}^2)$, while

$$\begin{aligned} \sup _{0<r_1, r_2<1} \frac{1}{|{\mathbb {D}}^2_r|}\int _{{\mathbb {D}}^2_r} |\phi -\phi _{{\mathbb {D}}^2_r}| =\infty , \end{aligned}$$

where

$$\begin{aligned} {\mathbb {D}}^2_r:=\left\{ (z,w)\in {\mathbb {C}}^2: |z|<r_1, \, |w|<r_2 \right\} . \end{aligned}$$

The following lemma shows that Fact 1, 2 is no more true for general bidiscs.

Lemma 3.1.1

$f(x,y):=-\sqrt{(x+y)y}$ is convex on $(-\infty , 0)^2$ and increasing in each variable; moreover,

$$\begin{aligned} {\sup }_{\{x, y\le -1\}} \left[ f(x,y)-f(x-1, y-1)\right] =\infty . \end{aligned}$$

(3.2)

Proof

The first conclusion follows by a straightforward calculation. For (3.2) it suffices to note that

$$\begin{aligned} f(x,-1)-f(x-1, -2)=\frac{5-x}{\sqrt{6-2x}+\sqrt{1-x}} \rightarrow \infty \end{aligned}$$

as $x\rightarrow -\infty$. The proof is complete. $\square$

Let us first verify that $\phi \notin BUO_\mathrm{loc}({\mathbb {D}}^2, {\mathcal {P}})$.

Lemma 3.1.2

$\sup _{0<r_1, r_2<1} \sup _{{\mathbb {D}}^2_r}(\phi - \phi _{\mathbb D^2_r})=\infty$.

Proof

With $x=\log r_1$ and $y=\log r_2$, we get

$$\begin{aligned} {\sup }_{{\mathbb {D}}^2_r} \left( \phi - \phi _{{\mathbb {D}}^2_r}\right) = f(x,y)-\int _{-\infty }^0 \int _{-\infty }^0 f(x+t, y+s) \, de^{2t} d e^{2s}=: I(x,y). \end{aligned}$$

Integrate by parts with respect to t and s successively, we may write

$$\begin{aligned} I(x,y)=I_1 +I_2, \end{aligned}$$

where

$$\begin{aligned} I_1=\int _{-\infty }^0 \frac{x+2y+2s}{-4f(x+t, y+s)} \, de^{2s} \end{aligned}$$

and

$$\begin{aligned} I_2=\int _{(-\infty , 0)^2} \frac{y+s}{-4f(x+t, y+s)} \, de^{2t} d e^{2s}. \end{aligned}$$

Obviously, $I_2(x,-1)$ is bounded on $(-\infty , 0]$, but $I_1(x,-1)\rightarrow \infty$ as $x\rightarrow -\infty$, from which the assertion immediately follows. $\square$

Proof of Theorem 3.1

By Lemma 3.0.2 we have (still with $x=\log r_1,\,y=\log r_2$)

$$\begin{aligned} \phi _{{\mathbb {D}}^2_r}= & {} \int _{-\infty }^0 \int _{-\infty }^0 f(x+t, y+s) \, de^{2t} d e^{2s} \\\ge & {} f(x-1/2, y-1/2)= {\sup }_{{\mathbb {D}}^2_{e^{-1/2} r}}\,\phi , \end{aligned}$$

which yields

$$\begin{aligned} \frac{1}{|{\mathbb {D}}^2_r|}\int _{{\mathbb {D}}^2_r} |\phi -\phi _{\mathbb D^2_r}|\ge & {} \frac{1}{|{\mathbb {D}}^2_r|}\int _{{\mathbb {D}}^2_{e^{-1/2} r}} \left( {\sup }_{{\mathbb {D}}^2_{e^{-1/2} r}}\,\phi - \phi \right) \\= & {} e^{-2} \left( { \sup }_{{\mathbb {D}}^2_{e^{-1/2} r}}\,\phi - \phi _{{\mathbb {D}}^2_{e^{-1/2} r}} \right) . \end{aligned}$$

By a similar argument as Lemma 3.1.2, we conclude the proof of Theorem 3.1. $\square$

3.4 Lelong class

In this section we shall prove the third part of Theorem 1.1. The key ingredient is the following counterpart of Lemma 3.0.3.

Lemma 3.1.3

Let $g(t)=g(t_1, \ldots , t_n)$ be a convex function on $\mathbb R^n$ which is increasing in each variable. Assume that

$$\begin{aligned} g(t) \le \max _{1\le j\le n}\{\log (1+e^{t_j})\}, \quad \forall \ t\in {\mathbb {R}}^n. \end{aligned}$$

Then for every $M>0$ we have

$$\begin{aligned} {\sup }_{t\in {\mathbb {R}}^n} \left[ g(t)-g(t-M)\right] \le M, \end{aligned}$$

where $t-M:=(t_1-M, \ldots , t_n-M)$.

