TITLE: Example of Calabi-Yau 3-fold fibered by both K3 surface and abelian surface?
QUESTION [3 upvotes]: I am looking for a compact Calabi-Yau 3-fold which is fibered by both K3 surface and abelian surface (with possibly singular fibers). Are there any examples?

REPLY [8 votes]: You can find such an example (which gives you much more indeed) in the following paper:
Oguiso, Keiji. On algebraic fiber space structures on a Calabi-Yau 3-fold. With an appendix by Noboru Nakayama. Internat. J. Math. 4 (1993), no. 3, 439-465.
It is the content of Theorem 4.9 on page 452. It says that there exists a Calabi-Yau $3$-fold which admits a Type $II_0$ fibration and infinitely many different fibrations of each of the three types $I_+$, $I_0$ and $II_+$.
In his notation, roughly speaking, these fibration types are respectively:

$I_0$, fibration on $\mathbb P^1$, general fiber an abelian surface.
$I_+$, fibration on $\mathbb P^1$, general fiber a $K3$ surface.
$II_0$ and $II_+$, fibrations on a rational surface, general fiber an elliptic curve (they are distinguished by the intersection of the second Chern class of $X$ with the class of the divisor defining the fibration).

This $X$ is constructed as a certain blow-up of the quotient $\overline X$ of $Y=S\times G$ by a canonically defined involution, where $G$ is an elliptic curve and $S$ a $K3$ surface with an involution whose fixed points locus is a union of rational curves. I refer to the original paper for more details!
Finally, notice that for Oguiso $X$ is a Calabi-Yau $3$-fold in the most restrictive way, that is $K_X$ is trivial and $X$ is simply connected.

REPLY [0 votes]: Let me give a construction which is a bit simpler than the construction suggested by Jason.
Let $S$ be an elliptically fibered K3 with a section. Let $C$ be an elliptic curve. Then take just $X = S \times C$. Then the fibers of $X \to S \to P^1$ are products of two elliptic curves, while the fibers of $X \to C$ are K3 surfaces.


TITLE: How to prove this inequality of bounded complex number.?
QUESTION [0 upvotes]: $|a_1\cdots a_m - b_1\cdots b_m|\leq \sum\limits_{j=1}^m|a_i-b_i|$
Where $|a_i|\leq 1, |b_i|\leq 1$, and they are any complex number satisfying the condition.

REPLY [0 votes]: By induction on $m$.
For $m = 2$ we have:
$$\begin{aligned}
\vert a_1 a_2 - b_1 b_2 \vert &= \vert a_1 a_2 -a_2 b_1 + a_2b_1 -b_1b_2 \vert\\
&\le \vert a_2 \vert \vert a_1 - b_1 \vert+ \vert b_1 \vert \vert a_2 - b_2 \vert\\
&\le \vert a_1 - b_1 \vert + \vert a_2 - b_2 \vert
\end{aligned}$$
Induction step from $m$ to $m+1$ follows from the case $m=2$, substituting in previous proof $a_1 \dots a_m \to a_1$, $b_1 \dots b_m \to b_1$, $a_{m+1} \to a_2$ and $b_{m+1} \to b_2$.


TITLE: Integrating Poisson groups
QUESTION [1 upvotes]: Recall that a symplectic groupoid (http://projecteuclid.org/euclid.bams/1183553676) is a Lie groupoid $\mathcal{G}\rightrightarrows X$ together with a symplectic structure on $\mathcal{G}$ multiplicative with respect to the groupoid multilpication and such that the unit map $X\rightarrow \mathcal{G}$ is Lagrangian.
It is known that some Poisson manifolds $X$ can be integrated to symplectic groupoids $\mathcal{G}\rightrightarrows X$. Conversely, given a symplectic groupoid $\mathcal{G}\rightrightarrows X$ one can uniquely recover a Poisson structure on $X$ such that $p_1\colon\mathcal{G}\rightarrow X$ is Poisson and $p_2\colon\mathcal{G}\rightarrow X$ is anti-Poisson.
What is an object integrating a Poisson-Lie group $G$?
Clearly, this should give rise to a symplectic groupoid $\mathcal{G}\rightrightarrows G$ together with three Lagrangians in $\mathcal{G}\times\mathcal{G}\times\overline{\mathcal{G}}$ (here bar means the manifold with the opposite symplectic structure) lifting the two projections and the multiplication, and a Lagrangian $\mathcal{G}\times_G\mathcal{G}\rightarrow \mathcal{G}\times\mathcal{G}\times\overline{\mathcal{G}}$ from the definition of a symplectic groupoid.
I would guess that the corresponding object should be a symplectic 2-groupoid $\mathcal{G}_2\stackrel{\rightarrow}\rightrightarrows G\rightrightarrows \mathrm{pt}$, but I haven't figured out how to get the 4 Lagrangians.
Note that there is a known integration of doubles of Poisson groups to double symplectic groupoids and hence to symplectic 2-groupoids (http://arxiv.org/abs/1012.4103), but the one-simplices of the latter are something like $G\times G^\ast$ instead of just $G$.

REPLY [3 votes]: I think this question was already answered in the comments, but I just wanted to point out Jiang Hua Lu's thesis (which can be found on her website http://hkumath.hku.hk/~jhlu/publications.html) talks about this in some detail. In section 4.2, given a Poisson Lie group $G$, she constructs a double symplectic groupoid $\Gamma$ over $G$ and $G^*$, which simultaneously integrates the Poisson structure on both $G$ and the dual Poisson Lie group $\bar G^*$ (the Poisson structure on $G^*$ is negated).
It's perhaps interesting to note that $\Gamma$ has a natural interpretation as a moduli space of flat connections (Pavol Severa does this in his paper `Moduli spaces of flat connections and Morita equivalence of quantum tori' http://arxiv.org/abs/1106.1366). His description gives a nice TFT interpretation of the algebraic structures on $\Gamma$.


TITLE: It's confusing to calculate Euler characteristic of this surface
QUESTION [5 upvotes]: This pic below is an exploded view of a cone.

I'm trying to calculate the Euler characteristic of the surface made from the fragment $M$, i.e.,

At first I thought the Euler characteristic is 0, but the one who made this question says it is actually 1.
And this is actually a part of an exercise to find the total geodesic curvature of $\partial M$. I tried to figure it out using the Gauss-Bonnet theorem.

I'm sorry that the question is not clear.
$M$ is the region between top circle of cone and closed geodesic meeting the top circle of cone at one point.

REPLY [0 votes]: If I'm reading that picture correctly, the Euler characteristic shouldn't exist.
The subset $M$ of the cone looks like it started as a topological disk facing left, and is curving to the right to connect back with itself, and will become a topological cylinder (a sphere with 2 disks removed).
A disk has $\chi=1$ , and a cylinder has $\chi=0$ . The surface $M$ is between those.
Nevertheless, Euler's formula can be applied to a polygonalization of $M$, formed as a square pyramid with 2 opposite triangles removed; the peak of the pyramid is the singular point.
The pyramid has $5$ vertices, $8$ edges, and $3$ faces. Thus $\chi=5-8+3=0$.


TITLE: Simplifying inner product of the A-norm terms(part of proof)
QUESTION [0 upvotes]: I am studying a proof for convergence of iterative algorithm (page 145 of this link )and I don't understand how it is simplified at some point. Here is the proof:
Assuming A is symmetric positive definite matrix, and $d_{k+1} = d_k - \alpha_kr_k$, Where $d_k \in \mathbb{R}^n$ and $A \in \mathbb{R}^{n\times n}$. We also know:
$r_i =Ad_i\quad $ or  $\quad d_i =A^{-1}r_i$ $\quad\quad\quad\forall  i$
By expanding the square of $A-\text{norm}$ of $d_{k+1}$ we have:
$\| d_{k+1}\|_A^{2}=(d_k-\alpha_k r_k,r_k) = (A^{-1}r_k,r_k)-\alpha_k(r_k,r_k)$
$\quad\quad\quad = \|d_k\|_A^{2}(1-\frac{(r_k,r_k)}{(r_k,A^{-1}r_k)} \times \frac{(r_k,r_k)}{(r_k,A^{-1}r_k)})$
My question :
I am not sure if if this helps but I can simplify  $\| d_{k+1}\|_A^{2}$ as follows :
$\| d_{k+1}\|_A^{2}= \| d_{k}\|_A^{2}(1-\frac{\alpha_k}{\|d_k\|^2}(r_k,r_k))= \| d_{k}\|_A^{2}(1 +  \frac{(A^{-1}r_{k+1},r_k)}{(r_{k},A^{-1}r_k)}   -\frac{(A^{-1}r_{k},r_k)}{(r_{k},A^{-1}r_k)}  )$
I cannot simplify it further to obtain:
$$ \|d_k\|_A^{2}(1-\frac{(r_k,r_k)}{(r_k,Ar_k)} \times \frac{(r_k,r_k)}{(r_k,A^{-1}r_k)})$$
Could anyone help please?
Thanks in advance.

REPLY [1 votes]: From the link you've provided, you've correctly transcribed this part of the proof:
$\| d_{k+1}\|_A^{2}=(d_k-\alpha_k r_k,r_k) = (A^{-1}r_k,r_k)-\alpha_k(r_k,r_k).$
I'm going to assume that you're okay up to here and that this is your starting point. Since $(A^{-1}r_k,r_k)= (d_k,r_k) = (d_k,Ad_k)= \|d_k\|_A^{2}$, we then have that
$$(A^{-1}r_k,r_k) -\alpha_k(r_k,r_k)=(A^{-1}r_k,r_k)(1-\frac{\alpha_k(r_k,r_k)}{(r_k,A^{-1}r_k)})=\|d_k\|_A^{2}(1-\frac{\alpha_k(r_k,r_k)}{(r_k,A^{-1}r_k)}).$$
Your desired expression then follows from the fact that $\alpha_k=\frac{(r_k,r_k)}{(r_k,Ar_k)}$ for the steepest descent algorithm.


text:
\begin{document}
\setlength{\baselineskip}{1.2\baselineskip}
\title[Gradient Estimates]{Gradient Estimates of Mean Curvature Equations with Neumann Boundary Condition}

\author{Xi-Nan Ma}
\address{Department of Mathematics\\
         University of Science and Technology of China\\
         Hefei Anhui 230026 CHINA}
         \email{xinan@ustc.edu.cn}
\author{Jinju Xu}
\address{Department of Mathematics\\
         University of Science and Technology of China\\
         Hefei Anhui 230026 CHINA}
         \email{july25@mail.ustc.edu.cn}

\thanks{Research of the first author was supported by NSFC  No.11125105 and Wu Wen-Tsun Key Laboratory of Mathematics. }
\thanks{2010 Mathematics Subject Classification: Primary 35B45; Secondary 35J92, 35B50}

\maketitle

\begin{abstract}
In this paper, we use the maximum principle to get the gradient estimate for the solutions of the prescribed  mean curvature equation with Neumann boundary value problem, which gives a positive answer for the question raised by Lieberman \cite{Lieb13} in page 360.
As a consequence, we obtain the corresponding existence theorem for a class of mean curvature equations.
\end{abstract}

\section{Introduction}
Gradient estimate for the prescribed mean curvature equation has been extensively studied. The interior gradient estimate, for the minimal surface equation was obtained in the case of two variables by Finn \cite{FN54}. Bombieri, De Giorgi and M.Miranda \cite{BGM69}  obtained the estimate for high dimension case. For the general mean curvature equation, such estimate had also been obtained by Ladyzhenskaya and Ural'tseva \cite{LU70}, Trudinger \cite{TR73} and Simon \cite{Sim76}. All their methods were test function argument and a resulting Sobolev inequality. In 1983, Korevaar \cite{Kor86} introduced the normal variation technique and got the maximum principle proof for the interior gradient estimate on the minimal surface equation. Wang \cite{Wang98} gave a new proof for the mean curvature equation via  standard Bernstein technique.  The Dirichlet problem for the prescribed mean curvature equation had  been  studied by Jenkins-Serrin \cite{JS68} and Serrin \cite{S69}. More detailed history could be found in Gilbarg and Trudinger \cite{GT01}.

  For the  mean curvature equation with prescribed contact angle boundary value problem,  Ural'tseva \cite{Ur73} first got the boundary gradient estimates and the corresponding existence theorem. At the same time, Simon-Spruck \cite{SS76} and Gerhardt \cite{Ger76} also obtained existence theorem on the positive gravity case. For more general quasilinear divergence structure equation with conormal derivative boundary value problem, Lieberman \cite{Lie83}  gave the gradient estimate. They obtained these estimates also via test function technique.

