In this section, by constructing a global auxiliary function, we can reduce the global estimates to the boundary. On the boundary, in order to get the mixed tangential normal second order derivatives of *u*, we apply the tangential operator to the boundary value condition (2). A bound of double normal derivative of *u* on *∂*Ω is obtained by constructing a suitable auxiliary function. We use the key trick of [14] to establish the global second order derivative estimates of solutions for the degenerate Monge–Ampère type equations with the Neumann boundary value condition. For the argument below, we assume that the functions *φ*, *ν* are smoothly extended to Ω̄. The constant *C* in this section changes from line to line.

*Mixed tangential normal derivative estimate on* *∂*Ω. We introduce the tangential gradient operator \(\delta =(\delta _{1}, \delta _{2}, \ldots , \delta _{n})\), where \(\delta _{i}=\sum_{j=1}^{n}(\delta _{ij}-\nu _{i}\nu _{j})D_{j}\) for \(i=1, \ldots , n\). Applying this tangential operator to boundary condition (2), we have

$$ (D_{k}u) (\delta _{i}\nu _{k})+\nu _{k}(\delta _{i}D_{k}u)= \delta _{i} \varphi\quad \text{on } \partial \Omega . $$

(26)

If *τ* is a direction tangential to *∂*Ω at any point \(y\in \partial \Omega \), we have

$$ \begin{aligned} D_{\tau \nu }u(y)&=\sum _{i, k=1}^{n}\tau _{i}\nu _{k}D_{ik}u= \sum_{i, k=1}^{n} \tau _{i}\nu _{k}D_{ik}u-\sum _{i, j, k=1}^{n}\tau _{i} \nu _{k}\nu _{i}\nu _{j}D_{jk}u \\ &=\sum_{i, j, k=1}^{n}\tau _{i} \nu _{k}(\delta _{ij}-\nu _{i}\nu _{j})D_{jk}u= \sum_{i, k=1}^{n} \tau _{i}\nu _{k}\delta _{i}D_{k}u \\ &=\sum_{i, k=1}\tau _{i}\delta _{i}\varphi -\tau _{i}(\delta _{i} \nu _{k})D_{k}u=\sum_{i, k=1} \tau _{i}D_{i}\varphi -\tau _{i}(D_{i} \nu _{k})D_{k}u, \end{aligned} $$

(27)

where the second equality and the last equality are both valid using the fact that \(\tau \cdot \nu =0\), and the fifth equality holds by (26). Hence, from (27) we obtain

$$ \vert D_{\tau \nu }u \vert \leq C $$

(28)

on *∂*Ω, where the constant *C* depends on \(\sup_{\bar{\Omega}} |Du|\) and Ω.

*Double normal derivative estimate on* *∂*Ω. We may consider any boundary point. Without loss of generality, we may take it to be the origin and the \(x_{n}\)-axis in the direction of the interior normal. From the uniform *A*-convexity of Ω and the regularity of *A*, there exists a \(C^{2}\) defining function in Ω̄ satisfying

$$ \phi =0 \quad \text{on } \partial \Omega ,\qquad D_{\nu } \phi =-1 \quad \text{on } \partial \Omega ,\quad \text{and}\quad \phi < 0\quad \text{in } \Omega , $$

(29)

together with the inequality

$$ \bigl\{ D_{ij}\phi -A_{ij}^{k}(x, Du)D_{k}\phi \bigr\} _{n\times n}\geq \delta _{0}I, $$

(30)

in a neighborhood \(\mathcal{N}\) of *∂*Ω, whenever \(D_{\nu }u\geq \varphi (x)\), where \(\delta _{0}\) is a positive constant and *I* denotes the identity matrix. We employ the auxiliary function

$$ w=\pm \bigl(D_{\nu }u-\varphi (x)\bigr)+\beta \phi $$

(31)

in

$$ \Omega _{d_{0}}:=\bigl\{ x\in \Omega \mid \operatorname{dist}(x, \partial \Omega )< d_{0} \bigr\} , $$

where *β* and \(d_{0}\) are two positive constants to be determined. By the continuity of \(D_{\nu }\phi \) and \(D_{\nu }\phi =-1\) on *∂*Ω, there exists a small constant \(d_{0}\) such that \(D_{\nu }\phi \leq -\frac{1}{2}\) in \(\Omega _{d_{0}}\).

It follows from (16) that

$$ \begin{aligned} Lw&=\pm \tilde{u}^{ij} \bigl[D_{ij}(\nu _{k}D_{k}u)-A_{ij}^{l}D_{l}( \nu _{k}D_{k}u)\bigr] \mp L\varphi +\beta L\phi \\ &=\pm \tilde{u}^{ij}\bigl[(D_{ij}\nu _{k})D_{k}u+2D_{j}\nu _{k}D_{ik}u-A_{ij}^{l}(D_{l} \nu _{k})D_{k}u+\nu _{k}A_{ij, k} \bigr] \\ &\quad {} +\nu _{k}\tilde{B}_{p_{k}} \mp L\varphi +\beta L\phi \\ &\geq \beta L\phi -C\mathcal{T}-C-CB^{-\frac{1}{n-1}} \\ &\geq \beta L\phi -C\mathcal{T}-CB^{-\frac{1}{n-1}}, \end{aligned} $$

(32)

where \(\mathcal{T}=\sum_{i=1}^{n} \tilde{u}^{ii}\), the first inequality is obtained by using (20) in Lemma 2.1, the second inequality is obtained by using the fact that

$$ \frac{1}{n} \mathcal{T} \ge \Biggl(\prod _{i=1}^{n} \tilde{u}^{ii} \Biggr)^{\frac{1}{n}}= \Biggl(\prod_{i=1}^{n} \tilde{u}_{ii} \Biggr)^{- \frac{1}{n}}=B^{-\frac{1}{n}}\ge C. $$

(33)

By a direct calculation, we obtain

$$ \begin{aligned} L\phi &=\tilde{u}^{ij} \bigl[D_{ij}\phi -A_{ij}^{k}(x, Du)D_{k} \phi \bigr]-\tilde{B}_{p_{k}}D_{k} \phi \\ &\geq \delta _{0}\mathcal{T} -\tilde{B}_{p_{k}}D_{k} \phi \\ &\geq \delta _{0}\mathcal{T} -CB^{-\frac{1}{2(n-1)}}, \end{aligned} $$

(34)

where the first and second inequalities are established by using (34) and (6), respectively. Therefore, from (32), (33), and (34), we have

$$ Lw\geq (\beta \delta _{0}-C)\mathcal{T} -CB^{-\frac{1}{n-1}}-C\beta B^{- \frac{1}{2(n-1)}}. $$