Proof

For fixed t, we consider the following convex increasing function

$$\begin{aligned} f(s):=g(t_1+s, \ldots , t_n+s) \end{aligned}$$

on ${\mathbb {R}}$. Convexity of f gives

$$\begin{aligned} \frac{f(0)-f(-M)}{M} \le \lim _{s\rightarrow \infty } \frac{f(s)-f(0)}{s}. \end{aligned}$$

By the assumption, we have

$$\begin{aligned} f(0) \le f(s) \le \max _{1\le j\le n}\{\log (1+e^{t_j} e^s)\} \end{aligned}$$

for every $s\ge 0$, so that

$$\begin{aligned} \lim _{s\rightarrow \infty } \frac{f(s)-f(0)}{s}\le 1. \end{aligned}$$

The proof is complete. $\square$

Proof of the third part of Theorem 1.1

Again for any polydisc

$$\begin{aligned} P:=\left\{ z\in {\mathbb {C}}^n: |z_j-{\hat{z}}_j|<r_j, \ 1\le j\le n\right\} , \end{aligned}$$

we may write

$$\begin{aligned} UO_{P} (\phi )={\sup }_P\, \phi -\phi _P =I_1+I_2, \end{aligned}$$

where

$$\begin{aligned} I_1:={\sup }_P\, \phi -\phi _{\partial P}, \quad I_2:=\phi _{\partial P}-\phi _{P}. \end{aligned}$$

By Lemma 3.0.4 we have

$$\begin{aligned} I_1\le 3^n \left( {\sup }_P\, \phi -{\sup }_{\frac{1}{2} P}\, \phi \right) . \end{aligned}$$

Put

$$\begin{aligned} P_t:=\left\{ z\in {\mathbb {C}}^n: |z_j-{\hat{z}}_j|<e^{t_j}r_j, \ 1\le j\le n\right\} \end{aligned}$$

and $f_1(t):=\sup _{P_t} \phi$. Since $\phi \in {{\mathcal {L}}}$, we know that for some constant $c_1\gg 1$ the function $f_1-c_1$ satisfies the assumption in Lemma 3.1.3, so that

$$\begin{aligned} {\sup }_{P}\, \phi -{\sup }_{\frac{1}{2} P}\,\phi =f_1(0)-f_1(-\log 2) \le \log 2, \end{aligned}$$

which in turn implies

$$\begin{aligned} I_1\le 3^n \log 2. \end{aligned}$$

Moreover, we infer from Lemma 3.0.5 that

$$\begin{aligned} I_2\le f(0)-f(-1/2, \ldots , -1/2), \quad f(t):=\phi _{\partial P_t}. \end{aligned}$$

Applying Lemma 3.1.3 in a similar way as above, we have

$$\begin{aligned} I_2 \le 1/2. \end{aligned}$$

Thus

$$\begin{aligned} UO_{P} (\phi ) \le 3^n \log 2+ 1/2 < 3^n, \end{aligned}$$

which finishes the proof. $\square$

4 Proof of Theorem 1.2

The starting point is the following

Definition 4.0.1

($\gamma$-constant) We shall define the constant $\gamma$ as the BUO norm of $\log |z|$ on ${\mathbb {C}}$ with respect to all line segments. More precisely,

$$\begin{aligned} \gamma :=\sup _{a\ne b\in {\mathbb {C}}} UO_{[a,b]} (\log |z|), \end{aligned}$$

where [a, b] denotes the line segment connecting a and b, and the upper oscillation is defined by

$$\begin{aligned} UO_{[a,b]} (\log |z|) :=\left( {\sup }_{0\le t\le 1}\,\log |a(1-t)+bt| \right) - \int _{0}^1 \log |a(1-t)+bt| \, dt. \end{aligned}$$

The key step is to show the following

Lemma 4.0.4

$1<\gamma <2$ is determined by

$$\begin{aligned} \gamma +\log (\gamma -1)=0. \end{aligned}$$

(4.1)

Proof

For each pair $a, b \in {\mathbb {C}}$, we shall compute

$$\begin{aligned} UO_{[a,b]} (\log |z|)={\sup }_{[a,b]}\, \log |z|-(\log |z|)_{[a,b]}. \end{aligned}$$

