On some Schwarz type inequalities

Abstract

First, we establish some Schwarz type inequalities for mappings with bounded Laplacian, then we obtain boundary versions of the Schwarz lemma.

1 Introduction and preliminaries

Motivated by the role of the Schwarz lemma in complex analysis and numerous fundamental results, see for instance [16, 19] and references therein, in 2016, the first author [1](a) has posted on ResearchGate the project “Schwarz lemma, the Carathéodory and Kobayashi Metrics and Applications in Complex Analysis”.¹ Various discussions regarding the subject can also be found in the Q&A section on ResearchGate under the question “What are the most recent versions of the Schwarz lemma?” [1](b).² In this project and in [16], cf. also [13], we developed the method related to holomorphic mappings with strip codomain (we refer to this method as the approach via the Schwarz–Pick lemma for holomorphic maps from the unit disc into a strip). It is worth mentioning that the Schwarz lemma has been generalized in various directions; see [2, 4, 7, 8, 13, 14, 18, 21] and the references therein.

Recently Wang and Zhu [20] and Chen and Kalaj [5] have studied boundary Schwarz lemma for solutions of Poisson’s equation. They improved Heinz’s theorem [10] and Theorem A below. We found that Theorem A is a forgotten result of Hethcote [11], published in 1977.

Note that previously Burgeth [3] improved the above result of Heinz and Theorem A for real-valued functions (it is easy to extend his result for complex-valued functions; see below) by removing the assumption $f(0)=0$ but it is overlooked in the literature. Recently, Mateljević and Sveltik [18] proved a Schwarz lemma for real harmonic functions with values in $(-1,1)$ using a completely different approach than Burgeth [3] and showed that the inequalities obtained are sharp.

In this paper, we further develop the method initiated in [18]. More precisely, we show that, if $\mathbb{U}$ denotes the open unit disc and $f: \mathbb {U}\to(-1,1)$, $f \in C^{2}(\mathbb{U})$ and is continuous on $\overline{\mathbb{U}}$, and $|\Delta f|\leq c$ on $\mathbb{U}$ for some $c>0$, then the mapping $u=f \pm \frac{c}{4} (1- |z|^{2})$ is subharmonic or superharmonic and we estimate the harmonic function $P[u^{*}]$; see Theorem 2. Next, we extend the previous result to complex-valued functions; see Corollary 1. As an application, we provide an elementary proof of a theorem of Chen and Kalaj [5] giving an estimate of the solutions of the Poisson equations. Finally, we establish Schwarz lemmas at the boundary for solutions of $|\Delta f| \leq c$. Our results are generalizations of Theorem 1.1 [20] and Theorem 2 [5].

The proofs are mainly based on two ingredients, the first of which is a sharp Schwarz lemma for real harmonic functions with values in $(-1,1)$, see Theorem B, and the second is the principle of harmonic majoration, which is a consequence of the maximum principle for subharmonic functions.

1.1 Notations and background

In this paper $\mathbb{T}$ denotes the unit circle.

Recall that a real-valued function u, defined in an open subset D of the complex plane $\mathbb{C}$, is harmonic if it satisfies Laplace’s equation $\Delta u =0$ on D.

A real-valued function $u \in\mathcal{C}^{2}(D)$ is called subharmonic if $\Delta u(z) \geq0$ for all $z\in D$.

Let P be the Poisson kernel, i.e., the function

$$P\bigl(z,e^{i\theta}\bigr)=\frac{1- \vert z \vert ^{2}}{ \vert z-e^{i\theta} \vert ^{2}}, $$

and let G be the Green function on the unit disc, i.e., the function

$$G(z,w)=\frac{1}{2\pi} \log \biggl\vert \frac{1-z\overline{w}}{z-w} \biggr\vert , \quad z,w\in\mathbb{U}, z \neq w. $$

Let $\phi\in L^{1}(\mathbb{T})$ be an integrable function on the unit circle. Then the function $P[\phi]$ given by

$$P[\phi](z)=\frac{1}{2\pi} \int_{0}^{2\pi} P\bigl(z,e^{i\theta}\bigr) \phi \bigl(e^{i\theta}\bigr) \,d\theta $$

is harmonic in $\mathbb {U}$ and has a radial limit that agrees with ϕ almost everywhere on $\mathbb{T}$.

