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 f is indeed a diffeomorphism, or what is the same its Jacobian \(J_f:=|\partial f|^2-|\bar{\partial }f|^2=|g'(z)|^2-|h'(z)|^2\ne 0\). 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 lemma states that if f is a holomorphic mapping of the unit disk \(\mathbf {U}\) into, itself such that \(f(0)=0\) then \(|f(z)|\le |z|\).

Its counter-part for harmonic mappings states the following ([2, Sect. 4.6]). Let f be a complex-valued function harmonic in the unit disk \(\mathbf {U}\) into itself, with \(f (0) = 0\). Then

$$\begin{aligned} | f (z)| \le \frac{ 4}{\pi }\arctan |z|, \end{aligned}$$

and this inequality is sharp for each point \(z \in \mathbf {U}\). Furthermore, the bound is sharp everywhere (but is attained only at the origin) for univalent harmonic mappings f of \(\mathbf {U}\) onto itself with \(f (0) = 0.\)

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

$$\begin{aligned} |f'(z)|\le \frac{1-|f(z)|^2}{1-|z|^2}. \end{aligned}$$
(1.1)

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:

$$\begin{aligned} |f'(z)|\le \frac{1}{1-|z|^2} \end{aligned}$$
(1.2)

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

$$\begin{aligned} |\partial f(z)|\le \frac{R}{1-|z|^2},\ \ \ z\in \mathbf {U} \end{aligned}$$
(2.1)

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

$$\begin{aligned} F(z):=e^{-i\theta }f\left( \frac{z+a}{1+z\bar{a}}\right) =R\left( \int _0^z\frac{ dt}{1+t^2\mu (t)}+\overline{\int _0^z\frac{\mu (t)dt}{1+t^2\mu (t)}}\right) . \end{aligned}$$
(2.2)

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 \(R=1\) and \(|\mu |_{\infty }=k<1\), then the mapping F is \(K=\frac{1+k}{1-k}\) bi-Lipschitz, and K—quasi-conformal.

Proof

We have that

$$\begin{aligned} F_z(z)=\frac{1}{1+z^2\mu (z)} \end{aligned}$$

and

$$\begin{aligned} \overline{F_{\bar{z}}(z)}= \frac{\mu (z)}{1+z^2\mu (z)}. \end{aligned}$$

Thus

$$\begin{aligned} \frac{1-k}{1+k}\le |F_z|-|F_{\bar{z}}|:=|lF|\le |dF|:=|F_z|+|F_{\bar{z}}|\le \frac{1+k}{1-k}. \end{aligned}$$

Thus, F is K—bi-Lipschitz. Furthermore, we have

$$\begin{aligned} \frac{|F_{\bar{z}}(z)|}{|F_z(z)|}=|\mu (z)|\le k, \end{aligned}$$

and so

$$\begin{aligned} \frac{(|F_z|+|F_{\bar{z}}|)^2}{|F_z|^2-|F_{\bar{z}}|^2}=\frac{|F_z|+|F_{\bar{z}}|}{|F_z|-|F_{\bar{z}}|}\le \frac{1+k}{1-k}=K. \end{aligned}$$

Therefore, f is K—quasi-conformal. \(\square \)

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 a disk.

Proof of Corollary 2.3

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

$$\begin{aligned} F(z):=e^{-i\theta }f\left( \frac{z+a}{1+z\bar{a}}\right) =R\left( \int _0^z\left( 1-t^2 h'(t)\right) dt+\overline{h(z)}\right) \end{aligned}$$
(2.3)

where \(h(z)=\sum _{k=0}^\infty a_k z^k\) is defined on the unit disk and satisfies the condition:

$$\begin{aligned} \frac{|h'(z)|}{|1-z^2 h'(z)|}< 1, \ \ z\in \mathbf {U}. \end{aligned}$$
(2.4)

Further

$$\begin{aligned} J_F(z)=|1-z^2 h'(z)|^2-|h'(z)|^2. \end{aligned}$$

Since

$$\begin{aligned} h'(z)=\sum _{k=0}^\infty b_k z^k=\sum _{k=0}^\infty (k+1)a_{k+1} z^{k}, \end{aligned}$$

and

$$\begin{aligned} 1-z^2 h'(z)=\sum _{k=0}^\infty c_k z^k=1-\sum _{k=2}^\infty (k-1)a_{k-1} z^{k}, \end{aligned}$$

it follows that

$$\begin{aligned} \begin{aligned} |f(\mathbf {U})|&=\int _{\mathbf {U}} J_F(z) dxdy\\ {}&=\int _0^{2\pi }\int _0^1 r J_F(re^{it})drdt \\ {}&= 2\pi R^2 \sum _{k=0}^\infty \frac{|b_k|^2}{2k+2}-2\pi R^2 \sum _{k=0}^\infty \frac{|c_k|^2}{2k+2} \\ {}&=2\pi R^2\left( 1+ \sum _{k=2}^\infty \frac{|(k-1)a_{k-1} |^2}{2k+2}-\sum _{k=0}^\infty \frac{|(k+1)a_{k+1}|^2}{2k+2}\right) \\ {}&=2\pi R^2\left( 1+ \sum _{k=0}^\infty \frac{|(k+1)a_{k+1} |^2}{2k+6}-\sum _{k=0}^\infty \frac{|(k+1)a_{k+1}|^2}{2k+2}\right) \\ {}&=\pi R^2\left( 1- 2\sum _{k=0}^\infty \frac{(1+k)|a_{k+1}|^2}{(3+k)}\right) . \end{aligned} \end{aligned}$$

If \(R=1\), this implies that \(\Omega =\mathbf {U}+a_0\) if and only if \(h\equiv a_0\). This concludes the proof. \(\square \)

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.

