# Introduction

In this article, we prove the classical Aleksandrov maximum principle for the second-order elliptic operator in non-divergence form: $\mathcal{L}u=a^{ij}D_{ij} u + b^i D_i u +cu,$ where $A=[a^{ij}]$ is elliptic in $\Omega$ in the sense that $a^{ij}$ is positive on $\Omega$ and $c\leq 0$ on $\Omega$. We further assume that eigenvalues of $A$ are greater than $\lambda>0$ and less than $\Lambda$. Then the geometric mean $D^*$ of eigenvalues of $A$ satisfies $0<\lambda\leq D^* \leq\Lambda$. We further assume that $\mathbf{b} /D^* \in L_{n}(\Omega)$. In 1963, Aleksandrov  proved the following theorem, whose proof was refined by Bakel'man . A similar result was shown by Pucci  in 1966.
Theorem 1. Let $\Omega$ be a bounded domain in $\mathbb{R}^n$ and $u\in W^{2}_{n,loc}(\Omega)\cap C(\overline{\Omega})$ satisfy $\mathcal{L}u\geq f$ in $\Omega$. Then $\sup_\Omega u \leq \sup_{\partial \Omega} u^+ + C\Vert{f/D^*}\Vert_{ L_{n}(\Omega)},$ where $C$ is a constant depending only on $n$, $\mathrm{diam}\, \Omega$, and $\Vert{\mathbf{b}/D^*}\Vert_{ L_{n}(\Omega)}$.
One may relax the assumption $u\in W^{2}_{n,loc}(\Omega)$ to $u\in W^{2}_{p}(\Omega)$, $p<n$. However, the following example suggests that this is impossible in general.
Example. Let $a^{ij} =\delta_{ij} + b \frac{x_i x_j}{|x|^2},\quad b=-1+\frac{n-1}{1-\lambda}, \quad 0<\alpha<1,$ where $n>2(2-\alpha)>2$. Then $u_1(x)=1$ and $u_2(x)=|x|^\alpha$ are in $W^{2}_{2}(B)$ and agreeing on $\partial B$, where $B$ is the unit ball. However, both functions are solutions of $a^{ij}D_{ij}u=0\quad \text{in } B_1.$ This example suggests that there cannot be an Alexandrov type estimate for $\mathcal{L}=a^{ij}D_{ij}$.
The example does not mean that we cannot expect Alexandrov type estimate $W^{2}_{p}$ for any $p<n$. Recently, Krylov  proved that there exists $n/2 <n_0 <n$ depending only on $\lambda$, $\Lambda$, $n$, $\mathrm{diam}\,\Omega$, and $\mathbf{b}$ such that if $n_0\leq p<\infty$, then the following estimate $u(x)\leq \sup_{\partial \Omega} u^+ + C \Vert{(\mathcal{L}u-cu)_{-}}\Vert_{L_{p}(\Omega)}\quad \text{for all } x\in \Omega$ holds for all $u\in W^{2}_{p,loc}(\Omega) \cap C(\overline{\Omega})$. This result was obtained by the probabilistic method. Without using probability theory, another proof was obtained by Dong-Krylov  by using the result of Byun-Lee-Palagachev  on Hessian estimates of fully nonlinear equations. The novelty of this result is to allow Morrey drifts $\mathbf{b}$ although the domain is restricted to $B_R$.

Here I do not claim that the proof of this theorem is not original. Mostly, I follow the proof given in the textbook of Qing Han  and Gilbarg-Trudinger .

# Proof of the Alexandrov maximum principle

Let $\Omega$ be a bounded domain in $\mathbb{R}^n$ and let $u\in C(\Omega)$. We define a subset $\Gamma^+$ of $\Omega$ as follows: $y\in\Gamma^+$ if and only if there exists $p=p(y)\in\mathbb{R}^n$ such that $u(x)\leq u(y)+ p\cdot(y-x)$ for all $x\in\Omega$. We call $\Gamma^+$ the upper contact set of $u$. Similarly, we can define the lower contact set of $u$. The upper contact set is the set of all points whose tangent hyperplane lies above the graph of $u$.
