Bounds on the growth of subharmonic frequently oscillating functions

Abstract

We present a Phragmén–Lindelöf type theorem with a flavour of Nevanlinna’s theorem for subharmonic functions with frequent oscillations between zero and one. We use a technique inspired by a paper of Jones and Makarov.

Appendices

Appendices

1.1 Appendix A: A proof for Claim 1.2

The proof we present here is a concatenation of two inequalities. To state and prove these inequalities we will need the following definitions: Define the function

$$\begin{aligned} k_d(t):= {\left\{ \begin{array}{ll} \log \left( {t}\right) ,&{}\quad d=2\\ \frac{-1}{t^{d-2}},&{}\quad d\ge 3 \end{array}\right. }. \end{aligned}$$

Following [3], for every measure $\nu$ we let

$$\begin{aligned} p_\nu (x):= & {} \int _{\mathbb R^d}^{}{k_d\left( {\left| {x-y}\right| }\right) }d{\nu }(y)\\ I(\nu ):= & {} \int _{\mathbb R^d}^{}{p_\nu (x)}d{\nu }(x)=\int _{\mathbb R^d}^{}{\int _{\mathbb R^d}^{}{k_d\left( {\left| {x-y}\right| }\right) }d{\nu }(y)}d{\nu }(x). \end{aligned}$$

The first is called the potential of the measure $\nu$ and the second is called the energy of the measure $\nu$. It is known that if the set E is compact, then there exists a probability measure $\nu _0$ so that

$$\begin{aligned} I(\nu _0)=\underset{\nu }{\sup }\; I(\nu ) \end{aligned}$$

where the supremum is takes over all the Borel probability measures $\nu$ which are supported on E. The measure $\nu _0$ is called the equilibrium measure of E. For more information see, for example, Theorem 5.4 on page 209 of [6].

We are now ready to state these inequalities:

Claim 4.1

Let $E\subset B\left( {0,\frac{1}{2}}\right)$ be a compact set with $\lambda _{d-1}(E)>0$. If $\nu _0$ is the equilibrium measure of E, then

$$\begin{aligned} \lambda _{d-1}(E)\lesssim _d\; \frac{-1}{I(\nu _0)}. \end{aligned}$$

The proof of this claim is a variation of the proof on p.94 in [3]. For the sake of completeness, and since the statement is not exactly the same, we bring here the full proof.

We will use the following result:

Lemma 4.2

(Frostman’s lemma) Let $\varphi$ be a gauge function, i.e a positive, increasing function on $[0,\infty )$ with $\varphi (0) = 0$. Let $K\subset \mathbb R^d$ be a compact set with positive $\varphi$-Hausdorff content, $\lambda _\varphi (K) > 0$. Then there is a positive Borel measure $\mu$ on K satisfying that for every ball of radius r, B, $\mu (B)\le C_d\cdot \varphi (r)$ while $\mu (K)\ge \lambda _\varphi (K)$. The constant $C_d$ depends on the dimension alone.

This lemma, and its proof can be found for example as Lemma 3.1.1, p.83 in [3], or as Theorem 1 on p.7 in [5].

Proof of Claim 4.1

Let $\mu$ be the measure we obtain by using Lemma 4.2 with $\varphi (t)=t^{d-1}$ and $K=E$. We begin by bounding from bellow the potential of the measure $\mu$. Then since $-k_d$ is a monotone decreasing positive function on [0, 1] and $diam(E)\le 1$

