text:
\section{Preliminaries and Main Results}

Let $\mathbb{C}$ denote the complex plane and $\mathbb{D}$ the open unit disk in $\mathbb{C}$. Let $\mathbb{T}=\partial\mathbb{D}$ be the boundary of $\mathbb{D}$, and $\overline{\mathbb{D}} = {\mathbb D} \cup \mathbb{T}$, the closure of $\mathbb{D}$. Furthermore, we denote by $\mathcal{C}^m(\Omega)$ the set of all complex-valued $m-$times continuously differentiable functions from $\Omega$ into $\mathbb{C}$, where
$\Omega$ stands for a domain of $\mathbb{C}$ and $m \in \mathbb{N}$. In particular, $\mathcal{C}(\Omega):= \mathcal{C}^0(\Omega)$ denotes the set of all continuous functions in $\Omega$.\\
For a real $2 \times 2$ matrix $A$, we use the matrix norm
$$\| A\|=\sup\lbrace|Az| : |z|=1\rbrace,$$
and the matrix function
$$\lambda(A)=\inf\lbrace |Az| : |z|=1 \rbrace .$$
For $z = x+iy \in {\mathbb C}$, the formal derivative of a complex-valued function $\Phi = u+iv$
is given by\\
$$D_\Phi=\begin{pmatrix}
u_x && u_y\\
v_x && v_y
\end{pmatrix},$$
so that
$$\| D_\Phi\|=|\Phi_z|+|\Phi_{\overline{z}}|\ \ \ \text{and}\ \ \ \lambda(D_\Phi)=\big||\Phi_z|-|\Phi_{\overline{z}}|\big|,$$
where
$$\Phi_z=\frac{1}{2}(\Phi_x-i\Phi_y)\ \ \ \text{and}\ \ \ \ \Phi_{\overline{z}}=\frac{1}{2}(\Phi_x+i\Phi_y).$$
We use
$$J_\Phi := \det D_\Phi = |\Phi_z|^2-|\Phi_{\overline{z}}|^2.$$

The main  objective of this paper is to establish a Schwarz-type lemma for  the solutions to the following inhomogeneous
biharmonic Dirichlet problem (briefly, IBDP):
\begin{equation}\label{ibdp}
\begin{cases}
\Delta^2 \Phi = g \ \ \  \text{in }\ \ {\mathbb D},\\
\Phi\  = f \ \ \  \ \ \text{on}\ \ \mathbb{T},\\
\partial_n \Phi =h \ \ \ \ \text{on}\ \  \mathbb{T}.
\end{cases}
\end{equation}
where $$\displaystyle\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2},$$ denotes the standard Laplacian and $\partial_n$ denotes the differentiation in the inward normal direction, $g \in \mathcal{C}(\overline{{\mathbb D}})$ and the boundary data $f$ and $h$ $\in\mathcal{C}(\mathbb{T})$.\\

We would like to mention that in \cite{Chel} and \cite{ChenZu}, the authors  have considered similar inhomogeneous biharmonic equations but with different boundaries conditions.\\

In order to state our main results, we introduce some necessary terminologies.
For $z,w \in {\mathbb D}$, let
$$G(z,w) = |z -w|^2\log\bigg|\frac{1 - z\overline{w}}{z - w}\bigg|^2- (1 - |z|^2)(1 - |w|^2),$$
and
$$P(z) = \frac{1-|z|^2}{|1 - z|^2}.$$
denote the biharmonic Green function and the harmonic Poisson kernel, respectively.\\
\\
By \cite[Theorem 1.1]{Li}, we see that all solutions
of IBDP  (\ref{ibdp}) are given by

\begin{equation}
\Phi(z) = F_0[f](z) + H_0[h](z)-G[g](z),\nonumber
\end{equation}

\noindent where
$$F_0[f](z) =\frac{1}{2\pi}\int^{2\pi}_0 F_0(ze^{-i\theta})f(e^{i\theta}) d\theta,\quad H_0[h](z) =
\frac{1}{2\pi}\int^{2\pi}_0 H_0(ze^{-i\theta})h(e^{i\theta}) d\theta,$$
$$\text{and}\ \ \ G[g](z)=\frac{1}{16}\int_{{\mathbb D}}G(z,\omega)g(\omega)dA(\omega)$$
where $dA(\omega)$ denotes the Lebesgue area measure in ${\mathbb D}$.
Here the kernels $H_0$ and $F_0$ are given by
$$F_0(z) = H_0(z) + K_2(z),$$
$$ H_0(z) =\displaystyle\frac{1}{2}(1-|z|^2)P(z),$$
$$K_2(z)=\frac{1}{2}\frac{(1-|z|^2)^3}{|1 - z|^4}.$$

Thus, the solutions of the equation (\ref{ibdp}) are given by

$$\Phi(z)=\frac{1}{2} (1-|z|^2)P[f+h](z)+K_2[f](z)-G[g](z). $$

Obviously  $P[f+h]$ is a bounded harmonic function, and  Heinz \cite{Hei} proved the following result, which is called the Schwarz lemma for planar
harmonic functions: If $\Phi$ is a harmonic mapping from ${\mathbb D}$ into itself with $\Phi(0) = 0$, then for $z \in {\mathbb D}$,
\begin{equation*}
|\Phi(z)|\leq \frac{4}{\pi} \arctan |z|.
\end{equation*}

Hethcote \cite{heth} and  Pavlovi\'c \cite[Theorem 3.6.1]{Pav} improved Heinz's result,  by removing the assumption $\Phi(0)=0$, and proved the following.

\begin{Thm}
	 Let $\Phi:{\mathbb D} \to  {\mathbb D}$ be a harmonic function from the unit disc to itself, then
\begin{equation} \label{harm}
\bigg|\Phi(z)-\frac{1-|z|^2}{1+|z|^2} \Phi(0)\bigg|\leq \frac{4}{\pi} \arctan|z|,\ \ \ \ \ \ \ \ z\in {\mathbb D}.
\end{equation}

\end{Thm}

\noindent Remark that $K_2[f]$ is a bounded $T_2$-harmonic which is a special type of biharmonic functions.  So naturally our first aim is to study the class of  $T_2$-harmonic functions.  Let us recall first  the definition of  $T_\alpha$-harmonic functions.
\begin{defn}\cite{Olf1}
Let $\alpha \in {\mathbb R}$, and let $f\in \mathcal{C}^2({\mathbb D})$. We say that $f$ is $T_\alpha$-harmonic if $f$ satisfies
 $$T_\alpha(f)=0\quad \text{in} \ {\mathbb D},$$
  \end{defn}
 where the $T_\alpha$-Laplacian operator is defined by
 $$T_\alpha=-\frac{\alpha^2}{4}(1-|z|^2)^{-(\alpha+1)}+\frac{1}{2}L_\alpha+\frac{1}{2}\overline{L_\alpha},$$
 with the weighted Laplacian operator $L_\alpha$ is defined by
 $$L_\alpha=\frac{\partial}{\partial\overline{z}}(1 - |z|^2)^{-\alpha}\frac{\partial}{\partial z}.$$

 \begin{rem}
 	Let $f$ be a $T_\alpha$-harmonic function.
 	\begin{enumerate}
 		\item If $\alpha=0$, then $f$ is harmonic.
 		\item If $\alpha=2(n-1)$, then  $f$ is $n$-harmonic,  where $n\in \{1,2,\ldots,\}$, see \cite{Abd,borh,Olf,Olf1}.
 	\end{enumerate}
 \end{rem}

 The following result is the homogeneous  expansion of $T_\alpha$-harmonic functions.

 \begin{Thm}\cite{Olf}
 	Let $\alpha\in {\mathbb R}$ and $f\in \mathcal{C}^2({\mathbb D})$. Then $f$ is $T_\alpha$-harmonic if and only if it has a series expansion of the form
 	\begin{equation}\label{expansion}
 	f(z)=\sum_{k=0}^\infty c_k F(-\frac{\alpha}{2},k-\frac{\alpha}{2};k+1;|z|^2)z^k+
 	\sum_{k=1}^\infty c_{-k} F(-\frac{\alpha}{2},k-\frac{\alpha}{2};k+1;|z|^2)\overline{z}^k  \end{equation}
 	for some sequence $\{c_k\}$ of complex numbers satisfying
 	$$\limsup_{|k|\to \infty} |c_k|^{\frac{1}{|k|}}\leq 1.$$
 \end{Thm}
$F$ is the {\it hypergeometric} function defined by the power series
$$F(a,b;c;x)=\sum_{n=0}^\infty \frac{(a)_n(b)_n}{(c)_n} \frac{x^n}{n!}, \quad |x|<1,$$
for $a,b,c\in {\mathbb R}$, with $c\not= 0,-1,-2,\ldots.,$ where $(a)_0=1$ and $(a)_n=a(a+1)\ldots (a+n-1)$ for $n=1,2,\ldots$ are the {\it Pochhammer} symbols.\\