Remark. (a) Suppose that $u$ is differentiable on $\Omega$. Then $y\in \Gamma^+ \quad \text{if and only if } u(x)\leq u(y)+ \nabla u(y)\cdot (y-x)\quad \text{for all } x\in \Omega.$ Since $\Omega$ is open, there exists $\delta>0$ such that $B_\delta (x)\subset \Omega$. Note that for $0<h<\delta$ and $x=y+he_j$, we have $u(y+he_j)-u(y)\leq p_j h.$ Since $u$ is differentiable, it follows that $D_ju(y)\leq p_j$. Similarly, we also have $p_j \leq D_j u(y)$. Hence, $p=\nabla u(y)$.
(b) If $u$ is twice differentiable in $\Omega$, then $-\nabla^2 u$ is positive semidefinite on $\Gamma^+$. Note that $\Gamma^+$ is relatively closed in $\Omega$. We write $\Gamma^+ = C_1 \cup C_2 \cup \cdots \cup C_k,$ where $C_i$ is a connected component of $\Gamma^+$. Hence we may assume that $\Gamma^+=\Omega$. Then one can easily show that $u$ is concave on $\Omega$. Define $v(t)=u(x+t\xi).$ Then $v$ is concave around $0$ and $v''(0)\leq 0$. By the chain rule, $v'(t)= u_{x_i} (x+t\xi) \xi_i$ and $v''(t)=u_{x_i x_j}(x+t\xi) \xi_i \xi_j$ for $\xi \in \mathbb{R}^n$. By plugging $t=0$, we see that $\nabla^2 u$ is nonpositive on $\Gamma^+$.
Lemma 2. Let $\Omega$ be a bounded domain in $\mathbb{R}^n$ and $g\in L_{1,loc}(\mathbb{R}^n)$ be nonnegative. Then for any $u\in C^2(\Omega)\cap C(\overline{\Omega})$, we have $\int_{B_{M/4}} g(p)\, dp \leq \int_{\Gamma^+} g(\nabla u) |\det (\nabla^2 u)|\, dx,$ where $\Gamma^+$ is the upper contact set of $u$, $M=\sup_\Omega u-\max_{\partial \Omega} u^+$, and $d= \mathrm{diam}\, \Omega$.
Proof. By replacing $u$ with $u-\sup_{\partial \Omega} u^+$, it suffices to assume that $u\leq 0$ on $\partial\Omega$. Also, we may assume that $M>0$. Set $\Omega^+ = \{u>0\}$. Then area formula gives $\int_{\nabla u(\Gamma^+\cap \Omega^+)} g(p) \, dp\leq \int_{\Gamma^+ \cap \Omega^+} g(\nabla u)| \det (\nabla^2 u)|\, dx.$ We show that $B_{M/d}\subset \nabla u(\Gamma^+\cap \Omega^+)$. By translation, we may assume that $u$ attains the maximum at $0$. Let $a\in B_{M/d}$ define $u_a(x)=u(x)-a\cdot x.$ Since $u\leq 0$ on $\partial\Omega$ and $a\in B_{M/d}$, it follows that $u_a(x)\leq -a\cdot x<M$ for all $x\in \partial\Omega$. Hence there exists a point $x_a \in \Omega$ such that $u_a$ attains its maximum at $x_a$. Also, $u(x_a)-a\cdot x_a =u_a(x_a)\geq M,$ i.e., $u(x_a)\geq M + a\cdot x_a >0.$ Moreover, $u(x)-a\cdot x =u_a(x) \leq u_a(x_a).$ Hence $u(x)\leq u(x_a)+a\cdot (x-x_a),$ which proves that $x_a \in \Gamma^+$ and $a=\nabla u(x_a)$. This proves the desired assertion.
