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For a $C^2$ solution $u$ of the minimal surface equation for a graph on the closure of a bounded domain,Write $a^{ij}(Du)u_{ij}=0$ with $a^{ij}(p)={\delta }_{ij}-p_i{p}_j/(1+|p{|}^2)$. Its differentiated operator $\mathcal{L}=a^{ij}{D}_{ij}+b^{\ell }{D}_{\ell }$, where $b^{\ell }=(\partial a^{ij}/\partial p_{\ell })u_{ij}$, satisfiesThe weak maximum principle for elliptic operators gives the result. Interior Schauder estimates justify differentiating a solution initially assumed only $C^2$.

For a harmonic function $w$ smooth inside a bounded domain and $C^1$ on its closure, $\Delta |Dw{|}^2=2|D^{2}w{|}^2\ge 0$. The weak maximum principle for elliptic operators and continuity of the gradient imply ${\sup }_{\Omega }|Dw|={\sup }_{\partial \Omega }|Dw|$.

Let $h:\mathbb{R}\to \mathbb{R}$ be continuous, odd, strictly increasing and onto. Suppose $u\in C^2({\mathbb{R}}^d)$ tends to zero at infinity and satisfies $-\Delta u+h(u)=f$, with bounded $f$. At a positive maximum $m$ of $u$, $-\Delta u\ge 0$, so $h(m)\le f\le \Vert f{\Vert }_{\infty }$. Apply the same argument to $-u$ to obtain the displayed bound, using the maximum-point argument of the weak maximum principle for elliptic operators. For $h(s)=s+\sin s$, strict monotonicity follows by integrating $h^{\prime }=1+\cos s$, whose zeros are isolated. Thus $\Vert u{\Vert }_{\infty }\le h^{-1}(\Vert f{\Vert }_{\infty })\le \Vert f{\Vert }_{\infty }+1$. The stronger bound by $\Vert f{\Vert }_{\infty }$ holds when that norm is at most $\pi$, but need not hold for larger sources.

For smooth non-increasing $V$ and an ordered subsolution and supersolution, begin with $u_0={\phi }^{-}$ and solve successive Poisson equations with forcing $V(u_k)$ and the fixed boundary data. The weak maximum principle for elliptic operators applied to the negatives of consecutive differences givesThe global Schauder estimate and the Hölder interpolation inequality bound the iterates uniformly in $C^{2,\alpha }$. Their bounded monotone sequence converges, and compact embedding of Hölder spaces permits passing to the equation. The uniform Hölder seminorm bounds retain $C^{2,\alpha }$ regularity of the limit.

With the convention $Lu\ge 0$ for a classical subsolution, the weak maximum principle for elliptic operators isHere $L$ is uniformly elliptic with bounded drift, has no zeroth-order term, and the domain is bounded. Equivalently, $Lu\le 0$ gives ${\inf }_{\Omega }u\ge {\min }_{\partial \Omega }u$.

One can see why measurable coefficients suffice for the pointwise classical statement by adding $\epsilon e^{\kappa x_1}$. Choose $\kappa$ large enough that $\lambda \kappa >\Vert b_1{\Vert }_{\infty }$; then $L(e^{\kappa x_1})=e^{\kappa x_1}({\kappa }^2{a}_{11}+\kappa b_1)>0$. The perturbed subsolution cannot attain a maximum inside: its gradient would be zero and its Hessian negative semidefinite, forcing its elliptic second-order term to be nonpositive. Its maximum is therefore on the boundary. Letting $\epsilon \downarrow 0$ proves the stated estimate. Only the symmetric part of $a$ acts on the Hessian, and ellipticity gives its positive definiteness.

Put $F={\sup }_{\Omega }{f}^{+}\ge 0$ and $m={\inf }_{\partial \Omega }u$, both finite by continuity on the compact closure. Set $z=Fv+m-u$. Constants are annihilated by $L$, soOn the boundary, $z\le m-u\le 0$, using $v\le 0$ and $F\ge 0$. The weak maximum principle for elliptic operators therefore gives $z\le 0$ inside as well. Rearrangement provesThe nonnegative multiplier is essential when multiplying the barrier inequality.