  Spruck \cite{Sp75} used the maximum principle to obtain boundary gradient estimate in two dimension for the positive gravity capillary problems.  Korevaar \cite{Kor88} generalized his normal variation technique and got the gradient estimates for the positive gravity case in high dimension case. In \cite{Lieb84, Lieb87}, Lieberman developed the maximum principle approach on the boundary gradient estimates to the quasilinear elliptic equation with oblique derivative boundary value problem, and in \cite{Lieb88} he got the maximum principle proof for the gradient estimates on the  general quasilinear elliptic equation with capillary boundary value problems.

  In a recent book written by Lieberman (\cite{Lieb13}, in page 360), he posed the following question, how to get the gradient estimates for the mean curvature equation with Neumann boundary value problem. In this paper we use the technique developed by Spruck \cite{Sp75}, Lieberman \cite{Lieb88} and Wang \cite{Wang98} to get a positive answer. As a consequence, we obtain an existence theorem for a class of mean curvature equations with Neumann boundary value problem.

We first consider the boundary gradient estimates for the mean curvature equation with Neumann boundary value problem.
Now let's state our main gradient estimates.
\begin{Thm}\label{Thm1.1}
Suppose $u\in C^{2}(\overline\Omega)\bigcap C^{3}(\Omega)$ is a bounded  solution for the following boundary value problem
\begin{align}
 \texttt{div}(\frac{Du}{\sqrt{1+|Du|^2}}) =&f(x, u)   \quad\text{in}\quad \Omega, \label{1.1}\\
              \frac{\partial u}{\partial \gamma} = &\psi(x, u)  \quad\text{on} \quad\partial \Omega,\label{1.2}
\end{align}
where $\Omega \subset \mathbb R^n $ is a bounded domain, $n\geq 2$, $\partial \Omega \in C^{3}$, $\gamma$ is the inward unit normal to $\partial\Omega $.

We assume $f(x,z) \in C^{1}(\overline\Omega\times [-M_0, M_0])$ and $\psi(x,z) \in C^{3}(\overline\Omega\times [-M_0, M_0])$,
and there exist positive constants $M_0, L_1, L_2$ such that
\begin{align}
|u|\leq& M_0\quad in\quad\overline\Omega,\label{1.3}\\
f_z(x,z)\geq &0 \,\,\,\,\quad in\quad \overline\Omega\times[-M_0, M_0],\label{1.4}\\
|f(x,z)|+|f_{x}(x,z)|\leq & L_1 \quad in\quad \overline\Omega\times[-M_0, M_0],\label{1.5}\\
|\psi(x,z)|_{C^3(\overline\Omega\times[-M_0, M_0])}\leq&  L_2.\label{1.6}
\end{align}

Then there exists a small positive constant
$\mu_0$ such that we have the following estimate
$$\sup_{\overline\Omega_{\mu_0}}|Du|\leq \max\{M_1, M_2\},$$
where $M_1$ is a positive constant depending only on $n, \mu_0, M_0, L_1$, which is from the interior gradient estimates;
$M_2$ is  a positive constant depending only on $n, \Omega, \mu_0, M_0, L_1, L_2$, and $d(x) =\texttt{dist}(x, \partial\Omega), \Omega_{\mu_0} = \{x \in \Omega : d(x)<\mu_0\}.$
\end{Thm}
 As we stated before, there is a standard interior gradient estimates for the mean curvature equation.

\begin{Rem}[\cite{GT01}]\label{Rem1.1}
If $u\in C^{3}(\Omega)$ is a bounded solution for the equation \eqref{1.1}  with \eqref{1.3}, and if $f \in C^{1}(\overline\Omega \times [-M_0, M_0])$ satisfies the conditions \eqref{1.4}-\eqref{1.5}, then for any subdomain
$\Omega'\subset\subset\Omega$, we have
$$\sup_{\Omega'}|Du|\leq M_1,$$
where $M_1$ is a positive constant depending only on $n,  M_0, \texttt{dist} (\Omega', \partial\Omega), L_1$.
\end{Rem}

From the standard bounded estimates for the prescribed mean curvature equation in Concus-Finn \cite{CF74} ( see also Spruck \cite{Sp75}), we can get the following existence theorem for the Neumann boundary value problem of mean curvature equation.

\begin{Thm}\label{Thm1.2}
Let $\Omega \subset\mathbb R^n $ be a bounded domain, $n\geq 2$, $\partial \Omega\in C^{3}$, $\gamma$ is the inward unit normal to $\partial\Omega $.
If  $\psi \in C^{3}(\overline\Omega)$,  is a given function, then the following boundary value problem
\begin{align}
 \texttt{div}(\frac{Du}{\sqrt{1+|Du|^2}}) =& u   \quad\text{in}\quad \Omega, \label{1.7}\\
              \frac{\partial u}{\partial \gamma} = &\psi(x)  \quad\text{on} \quad\partial \Omega,\label{1.8}
\end{align}
 exists a unique solution $ u \in C^2(\overline\Omega)$.
\end{Thm}

The rest of the paper is organized as follows. In section 2, we first give the definitions and some notations. We prove the main Theorem~\ref{Thm1.1} in section 3 under the help of one lemma. In section 4, we give the proof of  Theorem~\ref{Thm1.2}.

\section{PRELIMINARIES}
We denote by $\Omega$ a bounded  domain in $\mathbb{R}^n$, $n\geq 2$,  $\partial \Omega\in C^{3}$,   set
\begin{align*}
 d(x)=\texttt{dist}(x,\partial \Omega),
 \end{align*}
 and
\begin{align*}
 \Omega_\mu=&\{{x\in\Omega:d(x)<\mu}\}.
 \end{align*}
Then it is well known that there exists a positive constant $\mu_{1}>0$ such that $d(x) \in C^3(\overline \Omega_{\mu_{1}})$. As in Simon-Spruck \cite{SS76} or Lieberman \cite{Lieb13} in page 331,  we can take $\gamma= D d$ in $\Omega_{\mu_{1}}$ and note that  $\gamma$ is a $C^2(\overline \Omega_{\mu_{1}})$ vector field. As mentioned in \cite{Lieb88} and the book \cite{Lieb13}, we also have the following formulas

\begin{align}\label{2.1}
\begin{split}
|D\gamma|+|D^2\gamma|\leq& C(n,\Omega) \quad\text{in}\quad \Omega_{\mu_{1}},\\
 \sum_{1\leq i\leq n}\gamma^iD_j\gamma^i=0,  \sum_{1\leq i\leq n}\gamma^iD_i\gamma^j=&0, \,|\gamma|=1 \quad\text{in} \quad\Omega_{\mu_{1}}.
\end{split}
\end{align}
As in \cite{Lieb13}, we define
 \begin{align}\label{2.2}
\begin{split}
c^{ij}=&\delta_{ij}-\gamma^i\gamma^j  \quad \text{in} \quad \Omega_{\mu_{1}},
\end{split}
\end{align}
 and for a vector $\zeta \in R^n$, we write $\zeta'$ for the vector with $i-$th component $ \sum_{1\leq j\leq n}c^{ij}\zeta_j$. So
 \begin{align}\label{2.3}
\begin{split}
|D'u|^2=& \sum_{1\leq i,j\leq n}c^{ij}u_iu_j.
\end{split}
\end{align}
Let
\begin{align}\label{2.4}
\begin{split}
a^{ij}(D u)=v^2\delta_{ij}-u_iu_j, \quad
v=(1+|D u|^2)^{\frac{1}{2}}.
\end{split}
\end{align}

Then the equations \eqref{1.1}, \eqref{1.2} are equivalent to the following boundary value problem
\begin{align}\label{2.5}
\sum_{i,j=1}^n a^{ij}u_{ij}=&f(x,u)v^3 \quad \text{in}\,\Omega,\\
u_{\gamma}=& \psi(x, u) \quad \text{on}\,\partial\Omega.\label{2.6}
\end{align}

\section{Proof of Theorem~\ref{Thm1.1} }

Now we begin to prove Theorem~\ref{Thm1.1}, as  mentioned in introduction, using the technique developed by Spruck \cite{Sp75}, Lieberman \cite{Lieb88} and Wang \cite{Wang98}. We shall choose an auxiliary function which contains  $|Du|^2$ and other lower order terms. Then we use the maximum principle for this auxiliary function in $\overline\Omega_{\mu_0}, 0<\mu_0<\mu_1$. At last,  we get our estimates.

{\em Proof of  Theorem~\ref{Thm1.1}.}

Setting $w=u-\psi(x,u)d$,  we choose the following auxiliary function
$$\Phi(x)=\log|Dw|^2e^{1+M_0+u}e^{\alpha_0 d}, \quad  x \in \overline \Omega_{\mu_{0}},$$ where $\alpha_0=|\psi|_{C^0(\overline\Omega\times[-M_0, M_0])}+C_0+1 $, $C_0$ is  a positive constant depending only on $n,\Omega$.

In order to simplify the computation, we let
\begin{align}\label{3.0}
 \varphi(x)= \log \Phi(x) =\log\log|Dw|^2+h(u)+g(d).
 \end{align}

In our case, we take
\begin{align}\label{3.1}
 h(u)=1+M_0+u,\quad g(d)=\alpha_0 d.
 \end{align}

 We assume that
$\varphi(x)$ attains its maximum at $x_0 \in \overline \Omega_{\mu_{0}}$, where $0<\mu_0<\mu_1$ is a sufficiently small number which we shall decide it  later.

Now we divide three cases to complete the proof of  Theorem~\ref{Thm1.1}.

Case I: If $\varphi(x)$ attains its maximum at $x_0 \in \partial\Omega$, then we shall use the Hopf Lemma to get the bound of $|Du|(x_0)$.

Case II: If $\varphi(x)$ attains its maximum at $x_0 \in\partial\Omega_{\mu_0}\bigcap\Omega$, then we shall get the estimates via the standard interior gradient bound \cite{GT01}.

Case III:  If $\varphi(x)$ attains its maximum at $x_0 \in \Omega_{\mu_0}$, in this case for the sufficiently small constant $\mu_0>0$,  then we can use the maximum principle to get the bound of $|Du|(x_0)$.

Now  all computations work at the point $x_0$.

{\bf Case 1.} If  $\varphi(x)$ attains its maximum at $x_0\in \partial \Omega$, we shall get the bound of $|Du|(x_0)$.

We differentiate $\varphi$ along the normal direction.

\begin{align}\label{3.2}
\begin{split}
\frac{\partial\varphi}{\partial\gamma}=&\frac{\sum_{1\leq i\leq n}(|Dw|^2)_i\gamma^i}{|Dw|^2\log|Dw|^2}+h'u_{\gamma}+g'.
\end{split}
\end{align}

Since
\begin{align}
w_i=&u_i-\psi_u u_i d -\psi_{x_i} d-\psi\gamma^i,\label{3.3}\\
|Dw|^2=&|D'w|^2+w^2_\gamma,\label{3.4}
\end{align}

we have
\begin{align}
w_\gamma=&u_\gamma-\psi_uu_\gamma d-\sum_{1\leq i\leq n}\psi_{x_i}\gamma^i d-\psi=0\quad\text{on}\quad\partial\Omega,\label{3.5}\\
(|Dw|^2)_i=&(|D'w|^2)_i\quad\text{on}\quad\partial\Omega.\label{3.6}
\end{align}

Applying  \eqref{2.1}, \eqref{2.3} and \eqref{3.6}, it follows that
\begin{align}\label{3.7}
\begin{split}
\sum_{1\leq i\leq n}(|Dw|^2)_i\gamma^i=&\sum_{1\leq i\leq n}(|D'w|^2)_i\gamma^i\\
=&2\sum_{1\leq i,k,l\leq n}c^{kl}w_{ki}w_l\gamma^i\\
=&2\sum_{1\leq i,k,l\leq n}c^{kl}u_{ki}u_l\gamma^i-2\sum_{1\leq k,l\leq n}c^{kl}u_lD_k\psi,
\end{split}
\end{align}
where
\begin{align*}
D_k\psi= \psi_{x_k} + \psi_u u_k.
\end{align*}

Differentiating \eqref{2.6} with respect to tangential direction,   we have
\begin{align}\label{3.8}
\sum_{1\leq k\leq n}c^{kl}(u_{\gamma})_k=&\sum_{1\leq k\leq n}c^{kl}D_k\psi.
\end{align}

