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For a weak solution of a uniformly elliptic homogeneous divergence-form elliptic operator, testing with ${\eta }^2(u-c)$ and choosing the gradient of the cutoff function in $B_{2R}\setminus B_R$ gives ${\int }_{B_R}|\nabla u{|}^2\le C{R}^{-2}{\int }_{B_{2R}\setminus B_R}|u-c{|}^2$. The arbitrary constant $c$ can be chosen as an annular average, which makes the estimate suitable for a hole-filling argument.

For a $C^1$ flux $F$ and a weak solution of $\mathrm{div}F(Du)=0$, discrete integration by parts and the fundamental theorem of calculus along a line segment show that $w_h={\delta }_{\ell ,-h}u$ solves $\mathrm{div}(A_{h}D{w}_h)=0$, with $p_{-}=Du(x-h{e}_{\ell })$ and $p_{+}=Du(x)$. Coercive quadratic-form bounds and operator norm bounds for $DF$ on all segments between these gradients pass to $A_h$. This produces a divergence-form elliptic operator without initially assuming second partial derivatives of $u$.

For a uniformly elliptic divergence-form elliptic operator with lower-order coefficients in the supercritical Lebesgue spaces, truncation tests and Moser iteration bound the positive part of a weak subsolution on an interior ball by its Lp norm on a larger ball and the forcing norms. In particular, every weak subsolution of the homogeneous equation is locally bounded above. This is the preliminary estimate needed when a strong maximum principle is phrased using essential suprema.

On a nonempty bounded domain where the Poincaré inequality holds, no $a\in H_0^1(\Omega )$ can satisfy $a\ge a_{-}>0$ almost everywhere. For $g(s)=\min (\max (s,0),a_{-})$, Lipschitz truncation preserves zero-boundary Sobolev spaces, so $g(a)\in H_0^1(\Omega )$. But $g(a)$ would be the nonzero constant $a_{-}$, whose zero gradient contradicts the Poincaré inequality. In a uniformly elliptic divergence-form elliptic operator, the homogeneous boundary condition belongs to the unknown and test functions, not to a coefficient bounded below by a positive constant.

Write $R=\Lambda /\lambda$. The coefficient is measurable, so interpret the divergence-form elliptic operator weakly: $\int A{u}^{\prime }{\varphi }^{\prime }=0$ for compactly supported $H^1$ tests. Testing with $\varphi ={\zeta }^{2}u$ givesDivide if the first integral is positive; the zero case is immediate. Since $\zeta =1$ on the inner interval,This Caccioppoli inequality applies in particular to the smooth solutions in the question. No derivative of $A$ is taken.

There is a genuine conflict in the printed coefficient assumptions: no uniformly positive coefficient in a zero-boundary Sobolev space exists. A function $a\in H_0^1(\Omega )$ cannot also obey $a\ge a_{-}>0$ almost everywhere. Indeed, the Lipschitz truncation $g(s)=\min (\max (s,0),a_{-})$ has $g(0)=0$, so Lipschitz truncation preserves zero-boundary Sobolev spaces, giving $g(a)\in H_0^1(\Omega )$. But $g(a)$ would be the nonzero constant $a_{-}$. Its zero gradient contradicts the Poincaré inequality in the zero-boundary Sobolev space. Thus the literal coefficient class is empty.

For the meaningful uniformly elliptic problem, take $a\in L^{\infty }(\Omega )$ with $0<a_{-}\le a\le a_{+}$, and impose the Dirichlet boundary condition on the unknown and test functions: $V=H_0^1((0,1{)}^2)$. Additional $H^1$ regularity of $a$ is harmless, but a zero trace for $a$ must be removed. The divergence-form elliptic operator is $L=-\mathrm{div}(a\nabla )$. Integration by parts defines the symmetric bounded bilinear formIn particular,Here $2{\pi }^2$ is the first Dirichlet Laplacian eigenvalue on the unit square; the Poincaré inequality follows, for example, by applying the one-dimensional inequality in each coordinate and adding. Thus $B$ is coercive in the gradient norm on $V$, and the Dirichlet realization of $L$ is a positive definite symmetric operator. On its operator domain, $⟨Lv,v⟩=B(v,v)$.

The required functional and weak equation areBecause $f\in L^2$, the right side is a bounded linear functional by the Cauchy-Schwarz inequality and Poincaré inequality. The Lax-Milgram theorem gives a unique weak solution, and part (b) proves that it uniquely minimizes $I$. These formulas prove the intended conclusion under the repaired coefficient hypothesis; under the literal hypothesis there is no coefficient to which the conclusion can be applied.

Put $\alpha =1-n/q>0$ andFirst suppose $A<\infty$. For an interior ball $B_R$, with $R\le 1$, write $M_R={\mathrm{esssup}}_{B_R}u$, $m_R={\mathrm{essinf}}_{B_R}u$, and $\omega (R)=M_R-m_R$. Constant shifts of a divergence-form elliptic operator change its forcing. The two nonnegative weak supersolutions $v=M_R-u$ and $w=u-m_R$ satisfy, respectively,Their Weak Harnack inequalities on $B_{R/2}$ therefore have forcing errors bounded by $CN{R}^{\alpha }$. Indeed $|M_R|,|m_R|\le A$, the new divergence forcing is bounded by $\Vert F{\Vert }_q+A\Vert b{\Vert }_q$, and the scalar forcing by $\Vert g{\Vert }_{q/2}+A\Vert d{\Vert }_{q/2}$; also $R^{2\alpha }\le R^{\alpha }$.

