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. □