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For a uniformly elliptic operator $Lu=a^{ij}(x)D_{ij}u$ with continuous coefficients and $W⋐U$,Uniform continuity lets one freeze $a^{ij}$ on sufficiently small open balls and use the perturbation of a constant-coefficient elliptic second-derivative estimate. A partition of unity and cutoff functions reduce the global interior estimate to finitely many such balls; derivatives of the cutoffs produce only the displayed lower-order norm.

For a fixed bounded measurable uniformly elliptic operator in divergence form, solutions bounded by one have uniform local Caccioppoli inequalities and Hölder continuity estimates. A diagonal subsequence argument yields a local weak $W^{1,2}$ limit, while the Arzelà-Ascoli theorem gives uniform convergence on a fixed smaller closed Euclidean ball. The weak equation passes to the limit because the coefficients are fixed and bounded. A global supremum normalization does not force the interior limit to be nonzero.

For a fixed uniformly elliptic operator with $C^{0,\mu }$ coefficients on a bounded smooth connected domain, nonnegative solutions with zero Dirichlet boundary condition satisfy the displayed estimate when ${\Omega }^{\prime }$ is nonempty and compactly contained in $\Omega$. If it failed, normalize the global supremum to one. The forcing and the interior infimum then tend to zero. The global Schauder estimate and compact embedding of Hölder spaces produce a nonzero homogeneous limit with an interior zero, contradicting the strong minimum principle for elliptic operators. An arbitrary bounded zeroth-order coefficient is allowed: replace $c$ by $\min (c,0)$ for the positivity argument. Keeping global boundary control prevents the normalized maximum from disappearing in the limit.

On the unit ball take the uniformly elliptic operator $L=\Delta -2n$ and the smooth function $u(x)=-1-|x{|}^2$. The zeroth-order coefficient is negative andNeverthelessThis is consistent with the weak maximum principle, whose upper bound here is ${\sup }_{\partial B_1}{u}^{+}=0$. At a negative interior maximum the term $cu$ can be positive; this is why a negative boundary maximum cannot replace its positive part. If $c=0$, subtracting the boundary maximum preserves the differential inequality and gives the stronger bound. That subtraction is unavailable for general $c\le 0$.

The Hopf boundary point lemma requires an interior sphere condition, strictness of the boundary maximum, and quantitative control of the uniformly elliptic operator near the contact point. Assume that $y\in \partial \Omega$, that $B_R(z)\subset \Omega$ and $\overline{B_R(z)}\setminus \{y\}\subset \Omega$, and that $|y-z|=R$. After shrinking the tangent open ball, we may suppose that the coefficients are bounded there and that, for constants $0<\lambda \le \Lambda$ and $B<\infty$,The coefficient matrix may be replaced by its symmetric matrix part because the Hessian matrix is symmetric. Assume also that $u(x)<u(y)$ throughout the tangent open ball. In the usual statement, one assumes $u(x)<u(y)$ throughout $\Omega$; when $\Omega$ is connected, the strong maximum principle for elliptic operators supplies this strictness from a nonconstant solution attaining its global maximum at $y$.

Put $\mathbf{n}=(y-z)/R$, the outward unit normal of the tangent open ball. With only the given continuity at $y$, the precise conclusion isIf the outward normal derivative exists, it is strictly positive. In particular, if $u$ is continuously differentiable up to $y$ and the domain has this outward unit normal, then ${\partial }_{\mathbf{n}}u(y)>0$. The stated $C^0$ boundary regularity alone does not assert existence of a normal derivative.

For the proof, use the annulus $A=\{R/2<|x-z|<R\}$ and the barrier for the Dirichlet problemWriting $q=x-z$, direct differentiation givesChoose $k>2(n\Lambda +BR)/(\lambda R^2)$, so that $Lv>0$ on $A$. On the inner sphere, compactness and the strict maximum give $m={\min }_{|x-z|=R/2}(u(y)-u(x))>0$. Choose $\epsilon >0$ with $\epsilon {\max }_{|x-z|=R/2}v\le m$. On the outer sphere $v=0$ and $u-u(y)\le 0$, including at $y$ by continuity. Consequently $w=u-u(y)+\epsilon v$ has $Lw\ge 0$ and nonpositive boundary values. The allowed comparison principle, or the weak maximum principle for elliptic operators, yields $w\le 0$ in $A$.