We refer the reader to \cite{Olf}, where Olofsson gave a Poisson type  integral representation theorem for $T_\alpha$-harmonic mappings, for $\alpha>-1$.\\

It is well known that the Schwarz lemma is one of the most influential results  in many branches of mathematical research for more than a hundred years. We refer the reader to \cite{Bur,Chel,Kal1,Kra,Liu,Mate} for generalizations and applications of this lemma.\\

In this context, we establish  a Schwarz type  lemma for  $T_2$-harmonic functions.

\begin{thm}\label{theo1}
Let $u:{\mathbb D}\longrightarrow {\mathbb D}$ be a  $T_2$-harmonic function, then
\begin{equation}\label{t2}
\bigg|u(z)-\frac{(1-|z|^2)^3}{(1+|z|^2)^2}u(0)\bigg|\leq \frac{2}{\pi}\Bigg[(1+|z|^2)\arctan |z|+ \frac{|z|(1-|z|^2)}{1+|z|^2}\Bigg],\nonumber
\end{equation}
for all $z\in {\mathbb D}$.
\end{thm}

Next, we prove a Schwarz-Pick lemma for $T_2$-harmonic functions.
\begin{thm}\label{theo2}
	Let $u:{\mathbb D}\longrightarrow {\mathbb D}$ be a  $T_2$-harmonic function, then
		\begin{equation}\label{theoe2}
	\| D_u(z)\| \leq \frac{(2+5|z|)(1+|z|^2)}{1-|z|^2}, \text{ for all } z\in {\mathbb D}.
	\end{equation}
	\begin{equation}\label{grad0}
	\| D_u(0)\|\leq \frac{4}{\pi}.
	\end{equation}
	Moreover, the inequality (\ref{grad0}) is sharp.
\end{thm}

Next we establish a Landau type theorem for $T_2$-harmonic functions
\begin{thm}\label{landau}
	Let $u\in \mathcal{C}^2({\mathbb D})$ be a $T_2$-harmonic function satisfying $u(0)=|J_u(0)|-1=0$ and $\sup_{z\in {\mathbb D}}|f(z)|\leq M$, where $M>0$ and $J_u$ is the Jacobian of $u$. Then $u$ is univalent on $D_{r_0}$, where $r_0$ satisfies the following equation
	$$\frac{\pi}{4M}-\frac{4}{\pi}\frac{M r_0}{(1-r_0)^3}\big(-r_0^3+3r_0^2-3r_0+3 \big)=0.$$
	Moreover, $u({\mathbb D}_{r_0})$ contains a univalent disk $D_{R_0}$ with
		$$R_0 \geq  \frac{4M r_0^2}{\pi(1-r_0)^3}\big[-\frac{4}{5}r_0^3+\frac{9}{4}r_0^2-2r_0+\frac{3}{2}\big]. $$
\end{thm}

Now we are in the position to prove our main results
\begin{thm}\label{theo3}
Let $g\in \mathcal{C}(\overline{{\mathbb D}})$, $f, h \in \mathcal{C}(\mathbb{T})$ and suppose that $\Phi\in \mathcal{C}^4({\mathbb D}) \cap \mathcal{C}(\overline{{\mathbb D}})$
satisfies (\ref{ibdp}). Then for $z \in {\mathbb D}$,
\begin{eqnarray}\label{sch}
\bigg| \Phi(z)-\frac{1}{2}\frac{(1-|z|^2)^3}{(1+|z|^2)^2}P[f](0)-\frac{1}{2}\frac{(1-|z|^2)^2}{1+|z|^2}P[f+h](0)\bigg|&\leq & \Big[\frac{2}{\pi}(1-|z|^2)\arctan|z|\Big]\| f+h\|_\infty\nonumber\\
&+&\frac{2}{\pi}\bigg[(1+|z|^2)\arctan |z|+|z|\frac{1-|z|^2}{1+|z|^2}\bigg]\| f\|_\infty\nonumber\\
&+&\frac{(1-|z|^2)^2}{64}\,\| g\|_\infty.
\end{eqnarray}
where $\| f \|_\infty= \displaystyle\sup_{\zeta\in \mathbb{T}}|f(\zeta)|$, $\displaystyle\| f+h \|_\infty= \sup_{\zeta\in \mathbb{T}}|f(\zeta)+h(\zeta)|$ and  $\displaystyle\| g \|_\infty= \sup_{\zeta\in \mathbb{D}}|g(\zeta)|$.
\end{thm}

\begin{normalsize}
\textbf{Remark 1.2}
\end{normalsize}
Under the hypothesis of Theorem \ref{theo3}, if $g\equiv 0$, then $\Phi$ is biharmonic mapping and the estimate (\ref{sch}) can be written in the following form
\begin{eqnarray*}
\bigg| \Phi(z)-\frac{1}{2}\frac{(1-|z|^2)^3}{(1+|z|^2)^2}P[f](0)-\frac{1}{2}\frac{(1-|z|^2)^2}{1+|z|^2}P[f+h](0)\bigg|&\leq & \Big[\frac{2}{\pi}(1-|z|^2)\arctan|z|\Big]\| f+h\|_\infty \\
&+&\frac{2}{\pi}\bigg[(1+\!|z|^2)\arctan |z|+|z|\frac{1-|z|^2}{1+|z|^2}\bigg]\| f\|_\infty.
\end{eqnarray*}

\begin{thm}\label{theo4}
Let $g\in \mathcal{C}(\overline{{\mathbb D}})$, $f$ and $h \in \mathcal{C} (\mathbb{T})$. Suppose that $\Phi\in \mathcal{C}^4({\mathbb D})$ is  satisfying (\ref{ibdp}). Then for all $z\in {\mathbb D}$
\begin{equation}
\| D_\Phi (z)\| \leq  \frac{2+5|z|}{1-|z|^2}(1+|z|^2)\| f\|_\infty
+(\frac{2}{\pi} +|z| )\| f+h\|_\infty +\frac{23}{48}\| g\|_\infty.
\end{equation}
Moreover at $z=0$, we have
\begin{equation}
\| D_\Phi (0)\| \leq  \frac{4}{\pi}\| f\|_\infty
+\frac{2}{\pi}\| f+h\|_\infty +\frac{23}{48}\| g\|_\infty.
\end{equation}
\end{thm}
The classical Schwarz lemma at the boundary is as follows.

\begin{Thm}
Suppose $f:{\mathbb D}\longrightarrow{\mathbb D}$ is a holomorphic function with $f(0)=0$, and further, $f$ is analytic at $z=1$ with $f(1)=1$. Then, the following two conditions hold:
\begin{enumerate}
\item[(a)] $f'(1)\geq 1;$
\item[(b)]  $f'(1)=1$ if and only if $f(z)=z.$
\end{enumerate}
\end{Thm}

Which is known as the Schwarz lemma on the boundary, and its generalizations have important applications in geometric theory of functions (see, \cite{Gol,Lav,Pom}). Among the recent papers devoted to this subject, for example, Burns and Krantz \cite{Bur}, Krantz \cite{Kra}, Liu and Tang \cite{Liu} explored many versions of the Schwarz lemma at the boundary point of holomorphic functions, Dubinin also  applied this latter for algebraic polynomials and rational functions (see \cite{Dub,Dub1}). In the present paper, we refine the Schwarz type
lemma at the boundary for $\Phi$ satisfies (\ref{ibdp}) as an application of Theorem \ref{theo3}.
\begin{thm}\label{theo5}
Suppose that $\Phi\in \mathcal{C}^4({\mathbb D})\cap\mathcal{C}(\overline{{\mathbb D}})$ satisfies (\ref{ibdp}), where $g\in \mathcal{C}(\overline{{\mathbb D}})$ and $f$, $h\in \mathcal{C}(\mathbb{T})$ such that $\|f\|_\infty \leq 1$, and  $\| f+h\|_\infty\leq 1$.
If $\displaystyle\lim_{r\to 1} |\Phi(r\eta)|=1$  for  $\eta\in \mathbb{T}$, then
\begin{equation}
\liminf_{r\to 1}\frac{|\Phi(\eta)-\Phi(r\eta)|}{1-r}\geq 1-\| f+h\|_\infty.\nonumber
\end{equation}
In particular if $\|f+h\|_\infty=0$,  then
\begin{equation*}
\liminf_{r\to 1}\frac{|\Phi(\eta)-\Phi(r\eta)|}{1-r}\geq 1,
\end{equation*}
and this estimate is sharp.
\end{thm}