(35)

We decompose \(\Omega _{d_{0}}=\Omega _{d_{0}}^{1} \cup \Omega _{d_{0}}^{2}\), where

$$ \Omega _{d_{0}}^{1}=\Omega _{d_{0}} \cap \Biggl\{ \sum_{i=1}^{n} \tilde{u}_{ii} \geq n \Biggr\} $$

and

$$ \Omega _{d_{0}}^{2}=\Omega _{d_{0}} \cap \Biggl\{ \sum_{i=1}^{n} \tilde{u}_{ii} < n \Biggr\} . $$

If \(\max_{\bar {\Omega }_{d_{0}}}w=w(\bar {x})\) for some \(\bar {x}\in \bar {\Omega }_{d_{0}}\). The proof of the double normal derivative estimate of *u* on *∂*Ω splits into three stages according to whether \(\bar {x}\in \Omega _{d_{0}}^{1}\) or \(\bar {x}\in \Omega _{d_{0}}^{2}\), or neither.

*Case 1.* \(\bar {x}\in \Omega _{d_{0}}^{1}\). Since \(\max_{\bar {\Omega }_{d_{0}}}w=w(\bar {x})\) for \(\bar {x}\in \Omega _{d_{0}}^{1}\), then we have

$$ Lw(\bar {x}) \leq 0. $$

(36)

We may assume that \(\tilde{u}_{ij}(\bar {x})\) is diagonal. Without loss of generality, we may assume that \(\tilde{u}_{11}(\bar {x}) \geq 1\). At the point *x̄*, we have

$$ \begin{aligned} \mathcal{T} &\geq \sum _{i=2}^{n}\tilde{u}^{ii} \\ &\geq (n-1) \Biggl(\prod_{i=2}^{n} \tilde{u}^{ii}\Biggr)^{\frac{1}{n-1}} \\ &=(n-1) \Biggl(\prod_{i=1}^{n} \tilde{u}^{ii}\Biggr)^{\frac{1}{n-1}}(\tilde{u}_{11})^{ \frac{1}{n-1}} \\ &\geq (n-1)B^{-\frac{1}{n-1}}, \end{aligned} $$

(37)

where \(\tilde{u}_{11}(\bar{x})\ge 1\) is used in the last inequality. At *x̄*, we can assume that \(\sqrt{\mathcal{T}}\geq \frac{3C}{\delta _{0}\sqrt{n-1}}\), otherwise we have already got the global second order derivative estimates. Therefore at *x̄*, by (35), we have

$$ \begin{aligned} Lw &\geq \biggl(\frac{\beta \delta _{0}}{3}-C\biggr) \mathcal{T}+ \frac{\beta \delta _{0}}{3}\mathcal{T}-CB^{-\frac{1}{n-1}}+ \frac{\beta \delta _{0}}{3}\mathcal{T}-C\beta B^{-\frac{1}{2(n-1)}} \\ &\geq \biggl(\frac{\beta \delta _{0}}{3}-C\biggr)\mathcal{T}+\biggl( \frac{\beta \delta _{0}}{3}-\frac{C}{n-1}\biggr)\mathcal{T}+\biggl( \frac{\beta \delta _{0}}{3}\sqrt{\mathcal{T}}- \frac{C\beta }{\sqrt{n-1}}\biggr)\sqrt{ \mathcal{T}} \\ & >0, \end{aligned} $$

(38)

where we choose \(\beta \geq \frac{3C}{\delta _{0}}+1\) such that the last inequality holds. It is a contradiction with (36). Therefore, \(\bar {x}\notin \Omega _{d_{0}}^{1}\).

*Case 2.* \(\bar {x}\in \Omega _{d_{0}}^{2}\). For the unit inner normal vector \(\nu =(\nu _{1}, \ldots , \nu _{n})\), we must have

$$ 0\leq \vert \nu _{i} \vert \leq 1\quad \text{for } i=1, \ldots , n. $$

(39)

Since \(\max_{\bar {\Omega }_{d_{0}}^{2}}w=w(\bar {x})\), \(\bar {x}=(\bar {x}_{1}, \ldots , \bar {x}_{n})\in \Omega _{d_{0}}^{2}\), we obtain

$$ w_{\nu }(\bar {x})=0. $$

(40)

Therefore, we have at *x̄*

$$ \begin{aligned} 0&=w_{\nu }=\pm (D_{\nu \nu }u-D_{\nu } \varphi )+\beta D_{ \nu }\phi \\ &=\pm \Biggl(\sum_{i, j}^{n}\nu _{i}\nu _{j} D_{ij}u\Biggr) \mp D_{\nu }\varphi + \beta D_{\nu }\phi \\ &\leq n^{2} \sum_{i=1}^{n} \tilde{u}_{ii} \mp D_{\nu } \varphi + \beta D_{\nu }\phi . \end{aligned} $$

(41)

Choose \(\beta \geq 2\max |D\varphi |+4n^{3}\) such that \(\sum_{i=1}^{n}\tilde{u}_{ii}\geq 2n\), which is a contradiction with \(\sum_{i=1}^{n}\tilde{u}_{ii} < n\). Therefore, \(\bar {x}\notin \Omega _{d_{0}}^{2}\).

*Case 3.* \(\bar {x}\in \partial \Omega _{d_{0}}\). On \(\partial \Omega _{d_{0}}\cup \Omega \), we have

$$ w=\pm (D_{\nu }u-\varphi )+\beta \phi \leq 0, $$

(42)

where *β* is chosen large enough such that the last inequality holds. On *∂*Ω, we have

$$ w=\pm \bigl(D_{\nu }u-\varphi (x)\bigr)+\beta \phi = 0. $$

(43)

Therefore, for all \(x \in \bar {\Omega }_{d_{0}}\), we have

Since \(w(0)=0\), we have

$$ D_{\nu }w(0) \leq 0, $$

(45)

which implies

$$ \vert D_{\nu \nu }u \vert \leq C\quad \text{on } \partial \Omega . $$

(46)

### Remark 6

In the above proof, we only use (30) in a neighborhood \(\mathcal{N}\) of *∂*Ω whenever \(D_{\nu }u\geq \varphi (x)\). If \(D_{\nu }u< \varphi (x)\) in a neighborhood \(\mathcal{N}_{0}\) of some boundary point \(x_{0}\in \partial \Omega \), from the boundary condition, we directly get an upper bound \(D_{\nu \nu }u(x_{0})\le D_{\nu } \varphi (x_{0})\). While the lower bound \(D_{\nu \nu }u \ge \sum_{i,j}^{n} A_{ij} \nu _{i}\nu _{j}\) at \(x_{0}\) can be derived from the ellipticity.