It follows that
\begin{align}\label{3.9}
\sum_{1\leq i, k\leq n}c^{kl}u_{ik}\gamma^i=&-\sum_{1\leq i,k\leq n}c^{kl}u_i(\gamma^i)_k+\sum_{1\leq k\leq n}c^{kl}D_k\psi.
\end{align}

Inserting \eqref{3.9} into \eqref{3.7} and combining \eqref{2.6}, \eqref{3.2}, we have
\begin{align}\label{3.10}
\begin{split}
|Dw|^2\log|Dw|^2\frac{\partial\varphi}{\partial\gamma}(x_0)
=&(g'(0)+h'\psi)|Dw|^2\log|Dw|^2-2\sum_{1\leq i,k,l\leq n}c^{kl}u_iu_l(\gamma^i)_k.
\end{split}
\end{align}

From \eqref{3.3}, we obtain
\begin{align}\label{3.11}
|Dw|^2=&|Du|^2-\psi^2\quad\text{on}\quad\partial\Omega.
\end{align}

Assume $|Du|(x_0)\ge \sqrt{100+2|\psi|^2_{C^0(\overline\Omega\times[-M_0, M_0])}}$, otherwise we get the estimates. At $x_0$,   we have
\begin{align}
\frac{1}{2}|Du|^2\leq&|Dw|^2\leq |Du|^2,\label{3.12}\\
|Dw|^2\ge&50.\label{3.13}
\end{align}

Inserting  \eqref{3.12} and \eqref{3.13} into \eqref{3.10}, we have
\begin{align}\label{3.14}
\begin{split}
|Dw|^2\log|Dw|^2\frac{\partial\varphi}{\partial\gamma}(x_0)
\geq&(\alpha_0-|\psi|_{C^0(\overline\Omega\times[-M_0, M_0])}-C_0)|Dw|^2\log|Dw|^2\\
=&|Dw|^2\log|Dw|^2\\
>&0.
\end{split}
\end{align}
On the other hand, from Hopf Lemma, we have
$$\frac{\partial\varphi}{\partial\gamma}(x_0)\leq 0,$$
it is a contradiction to \eqref{3.14}.

Then we have
\begin{align}\label{3.15}
|Du|(x_0)\leq\sqrt{100+2|\psi|^2_{C^0(\overline\Omega\times[-M_0, M_0])}}.
\end{align}

{\bf Case 2.} $ x_0\in \partial\Omega_{\mu_{0}}\bigcap\Omega$. This is due to interior gradient estimates. From Remark~\ref{Rem1.1}, we have
 \begin{align}\label{3.16}
\sup_{\partial\Omega_{\mu_0}\bigcap\Omega}|Du|\leq \tilde{M}_1.
\end{align}
where $\tilde{M}_1$ is a positive constant depending only on $n, M_0, \mu_0, L_1$.\par

{\bf Case 3.} $x_0\in\Omega_{\mu_{0}}$. \par

In this case, $x_0$ is a critical point of $\varphi$. We choose the normal coordinate at $x_0$, by rotating the coordinate system
suitably, we may assume that $u_i(x_0)=0,\,2\leq i\leq n$ and $u_1(x_0)=|Du|>0$. And we can further assume that the matrix $(u_{ij}(x_0))(2\leq i,j\leq n)$ is diagonal. Let $$\mu_2 \le \frac{1}{10
0L_2}$$ such that
 \begin{align}\label{3.16a}
 |\psi_u| \mu_2 \le \frac{1}{100
}, \quad \text{then}\quad \frac{99}{100}\le 1-\psi_u \mu_2 \le
 \frac{101}{100}.
\end{align}

 We can choose $$ \mu_0=\frac{1}{2} \min\{ \mu_1,\mu_2,  1 \}.$$

In order to simplify the calculations, we let
\begin{align*}
w=&u-G,\quad G=\psi(x,u)d.
\end{align*}
Then we have
\begin{align}
w_k=(1-G_u)u_k-G_{x_k}.\label{3wk}
\end{align}
Since at $x_0$,
\begin{align}
|Dw|^2=&w_1^2+\sum_{2\leq i\leq n}w_i^2,\label{3Aa}\\
w_i=&-G_{x_i}=-\psi_{x_i}d-\psi\gamma^i,\quad i=2,\ldots,n,\label{3wi2}\\
w_1=&(1-G_u)u_1-G_{x_1}=(1-G_u)u_1 -\psi_{x_1}d-\psi\gamma^1.\label{3wi2a}
\end{align}
So from the above relation,  at $x_0$, we can assume
\begin{align}
u_1 =  |Du|(x_0)\ge \sqrt{3000(1+ |\psi|^2_{C^1(
\overline\Omega\times[-M_0, M_0])})},\label{3wk}
\end{align}
then
\begin{align}\label{3.18}
\frac{9}{10}u_1^2\leq|Dw|^2\leq \frac{11}{10}u_1^2, \quad \frac{9}{10}u_1^2\leq w_1^2\leq \frac{11}{10}u_1^2
\end{align}
and by
 the choice of $\mu_0$ and  \eqref{3.16a}, we have
\begin{align}
\frac{99}{100}
\le 1-G_u \le\frac {101}{100}.
\label{wi2aa}
\end{align}

From the above choice, we shall prove Theorem~\ref{Thm1.1} with three steps, as we mentioned before, all the calculations will be done at the fixed point $x_0$.

{\bf Step 1:} We first get the formula \eqref{3aijvarphiijb}.\par

Taking the first  derivatives of $\varphi$,
\begin{align}\label{3varphii}
\varphi_i=&\frac{(|Dw|^2)_i}{|Dw|^2\log|Dw|^2}+h'u_i+g'\gamma^i.
\end{align}
From $\varphi_i(x_0)=0$,  we have
\begin{align}\label{3varphii=0}
(|Dw|^2)_i=-|Dw|^2\log|Dw|^2(h'u_i+g'\gamma^i).
\end{align}
Take the derivatives again for  $\varphi_i$,
\begin{align}\label{3varphiija}
\begin{split}
\varphi_{ij}
=&\frac{(|Dw|^2)_{ij}}{|Dw|^2\log|Dw|^2}-(1+\log|Dw|^2)\frac{(|Dw|^2)_i(|Dw|^2)_j}{(|Dw|^2\log|Dw|^2)^2}\\&
+h'u_{ij}
+h''u_iu_j+g''\gamma^i\gamma^j+g'(\gamma^i)_j.
\end{split}
\end{align}
Using \eqref{3varphii=0}, it follows that
\begin{align}\label{3varphiijb}
\begin{split}
\varphi_{ij}
=&\frac{(|Dw|^2)_{ij}}{|Dw|^2\log|Dw|^2}+h'u_{ij}+\big[h''-(1+\log|Dw|^2)h'^2\big]u_iu_j\\&
+\big[g''-(1+\log|Dw|^2)g'^2\big]\gamma^i\gamma^j
-(1+\log|Dw|^2)h'g'(\gamma^iu_j+\gamma^ju_i)+g'(\gamma^i)_j.
\end{split}
\end{align}
Then we get
\begin{align}\label{3aijvarphiija}
\begin{split}
0\geq \sum_{1\leq i,j\leq n}a^{ij}\varphi_{ij}
=:&I_1+I_2,
\end{split}
\end{align}
where
\begin{align}\label{3I1a}
I_1=\frac{1}{|Dw|^2\log|Dw|^2}\sum_{1\leq i,j\leq n}a^{ij}(|Dw|^2)_{ij},
\end{align}
and
\begin{align}
I_2=&\sum_{1\leq i,j\leq n}a^{ij}\bigg\{h' u_{ij}+\big[h''-(1+\log|Dw|^2)h'^2\big]u_iu_j+\big[g''-(1+\log|Dw|^2)g'^2\big]\gamma^i\gamma^j\notag\\
&\qquad\qquad\quad-2(1+\log|Dw|^2)h'g'\gamma^iu_j+g'(\gamma^i)_j\bigg\}.\label{3I2a}
\end{align}
From the choice of the coordinate, we have
\begin{align}\label{3aij}
a^{11}=1,\, a^{ii}=v^2=1+u_1^2 \,\,(2\leq i\leq n ), a^{ij}=0\,\, (i\neq j,1\leq i,j\leq n).
\end{align}

Now we first treat $I_2$.

From the choice of the coordinate and the equations \eqref{2.5}, \eqref{3aij}, we have
\begin{align}
I_2
=&h'f v^3-h'^2u_1^2\log|Dw|^2+(h''-h'^2)u_1^2+\big[g''-(1+\log|Dw|^2)g'^2\big]\sum_{1\leq i\leq n}a^{ii}(\gamma^i)^2\notag\\
&-2(1+\log|Dw|^2)h'g'\gamma^1u_1+g'\sum_{1\leq i\leq n}a^{ii}(\gamma^i)_i\notag\\
=&h'f v^3-(h'^2+c^{11}g'^2) u_1^2\log|Dw|^2+\big[h''-h'^2+c^{11}(g''-g'^2)+g'\sum_{2\leq i\leq n}(\gamma^i)_i\big]u_1^2\notag\\
&-2h'g'\gamma^1 u_1\log|Dw|^2-2h'g'\gamma^1 u_1+g''-g'^2
+g'\sum_{1\leq i\leq n}(\gamma^i)_i.\label{3I2b}
\end{align}
Hence from \eqref{3.1}, we have
\begin{align}
I_2
=&f v^3-(1+c^{11}\alpha_0^2) u_1^2\log|Dw|^2+\big[-1-c^{11}\alpha_0^2+\alpha_0\sum_{2\leq i\leq n}(\gamma^i)_i\big] u_1^2\notag\\
&-2\alpha_0\gamma^1 u_1\log|Dw|^2-2\alpha_0\gamma^1u_1-\alpha_0^2
+\alpha_0\sum_{1\leq i\leq n}(\gamma^i)_i\notag\\
\geq&f v^3-(1+c^{11}\alpha_0^2) u_1^2\log|Dw|^2-C_1u_1^2.\label{3I2c}
\end{align}
here  $C_1$  is a positive constant  depending  only on $n, \Omega, M_0, \mu_0, L_2$.\par

Next, we calculate $I_1$ and get the formula \eqref{3I1c}. \par

Taking the first  derivatives of $|Dw|^2$, we have
\begin{align}\label{3|Dw|^2i}
\begin{split}
(|Dw|^2)_i=&2\sum_{1\leq k\leq n}w_kw_{ki}.
\end{split}
\end{align}
Taking the derivatives of $|Dw|^2$ once more, we have
\begin{align}\label{3|Dw|^2ij}
\begin{split}
(|Dw|^2)_{ij}=&2\sum_{1\leq k\leq n}w_kw_{kij}+2\sum_{1\leq k\leq n}w_{ki}w_{kj}.
\end{split}
\end{align}
By \eqref{3I1a} and \eqref{3|Dw|^2ij}, we can rewrite $I_1$ as
\begin{align}
I_1
=&\frac{1}{|Dw|^2\log|Dw|^2}\big[2\sum_{1\leq i,j,k\leq n}w_ka^{ij}w_{ijk}+2\sum_{1\leq i,j,k\leq n}a^{ij}w_{ki}w_{kj}\big]\notag\\
=:&\frac{1}{|Dw|^2\log|Dw|^2}\big[I_{11}+I_{12}\big].\label{3I1b}
\end{align}
In the following, we shall deal with $I_{11}$ and $I_{12}$ respectively. \par

For the term $I_{11}$. As we have
let
\begin{align}
w=&u-G,\quad G=\psi(x,u)d.\label{3w}
\end{align}
Then we have
\begin{align}
w_k=&(1-G_u)u_k-G_{x_k},\notag\\
w_{ki}=&(1-G_u)u_{ki}-G_{uu}u_ku_i-G_{ux_i}u_k-G_{ux_k}u_i-G_{x_kx_i},\label{3wki}\\
w_{kij}=&(1-G_u)u_{kij}-G_{uu}(u_{ki}u_j+u_{kj}u_i+u_{ij}u_k)\notag\\
&-G_{ux_i}u_{kj}-G_{ux_j}u_{ki}-G_{ux_k}u_{ij}\notag\\
&-G_{uuu}u_ku_iu_j-G_{uux_i}u_ju_k-G_{uux_j}u_iu_k-G_{uux_k}u_iu_j\notag\\
&-G_{ux_ix_j}u_k-G_{ux_kx_j}u_i-G_{ux_ix_k}u_j-G_{x_ix_jx_k}.\label{3wkij}
\end{align}