The weak maximum principle for elliptic operators gives the clipped boundary boundsIf $c=0$, constants solve the equation, and applying these bounds after subtracting the boundary maximum or minimum gives the sharper conclusion that both extrema occur on the boundary. For $c<0$, the clipping by zero is necessary in the general statement.

To prove the upper bound, write $B=\Vert b_1{\Vert }_{\infty }$ and $C_0=\Vert c{\Vert }_{\infty }$, and choose $\alpha >0$ large enough that $\lambda {\alpha }^2-B\alpha -C_0>0$. For $\varphi (x)=e^{\alpha x_1}$, uniform ellipticity givesPut $u_{\epsilon }=u+\epsilon \varphi$. Then $L{u}_{\epsilon }>0$. If $u_{\epsilon }$ had a positive maximum at an interior point, its gradient would vanish there and its Hessian matrix would be negative semidefinite. Only the symmetric part of $a_{ij}$ contracts with this Hessian, and its positive definiteness implies $a_{ij}{D}_{ij}{u}_{\epsilon }\le 0$. Also $c{u}_{\epsilon }\le 0$. These facts would give $L{u}_{\epsilon }\le 0$, a contradiction.

Since $\overline{\Omega }$ is compact and $u_{\epsilon }$ is continuous on it, its maximum exists. It is therefore either nonpositive or attained on the boundary, proving ${\sup }_{\Omega }{u}_{\epsilon }\le \max \{0,{\sup }_{\partial \Omega }{u}_{\epsilon }\}$. The exponential is bounded on the bounded domain; letting $\epsilon \downarrow 0$ proves the upper estimate. Apply the same argument to $-u$ for the lower estimate. This is the exponential perturbation proof of the weak maximum principle with drift and requires no continuity of the coefficient functions.

For $\eta >0$, use the weak maximum principle for elliptic operators perturbation ${\varphi }_{\eta }(x)=\varphi (x)+\eta |x{|}^2$. Its Laplacian is at least $4\eta$, so the preceding argument puts its maximum on the boundary. If $R^2={\max }_{\bar{\Omega }}|x{|}^2$, then, for any $x\in \bar{\Omega }$,Letting $\eta \downarrow 0$ gives $\varphi (x)\le {\max }_{\partial \Omega }\varphi$. The boundary is compact, so this maximum is attained. This proves the weak weak maximum principle for elliptic operators even when the Laplacian may vanish.

Apply the weak weak maximum principle for elliptic operators to the harmonic function $\varphi$ and to $-\varphi$. Their boundary maxima are both zero, giving $\varphi \le 0$ and $\varphi \ge 0$ on the closure. Consequently $\varphi \equiv 0$.

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.

For the corrected bounded-domain problem, the interior theorem for the Perron method for the Dirichlet problem gives a classical solution of the Poisson equation, and the power barrier for an exterior cone argument gives its continuous extension with the required Dirichlet boundary condition. Hence existence holds in $C^2(\Omega )\cap C^0(\overline{\Omega })$.

If $u_1,u_2$ are two such classical solutions, then $w=u_1-u_2$ is a harmonic function with zero boundary trace. The weak maximum principle for elliptic operators, applied to $w$ and $-w$ on the bounded domain, gives $w\le 0$ and $-w\le 0$. ConsequentlyThere is exactly one solution after adding boundedness of $\Omega$. For the printed unbounded-domain version, the half-plane counterexample in part (i) disproves uniqueness; an additional condition at infinity or a suitable restricted solution class would instead have to be specified.

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 $L=\Delta +{\lambda }_1$.

Apply the weak maximum principle for elliptic operators to $q=|Dw{|}^2$. The product rule and the equation for the harmonic function giveThe function $q$ is twice differentiable inside and continuous on the closure, by the prescribed regularity. Hence ${\sup }_{\Omega }q\le {\sup }_{\partial \Omega }q$. Conversely, each boundary value of $q$ is a limit of values at interior points, so the reverse inequality holds for the suprema. Taking square roots proves the gradient maximum principle for harmonic functions:This argument controls the length of the entire gradient, without comparing its separate components.