Since $\log |z|$ is $S^1$-invariant, by a rotation of z, we may assume that

$$\begin{aligned} b \in {\mathbb {R}},\quad b>|a|. \end{aligned}$$

Thus

$$\begin{aligned} {\sup }_{[a,b]}\, \log |z|=\log b \end{aligned}$$

is independent of a. Since

$$\begin{aligned} (\log |z|)_{[a,b]}= & {} \int _{0}^1 \log |a(1-t)+bt| \, dt\\\ge & {} \int _{0}^1 \log |\mathrm{Re\,}a\cdot (1-t)+bt| \, dt\\= & {} (\log |z|)_{[\mathrm{Re\,}a,b]} \end{aligned}$$

with equality holds if and only if $a\in {\mathbb {R}}$. Thus it suffices to verify (4.1) for

$$\begin{aligned} a, b\in {\mathbb {R}}, \quad |a|<b. \end{aligned}$$

Consider $\log |z|-\log b$ instead of $\log |z|$, one may further assume that

$$\begin{aligned} b=1, \quad -1<a<1, \end{aligned}$$

which implies

$$\begin{aligned} UO_{[a,1]} (\log |z|)= \log 1- (\log |z|)_{[a, 1]}=-(\log |z|)_{[a, 1]}. \end{aligned}$$

We divide into two cases. (i) $0\le a<1$. Then we have

$$\begin{aligned} -(\log |z|)_{[a, 1]}=\frac{-1}{1-a} \int _{a}^1 \log x \, dx=\frac{a\log a}{1-a}+1 \le 1. \end{aligned}$$

(ii) $-1<a<0$. Then we have

$$\begin{aligned} -(\log |z|)_{[a, 1]}=\frac{-1}{1-a} \int _{a}^1 \log |x| \, dx=\frac{a\log (-a)}{1-a}+1 >1. \end{aligned}$$

Thus

$$\begin{aligned} \gamma =\sup _{-1<a<0} \frac{a\log (-a)}{1-a}+1. \end{aligned}$$

It suffices to verify that $\gamma$ satisfies (4.1). To see this, put

$$\begin{aligned} t^{-1}:=1-a \in (1,2) \end{aligned}$$

and write

$$\begin{aligned} \frac{a\log (-a)}{1-a}= (1-t)\log t-(1-t)\log (1-t)=:f(t). \end{aligned}$$

Since

$$\begin{aligned} f'(t)=t^{-1}-\log t +\log (1-t), \end{aligned}$$

it follows that $f'(t)=0$ if and only if

$$\begin{aligned} t^{-1}=\log \frac{1}{t^{-1}-1}, \end{aligned}$$

i.e.,

$$\begin{aligned} 1-a+\log (-a)=0. \end{aligned}$$

Thus we have

$$\begin{aligned} \gamma =\sup _{-1<a<0} \frac{a\log (-a)}{1-a}+1= \frac{a_0\log (-a_0)}{1-a_0}+1, \end{aligned}$$

where $a_0$ is determined by

$$\begin{aligned} 1-a_0+\log (-a_0)=0, \end{aligned}$$

(4.2)

which gives

$$\begin{aligned} \gamma =1-a_0 \in (1,2). \end{aligned}$$

It is clear that (4.2) is equivalent to (4.1). $\square$

Since a translation of a line segment is still a line segment, we know that $\log |z-z_0|$ and $\log |z|$ have the same line segment BUO norm. This fact can be used to estimate the line segment BUO norm of $\log |p|$ for general polynomials p. In fact, if we write

$$\begin{aligned} p=a_0(z-a_1)^{n_1} \cdots (z-a_k)^{n_k}, \end{aligned}$$

then

$$\begin{aligned} {\sup }_{[a,b]}\, \log |p| \le \log |a_0|+ \sum _{j=1}^k n_j {\sup }_{[a,b]}\, \log |z-a_j| \end{aligned}$$

and

$$\begin{aligned} (\log |p|)_{[a,b]}= \log |a_0|+ \sum _{j=1}^k n_j (\log |z-a_j|)_{[a,b]}. \end{aligned}$$

Thus

$$\begin{aligned} UO_{[a,b]} (\log |p|):= & {} {\sup }_{[a,b]}\, \log |p|-(\log |p|)_{[a,b]}\\\le & {} \sum _{j=1}^k n_j UO_{[a,b]}(\log |z-a_j|). \end{aligned}$$

This combined with the fact $UO_{[a,b]}(\log |z-a_j|) \le \gamma$ gives

$$\begin{aligned} UO_{[a,b]} (\log |p|) \le \gamma \cdot \deg p. \end{aligned}$$

(4.3)

for all polynomials p and all $a,b\in {\mathbb {C}}$.