$$\begin{aligned} -p_\mu (x):= & {} \int _{E}^{}{-k_d(\left| {x-y}\right| )}d{\mu }(y) \\\le & {} \underset{{n} ={0}}{\overset{\infty }{\sum }}\;\int _{\left| {x-y}\right|<2^{-n-1}}^{\left| {x-y}\right|<2^{-n}}{-k_d(\left| {x-y}\right| )}d{\mu }(y)\\\le & {} \underset{{n} ={0}}{\overset{\infty }{\sum }}\;\left\{ -k_d\left( {2^{-n-1}}\right) \left[ \mu \left( {\left\{ {\left| {x-y}\right|<2^{-n}}\right\} }\right) -\mu \left( {\left\{ {\left| {x-y}\right| <2^{-n-1}}\right\} }\right) \right] \right\} \\\le & {} C_d\underset{{n} ={0}}{\overset{\infty }{\sum }}\;2^{-n(d-1)} \left( {-k_d\left( {2^{-(n+1)}}\right) }\right) \\= & {} 2^{d-1}\cdot C_d\underset{{n} ={0}}{\overset{\infty }{\sum }}\;2^{-(n+1)(d-1)}\left( { -k_d\left( {2^{-(n+1)}}\right) }\right) \\\le & {} 2^{d-1}\cdot C_d\underset{{n} ={1}}{\overset{\infty }{\sum }}\;2^{-n(d-1)}\left( { -k_d\left( {2^{-n}}\right) }\right) \\= & {} {\left\{ \begin{array}{ll} 2^{d-1}\cdot C_d\underset{{n} ={1}}{\overset{\infty }{\sum }}\;\log \left( {2^n}\right) 2^{-n},&{} \quad d=2\\ 2^{d-1}\cdot C_d\underset{{n} ={1}}{\overset{\infty }{\sum }}\;\frac{2^{n(d-2)}}{2^{n(d-1)}},&{}\quad d\ge 3 \end{array}\right. } \end{aligned}$$

both series are convergent and bounded by 2. We conclude that $-p_\mu (x)\le 2^dC_d$. Next, define the measure $d\nu =\frac{d\mu }{\mu (E)}$. Then $\nu$ is a probability measure supported on E, while

$$\begin{aligned} I(\nu )=\frac{1}{\mu (E)^2}I(\mu )=\frac{-1}{\mu (E)^2}\int _{\mathbb R^d}^{}{-p_\mu (x)}d{\mu }(x)\ge \frac{-2^dC_d}{\mu (E)}\ge \frac{-2^dC_d}{\lambda _{d-1}(E)}. \end{aligned}$$

By definition of equilibrium measure we see that

$$\begin{aligned} I(\nu _0)=\underset{\mu }{\sup }\; I(\mu )\ge I(\nu )\ge \frac{-2^dC_d}{\lambda _{d-1}(E)}\Rightarrow \lambda _{d-1}(E)\le \frac{-2^dC_d}{I(\nu _0)}, \end{aligned}$$

as $I(\nu _0)<0$, concluding our proof. $\square$

Claim 4.3

Let $E\subset B\left( {0,\frac{1}{2}}\right)$ be a compact set with $\left| {I(\nu _0)}\right| <\infty$. Then

$$\begin{aligned} \omega (0,E;B(0,1){\setminus } E) > rsim _d\; \frac{-1}{I(\nu _0)}. \end{aligned}$$

Proof

Define the function $w:B(0,1)\rightarrow \mathbb R$ by

$$\begin{aligned} w(x)= {\left\{ \begin{array}{ll} \omega \left( {x,E,B\left( {0,1}\right) {\setminus } E}\right) &{}\quad \xi \in B\left( {0,1}\right) {\setminus } E\\ 1 &{}\quad \text {otherwise} \end{array}\right. }. \end{aligned}$$

This function is super-harmonic, and by Poisson-Jensen formula for every $x\in B\left( {0,1}\right)$, since $w|_{\partial B(0,1)}\equiv 0$,

$$\begin{aligned} -w(x)= -\int _{B\left( {0,1}\right) }^{}{G_B\left( {y,x}\right) }d{\mu }_w(y), \end{aligned}$$

(⋆)

where $G_B$ is Greens’s function on $B=B\left( {0,1}\right)$, and $\mu _w$ the Reisz measure of w, defined by $d\mu _w=\frac{\Delta w}{\sigma (\partial B(0,1))}dm_d$, where $\sigma _d$ is the surface measure in $\mathbb R^d$. As $E\subseteq B\left( {0,\frac{1}{2}}\right)$, there exists a constant $c_d>0$ so that $G_B(0,y)>c_d$ for every $y\in E$, implying that

$$\begin{aligned} \omega \left( {0,E,B\left( {0,1}\right) {\setminus } E}\right)&=w\left( {0}\right) \overset{\text {by} \, (\star )}{=}\int _{E}^{}{G_B\left( {0,y}\right) }d{\mu }_w\left( {y}\right) \\&\ge c_d\cdot \int _{E}^{}{}d{\mu }_w\left( {y}\right) =c_d\cdot \left| \left| {\mu _w}\right| \right| _{}. \end{aligned}$$