For $g \in\mathcal{C}(\overline{\mathbb {U}})$, let

$$G[g](z)= \int_{\mathbb{U}} G(z,w) g(w) \,dm(w), $$

$|z|<1$ and let $dm(w)$ denote the Lebesgue measure in $\mathbb{U}$.

If we consider the function

$$u(z):=P[\phi](z)-G[g](z), $$

then u satisfies the Poisson equation

$$\textstyle\begin{cases} \Delta u=g \quad\mbox{on the disc }\mathbb{U},\\ \lim_{r\to1^{-}}u(re^{i\theta})=\phi(e^{i\theta}) \quad\mbox{a.e. on the circle.} \end{cases} $$

One can easily see that the previous equation has a non-unique solution. Indeed, the Poisson kernel $P(z)=\frac {1-|z|^{2}}{|1-z|^{2}}$ is a harmonic function on the unit disc and $\lim_{r\to1^{-}}P(re^{i\theta})=0$ a.e., but $P \neq0$.

It is well known that, if ϕ is continuous on the unit circle, then the harmonic function $P[\phi]$ extends continuously on $\mathbb {T}$ and equals ϕ on $\mathbb{T}$; see Hörmander [12].

The following is a consequence of the maximum principle for subharmonic functions.

Theorem

(Harmonic majoration)

Letube a subharmonic function in$\mathcal{C}^{2}(\mathbb{U})\cap \mathcal{C}(\overline{\mathbb {U}})$. Then

$$u \leq P[u\vert_{\mathbb{T}}] \quad\textit{on } \mathbb{U}. $$

2 The Schwarz lemma for harmonic functions

In [10], Heinz proved that, if f is a harmonic mapping f from the unit disc into itself such that $f(0)=0$, then

$$\bigl\vert f(z) \bigr\vert \leq\frac{4}{\pi} \arctan \vert z \vert . $$

Moreover, this inequality is sharp for each point $z\in\mathbb{U}$.

This inequality for functions from the unit disk to unit ball of $\mathbb{C}^{n}$ are discussed in [9] to establish Landau’s theorem for p-harmonic mappings in several variables.

Later, in 1977, Hethcote [11] improved the above result of Heinz by removing the assumption $f(0)=0$ and showed the following.

Theorem A

([11])

Iffis a harmonic mapping from the unit disc into itself, then

$$\biggl\vert f(z) -\frac{1 - \vert z \vert ^{2}}{1 + \vert z \vert ^{2}} f(0) \biggr\vert \leq \frac {4}{\pi} \arctan \vert z \vert $$

holds for all$z\in\mathbb{U}$.

We remark that the estimate of Theorem A cannot be sharp for all values z in the unit disc.

Recently, Mateljević and Sveltik [18] proved a Schwarz lemma for real harmonic functions with values in $(-1,1)$ using a completely different approach from Burgeth [3].

Theorem B

([18])

Let$u:\mathbb{U}\to(-1,1)$be a harmonic function such that$u(0)=b$. Then

$$m_{b}\bigl( \vert z \vert \bigr) \leq u(z) \leq M_{b}\bigl( \vert z \vert \bigr) \quad \textit{for all } z\in\mathbb{U}. $$

Moreover, this inequality is sharp for each$z\in\mathbb{U}$, where$M_{b}(r):= \frac{4}{\pi} \arctan \frac{a+r}{1+ar}$, $m_{b}(r):=\frac {4}{\pi} \arctan\frac{a-r}{1-ar}$, and$a= \tan\frac{b\pi}{4}$.

Clearly Theorem B improves Theorem A for real harmonic functions, as one can check the following elementary proposition.

Proposition 2.1

Letbbe in$(-1,1)$and$r\in[0,1)$. Then

• (1) $M_{b}(r) \leq \frac{1-r^{2}}{1+r^{2}}b+ \frac{4}{\pi} \arctan r =:A_{b}(r)$and$m_{b}(r) \geq \frac{1-r^{2}}{1+r^{2}}b- \frac{4}{\pi}\arctan r$.

