[proofplan] We first prove the result under the stronger hypothesis $Lu < 0$ (strict inequality), where a direct contradiction at any [interior](/page/Interior) non-negative maximum shows the maximum must occur on the [boundary](/page/Boundary). We then reduce the general case $Lu \le 0$ to the strict case by adding a small perturbation $\varepsilon v$ with $v(x) = e^{\lambda x_1}$, chosen so that $Lv < 0$ everywhere. Taking $\varepsilon \to 0$ recovers the result for the original $u$. [/proofplan] [step:Rule out interior non-negative maxima when $Lu < 0$ strictly] Assume $Lu < 0$ in $U$. Suppose for contradiction there exists $x_0 \in U$ with $u(x_0) = \max_{\bar{U}} u \ge 0$. Since $x_0$ is an interior maximum of $u \in C^2(U)$, the first-order necessary condition gives $\nabla u(x_0) = 0$, and the second-order condition gives that the Hessian $D^2 u(x_0)$ is negative semi-definite. [guided] Why does $Lu(x_0) < 0$ lead to a contradiction at an interior maximum? The operator $L$ evaluated at $x_0$ decomposes into three contributions: \begin{align*} Lu(x_0) = \underbrace{-\sum_{i,j} a_{ij}(x_0)\, \partial_{x_i x_j} u(x_0)}_{\text{second-order}} + \underbrace{\sum_i b_i(x_0)\, \partial_{x_i} u(x_0)}_{\text{first-order}} + \underbrace{c(x_0)\, u(x_0)}_{\text{zeroth-order}}. \end{align*} At the interior maximum $x_0$, the first-order necessary condition gives $\nabla u(x_0) = 0$, which kills the first-order term entirely: \begin{align*} \sum_i b_i(x_0)\, \partial_{x_i} u(x_0) = 0. \end{align*} The second-order necessary condition gives that $D^2 u(x_0)$ is negative semi-definite. Since $(a_{ij}(x_0))$ is symmetric and positive definite by the ellipticity hypothesis, diagonalizing via an orthogonal matrix $Q$ with $Q^\top (a_{ij}) Q = \operatorname{diag}(\lambda_1, \dots, \lambda_n)$ where each $\lambda_k > 0$ gives: \begin{align*} \sum_{i,j} a_{ij}(x_0)\, \partial_{x_i x_j} u(x_0) = \sum_{k=1}^n \lambda_k\, \partial_{y_k y_k} \tilde{u}(0) \le 0, \end{align*} since the trace of a product of a positive-definite matrix with a negative semi-definite matrix is non-positive. The minus sign in $-\sum a_{ij} \partial_{x_i x_j} u$ therefore makes the second-order term non-negative: \begin{align*} -\sum_{i,j} a_{ij}(x_0)\, \partial_{x_i x_j} u(x_0) \ge 0. \end{align*} For the zeroth-order term, the hypotheses give $c(x_0) \ge 0$ and the assumption $u(x_0) \ge 0$ (since $x_0$ is a non-negative maximum), so $c(x_0)\, u(x_0) \ge 0$. Combining all three: \begin{align*} Lu(x_0) \ge 0 + 0 + 0 = 0, \end{align*} which contradicts the hypothesis $Lu(x_0) < 0$. Therefore no interior point can achieve a non-negative maximum. [/guided] [/step] [step:Evaluate $Lu(x_0)$ using ellipticity and the maximum conditions] Since $A(x_0) = (a_{ij}(x_0))$ is symmetric and positive definite, there exists an orthogonal matrix $Q$ such that $Q^\top A(x_0) Q = \operatorname{diag}(\lambda_1, \dots, \lambda_n)$ with each $\lambda_k > 0$. In the rotated coordinates $y = Q^\top(x - x_0)$, writing $\tilde{u}(y) = u(x_0 + Qy)$: \begin{align*} \sum_{i,j=1}^n a_{ij}(x_0)\, \partial_{x_i x_j} u(x_0) = \sum_{k=1}^n \lambda_k\, \partial_{y_k y_k} \tilde{u}(0). \end{align*} Since $D^2 u(x_0)$ is negative semi-definite, so is $D^2 \tilde{u}(0)$, and in particular $\partial_{y_k y_k} \tilde{u}(0) \le 0$ for each $k$. Since $\lambda_k > 0$: \begin{align*} \sum_{i,j=1}^n a_{ij}(x_0)\, \partial_{x_i x_j} u(x_0) = \sum_{k=1}^n \lambda_k\, \partial_{y_k y_k} \tilde{u}(0) \le 0. \end{align*} Therefore: \begin{align*} Lu(x_0) = \underbrace{-\sum_{i,j=1}^n a_{ij}(x_0)\, \partial_{x_i x_j} u(x_0)}_{\ge\, 0} + \underbrace{\sum_{i=1}^n