So from the choice of the coordinate and the equations \eqref{2.5}, \eqref{3aij}, we have
\begin{align}
\sum_{1\leq i,j,k\leq n}w_ka^{ij}w_{ijk}=&\sum_{1\leq i,j,k\leq n}\big[(1-G_u)w_ka^{ij}u_{ijk}-G_{uu}w_ku_ka^{ij}u_{ij}
-2G_{uu}w_ka^{ij}u_{ki}u_j\notag\\
&\qquad\qquad-w_kG_{ux_k}a^{ij}u_{ij}-2w_ka^{ij}G_{ux_i}u_{kj}-G_{uuu}w_ku_ka^{ij}u_iu_j\notag\\
&\qquad\qquad-2w_ku_ka^{ij}G_{uux_i}u_j-w_kG_{uux_k}a^{ij}u_iu_j
-w_ku_ka^{ij}G_{ux_ix_j}\notag\\
&\qquad\qquad-2w_ka^{ij}G_{ux_ix_k}u_j-w_ka^{ij}G_{x_ix_jx_k}\big]\notag\\
=&(1-G_u)\sum_{1\leq i,j,k\leq n}w_ka^{ij}u_{ijk}-2(G_{uu}u_1+G_{ux_1})\sum_{1\leq k\leq n}w_ku_{k1}\notag\\
&-2v^2\sum_{2\leq i\leq n}G_{ux_i}\sum_{1\leq k\leq n}w_ku_{ki}-G_{uu}fv^3u_1w_1
-G_{uuu}u_1^3w_1\notag\\&
-v^2u_1w_1\sum_{2\leq i\leq n}G_{ux_ix_i}-fv^3\sum_{1\leq k\leq n}w_kG_{ux_k}-2G_{uux_1}u_1^2w_1\notag\\&-u_1^2\sum_{1\leq k\leq n}w_kG_{uux_k}
-v^2\sum_{1\leq k\leq n}w_k\sum_{2\leq i\leq n}G_{x_ix_ix_k}-G_{ux_1x_1}u_1w_1\notag\\
&-2u_1\sum_{1\leq k\leq n}G_{ux_1x_k}w_k-\sum_{1\leq k\leq n}G_{x_1x_1x_k}w_k.\label{3wkaijwkij}
\end{align}
By the equation \eqref{2.5}, we have
\begin{align}\label{3u11a}
\begin{split}
u_{11}=&fv^3-v^2\sum_{2\leq i\leq n}u_{ii},
\end{split}
\end{align}
and
\begin{align}\label{3Delta u}
\begin{split}
\Delta u=&fv+\frac{u_1^2}{v^2}u_{11}.
\end{split}
\end{align}
Differentiating \eqref{2.5}, we have
\begin{align}\label{3aijuijka}
\begin{split}
\sum_{1\leq i,j\leq n}a^{ij}u_{ijk}=&-\sum_{1\leq i,j,l\leq n}a^{ij}_{p_l}u_{lk}u_{ij}+v^3D_{k}f+3fv^2v_k.
\end{split}
\end{align}
From \eqref{2.4}, we have
\begin{align}\label{3aijpl}
\begin{split}
a^{ij}_{p_l}=&2u_l\delta_{ij}-\delta_{il}u_j-\delta_{jl}u_i.
\end{split}
\end{align}
By the definition of $v$, we have
\begin{align}\label{3vvk}
\begin{split}
v v_k=&u_1u_{1k}.
\end{split}
\end{align}
Since
\begin{align}\label{3Dkf}
\begin{split}
D_kf=&f_uu_k+f_{x_k},
\end{split}
\end{align}
 from \eqref{3aijpl}, \eqref{3u11a} and \eqref{3vvk}, we have
\begin{align}\label{3aijuijkb}
\begin{split}
\sum_{1\leq i,j\leq n}a^{ij}u_{ijk}
=&-2u_1u_{1k}\Delta u+2u_1\sum_{1\leq i\leq n}u_{1i}u_{ik}+v^3D_{k}f+3fvu_1u_{1k},\\
=&\frac{2u_1}{v^2}u_{11}u_{1k}+2u_1\sum_{2\leq i\leq n}u_{1i}u_{ik}+v^3D_{k}f+fvu_1u_{1k}\\
=&\frac{2u_1}{v^2}u_{11}u_{1k}+2u_1\sum_{2\leq i\leq n}u_{1i}u_{ik}+f_uv^3u_k+f_{x_k}v^3+fvu_1u_{1k}.
\end{split}
\end{align}
Inserting \eqref{3aijuijkb} into \eqref{3wkaijwkij}, we rewrite $I_{11}$ as
\begin{align}\label{3I11a}
I_{11}
=&4(1-G_u)\frac{u_1}{v^2} u_{11}\sum_{1\leq k\leq n}w_ku_{k1}+4(1-G_u)u_1\sum_{2\leq i\leq n}u_{1i}\sum_{1\leq k\leq n}w_ku_{ki}\notag\\
&+[2(1-G_u)fu_1v-4G_{uu}u_1-4G_{ux_1}]\sum_{1\leq k\leq n}w_ku_{k1}-4v^2\sum_{2\leq i\leq n}G_{ux_i}\sum_{1\leq k\leq n}w_ku_{ki}\notag\\
&+2(1-G_u)f_uv^3u_1w_1-2G_{uu}fv^3u_1w_1+2(1-G_u)v^3\sum_{1\leq k\leq n}f_{x_k}w_k\notag\\&-2fv^3\sum_{1\leq k\leq n}G_{ux_k}w_k
-2G_{uuu}u_1^3w_1-2v^2u_1w_1\sum_{2\leq i\leq n}G_{ux_ix_i}-4G_{uux_1}u_1^2w_1\notag\\&-2u_1^2\sum_{1\leq k\leq n}w_kG_{uux_k}
-2v^2\sum_{1\leq k\leq n}w_k\sum_{2\leq i\leq n}G_{x_ix_ix_k}-2G_{ux_1x_1}u_1w_1\notag\\
&-4u_1\sum_{1\leq k\leq n}G_{ux_1x_k}w_k-2\sum_{1\leq k\leq n}G_{x_1x_1x_k}w_k.
\end{align}

For the term $I_{12}$: applying \eqref{3aij} and \eqref{3wki}, we have
\begin{align*}
I_{12}
=&2\sum_{1\leq i,k\leq n}a^{ii}w^2_{ki}\notag\\
=&2\sum_{1\leq k\leq n}w^2_{k1}+2v^2\sum_{1\leq k\leq n}\sum_{2\leq i\leq n}w^2_{ki}\notag\\
=&2w_{11}^2+2(v^2+1)\sum_{2\leq i\leq n}w^2_{1i}+2v^2\sum_{2\leq i,k\leq n}w^2_{ki}\notag\\
=&2[(1-G_u)u_{11}-(G_{uu}u_1^2+2G_{ux_1}u_1+G_{x_1x_1})]^2\notag\\
&+2(v^2+1)\sum_{2\leq i\leq n}[(1-G_u)u_{1i}-(G_{ux_i}u_1+G_{x_1x_i})]^2\notag\\
&+2v^2\sum_{2\leq i,k\leq n}[(1-G_u)u_{ki}-G_{x_kx_i}]^2\notag\\
=&2(1-G_u)^2u_{11}^2+2(1-G_u)^2(v^2+1)\sum_{2\leq i\leq n}u_{1i}^2+2(1-G_u)^2v^2\sum_{2\leq i\leq n}u_{ii}^2\notag\\
\end{align*}
\begin{align}
&-4(1-G_u)(G_{uu}u_1^2+2G_{ux_1}u_1+G_{x_1x_1})u_{11}\notag\\
&-4(1-G_u)(v^2+1)\sum_{2\leq i\leq n}(G_{ux_i}u_1+G_{x_1x_i})u_{1i}\notag\\
&-4(1-G_u)v^2\sum_{2\leq i\leq n}G_{x_ix_i}u_{ii}+2(G_{uu}u_1^2+2G_{ux_1}u_1+G_{x_1x_1})^2\notag\\
&+2(v^2+1)\sum_{2\leq i\leq n}(G_{ux_i}u_1+G_{x_1x_i})^2
+2v^2\sum_{2\leq i,k\leq n}G_{x_kx_i}.\label{3I12a}
\end{align}
Combining  \eqref{3I11a}, \eqref{3I12a}, it follows that
\begin{align}
I_1=&\frac{1}{|Dw|^2\log|Dw|^2}\bigg[2(1-G_u)^2u_{11}^2+2(1-G_u)^2(v^2+1)\sum_{2\leq i\leq n}u_{1i}^2\notag\\&+2(1-G_u)^2v^2\sum_{2\leq i\leq n}u_{ii}^2
+4(1-G_u)\frac{u_1}{v^2} u_{11}\sum_{1\leq k\leq n}w_ku_{k1}\notag\\&+4(1-G_u)u_1\sum_{2\leq i\leq n}u_{1i}\sum_{1\leq k\leq n}w_ku_{ki}\notag\\&
+[2(1-G_u)fu_1v-4G_{uu}u_1-4G_{ux_1}]\sum_{1\leq k\leq n}w_ku_{k1}\notag\\&-4v^2\sum_{2\leq i\leq n}G_{ux_i}\sum_{1\leq k\leq n}w_ku_{ki}\notag\\
&-4(1-G_u)(G_{uu}u_1^2+2G_{ux_1}u_1+G_{x_1x_1})u_{11}\notag\\
&-4(1-G_u)(v^2+1)\sum_{2\leq i\leq n}(G_{ux_i}u_1+G_{x_1x_i})u_{1i}\notag\\
&-4(1-G_u)v^2\sum_{2\leq i\leq n}G_{x_ix_i}u_{ii}+2(1-G_u)f_uv^3u_1w_1-2G_{uu}fv^3u_1w_1\notag\\
&+2(G_{uu}u_1^2+2G_{ux_1}u_1+G_{x_1x_1})^2+2(v^2+1)\sum_{2\leq i\leq n}(G_{ux_i}u_1+G_{x_1x_i})^2\notag\\
&+2v^2\sum_{2\leq i,k\leq n}G_{x_kx_i}+2(1-G_u)v^3\sum_{1\leq k\leq n}f_{x_k}w_k-2fv^3\sum_{1\leq k\leq n}G_{ux_k}w_k\notag\\&
-2G_{uuu}u_1^3w_1-2v^2u_1w_1\sum_{2\leq i\leq n}G_{ux_ix_i}-4G_{uux_1}u_1^2w_1-2u_1^2\sum_{1\leq k\leq n}w_kG_{uux_k}\notag\\&
-2v^2\sum_{1\leq k\leq n}w_k\sum_{2\leq i\leq n}G_{x_ix_ix_k}-2G_{ux_1x_1}u_1w_1\notag\\
&-4u_1\sum_{1\leq k\leq n}G_{ux_1x_k}w_k-2\sum_{1\leq k\leq n}G_{x_1x_1x_k}w_k\bigg].\label{3I1c}
\end{align}
Inserting \eqref{3I1c} and \eqref{3I2b} into \eqref{3aijvarphiija}, we  can obtain the following formula
\begin{align}\label{3aijvarphiijb}
\begin{split}
0\geq \sum_{1\leq i,j\leq n}a^{ij}\varphi_{ij}=: Q_1+Q_2+Q_3,
\end{split}
\end{align}
where $Q_1$ contains all the quadratic terms of $u_{ij}$; $Q_2$  is the term which contains all linear terms of $u_{ij}$;  and   the remaining terms are denoted by $Q_3$.
Then we have
\begin{align}\label{3Q1a}
Q_1=&\frac{1}{|Dw|^2\log|Dw|^2}\bigg[ 2(1-G_u)^2u_{11}^2+2(1-G_u)^2(v^2+1)\sum_{2\leq i\leq n}u_{1i}^2\notag\\&
+4(1-G_u)\frac{u_1}{v^2} u_{11}\sum_{1\leq k\leq n}w_ku_{k1}
+4(1-G_u)u_1\sum_{2\leq i\leq n}u_{1i}\sum_{1\leq k\leq n}w_ku_{ki}\notag\\&
+2(1-G_u)^2v^2\sum_{2\leq i\leq n}u_{ii}^2\bigg];
\end{align}
The linear terms of $u_{ij}$ are
\begin{align}\label{3Q2a}
Q_2=&\frac{1}{|Dw|^2\log|Dw|^2}\bigg[\big[2(1-G_u)fu_1v-4G_{uu}u_1-4G_{ux_1}\big]\sum_{1\leq k\leq n}w_ku_{k1}\notag\\&
-4v^2\sum_{2\leq i\leq n}G_{ux_i}\sum_{1\leq k\leq n}w_ku_{ki}-4(1-G_u)(G_{uu}u_1^2+2G_{ux_1}u_1+G_{x_1x_1})u_{11}\notag\\
&-4(1-G_u)(v^2+1)\sum_{2\leq i\leq n}(G_{ux_i}u_1+G_{x_1x_i})u_{1i}-4(1-G_u)v^2\sum_{2\leq i\leq n}G_{x_ix_i}u_{ii}\bigg];
\end{align}
and  the
remaining terms are
\begin{align}
Q_3=&I_2+\frac{1}{|Dw|^2\log|Dw|^2}\bigg[2(1-G_u)f_uv^3u_1w_1-2G_{uu}fv^3u_1w_1\notag\\
&+2(G_{uu}u_1^2+2G_{ux_1}u_1+G_{x_1x_1})^2+2(v^2+1)\sum_{2\leq i\leq n}(G_{ux_i}u_1+G_{x_1x_i})^2\notag\\
&+2v^2\sum_{2\leq i,k\leq n}G_{x_kx_i}+2(1-G_u)v^3\sum_{1\leq k\leq n}f_{x_k}w_k-2fv^3\sum_{1\leq k\leq n}G_{ux_k}w_k\notag\\&
-2G_{uuu}u_1^3w_1-2v^2u_1w_1\sum_{2\leq i\leq n}G_{ux_ix_i}-4G_{uux_1}u_1^2w_1-2u_1^2\sum_{1\leq k\leq n}w_kG_{uux_k}\notag\\&
-2v^2\sum_{1\leq k\leq n}w_k\sum_{2\leq i\leq n}G_{x_ix_ix_k}-2G_{ux_1x_1}u_1w_1\notag\\
&-4u_1\sum_{1\leq k\leq n}G_{ux_1x_k}w_k-2\sum_{1\leq k\leq n}G_{x_1x_1x_k}w_k\bigg].\label{3Q3a}
\end{align}
From the estimate on $I_2$ in \eqref{3I2c}, we have
\begin{align}\label{3Q3b}
Q_3\geq&f v^3-\frac{2fG_{uu}}{|Dw|^2\log|Dw|^2}v^3u_1w_1-(1+c^{11}\alpha_0^2) u_1^2\log|Dw|^2-C_2u_1^2,
\end{align}
in the computation of $Q_3$, we use the relation $f_u\geq 0$, where $C_2$ is a positive constant which depends only on $n, \Omega, M_0, \mu_0, L_1, L_2$.