Based on the mixed tangential normal derivative estimate and the double normal derivative estimate on *∂*Ω, we now use the method in [15] to prove Theorem 1.1, which is a modification of the original method in [14].

### Proof of Theorem 1.1

We modify the elliptic subsolution \(\underline{u}\) by adding a perturbation function *aϕ*, where *a* is a small positive constant and *ϕ* is the defining function of the domain Ω satisfying (29). Note that the function is still uniformly elliptic in Ω if *a* is a sufficiently small. By a direct computation, we have

on *∂*Ω. We define a function with the form with a positive constant *K*. Replacing \(\underline{u}\) in Lemma 2.2 with , from (24), we have

$$ L \Phi \geq \epsilon _{1} \mathcal{T}-C \bigl( B^{-\frac{1}{2(n-1)}} + 1 \bigr)\quad \text{in } \Omega . $$

(48)

By a direct calculation, we also have

$$ D_{\nu }\Phi \geq Ka\inf \bigl(e^{K(\underline{u}-u)}\bigr)>0\quad \text{on } \partial \Omega . $$

(49)

We employ the auxiliary function

$$ G(x, \xi )=\bigl(\tilde{u}_{\xi \xi }-v'(x, \xi )\bigr)e^{\frac{\alpha }{2}|Du|^{2}+ \kappa \Phi } $$

(50)

in \(\bar {\Omega }\times \mathbb{R}^{n}\), where \(|\xi |=1\), *α* and *κ* are positive constants to be determined, \(v'\) is given by

$$ v'(x, \xi )=2(\xi \cdot \nu )\xi '_{i}\bigl(D_{i}\varphi (x) - D_{k}uD_{i} \nu _{k}-A_{ij}\nu _{j}\bigr), $$

(51)

and \(\xi '\) is given by

$$ \xi '=\xi -(\xi \cdot \nu )\nu . $$

(52)

Assume that *G* attains its maximum at \(x_{0}\in \Omega \) and \(\xi =\xi _{0}\). Let

$$ H(x,\xi )=\log G(x,\xi )=\log \bigl(\tilde{u}_{\xi \xi }-v' \bigr)+ \frac{\alpha }{2} \vert Du \vert ^{2}+\kappa \Phi , $$

(53)

then the function *H* attains its maximum at \(x_{0}\in \Omega \) and \(\xi =\xi _{0}\). From now on, all the calculations will be made at the point \(x=x_{0}\) and \(\xi =\xi _{0}\) unless otherwise specified. At \(x_{0}\), we obtain

$$\begin{aligned}& 0=D_{i} H= \frac{D_{i}(\tilde{u}_{\xi \xi }-v')}{\tilde{u}_{\xi \xi }-v'}+\alpha D_{k}uD_{ik}u+ \kappa D_{i}{\Phi },\quad \text{for } i=1,\ldots, n, \end{aligned}$$

(54)

$$\begin{aligned}& \begin{aligned} 0 \geq{}& D_{ij}H \\ ={}& \frac{D_{ij}(\tilde{u}_{\xi \xi }-v')}{\tilde{u}_{\xi \xi }-v'}- \frac{D_{i}(\tilde{u}_{\xi \xi }-v')D_{j}(\tilde{u}_{\xi \xi }-v')}{(\tilde{u}_{\xi \xi }-v')^{2}} \\ &{}+\alpha (D_{ik}uD_{jk}u+D_{k}uD_{ijk}u)+ \kappa D_{ij}\Phi . \end{aligned} \end{aligned}$$

(55)

Therefore, at \(x_{0}\) we have

$$ \begin{aligned} 0 \geq{}& LH \\ =&\frac{1}{\tilde{u}_{\xi \xi }-v'}L \bigl(\tilde{u}_{ \xi \xi }-v'\bigr)-\frac{1}{(\tilde{u}_{\xi \xi }-v')^{2}} \tilde{u}^{ij}D_{i}\bigl( \tilde{u}_{\xi \xi }-v' \bigr)D_{j}\bigl(\tilde{u}_{\xi \xi }-v'\bigr) \\ &{}+\alpha \tilde{u}^{ij}D_{ik}uD_{jk}u+ \alpha D_{k}uLu_{k}+\kappa L \Phi . \end{aligned} $$

(56)

By a direct computation and using (17)–(19), we have

$$ \begin{aligned} Lu_{\xi \xi }\ge {}& \tilde{u}^{ik}\tilde{u}^{jl}D_{\xi } \tilde{u}_{ij}D_{\xi }\tilde{u}_{kl}+ \tilde{u}^{ij}D_{p_{k}p_{l}}A_{ij}D_{ \xi }u_{k}D_{\xi }u_{l} \\ &{}-C\bigl[(1+\tilde{u}_{ii})\mathcal{T}+(\tilde{u}_{ii})^{2} \bigr]-C(1+\tilde{u}_{ii})B^{- \frac{1}{n-1}} \\ \ge {}&\tilde{u}^{ik}\tilde{u}^{jl}D_{\xi } \tilde{u}_{ij}D_{\xi } \tilde{u}_{kl}-C\bigl[(1+ \tilde{u}_{ii})\mathcal{T} +(\tilde{u}_{ii})^{2} \bigr]-C(1+ \tilde{u}_{ii}) B^{-\frac{1}{n-1}}, \end{aligned} $$

(57)

where the first inequality is established by (21) in Lemma 2.1, and the second inequality is obtained by using the A3w condition. Note that the term \(\sum_{k=1}^{n} \tilde{B}_{p_{k}}D_{ii}u_{k}\) in (21) does not appear on the right-hand side of the first inequality in (57) since it can be subtracted using the definition of the linearized operator *L* in (23). By a direct computation, we obtain

$$ \begin{aligned} \vert LA_{\xi \xi } \vert & \leq C\bigl[(1+\tilde{u}_{ii})\mathcal{T}+ \tilde{u}_{ii} \bigr]+ \biggl\vert \frac{B_{p_{l}}}{B}A_{\xi \xi ,l} \biggr\vert \\ &\leq C\bigl[(1+\tilde{u}_{ii})\mathcal{T}+\tilde{u}_{ii} \bigr]+C(1+\tilde{u}_{ii}) B^{-\frac{1}{n-1}}, \end{aligned} $$

(58)

where (20) in Lemma 2.1 is used in the second inequality. It follows from (16) and (20) in Lemma 2.1 that

$$ \begin{aligned} \bigl\vert Lv' \bigr\vert &\leq C\mathcal{T}+C \bigl\vert L(u_{k}) \bigr\vert \\ &\leq C \mathcal{T}+C B^{-\frac{1}{n-1}}. \end{aligned} $$

(59)

Note that there is no \(1+\tilde{u}_{ii}\) in the coefficient of \(B^{-\frac{1}{n-1}}\) in (59) since the term \(\tilde{B}_{p_{l}}D_{l}u_{k}\) is already subtracted in \(L(u_{k})\).