Denote by $\mathcal{H}({\mathbb D})$ the set of all holomorphic functions $\Phi$ in ${\mathbb D}$ satisfying the standard normalization: $\Phi(0) = \Phi'(0)-1 = 0$. In the early 20th century, Landau \cite{Lan} showed
that there is a constant $r > 0$, independent of elements in $\mathcal{H}({\mathbb D})$, such that $\Phi({\mathbb D})$ contains a disk of radius $r$. Later, the Landau theorem has become an important
tool in geometric function theory.  To establish analogs of the Landau type theorem for more general classes of functions, it is necessary to
restrict our focus on certain subclasses (cf. \cite{Abd}, \cite{Boc}, \cite{Bonk}, \cite{Che1}, \cite{Che2},
\cite{Chep}, \cite{Chev}). \\
\\
For convenience, we make a notational convention: for $g\in\mathcal{C}(\overline{{\mathbb D}})$ and $h \in \mathcal{C}(\mathbb{T})$, let
$\mathcal{BF}_{g,h}(\overline{{\mathbb D}})$ denote the class of all complex-valued functions $\Phi\in \mathcal{C}^4({\mathbb D})\cap \mathcal{C}(\overline{{\mathbb D}})$ satisfying
(\ref{ibdp}) with the normalization $\Phi(0) = J_\Phi(0) - 1 = 0$.\\

We establish the following Landau-type theorem for $\Phi \in \mathcal{BF}_{g,h}(\overline{{\mathbb D}})$. In particular, if $g \equiv 0$, then $\Phi \in \mathcal{BF}_{g,h}(\overline{{\mathbb D}})$  is biharmonic. In this
sense, the following result is a generalization of \cite[Theorem 1]{Abd} and \cite[Theorem 2]{Abd}.
\begin{thm}\label{theo6}
Suppose that $M_1>0$, $M_2>0$ and $M_3>0$ are constants, and suppose that
$\Phi \in \mathcal{BF}_{g,h}(\overline{{\mathbb D}})$ satisfies the following conditions:
$$\sup_{z\in \mathbb{T}} |f(z)|\leq M_1,\ \ \ \sup_{z\in \mathbb{T}} |f(z)+h(z)|\leq M_2,\ \ \text{ and }\sup_{z\in {\mathbb D}} |g(z)|\leq M_3.$$
Then $\Phi$ is univalent in ${\mathbb D}_{r_0}$, and $\Phi({\mathbb D}_{r_0})$ contains a univalent disk ${\mathbb D}_{R_0}$,where $r_0$ satisfies
the following equation:
$$(\frac{4}{\pi} M_1+\frac{2}{\pi}M_2+\frac{23}{48}M_3)\sigma(r_0)=1,$$
with
\begin{eqnarray*}
\sigma(|z|)&:=&(M_1+M_2+\frac{101}{120} M_3)|z|+\frac{2M_2|z|}{\pi}\bigg[\frac{(2-|z|)(1+|z|^2)}{(1-|z|)^2}+|z| \bigg]\\
&+&\frac{4M_1|z|}{\pi(1-|z|)^3}\big(-|z|^3+3|z|^2-3|z|+3\big).
\end{eqnarray*}

and
$$R_0\geq \frac{r_0}{\frac{8}{\pi} M_1+\frac{4}{\pi}M_2+\frac{23}{24}M_3}. $$

\end{thm}
\section{Schwarz and Landau Type Lemmas for $T_2$-harmonic functions}
\subsection{Schwarz Type Lemma for $T_2$-harmonic functions}\hfill

The main purpose of this section is to prove Schwarz type lemma   for $T_2$-harmonic functions.

\begin{proof}[Proof of Theorem \ref{theo1} ]
Let $0\leq r=|z|<1$. As $u$ is a $T_2$-harmonic function, then
$$ u(z)=\frac{1}{2\pi}\int_0^{2\pi}\frac{1}{2}\frac{(1-r^2)^3}{|1-ze^{-i\theta}|^4}u^*(e^{i\theta})d\theta,$$
where $u^*\in L^\infty(\mathbb{T})$. Thus
\begin{eqnarray*}
\bigg|u(z)-\frac{(1-r^2)^3}{(1+r^2)^2}u(0)\bigg|&\leq &\frac{1}{4\pi}\int_0^{2\pi}\bigg|\frac{(1-r^2)^3}{(1+r^2-2r\cos\theta)^2}-\frac{(1-r^2)^3}{(1+r^2)^2}\bigg|d\theta \\
&=&\frac{1}{2\pi}\int_0^{\pi}\bigg|\frac{(1-r^2)^3}{(1+r^2-2r\cos\theta)^2}-\frac{(1-r^2)^3}{(1+r^2)^2}\bigg|d\theta\\
&=&\frac{1}{2\pi}\bigg[\int_0^{\pi/2}\frac{(1-r^2)^3}{(1+r^2-2r\cos\theta)^2}-\frac{(1-r^2)^3}{(1+r^2)^2}d\theta\\
&-&\int_{\pi/2}^{\pi}\frac{(1-r^2)^3}{(1+r^2-2r\cos\theta)^2}-\frac{(1-r^2)^3}{(1+r^2)^2}d\theta\bigg]\\
&=&\frac{1}{2\pi}\bigg[\int_0^{\pi/2}\frac{(1-r^2)^3}{(1+r^2-2r\cos\theta)^2}d\theta -\int_{\pi/2}^{\pi}\frac{(1-r^2)^3}{(1+r^2-2r\cos\theta)^2}d\theta\bigg]\\
&=&\frac{1}{2\pi}\bigg[2J(\pi/2)-J(\pi)\bigg],
\end{eqnarray*}

where

$$ J(\theta):=\int_0^{\theta}\frac{(1-r^2)^3}{(1+r^2-2r\cos\varphi)^2}d\varphi.$$
Easy but tedious computations show that
\begin{lem}\label{lem1}
For $0\leq\theta<\pi$, and $r\in[0,1)$, we have
$$J(\theta)=\frac{2r(1-r^2)\sin\theta}{1+r^2-2r\cos\theta}+2(1+r^2)\arctan\bigg(\frac{(1+r)\tan\theta/2}{1-r}\bigg),$$
and  $\displaystyle J(\pi)= \lim_{\theta \to \pi} J(\theta)= \pi(1+r^2).$
\end{lem}

Then by Lemma \ref{lem1}, and using the fact that $ \displaystyle \arctan(\frac{1+r}{1-r})-\frac{\pi}{4}=\arctan r$,  we have
\begin{eqnarray*}
\bigg|u(z)-\frac{(1-r^2)^3}{(1+r^2)^2}u(0)\bigg|&\leq &\frac{1}{2\pi}\bigg[2J(\pi/2)-J(\pi)\bigg]\\
&=&\frac{1}{2\pi}\bigg[4\frac{r(1-r^2)}{1+r^2}+4(1+r^2)\arctan(\frac{1+r}{1-r})-\pi(1+r^2)\bigg]\\
&=&\frac{2}{\pi}\bigg[\frac{r(1-r^2)}{1+r^2}+(1+r^2)\arctan r\bigg].
\end{eqnarray*}
\end{proof}

To prove Theorem \ref{theo2}, we need the following lemma

\begin{Lem}{\cite{Li}}\label{theod}
For any $z\in {\mathbb D}$, we have
$$I_\alpha(z):=\frac{1}{2\pi}\int^{2\pi}_0\frac{d\theta}{|1-ze^{i\theta}|^{2\alpha}}=\sum^{+\infty}_{n=0}\bigg(\frac{\Gamma(n+\alpha)}{n!\Gamma(\alpha)}\bigg)^2|z|^{2n},$$
where $\alpha >0$ and $\Gamma$ denotes the Gamma function.
\end{Lem}
Thus
\begin{equation}\label{j2}
I_2(z)=\sum_{n=0}^\infty (n+1)^2|z|^{2n}=\frac{1+|z|^2}{(1-|z|^2)^3}.
\end{equation}

\begin{proof}[Proof of Theorem \ref{theo2}]