{\bf Step 2:} In this step we shall treat the terms $Q_1, Q_2$ , using the first order derivative condition
 $$\varphi_i(x_0)=0,$$
  and let
  \begin{align}\label{3A}
A=|Dw|^2\log|Dw|^2.
\end{align}

By \eqref{3varphii=0}  and  \eqref{3|Dw|^2i},  we have
\begin{align}\label{3wkwki=1}
\begin{split}
\sum_{1\leq k\leq n}w_kw_{ki}=&-\frac{h'}{2}|Dw|^2\log|Dw|^2u_i-\frac{g'\gamma^i}{2}|Dw|^2\log|Dw|^2\\
=&-\frac{h'}{2}Au_i-\frac{g'\gamma^i}{2}A,\qquad i=1,2,\ldots,n.
\end{split}
\end{align}

Putting \eqref{3wki} into \eqref{3wkwki=1},  we get
\begin{align}\label{3wkuki=1}
\begin{split}
(1-G_u)\sum_{1\leq k\leq n} w_ku_{ki}
=&-\frac{h'}{2}Au_i-\frac{g'\gamma^i}{2}A+G_{uu}\sum_{1\leq k\leq n} w_ku_ku_i+\sum_{1\leq k\leq n} w_ku_kG_{ux_i}\\
&+\sum_{1\leq k\leq n} w_kG_{ux_k}u_i+\sum_{1\leq k\leq n} w_kG_{x_kx_i}\\
=&-\frac{h'}{2}Au_i-\frac{g'\gamma^i}{2}A+G_{uu}w_1u_1u_i+ w_1u_1G_{ux_i}\\
&+\sum_{1\leq k\leq n} w_kG_{ux_k}u_i+\sum_{1\leq k\leq n} w_kG_{x_kx_i},\quad i=1,2,\ldots,n.
\end{split}
\end{align}

By \eqref{3wkuki=1}, we have
\begin{align}\label{3wkuk1}
\begin{split}
(1-G_u)\sum_{1\leq k\leq n} w_ku_{k1}
=&-\frac{h'}{2}Au_1-\frac{g'\gamma^1}{2}A+G_{uu}u_1^2w_1+G_{ux_1}w_1u_1\\
&+\sum_{1\leq k\leq n} w_kG_{ux_k}u_1+\sum_{1\leq k\leq n} w_kG_{x_kx_1},
\end{split}
\end{align}
and
\begin{align}\label{3wkuki=2}
\begin{split}
(1-G_u)\sum_{1\leq k\leq n} w_ku_{ki}
=&-\frac{g'\gamma^i}{2}A+G_{ux_i}w_1u_1+\sum_{1\leq k\leq n} w_kG_{x_kx_i},\quad i=2,\ldots,n.
\end{split}
\end{align}

Through \eqref{3wkuki=2}  and the choice of the coordinate at $x_0$, we have
\begin{align}\label{3u1i}
\begin{split}
u_{1i}
=&-\frac{1}{w_1}w_iu_{ii}-\frac{g'\gamma^i}{2(1-G_u)}\frac{A}{w_1}+\frac{G_{ux_i}}{1-G_u}u_1+\frac{1}{(1-G_u)w_1}\sum_{1\leq k\leq n} w_kG_{x_kx_i},\\
&\hspace{200pt} i=2,3,\ldots,n.
\end{split}
\end{align}
Using \eqref{3wkuk1}  and \eqref{3u1i},  it follows that
\begin{align}\label{3u11b}
\begin{split}
u_{11}
=&\sum_{2\leq i\leq n}\frac{w_i^2}{w_1^2}u_{ii}-\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}+\frac{G_{uu}}{1-G_u}u_1^2
-\frac{g'\gamma^1}{2(1-G_u)}\frac{A}{w_1}\\
&+\frac{G_{ux_1}}{1-G_u}u_1+\frac{u_1}{(1-G_u)w_1}\sum_{1\leq k\leq n} w_kG_{ux_k}+\frac{g'}{2(1-G_u)}\sum_{2\leq i\leq n}w_i\gamma^i\frac{A}{w_1^2}\\
&-\frac{u_1}{(1-G_u)w_1}\sum_{2\leq i\leq n}G_{ux_i}w_i+\frac{1}{(1-G_u)w_1}\sum_{1\leq k\leq n} w_kG_{x_kx_1}\\&
-\frac{1}{(1-G_u)w_1^2}\sum_{1\leq k,i\leq n} w_iw_kG_{x_kx_i}\\
=&:\sum_{2\leq i\leq n}\frac{w_i^2}{w_1^2}u_{ii}-\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}+\frac{D}{1-G_u},
\end{split}
\end{align}
 where we have let
 \begin{align}\label{3Da}
\begin{split}
D=&G_{uu}u_1^2
-\frac{g'\gamma^1}{2}\frac{A}{w_1}+G_{ux_1}u_1+\frac{u_1}{w_1}\sum_{1\leq k\leq n} w_kG_{ux_k}+\frac{g'}{2}\sum_{2\leq i\leq n}w_i\gamma^i\frac{A}{w_1^2}\\
&-\frac{u_1}{w_1}\sum_{2\leq i\leq n}G_{ux_i}w_i+\frac{1}{w_1}\sum_{1\leq k\leq n} w_kG_{x_kx_1}
-\frac{1}{w_1^2}\sum_{1\leq k,i\leq n} w_iw_kG_{x_kx_i}.
\end{split}
\end{align}
It follows that
\begin{align}\label{3Db}
\begin{split}
|D|
\le C_3 u_1^2.
\end{split}
\end{align}

By \eqref{3u11a}  and \eqref{3u11b},  we have
\begin{align}\label{3uii=2}
\begin{split}
\sum_{2\leq i\leq n}(v^2+\frac{w_i^2}{w_1^2})u_{ii}
=f v^3+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}.
\end{split}
\end{align}

Now we use the formulas \eqref{3wkuk1}-\eqref{3u11b}  to treat  each term in $Q_1, Q_2$.  At first, we treat  the first four terms of $Q_1$ in \eqref{3Q1a}, and get \eqref{3u11^2}-\eqref{3u1iwkuki}.

By \eqref{3u11b}, we have
\begin{align}\label{3u11^2}
2(1-G_u)^2u_{11}^2=&2(1-G_u)^2\big[\sum_{2\leq i\leq n}\frac{w_i^2}{w_1^2}u_{ii}-\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}+\frac{D}{1-G_u}\big]^2\notag\\
=&2(1-G_u)^2(\sum_{2\leq i\leq n}\frac{w_i^2}{w_1^2}u_{ii})^2-2(1-G_u)(h'\frac{Au_1}{w_1^3}-2\frac{D}{w_1^2})\sum_{2\leq i\leq n}w_i^2u_{ii}\notag\\
&+\frac{h'^2}{2}\frac{A^2u_1^2}{w_1^2}-2h'\frac{Au_1}{w_1}D+2D^2
\notag\\
=&2(1-G_u)^2(\sum_{2\leq i\leq n}\frac{w_i^2}{w_1^2}u_{ii})^2-2(1-G_u)(h'\frac{Au_1}{w_1^3}-2\frac{D}{w_1^2})\sum_{2\leq i\leq n}w_i^2u_{ii}\notag\\
&+\frac{h'^2}{2}\frac{A^2u_1^2}{w_1^2}+AO(u_1^2).
\end{align}

By \eqref{3u1i}, we have
\begin{align}\label{3u1i^2}
&2(1-G_u)^2(v^2+1)\sum_{2\leq i\leq n}u_{1i}^2\notag\\
=&2(v^2+1)\sum_{2\leq i\leq n}\big[(1-G_u)\frac{w_i}{w_1}u_{ii}+\frac{g'\gamma^i}{2}\frac{A}{w_1}-G_{ux_i}u_1-\frac{1}{w_1}\sum_{1\leq k\leq n} w_kG_{x_kx_i}\big]^2\notag\\
=&2(1-G_u)^2(v^2+1)\sum_{2\leq i\leq n}\frac{w_i^2}{w_1^2}u_{ii}^2+2(1-G_u)g'A\frac{v^2+1}{w_1^2}\sum_{2\leq i\leq n}w_i\gamma^iu_{ii}\notag\\
&-4(1-G_u)\frac{v^2+1}{w_1^2}\sum_{2\leq i\leq n}(G_{ux_i}u_1w_1+\sum_{1\leq k\leq n} w_kG_{x_kx_i})w_iu_{ii}\notag\\
&+\frac{c^{11}}{2}g'^2\frac{v^2+1}{w_1^2}A^2-2g'A\frac{v^2+1}{w_1^2}\sum_{2\leq i\leq n}(G_{ux_i}u_1w_1+\sum_{1\leq k\leq n} w_kG_{x_kx_i})\gamma^{i}\notag\\
&+\frac{2(v^2+1)}{w_1^2}\sum_{2\leq i\leq n}(G_{ux_i}u_1w_1+\sum_{1\leq k\leq n} w_kG_{x_kx_i})^2
\notag\\
=&2(1-G_u)^2(v^2+1)\sum_{2\leq i\leq n}\frac{w_i^2}{w_1^2}u_{ii}^2+2(1-G_u)g'A\frac{v^2+1}{w_1^2}\sum_{2\leq i\leq n}w_i\gamma^iu_{ii}\notag\\
&-4(1-G_u)\frac{v^2+1}{w_1^2}\sum_{2\leq i\leq n}(G_{ux_i}u_1w_1+\sum_{1\leq k\leq n} w_kG_{x_kx_i})w_iu_{ii}\notag\\
&+\frac{c^{11}}{2}g'^2\frac{v^2+1}{w_1^2}A^2+AO(u_1^2).
\end{align}