Hence, combining (57), (58), and (59), we have

$$ L\bigl(\tilde{u}_{\xi \xi }-v'\bigr)\geq \tilde{u}^{ik}\tilde{u}^{jl}D_{\xi } \tilde{u}_{ij}D_{\xi }\tilde{u}_{kl}-C\bigl[(1+ \tilde{u}_{ii})\mathcal{T}+ \tilde{u}_{ii}^{2} \bigr]-C(1+\tilde{u}_{ii})B^{-\frac{1}{n-1}}. $$

(60)

By Cauchy’s inequality, we obtain

$$ \tilde{u}^{ij}D_{i}\bigl( \tilde{u}_{\xi \xi }-v'\bigr)D_{j}\bigl( \tilde{u}_{\xi \xi }-v'\bigr) \leq (1+\theta ) \tilde{u}^{ij}D_{i}\tilde{u}_{\xi \xi }D_{j} \tilde{u}_{ \xi \xi }+C(\theta )\tilde{u}^{ij}D_{i} v'D_{j} v' $$

(61)

for any \(\theta > 0\), where \(C(\theta )\) is a positive constant depending on *θ*. Inserting (48), (60), and (61) into (56), by calculations, we get

$$ \begin{aligned} 0 \geq {}& LH \\ \geq {}& \frac{1}{\tilde{u}_{\xi \xi }-v'}\tilde{u}^{ik}\tilde{u}^{jl}D_{ \xi } \tilde{u}_{ij}D_{\xi }\tilde{u}_{kl}- \frac{1+\theta }{(\tilde{u}_{\xi \xi }-v')^{2}}\tilde{u}^{ij}D_{i} \tilde{u}_{\xi \xi }D_{j} \tilde{u}_{\xi \xi } \\ &{}-C\frac{1}{\tilde{u}_{\xi \xi }-v'}\bigl\{ \bigl[(1+\tilde{u}_{ii})\mathcal{T}+( \tilde{u}_{ii})^{2}\bigr]+C(1+\tilde{u}_{ii})B^{-\frac{1}{n-1}} \bigr\} \\ &{}-\frac{C(\theta )}{(\tilde{u}_{\xi \xi }-v')^{2}}\tilde{u}^{ij}D_{i} v' D_{j} v'-C\alpha \bigl(B^{-\frac{1}{n-1}}+\mathcal{T}\bigr) \\ &{}+\alpha \sum_{i=1}^{n} \tilde{u}_{ii}+\kappa \bigl[\epsilon _{1} \mathcal{T}-C \bigl( B^{-\frac{1}{2(n-1)}} + 1 \bigr) \bigr]. \end{aligned} $$

(62)

Next, we shall deal with the terms on the right-hand side of (62). Without loss of generality, we assume that \(\{\tilde{u}_{ij}\}\) is diagonal at \(x_{0}\) with the maximum eigenvalue \(\tilde{u}_{11}\). We can always assume that \(\tilde{u}_{11}>1\) and is as large as we want; otherwise we are done. Since \(v'\) is bounded, \(\tilde{u}_{11}\) and \(\tilde{u}_{\xi \xi }\) are comparable in the sense that, for any \(\theta >0\), there exists a further constant \(C(\theta )\) such that

$$ \bigl\vert \tilde{u}_{11}-\tilde{u}_{\xi \xi }+v' \bigr\vert < \theta \tilde{u}_{11} $$

(63)

if \(\tilde{u}_{11}> C(\theta )\). Therefore, using (63), we have

$$ \begin{aligned} &\frac{1}{\tilde{u}_{\xi \xi }-v'} \bigl[(1+ \tilde{u}_{ii}) \mathcal{T}+(\tilde{u}_{ii})^{2} \bigr] \\ &\quad \leq \frac{1}{(1-\theta )\tilde{u}_{11}} \bigl[(1+\tilde{u}_{ii}) \mathcal{T}+( \tilde{u}_{ii})^{2}\bigr] \\ &\quad \leq C \Biggl(\mathcal{T}+\sum_{i=1}^{n} \tilde{u}_{ii} \Biggr) \end{aligned} $$

(64)

for some constant *C* if \(\theta \in (0, 1/2)\).

We shall treat the first two terms on the right-hand side of (62). For this purpose, we will make a more detailed calculations than [15] since the equality \(D_{i}A_{\xi \xi }=D_{\xi }A_{i\xi }\) does not hold in general. Because in (62) the vector *ξ* is not equal to \(e_{1}\), the situation here is different from the formula below (24) in [21]. Next, we divide the discussions into the two cases (a) and (b) assumed in the statement of this theorem.

In case (a), \(A_{ij}=f_{ij}(x,u_{i})\delta _{ij}\). We define

$$ \mathcal{P} :=\frac{1}{\tilde{u}_{11}}\tilde{u}^{ik} \tilde{u}^{jl}D_{ \xi }\tilde{u}_{ij}D_{\xi } \tilde{u}_{kl}-\frac{1-2\theta }{1-\theta } \frac{1}{\tilde{u}_{11}^{2}} \tilde{u}^{ij}D_{i}\tilde{u}_{\xi \xi }D_{j} \tilde{u}_{\xi \xi }, $$

(65)

we shall get a lower bound of the quantity \(\mathcal{P}\) in terms of \(\mathcal{T}\). We define the matrix

$$ \{a_{ik}\}:=\bigl\{ \tilde{u}^{jl}D_{\xi } \tilde{u}_{ij}D_{\xi }\tilde{u}_{kl} \bigr\} $$

(66)

and let Λ be its maximum eigenvalue. For the first term of \(\mathcal{P}\), we have

$$ \frac{1}{\tilde{u}_{11}}\tilde{u}^{ik} \tilde{u}^{jl}D_{\xi }\tilde{u}_{ij}D_{ \xi } \tilde{u}_{kl}= \frac{\operatorname{trace}\{\tilde{u}^{ik}a_{kj}\}}{\tilde{u}_{11}} \geq \frac{\tilde{u}^{11}\Lambda }{\tilde{u}_{11}} = \frac{\Lambda }{\tilde{u}_{11}^{2}}. $$