Corollary 3. Let $A=\{a^{ij}\}$ be a positive definite matrix. Then we have $\det (-\nabla^2 u)\leq \frac{1}{D}\left(\frac{-a^{ij} D_{ij} u}{n}\right)^n \quad \text{on } \Gamma^+,$ where $D$ is the determinant of $A$. Thus, we have $\int_{B_{\tilde{M}}}g\, dx\leq \int_{\Gamma^+ \cap \{u>0\}} g(\nabla u)\left(-\frac{a^{ij} D_{ij} u}{n D^*}\right)^n \, dx,$ where $\tilde{M}=(\sup_\Omega u^+-\sup_{\partial\Omega } u^+)/d$.
Proof. By the spectral theorem, we can easily show the following inequality for two positive semidefinite matrices: $(\det A) (\det B) \leq \left(\frac{\mathrm{Tr}(AB)}{n}\right)^n.$ Note that on $\Gamma^+$, $-\nabla^2 u$ is positive semidefinite. Hence it follows that $\det (-\nabla^2 u) \leq \frac{1}{D}\left(-\frac{a^{ij} D_{ij}u}{n}\right)^n\quad \text{on } \Gamma^+.$ Hence $\int_{B_{\tilde{M}}} g\, dx\leq \int_{\Gamma^+\cap \{u>0\}} g(\nabla u) \frac{1}{D}\left(-\frac{a^{ij}D_{ij}u}{n} \right)^n \, dx.$ This completes the proof.

Now we are ready to prove Theorem 1.

Proof of Theorem 1. We first assume that $u\in C^2(\Omega)\cap C(\overline{\Omega})$. For $\mu>0$, set $g(p)=(|p|^n+\mu^n)^{-1}$. Then $g\in L_{1,loc}(\mathbb{R}^n)$. On $\Omega^+=\{ u>0\}$, we have
\begin{align} -a^{ij}D_{ij}u&\leq b^i D_i u+cu-f\nonumber \\ &\leq b^i D_i u-f\nonumber \\ &\leq |\mathbf{b}||\nabla u| +f^-\nonumber\\ &\leq |\mathbf{b}||\nabla u \cdot 1 + \frac{f^-}{\mu} \mu \cdot 1\nonumber \\ &\leq \left(|\mathbf{b}|^n + \left(\frac{f^-}{\mu} \right)^n \right)^{1/n}\left(|\nabla u|^n +\mu^n\right)^{1/n} \cdot (1+1)^{(n-2)/n},\nonumber \end{align}
that is, $(-a^{ij}D_{ij} u)^n \leq \left(|\mathbf{b}|^n + \left(\frac{f^-}{\mu} \right)^n \right)(|\nabla u|^n + \mu^n)2^{n-2}$ on $\Omega^+$. Then it follows from Lemma 2 that $\int_{B_{\tilde{M}}} g\, dx\leq c \int_{\Gamma^+\cap \{u>0\}} \frac{|\mathbf{b}|^n + (f^-/\mu)^n}{(D^*)^n}\, dx.$ On the other hand, using a change of variable formula, we have $\int_{B_{\tilde{M}}} g\, dx =\frac{\omega_n}{n} \log \left(\frac{\tilde{M}^n}{\mu^n} +1\right).$ Therefore, by taking exponential, we finally get $\tilde{M}^n \leq \mu^n\left[\exp \left(\frac{2^{n-2}}{\omega_n n^n} \left[\Vert{\mathbf{b}/D^*}\Vert_{ L_{n}(\Gamma^+\cap \Omega^+)}^n+\mu^{-n} \Vert{f^-/D^*}\Vert_{ L_{n}(\Gamma^+\cap \Omega^+)} \right]\right)-1 \right].$ If $f=0$, we let $\mu\rightarrow 0+$. If $f\neq 0$, we take $\mu = \Vert{f^-/D^*}\Vert_{ L_{n}(\Gamma^+\cap \Omega^+)}$. Then we finally get $\sup_{\Omega} u \leq \sup_{\partial \Omega} u^+ +C\Vert{f^-/D^*}\Vert_{ L_{n}(\Gamma^+\cap \Omega^+)}$ for some constant $C=C(\Vert{\mathbf{b}}\Vert_{ L_{n}(\Omega)},n, \mathrm{diam}\, \Omega)>0$.
