A sharp inequality for harmonic diffeomorphisms of the unit disk

TL;DR

This paper extends the Schwarz-Pick inequality to harmonic mappings between the unit disk and Jordan domains with specified perimeter, identifying extremals as harmonic diffeomorphisms solving a second-order Beltrami equation.

Contribution

It introduces a new inequality for harmonic diffeomorphisms of the unit disk into Jordan domains with fixed perimeter, highlighting extremals as solutions to a specific Beltrami equation.

Findings

Extended Schwarz-Pick inequality for harmonic maps

Identified extremals as harmonic diffeomorphisms solving a second-order Beltrami equation

Applicable to convex Jordan domains with given perimeter

Abstract

We extend the classical Schwarz-Pick inequality to the class of harmonic mappings between the unit disk and a Jordan domain with given perimeter. It is intriguing that the extremals in this case are certain harmonic diffeomorphisms between the unit disk and a convex domain that solve the Beltrami equation of second order.

Equations

$$|f^{\prime }(z)|⩽\frac{1-|f(z){|}^2}{1-|z{|}^2}.$$

$$|f^{\prime }(z)|⩽\frac{1}{1-|z{|}^2}$$

$$|\partial f(z)|⩽\frac{R}{1-|z{|}^2},z\in \mathbf{U}$$

$$F(z):=e^{-i\theta }f\left(\frac{z+a}{1+z\bar{a}}\right)=R\left({\int }_0^z\frac{dt}{1+t^2\mu (t)}+\overline{{\int }_0^z\frac{\mu (t)dt}{1+t^2\mu (t)}}\right).$$

$$f_z(z)=\frac{1}{1+z^2\mu (z)}$$

$$\overline{f_{\bar{z}}(z)}=\frac{\mu (z)}{1+z^2\mu (z)}.$$

$$\frac{1-k}{1+k}⩽|F_z|-|F_{\bar{z}}|=|lF|⩽|dF|=|F_z|+|F_{\bar{z}}|⩽\frac{1+k}{1-k}.$$

$$F(z):=e^{-i\theta }f\left(\frac{z+a}{1+z\bar{a}}\right)=R\left({\int }_0^z\left(1-t^2{h}^{\prime }(t)\right)dt+\overline{h(z)}\right)$$

$$\frac{|h^{\prime }(z)|}{|1-z^2{h}^{\prime }(z)|}<1,z\in \mathbf{U}.$$

$$|f(\mathbf{U})|=R^2\pi \left(1-\sum _{k=0}^{\infty }\frac{|a_k{|}^2}{(n+1)(n+2)}\right).$$

$$\left\{\begin{array}[]{ll}\overline{f_{\bar{z}}(z)}=\mu(z)f_{z}(z),&\hbox{}\\ f_{z}(0)=R,&\hbox{}\\ f(0)=0.&\hbox{}\end{array}\right.$$

$$\left\{\begin{array}[]{ll}\overline{g_{\bar{z}}(z)}=\mu(z)g_{z}(z),&\hbox{}\\ g_{z}(a)=\frac{R}{1-|a|^{2}},&\hbox{}\\ g(a)=0,&\hbox{}\end{array}\right.$$

$$g(z)=e^{i\theta }f(\frac{z-a}{1-z\bar{a}})$$

$${\partial }_t\left(g(z)+\overline{h(z)}\right)=i{g}^{\prime }(z)z+\overline{i{h}^{\prime }(z)z}$$

$$|i{g}^{\prime }(z)z+\overline{i{h}^{\prime }(z)z}|=|g^{\prime }(z)-\overline{h^{\prime }(z)z^2}|.$$

$$2\pi ={\int }_{\mathbf{T}}\left|{\partial }_t\left(g(z)+\overline{h(z)}\right)\right||dz|={\int }_{\mathbf{T}}|g^{\prime }(z)-\overline{h^{\prime }(z)z^2}||dz|.$$

$$|g^{\prime }(0)|⩽\frac{1}{2\pi }{\int }_{\mathbf{T}}|g^{\prime }(z)-\overline{h^{\prime }(z)z^2}||dz|.$$