• (2) The mapping$b \mapsto M_{b}(r)$is increasing on$(-1,1)$.

Using a standard rotation, we can extend Theorem B for complex harmonic functions from the unit disc into itself.

Theorem 1

Let$f: \mathbb {U}\to \mathbb {U}$be a harmonic function from the unit disc into itself. Then

$$\bigl\vert f(z) \bigr\vert \leq M_{ \vert f(0) \vert }\bigl( \vert z \vert \bigr) $$

holds for all$z\in\mathbb{U}$.

Proof

Fix $z_{0}$ in the unit disc and choose unimodular λ such that $\lambda f(z_{0})=|f(z_{0})|$.

Define $u(z)=\Re(\lambda f(z))$.

Hence, using Theorem B, we get

$$\bigl\vert f(z_{0}) \bigr\vert = u(z_{0}) \leq M_{u(0)}\bigl( \vert z_{0} \vert \bigr) \leq M_{ \vert f(0) \vert } \bigl( \vert z_{0} \vert \bigr), $$

as the mapping $b\mapsto M_{b}(|z_{0}|)$ is increasing. □

3 Schwarz lemma for mappings with bounded Laplacian

The following theorem is our main result of this section.

Theorem 2

Letfbe a$C^{2}(\mathbb{U})$real-valued function, continuous on$\overline{\mathbb {U}}$and$f^{*}=f\vert_{\mathbb{T}}$. Let$b=P[f^{*}](0)$, $c\in\mathbb{R}$andKbe a positive number such that$K \geq \|P[f^{*}]\|_{\infty}$.

• (i) Iffsatisfies$\Delta f\geq- c$, then

  $$f(z)\leq K M_{b/K}\bigl( \vert z \vert \bigr) + \frac{c}{4} \bigl(1- \vert z \vert ^{2}\bigr) $$

  holds for all$z\in\mathbb{U}$.

• (ii) Iffsatisfies$\Delta f\leq c$, then

  $$f(z) \geq K m_{b/K}\bigl( \vert z \vert \bigr) - \frac{c}{4} \bigl(1- \vert z \vert ^{2}\bigr) $$

  holds for all$z\in\mathbb{U}$.

Proof

(i) Define $f^{0}(z)= f(z) + \frac{c}{4} (|z|^{2}-1)$, and set $P[f^{*}](0)=b$. Then $f^{0}$ is subharmonic and $f^{0} \leq P[f^{*}]$. As $\frac{1}{K} P[f^{*}]$ is a real harmonic function with codomain $(-1,1)$, by Theorem B, we obtain $P[f^{*}](z) \leq K M_{b/K}(|z|)$. Thus

$$f(z)\leq K M_{b/K}\bigl( \vert z \vert \bigr) + \frac{c}{4} \bigl(1- \vert z \vert ^{2}\bigr),\quad \mbox{for all } z\in\mathbb{U}. $$

(ii) If f satisfies $\Delta f\leq c$, then define $f_{0}(z)= f(z) - \frac{c}{4} (|z|^{2}-1)$, and set $P[f^{*}](0)=b$. In a similar way, we show that the inequality

$$f(z) \geq K m_{b/K}\bigl( \vert z \vert \bigr) - \frac{c}{4} \bigl(1- \vert z \vert ^{2}\bigr) $$

holds for all $z\in\mathbb{U}$. □

For complex-valued functions with bounded Laplacian from the unit disc into itself, we prove the following.

Corollary 1

Suppose that$f: \mathbb {U}\to \mathbb {U}$, $f \in C^{2}(\mathbb{U})$and continuous on$\overline{\mathbb{U}}$, and$|\Delta f|\leq c$on$\mathbb{U}$for some$c>0$. Then

$$\bigl\vert f(z) \bigr\vert \leq M_{b}\bigl( \vert z \vert \bigr) + \frac{c}{4} \bigl(1- \vert z \vert ^{2}\bigr) $$

holds for all$z\in\mathbb{U}$, where$b=|P[f^{*}](0)|$.