As $u$  is a bounded $T_2$-harmonic function, then
$$ u(z)= \frac{1}{2\pi}\int_0^{2\pi} K_2(ze^{-i\theta})u^*(e^{i\theta})d\theta,$$
where $u^*\in L^\infty(\mathbb{T})$. Elementary computations show that

\begin{eqnarray}\label{uz0}
u_z(z)&=&\frac{1}{2\pi}\int^{2\pi}_0 [K_2(ze^{-i\theta})]_z u^*(e^{i\theta})d\theta\nonumber \\
&=&
 \frac{1}{2\pi}\int^{2\pi}_0\frac{(1-|z|^2)^2[2e^{-i\theta}(1-|z|^2)-3\overline{z}(1-ze^{-i\theta})]}{2(1-\overline{z}e^{i\theta})^2(1-ze^{-i\theta})^3}u^*(e^{i\theta})d\theta.
\end{eqnarray}
Hence
\begin{eqnarray*}
|u_z(z)|
&\leq & \frac{(1-|z|^2)^2}{2}\bigg[\frac{1}{2\pi}\int^{2\pi}_0\bigg|\frac{2e^{-i\theta}(1-|z|^2)}{(1-\overline{z}e^{i\theta})^2(1-ze^{-i\theta})^3}\bigg|d\theta +\frac{1}{2\pi}\int^{2\pi}_0\bigg|\frac{3\overline{z}(1-ze^{-i\theta})}{(1-\overline{z}e^{i\theta})^2(1-ze^{-i\theta})^3}\bigg|d\theta\bigg]\\
&\leq & \frac{(1-|z|^2)^2}{2}\bigg[2(1-|z|^2)I_{5/2}(z)+3|z|I_2(z)\bigg]\\
&\leq & \frac{(1-|z|^2)^2}{2}(2+5|z|)I_2(z).
\end{eqnarray*}
The last inequality follows from the following estimate  $\displaystyle I_{5/2}(z) \leq \frac{I_2(z)}{1-|z|}$.\\
Using the explicit expression of $I_2$,  see (\ref{j2}),  we obtain

\begin{equation}
 |u_z(z)|\leq  \frac{(2+5|z|)(1+|z|^2)}{2(1-|z|^2)}.\nonumber
  \end{equation}
Similarly,
\begin{equation}
|u_{\overline{z}}(z)| \leq  \frac{(2+5|z|)(1+|z|^2)}{2(1-|z|^2)}.\nonumber
\end{equation}
Thus
$$
	\| D_u(z)\| =|u_z(z)|+|u_{\overline{z}}(z)|\leq  \frac{(2+5|z|)(1+|z|^2)}{1-|z|^2}.$$

Next let us show the estimate (\ref{grad0}). From Theorem \ref{theo1}, we deduce that near $0$
\begin{equation}\label{gradient0}
 |u(z)-u(0)| \leq \frac{4}{\pi}|z|+O(|z|^2).
\end{equation}
Indeed, near 0,  $$  (1+r^2)\arctan r+ r\frac{1-r^2}{1+r^2}=2r+O(r^2), \text{ and }$$
$$\frac{(1-r^2)^3}{(1+r^2)^2}= 1+O(r^2).$$
Hence from (\ref{gradient0}), we get
$$\|D_u(0)\| \leq \frac{4}{\pi}.$$

To show that the last estimate is sharp. Let us consider the $T_2$-harmonic mapping defined by
$$U(z)=K_2[\chi_{{\mathbb T}^+}- \chi_{{\mathbb T}^-}](z),$$
where $\chi_{{\mathbb T}^+}$ (resp. $\chi_{{\mathbb T}^-}$) denotes the characteristic function of the upper (resp. lower) unit circle ${\mathbb T}$.

By  (\ref{uz0}), we have
$$U_z(0)= \frac{1}{2\pi}  \int_0^{2\pi} e^{-i\theta}
(\chi_{{\mathbb T}^+}- \chi_{{\mathbb T}^-})(\theta) d\theta=\frac{1}{2\pi}  \int_0^{\pi} e^{-i\theta}d\theta-\frac{1}{2\pi}  \int_\pi^{2\pi} e^{-i\theta}d\theta=-\frac{2i}{\pi}.$$
Hence
$$|\nabla U(0)|=2|U_z(0)|=\frac{4}{\pi}.$$
\end{proof}

\subsection{Landau type theorem for $T_2$- harmonic functions}\hfill

We will use the following theorem provides some estimates on the coefficients of $T_\alpha$-harmonic mappings.

\begin{Thm}\cite{Chev}
	For $\alpha>-1$, let $u\in \mathcal{C}^2({\mathbb D})$ be a $T_\alpha$-harmonic function with the series expansion of the form (\ref{expansion}) and $\sup_{z\in {\mathbb D}}|u(z)| \leq M$, where $M>0$. Then, for $k\in \{1,2,\ldots\}$,
	\begin{equation}\label{2.6}
	\left|c_kF(-\frac{\alpha}{2},k-\frac{\alpha}{2};k+1;1)\right|+\left|c_{-k}F(-\frac{\alpha}{2},k-\frac{\alpha}{2};k+1;1)\right| \leq \frac{4M}{\pi},
	\end{equation}
	and
	\begin{equation}\label{c0}
	|c_0F(-\frac{\alpha}{2},-\frac{\alpha}{2};1;1)|\leq M.
	\end{equation}
\end{Thm}

\begin{proof}[Proof of Theorem \ref{landau}]

	As $u$ is $T_2$-harmonic function on the unit disk with $u(0)=0$, then by plugging $\alpha=2$ in  (\ref{expansion}), we have $c_0=0$ and
	$$ u(z)=\sum_{k=1}^\infty c_k F(-1,k-1;k+1;|z|^2)z^k+
	\sum_{k=1}^\infty c_{-k} F(-1,k-1;k+1;|z|^2)\overline{z}^k.$$

An elementary computation shows that the function $\displaystyle F(-1,k-1;k+1;\bigcdot)$ is given by
\begin{equation}\label{26}
F(-1,k-1;k+1;w)=1-\frac{k-1}{k+1}w, \quad w\in [0,1).\\
\end{equation}

Clearly, the function $w \mapsto F(-1,k-1;k+1;w)$ is decreasing on $[0,1]$ for $k \geq 1$. Thus

\begin{equation} \label{2.7}
\frac{2}{k+1} \leq F(-1,k-1;k+1;w) \leq 1 \quad  \text { for } k\geq 1 \text{ and } w\in [0,1].\\
\end{equation}

Combining  (\ref{2.6})  for $\alpha=2$ and (\ref{2.7}), we obtain

\begin{equation}\label{ck}
|c_k|+|c_{-k}|\leq \frac{2M}{\pi}(k+1) \quad \text{ for } k\geq 1.
\end{equation}

Using (\ref{26}), we see that the mapping $u$ is given by
\begin{equation}
u(z)=\sum_{k=1}^\infty c_kz^k+c_{-k}\overline{z}^k -\sum_{k=1}^\infty \frac{k-1}{k+1}\bigg(c_kz^{k+1}\overline{z}+c_{-k}\overline{z}^{k+1}z\bigg).\nonumber
\end{equation}

Therefore

\begin{equation}\label{uz}
u_z(z)-u_z(0)= \sum_{k=2}^\infty kc_kz^{k-1}-\sum_{k=2}^\infty \frac{k-1}{k+1}\bigg( (k+1)c_kz^k\overline{z}+c_{-k}\overline{z}^{k+1} \bigg).
\end{equation}

Similarly

\begin{equation}\label{uzbar}
u_{\overline{z}}(z)-u_{\overline{z}}(0)= \sum_{k=2}^\infty kc_{-k}\overline{z}^{k-1}-\sum_{k=2}^\infty \frac{k-1}{k+1}\bigg( (k+1)c_{-k}\overline{z}^k z+c_{k}z^{k+1} \bigg).
\end{equation}