Along the inward radius this givesThis proves the Hopf boundary point lemma. There is no sign restriction on $u(y)$ here, because the operator has no zeroth-order term.

Write the minimal surface equation for a graph as $\mathrm{div}F(Du)=0$ in its weak formulation, where $F(p)=p/\sqrt{1+|p{|}^2}$. For negative increments use the same difference quotient convention, soFor $\zeta \in C_c^1({\Omega }^{\prime \prime })$, the function ${\delta }_{\ell ,h}\zeta$ is an admissible test function in $\Omega$. Commuting the first partial derivative with translation, then applying discrete integration by parts, givesThe fundamental theorem of calculus along a line segment gives the averaged linearization of a nonlinear divergence-form equationThus the backward difference quotient solves a linear equation in divergence form:To verify the ellipticity of the minimal surface flux, computeand henceThe eigenvalue parallel to $p$ is $(1+|p{|}^2{)}^{-3/2}$; every orthogonal eigenvalue is $(1+|p{|}^2{)}^{-1/2}$. The same lower and upper bounds pass to the averaged symmetric matrix $A_h$ whenever both endpoint gradients have norm at most $M$:The printed $C^1(\Omega )$ hypothesis supplies such an $M$ on each relatively compact subset, uniformly for sufficiently small $h$. It therefore establishes a locally uniformly elliptic operator. A single lower bound on all of ${\Omega }^{\prime \prime }$ additionally requires bounded gradients there and at the shifted points; this is automatic for bounded $\Omega$ and fixed $h$, but is not supplied on an arbitrary open set.

There is no need to assume that the original zeroth-order coefficient is nonpositive. Set $c_{+}=\max (c,0)$ and $\tilde{c}=\min (c,0)$, and use the uniformly elliptic operatorBecause $u\ge 0$ and $f\le 0$,Now $\tilde{c}\le 0$, so the strong minimum principle for elliptic operators applies. If $u$ vanished at an interior point of the connected domain, this principle would force $u$ to vanish identically. The given nontriviality therefore provesEvery nontrivial nonnegative solution is strictly positive inside. This uses the nondivergence-form strong minimum principle for elliptic operators; the divergence-form Harnack inequality for uniformly elliptic divergence-form equations in part (a) is not being applied to this operator.

Positivity and continuity of $g$ give $\mu ={\min }_{[1/2,2]}g>0$ and a finite maximum $G$. Hence, for every $x$ in the shell and every $\xi \in {\mathbb{R}}^3$,This is the uniformly elliptic operator condition. Positivity at separate points without a compactness/continuity lower bound would not alone supply a common ellipticity constant.

Assume real bounded coefficients, with a principal part defining a uniformly elliptic operator, say $a^{ij}{\xi }_i{\xi }_j\ge \lambda |\xi {|}^2$ for some $\lambda >0$, and assume $c\le 0$. No smoothness of the boundary is needed for this classical weak maximum principle for elliptic operators. Its conclusion isIf $c=0$, the zero may be omitted. Only the symmetric part of the principal coefficients contributes to the Hessian matrix contraction.

For a strict inequality $Lu>0$, a positive interior maximum is impossible: there $Du=0$, $D^{2}u$ is negative semidefinite, and $cu\le 0$, so $Lu\le 0$. To reduce the non-strict case to this one, let $q(x)=e^{\kappa x_1}$. Boundedness of $b^1,c$ permits choosing $\kappa$ so large thatThus $L(u+\epsilon q)>0$. The preceding maximum argument and continuity on the compact closure giveLet $\epsilon \downarrow 0$; the exponential is bounded on the bounded domain. This is the exponential perturbation proof of the weak maximum principle with drift, with the zeroth-order term included. For $c=0$, any interior maximum, of either sign, is impossible for the perturbed function, giving the sharper boundary bound. The sign restriction on $c$ is essential: a positive first Dirichlet Laplacian eigenfunction violates the conclusion for $L=\Delta +{\lambda }_1$.