Proof

Fix $z_{0}$ in the unit disc and choose λ such that $\lambda f(z_{0})=|f(z_{0})|$. Define $u(z)=\Re(\lambda f(z))$ (we say that u is a real-valued harmonic associated to complex-valued harmonic f at $z_{0}$). We have $\Delta u= \Re(\lambda\Delta f)$. As u is a real function with codomain $(-1,1)$ satisfying $|\Delta u|\leq c$, by Theorem 2, we get

$$\bigl\vert u(z) \bigr\vert \leq M_{b_{1}}\bigl( \vert z \vert \bigr) + \frac{c}{4} \bigl(1- \vert z \vert ^{2}\bigr), \quad\mbox{where } b_{1}=P\bigl[u^{*}\bigr](0). $$

We have $b_{1}=P[u^{*}](0)=\Re(\lambda P[ f^{*}](0))\leq|P[f^{*}](0)|$. Hence

$$\bigl\vert f(z_{0}) \bigr\vert \leq M_{b}\bigl( \vert z_{0} \vert \bigr) + \frac{c}{4} \bigl(1- \vert z_{0} \vert ^{2}\bigr), $$

where $b=|P[f^{*}](0)|$, as the mapping $b\mapsto M_{b}(|z_{0}|)$ is increasing. □

Under the conditions of the previous theorem and using Proposition 2.1 we obtain

$$ \biggl\vert f(z)- \frac{1- \vert z \vert ^{2}}{1+ \vert z \vert ^{2}}P\bigl[f^{*}\bigr](0) \biggr\vert \leq\frac {4}{\pi} \arctan \vert z \vert + \frac{c}{4} \bigl(1- \vert z \vert ^{2}\bigr). $$

(3.1)

3.1 Applications

For a given continuous function $g: G\rightarrow\mathbb{C}$, Chen and Kalaj [5] established some Schwarz type Lemmas for mappings f in G satisfying the Poisson equation $\Delta f=g$, where G is a subset of the complex plane $\mathbb{C}$. Then they applied these results to obtain a Landau type theorem, which is a partial answer to the open problem in [6].

We provide a different and an elementary proof of Theorem C, giving a Schwarz type lemma for mappings satisfying Poisson’s equations.

Theorem C

([5])

Let$g\in C(\overline{\mathbb{U}})$and$\phi\in C(\mathbb{T})$. If a complex-valued functionfsatisfies$\Delta f = g$in$\mathbb{U}$and$f = \phi$in$\mathbb{T}$, then for$z\in\mathbb{U}$

$$ \biggl\vert f(z)- P[\phi](0) \frac{1- \vert z \vert ^{2}}{1+ \vert z \vert ^{2}} \biggr\vert \leq\frac {4}{\pi} \bigl\Vert P[\phi] \bigr\Vert _{\infty}\arctan \vert z \vert + \frac{1}{4} \| g \| _{\infty}\bigl(1- \vert z \vert ^{2}\bigr), $$

(3.2)

where$\|P[\phi]\|_{\infty}=\sup_{z\in\mathbb{U}}|P[\phi](z)|$and$\|g\|_{\infty}=\sup_{z\in\mathbb{U}}|g(z)|$.

Now we show that Theorem 2 implies Theorem C.

We will consider first the case when f is a real-valued $C^{2}(\mathbb{U})$ function, continuous on $\overline{\mathbb {U}}$, satisfying $\Delta f=g$ and $f^{*}=\phi$. Let $K := \|P[\phi]\|_{\infty}$. By Theorem 2, we have

$$m_{b/K}\bigl( \vert z \vert \bigr) K - \Vert g \Vert _{\infty}\frac{(1- \vert z \vert ^{2})}{4} \leq f(z)\leq M_{b/K}\bigl( \vert z \vert \bigr) K + \Vert g \Vert _{\infty}\frac{(1- \vert z \vert ^{2})}{4}, $$

where $b=P[\phi](0)$. Using Proposition 2.1(1), we get

$$M_{b/K} \bigl( \vert z \vert \bigr)K \leq\frac{1- \vert z \vert ^{2}}{1+ \vert z \vert ^{2}}b+ \frac{4K}{\pi} \arctan \vert z \vert $$

and

$$m_{b/K}\bigl( \vert z \vert \bigr)K \geq\frac{1- \vert z \vert ^{2}}{1+ \vert z \vert ^{2}}b- \frac{4K}{\pi} \arctan \vert z \vert . $$