(⋆⋆)

Next, let $\nu$ denote the equilibrium measure of the set E, which is compact. For every measure $\mu$ supported on E, by Frostman’s theorem:

$$\begin{aligned} \begin{aligned} \left| \left| {\mu }\right| \right| _{}&= \mu (E)=\frac{-1}{I(\nu )}\int _{\mathbb R^d}^{}{\int _{\mathbb R^d}^{}{-k_d(\left| {x-y}\right| )}d{\nu }(x)}d{\mu }(x)\\& > rsim _d \frac{-1}{I(\nu )}\int _{E}^{}{\int _{E}^{}{G_B\left( {x,y}\right) }d{\mu }(x)}d{\nu }(y), \end{aligned} \end{aligned}$$

for $G_B$ Green’s function on $B=B\left( {0,1}\right)$.

In particular, for $\mu =\mu _w$ we get that

$$\begin{aligned} \begin{aligned} \left| \left| {\mu _w}\right| \right| _{}&\ge \frac{-1}{I(\nu )}\int _{E}^{}{\int _{E}^{}{G_B\left( {x,y}\right) }d{\mu }_w(y)}d{\nu }(x)\overset{\text {by} \, (\star )}{=}\frac{-1}{I(\nu )}\int _{E}^{}{w\left( {x}\right) }d{\nu }(x) \\&\overset{(a)}{=}\frac{-1}{I(\nu )}\int _{E}^{}{}d{\nu }(y)\overset{(b)}{=}\frac{-1}{I(\nu )}, \end{aligned} \end{aligned}$$

where (a) is since $w|_E\equiv 1$, and (b) is since $\nu$ is a probability measure supported on E.

Combining this with (⋆⋆), we see that

$$\begin{aligned} \omega \left( {0,E,B\left( {0,1}\right) {\setminus } E}\right) > rsim _d\;\left| \left| {\mu _w}\right| \right| _{}\ge \frac{-1}{I(\nu )}, \end{aligned}$$

concluding the proof. $\square$

Combining these two claims we see that

$$\begin{aligned} \omega (0,E; B(0,1){\setminus } E)\ge \frac{-C_1}{I(\nu )}\ge C_1\cdot C_2\cdot \lambda _{d-1}(E), \end{aligned}$$

concluding the proof of Claim 1.2.

1.2 Appendix B: Another proof of the main lemma

It is possible to prove the Main Lemma, Lemma 2.1, by using the Main Lemma in [7] with out repeating Jones and Makarov’s ingenious idea. This is done using the following lemma, suggested to the author by M.Sodin:

Lemma 4.4

Let u be a subharmonic function defined in a neighbourhood of $Q_0:=\left[ - N, N\right] ^d$ for some $N\gg 1$ and define $Q=\left[ -\frac{N}{4\sqrt{d}},\frac{N}{4\sqrt{d}}\right] ^d$. Let $E\subseteq Z_u \cap Q$ be a closed set. Then for every basic cube $I\subset Q$ satisfying that $\lambda _{d-1}(E\cap I)>\varepsilon _d>0$

$$\begin{aligned} \frac{M_u(I)}{M_u(Q_0)}\lesssim \frac{\omega (I^*)}{\omega (E)}, \end{aligned}$$

where $\omega (\cdot )=\omega (\infty ,\;\cdot \;;\mathbb R^d{\setminus } E)$ and $I^*$ is a cube concentric with I having edge length which depends on the dimension alone.