Now we may conclude the proof of Theorem 1.2 as follows. Since A is compact, we may choose $z_0\in A$ such that

$$\begin{aligned} |p(z_0)|={\sup }_{z\in A}\, |p(z)|. \end{aligned}$$

For every ray (half line), say L, starting from $z_0$, we see that $A\cap L$ is a line segment in view of convexity of A. Let $L_{{\mathbb {C}}}$ be the complex line containing L. Apply (4.3) to $p|_{L_{{\mathbb {C}}}}$, we have

$$\begin{aligned} UO_{A\cap L} (\log |p|)= UO_{A\cap L} (\log |p|_{L_{{\mathbb {C}}}}|) \le \gamma \deg p|_{L_{{\mathbb {C}}}}\le \gamma \deg p, \end{aligned}$$

which gives

$$\begin{aligned} UO_{A} (\log |p|) \le \gamma \deg p \end{aligned}$$

since $UO_{A} (\log |p|)$ is a certain average of $UO_{A\cap L} (\log |p|)$ for all L starting from $z_0$: in fact, since $z_0$ is a maximum point of $\log |p|$ on A and L contains $z_0$, we always have

$$\begin{aligned} \sup _{A\cap L} \log |p|= \log |p(z_0)|, \end{aligned}$$

together with (4.3) it gives

$$\begin{aligned} \gamma \cdot \deg p\ge UO_{A\cap L} (\log |p|) = \log |p(z_0)| - \frac{1}{|A\cap L|} \int _{A\cap L} \log |p|. \end{aligned}$$

(4.4)

Thus

$$\begin{aligned} \int _A \log |p|= & {} \int _{ S_{2n-1}} \int _{A\cap L} \log |p| \, d\mu (L)\\\ge & {} (\log |p(z_0)|-\gamma \cdot \deg p)\int _{ S_{2n-1}} |A\cap L| \, d\mu (L)\\= & {} (\log |p(z_0)|-\gamma \cdot \deg p) |A|, \end{aligned}$$

where $d\mu$ is a certain measure on the unit sphere $S_{2n-1}$ and we identify the set of rays L starting from $z_0$ with $S_{2n-1}$. Notice that the above inequality gives

$$\begin{aligned} \gamma \cdot \deg p \ge \log |p(z_0)| - \frac{1}{|A|} \int _A \log |p|=UO_A (\log |p|), \end{aligned}$$

from which the assertion immediately follows.

5 Proof of Theorem 1.3

The starting point is the following

Proposition 5.0.1

(John–Nirenberg inequality) Suppose $\phi \in PSH(\Omega )$ and $\Omega _0\subset \subset \Omega$ is open. For each $a=(a_1,\ldots ,a_n)$ with all $a_j>0$ there exists $\varepsilon _0=\varepsilon (a,\phi ,\Omega _0,\Omega ) >0$ such that

$$\begin{aligned} \sup _{ P_{r^a}(\hat{z})\subset \Omega _0} \frac{1}{|P_{r^a}(\hat{z})|}\int _{P_{r^a}(\hat{z})} e^{-\varepsilon (\phi -\sup _{P_{r^a}(\hat{z})} \phi )} <\infty , \end{aligned}$$

for every $\varepsilon \le \varepsilon _0$. Here

$$\begin{aligned} P_{r^a}(\hat{z})=\{z\in {\mathbb {C}}^n: |z_j-\hat{z}_j|\le r^{a_j}\}. \end{aligned}$$

Although the argument is fairly standard, we will provide a proof in Appendix, because the result cannot be found in literature explicitly.

Lemma 5.0.5

Let $\psi$ be a psh function on $\Omega$ which satisfies $\sup _{\Omega } \psi <\infty$ and $\int _\Omega e^{-\psi } <\infty$. Suppose $\Omega$ is circular, i.e., $\zeta z\in \Omega$ for every $\zeta \in {\mathbb {C}}$, $|\zeta |\le 1$, and $z\in \Omega$. Then

$$\begin{aligned} \left( \frac{1}{|\Omega |} \int _{\Omega } e^{-(\psi -\sup _{\Omega } \psi )} \right) ^{-1} \le K_{\psi , \,\Omega }(0) \cdot |\Omega |\cdot e^{-\sup _\Omega \psi } \le 1. \end{aligned}$$

(5.1)