(67)

For the second term of \(\mathcal{P}\), using \(\{\tilde{u}_{ij}\}: =\{u_{ij}-A_{ij}\}\), we have

$$ \begin{aligned} & \frac{1-2\theta }{1-\theta } \frac{1}{\tilde{u}_{11}^{2}} \tilde{u}^{ij}D_{i} \tilde{u}_{\xi \xi }D_{j}\tilde{u}_{\xi \xi } \\ &\quad =\frac{1-2\theta }{1-\theta }\frac{1}{\tilde{u}_{11}^{2}}\tilde{u}^{ij}(D_{\xi }{u_{ki}}-D_{i}A_{k\xi }) (D_{\xi }{u_{lj}}-D_{i}A_{l\xi }) \xi _{k}\xi _{l} \\ &\quad =\frac{1-2\theta }{1-\theta }\frac{1}{\tilde{u}_{11}^{2}}\tilde{u}^{ij} \bigl[D_{\xi }(u_{ik}-A_{ik})+(D_{\xi }A_{ki}-D_{i}A_{k\xi }) \bigr] \\ &\qquad {} \cdot \bigl[D_{\xi }(u_{jl}-A_{jl}) +(D_{\xi }A_{jl}-D_{j}A_{l\xi }) \bigr]\xi _{k} \xi _{l} \\ &\quad =\frac{1-2\theta }{1-\theta }\frac{1}{\tilde{u}_{11}^{2}}\tilde{u}^{ij}D_{\xi } \tilde{u}_{i \xi }D_{\xi }\tilde{u}_{j\xi }+ \frac{1-2\theta }{1-\theta }\frac{2}{\tilde{u}_{11}^{2}}\tilde{u}^{ij}D_{\xi } \tilde{u}_{i \xi }(D_{\xi }A_{j\xi }-D_{j}A_{\xi \xi }) \\ &\qquad {}+\frac{1-2\theta }{1-\theta }\frac{1}{\tilde{u}_{11}^{2}}\tilde{u}^{ij}(D_{\xi }A_{i \xi }-D_{i}A_{\xi \xi }) (D_{\xi }A_{j\xi }-D_{j}A_{\xi \xi }) \\ &\quad \leq \frac{1}{\tilde{u}_{11}^{2}}\tilde{u}^{ij}D_{\xi } \tilde{u}_{i \xi }D_{\xi }\tilde{u}_{j\xi }+ \frac{1-2\theta }{\theta } \frac{1}{\tilde{u}_{11}^{2}} \tilde{u}^{ij}(D_{\xi }A_{i \xi }-D_{i}A_{ \xi \xi }) (D_{\xi }A_{j\xi }-D_{j}A_{\xi \xi }), \end{aligned} $$

(68)

where the last inequality is valid by using Cauchy’s inequality. We now calculate the terms \(D_{i}A_{\xi \xi }\) and \(D_{\xi }A_{i\xi }\). For general \(A(x,Du)\), we have

$$ D_{i}A_{\xi \xi } = D_{x_{i}}A_{\xi \xi } + (D_{p_{k}}A_{\xi \xi }) u_{ki},\quad {\forall } i=1, \ldots , n. $$

(69)

For \(A_{ij}=f_{ij}(x,u_{i})\delta _{ij}\), we have

$$ D_{\xi }A_{i\xi } = (D_{x_{k}}f_{ii}) \xi _{i}\xi _{k} + (D_{p_{i}}f_{ii}) \xi _{i} u_{i\xi }, \quad {\forall } i=1, \ldots , n. $$

(70)

Substituting (69) and (70) into the last term of (68), we have

$$ \begin{aligned} & \frac{1-2\theta }{\theta } \frac{1}{\tilde{u}_{11}^{2}} \tilde{u}^{ij} (D_{\xi }A_{i \xi }-D_{i}A_{\xi \xi }) (D_{\xi }A_{j\xi }-D_{j}A_{ \xi \xi }) \\ &\quad = \frac{1-2\theta }{\theta }\frac{1}{\tilde{u}_{11}^{2}} \tilde{u}^{ij} (D_{\xi }A_{i \xi }D_{\xi }A_{j\xi } - D_{\xi }A_{i \xi }D_{j}A_{\xi \xi } -D_{i}A_{ \xi \xi }D_{\xi }A_{j\xi } + D_{i}A_{\xi \xi }D_{j}A_{\xi \xi }) \\ &\quad \le \frac{1-2\theta }{\theta }\frac{C}{\tilde{u}_{11}^{2}} ( \mathcal{T}+ \tilde{u}_{11} ) \\ &\quad \le C(\mathcal{T} + 1), \end{aligned} $$

(71)

where \(\tilde{u}^{ij}\tilde{u}_{jk}=\delta _{ik}\) and \(\{\tilde{u}_{ij}\}: =\{u_{ij}-A_{ij}\}\) are used in the first inequality, and \(\tilde{u}_{11}\geq \frac{1-2\theta }{\theta }\) is assumed in the last inequality. Combining (65), (67), (68), and (71), we now get the lower bound of \(\mathcal{P}\) in terms of \(\mathcal{T}\),

$$ \mathcal{P}\ge \frac{1}{\tilde{u}_{11}^{2}}(\Lambda - a_{kl}\xi _{k} \xi _{l}) -C (\mathcal{T}+1) \ge - C (\mathcal{T}+1), $$

(72)

where the definitions of the matrix \(\{a_{ik}\}\) and its maximum eigenvalue Λ are used successively. On the other hand, it follows from \(D_{i}H=0\) in (54) that

$$ \begin{aligned} \tilde{u}^{ij}D_{i} \tilde{u}_{\xi \xi }D_{j} \tilde{u}_{ \xi \xi } &\leq 2 \tilde{u}^{ii}\bigl[ \bigl\vert D_{i}v' \bigr\vert ^{2}+\bigl(\tilde{u}_{\xi \xi }-v' \bigr)^{2}( \alpha D_{k}uD_{ik}u+\kappa D_{i}\Phi )^{2}\bigr] \\ &\leq 2\tilde{u}^{ii} \bigl\vert D_{i}v' \bigr\vert ^{2}+C\bigl(\tilde{u}_{\xi \xi }-v' \bigr)^{2} \Biggl(\alpha ^{2} \sum _{i=1}^{n}\tilde{u}_{ii} +\kappa ^{2} \mathcal{T} \Biggr). \end{aligned} $$