Applying (\ref{uz}), (\ref{uzbar}) and (\ref{ck}), we obtain
\begin{eqnarray}
& & |u_z(z)-u_z(0) |+| u_{\overline{z}}(z)-u_{\overline{z}}(0)|\nonumber \\
&\leq& \sum_{k=2}^\infty  k \left(|c_k|+|c_{-k}|\right)|z|^{k-1}  \nonumber\\
&+&  \sum_{k=2}^\infty (k-1)\left(|c_k|+|c_{-k}| \right)|z|^{k+1}\nonumber\\
&+&  \sum_{k=2}^\infty \frac{k-1}{k+1}\left(|c_k|+|c_{-k}| \right)|z|^{k+1}\nonumber\\
&\leq& \frac{2M}{\pi}\bigg(\sum_{k=2}^\infty k(k+1)|z|^{k-1} +\sum_{k=2}^\infty (k-1)(k+1)|z|^{k+1}+\sum_{k=2}^\infty (k-1)|z|^{k+1}  \bigg)\nonumber\\
&=& \frac{2M|z|}{\pi(1-|z|)^2}\bigg( 2\frac {|z|^2-3|z|+3}{1-|z|}+\frac{|z|^2(3-|z|)}{1-|z|}+|z|^2\bigg).\nonumber
\end{eqnarray}

That is
\begin{equation}\label{uzzz}
 |u_z(z)-u_z(0) |+| u_{\overline{z}}(z)-u_{\overline{z}}(0)|
\leq \frac{4M|z|}{\pi(1-|z|)^3}\big(-|z|^3+3|z|^2-3|z|+3\big).
\end{equation}
Moreover, the RHS of (\ref{uzzz}) is strictly increasing as a function of $|z|$.\\

Applying Theorem \ref{theo2} (\ref{grad0}), we get
\begin{equation}
1=J_u(0)=\|D_u(0)\| \lambda(D_u(0)) \leq \frac{4M}{\pi}\lambda(D_u(0)),\nonumber
\end{equation}
which gives that
\begin{equation}\label{ldu}
 \lambda\left(D_u(0)\right) \geq \frac{\pi}{4M}.
\end{equation}
We will show that $u$ is univalent in ${\mathbb D}_{r_0}$, where $r_0$ satisfies the following equation
	$$\frac{\pi}{4M}-\frac{4}{\pi}\frac{M r_0}{(1-r_0)^3}\big(-r_0^3+3r_0^2-3r_0+3 \big)=0.$$
Indeed, let $z_1$ and $z_2$ be two distinct points in $D_{r_0}$ and let $[z_1,z_2]$ denote the line segment from $z_1$ to $z_2$.\\

By (\ref{uzzz}), (\ref{ldu}) we have

\begin{eqnarray*}
	|u(z_1)-u(z_2)|&=&\bigg|\int_{[z_1,z_2]}u_z(z)\,dz+u_{\overline{z}}(z)\,d\overline{z}\bigg|\\
	&\geq & \bigg|\int_{[z_1,z_2]}u_z(0)\,dz+u_{\overline{z}}(0)\,d\overline{z}\bigg|\\
	&-&\bigg|\int_{[z_1,z_2]}(u_z(z)-u_z(0))\,dz+(u_{\overline{z}}(z)-u_{\overline{z}}(0))\,d\overline{z}\bigg|\\
	&\geq &\lambda( D_u (0))|z_1- z_2|\\
	&-&\int_{[z_1,z_2]}(|u_z(z)-u_z(0)|+|u_{\overline{z}}(z)-u_{\overline{z}}(0)|)\,|dz|\\
	&>& |z_2-z_1|\bigg\{ \frac{\pi}{4M}-\frac{4}{\pi}\frac{M r_0}{(1-r_0)^3}\big(-r_0^3+3r_0^2-3r_0+3 \big)\bigg\}\\
	&=&0.
\end{eqnarray*}

This implies that $u$ is univalent on $D_{r_0}$.\\

let $\xi=r_0e^{i\theta}\in \partial{\mathbb D}_{r_0}$. Then we infer from (\ref{uzzz}) that
\begin{eqnarray*}
	|u(\xi)-u(0)|&\geq  &\lambda( D_\Phi (0))r_0
	-\int_{[0,\xi]}(|u_z(z)-u_z(0)|+|u_{\overline{z}}(z)-u_{\overline{z}}(0)|)|dz|\\
	&\geq& \frac{\pi r_0}{4M}-\frac{4M}{\pi}\frac{r_0^2}{(1-r_0)^3}\int_0^1(  -r_0^3t^4+3r_0^2t^3-3r_0t^2+3t )dt\\
	&=& \frac{\pi r_0}{4M}-\frac{4M}{\pi}\frac{r_0^2}{(1-r_0)^3}\bigg(-\frac{r_0^3}{5}+\frac{3}{4}r_0^2-r_0+\frac{3}{2} \bigg)\\
	&=& \frac{4  Mr_0^2}{\pi(1-r_0)^3}\bigg(-r_0^3+3r_0^2-3r_0+3 -(-\frac{r_0^3}{5}+\frac{3}{4}r_0^2-r_0+\frac{3}{2})\bigg)\\
	&=& \frac{4 M r_0^2}{\pi(1-r_0)^3}\big(-\frac{4}{5}r_0^3+\frac{9}{4}r_0^2-2r_0+\frac{3}{2}\big).\\
\end{eqnarray*}
Hence $u(D_{r_0})$ contains an univalent disk $D_{R_0}$ with
$$R_0 \geq  \frac{4M r_0^2}{\pi(1-r_0)^3}\big(-\frac{4}{5}r_0^3+\frac{9}{4}r_0^2-2r_0+\frac{3}{2}\big). $$
\end{proof}

 \section{Schwarz-Type Lemmas for Solutions to Inhomogeneous Biharmonic Equations}
\begin{proof}[Proof of Theorem \ref{theo3}]

The solution of (\ref{ibdp}) can be written in the following form
\begin{equation*}
\Phi(z)=\frac{1}{2}(1-|z|^2)P[f+h](z)+K_2[f](z)-G[g](z).\label{phi1}
\end{equation*}
As $z\longmapsto K_2[f](z) $ is $T_2$-harmonic function, then by Theorem \ref{theo1}, we have
\begin{equation}\label{31}
\bigg|K_2[f](z)-\frac{(1-|z|^2)^3}{(1+|z|^2)^2}K_2[f](0)\bigg| \leq  \frac{2}{\pi}\bigg[(1+|z|^2)\arctan |z|+ \frac{|z|(1-|z|^2)}{1+|z|^2}\bigg]\| f\|_\infty .
\end{equation}
On the other hand, using  the  estimate (\ref{harm}) for the harmonic mapping $P[f+h]$, we get
\begin{equation}\label{32}
\bigg|P[f+h](z)-\frac{1-|z|^2}{1+|z|^2}P[f+h](0)\bigg|\leq \frac{4}{\pi}\arctan|z|\,\| f+h\|_\infty.
\end{equation}

\noindent Using \cite[inequality 2.3]{Chel}, we obtain
\begin{equation}\label{33}
|G[g](z)| \leq \frac{(1-|z|^2)^2}{64}\| g\|_\infty .
\end{equation}

\noindent Finally as $\displaystyle  K_2[f](0)=\frac{1}{2}P[f](0)$, then if follows from (\ref{31}--\ref{33}) that
\begin{eqnarray*}
\bigg| \Phi(z)-\frac{1}{2}\frac{(1-|z|^2)^3}{(1+|z|^2)^2}P[f](0)-\frac{1}{2}\frac{(1-|z|^2)^2}{1+|z|^2}P[f+h](0)\bigg|&\leq &\frac{2}{\pi}(1-|z|^2)\arctan|z|\| f+h\|_\infty\nonumber\\
&+&\frac{2}{\pi}\Bigg[(|z|^2+1)\arctan |z|+\frac{|z|(1-|z|^2)}{1+|z|^2}\Bigg]\| f\|_\infty\nonumber\\
&+&\frac{(1-|z|^2)^2}{64}\|g\|_\infty.
\end{eqnarray*}
Hence, the proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{theo4}]
	The solution of (\ref{ibdp}) can be written in the following form
	 \begin{eqnarray*}
 \Phi(z)&=&\frac{1}{2} (1-|z|^2)P[f+h](z)+ K_2[f](z) - G[g](z).
\end{eqnarray*}
Thus
\begin{eqnarray}
 \Phi_z(z)&=&\frac{1}{2} \bigg[(1-|z|^2)[P[f+h](z)]_z-\overline{z} P[f+h](z)\bigg]+ K_2[f]_z(z)-G[g]_z(z), \nonumber
\end{eqnarray}
\begin{eqnarray}
 \Phi_{\overline{z}}(z)&=&\frac{1}{2} \bigg[(1-|z|^2)[P[f+h](z)]_{\overline{z}}-z P[f+h](z)\bigg]+K_2[f]_{\overline{z}}(z)-G[g]_{\overline{z}}(z).\nonumber
 \end{eqnarray}
 Therefore
$$ \| D_\Phi (z)\| \leq  \frac{1}{2}(1-|z|^2)\| D_{P[f+h]} (z)\| +|z||P[f+h](z)| +\| D_{K_2[f]}(z)\|+\| D_{G[g]}(z)\|.$$