Proof

The extremal property of the Bergman kernel implies that

$$\begin{aligned} K_{\psi , \,\Omega }(0) \ge \frac{1}{\int _{{\mathbb {D}}_r} e^{-\psi }} \end{aligned}$$

and the first inequality in (5.1) holds. On the other hand, as $\Omega$ is circular, it is easy to verify that

$$\begin{aligned} f(0)=\frac{1}{|\Omega |} \int _{\Omega } f \end{aligned}$$

for all $f\in {{\mathcal {O}}}(\Omega )$. Thus we have

$$\begin{aligned} |f(0)|^2 = \left| \,\frac{1}{|\Omega |} \int _{\Omega } f\,\right| ^2 \le \frac{1}{|\Omega |} \int _{\Omega } |f|^2e^{-\psi } \cdot e^{\sup _\Omega \psi }, \end{aligned}$$

so that the second inequality in (5.1) also holds. $\square$

Proof of Theorem 1.3

Since

$$\begin{aligned} \int _{{\mathbb {D}}^n} |f|^2e^{-\varepsilon \phi ^t} =|t_1\cdots t_n|^{-2}\int _{{\mathbb {D}}_t^n} |f|^2e^{-\varepsilon \phi },\ \ \ \forall \,f\in {{\mathcal {O}}}(\Omega ), \end{aligned}$$

it follows that

$$\begin{aligned} K_{\varepsilon \phi ^t, \,{\mathbb {D}}^n}(z)=\frac{|\mathbb D^n_t|}{|{\mathbb {D}}^n|} \cdot K_{\varepsilon \phi , \,\mathbb D^n_t}(z), \end{aligned}$$

(5.2)

where

$$\begin{aligned} {\mathbb {D}}^n_t:=\left\{ z\in {{\mathbb {C}}}^n: |z_j|<|t_j|, \ 1\le j\le n \right\} . \end{aligned}$$

Thus we have

$$\begin{aligned} F(\varepsilon \phi )(t,0)=\log (|{\mathbb {D}}^n_t|\cdot K_{\varepsilon \phi , \,{\mathbb {D}}^n_t}(0))-n \log \pi . \end{aligned}$$

This combined with Lemma 5.0.5 gives

$$\begin{aligned} -\log \left( \frac{1}{|{\mathbb {D}}^n_t|} \int _{{\mathbb {D}}^n_t} e^{-\varepsilon (\phi -\sup _{{\mathbb {D}}^n_t} \phi )}\right) -n \log \pi \le F(\varepsilon \phi )(t,0) - \varepsilon \, {\sup }_{\mathbb D^n_t}\, \phi \le -n \log \pi . \end{aligned}$$

By Proposition 5.0.1, we conclude the proof. $\square$

6 Proof of Theorem 1.4

Recall that

$$\begin{aligned} \phi ^t(z):=\phi (t_1z_1,\ldots , t_nz_n). \end{aligned}$$

By Proposition 2.2 in [3], we have

$$\begin{aligned} \frac{\partial }{\partial t_j} K_{\phi ^t, \,\mathbb D^n}(0)=\int _{{\mathbb {D}}^n} \frac{\partial \phi ^t}{\partial t_j}\,|K_{\phi ^t, \,{\mathbb {D}}^n}(z,0)|^2 e^{-\phi ^t}, \end{aligned}$$

(6.1)

where $K_{\phi ^t, \,{\mathbb {D}}^n}(z,0)$ satisfies the following reproducing property

$$\begin{aligned} f(0)=\int _{{\mathbb {D}}^n } f(z)\overline{K_{\phi ^t, \,\mathbb D^n}(z,0)}e^{-\phi ^t} \end{aligned}$$

for all $L^2$ holomorphic functions f on ${\mathbb {D}}^n$. In particular, if $f=zK_{\phi ^t, \,{\mathbb {D}}^n}(z,0)$ then

$$\begin{aligned} 0=\int _{{\mathbb {D}}^n} z\cdot |K_{\phi ^t, \,{\mathbb {D}}^n}(z,0)|^2 e^{-\phi ^t}, \end{aligned}$$

and since $\frac{\partial \phi ^t}{\partial t_j}|_{t=0}=z_j\phi _{z_j}(0),$ we get

$$\begin{aligned} \left. \int _{{\mathbb {D}}^n} \frac{\partial \phi ^t}{\partial t_j}\right| _{t=0} |K_{\phi ^t, \,{\mathbb {D}}^n}(z,0)|^2 e^{-\phi ^t}=0 \end{aligned}$$

for all $t\in {\mathbb {D}}^n$. Thus we may write (6.1) as

$$\begin{aligned} \frac{\partial }{\partial t_j} K_{\phi ^t, \,{\mathbb {D}}^n}(0) =\int _{{\mathbb {D}}^n} \left( \frac{\partial \phi ^t}{\partial t_j}-\left. \frac{\partial \phi ^t}{\partial t_j}\right| _{t=0}\right) |K_{\phi ^t, \,{\mathbb {D}}^n}(z,0)|^2 e^{-\phi ^t}. \end{aligned}$$

In particular,

$$\begin{aligned} \left. \frac{\partial }{\partial t_j} K_{\phi ^t, \,\mathbb D^n}(0)\right| _{t=0}=0. \end{aligned}$$