Assume an interior sphere condition at $y$, and assume the coefficients are bounded and the principal part defines a uniformly elliptic operator near its tangent Euclidean ball. Shrinking the Euclidean ball while keeping it tangent if necessary, arrange that $\overline{B_R(z)}\setminus \{y\}\subset \Omega$. Write $\nu =(y-z)/R$, the outward radial direction. The conclusion of the Hopf boundary point lemma in the regularity stated in the question isIf the one-sided normal derivative exists, this says ${\partial }_{\nu }u(y)>0$. For a $C^1$ boundary its outward normal agrees with the tangent Euclidean ball's radial normal. Continuity alone does not guarantee that the derivative exists, so its existence must be added when asserting a finite derivative.

On the annulus $R/2<r=|x-z|<R$, use the barrierThe operator has no zeroth-order term, and differentiation givesfor sufficiently large $k$: the quadratic-in-$k$ term is at least $k^2\lambda R^2$, while the remaining terms are bounded multiples of $k$.

On the inner sphere, compactness and strict inequality give a positive minimum of $u(y)-u(x)$. Choose $\epsilon >0$ small enough that $u-u(y)+\epsilon h\le 0$ there. The same inequality holds on the outer sphere since $h=0$. The weak maximum principle for elliptic operators yields $u-u(y)+\epsilon h\le 0$ throughout the annulus. Along the inward radius,This proves the lemma. The interior sphere condition supplies precisely the geometry needed for the barrier; no global boundary regularity is used.

The principal coefficients satisfy the positive uniformly elliptic operator condition if there is $\theta >0$, independent of $x$, such thatThe smooth coefficients on the compact closure also give a uniform upper bound $\sum a^{ij}{\xi }_i{\xi }_j\le \Theta |\xi {|}^2$. The first-order and zeroth-order coefficients do not enter uniform ellipticity.

Integration by parts gives the minimal surface equation for a graph in nondivergence form:Its eigenvalues are between $(1+|p{|}^2{)}^{-1}$ and one. On each relatively compact subdomain, $Du$ is bounded, so it is a uniformly elliptic operator. Since $u\in C^2$, its coefficient functions are locally Lipschitz. The interior Schauder estimate gives $C_{\mathrm{loc}}^{2,\beta }$ regularity for every $0<\beta <1$; differentiating and repeatedly applying regularity estimates gives $\boxed{u\in C^{\infty }(\Omega )}$.

Differentiate in direction $k$. Each $u_k$ satisfies $\mathcal{L}{u}_k=0$, withFor $v=|Du{|}^2$ this yieldsIf $u\in C^2(\overline{\Omega })$, the principal coefficients are uniformly elliptic globally and the drift is bounded. The weak maximum principle for elliptic operators bounds $v$ by its boundary maximum. Boundary values are limits of interior values by continuity, proving the gradient maximum principle for a minimal graph:

The cone condition implies $u(\lambda x)=\lambda u(x)$, so $Du(\lambda x)=Du(x)$. The gradient is homogeneous of degree zero, and $v=|Du{|}^2$ attains its global maximum on the compact unit sphere. For $n\ge 2$ the punctured space is connected. Part (b) gives $\mathcal{L}v\ge 0$ for a locally uniformly elliptic operator with no zeroth-order term. The strong maximum principle for elliptic operators makes $v$ constant. HencePositive definiteness gives $D^{2}u=0$. Thus $u=a\cdot x+b$, and degree-one homogeneity forces $b=0$: $\boxed{u(x)=a\cdot x}$. This proves that a minimal graphical cone is a plane. Strictly, the graph restricted to $x\ne 0$ is the plane with its origin omitted; its closure is the full plane through the origin.

Extend $u$ by zero to ${\mathbb{R}}^n$; its compact support inside the ball makes the extension smooth. The Fourier transform of a derivative givesThe assumption says that $L_0$ is a uniformly elliptic operator, soMoreover,The Plancherel theorem therefore yieldsAll integrands vanish outside the original support where appropriate, so

Uniform ellipticity means that there is $\lambda >0$ such that the symmetric part of the principal coefficient matrix satisfiesfor every $x\in \Omega$ and $\xi \in {\mathbb{R}}^n$. This is the defining coercive bound for a uniformly elliptic operator.