By Colonna \cite{Col}, we have

\begin{equation}\label{colo}
\| D_{P[f+h]}(z) \| \leq \frac{4}{\pi} \frac{1}{1-|z|^2}\|f+h\|_\infty.
\end{equation}
It follows from  \cite[Lemma 2.5]{Li},  that
\begin{equation}\label{25}
\| D_{G[g]}(z)\| \leq \frac{23}{48}\|g\|_\infty,
\end{equation}
since
$$ \displaystyle \int_{\mathbb D} |G_z(z,\omega) g(\omega)|dA(\omega) \leq \frac{23}{6} \|g\|_\infty \text{ and } \int_{\mathbb D} |G_{\overline{z}}(z,\omega) g(\omega)|dA(\omega) \leq \frac{23}{6} \|g\|_\infty.$$
Therefore, combining  (\ref{colo})-(\ref{25}), we obtain
\begin{eqnarray*}
\| D_\Phi (z)\| &\leq & \frac{1}{2}(1-|z|^2)\| D_{P[f+h]} (z)\| +|z||P[f+h](z)| +\| D_{K_2[f]}(z)\|+\| D_{G[g]}(z)\|\\
&\leq &\frac{2}{\pi}\| P[f+h]\|_\infty +|z| \| f+h\|_\infty + \frac{(2+5|z|)(1+|z|^2)}{1-|z|^2} \| f\|_\infty+\frac{23}{48}\| g\|_\infty \\
&\leq &(\frac{2}{\pi} +|z|)\| f+h\|_\infty +\frac{(2+5|z|)(1+|z|^2)}{1-|z|^2} \|f\|_\infty+\frac{23}{48}\| g\|_\infty.
\end{eqnarray*}
\end{proof}

\begin{proof}[Proof of  Theorem \ref{theo5}]
Suppose that $|z|=r$,
it follows from Theorem \ref{theo3} that
\begin{eqnarray*}
|\Phi(\eta)-\Phi(r\eta)|&\geq &1-\frac{1}{2}(1-r^2) \| f+h\|_\infty -\frac{2}{\pi}\bigg[(r^2+1)\arctan r+\frac{r(1-r^2)}{1+r^2}\bigg]\\
&-&\frac{\| g\|_\infty (1-r^2)^2}{64}-\frac{1}{2}\frac{(1-r^2)^3}{(1+r^2)^2}|P[f](0)| -\frac{1}{2}\frac{(1-r^2)^2}{1+r^2}|P[f+h](0)|
\end{eqnarray*}
Divide by $1-r$ and used the Hospital rule, we obtain
\begin{eqnarray*}
\liminf_{r\to 1}\frac{|\Phi(\eta)-\Phi(r\eta)|}{1-r}&\geq &\lim_{r\to 1}\frac{1-\frac{2}{\pi}(r^2+1)\arctan r}{1-r}-\lim_{r\to 1}\frac{2}{\pi}\frac{r(1-r^2)}{(1-r)(1+r^2)}\\
&-&\frac{1}{2}\lim_{r\to 1}(1+r)\| f+h\|_\infty \\
&=&[\varphi'(r)]_{r=1}-\frac{2}{\pi}-\| f+h\|_\infty,
\end{eqnarray*}
where $$\varphi(r)=\frac{2}{\pi}(r^2+1)\arctan r.$$
Then $$\varphi'(r)=\frac{2}{\pi}\big[2r\arctan r+1\big], \text{ so  } \varphi'(1)=1+\frac{2}{\pi}.$$
Hence
$$\liminf_{r\longrightarrow 1}\frac{|\Phi(\eta)-\Phi(r\eta)|}{1-r}\geq 1-\| f+h\|_\infty.$$
\end{proof}

\section{A Landau-Type Theorem for Solutions to Inhomogeneous Biharmonic Equations }

 First, let us recall the following result.\\
\\
\begin{Thm}[\cite{Chep}, Lemma 1]
Suppose $f$ is a harmonic mapping of ${\mathbb D}$ into ${\mathbb C}$ such that $|f(z)|\leq M$ for all $z\in {\mathbb D}$ and
$$f(z)=\sum_{n=0}^\infty a_n z^n+\sum_{n=1}^\infty \overline{b}_n \overline{z}^n.$$
Then $|a_0|\leq M$ and for all $n\geq 1,$
$$|a_n|+|b_n|\leq \frac{4M}{\pi}.$$
\end{Thm}

\begin{proof}[Proof of Theorem \ref{theo6}]
	The solution of (\ref{ibdp}) can be written in the following form
$$\Phi(z)=H_0[f+h](z)+K_2[f](z)-G[g](z). $$

where
\begin{equation}\label{h0}
H_0[f+h](z)= \frac{1}{2} (1-|z|^2)P[f+h](z).
\end{equation}
Since $P[f+h](z)$ is harmonic in ${\mathbb D}$,  we have
$$H_0[f+h](z)=\sum_{n=0}^\infty a_n z^n+\sum_{n=1}^\infty \overline{b}_n \overline{z}^n$$
Since $|P[f+h](z)|\leq M_2$ for all $z\in {\mathbb D}$, by Theorem F, we have
\begin{equation} \label{est}
|a_n|+|b_n|\leq \frac{4M_2}{\pi}  \quad \text{ for } n\geq 1.
\end{equation}
By (\ref{h0}) and (\ref{est}),  we have
$$[H_0[f+h](z)]_z=\frac{1}{2}(1-|z|^2)P[f+h]_z(z)-\frac{1}{2}\overline{z}P[f+h](z),$$
and
$$[H_0[f+h](z)]_{\overline{z}}=\frac{1}{2}(1-|z|^2)P[f+h]_{\overline{z}}(z)-\frac{1}{2}z P[f+h](z).$$
Thus
\begin{eqnarray}
& & \bigg|H_0[f+h]_z(z)-H_0[f+h]_z(0)\bigg|+\bigg|H_0[f+h]_{\overline{z}}(z)-H_0[f+h]_{\overline{z}}(0)\bigg|\nonumber  \\
&\leq&|z| |P[f+h](z)|+\frac{1}{2} \bigg( \big| P[f+h]_z(z)-P[f+h]_z(0)\big|+P[f+h]_{\overline{z}}(z)-P[f+h]_{\overline{z}}(0)\big| \bigg)\nonumber\\
&+&\frac{1}{2}|z|^2\bigg( \big| P[f+h]_z(z)\big|+\big|P[f+h]_{\overline{z}}(z)\big| \bigg)\nonumber\\
&\leq& M_2|z|+\frac{1}{2}\sum_{n\geq 2}n(|a_n|+|b_n|) |z|^{n-1} +\frac{1}{2}|z|^2 \sum_{n\geq 1}n(|a_n|+|b_n|) |z|^{n-1}  \nonumber\\
&\leq& M_2|z| +\frac{1}{2}(1+|z|^2) \sum_{n\geq 2}n(|a_n|+|b_n|) |z|^{n-1}+\frac{1}{2}|z|^2(|a_1|+|b_1|) \nonumber\\
&\leq & M_2|z|+\frac{2M_2|z|}{\pi}\bigg[\frac{(2-|z|)(1+|z|^2)}{(1-|z|)^2}+|z| \bigg].\label{h0z}
\end{eqnarray}

Since $K_2$ is $T_2$-harmonic, then
	$$ K_2(z)=\sum_{k=0}^\infty c_k F(-1,k-1;k+1;|z|^2)z^k+
\sum_{k=1}^\infty c_{-k} F(-1,k-1;k+1;|z|^2)\overline{z}^k.$$
Let us denote $$K_2^0[f](z):=K_2(z)-c_0F(-1,-1;1;|z|^2)=K_2[f](z)-c_0(1+|z|^2). $$
Hence
$$ K_2[f]=K_2^0[f](z)+c_0(1+|z|^2).$$

\begin{eqnarray}
& & \bigg|K_2[f]_z(z)-K_2[f]_z(0)\bigg|+\bigg|K_2[f]_{\overline{z}}(z)-K_2[f]_{\overline{z}}(0)\bigg|\nonumber  \\
&\leq&\bigg|K_2^0[f]_z(z)-K_2^0[f]_z(0)\bigg|+\bigg|K_2^0[f]_{\overline{z}}(z)-K_2^0[f]_{\overline{z}}(0)\bigg|+2|c_0||z| \nonumber
\end{eqnarray}
By (\ref{c0}), we have
\begin{equation}
2|c_0| \leq M_1. \nonumber
\end{equation}
On the other hand, as $K_2^0(f)$ is a $T_2$-harmonic function with $K_2^0(0)=0$, it yields
\begin{equation} |K_2^0[f]_z(z)-K_2^0[f]_z(0) |+| K_2^0[f]_{\overline{z}}(z)-K_2^0[f]_{\overline{z}}(0)|
\leq \frac{4M_1|z|}{\pi(1-|z|)^3}\big(-|z|^3+3|z|^2-3|z|+3\big).\nonumber
\end{equation}