Thus we can further write (6.1) as

$$\begin{aligned} \frac{\partial }{\partial t_j} K_{\phi ^t, \,\mathbb D^n}(0)-\left. \frac{\partial }{\partial t_j} K_{\phi ^t, \,\mathbb D^n}(0)\right| _{t=0}=\int _{{\mathbb {D}}^n} \left( \frac{\partial \phi ^t}{\partial t_j}-\left. \frac{\partial \phi ^t}{\partial t_j}\right| _{t=0}\right) |K_{\phi ^t, \,{\mathbb {D}}^n}(z,0)|^2 e^{-\phi ^t}, \end{aligned}$$

which implies

$$\begin{aligned} \left. \frac{\partial ^2}{\partial t_j \partial {\bar{t}}_k} K_{\phi ^t, \,{\mathbb {D}}^n}(0)\right| _{t=0}=\int _{{\mathbb {D}}^n} \left. \frac{\partial ^2 \phi ^t}{\partial t_j\partial \bar{t}_k}\right| _{t=0}\cdot |K_{\phi (0), \,{\mathbb {D}}^n}(z,0)|^2 e^{-\phi (0)}. \end{aligned}$$

Since

$$\begin{aligned} K_{\phi (0), \,{\mathbb {D}}^n}(z,0)=\frac{e^{\phi (0)}}{\pi ^n} \end{aligned}$$

and

$$\begin{aligned} \left. \frac{\partial ^2 \phi ^t}{\partial t_j\partial \bar{t}_k}\right| _{t=0}=z_j{\bar{z}}_k \phi _{z_j {\bar{z}}_k}(0), \end{aligned}$$

we get

$$\begin{aligned} \left. \frac{\partial ^2}{\partial t_j \partial {\bar{t}}_k} K_{\phi ^t, \,{\mathbb {D}}^n}(0)\right| _{t=0}=\frac{e^{\phi (0)}\phi _{z_j \bar{z}_k}(0)}{\pi ^{2n}}\int _{{\mathbb {D}}^n} z_j {\bar{z}}_k. \end{aligned}$$

Notice that

$$\begin{aligned} \int _{{\mathbb {D}}^n} z_j {\bar{z}}_k= {\left\{ \begin{array}{ll} \begin{array}{ll} \pi ^{n}/2 &{}\quad \text {if}\ \, j=k\\ 0 &{}\quad \text {if}\ \, j\ne k, \end{array} \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \left. \frac{\partial ^2}{\partial t_j \partial {\bar{t}}_k} K_{\phi ^t, \,{\mathbb {D}}^n}(0)\right| _{t=0} = K_{\phi ^t, \,{\mathbb {D}}^n}(0)\cdot \left. \frac{\partial ^2}{\partial t_j \partial {\bar{t}}_k} \log K_{\phi ^t, \,{\mathbb {D}}^n}(0)\right| _{t=0}, \end{aligned}$$

our assertion follows.

Appendix

In this section we provide a proof of Proposition 5.0.1. Let us first recall a few basic facts in real-variable theory, by following Stein [17]. A quasi-distance defined on ${{\mathbb {R}}}^m$ means a nonnegative continuous function $\rho$ on ${{\mathbb {R}}}^m\times {\mathbb {R}}^m$ for which there exists a constant $c>0$ such that

• 1. $\rho (x,y)=0$ iff $x=y$;

• 2. $\rho (x,y)\le c\rho (y,x)$;

• 3. $\rho (x,y)\le c(\rho (x,z)+\rho (y,z))$.

Given such a $\rho$, we define “balls”

$$\begin{aligned} B(x,r):=\{y:\rho (y,x)<r\},\quad r>0. \end{aligned}$$

One can verify that there exists a constant $c_1>1$ such that for all x, y and r,

$$\begin{aligned} B(x,r)\cap B(y,r)\ne \emptyset \ \ \ \Rightarrow \ \ \ B(y,r)\subset B(x,c_1r). \end{aligned}$$

(7.1)

In the case of Proposition 5.0.1, we define

$$\begin{aligned} \rho (z,w)=\max _k |z_k-w_k|^{1/a_k},\quad z,w\in {{\mathbb {C}}}^n. \end{aligned}$$

It is easy to verify that $\rho$ is a quasi-distance on ${\mathbb C}^n$ and

$$\begin{aligned} B(\hat{z},r)=P_{r^a}(\hat{z}),\quad \hat{z}\in {\mathbb {C}}^n,\, r>0. \end{aligned}$$

Besides (7.1), the following properties also hold for $B(\hat{z},r)$:

$$\begin{aligned}&|B(\hat{z},c_1r)|\le c_1^{2\sum _k a_k} \cdot |B(\hat{z},r)|=:c_2\, |B(\hat{z},r)|; \end{aligned}$$