$$\partial F(0)=f^{\prime }(a)m^{\prime }(0)=\partial f(a)(1-|a{|}^2).$$

$$|\partial f(a)|⩽\frac{1}{1-|a{|}^2}.$$

$$|g^{\prime }(0)|=\frac{1}{2\pi r}{\int }_{\mathbf{rT}}|g^{\prime }(z)-\overline{h^{\prime }(z)z^2}||dz|,$$

$$|g^{\prime }(0)|=\frac{1}{2\pi }{\int }_{\mathbf{T}}|g^{\prime }(zr)-\overline{h^{\prime }(zr)r^2{z}^2}||dz|.$$

$$\frac{1}{2\pi }{\int }_{\mathbf{T}}|g^{\prime }(zr)-\overline{h^{\prime }(rz)r^2{z}^2}||dz|-|g^{\prime }(0)|\equiv 0.$$

$${\int }_{\mathbf{U}}\phi (z)d\mu (z)={\int }_{\mathbf{U}}u\Delta \phi (z)dm(z),\phi \in C_0^2(\mathbf{U}).$$

$$d\mu =\Delta udm.$$

$$\frac{1}{2\pi }{\int }_{\mathbf{T}}u(rz)|dz|-u(0)=\frac{1}{2\pi }{\int }_{|z|<r}\log \frac{r}{|z|}d\mu (z)$$

$$u(z)=|g^{\prime }(z)-\overline{h^{\prime }(z)z^2}|$$

$$\frac{1}{2\pi }{\int }_{|z|<r}\log \frac{r}{|z|}d\mu (z)\equiv 0.$$

$$\Delta u=u|\nabla \theta {|}^2=0.$$

$$e^{-i\theta }(g^{\prime }(z)-\overline{h^{\prime }(z)z^2})=G(z)+\overline{H(z)},$$

$$G(z)=e^{-i\theta }{g}^{\prime }(z)$$

Full text

1. Introduction

Let $\mathbf{U}$ be the unit disk in the complex plane $\mathbf{C}$ and denote by $\mathbf{T}$ its boundary. A harmonic mapping $f$ of the unit disk into the complex plane can be written by $f(z)=g(z)+\overline{h(z)}$ where $g$ and $h$ are holomorphic functions defined on the unit disk. Two of essential properties of harmonic mappings are given by Lewy theorem, and Rado-Kneser-Choquet theorem. Lewy theorem states that a injective harmonic mapping is indeed a diffeomorphism. Rado-Kneser-Choquet theorem states that a Poisson extension of a homeomorphism of the unit circle $\mathbf{T}$ onto a convex Jordan curve $\gamma$ is a diffeomorphism on the unit disk onto the inner part of $\gamma$. For those and many more important properties of harmonic mappings we refer to the book of Duren [2].

The standard Schwarz-Pick lemma for holomorphic mappings states that every holomorphic mapping $f$ of the unit disk onto itself satisfies the inequality

[TABLE]

If the equality is attained in (1.1) for a fixed $z=a\in\mathbf{U}$, then $f$ is a Möbius transformation of the unit disk.

It follows from (1.1) the weaker inequality

[TABLE]

with the equality in (1.2) for some fixed $z=a$ if and only if $f(z)=e^{it}\frac{z-a}{1-z\bar{a}}$. We will extend this result to harmonic mappings.

2. Main result

Theorem 2.1**.**

If $f$ is a harmonic orientation preserving diffeomorphism of the unit disk $\mathbf{U}$ onto a Jordan domain $\Omega$ with rectifiable boundary of length $2\pi R$, then the sharp inequality

[TABLE]

holds. If the equality in (2.1) is attained for some $a$, then $\Omega$ is convex and there is a holomorphic function $\mu:\mathbf{U}\to\mathbf{U}$ and a constant $\theta\in[0,2\pi]$ such that

[TABLE]

Moreover every function $f$ defined by (2.2), is a harmonic diffeomorphism and maps the unit disk to a Jordan domain bounded by a convex curve of length $2\pi R$ and the inequality (2.1) is attained for $z=a$.