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 global Schauder estimate on a bounded $C^{2,\alpha }$ domain isHere $C$ depends on $n,\alpha$, the domain's boundary regularity and $\Vert c{\Vert }_{C^{0,\alpha }}$, but not on the particular solution or data. The $C^0$ term is needed before uniqueness has been established. The given $\psi$ is an extension of the boundary data to the closure.

For $c\le 0$, choose $d$ with $\Omega \subset \{|x_1|<d\}$ and set $M={\sup }_{\partial \Omega }|\psi |$, $F={\sup }_{\Omega }|f|$. The hinted functionis nonnegative, dominates $|\psi |$ on the boundary, and satisfiessince $x_1+d>0$. Therefore $(\Delta +c)(u-b)=f-(\Delta +c)b\ge 0$ and $(\Delta +c)(-u-b)=-f-(\Delta +c)b\ge 0$. Both functions are nonpositive on the boundary. The weak maximum principle for elliptic operators gives $|u|\le b$, hence the supremum norm barrier for an elliptic Dirichlet problem:Unlike the full Schauder estimate, this last constant depends only on a slab containing $\Omega$, and is independent of how negative $c$ is.

Assume uniform ellipticity $a^{ij}{\xi }_i{\xi }_j\ge \lambda |\xi {|}^2$ with $\lambda >0$, bounded coefficients, and $c\le 0$. The weak maximum principle for elliptic operators isWhen $c=0$, zero can be omitted. The sign restriction is necessary, as shown by the failure of the weak maximum principle with a positive zeroth-order coefficient.

Let $M=\max (0,{\sup }_{\partial \Omega }u)$ and $w=u-M$, so $Lw=Lu-cM\ge 0$ and $w\le 0$ on the boundary. For sufficiently large $k$, $q=e^{k{x}_1}$ satisfiesIf $w$ is positive somewhere, $w+\epsilon q$ retains a positive interior maximum above all its boundary values for small $\epsilon >0$. At that point its gradient is zero and its Hessian matrix is negative semidefinite; ellipticity and $c\le 0$ give $L(w+\epsilon q)\le 0$, contradicting its strict positivity. This exponential perturbation proof of the weak maximum principle with drift also works with $M={\sup }_{\partial \Omega }u$ of either sign when $c=0$.

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:

As printed, the hypotheses are insufficient. They order the barriers only on the boundary; the intended monotone iteration for a semilinear elliptic equation needs an ordered subsolution and supersolution throughout the domain. On $B_1$, take a positive first Dirichlet Laplacian eigenfunction $e_1$ with $-\Delta e_1={\lambda }_1{e}_1$, and setBoth barriers satisfy $Q{\phi }^{\pm }=0$ and vanish on the boundary, but ${\phi }^{-}>{\phi }^{+}$ inside. No function lies between them. This also disproves the unconditional conclusions in (b) and (d).

With the intended extra hypothesis ${\phi }^{-}\le {\phi }^{+}$ on $\overline{\Omega }$, solvability of the Poisson equation and the Schauder estimate give a unique $u_1\in C^{2,\alpha }$ with forcing $V({\phi }^{-})$ and boundary data $\psi$. Smooth $V$ composed with the $C^{0,\alpha }$ barrier has the required forcing regularity. The barrier inequalities giveBoth differences have nonnegative boundary values. The weak maximum principle for elliptic operators applied to their negatives gives $\boxed{{\phi }^{-}\le u_1\le {\phi }^{+}}$ under the corrected hypothesis.

Use the interior ordered subsolution and supersolution correction from part (a). Recursively solve the Poisson equation $\Delta u_{k+1}=V(u_k)$ with boundary value $\psi$, starting from $u_0={\phi }^{-}$. Part (a) gives $u_0\le u_1\le {\phi }^{+}$. If $u_{k-1}\le u_k\le {\phi }^{+}$, thenThe first difference has zero boundary values, the second nonnegative values. The weak maximum principle for elliptic operators applied to their negatives gives $u_k\le u_{k+1}\le {\phi }^{+}$. Induction proves the bounded monotone sequenceWithout interior ordering, the counterexample in part (a) disproves the claim.