(7.2)

$$\begin{aligned}&{\bigcap }_r \, \overline{B}(\hat{z},r)=\{\hat{z}\}\quad {\text {and}}\quad {\bigcup }_r\, B(\hat{z},r)={\mathbb {C}}^n; \end{aligned}$$

(7.3)

$$\begin{aligned}&\text {For each open set } U \text { and each }r>0, \text { the function } \hat{z}\mapsto |B(\hat{z},r)\cap U| \text { is continuous.}\nonumber \\ \end{aligned}$$

(7.4)

Fix a pair of positive constants $c^*$ and $c^{**}$ with $1<c^*<c^{**}$. For $B=B(\hat{z},r)$ we define $B^*=B(\hat{z},c^*r)$ and $B^{**}=B(\hat{z},c^{**} r)$. Then we have

Lemma 7.0.6

(cf. [17, p. 15–16]) Choose $c^*=4c_1^2$ and $c^{**}=16c_1^2$. Given a closed nonempty set $F\subset {{\mathbb {C}}}^n$, there exists a collection of balls $\{B_k\}$ such that

• (1) The $B_k$ are pairwise disjoint;

• (2) $\bigcup _k B_k^*= F^c:={\mathbb {C}}^n\backslash F$;

• (3) $B_k^{**}\cap F\ne \emptyset$ for each k.

Proposition 7.0.2

(Calderón–Zygmund decomposition) Let $B_0$ be a ball in ${\mathbb {C}}^n$ and $f\in L^1(B_0)$. There is a constant $c=c(c_1,c_2)>0$ such that given a positive number $\alpha$, there exists a sequences of balls $\{B_k^*\}$ in $B_0$ such that

• (1) $|f(z)|\le \alpha$, for a.e. $z\in B_0\backslash \bigcup _k B_k^*$;

• (2) $\int _{B_k^*} |f|\le c\alpha |B_k^*|,$ for each k;

• (3) $\sum _k |B_k^*| \le \frac{c}{\alpha } \int _{B_0} |f|$.

Proof

We extend f to an integrable function on ${{\mathbb {C}}}^n$ by setting $f=0$ outside $B_0$. Recall the following two types of Hardy–Littlewood maximal functions:

$$\begin{aligned}&Mf(z):=\sup _{r>0}\, \frac{1}{|B(z,r)|} \int _{B(z,r)} |f|, \\&\widetilde{M}f(z):=\sup _{z\in B}\, \frac{1}{|B|} \int _B |f| \end{aligned}$$

where the supremum is taken over all balls B containing z. The relationship between Mf and $\widetilde{M}f$ is as follows:

$$\begin{aligned} Mf \le \widetilde{M}f\le c_2\, Mf. \end{aligned}$$

(7.5)

Notice that

$$\begin{aligned} E_\alpha :=\left\{ z\in B_0: \widetilde{M}f(z)>\alpha \right\} \end{aligned}$$

is an open set since $\widetilde{M}f$ is lower semicontinuous, and

$$\begin{aligned} |E_\alpha |\le \frac{c}{\alpha } \int _{{\mathbb {C}}^n}|f| = \frac{c}{\alpha } \int _{B_0}|f| \end{aligned}$$

in view of (7.5) and [17, p. 13, Theorem 1]. Here and in what follows c will denote a generic positive constant depending only on $c_1,c_2$. With $F:={\mathbb {C}}^n\backslash E_\alpha$ we choose balls $\{B_k\}$, $\{B_k^*\}$ and $\{B_k^{**}\}$ according to Lemma 7.0.6. Then we have

$$\begin{aligned} {\sum }_k |B_k^*| \le c \, {\sum }_k |B_k| \le c\,|E_\alpha | \le \frac{c}{\alpha } \int _{B_0} |f|. \end{aligned}$$

Since $B_k^{**}\cap F\ne \emptyset$ for each k, we have

$$\begin{aligned} \int _{B_k^*} |f|\le \int _{B_k^{**}} |f| \le \alpha |B_k^{**}| \le c\alpha |B_k^*|. \end{aligned}$$

Finally, by (7.5) and [17, p. 13, Corollary], we know that $|f(z)|\le \widetilde{M}f(z)$ for a.e. z, from which (1) immediately follows. $\square$

Proof of Proposition 5.0.1

By Theorem 1.1, we know that

$$\begin{aligned} M:=\sup _{B(\hat{z},r)\subset \Omega _0} \frac{1}{|B(\hat{z},r)|} \int _{B(\hat{z},r)} |\phi -\phi _{B(\hat{z},r)}|<\infty . \end{aligned}$$