Hence, the following inequality:

$$\biggl\vert f(z)- \frac{1- \vert z \vert ^{2}}{1+ \vert z \vert ^{2}} P[\phi](0) \biggr\vert \leq \frac {4}{\pi} \bigl\Vert P[\phi] \bigr\Vert _{\infty}\arctan \vert z \vert + \frac{1}{4} \| g \| _{\infty}\bigl(1- \vert z \vert ^{2}\bigr) $$

holds for all $z\in\mathbb{U}$.

If f is a complex-valued function, we may consider $u=\Re(\lambda f)$, where λ is a complex number of modulus 1. Indeed, we have

$$u(z)- \frac{1- \vert z \vert ^{2}}{1+ \vert z \vert ^{2}} P\bigl[u^{*}\bigr](0)=\Re \biggl( \lambda \biggl(f(z)- \frac{1- \vert z \vert ^{2}}{1+ \vert z \vert ^{2}} P[\phi](0) \biggr) \biggr), $$

where $u^{*}=\Re(\lambda\phi)$ on $\mathbb{T}$. Now, one can choose λ such that

$$\biggl\vert f(z)- \frac{1- \vert z \vert ^{2}}{1+ \vert z \vert ^{2}} P[\phi](0) \biggr\vert =u(z)- \frac {1- \vert z \vert ^{2}}{1+ \vert z \vert ^{2}} P\bigl[u^{*}\bigr](0). $$

4 Boundary Schwarz lemmas

We establish Schwarz lemmas at the boundary for solutions of $|\Delta f| \leq c$. Our results are generalizations of Theorem 1.1 [20] and Theorem 2 [5].

Theorem 3

Suppose$f\in C^{2}(\mathbb {U})$, continuous on$\overline{\mathbb {U}}$with codomain$(-1,1)$, such that$\Delta f \geq-c$. Iffis differentiable at$z=1$with$f(1) =1$, then the following inequality holds:

$$f_{x}(1)\geq \frac{2}{\pi} \tan\frac{\pi}{4}(1-b) - \frac{c}{2}, $$

where

$$b=P\bigl[f^{*}\bigr](0). $$

Before giving the proof, one can easily show that

$$M_{b}'(r)=\frac{4}{\pi} \biggl[ \dfrac{1-a^{2}}{ (a^{2}+1 )r^{2}+4ar+a^{2}+1} \biggr]. $$

Hence

$$M_{b}'(1)= \frac{2}{\pi} \biggl[ \frac{1-a}{1+a} \biggr]=\frac{2}{\pi } \tan\frac{\pi}{4}(1-b), $$

as $a=\tan\frac{b\pi}{4}$.

Proof

Since f is differentiable at $z=1$, we know that

$$f(z)=1+f_{z}(1) (z-1)+f_{\bar{z}}(1) (\bar{z}-1)+ o\bigl( \vert z -1 \vert \bigr). $$

That is,

$$f_{x}(1)= \lim_{r \to1^{-}} \frac{f(r)-1}{r-1}. $$

On the other hand, Theorem 2(i) leads to

$$1-f(r) \geq1-M_{b}(r) -\frac{c}{4} \bigl(1- r^{2}\bigr). $$

Dividing by $(1-r)$ and letting $r\to1^{-}$, we get

$$ f_{x}(1) \geq M_{b}'(1)- \frac{c}{2}. $$

(4.1)

Thus

$$f_{x}(1) \geq \frac{2}{\pi} \tan\frac{\pi}{4}(1-b) -\frac{c}{2}. $$

 □

Corollary 2

Suppose$f\in C^{2}(\mathbb {U})$, with codomain$(-1,1)$, is continuous on$\overline{\mathbb {U}}$and is differentiable at$z=1$with$f(1) =1$.