To show that the theorem holds for $u\in W^{2}_{n,loc}(\Omega)\cap C(\overline{\Omega})$, we may assume that $|\mathbf{b}|/\lambda$ is bounded. We first assume that $\mathcal{L}$ is uniformly elliptic in $\Omega$ and choose a sequence $\{u_k\}$ in $C^2(\Omega)$ so that $u_k \rightarrow u$ in $W^{2}_{n,loc}(\Omega)$. For $\varepsilon>0$, we further assume that $u_k$ converges to $u$ in $W^{2}_{n}(\Omega_\varepsilon)$ and $u_m \leq \varepsilon + \sup_{\partial \Omega} u$ on $\partial\Omega_{\varepsilon}$ for some domain $\Omega_{\varepsilon}\subset \subset \Omega$. Write $a^{ij}D_{ij} u_m + b^i D_i u_m \geq f-a^{ij}D_{ij} (u-u_m)+b^i D_i (u-u_m)\quad \text{in } \Omega_\varepsilon.$ By the Aleksandrov maximum principle, we have \begin{align} \sup_{\Omega_\varepsilon} u_m &\leq \varepsilon + \sup_{\partial \Omega} u + C\Vert{f/D^*}\Vert_{ L_{n}(\Omega_\varepsilon)}\nonumber\\ &\quad+C\Vert{(a^{ij}D_{ij}(u-u_m)+b^i D_i (u-u_m))/D^*}\Vert_{ L_{n}(\Omega_\varepsilon)}. \nonumber \end{align} By letting $m\rightarrow \infty$, we get $\sup_{\Omega_\varepsilon} u \leq \varepsilon + \sup_{\partial \Omega} u^+ + C\Vert{f/D^*}\Vert_{ L_{n}(\Omega_\varepsilon)}.$ For $\eta>0$, set $\mathcal{L}_\eta = (\eta \Lambda) \Delta +\mathcal{L}.$ It is easy to see that $\mathcal{L}_\eta$ is uniformly elliptic. Since $\mathcal{L}_\eta u = (\eta \Lambda)\Delta u +\mathcal{L}u\geq (\eta \Lambda)\Delta u+f,$ it follows that \begin{align} \sup_{\Omega_\varepsilon} u&\leq \varepsilon + \sup_{\partial \Omega} u^+ + C\Vert{(\eta \Lambda)\Delta u/D^*_\eta}\Vert_{ L_{n}(\Omega_\varepsilon)}+\Vert{f/D^*_\eta}\Vert_{ L_{n}(\Omega_\varepsilon)}\nonumber\\ &\leq \varepsilon + \sup_{\partial \Omega} u^+ + C\Vert{(\eta \Lambda)\Delta u/D^*_\eta}\Vert_{ L_{n}(\Omega_\varepsilon)}+\Vert{f/D^*}\Vert_{ L_{n}(\Omega_\varepsilon)}\nonumber \end{align} where $D_\eta^*$ denotes the geometric mean of eigenvalues of the matrix $\eta\Lambda \delta_{ij} + a^{ij}$. By the dominated convergence theorem and $u\in W^{2}_{n}(\Omega_\varepsilon)$, we get $\sup_{\Omega_\varepsilon} u \leq \varepsilon + \sup_{\partial \Omega} u^++C\Vert{f/D^*}\Vert_{ L_{n}(\Omega)}.$ Letting $\varepsilon \rightarrow 0$, we get the desired conclusion. This completes the proof of Theorem 1.
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