By \eqref{3u11b} and \eqref{3wkuk1}, we have
\begin{align}\label{3u11wkuk1}
&4(1-G_u)\frac{u_1}{v^2} u_{11}\sum_{1\leq k\leq n}w_ku_{k1}\notag\\
=&\frac{4u_1}{v^2}\big[\sum_{2\leq i\leq n}\frac{w_i^2}{w_1^2}u_{ii}-\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}+\frac{D}{1-G_u}\big]
\big[-\frac{h'}{2}Au_1-\frac{g'\gamma^1}{2}A+G_{uu}u_1^2w_1\notag\\&+G_{ux_1}w_1u_1
+\sum_{1\leq k\leq n} w_kG_{ux_k}u_1+\sum_{1\leq k\leq n} w_kG_{x_kx_1}\big]\notag\\
=&\big[-2h'\frac{Au_1^2}{v^2w_1^2}+4G_{uu}\frac{u_1^3}{v^2w_1}-2g'\gamma^1\frac{Au_1}{v^2w_1^2}+4G_{ux_1}\frac{u_1^2}{v^2w_1}
+4\sum_{1\leq k\leq n} w_kG_{ux_k}\frac{u_1^2}{v^2w_1^2}\notag\\&
+4\sum_{1\leq k\leq n} w_kG_{x_kx_1}\frac{u_1}{v^2w_1^2}\big]\sum_{2\leq i\leq n}w_i^2u_{ii}+\frac{h'^2}{1-G_u}\frac{A^2u_1^3}{v^2w_1}-\frac{2h'}{1-G_u}\frac{Au_1^2}{v^2}D\notag\\&
-\frac{4u_1}{v^2}[\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}][\frac{g'\gamma^1}{2}A-G_{uu}u_1^2w_1-G_{ux_1}w_1u_1\notag\\&
-\sum_{1\leq k\leq n} w_kG_{ux_k}u_1-\sum_{1\leq k\leq n} w_kG_{x_kx_1}]
\notag\\
=&\big[-2h'\frac{Au_1^2}{v^2w_1^2}+4G_{uu}\frac{u_1^3}{v^2w_1}-2g'\gamma^1\frac{Au_1}{v^2w_1^2}+4G_{ux_1}\frac{u_1^2}{v^2w_1}
+4\sum_{1\leq k\leq n} w_kG_{ux_k}\frac{u_1^2}{v^2w_1^2}\notag\\&
+4\sum_{1\leq k\leq n} w_kG_{x_kx_1}\frac{u_1}{v^2w_1^2}\big]\sum_{2\leq i\leq n}w_i^2u_{ii}+\frac{h'^2}{1-G_u}\frac{A^2u_1^3}{v^2w_1}+AO(u_1^2).
\end{align}

By \eqref{3u1i} and \eqref{3wkuki=2}, we have
\begin{align*}
&4(1-G_u)u_1\sum_{2\leq i\leq n}u_{1i}\sum_{1\leq k\leq n}w_ku_{ki}\notag\\
=&4u_1\sum_{2\leq i\leq n}\big[-\frac{g'\gamma^i}{2}A+G_{ux_i}w_1u_1+\sum_{1\leq k\leq n} w_kG_{x_kx_i}\big]\big[-\frac{1}{w_1}w_iu_{ii}-\frac{g'\gamma^i}{2(1-G_u)}\frac{A}{w_1}\notag\\&+\frac{G_{ux_i}}{1-G_u}u_1+\frac{1}{(1-G_u)w_1}\sum_{1\leq k\leq n} w_kG_{x_kx_i}\big]\notag\\
=&2g'\frac{Au_1}{w_1}\sum_{2\leq i\leq n}w_i\gamma^iu_{ii}-4u_1^2\sum_{2\leq i\leq n}w_iG_{ux_i}u_{ii}
-\frac{4u_1}{w_1}\sum_{1\leq k\leq n}w_k\sum_{2\leq i\leq n}G_{x_kx_i}w_iu_{ii}\notag\\
&
+\frac{c^{11}g'^2}{1-G_u}\frac{A^2u_1}{w_1}-\sum_{2\leq i\leq n}
\big[2g'\gamma^iAu_1-4G_{ux_i}w_1u_1^2-4u_1\sum_{1\leq k\leq n} w_kG_{x_kx_i}\big]\big[\frac{G_{ux_i}}{1-G_u}u_1\notag\\&+\frac{1}{(1-G_u)w_1}\sum_{1\leq k\leq n} w_kG_{x_kx_i}\big]-\frac{2g'}{1-G_u}\frac{Au_1}{w_1}\sum_{2\leq i\leq n}\big[G_{ux_i}u_1w_1+\sum_{1\leq k\leq n} w_kG_{x_kx_i}\big]\gamma^i
\notag\\
\end{align*}
\begin{align}
=&2g'\frac{Au_1}{w_1}\sum_{2\leq i\leq n}w_i\gamma^iu_{ii}-4u_1^2\sum_{2\leq i\leq n}w_iG_{ux_i}u_{ii}
-\frac{4u_1}{w_1}\sum_{1\leq k\leq n}w_k\sum_{2\leq i\leq n}G_{x_kx_i}w_iu_{ii}\notag\\&
+\frac{c^{11}g'^2}{1-G_u}\frac{A^2u_1}{w_1}+AO(u_1^2).\label{3u1iwkuki}
\end{align}

Now we  treat  the first four terms of $Q_2$ in \eqref{3Q2a}, and get \eqref{3wkuk1a}-\eqref{3u1ia}.

From \eqref{3wkuk1}, we get
\begin{align}\label{3wkuk1a}
&\big[2(1-G_u)fu_1v-4G_{uu}u_1-4G_{ux_1}\big]\sum_{1\leq k\leq n}w_ku_{k1}\notag\\
=&\frac{1}{1-G_u}\big[2(1-G_u)fu_1v-4G_{uu}u_1-4G_{ux_1}\big]\big[-\frac{h'}{2}Au_1+G_{uu}u_1^2w_1\notag\\&-\frac{g'\gamma^1}{2}A+G_{ux_1}w_1u_1
+\sum_{1\leq k\leq n} w_kG_{ux_k}u_1+\sum_{1\leq k\leq n} w_kG_{x_kx_1}\big]\notag\\
=&-h'fAvu_1^2+2G_{uu}fvu_1^3w_1-fg'\gamma^1Avu_1\notag\\
&+2fvu_1[G_{ux_1}w_1u_1
+\sum_{1\leq k\leq n} w_kG_{ux_k}u_1+\sum_{1\leq k\leq n} w_kG_{x_kx_1}]\notag\\
&-\frac{4}{1-G_u}(G_{uu}u_1+G_{ux_1})\big[-\frac{h'}{2}Au_1+G_{uu}u_1^2w_1\notag\\&-\frac{g'\gamma^1}{2}A+G_{ux_1}w_1u_1
+\sum_{1\leq k\leq n} w_kG_{ux_k}u_1+\sum_{1\leq k\leq n} w_kG_{x_kx_1}\big]
\notag\\
=&-h'fvu_1^2+2G_{uu}fvu_1^3w_1+AO(u_1^2).
\end{align}

From \eqref{3wkuki=2}, we have
\begin{align}\label{3wkuki=2a}
&-4v^2\sum_{2\leq i\leq n}G_{ux_i}\sum_{1\leq k\leq n}w_ku_{ki}\notag\\
=&-\frac{4v^2}{1-G_u}\sum_{2\leq i\leq n}G_{ux_i}\big[-\frac{g'\gamma^i}{2}A+G_{ux_i}w_1u_1+\sum_{1\leq k\leq n} w_kG_{x_kx_i}\big]\notag\\
=&\frac{2g'}{1-G_u}\sum_{2\leq i\leq n}G_{ux_i}\gamma^iAv^2-\frac{4}{1-G_u}\sum_{2\leq i\leq n}G^2_{ux_i}v^2u_1w_1\notag\\
&-\frac{4}{1-G_u}\sum_{2\leq i\leq n}G_{ux_i}\sum_{1\leq k\leq n} w_kG_{x_kx_i}v^2
\notag\\
=&AO(u_1^2).
\end{align}

From \eqref{3u11b}, we have
\begin{align}\label{3u11c}
&-4(1-G_u)(G_{uu}u_1^2+2G_{ux_1}u_1+G_{x_1x_1})u_{11}\notag\\
=&-4(G_{uu}u_1^2+2G_{ux_1}u_1+G_{x_1x_1})\big[(1-G_u)\sum_{2\leq i\leq n}\frac{w_i^2}{w_1^2}u_{ii}-\frac{h'}{2}\frac{Au_1}{w_1}+D\big]\notag\\
=&-4(1-G_u)(G_{uu}u_1^2+2G_{ux_1}u_1+G_{x_1x_1})\sum_{2\leq i\leq n}\frac{w_i^2}{w_1^2}u_{ii}\notag\\
&+4(G_{uu}u_1^2+2G_{ux_1}u_1+G_{x_1x_1})(\frac{h'}{2}\frac{Au_1}{w_1}-D)
\notag\\
=&-4(1-G_u)(G_{uu}u_1^2+2G_{ux_1}u_1+G_{x_1x_1})\sum_{2\leq i\leq n}\frac{w_i^2}{w_1^2}u_{ii}+AO(u_1^2).
\end{align}

From \eqref{3u1i}, we have
\begin{align}\label{3u1ia}
&-4(1-G_u)(v^2+1)\sum_{2\leq i\leq n}(G_{ux_i}u_1+G_{x_1x_i})u_{1i}\notag\\
=&4(v^2+1)\sum_{2\leq i\leq n}(G_{ux_i}u_1+G_{x_1x_i})\big[\frac{1-G_u}{w_1}w_iu_{ii}+\frac{g'\gamma^i}{2}\frac{A}{w_1}\notag\\&
-G_{ux_i}u_1-\frac{1}{w_1}\sum_{1\leq k\leq n} w_kG_{x_kx_i}\big]\notag\\
=&4(1-G_u)\frac{v^2+1}{w_1}\sum_{2\leq i\leq n}(G_{ux_i}u_1+G_{x_1x_i})w_iu_{ii}\notag\\
&+4(v^2+1)\sum_{2\leq i\leq n}(G_{ux_i}u_1+G_{x_1x_i})\big[\frac{g'\gamma^i}{2}\frac{A}{w_1}
-G_{ux_i}u_1-\frac{1}{w_1}\sum_{1\leq k\leq n} w_kG_{x_kx_i}\big]
\notag\\
=&4(1-G_u)\frac{v^2+1}{w_1}\sum_{2\leq i\leq n}(G_{ux_i}u_1+G_{x_1x_i})w_iu_{ii}+AO(u_1^2).
\end{align}