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 (Beltrami equation)

$$\begin{aligned} \left\{ \begin{array}{ll} \overline{f_{\bar{z}}(z)}=\mu (z) f_z(z), &{} \hbox {} \\ f_z(0)=R, &{} \hbox {} \\ f(0)=0, &{} \hbox {} \end{array} \right. \end{aligned}$$
(2.5)

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

Remark 2.6

If instead of boundary problem (2.5), we observe

$$\begin{aligned} \left\{ \begin{array}{ll} \overline{g_{\bar{z}}(z)}=\mu (z) g_z(z), &{} \hbox {} \\ g_z(a)=\frac{R}{1-|a|^2}, &{} \hbox {} \\ g(a)=0, &{} \hbox {} \end{array} \right. \end{aligned}$$
(2.6)

then the solution g is given by

$$\begin{aligned} g(z)=e^{i\theta }f\left( \frac{z-a}{1-z\bar{a}}\right) \end{aligned}$$

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 loss of generality that \(R=1\). Then, we have

$$\begin{aligned} \partial _t \left( {g(z)+\overline{h(z)}}\right) =ig'(z) z+\overline{i h'(z) z}\end{aligned}$$
(3.1)

Therefore, for \(z=e^{it}\)

$$\begin{aligned} |ig'(z) z+\overline{i h'(z) z}|=|g'(z)-\overline{h'(z) z^2}|. \end{aligned}$$

Thus

$$\begin{aligned} 2\pi =\int _{\mathbf {T}} \left| \partial _t \left( {g(z)+\overline{h(z)}}\right) \right| |dz|=\int _{\mathbf T}|g'(z)-\overline{h'(z) z^2}||dz|. \end{aligned}$$

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

$$\begin{aligned} |g'(0)|\le \frac{1}{2\pi }\int _{\mathbf T}|g'(z)-\overline{h'(z) z^2}||dz|. \end{aligned}$$

Thus, we have that \(|g'(0)|\le 1\). Now, if \(m(z)=\frac{z+a}{1+z \bar{a}}\), then \(m(0)=a\), and thus, \(F(z)=f(m(z))\) is a harmonic diffeomorphism of the unit disk onto itself. Furthermore

$$\begin{aligned} \partial F(0)=f'(a)m'(0)=\partial f(a)(1-|a|^2). \end{aligned}$$

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

$$\begin{aligned} |\partial f(a)|\le \frac{1}{1-|a|^2}. \end{aligned}$$

Assume now that the equality is attained for \(z=0\). Then

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

or what is the same

$$\begin{aligned} |g'(0)|= \frac{1}{2\pi }\int _{\mathbf {T}}|g'(zr)-\overline{h'(zr) r^2z^2}||dz|. \end{aligned}$$

Thus, for \(0\le r\le 1\), we have

$$\begin{aligned} \frac{1}{2\pi }\int _{\mathbf {T}}|g'(zr)-\overline{h'(rz) r^2z^2}||dz|-|g'(0)|\equiv 0. \end{aligned}$$
(3.2)

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

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

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

$$\begin{aligned} d\mu =\Delta u dm. \end{aligned}$$

\(\square \)

We need the following proposition.

Proposition 3.1

[5, Theorem 2.6 (Riesz representation theorem)]. If u is a subharmonic function defined on the unit disk then for \(r<1\), we have

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

where \(\mu \) is the Riesz measure of u.

By applying Proposition 3.1 to the subharmonic function

$$\begin{aligned} u(z)=|g'(z)-\overline{h'(z) z^2}| \end{aligned}$$

in view of (3.2), we obtain that

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

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

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

Therefore, \(\nabla \theta \equiv 0\), and hence, \(\theta =\mathrm {const}\).

