Let $Lu = -\sum_{ij}\partial_i(a^{ij}\partial_j u) + \sum_i b^i\partial_i u + cu$ with $L$ uniformly elliptic, and suppose $Lu \leq 0$ in $\Omega$ and $c \leq 0$.

Claim: $\max_{\overline{\Omega}} u = \max_{\partial\Omega} u$

Step 1: Assume Strict Inequality

Suppose for contradiction that $u$ attains its maximum $M$ at an interior point $x_0 \in \Omega$ and $M > \max_{\partial\Omega} u$.

Choose $\epsilon > 0$ small and define: $v(x) = u(x) + \epsilon|x - x_0|^2$

For small $\epsilon$, $v$ also attains an interior maximum at some point $x_\epsilon$ near $x_0$.

Step 2: Compute $Lv$ at the Maximum

At $x_\epsilon$, we have $\nabla v(x_\epsilon) = 0$ and the Hessian $D^2v \leq 0$ (since it's a maximum).

Computing: $Lv = Lu + \epsilon L(|x - x_0|^2)$

The second term is: $L(|x - x_0|^2) = -\sum_{ij}a^{ij}\partial_{ij}|x - x_0|^2 + \text{lower order}$ $= -2\sum_{ij}a^{ij}\delta_{ij} + \text{lower order} = -2\text{tr}(A) + \text{lower order}$

By uniform ellipticity, $\text{tr}(A) \geq n\lambda > 0$, so: $L(|x - x_0|^2) \leq -2n\lambda + C$

for some constant $C$ depending on bounds of lower-order terms.

Step 3: Derive Contradiction at Maximum

For small enough $\epsilon$, at the interior maximum $x_\epsilon$: $Lv(x_\epsilon) = Lu(x_\epsilon) + \epsilon L(|x - x_0|^2) < 0$

since $Lu \leq 0$ and the second term is negative for small $\epsilon$.

But at an interior maximum $x_\epsilon$ where $\nabla v = 0$ and $D^2v \leq 0$: $Lv = -\sum_{ij}a^{ij}\partial_{ij}v + \sum_i b^i\partial_i v + cv$

The first term is $-\sum a^{ij}(D^2v)_{ij} \geq 0$ by ellipticity and $D^2v \leq 0$. The second term vanishes since $\nabla v = 0$. The third term is $cv(x_\epsilon) \leq 0$ since $c \leq 0$ and $v > 0$ near the max.

Therefore $Lv(x_\epsilon) \geq 0$, contradicting $Lv(x_\epsilon) < 0$.

Step 4: Conclude

The contradiction shows $u$ cannot attain its maximum strictly inside $\Omega$. Hence: $\max_{\overline{\Omega}} u = \max_{\partial\Omega} u$