We treat the term $Q_1$ using the relations \eqref{3u11^2}-\eqref{3u1iwkuki} and  use the formulas \eqref{3wkuk1a}-\eqref{3u1ia}  to treat the term $Q_2$.
 By the formula on $Q_3$ in \eqref{3Q3a},  we can get the following new formula of \eqref{3aijvarphiijb},
\begin{align}\label{3aijvarphiijc}
0\geq \sum_{1\leq i,j\leq n}a^{ij}\varphi_{ij}=: J_1+J_2,
\end{align}
where $J_1$  only contains the terms with $u_{ii}$ , the other terms belong to $J_2$. We can write
\begin{align}\label{3J1a}
J_1=:\frac{1}{A}\big[J_{11}+J_{12}\big],
\end{align}
here $J_{11}$  contains the quadratic terms of $u_{ii}\,(i\geq2)$, and  $J_{12}$ is the term including linear terms of $u_{ii}\,(i\geq2)$.
It follows that
\begin{align}\label{3J11a}
J_{11}=&2(1-G_u)^2\big\{(\sum_{2\leq i\leq n}\frac{w_i^2}{w_1^2}u_{ii})^2+(v^2+1)\sum_{2\leq i\leq n}\frac{w_i^2}{w_1^2}u_{ii}^2+v^2\sum_{2\leq i\leq n}u_{ii}^2\big\}\notag\\
=&2(1-G_u)^2\big\{\sum_{2\leq i\leq n}\frac{w_i^4}{w_1^4}u_{ii}^2+2\sum_{2\leq i< j\leq n}\frac{w_i^2w_j^2}{w_1^4}u_{ii}u_{jj}
+\sum_{2\leq i\leq n}[v^2+(v^2+1)\frac{w_i^2}{w_1^2}]u_{ii}^2\big\}\notag\\
=&2(1-G_u)^2\big\{\sum_{2\leq i\leq n}(v^2+\frac{w_i^2}{w_1^2})(1+\frac{w_i^2}{w_1^2})u_{ii}^2+2\sum_{2\leq i< j\leq n}\frac{w_i^2w_j^2}{w_1^4}u_{ii}u_{jj}
\big\}\notag\\
=&2(1-G_u)^2\big\{\sum_{2\leq i\leq n}d_ie_iu_{ii}^2+2\sum_{2\leq i< j\leq n}\frac{w_i^2w_j^2}{w_1^4}u_{ii}u_{jj}
\big\},
\end{align}
where
\begin{align}
d_i=&v^2+\frac{w_i^2}{w_1^2}, \quad i=2,3,\ldots,n,\label{3di}\\
e_i=&1+\frac{w_i^2}{w_1^2},\quad i=2,3,\ldots,n.\label{3ei}
\end{align}
And
\begin{align}\label{3J12a}
J_{12}=&-2(1-G_u)(h'\frac{Au_1}{w_1^3}-2\frac{D}{w_1^2})\sum_{2\leq i\leq n}w_i^2u_{ii}
+2(1-G_u)g'A\frac{v^2+1}{w_1^2}\sum_{2\leq i\leq n}w_i\gamma^iu_{ii}\notag\\
&-4(1-G_u)\frac{v^2+1}{w_1^2}\sum_{2\leq i\leq n}(\sum_{1\leq k\leq n} w_kG_{x_kx_i}-G_{x_1x_i}w_1)w_iu_{ii}\notag\\
&-\big[2h'\frac{Au_1^2}{v^2w_1^2}-4G_{uu}\frac{u_1^3}{v^2w_1}+2g'\gamma^1\frac{Au_1}{v^2w_1^2}-4G_{ux_1}\frac{u_1^2}{v^2w_1}
-4\sum_{1\leq k\leq n} w_kG_{ux_k}\frac{u_1^2}{v^2w_1^2}\notag\\&
-4\sum_{1\leq k\leq n} w_kG_{x_kx_1}\frac{u_1}{v^2w_1^2}\big]\sum_{2\leq i\leq n}w_i^2u_{ii}+2g'\frac{Au_1}{w_1}\sum_{2\leq i\leq n}w_i\gamma^iu_{ii}
-4u_1^2\sum_{2\leq i\leq n}w_iG_{ux_i}u_{ii}\notag\\&
-\frac{4u_1}{w_1}\sum_{1\leq k\leq n}w_k\sum_{2\leq i\leq n}G_{x_kx_i}w_iu_{ii}
-4(1-G_u)v^2\sum_{2\leq i\leq n}G_{x_ix_i}u_{ii}\notag\\&-4(1-G_u)(G_{uu}u_1^2+2G_{ux_1}u_1+G_{x_1x_1})\sum_{2\leq i\leq n}\frac{w_i^2}{w_1^2}u_{ii}\notag\\
=:&\sum_{2\leq i\leq n}K_iu_{ii},
\end{align}
where
\begin{align}\label{3Kia}
K_i=&-2(1-G_u)(h'\frac{Au_1}{w_1^3}-2\frac{D}{w_1^2})w_i^2
+2(1-G_u)g'A\frac{v^2+1}{w_1^2}w_i\gamma^i\notag\\
&-4(1-G_u)\frac{v^2+1}{w_1^2}(G_{ux_i}u_1w_1+\sum_{1\leq k\leq n} w_kG_{x_kx_i})w_i\notag\\
&+\big[-2h'\frac{Au_1^2}{v^2w_1^2}+4G_{uu}\frac{u_1^3}{v^2w_1}-2g'\gamma^1\frac{Au_1}{v^2w_1}+4G_{ux_1}\frac{u_1^2}{v^2w_1}
+4\sum_{1\leq k\leq n} w_kG_{ux_k}\frac{u_1^2}{v^2w_1^2}\notag\\&
+\sum_{1\leq k\leq n} w_kG_{x_kx_1}\frac{u_1}{v^2w_1^2}\big]w_i^2
+2g'\frac{Au_1}{w_1}w_i\gamma^i-4u_1^2w_iG_{ux_i}
-\frac{4u_1}{w_1}\sum_{1\leq k\leq n}w_kG_{x_kx_i}w_i\notag\\&
-4(1-G_u)v^2G_{x_ix_i}-4(1-G_u)(G_{uu}u_1^2+2G_{ux_1}u_1+G_{x_1x_1})\frac{w_i^2}{w_1^2}\notag\\
&+4(1-G_u)\frac{v^2+1}{w_1}(G_{ux_i}u_1+G_{x_1x_i})w_i.
\end{align}
It follows that
\begin{align}\label{3Kib}
|K_i|\leq&C_4A,\quad i=2,\ldots,n.
\end{align}

We write other terms as $J_2$, then
\begin{align}\label{3J2a}
J_2=&Q_3-h'fvu_1^2+\frac{2fG_{uu}}{A}vu_1^3w_1+\frac{h'^2}{2}\frac{Au_1^2}{w_1^2}\notag\\
+&\frac{h'^2}{1-G_u}\frac{Au_1^3}{v^2w_1}
+\frac{c^{11}}{2}g'^2\frac{u_1^2}{w_1^2}A
+\frac{c^{11}g'^2}{1-G_u}\frac{Au_1}{w_1}+O(u_1^2).
\end{align}

So by the choice of $\mu_0$ and the formula on $Q_3$ in  \eqref{3Q3b},\eqref{3A} and \eqref{3.1}, we get the following estimate on $J_2$,
\begin{align}\label{3J2b}
J_2\geq&-(1+c^{11}\alpha_0^2) u_1^2\log|Dw|^2+\frac{h'^2}{2}\frac{Au_1^2}{w_1^2}+\frac{h'^2}{1-G_u}\frac{Au_1^3}{v^2w_1}\notag\\&
+\frac{c^{11}}{2}g'^2\frac{u_1^2}{w_1^2}A
+\frac{c^{11}g'^2}{1-G_u}\frac{Au_1}{w_1}-C_5u_1^2\notag\\
\geq&\frac{1}{4}(1+c^{11}\alpha_0^2) u_1^2\log|Dw|^2-C_6u_1^2,
\end{align}
where  $ C_4, C_5, C_6 $ are  positive constants which only depend on $n, \Omega, \mu_0, M_0, L_1, L_2$.

{\bf Step 3:}
In this step, we concentrate on $J_1$. We first treat the terms $J_{11}$ and $J_{12}$  and obtain the formula \eqref{3aijvarphiijd},
then we complete the proof of  Theorem~\ref{Thm1.1} through  Lemma~\ref{Lem4.5}.

 By \eqref{3uii=2}, we have
\begin{align}\label{3u22}
\begin{split}
u_{22}
=&-\frac{1}{d_2}\sum_{3\leq i\leq n}d_iu_{ii}
+\frac{1}{d_2}[f v^3+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}],
\end{split}
\end{align}

We first treat the term $J_{11}$:
using \eqref{3u22} to simplify \eqref{3J11a}, we get
\begin{align}\label{3J11b}
\begin{split}
J_{11}
=&2(1-G_u)^2\bigg\{d_2e_2u_{22}^2+\sum_{3\leq i\leq n}d_ie_iu_{ii}^2+2\frac{w_2^2}{w_1^2}u_{22}\sum_{3\leq j\leq n}\frac{w_j^2}{w_1^2}u_{jj}\\&
+2\sum_{3\leq i< j\leq n}\frac{w_i^2w_j^2}{w_1^4}u_{ii}u_{jj}
\bigg\}\\
=&2(1-G_u)^2\bigg\{\frac{e_2}{d_2}\big[-\sum_{3\leq i\leq n}d_iu_{ii}+f v^3+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}\big]^2
+\sum_{3\leq i\leq n}d_ie_iu_{ii}^2\\&+\frac{2w_2^2}{d_2w_1^2}\big[-\sum_{3\leq i\leq n}d_iu_{ii}+f v^3
+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}\big]\sum_{3\leq j\leq n}\frac{w_j^2}{w_1^2}u_{jj}\\&
+2\sum_{3\leq i< j\leq n}\frac{w_i^2w_j^2}{w_1^4}u_{ii}u_{jj}
\bigg\}\\
=&\frac{2(1-G_u)^2}{d_2}\bigg\{\sum_{3\leq i\leq n}\big[e_2d_i^2+e_id_id_2-2\frac{w_2^2w_i^2}{w_1^4}d_i\big]u_{ii}^2\\&
+2\sum_{3\leq i< j\leq n}\big[e_2d_id_j-\frac{w_2^2w_i^2}{w_1^4}d_j-\frac{w_2^2w_j^2}{w_1^4}d_i+\frac{w_i^2w_j^2}{w_1^4}\big]u_{ii}u_{jj}\\&
-2e_2\big[f v^3+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}\big]\sum_{3\leq i\leq n}d_iu_{ii}\\&
+2\frac{w_2^2}{w_1^4}\big[f v^3+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}\big]\sum_{3\leq i\leq n}w_i^2u_{ii}\\&
+e_2\big[f v^3+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}\big]^2\bigg\}.
\end{split}
\end{align}

 We can rewrite it as the following
\begin{align}\label{3J11bb}
\begin{split}
J_{11}
  =&:\frac{2(1-G_u)^2}{d_2}\bigg\{\sum_{3\leq i\leq n}b_{ii}u_{ii}^2+2\sum_{3\leq i< j\leq n}b_{ij}u_{ii}u_{jj}\\&-2e_2\big[f v^3+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}\big]\sum_{3\leq i\leq n}d_iu_{ii}\\&
+2\frac{w_2^2}{w_1^4}\big[f v^3+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}\big]\sum_{3\leq i\leq n}w_i^2u_{ii}\\&
+e_2\big[f v^3+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}\big]^2\bigg\},
\end{split}
\end{align}
where
\begin{align}\label{3bija}
\begin{split}
b_{ii}=&e_2d_i^2+e_id_id_2-2\frac{w_2^2w_i^2}{w_1^4}d_i\\

=&:2u_1^4+A_{1i}u_1^2+A_{2i},\quad i\geq 3\\
b_{ij}=&e_2d_id_j-\frac{w_2^2w_i^2}{w_1^4}d_j-\frac{w_2^2w_j^2}{w_1^4}d_i+\frac{w_i^2w_j^2}{w_1^4}\\
=&:u_1^4+G_{ij}u_1^2+\hat{G}_{ij},
\qquad\qquad\qquad\qquad i\neq j,\,i,j\geq 3,
\end{split}
\end{align}
and
\begin{align}\label{3A1i}
\begin{split}
A_{1i}=&4+(w_2^2+w_i^2)\frac{u_1^2}{w_1^2}, \\
A_{2i}=&2+(3w_2^2+5w_i^2)\frac{u_1^2}{w_1^2}
+w_i^2(w_2^2+w_i^2)\frac{u_1^2}{w_1^4}+\frac{2w_2^2+4w_i^2}{w_1^2}+\frac{2w_i^2(w_2^2+w_i^2)}{w_1^4}, \\
G_{ij}=&2+w_2^2\frac{u_1^2}{w_1^2}, \\
\hat{G}_{ij}=&1+(2w_2^2+w_i^2+w_j^2)\frac{u_1^2}{w_1^2}
+\frac{w_2^2+w_i^2+w_j^2}{w_1^2}+\frac{2w_i^2w_j^2}{w_1^4}-\frac{w_2^2w_i^2w_j^2}{w_1^6}.
\end{split}
\end{align}

Now we simplify the terms in $J_{12}$: by  \eqref{3u22},  we can rewrite \eqref{3J12a} as
\begin{align}\label{3J12b}
\begin{split}
J_{12}=&K_2u_{22}+\sum_{3\leq i\leq n}K_iu_{ii}\\
=&\frac{K_2}{d_2}\big[-\sum_{3\leq i\leq n}d_iu_{ii}
+f v^3+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}\big]+\sum_{3\leq i\leq n}K_iu_{ii}\\
=&\sum_{3\leq i\leq n}[K_i-\frac{K_2d_i}{d_2}]u_{ii}
+\frac{K_2}{d_2}[f v^3+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}].
\end{split}
\end{align}
Using  \eqref{3J11b} and \eqref{3J12b} to treat \eqref{3J1a},  we have
\begin{align}\label{3J1b}
\begin{split}
J_{1}=&\frac{2(1-G_u)^2}{Ad_2}\bigg\{\sum_{3\leq i\leq n}b_{ii}u_{ii}^2+2\sum_{3\leq i< j\leq n}b_{ij}u_{ii}u_{jj}\\&
+\sum_{3\leq i\leq n}\tilde{K}_iu_{ii}\big]+ R,
\end{split}
\end{align}
where
\begin{align}\label{3tildeKia}
\tilde{K}_i=&-2e_2\big[f v^3+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}\big]d_i\notag\\&
+2\frac{w_2^2}{w_1^4}\big[f v^3+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}\big]w_i^2+
K_id_2-K_2d_i.
\end{align}
We also have let
\begin{align*}
\begin{split}
R=&\frac{2e_2(1-G_u)^2}{Ad_2}\big[f v^3+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}\big]^2\\
+&\frac{K_2}{Ad_2}[f v^3+\frac{h'}{2(1-G_u)}\frac{Au_1}{w_1}-\frac{D}{1-G_u}].
\end{split}
\end{align*}

For $\tilde{K}_i$ and $R$, using the formulas on $D, K_i$ in \eqref{3Db}, \eqref{3Kib}; the formula of $A$ in \eqref{3A}; $ e_i; d_i$  in \eqref{3ei}-\eqref{3di},
and  $h(u), g(d)$ in \eqref{3.1}, we have the following estimates
\begin{align}
|\tilde{K}_i|\leq& C_7u_1^5,\quad i=3,\ldots,n;\label{3tildeKib}\\
|R|\leq &C_8\frac{u_1^2}{\log u_1}.\label{3R}
\end{align}

Now we use  Lemma~\ref{Lem4.5}, if there is a sufficiently large positive constant $C_9$ such that
\begin{align}
|Du|(x_0) \ge C_9,\label{3C9}
\end{align}
 then we have
\begin{align}\label{3J1c}
J_{1}
\geq &\frac{2}{Ad_2}({-C_{10} u_1^6}),\notag\\
\ge & -C_{11} u_1^2           ,
\end{align}
where we use the formulas  $d_2$ in \eqref{3di} and $A$ in \eqref{3A}.