On $B_{R/2}$ we have $v+w=\omega (R)$. Even if $p<1$, the elementary inequality $(s+t{)}^p\le C_p(s^p+t^p)$ impliesThe sum of the two essential infima is $\omega (R)-\omega (R/2)$. Enlarging a uniform constant $C_0>1$ giveswhere $\theta =1-C_0^{-1}\in (0,1)$. This is an oscillation decay estimate; inhomogeneous oscillation decay implies Hölder continuity. ChooseIterating on concentric balls, starting at a fixed admissible radius $R_{\ast }\le 1$, yieldsThe strict choice of $\mu$ bounds the geometric series, including the case in which the two original decay rates coincide. Monotonicity of oscillation extends this bound to intermediate radii.

Take $R_{\ast }$ uniformly smaller than the distance of ${\Omega }^{\prime }$ from the boundary of $\Omega$. The vanishing essential oscillation produces a continuous representative of $u$. A ball centred at either of two sufficiently close points of ${\Omega }^{\prime }$ contains both, so its oscillation bounds their difference by $CN|x-y{|}^{\mu }$. For separated points use $|u(x)-u(y)|\le 2A$. Thus the Hölder seminorm satisfiesThe constants depend only on the indicated domain separation, dimension, coefficient norms, uniform ellipticity bounds and $q$. If the global essential supremum is infinite, the displayed global estimate is vacuous; local boundedness of weak elliptic subsolutions on slightly larger compact subsets still makes the same argument prove interior Hölder continuity.

Let $M={\mathrm{esssup}}_{\Omega }u$. The standard local boundedness of weak elliptic subsolutions gives a finite upper bound on every relatively compact ball. In particular, the assumed equality of essential suprema makes $M$ finite. Equivalently, when using the maximum principle for an upper-bounded weak subsolution, this preliminary fact is already part of the hypothesis.

Set $v=M-u$. Because the divergence-form elliptic operator here annihilates constants, $v\ge 0$ and $Lv=-Lu\le 0$. The hypothesis says ${\mathrm{essinf}}_{B}v=0$. Cover the compact closure of $B$ by finitely many balls $D_j$ whose doubled balls lie in $\Omega$. At least one $D_j$ has essential infimum zero: otherwise the minimum of their finitely many positive lower bounds would be a positive lower bound on $B$. The Weak Harnack inequality with zero forcing now givesConsequently $v=0$ almost everywhere on $D_j$. If another admissible ball overlaps $D_j$ in an open set, its essential infimum of $v$ is also zero, and the same argument makes $v$ vanish there. Since a domain is connected, a chain of overlapping interior balls reaches every point of $\Omega$. This propagation of zeros by the weak Harnack inequality proves $u=M$ almost everywhere on $\Omega$, which is the claimed strong maximum principle. Notice why the absence of the $b,d$ terms matters: it is precisely what makes $L(M-u)=-Lu$.

For the divergence-form elliptic operator use the weak subsolution convention $\int a^{ij}{D}_{j}u{D}_i\phi \le 0$ for nonnegative test functions. Set $u_{\epsilon }=u+\epsilon$ and test with $\phi ={\eta }^2{u}_{\epsilon }^{\beta -1}$, where $\beta >1$. The Sobolev chain rule makes this test admissible; boundedness and the positive regularization control the derivative of its power. Ellipticity and the Cauchy-Schwarz inequality give, with $X=\int |Du{|}^2{u}_{\epsilon }^{\beta -2}{\eta }^2$ and $Y=\int u_{\epsilon }^{\beta }|D\eta {|}^2$,Division when $X>0$, and the trivial case $X=0$, establish the power Caccioppoli inequality. Letting $\epsilon \downarrow 0$ givesFor $1<\beta <2$ this limit uses the Fatou lemma; the gradient of a Sobolev function vanishes on a level set, so the weighted integrand is assigned zero on $\{u=0\}$. The same regularization justifies differentiating $u^{\beta /2}\eta$ in $W^{1,2}$.

The PDF bounds individual entries by $\Lambda$. Then ${\Lambda }_{\ast }\le n\Lambda$: dimensional factors are included in the ellipticity constants when the first estimate is written as $C(\lambda ,\Lambda )$. With literal entrywise bounds, the explicit valid constant displayed here is $4n^2{\Lambda }^2/{\lambda }^2$. If $\Lambda$ bounds the operator norm instead, it is $4{\Lambda }^2/{\lambda }^2$. This distinction does not alter any later conclusion, whose constants explicitly depend on $n$.

For $n\ge 3$, put $\sigma =n/(n-2)$. Apply the Sobolev inequality to $g=u^{\beta /2}\eta$ and useFor $\beta \ge p>1$, $\beta /(\beta -1)\le p/(p-1)$, so the resulting constant is uniform in $\beta$. Taking a cutoff equal to one on $B_r$, supported in $B_R$, with $|D\eta |\le C/(R-r)$, yieldsAt $R=1$ a compactly supported cutoff can still be chosen with this bound by leaving an outer margin; alternatively pass to the limit from smaller radii.

For Moser iteration, set ${\beta }_j=p{\sigma }^j$ and $r_j=1/2+2^{-j-1}$. Repeated use of the preceding estimate bounds the successive norms byThis product is finite becauseAs the exponents tend to infinity, their norms on the Euclidean ball of radius $1/2$ tend to its essential supremum. ThereforeThe power $\beta$ is independent of the exponents of Hölder continuity in earlier questions.