Therefore

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

is a real harmonic function. Here

$$\begin{aligned} G(z)=e^{-i\theta }g'(z) \end{aligned}$$

and

$$\begin{aligned} H(z)=-e^{i\theta }{h'(z) z^2} \end{aligned}$$

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

$$\begin{aligned} G(z)+\overline{H(z)}=\overline{G(z)}+H(z) \end{aligned}$$

or what is the same

$$\begin{aligned} G(z)-H(z)=\overline{G(z)-H(z)}. \end{aligned}$$

Thus, \(G(z)-H(z)\) is a real holomorphic function, and therefore, it is a constant function. Furthermore

$$\begin{aligned} e^{-i\theta }g'(z)+e^{i\theta }{h'(z) z^2}=G(z)-H(z)=G(0)-H(0)=e^{-i\theta }g'(0). \end{aligned}$$

Hence

$$\begin{aligned} G(z)+\overline{H(z)}=G(z)+\overline{G(z)}-e^{-i\theta }g'(0)=2\mathfrak {R}\left[ e^{-i\theta }g'(z)\right] -e^{-i\theta }g'(0). \end{aligned}$$

Assume without loss of the generality that \(\theta =0\) and \(g'(0)=1\). Then

$$\begin{aligned} g'(z)=1-{h'(z) z^2}. \end{aligned}$$
(3.3)

From (2.4), we infer that

$$\begin{aligned} (1-2\mathfrak {R}(h'(z) z^2))>|h'(z)|^2(1-|z|^2). \end{aligned}$$
(3.4)

Further for \(z=e^{it}\), from (3.1) and (3.3), we have

$$\begin{aligned} \partial _t f(z)=i z(1-2\mathfrak {R}(h'(z) z^2)).\end{aligned}$$
(3.5)

To get the representation (2.2), by Lewy theorem, we have that the holomorphic mapping \(\mu (z)=\frac{h'(z)}{g'(z)}\) maps the unit disk into itself. By (3.3), we deduce that

$$\begin{aligned} g(z)=\int _0^z\frac{ dt}{1+t^2\mu (t)} \end{aligned}$$

and

$$\begin{aligned} h(z)={\int _0^z\frac{\mu (t)dt}{1+t^2\mu (t)}}. \end{aligned}$$

It follows by (3.5) and (3.4) that \(\partial _t \arg \partial _t f(z)=1>0\), and this implies that the image of \(\mathbf {U}\) under f is a convex domain.

To prove that, every mapping f defined by (2.2) is a diffeomorphism of the unit disk onto a convex Jordan domain, we use Choquet–Kneser–Rado theorem. First of all, we have

$$\begin{aligned} \arg \partial _t F(z)= (\pi /2+t). \end{aligned}$$

Therefore

$$\begin{aligned} \partial _t \arg \partial _t F(z)=1>0 \end{aligned}$$

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

As

$$\begin{aligned} \frac{\partial _t F(z)}{|\partial _t F(z)|}=iz, \end{aligned}$$

if \(z_1, z_2\in \mathbf {T}\) with \(f(z_1)= f(z_2)\), then

$$\begin{aligned} \frac{\partial _t F(z_1)}{|\partial _t F(z_1)|}=\frac{\partial _t F(z_2)}{|\partial _t F(z_2)|} \end{aligned}$$

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 \(\Omega _n=\phi (r_n\mathbf {U})\), and let \(U_n=f^{-1}\Omega _n\). Let \(\phi _n:\mathbf {U}\rightarrow U_n\) 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 \(\Omega _n\). Furthermore, by subharmonic property of \(|\phi '(z)|\), we conclude that

$$\begin{aligned} R_n=|\partial \Omega _n|=\int _{\mathbf T}|\phi '(r_n z)|dz|< \int _{\mathbf T}|\phi '( z)|dz|=|\partial \Omega |=R=1. \end{aligned}$$

Then, we have that

$$\begin{aligned} |\partial f_n(z)|\le \frac{R_n}{1-|z|^2},\ \ \ z\in \mathbf {U}. \end{aligned}$$
(3.6)

As \(\phi _n\) converges in compacts to the identity mapping, and thus, \(\phi '_n\) 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. However, 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. \(\square \)

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

$$\begin{aligned} \partial _z F(z) = \frac{R}{1+z^{n+2}},\ \ \ \partial _{\bar{z}} F(z)=\frac{Rz^{n}}{1+z^{n+2}}. \end{aligned}$$

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

$$\begin{aligned} |DF|^2:=|F_z|^2+|F_{\bar{z}}|^2\ge \frac{R^2}{2} . \end{aligned}$$

Indeed, we have that

$$\begin{aligned} |DF|^2=R^2\frac{1+|\mu |^2}{|1+z^2\mu |^2}\ge \frac{R^2}{2}=\frac{L^2}{8\pi ^2}\ge \frac{\rho ^2}{2}. \end{aligned}$$

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

$$\begin{aligned} |DF|^2\ge \frac{\rho ^2}{2}.\end{aligned}$$
(3.7)

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

$$\begin{aligned} |Df|^2\ge \frac{\rho ^2}{16},\end{aligned}$$
(3.8)

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

$$\begin{aligned} |Df|^2\ge \frac{\rho ^2}{8},\end{aligned}$$
(3.9)

The last conjectured inequality is not proved. The gap between \(\frac{\rho ^2}{2}\) and \(\frac{\rho ^2}{8}\) in (3.7) and (3.9) appears as the mappings F are special extremal mappings which for the case of \(\Omega \) being the unit disk are just rotations.