b_i(x_0)\, \partial_{x_i} u(x_0)}_{=\, 0} + \underbrace{c(x_0)\, u(x_0)}_{\ge\, 0} \ge 0. \end{align*} This contradicts $Lu(x_0) < 0$. Therefore no interior point can be a non-negative maximum, and $\max_{\bar{U}} u \le \max_{\partial U} u^+$. [/step] [step:Reduce the general case $Lu \le 0$ to the strict case via an exponential perturbation] Now assume only $Lu \le 0$ in $U$. Define the auxiliary [function](/page/Function): \begin{align*} v: U &\to \mathbb{R} \\ x &\mapsto e^{\lambda x_1}, \end{align*} where $\lambda > 0$ is a constant to be chosen. Computing $Lv$: \begin{align*} Lv = e^{\lambda x_1}\bigl(-\lambda^2 a_{11}(x) + \lambda b_1(x) + c(x)\bigr). \end{align*} By uniform ellipticity, $a_{11}(x) \ge \theta > 0$ for all $x \in U$. Since $b_1$ and $c$ are bounded on $\bar{U}$, the expression $-\lambda^2 \theta + \lambda \|b_1\|_{L^\infty(U)} + \|c\|_{L^\infty(U)}$ is negative for all $\lambda$ sufficiently large. Fix such a $\lambda$, giving $Lv < 0$ in $U$. For any $\varepsilon > 0$, define $u_\varepsilon := u + \varepsilon v$. By linearity of $L$: \begin{align*} Lu_\varepsilon = Lu + \varepsilon Lv \le 0 + \varepsilon Lv < 0 \quad \text{in } U. \end{align*} By the strict case established above, $\max_{\bar{U}} u_\varepsilon \le \max_{\partial U} u_\varepsilon^+$. [/step] [step:Take $\varepsilon \to 0$ to obtain the result for $u$] For any $x \in \bar{U}$: \begin{align*} u(x) = u_\varepsilon(x) - \varepsilon v(x) \le u_\varepsilon(x) \le \max_{\partial U} u_\varepsilon^+. \end{align*} On $\partial U$, $u_\varepsilon^+(x) = \max(u(x) + \varepsilon e^{\lambda x_1}, 0) \le u^+(x) + \varepsilon e^{\lambda \operatorname{diam}(U)}$. Therefore: \begin{align*} \max_{\bar{U}} u \le \max_{\partial U} u^+ + \varepsilon\, e^{\lambda \operatorname{diam}(U)}. \end{align*} Since $\varepsilon > 0$ was arbitrary, taking $\varepsilon \to 0$ gives: \begin{align*} \max_{\bar{U}} u \le \max_{\partial U} u^+. \end{align*} [guided] The perturbation trick is a standard technique for passing from strict to non-strict inequalities. The function $v = e^{\lambda x_1}$ is chosen because it is strictly positive (so the perturbation $\varepsilon v$ does not introduce new zeros), and its derivatives have a specific structure. Computing $Lv$ explicitly: \begin{align*} Lv &= -a_{11}\, \lambda^2\, e^{\lambda x_1} + b_1\, \lambda\, e^{\lambda x_1} + c\, e^{\lambda x_1} \\ &= e^{\lambda x_1}\bigl(-\lambda^2 a_{11} + \lambda b_1 + c\bigr). \end{align*} The second [derivative](/page/Derivative) $\partial_{x_1 x_1} v = \lambda^2 e^{\lambda x_1}$ contributes the $-\lambda^2 a_{11}$ term, which grows quadratically in $\lambda$ and dominates the first-order term $\lambda b_1$ and zeroth-order term $c$ for $\lambda$ large. The key observation is that $Lv < 0$ requires only $a_{11} \ge \theta > 0$ (which follows from ellipticity applied with $\xi = e_1$) and boundedness of $b_1$ and $c$. Concretely, $Lv < 0$ holds whenever: \begin{align*} \lambda^2 \theta > \lambda \|b_1\|_{L^\infty(U)} + \|c\|_{L^\infty(U)}, \end{align*} which is satisfied for all $\lambda$ sufficiently large since the left side grows as $\lambda^2$ while the right side grows as $\lambda$. With $Lv < 0$ and $Lu \le 0$, the combination $Lu_\varepsilon = Lu + \varepsilon Lv < 0$ is strictly negative, placing us in the strict case already established. Taking $\varepsilon \to 0$ recovers the result because the perturbation $\varepsilon v$ vanishes uniformly on $\bar{U}$: \begin{align*} \|\varepsilon v\|_{L^\infty(\bar{U})} = \varepsilon\, e^{\lambda \max_{x \in \bar{U}} x_1} \to 0 \quad \text{as } \varepsilon \to 0. \end{align*} [/guided] [/step]