In light of this lemma, and using the Main Lemma in [7], we get another proof for Lemma 2.1. For completeness, we bring here a reformulation of the Main Lemma in [7]:

Lemma 4.5

(The Main Lemma in [7]) Let $E\subset B(0,1)$ be a compact set , and let $\omega (\cdot )=\omega (\infty ,\cdot ;\mathbb R^d{\setminus } E)$. We subdivide $[0,1]^d$ into $N^d$ cubes I with side length $\frac{1}{N}$ and denote by ${\mathcal {G}}$ the whole collection of cubes. We define a subset ${\mathcal {E}}\subset {\mathcal {G}}$ (“empty” squares), as follows:

$$\begin{aligned} I\in {\mathcal {E}} \text { if }\lambda _{d-1}(E\cap I)\le \frac{\varepsilon _d}{N^{d-1}}, \end{aligned}$$

where $\lambda _{d-1}$ is the $(d-1)$-dimensional dyadic Hausdorff content. If $\#{\mathcal {E}}\le c_d\cdot N^d$ for some absolute constant $c_d$, then for at least half of the cubes $I\in {\mathcal {G}}$ the following inequality holds:

$$\begin{aligned} \frac{\omega (I)}{\omega ([0,1]^d)}<\exp \left( {-\alpha _d\left( {\frac{N^d}{N+\#{\mathcal {E}}}\log ^d\left( {2+\frac{\#{\mathcal {E}}}{N}}\right) }\right) ^{\frac{1}{d-1}}}\right) , \end{aligned}$$

where $\alpha _d$ is an absolute constant.

Alternative proof of Lemma 2.1

Following Lemma 4.4, it is enough to bound $\omega (I^*)$ from above.

Let $I\in {\mathcal {G}}{\setminus }{\mathcal {E}}$ be a basic cube and let $I^*$ be the corresponding cube from Lemma 4.4. We rescale the set E by $\frac{1}{\ell (I^*)N}$ so that rescaling $I^*$, we get a cube of edge-length $\frac{1}{N}$, and $E\subset B(0,1)$. We may now apply the Main Lemma in [7], on the rescaled set, to conclude the proof. $\square$

1.3 The proof of Lemma 4.4

Let g denote Green’s function for the domain $\Omega :=\mathbb R^d{\setminus } E$ with pole at $\infty$. Since $E\subseteq Q\subset B\left( {0,\frac{N}{2}}\right)$ and Green’s functions satisfy the subordination principle, for every $\left| {x}\right| \ge N$

$$\begin{aligned} g(x)\ge g_{\mathbb R^d{\setminus } B\left( {0,\frac{N}{2}}\right) }(\infty ,x)\ge \underset{\left| {x}\right| {=}N}{\min }\; g_{\mathbb R^d{\setminus } B\left( {0,\frac{N}{2}}\right) }(\infty ,x){=}\underset{\left| {x}\right| =2}{\min }\; g_{\mathbb R^d{\setminus } B(0,1)}(\infty ,x):= c_1, \end{aligned}$$

where $c_1$ is a constant which depends on the dimension, but independent of N. By the maximum principle, for every subharmonic function, u, with $E\subset Z_u$,

$$\begin{aligned} \frac{u(x)}{M_u(Q_0)}\le 1\le \frac{g(x)}{c_1}, \end{aligned}$$

implying that

$$\begin{aligned} \frac{M_u(I)}{M_u(Q_0)}\lesssim M_g(I). \end{aligned}$$

To conclude the proof we shall bound $M_g(I)$ from above.

Let h be a subharmonic function and let I be a basic cube satisfying property (P2), and let $x_I\in \partial I$ be a point satisfying that $M_h(I)=h(x_I)$. By taking a square double in edge length, we may assume without loss of generality that $dist(x_I,E\cap I)>\frac{1}{3}$. We note that following Nevanlinna’s first fundamental theorem (see for example [6, Theorem 3.19]), for every $r>\sqrt{d}$

$$\begin{aligned} M_h\left( {B\left( {x_I,\sqrt{d}}\right) }\right) \le \frac{r^{d-2}(r+\sqrt{d})}{(r-\sqrt{d})^{d-1}}\left( {h(x_I)+\underset{y\in E\cap I}{\max }G_{B(x_I,r)}(x_I,y)\cdot \mu _h(B(x_I,r))}\right) , \end{aligned}$$

and $\mu _h$ is Riesz measure, defined by $d\mu _h(x)=\frac{\Delta h(x)}{\sigma _d(\partial B(0,1))}dm_d(x)$, for $\sigma _d$ the surface area measure in $\mathbb R^d$.