Corollary 2.2**.**

Under the conditions of Theorem 2.1, if $|\mu|_{\infty}=k<1$, then the mapping $F$ is $K=\frac{1+k}{1-k}$ bi-Lipschitz, and $K-$quasiconformal.

Proof.

We have that

[TABLE]

and

[TABLE]

Thus

[TABLE]

∎

Corollary 2.3**.**

If $\Omega=\mathbf{U}$, then the equality is attained in (2.1) for some $a$ if only if $f$ is a Möbius transformation of the unit disk onto itself.

Proof of Corollary 2.3.

Under conditions of Theorem 2.1 the function (2.2) can be written as

[TABLE]

where $h(z)=\sum_{k=0}^{\infty}a_{k}z^{k}$ is defined on the unit disk and satisfies the condition

[TABLE]

Moreover

[TABLE]

If $R=1$, this implies that $\Omega=\mathbf{U}$ if and only if $h\equiv 0$. ∎

By using the corresponding result in [1] and Theorem 2.1 we have

Corollary 2.4**.**

If as in (2.3), $F(z)=g+\overline{h}$, then $F(z)=g(z)-h(z)$ is univalent and convex in direction of real axis.

By using Theorem 2.1 we obtain

Corollary 2.5**.**

For every positive constant $R$ and every holomorphic function $\mu$ of the unit disk into itself, there is a unique convex Jordan domain $\Omega=\Omega_{\mu,R}$, with the perimeter $2\pi R$ , such that the initial boundary problem

[TABLE]

admits a unique univalent harmonic solution $f=f_{\mu,R}:\mathbf{U}\xrightarrow[]{{}_{\!\!\textnormal{onto}\!\!}}\Omega$.

Remark 2.6*.*

If instead of boundary problem (2.5) we observe

[TABLE]

then the solution $g$ is given by

[TABLE]

and thus $g(\mathbf{U})=e^{i\theta}\cdot\Omega_{\mu,R}$. Here $f$ is a solution of (2.5).

3. Proof of the main result

Proof of Theorem 2.1.

Assume first that $f(z)=g(z)+\overline{h(z)}$ has $C^{1}$ extension to the boundary and assume without loos of generality that $R=1$. Then we have

[TABLE]

So for $z=e^{it}$,

[TABLE]

Thus

[TABLE]

As $|g^{\prime}(z)-\overline{h^{\prime}(z)z^{2}}|$ is subharmonic, it follows that

[TABLE]

Thus we have that $|g^{\prime}(0)|\leqslant 1$. Now if $m(z)=\frac{z+a}{1+za}$, then $m(0)=a$, and thus $F(z)=f(m(z))$ is a harmonic diffeomorphism of the unit disk onto itself. Further,

[TABLE]

Therefore by applying the previous case to $F$ we obtain

[TABLE]

Assume now that the equality is attained for $z=0$. Then

[TABLE]

or what is the same

[TABLE]

Thus for $0\leqslant r\leqslant 1$ we have

[TABLE]

In order to continue recall the definition of the Riesz measure $\mu$ of a subharmonic function $u$. Namely there exists a unique positive Borel measure $\mu$ so that

[TABLE]

Here $dm$ is the Lebesgue measure defined on the complex plane $\mathbf{C}$. If $u\in C^{2}$, then

[TABLE]

Proposition 3.1**.**

[5, Theorem 4.5.1]** If $u$ is a subharmonic function defined on the unit disk then for $r<1$ we have

[TABLE]

where $\mu$ is the Riesz measure of $u$.

By applying Proposition 3.1 to the subharmonic function

[TABLE]

in view of (3.1) we obtain that

[TABLE]

Thus in particular we infer that $\mu=0$, or what is the same $\Delta u=0$. As $u=|w|$ where $w=|u|e^{i\theta}$ is harmonic, it follows that

[TABLE]

Therefore $\nabla\theta\equiv 0$, and therefore $\theta=\mathrm{const}$.