(73)

Combining (63), (72), and (73), we obtain

$$ \begin{aligned} &\frac{1}{\tilde{u}_{\xi \xi }-v'} \tilde{u}^{ik}\tilde{u}^{jl}D_{ \xi } \tilde{u}_{ij}D_{\xi }\tilde{u}_{kl} - \frac{1+\theta }{(\tilde{u}_{\xi \xi }-v')^{2}}\tilde{u}^{ij}D_{i} \tilde{u}_{\xi \xi }D_{j} \tilde{u}_{\xi \xi } \\ &\quad \geq \frac{1}{1-\theta } \biggl(\frac{1}{\tilde{u}_{11}}\tilde{u}^{ik} \tilde{u}^{jl}D_{\xi }\tilde{u}_{ij}D_{\xi } \tilde{u}_{kl} - \frac{1+\theta }{(1-\theta )\tilde{u}_{11}^{2}}\tilde{u}^{ij}D_{i} \tilde{u}_{\xi \xi }D_{j} \tilde{u}_{\xi \xi } \biggr) \\ &\quad \geq \frac{1}{1-\theta } \biggl(\mathcal{P}- \frac{3\theta }{(1-\theta )\tilde{u}_{11}^{2}} \tilde{u}^{ij}D_{i} \tilde{u}_{\xi \xi }D_{j} \tilde{u}_{\xi \xi } \biggr) \\ &\quad \geq -\frac{C(\mathcal{T}+1)}{1-\theta }- \frac{3\theta }{[(1-\theta )\tilde{u}_{11}]^{2}} \Biggl[2 \tilde{u}^{ii} \bigl\vert D_{i}v' \bigr\vert ^{2}+C\bigl( \tilde{u}_{\xi \xi }-v' \bigr)^{2} \Biggl(\alpha ^{2} \sum _{i=1}^{n} \tilde{u}_{ii} +\kappa ^{2}\mathcal{T} \Biggr) \Biggr] \\ &\quad \geq -C(\mathcal{T}+1)-C\theta \alpha ^{2}\sum _{i=1}^{n}\tilde{u}_{ii}-C \theta \kappa ^{2}\mathcal{T} \end{aligned} $$

(74)

for \(\theta \in (0,1/2)\), where *C* becomes a further constant in the last inequality. We can assume \(\tilde{u}_{11}\geq 1\), otherwise we have already obtained the desired estimate. Similar to (37), we also have

$$ B^{-\frac{1}{n-1}} \le \frac{\mathcal{T}}{n-1}. $$

(75)

For the last term in (62), using Cauchy’s inequality, we have

$$ \begin{aligned} & \epsilon _{1} \mathcal{T}-C \bigl( B^{- \frac{1}{2(n-1)}} + 1 \bigr) \\ &\quad \ge \epsilon _{1} \mathcal{T} - \epsilon B^{-\frac{1}{n-1}} - C( \epsilon ) \\ &\quad \ge \frac{\epsilon _{1}}{2} \mathcal{T} - C(\epsilon _{1}), \end{aligned} $$

(76)

where we take \(\epsilon = \frac{n-1}{2}\epsilon _{1}\) and use (75) in the second inequality.

Inserting (63), (64), (74), (75), and (76) into (62), we obtain, for \(\tilde{u}_{11}\ge \max \{C(\theta ), 1\}\),

$$ \begin{aligned} & \alpha \sum _{i=1}^{n}\tilde{u}_{ii}+ \frac{\kappa \epsilon _{1}}{2} \mathcal{T} \\ &\quad \leq C \Biggl\{ \bigl[1+\kappa C(\epsilon _{1})\bigr]+\bigl(1+{ \alpha }^{2} \theta \bigr) \sum_{i=1}^{n} \tilde{u}_{ii}+\bigl(1+\alpha +\kappa ^{2}\theta \bigr) \mathcal{T} \Biggr\} . \end{aligned} $$

(77)

By choosing \(\kappa \gg \alpha \gg 1\) and fixing a small positive \(\theta = 1/ \kappa ^{2}\), we can get from (77) that

$$ \sum_{i=1}^{n} \tilde{u}_{ii}(x_{0}) \le C, $$

(78)

which implies a corresponding estimate for \(|D^{2}u(x_{0})|\) in case (a).

In case (b), \(|D_{p_{k}}A_{ij}|<\delta \) for all \(i,j,k = 1, \ldots , n\), and sufficiently small *δ*. We define

$$ \mathcal{P}' :=\frac{1}{\tilde{u}_{11}} \tilde{u}^{ik}\tilde{u}^{jl}D_{ \xi } \tilde{u}_{ij}D_{\xi }\tilde{u}_{kl}- \frac{1}{\tilde{u}_{11}^{2}} \tilde{u}^{ij}D_{i} \tilde{u}_{\xi \xi }D_{j} \tilde{u}_{\xi \xi }. $$

(79)

Similarly to (68), we can get

$$ \begin{aligned} & \frac{1}{\tilde{u}_{11}^{2}} \tilde{u}^{ij}D_{i} \tilde{u}_{\xi \xi }D_{j} \tilde{u}_{\xi \xi } \\ &\quad = \frac{1}{\tilde{u}_{11}^{2}}\tilde{u}^{ij}D_{\xi } \tilde{u}_{i \xi }D_{\xi }\tilde{u}_{j\xi }+ \frac{2}{\tilde{u}_{11}^{2}}\tilde{u}^{ij}D_{\xi } \tilde{u}_{i \xi }(D_{\xi }A_{j\xi }-D_{j}A_{\xi \xi }) \\ &\qquad {}+\frac{1}{\tilde{u}_{11}^{2}}\tilde{u}^{ij}(D_{\xi }A_{i \xi }-D_{i}A_{ \xi \xi }) (D_{\xi }A_{j\xi }-D_{j}A_{\xi \xi }). \end{aligned} $$

(80)