Assume without loss of generality $M=1$. Fix a ball $B_0\subset \Omega _0$. It suffices to show

$$\begin{aligned} |\{z\in B_0: |\phi -\phi _{B_0}|>t\}|\le \mathrm{const}\cdot e^{-\varepsilon t}\,|B_0|,\quad t>0, \end{aligned}$$

(7.6)

for certain $\varepsilon \ll 1$. With c as Proposition 7.0.2 we choose

$$\begin{aligned} \alpha>c>1\ge \frac{1}{|B_0|}\int _{B_0} |\phi -\phi _{B_0}|. \end{aligned}$$

Applying Proposition 7.0.2 with $f= |\phi -\phi _{B_0}|,$ we have a sequence of balls $\{B_k^{(1)}\}$ in $B_0$ such that

$$\begin{aligned}&|\phi (z)-\phi _{B_0}|\le \alpha \quad \text {a.e.}\ \ \ z\in B_0\backslash {\bigcup }_k B_k^{(1)}, \\&{\sum }_k |B_k^{(1)}|\le \frac{c}{\alpha } \int _{B_0} |\phi -\phi _{B_0}|\le \frac{c}{\alpha }\,|B_0|, \end{aligned}$$

and

$$\begin{aligned} |\phi _{B_k^{(1)}}-\phi _{B_0}|\le \frac{1}{|B_k^{(1)}|} \int _{B_k^{(1)}} |\phi -\phi _{B_0}|\le c\alpha . \end{aligned}$$

Applying Proposition 7.0.2 with $f= |\phi -\phi _{B_k^{(1)}}|$ for each k, we obtain a sequence of balls $\{B_k^{(2)}\}$ in $\bigcup _k B_k^{(1)}$ such that

$$\begin{aligned} {\sum }_k |B_k^{(2)}|\le \frac{c}{\alpha } \,{\sum }_k \int _{B_k^{(1)}} |\phi -\phi _{B_k^{(1)}}|\le \frac{c}{\alpha } \,{\sum }_k |B_k^{(1)}| \le \frac{c^2}{\alpha ^2} \, |B_0| \end{aligned}$$

and

$$\begin{aligned} |\phi (z)-\phi _{B_k^{(1)}}|\le \alpha \quad \text {a.e.}\ \ \ z\in B_k^{(1)}\backslash {\bigcup }_k B^{(2)}_k, \end{aligned}$$

which in turn implies

$$\begin{aligned} |\phi (z)-\phi _{B_0}|\le 2\cdot c\alpha \quad \text {a.e.}\ \ \ z\in B_0\backslash {\bigcup }_k B^{(2)}_k. \end{aligned}$$

Continue this process. For each j there exists a sequence of balls $\{B_k^{(j)}\}$ in $\bigcup _k B_k^{(j-1)}$ such that

$$\begin{aligned}&{\sum }_k |B_k^{(j)}|\le \frac{c^j}{\alpha ^j} \, |B_0|, \\&|\phi (z)-\phi _{B_0}|\le j\cdot c\alpha \quad \text {a.e.}\ \ \ z\in B_0\backslash {\bigcup }_k B^{(j)}_k. \end{aligned}$$

Thus

$$\begin{aligned} |\{z\in B_0:|\phi -\phi _{B_0}|>j\cdot c\alpha \}|\le {\sum }_k |B_k^{(j)}|\le \frac{c^j}{\alpha ^j} \, |B_0|. \end{aligned}$$

For any t there exists an integer j such that $t\in [j\cdot c\alpha ,(j+1)\cdot c\alpha )$. It follows that

$$\begin{aligned} (c/\alpha )^j=(\alpha /c)e^{-(j+1)\log \alpha /c}\le (\alpha /c)e^{-\frac{\log \alpha /c}{c\alpha } \, t}, \end{aligned}$$

from which (7.6) immediately follows. Now we have

$$\begin{aligned} \frac{1}{|B_0|} \int _{B_0} e^{\varepsilon |\phi -\phi _{B_0}|} = \frac{1}{|B_0|} \int _{0}^\infty e^{\varepsilon t} |\{|\phi -\phi _{B_0}|>t\}| + \frac{ |\{\phi =\phi _{B_0}\}| }{|B_0|} \le \mathrm{const}+1, \end{aligned}$$

which gives

$$\begin{aligned} \frac{1}{|B_0|} \int _{B_0} e^{-\varepsilon (\phi -\sup _{B_0}\phi )} \le (\mathrm{const}+1) e^{\varepsilon (\sup _{B_0}\phi -\phi _{B_0})}, \end{aligned}$$

By Theorem 1.1, $\sup _{B_0}\{\sup _{B_0}\phi -\phi _{B_0}\} <\infty$, thus Proposition 5.0.1 follows. $\square$