So

[TABLE]

is a real harmonic function. Here

[TABLE]

and

[TABLE]

are analytic functions satisfying the condition $|H(z)|<|G(z)|$ in view of Lewy theorem. Thus

[TABLE]

or what is the same

[TABLE]

Thus $G(z)-H(z)$ is a real holomorphic function and therefore it is a constant function. Further

[TABLE]

Hence

[TABLE]

Assume without losing the generality that $\theta=0$ and $g^{\prime}(0)=1$. Then

[TABLE]

Further for $z=e^{it}$,

[TABLE]

and

[TABLE]

From (2.4), we infer that

[TABLE]

In order to get the representation (2.2), by Lewy theorem, we have that the holomorphic mapping $\mu(z)=\frac{h^{\prime}(z)}{g^{\prime}(z)}$ maps the unit disk into itself. By (3.2) we deduce that

[TABLE]

and

[TABLE]

In order to prove that, every mapping $f$ defined by (2.2) is a diffeomorphism we use Choquet-Kneser-Rado theorem. First of all

[TABLE]

Therefore

[TABLE]

which means that $F(\mathbf{T})$ is a convex curve.

As

[TABLE]

if $z_{1},z_{2}\in\mathbf{T}$ with $f(z_{1})=f(z_{2})$, then

[TABLE]

and so $z_{1}=z_{2}$. Thus by Choquet-Kneser-Rado theorem, $F$ is a diffeomorphism.

If $f$ is not $C^{1}$ up to the boundary, then we apply the approximating sequence. Let $\Omega$ be a fixed Jordan domain and assume that $\phi$ is a conformal mapping of the unit disk onto $\Omega$, with $\phi(0)=0$. For $r_{n}=\frac{n}{n+1}$, let $\Omega_{n}=\varphi(r_{n}\mathbf{U})$, and let $U_{n}=f^{-1}\Omega_{n}$. Let $\phi_{n}:\mathbf{U}\to U_{n}$ be a conformal mapping satisfying the condition $\phi_{n}(0)=0$. Then $f_{n}=f\circ\phi_{n}$ is a conformal mapping of the unit disk onto the Jordan domain $\Omega_{n}$. Further, by subharmonic property of $|\phi^{\prime}(z)|$ we conclude that

[TABLE]

Then we have that

[TABLE]

As $\phi_{n}$ converges in compacts to the identity mapping, and thus $\phi^{\prime}_{n}$ converges in compacts to the constant $1$, we conclude that the inequality (2.1) is true for non-smooth domains.

It remains to consider the equality statement in this case. But we know that $\partial\Omega$ is rectifiable if and only if $\partial_{t}f\in h_{1}(\mathbf{U})$. (See e.g. [4, Theorem 2.7]). Here $h_{1}$ stands for the Hardy class of harmonic mappings. Now the proof is just repetition of the previous approach, and we omit the details.

∎

Example 3.2**.**

If $\mu(z)=z^{n}$, then $F$ defined in (2.2), maps the unit disk to $n+2-$regular polygon of perimeter $2\pi R$ and centered at 0. Namely we have that

[TABLE]

The rest follows from the similar statement obtained by Duren in [2, p. 62].

Remark 3.3*.*

If $\mu$ is a holomorphic mapping of the unit disk onto itself and $F$ is defined by (2.2), then $F(0)=0$ and

[TABLE]

Indeed we have that

[TABLE]

Here $\rho=\mathrm{dist}(0,\partial\Omega)$. Thus we have the sharp inequality

[TABLE]

In [3] it is proved that we have the general inequality

[TABLE]

for every harmonic diffeomorphism of the unit disk onto a convex domain $\Omega$ with $f(0)=0$. Some examples suggest that the best inequality in this context is

[TABLE]

The last conjectured inequality is not proved. The gap between $\frac{\rho^{2}}{2}$ and $\frac{\rho^{2}}{8}$ in (3.4) and (3.6) appears as the mappings $F$ are special extremal mappings which for the case of $\Omega$ being the unit disk are just rotations.