If $r=\alpha \cdot \sqrt{d}$, then for every d fixed

$$\begin{aligned}&\frac{r^{d-2}(r+\sqrt{d})}{(r-\sqrt{d})^{d-1}} =\frac{\alpha ^{d-2}d^{\frac{d-2}{2}}\cdot \sqrt{d}(\alpha +1)}{d^{\frac{d-1}{2}}(\alpha -1)^{d-1}}=\frac{\alpha ^{d-2}(\alpha +1)}{(\alpha -1)^{d-1}}=\frac{1+\frac{1}{\alpha }}{\left( {1-\frac{1}{\alpha }}\right) ^{d-1}}\searrow 1, \\&\quad \text { as } \alpha \rightarrow \infty . \end{aligned}$$

We conclude that for every $\delta >0$ there exists $\alpha$ so that

$$\begin{aligned} M_h\left( {B\left( {x_I,\sqrt{d}}\right) }\right) \le \left( {1+\delta }\right) \left( {h(x_I)+\underset{y\in E\cap I}{\max }G_{B(x_I,\alpha \sqrt{d})}(x_I,y)\cdot \mu _h(B(x_I,\alpha \sqrt{d}))}\right) . \end{aligned}$$

(⋆)

Next, since the basic cube I satisfies that $\lambda _{d-1}(E\cap I)>\varepsilon _d>0$, following Observation 1.1, there exists $\delta _d>0$ for which

$$\begin{aligned} h(x_I)<M_h\left( {B\left( {x_I,\sqrt{d}}\right) }\right) (1-\delta _d). \end{aligned}$$

Combining this with (⋆), we note that if we choose $\alpha >0$ large enough (depending on the dimension d, and, in particular, on $\delta _d$)

$$\begin{aligned}&M_h\left( {B\left( {x_I,\sqrt{d}}\right) }\right) \\&\quad \le \left( {1+\frac{\delta _d}{2}}\right) \left( {h(x_I)+\underset{y\in E\cap I}{\max }G_{B(x_I,\alpha \sqrt{d})}(x_I,y)\cdot \mu _h(B(x_I,\alpha \cdot \sqrt{d}))}\right) \\&\quad \le \left( {1+\frac{\delta _d}{2}}\right) \left( {(1-\delta _d)M_h\left( {B\left( {x_I,\sqrt{d}}\right) }\right) +c_d\cdot \mu _h(B(x_I,\alpha \cdot \sqrt{d}))}\right) , \end{aligned}$$

for some constant $c_d$, which depends on the dimension alone.

This implies that if $I^*$ is a cube centred at $x_I$ with edge length $2\alpha \sqrt{d}$ then

$$\begin{aligned} M_h(I)\le M_h\left( {B\left( {x_I,\sqrt{d}}\right) }\right) \le \frac{4c_d}{\delta _d}\cdot \mu _h(B(x_I,\alpha \cdot \sqrt{d}))\le \frac{4c_d}{\delta _d}\cdot \mu _h(I^*). \end{aligned}$$

In particular, for $h=g$ we see that

$$\begin{aligned} M_g(I)\lesssim \mu _g(I^*). \end{aligned}$$

To conclude the proof, we use the fact that for every cube J:

$$\begin{aligned} \mu _g(J)=\sigma _d\left( {B(0,1)}\right) \cdot \omega (\infty ,J;\Omega ), \end{aligned}$$

for $\sigma _d$ the surface area in $\mathbb R^d$, implying that

$$\begin{aligned} \frac{M_u(I)}{M_u(Q_0)}\lesssim M_g(I)\lesssim \mu _g(I^*)\sim _d\omega (\infty ,I^*,\mathbb R^d\setminus E)\le \frac{\omega (I^*)}{\omega (E)}. \end{aligned}$$

$\square$