Thus
\begin{equation}\label{kef}
\bigg|K_2[f]_z(z)-K_2[f]_z(0)\bigg|+\bigg|K_2[f]_{\overline{z}}(z)-K_2[f]_{\overline{z}}(0)\bigg|\leq  \frac{4M_1|z|}{\pi(1-|z|)^3}\big(-|z|^3+3|z|^2-3|z|+3\big)+M_1|z|.
\end{equation}
Let
$$\psi_1(z)=\bigg|\frac{1}{16\pi}\int_{\mathbb D} g(\omega)(G_z(z,\omega)-G_z(0,\omega))dA(\omega)\bigg|$$
and
$$\psi_2(z)=\bigg|\frac{1}{16\pi}\int_{\mathbb D} g(\omega)(G_{\overline{z}}(z,\omega)-G_{\overline{z}}(0,\omega))dA(\omega)\bigg|.$$

Then by \cite[Inequality $(3.6)$]{Chel}, we have
\begin{equation}
\psi_1(z)\leq \bigg(\frac{1-|z|^2}{16}+\frac{43}{120}\bigg)\| g\|_\infty |z|,\label{psi1}
\end{equation}
and
$$\psi_2(z)\leq \bigg(\frac{1-|z|^2}{16}+\frac{43}{120}\bigg)\| g\|_\infty |z|.$$

Now, it follows from  (\ref{h0z}), (\ref{kef}) and (\ref{psi1})  that

\begin{eqnarray}
& &|\Phi_z (z)-\Phi_z (0)|+|\Phi_{\overline{z}} (z)-\Phi_{\overline{z}} (0)|\nonumber\\
&\leq & M_2|z|+\frac{2M_2|z|}{\pi}\bigg[\frac{(2-|z|)(1+|z|^2)}{(1-|z|)^2}+|z| \bigg]+\nonumber\\
&+& \frac{4M_1|z|}{\pi(1-|z|)^3}\big(-|z|^3+3|z|^2-3|z|+3\big)+M_1|z|+\psi_1(z)+\psi_2(z)\nonumber\\
&\leq & \sigma(z),\label{sigg}
\end{eqnarray}
where
$$\sigma(|z|):=(M_1+M_2+\frac{101}{120} M_3)|z|+\frac{2M_2|z|}{\pi}\bigg[\frac{(2-|z|)(1+|z|^2)}{(1-|z|)^2}+|z| \bigg]+\frac{4M_1|z|}{\pi(1-|z|)^3}\big(-|z|^3+3|z|^2-3|z|+3\big)$$
Remark that not only $\sigma(|z|)$ is increasing but also  $\displaystyle\frac{\sigma(|z|)}{|z|}$ is  increasing with respect to $|z|$ in $[0,1)$.\\

By Theorem \ref{theo4}, we obtain that
$$1=J_\Phi (0)=\| D_\Phi (0)\|\lambda( D_\Phi (0))\leq \lambda( D_\Phi (0))\bigg(\frac{4}{\pi}M_1+\frac{2}{\pi}M_2+\frac{23}{48}M_3\bigg)$$
yields
\begin{equation}\displaystyle
\lambda( D_\Phi (0))\geq \frac{1}{\frac{4}{\pi}M_1+\frac{2}{\pi}M_2+\frac{23}{48}M_3}.\label{lambd}
\end{equation}

We will prove $\Phi$ is univalent in ${\mathbb D}_{r_0}$, where $r_0$ satisfies the following
equation:
\begin{equation}
(\frac{4}{\pi} M_1+\frac{2}{\pi}M_2+\frac{23}{48}M_3)\sigma(r_0)=1.\label{eq0}
\end{equation}
We choose two points $z_1\neq z_2 \in {\mathbb D}_{r_0}$,
and by  (\ref{sigg})-(\ref{eq0}), we obtain
\begin{eqnarray*}
|\Phi(z_1)-\Phi(z_2)|&=&\bigg|\int_{[z_1,z_2]}\Phi_z(z)dz+\Phi_{\overline{z}}(z)d\overline{z}\bigg|\\
&\geq & \bigg|\int_{[z_1,z_2]}\Phi_z(0)dz+\Phi_{\overline{z}}(0)d\overline{z}\bigg|\\
&-&\bigg|\int_{[z_1,z_2]}(\Phi_z(z)-\Phi_z(0))dz+(\Phi_{\overline{z}}(z)-\Phi_{\overline{z}}(0))d\overline{z}\bigg|\\
&\geq &\lambda( D_\Phi (0))|z_1- z_2|\\
&-&\int_{[z_1,z_2]}(|\Phi_z(z)-\Phi_z(0)|+|\Phi_{\overline{z}}(z)-\Phi_{\overline{z}}(0)|)|dz|\\
&>& |z_1- z_2|\bigg(\frac{1}{\frac{4}{\pi} M_1+\frac{2}{\pi}M_2+\frac{23}{48}M_3}-\sigma(r_0)\bigg)\\
&=&0.
\end{eqnarray*}
Thus, from the arbitrariness of $z_1$ and $z_2$, the univalence of $\Phi$ follows.\\

Now, we will prove $\Phi({\mathbb D}_{r_0})$ contains an univalent disk ${\mathbb D}_{R_0}$
To reach this goal, let $\xi=r_0e^{i\theta}\in \partial{\mathbb D}_{r_0}$. As the mapping $\frac{\sigma(|z|)}{|z|}$ is increasing, we deduce

\begin{equation}
\int_{[0,\xi]} \sigma(|z|)|dz| \leq \frac{\sigma(r_0)r_0}{2}.\nonumber
\end{equation}
Therefore,

\begin{eqnarray*}
|\Phi(\xi)-\Phi(0)|&=&\bigg|\int_{[0,\xi]}\Phi_z(z)dz+\Phi_{\overline{z}}(z)d\overline{z}\bigg|\\
&\geq & \bigg|\int_{[0,\xi]}\Phi_z(0)dz+\Phi_{\overline{z}}(0)d\overline{z}\bigg|\\
&-&\bigg|\int_{[0,\xi]}(\Phi_z(z)-\Phi_z(0))dz+(\Phi_{\overline{z}}(z)-\Phi_{\overline{z}}(0))d\overline{z}\bigg|\\
&\geq &\lambda( D_\Phi (0))r_0 -\int_{[0,\xi]}\sigma(z)|dz|\\
&\geq&  \sigma(r_0)r_0-\frac{\sigma(r_0)r_0}{2}\\
&=&\frac{\sigma(r_0)r_0}{2}\\
&=&\frac{r_0}{\frac{8}{\pi} M_1+\frac{4}{\pi}M_2+\frac{23}{24}M_3}.
\end{eqnarray*}

Hence  $\Phi({\mathbb D}_{r_0})$ contains an univalent disk ${\mathbb D}_{R_0}$ with the
radius $R_0$ satisfying
$$R_0\geq \frac{r_0}{\frac{8}{\pi} M_1+\frac{4}{\pi}M_2+\frac{23}{24}M_3}. $$
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Elżbieta Zakrzewska z domu Zalewska (ur. 25 listopada 1962 w Jeleniej Górze) – polska polityk i samorządowiec, posłanka na Sejm VI kadencji, od 2018 burmistrz Kowar.

Życiorys
Córka Augusta i Anny. Ukończyła studia ekonomiczne na Wydziale Ekonomiki i Organizacji Produkcji Akademii Ekonomicznej we Wrocławiu, podyplomowo kształciła się w zakresie zarządzania. W latach 1990–1998 przez dwie kadencje pełniła funkcję wójta Mysłakowic, następnie przez dwa lata była zastępczynią dyrektora w Dolnośląskiej Regionalnej Kasie Chorych Oddziału Jelenia Góra. Od 2000 do 2008 zajmowała stanowisko dyrektora Samodzielnego Publicznego Zespołu Opieki Zdrowotnej w Bogatyni, następnie została dyrektorem Szpitala Specjalistycznego MSWiA w Jeleniej Górze.