Differentiate the equation for $v_R$ with respect to $x_k$. The derivative $w=D_k{v}_R$ is a weak solution ofwhereThe eigenvalue in the direction of $p$ is $(1+|p{|}^2{)}^{-3/2}$ and every orthogonal eigenvalue is $(1+|p{|}^2{)}^{-1/2}$. Thus a uniform bound on $|Dv|$ makes this a uniformly elliptic operator with constants independent of $R$.

The operator is strictly elliptic when its symmetric principal matrix $A(x)=(a^{ij}(x))$ is positive definite at every point:It is a uniformly elliptic operator when some $\lambda >0$ satisfiesfor every $x$ and $\xi$. Here the least eigenvalue of the continuous matrix $A(x)$ is a positive continuous function on the compact set ${\overline{B}}_1$. It therefore has a positive minimum, which supplies $\lambda$ and proves uniform ellipticity.

Testing the weak formulation with the constant function $1$ proves the necessary compatibility conditionAssume first that $U$ is connected and this condition holds. On the mean-zero Sobolev spaceuse the norm $\Vert v{\Vert }_{†}=\Vert Dv{\Vert }_{L^2(U)}$. To prove the needed Poincare-Wirtinger inequality, suppose it failed. There would be $v_k\in H_{†}^1(U)$ with $\Vert v_k{\Vert }_2=1$ and $\Vert D{v}_k{\Vert }_2\to 0$. The Rellich-Kondrachov compactness theorem gives a subsequence converging strongly in $L^2$ and weakly in $H^1$ to a function $v$. Its weak derivative vanishes, so connectedness makes $v$ constant; its zero mean makes it zero. This contradicts $\Vert v{\Vert }_2=1$. Hence $\Vert Dv{\Vert }_2$ is an equivalent Hilbert space norm on $H_{†}^1(U)$.

DefineBoundedness of $a^{ij}$ makes $B$ a bounded bilinear form, while uniform ellipticity givesso it is a coercive bilinear form. The Cauchy-Schwarz inequality and the Poincare-Wirtinger inequality make $\ell$ a bounded linear functional. The Lax-Milgram theorem supplies a unique $u\in H_{†}^1(U)$ satisfying $B(u,v)=\ell (v)$ for every mean-zero $v$. For arbitrary $v\in H^1(U)$, subtract its mean; the omitted constant contributes zero on both sides because $D1=0$ and ${\int }_{U}f=0$. Thus $u$ solves the original problem.

If two solutions exist, their difference $w$ satisfies $B(w,w)=0$, so uniform ellipticity gives $Dw=0$. It is therefore constant on $U$. The solution is unique up to an additive constant, and its mean-zero representative is unique. If $U$ is disconnected, the precise condition is ${\int }_{U_j}f=0$ on every connected component $U_j$, and one independent additive constant remains on each component.

For a locally uniformly elliptic operator $L=a^{ij}{D}_{ij}+b^i{D}_i+c$ with locally bounded coefficients and $c\le 0$, a $C^2$ function satisfying $Lu\ge 0$ on a connected domain cannot attain a nonnegative global maximum inside unless it is constant. If $c=0$, the maximum can have either sign. If the maximum set were proper, choose an interior ball in its complement tangent to that set. The Hopf boundary point lemma gives a nonzero normal derivative at the contact point, contradicting the vanishing gradient at an interior maximum.

The strong minimum principle for elliptic operators follows by applying this to $-u$. For nonnegative $u$ with $Lu\le 0$, replacing $c$ by $\min (c,0)$ retains the inequality, so the strict positivity-or-zero conclusion does not require an original sign restriction on $c$.

For $Lu=-{\partial }_j(a^{ij}{\partial }_{i}u)+b^i{\partial }_{i}u+cu$ with smooth coefficients on a bounded smooth domain, the weak formulation of $(L+\lambda )u=f$ and conormal Robin boundary condition $a^{ij}{\partial }_{i}u{\nu }_j+\beta u=0$ isThe bilinear form is bounded by the Sobolev trace theorem. If $a^{ij}$ satisfy the positive uniformly elliptic operator inequality, the multiplicative trace inequality on a bounded smooth domain and the Young inequality show that the form is coercive when $\lambda$ is sufficiently large, even if $\beta$ and $c$ have negative parts. The Lax-Milgram theorem then gives a unique weak solution for every $f\in L^2(U)$.