For the middle term on the right-hand side of (80), rather than using Cauchy’s inequality, it can be dealt with by using \(D_{\xi }\tilde{u}_{i\xi }=D_{i}u_{\xi \xi }-D_{\xi }A_{i \xi }\), (69), (70), (54), and \(|D_{p_{k}}A_{ij}|<\delta \), namely

$$\begin{aligned}& \frac{2}{\tilde{u}_{11}^{2}} \tilde{u}^{ij}D_{\xi }\tilde{u}_{i \xi }(D_{\xi }A_{j\xi }-D_{j}A_{\xi \xi }) \\& \quad = \frac{2}{\tilde{u}_{11}^{2}}\tilde{u}^{ij}\bigl[D_{i} \tilde{u}_{\xi \xi }-(D_{\xi }A_{i\xi }-D_{i}A_{\xi \xi }) \bigr](D_{\xi }A_{j\xi }-D_{j}A_{\xi \xi }) \\& \quad \le \frac{2}{\tilde{u}_{11}^{2}}\tilde{u}^{ij} D_{i} \tilde{u}_{\xi \xi } (D_{\xi }A_{j\xi }-D_{j}A_{\xi \xi }) \\& \quad = \frac{2}{\tilde{u}_{11}^{2}}\tilde{u}^{ij} \bigl[D_{i}v' - \bigl(\tilde{u}_{ \xi \xi }-v'\bigr) (\alpha D_{k}u D_{ik}u+\kappa D_{i} \Phi )\bigr] (D_{\xi }A_{j \xi }-D_{j}A_{\xi \xi }) \\& \quad \le C \bigl[ 1+ \delta (\alpha + \kappa \mathcal{T}) \bigr], \end{aligned}$$

(81)

provided \(\tilde{u}_{11} \ge 1\). We can analyze the last term on the right-hand side of (80) for general matrix *A* to get

$$ \frac{1}{\tilde{u}_{11}^{2}}\tilde{u}^{ij}(D_{\xi }A_{i \xi }-D_{i}A_{ \xi \xi }) (D_{\xi }A_{j\xi }-D_{j}A_{\xi \xi }) \le C (\mathcal{T}+1), $$

(82)

provided \(\tilde{u}_{11} \ge 1\). Combining (79), (80), (81), and (82), we obtain

$$ \begin{aligned} \mathcal{P}'&\ge \frac{1}{\tilde{u}_{11}^{2}}(\Lambda - a_{kl} \xi _{k}\xi _{l}) -C \bigl[1+\mathcal{T} + \delta (\alpha + \kappa \mathcal{T}) \bigr] \\ &\ge -C \bigl[1+\mathcal{T} + \delta (\alpha + \kappa \mathcal{T})\bigr]. \end{aligned} $$

(83)

Combining (63), (83), and (73), we obtain

$$ \begin{aligned} &\frac{1}{\tilde{u}_{\xi \xi }-v'} \tilde{u}^{ik}\tilde{u}^{jl}D_{ \xi } \tilde{u}_{ij}D_{\xi }\tilde{u}_{kl} - \frac{1+\theta }{(\tilde{u}_{\xi \xi }-v')^{2}}\tilde{u}^{ij}D_{i} \tilde{u}_{\xi \xi }D_{j} \tilde{u}_{\xi \xi } \\ &\quad \geq \frac{1}{1-\theta } \biggl(\frac{1}{\tilde{u}_{11}}\tilde{u}^{ik} \tilde{u}^{jl}D_{\xi }\tilde{u}_{ij}D_{\xi } \tilde{u}_{kl} - \frac{1+\theta }{(1-\theta )\tilde{u}_{11}^{2}}\tilde{u}^{ij}D_{i} \tilde{u}_{\xi \xi }D_{j} \tilde{u}_{\xi \xi } \biggr) \\ &\quad \geq \frac{1}{1-\theta } \biggl(\mathcal{P}'- \frac{2\theta }{(1-\theta )\tilde{u}_{11}^{2}}\tilde{u}^{ij}D_{i} \tilde{u}_{\xi \xi }D_{j} \tilde{u}_{\xi \xi } \biggr) \\ &\quad \geq -C \bigl[1+\mathcal{T} + \delta (\alpha + \kappa \mathcal{T})\bigr]-C \theta \alpha ^{2}\sum_{i=1}^{n} \tilde{u}_{ii}-C\theta \kappa ^{2} \mathcal{T} \end{aligned} $$

(84)

for \(\theta \in (0,1/2)\). Using the estimate in (83) and deducing as in case (a), for \(\tilde{u}_{11}\ge \max \{C(\theta ), 1\}\), we obtain in place of (77)

$$ \begin{aligned} & \alpha \sum _{i=1}^{n}\tilde{u}_{ii}+ \frac{\kappa \epsilon _{1}}{2} \mathcal{T} \\ &\quad \leq C \Biggl\{ \bigl[1+\kappa C(\epsilon _{1})+\delta \alpha \bigr]+\bigl(1+{ \alpha }^{2} \theta \bigr)\sum _{i=1}^{n}\tilde{u}_{ii}+\bigl(1+ \alpha +\kappa ^{2} \theta +\kappa \delta \bigr)\mathcal{T} \Biggr\} . \end{aligned} $$

(85)

By choosing \(\alpha = 2C +1\) and \(\kappa =\frac{2[C(\alpha +3)+1]}{\epsilon _{1}}\) successively and fixing the positive constants \(\theta = 1/ \kappa ^{2}\) and \(\delta =1/ \kappa \), we can get from (85) that (78) holds again, which implies a corresponding estimate for \(|D^{2}u(x_{0})|\) in case (b).

Next, we consider the case \(x_{0}\in \partial \Omega \), namely the function *G* in (50) attains its maximum over Ω̄ at \(x_{0}\in \partial \Omega \) and a unit vector *ξ*. The estimation of the rest of the Hessian \(D^{2}u\) splits into two stages according to a different direction of *ξ*.

*Case (i)*. *ξ* tangential. Since \(v'(x_{0}, \xi )=0\), at \(x_{0}\) we obtain

$$ \begin{aligned} 0 &\geq D_{\nu }G \\ &=D_{\nu } \bigl\{ \bigl[\tilde{u}_{\xi \xi }-v'(x, \xi )\bigr]e^{\frac{\alpha }{2} \vert Du \vert ^{2}+ \kappa \Phi } \bigr\} \\ &=e^{\frac{\alpha }{2} \vert Du \vert ^{2}+\kappa \Phi } \biggl\{ \bigl(\tilde{u}_{\xi \xi }-v'(x, \xi )\bigr)D_{\nu } \biggl(\frac{\alpha }{2} \vert Du \vert ^{2}+\kappa \Phi \biggr)+D_{ \nu }\bigl( \tilde{u}_{\xi \xi }-v'(x, \xi )\bigr) \biggr\} \\ &=e^{\frac{\alpha }{2} \vert Du \vert ^{2}+\kappa \Phi } \bigl\{ \bigl[\alpha D_{k}uD_{ \nu }(D_{k}u)+ \kappa D_{\nu }\Phi \bigr]\tilde{u}_{\xi \xi }+D_{\nu }u_{\xi \xi }-D_{\nu } \bigl(A_{\xi \xi }+v'\bigr) \bigr\} \\ &=e^{\frac{\alpha }{2} \vert Du \vert ^{2}+\kappa \Phi } \bigl\{ \bigl[\kappa D_{\nu } \Phi +\alpha D_{k}u(\varphi _{k}-D_{i}uD_{k} \nu _{i})\bigr]\tilde{u}_{\xi \xi } \\ &\quad {} +D_{\nu }u_{\xi \xi }-D_{\nu }\bigl(A_{\xi \xi }+v' \bigr) \bigr\} \\ &\geq e^{\frac{\alpha }{2} \vert Du \vert ^{2}+\kappa \Phi } \bigl\{ (\kappa c_{0}- \alpha M) \tilde{u}_{\xi \xi }+D_{\nu }u_{\xi \xi }-D_{\nu } \bigl(A_{\xi \xi }+v'\bigr) \bigr\} , \end{aligned} $$