W 2006 w wyborach samorządowych wybrano ją na radną sejmiku dolnośląskiego z ramienia Lewicy i Demokratów. Również z listy LiD kandydowała bezskutecznie w wyborach w 2007 do Sejmu w okręgu legnickim. Pełniła funkcję wiceprzewodniczącej rady krajowej Unii Pracy. W sejmiku w 2008 przystąpiła do klubu radnych Platformy Obywatelskiej.

Po śmierci Jerzego Szmajdzińskiego w katastrofie polskiego Tu-154 w Smoleńsku uzyskała uprawnienie do objęcia po nim mandatu, na co wyraziła zgodę, deklarując akces do klubu poselskiego Lewica (we wrześniu 2010 klub został przemianowany na klub Sojuszu Lewicy Demokratycznej). 7 maja 2010 złożyła ślubowanie. W latach 2010–2013 była jedną z wiceprzewodniczących UP, następnie ponownie została jedną z wiceszefów rady krajowej tej partii. W 2011 nie uzyskała poselskiej reelekcji, startując z listy SLD (w ramach porozumienia UP z tą partią). W 2014 bez powodzenia kandydowała do sejmiku województwa z listy koalicyjnego komitetu SLD Lewica Razem.

W marcu 2017, po rezygnacji Marka Pola, została przewodniczącą rady krajowej UP. W 2018 została wybrana na stanowisko burmistrza miasta Kowary.

Odznaczona Srebrnym Krzyżem Zasługi (1998).

Przypisy

Bibliografia

Absolwenci Uniwersytetu Ekonomicznego we Wrocławiu
Ludzie urodzeni w Jeleniej Górze
Ludzie związani z Bogatynią
Odznaczeni Srebrnym Krzyżem Zasługi
Politycy Unii Pracy
Polscy burmistrzowie
Polscy wójtowie
Radni sejmiku województwa dolnośląskiego
Posłowie z okręgu Legnica
Urodzeni w 1962

redpajama_set_name: RedPajamaWikipedia


When Susan Mills began taking piano lessons in the second grade, her father would play the role of chauffeur, waiting in the car while she honed her craft. It was a weekly ritual that would continue until Susan was old enough to drive herself, but her father didn't mind. Both he and Susan's mother were music lovers as well. Her father played trombone in a band, while her mother had only to hear a tune once before picking it out on the piano herself.
"Both of them had perfect pitch, I think," said Susan, laughing.
In 2014, Susan moved from Washington D.C. to Shepherdstown following her retirement from a 46-year career as an award-winning producer with CBS News and The MacNeil/Lehrer News Hour (PBS).
While Susan continued to play piano through early adulthood, even serving as accompanist for her high school's senior choir, her interest in music had always included singing as well. Upon moving to Shepherdstown, she became intrigued by the Masterworks Chorale, the Eastern Panhandle's premiere vocal ensemble comprised of Shepherd University vocal students and members from the surrounding community. She auditioned and joined the ensemble in 2015.
Susan's humble observations have led her to establish the Susan L. Mills Music Scholarship in Honor of Camille and James Mills, acknowledging both her parents' support of her own musical interests, as well as her admiration for Shepherd music students and faculty. Designated in support of music majors, the deferred gift will be endowed through a bequest. As a result, Susan joins the Joseph P. McMurran Society, a special group of donors who have chosen to share their lifelong financial achievements with Shepherd University through estate and other planned gifts.


Enforcement agencies have a relatively good understanding of how to measure the loss of price competition caused by merger. However, when firms compete in multiple dimensions, merger effects are not well understood. In this paper, we study mergers in industries where firms compete by setting both price and promotion, and ask what happens if we mistakenly assume that price is the only dimension of competition. To answer the question, we build a structural model of the super-premium ice cream industry, where a 2003 merger between Häagen-Dazs and Dreyer's was challenged by the Federal Trade Commission. A structural merger model that ignores promotional competition under-predicts the price effects of a merger in this industry (5% instead of 12%). About three-fourths of the difference can be attributed to estimation bias (estimated demand is too elastic), with the remainder due to extrapolation bias from assuming post-merger promotional activity stays constant (instead it declines by 31%).


Apparently I am not the only one who wants to mod a Kinesis (Advantage, Pro, Classic..) to QMK. That's why I'm going to offer "modding kits" for the Kinesis ergonomic board lineup!
The MOQ is 9 kits - I doubt that there is high demand for modding-kits, but in case I am fine with ordering more!
if I can order all necessary parts on 01.11., delivery til January or even earlier should be possible!
What do I have to do with the kit?
in theory I can offer to mod your Kinesis for you: however I doubt shipping the board twice is worthwile..
I have to thank u/tobsn as he helped me format this post and even fixed some mistakes. thank you!!


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text:
Tainan City Council History Museum (Former Tainan Prefectural Council)(台南市議政史料館(原台南州會))
Walking from the Museum of Taiwan Literature towards Zhongzheng Road you will see a green bridge connecting the museum to another building. This is the Tainan City Council History Museum(台南市議政史料館), and the former Tainan Prefectural Council building(原台南州會). From behind you can still see the original appearance of the Tainan Prefectural Council building. After World War II, it became the seat of Tainan City Council in the 1950s. Later, the council moved to Anping, and from 1998 the building was left empty for a period. It was used for a while by the Photography Culture Museum, and now it has become the Tainan City Council History Museum, displaying a wealth of valuable literary and historical material and photographs.
Address West Central Dist., Tainan City Taiwán, R.O.C
Category Historic Sites、Culture
Open Hours (Under renovation, not open to the public)
Tuesday to Friday 13:00~21:00, Saturday to Sunday 10:00~19:00 (Closed on Monday), free entry
Attractions 210
Accommodation 273
Baoen Hall(報恩堂)
Former Tainan He Tong Building(Tainan City Fire Museum)(原台南合同廳舍)(消防史料館)
Chongqing Temple(重慶寺)
Tainan branch of Land Bank of Taiwan (Former Tainan Branch of Nippon Kangyo Bank)(土地銀行台南分行(原日本勸業銀行台南支店))

redpajama_set_name: RedPajamaCommonCrawl


Age is a major risk factor for AMD development, with people over the age of 50 at a higher risk. Smoking is another risk factor, as is heredity. If you have any of these risk factors, it's advised to have regular eye exams, which allow for early detection and treatment of AMD.
There are three stages of AMD: early, intermediate and late. In early AMD, you may not notice any changes to your vision. However, during a comprehensive eye exam, your eye doctor may detect a medium-sized drusen, a yellow waste deposit seen beneath the retina.
Similarly, intermediate AMD is generally asymptomatic. That said, large drusen and pigment changes in the retina can be detected through an eye exam. Late AMD would present with large drusen and noticeable macular damage. Signs of vision loss are apparent as well.
Currently, there's no proven cure for AMD. That said, there are things you can do to slow down its progression. For one, make sure to have a comprehensive eye exam at least once a year. Eye exams are not only recommended for those who wear eyeglasses. They are also helpful in detecting and managing various eye conditions, like AMD. Undergoing one regularly can help us monitor any changes in your condition.
Eating foods rich in antioxidants and omega-3 fatty acids can help as well. We may also recommend taking certain nutritional supplements that can help prevent the disease from advancing. Exercising regularly and quitting smoking can help maintain your eye health as well.
For more information on AMD or other eye care services, call us at (352) 504-4560. We serve Oxford and nearby FL areas.


Canadian Friends of Finland supports: Portrait Concert of Tomi Räisänen at Heliconian Hall: junctQín is joined by Finnish-Canadian pianist Heidi Saario and guitarist Rob MacDonald in concert. The music of Finnish composer Tomi Räisänen will be featured, with the composer in attendance, including a work for guitar and four balloons and a world premiere for piano 6 hands.
You are cordially invited to attend The Annual General Meeting of the Canadian Friends of Finland and The Annual Varpu Lindström Memorial Lecture Program: 7 PM: AGM Agenda as required by CFF by-laws. 8 PM: LECTURE "Thank you, Varpu" Evgeni Efremkin, PhD Historian and Entrepreneur, Varpu's former student. 8:45 PM Refreshments and Mingling Please register by phone message at 416-730-8350 or by email .