• (i) If$\Delta f \geq-c$, then

  $$f_{x}(1) \geq \frac{2}{\pi}-b-\frac{c}{2}. $$

• (ii) If$|\Delta f| \leq c$and$f(0)=0$, then$|b| \leq\frac {c}{4}$and

  $$f_{x}(1) \geq\frac{2}{\pi} -\frac{3}{4}c , $$

where$b=P[f^{*}](0)$.

Proof

(i) Using the inequality $M_{b} \leq A_{b}$ from Proposition 2.1 and $M_{b}(1)=A_{b}(1)=1$, we get

$$M_{b}'(1) \geq A_{b}'(1) = \frac{2}{\pi}-b. $$

(ii) The estimate $|b| \leq\frac{c}{4}$ follows directly from Theorem 2 using the assumption ${f(0)=0}$. □

Remark 4

One can also prove directly that $M_{b}'(1) \geq A_{b}'(1)$, that is,

$$ \frac{2}{\pi} \tan\frac{\pi}{4}(1-b) \geq \frac{2}{\pi}-b \quad\mbox{for } b\in[0,1). $$

(4.2)

Using the convexity of the tangent function, we get

$$\tan x \geq2\biggl(x-\frac{\pi}{4}\biggr)+1 \quad\mbox{for } x\in[0,\pi/2). $$

For $b\in[0,1)$, let us substitute x by $\frac{\pi}{4}(1-b)$, we obtain

$$\frac{2}{\pi} \tan\frac{\pi}{4}(1-b) \geq\frac{2}{\pi}-b. $$

The following theorem is a generalization of Theorem 2 in [5] where the authors proved a Schwarz lemma on the boundary for a function f satisfying $\Delta f=g$ and under the assumption $f(0)=0$.

Theorem 5

Suppose that$f\in\mathcal{C}^{2}(\mathbb {U})\cap\mathcal{C}(\overline{\mathbb {U}})$is a function of$\mathbb {U}$into$\mathbb {U}$satisfying$|\Delta f| \leq c$, where$0\leq c<\frac{4}{\pi} \tan\frac{\pi}{4}(1-b)$. If, for some$\xi\in\mathbb{T}$, $\lim_{r \to1^{-}} |f(r\xi)|=1$, then

$$\liminf_{r \to1^{-}} \frac{ \vert f(\xi)-f(r\xi) \vert }{1-r} \geq \frac {2}{\pi} \tan\frac{\pi}{4}(1-b) -\frac{c}{2}, $$

where$b=|P[f^{*}](0)|$.

If, in addition, we assume that$f(0)=0$, then

$$\vert b \vert \leq\frac{c}{4} $$

and

$$\liminf_{r \to1^{-}} \frac{ \vert f(\xi)-f(r\xi) \vert }{1-r} \geq \frac{2}{\pi } -\frac{3}{4}c. $$

Proof

Using Corollary 1, we have

$$\bigl\vert f(\xi)-f(r\xi) \bigr\vert \geq1- \bigl\vert f(r\xi) \bigr\vert \geq1-M_{b}(r)-\frac{c}{4} \bigl(1-r^{2} \bigr). $$

Thus

$$\liminf_{r \to1^{-}} \frac{ \vert f(\xi)-f(r\xi) \vert }{1-r} \geq\lim _{r \to 1^{-}} \frac{ 1-M_{b}(r)-\frac{c}{4} (1-r^{2})}{1-r}=M_{b}'(1)- \frac{c}{2}. $$

The conclusion follows as $M_{b}'(1)= \frac{2}{\pi} \tan\frac{\pi }{4}(1-b)$.

If in addition, we assume that $f(0)=0$, using the inequality (3.1), we obtain $|b|<\frac{c}{4}$. Hence

$$\liminf_{r \to1^{-}} \frac{ \vert f(\xi)-f(r\xi) \vert }{1-r} \geq \frac {2}{\pi} \tan\frac{\pi}{4}(1-b)-\frac{c}{2} \geq \frac{2}{\pi }-b-\frac{c}{2}\geq\frac{2}{\pi}- \frac{3}{4}c. $$

The second estimate follows from the inequality (4.2). □