(86)

where \(c_{0}=Ka\inf (e^{K(\underline{u}-u)})\), \(M=\max_{x\in \partial \Omega }|D_{k}u(\varphi _{k}-D_{i}uD_{k} \nu _{i})|\). The above inequality gives a relationship between \(\tilde{u}_{\xi \xi }(x_{0})\) and \(D_{\nu }u_{\xi \xi }(x_{0})\) at \(x_{0}\), namely

$$ D_{\nu }u_{\xi \xi }\leq -(\kappa c_{0}-\alpha M) \tilde{u}_{\xi \xi }+D_{ \nu }\bigl(A_{\xi \xi }+v' \bigr). $$

(87)

On the other hand, by tangential differentiating the boundary condition twice, we obtain

$$ (D_{k}u)\delta _{i} \delta _{j}\nu _{k}+(\delta _{i} D_{k}u)\delta _{j} \nu _{k}+(\delta _{j}D_{k}u) \delta _{i}\nu _{k}+\nu _{k}\delta _{i} \delta _{j}D_{k}u=\delta _{i}\delta _{j}\varphi\quad \mbox{on } \partial \Omega . $$

(88)

Hence, for the tangential direction *ξ* at \(x_{0}\), we have

$$ \begin{aligned} D_{\nu }u_{\xi \xi }&\geq -2(\delta _{i} \nu _{k})D_{jk}u \xi _{i}\xi _{j}+(\delta _{i}\nu _{j})\xi _{i}\xi _{j}D_{\nu \nu }u-C \\ &\geq -2(\delta _{i} \nu _{k})D_{jk}u\xi _{i}\xi _{j}-C, \end{aligned} $$

(89)

where the double normal derivative estimate (46) on *∂*Ω is used in the second inequality. Thus, at \(x_{0}\) we have

$$ \begin{aligned} (\kappa c_{0}-\alpha M) \tilde{u}_{\xi \xi }&\leq 2( \delta _{i} \nu _{k})D_{jk}u\xi _{i}\xi _{j}+D_{\nu }\bigl(A_{\xi \xi }+v' \bigr)+C \\ &\leq 2(\delta _{i} \nu _{k})D_{jk}u\xi _{i}\xi _{j}+C \vert DD_{\nu }u \vert +C \\ &\leq 2(\delta _{i} \nu _{k})D_{jk}u\xi _{i}\xi _{j}+C, \end{aligned} $$

(90)

where the mixed tangential normal derivative estimate (28) and the double normal derivative estimate (46) are used to obtain the last inequality, and the constant *C* changes from line to line. Without loss of generality, we can assume the normal at \(x_{0}\) to be \(\nu =(0,\ldots ,1)\), and we can assume \(\{\tilde{u}_{ij}(x_{0})\}_{i, j< n}\) is diagonal with maximum eigenvalue \(\tilde{u}_{11}(x_{0})>1\). Then, at \(x_{0}\) we obtain

$$ (\kappa c_{0}-\alpha M)\tilde{u}_{\xi \xi } \leq C(\tilde{u}_{11}+1). $$

(91)

Since \(G(x_{0}, e_{1})\le G(x_{0}, \xi )\), we have

$$ \tilde{u}_{11}(x_{0}) \le \tilde{u}_{\xi \xi }(x_{0})+v'(x_{0}, e_{1})-v'(x_{0}, \xi ). $$

(92)

Combining (91) and (92) and choosing \(\kappa \geq \frac{2}{c_{0}}(\alpha M+C)\), we obtain

$$ u_{\xi \xi }(x_{0})\leq C. $$

(93)

*Case (ii)*. *ξ* non-tangential. We write *ξ* as

$$ \xi =\alpha \tau +\beta \nu , $$

(94)

where \(\alpha =\xi \cdot \tau \), \(|\tau |=1\), \(\tau \cdot \nu =0\), \(\beta =\xi \cdot \nu \neq 0\), and

$$ \alpha ^{2}+\beta ^{2}=1. $$

(95)

Therefore, at \(x_{0}\) we have

$$ \begin{aligned} D_{\xi \xi }\tilde{u}&=\alpha ^{2} \tilde{u}_{\tau \tau }+ \beta ^{2}\tilde{u}_{\nu \nu }+2 \alpha \beta \tilde{u}_{\tau \nu } \\ &=\alpha ^{2}\tilde{u}_{\tau \tau }+\beta ^{2} \tilde{u}_{\nu \nu }+v'(x, \xi ), \end{aligned} $$

(96)

where the definition of \(v'\) in (51) is used. Since \(G(x_{0},\tau ) \leq G(x_{0},\xi )\), we obtain

$$ \begin{aligned} G(x_{0},\xi )&=\alpha ^{2}G(x_{0},\tau )+\beta ^{2}G(x_{0}, \nu ) \\ &\leq \alpha ^{2} G(x_{0},\xi )+\beta ^{2}G(x_{0},\nu ). \end{aligned} $$

(97)

Using (95) in (97), we get

$$ G(x_{0}, \xi )\leq G(x_{0},\nu ), $$

(98)

which implies

$$ D_{\xi \xi }u(x_{0})\leq C+D_{\nu \nu }u(x_{0}) \leq C, $$

(99)

where the double normal derivative estimate (46) on *∂*Ω is used again.

Therefore, we can conclude from the two cases (i) and (ii) that if *G* attains its global maximum at \(x_{0}\in \partial \Omega \) and a unit vector *ξ*, then \(D_{\xi \xi }u(x_{0})\) is bounded from above. Taking (78) into account, we can derive estimate (11). Hence, Theorem 1.1 is proved. □