Using the estimates on $J_1$ in \eqref{3J1c} and  $J_2$ in \eqref{3J2b}, from  \eqref{3aijvarphiijc} we obtain
\begin{align}\label{3aijvarphiijd}
0\geq &\sum_{1\leq i,j\leq n}a^{ij}\varphi_{ij}\notag\\
 \geq&\frac{1}{4}(1+c^{11}\alpha_0^2) u_1^2\log|Dw|^2-C_{12}u_1^2\notag\\
 \geq&\frac{1}{4} u_1^2\log|Dw|^2-C_{12}u_1^2,
\end{align}

 There exists a positive constant
$C_{13}$ such that
 \begin{align}\label{3C13}
 |Du|(x_0)\leq C_{13}.
 \end{align}

 So from  Case 1, Case 2,  and \eqref{3C13}, we have
 $$|Du|(x_0)\leq C_{14}, \quad \quad x_0\in\Omega_{\mu_0}\bigcup\partial\Omega.$$

Since $\varphi(x)\leq\varphi(x_0),\quad \text{for} \quad x\in \Omega_{\mu_0}$, there exists $M_2$ such that
\begin{align}\label{3M2}
|Du|(x)\leq M_2, \quad in\quad\Omega_{\mu_0}\bigcup\partial\Omega,
\end{align}
where~$M_2$ depends only on ~$n, \Omega, \mu_0, M_0,  L_1, L_2$.

So at last we get the following estimate
$$\sup_{\overline\Omega_{\mu_0}}|Du|\leq \max\{M_1, M_2\},$$
where the positive constant  ~$ M_1$ depends only on $n, \mu_0, M_0, L_1$; and $ M_2$ depends only on ~$n, \Omega, \mu_0, M_0, L_1, L_2$.

 So  we  complete the proof of Theorem~\ref{Thm1.1}.\qed

   Now we
 prove the main Lemma~\ref{Lem4.5}  which was used to estimate  $J_1$ defined in  \eqref{3J1c}.

\begin{Lem}\label{Lem4.5}
We define $(b_{ij})$  as in \eqref{3bija}; $d_i ,e_i$ defined as in \eqref{3di}-\eqref{3ei}; $A_{1i}, A_{2i}, G_{ij}, \hat{G}_{ij}$ defined as in \eqref{3A1i}.
We study the following quadratic form
\begin{align}\label{4.6}
\begin{split}
Q(x_3,x_4,\ldots,x_n)=&\sum_{3\leq i\leq n}b_{ii}x_i^2
+2\sum_{3\leq i< j\leq n}b_{ij}x_i x_j
+\sum_{3\leq i\leq n}\tilde{K}_i x_i,
\end{split}
\end{align}
where $\tilde{K}_i$ defined in  \eqref{3tildeKia} and we have the estimate  \eqref{3tildeKib} for $\tilde{K}_i$.
Then there exists a sufficiently large positive constant $C_{15}$ which depends only on  $n, \Omega, \mu_0, M_0, L_1, L_2,$ such that if
\begin{align}\label{4C15}
|Du|(x_0)=u_1(x_0) \ge C_{15},
\end{align}
then  the followings hold.

(I): The matrix $(b_{ij})$ is positive definite since the matrix  ~$(b^1_{ij})= (1+\delta_{ij})$ is positive definite.\par

(II): We have
\begin{align}\label{4.7}
\begin{split}
Q(x_3,x_4,\ldots,x_n)
\geq &-C_{16} u_1^6,
\end{split}
\end{align}
where positive constant $C_{16}$ also depends only on  $n, \Omega, \mu_0, M_0, L_1, L_2.$

\end{Lem}
{\bf Proof:} Let $$ B=(b_{ij})=B_1+B_2, B_1=u_1^4(b^1_{ij}), B_2=(O(u^2_1)\delta_{ij}).$$
We first  prove  (I):
\begin{align}\label{4.8}
\begin{split}
\sigma_k(B)=&\sigma_k(B_1+B_2)\\
=&\sigma_k(B_1)+\sigma_k(B_1, B_1, \ldots, B_1, B_2)\\&+\cdots+\sigma_k(B_1, B_2,\ldots, B_2, B_2)+\sigma_k(B_2)\\
=&u_1^{4k}\sigma_k(b^1_{ij})+O(u_1^{4k-2}),
\end{split}
\end{align}
so if $u_1$ is sufficiently large, then $\sigma_k(B)>0\Longleftrightarrow\sigma_k(b^1_{ij})>0$.

Now we prove   (II):
Since  ~$B_1=u_1^4(b^1_{ij})_{3\leq i,j\leq n}$  is positive definite, from the argument in (I), we get
\begin{align}\label{4.9}
B^{-1}=(B_1+B_2)^{-1}=B_1^{-1}(I+B_1^{-1}B_2)^{-1}=u_1^{-4}(b^1_{ij})^{-1}\big(1+o(1)\big).
\end{align}
Then we have
\begin{align}\label{4.10}
\begin{split}
(b^1_{ij})^{-1}=&\left(\begin{array}{cccc}2&1&\cdots &1\\
1&2&\cdots &1\\
\vdots & \vdots & \vdots &\vdots \\
1&1&\cdots& 2
\end{array}\right)^{-1}
=\frac{1}{n-1}\left(\begin{array}{cccc}
n-2&-1&\cdots & -1\\
-1&n-2&\cdots & -1\\
\vdots & \vdots & \vdots &\vdots \\
-1&-1&\cdots & n-2
\end{array}\right).
\end{split}
\end{align}

Now we solve the following linear  algebra equation
\begin{align}\label{4.11}
\frac{\partial Q}{\partial x_k}=0,\quad k=3,4,\ldots,n.
\end{align}
 We assume $(\bar{x}_{3},\bar{x}_{4},\ldots,\bar{x}_{n})$ is the extreme point of the quadratic form $Q(x_3,x_4,\ldots,x_n)$.
From the definition of $b_{ij}, \tilde{K}_i$ in ~\eqref{3bija}, ~\eqref{3tildeKia} and the estimate for $\tilde{K}_i$ in ~\eqref{3tildeKib}, using the formulas \eqref{4.9} and \eqref{4.10}, it follows that
\begin{align}\label{4.12}
\begin{split}
\left(\begin{array}{c}
\bar{x}_{3}\\
\bar{x}_{4}\\
\vdots  \\
\bar{x}_{n}
\end{array}\right)=& O(u_1^5)B^{-1}
\left(\begin{array}{c}
1\\
1\\
\vdots\\
1
\end{array}\right)
=O(u_1)
\left(\begin{array}{c}
1\\
1\\
\vdots\\
1
\end{array}\right).
\end{split}
\end{align}

It follows that we have the following minimum of the quadratic $Q$,
\begin{align}\label{4.14}
\begin{split}
Q(\bar{x}_3, \bar{x}_4,\ldots, \bar{x}_n)
\geq&
-C_{17}u_1^6.
\end{split}
\end{align}

 In this computation, the bounds in the coefficient on $ O(u_1^5), O(u_1)$ depend only on $n, \Omega,  M_0, \mu_0,  L_1, L_2$. Thus we complete this proof.\qed

\section{ The proof of Theorem \ref{Thm1.2}  }

In this section we first prove the Theorem~\ref{Thm1.2}.

  In the proof of the existence theorem for the Neumann boundary value problem, we need the a priori estimates. For the $C^0$ estimates we use the methods introduced by ~Concus-Finn\cite{CF74} and ~Spruck\cite{Sp75}. As in  Simon-Spruck\cite{SS76}, we use the continuity method to complete the proof of theorem ~\ref{Thm1.2}

{\em Proof of Theorem~\ref{Thm1.2}:}

We consider the following family on the mean curvature equation with Neumann boundary value problem:
\begin{align}
\texttt{div}(\frac{Du}{\sqrt{1+|Du|^2}}) =&u   \quad\text{in}\quad\Omega,\label{5.1} \\
              \frac{\partial u}{\partial \gamma} =&\tau\psi(x)  \quad\text{on}\quad \partial \Omega,\label{5.2}
\end{align}
where $\tau\in[0,1]$.

For
$\tau=0$, then  $u=0$ is the unique solution. And we need to find the solution for
$\tau=1.$
By the standard existence  theorem \cite{Ur73, LU68}, as in Simon-Spruck \cite{SS76}, if we can get the a priori estimates for the $C^2(\overline\Omega)$ solution of the equation  \eqref{5.1} and \eqref{5.2}
\begin{align}
\sup_{\Omega}|u|\leq& \hat{K}_1,\label{5.3}\\
\sup_{\Omega}|Du|\leq& \hat{K}_2,\label{5.4}
\end{align}
where ~$\hat{K}_1, \hat{K}_2$  are independent of~$\tau$.
Then we can get the existence theorem. From the interior gradient estimates and our boundary gradient estimates, we only need get the $C^0$ estimates for the solution $u$ in  \eqref{5.1} and \eqref{5.2}.

In the paper by  Spruck\cite{Sp75}, he used the comparison theorem developed by Concus-Finn\cite{CF74} to get the $C^0$ estimates for the mean curvature equation with prescribed contact angle boundary value problem. In our case, his proof is still true, so we complete the proof of Theorem \ref{Thm1.2}.\qed

\begin{Rem}\label{4Rem2}
In X.N. Ma  and J.J. Xu \cite{MX14}, we have generalized the boundary gradient estimates to Hessian equation and the higher order curvature equation with Neumann boundary value and the capillary boundary value problem. And  we can reduce the condition on $\psi \in C^3$ to
$\psi \in  C^2$, then we could get the similar results  on Theorem 1.1 and Theorem 1.3.
\end{Rem}

      {\bf
Acknowledgments:}
The authors would like to
thank Qiu Guohuan for
 pointing
a flaw in our first version of this paper.

config: arxiv


TITLE: mapping from S to T and T to U such that S to U is injective.
QUESTION [0 upvotes]: Let S, T, U be three non-void sets and $f : S \rightarrow T, g : T \rightarrow U$ be so that $g o f : S \rightarrow U$ is injective. Then should $g$ and $f$ both be injective?
I think that $g$ should be injective but I'm not sure about $f$

REPLY [0 votes]: No, $g$ need not be injective, but $f$ does.
Indeed, if $f(a)=f(b)$, then $g(f(a))=g(f(b))$; because $g\circ f$ is injective, we conclude that $a=b$.
For a counterexample, consider $S=\{0\}$, $T=\{1,2\}$ and $U=\{3\}$. Define $f(0)=1$ and let $g$ be the unique map $g\colon T\to U$. Then $g\circ f$ is injective (because the domain has a single element), but $g$ isn't.

The same example shows that if $g\circ f$ is surjective, then $f$ need not be surjective.