A k-Sperner family meets every maximal chain in a Boolean lattice in at most $k$ members. The random-chain calculation in part (a) therefore gives the generalized LYM inequality $L(\mathcal{A})\le k$. Define the proportion of rank $i$ that is present byThen ${\sum }_i{x}_i=L(\mathcal{A})\le k$ and $w(\mathcal{A})={\sum }_{i}u(i)x_i$. This reduces the weighted theorem for k-Sperner families to allocating at most $k$ units of mass among ranks, each of capacity one.

To handle ties explicitly, let $t>0$ be the $k$th largest value among the $u(i)$, let $H=\{i:u(i)>t\}$, and put $h=|H|<k$. ThenThe last expression is exactly the sum of the $k$ largest rank weights, with multiplicity. Writing those values in descending order as $u_{(1)},\dots ,u_{(n+1)}$, the answer is

Choose a Uniformly random maximal chain in a Boolean lattice by taking a uniformly random permutation of $X$ and successively adjoining its elements. A fixed $r$-element set occurs on this maximal chain in a Boolean lattice with probability $r!(n-r)!/n!={\left(\genfrac{}{}{0ex}{}{n}{r}\right)}^{-1}$. Because an antichain meets every such chain in a partially ordered set at most once, linearity of expectation gives the LYM inequality:Here $L(\mathcal{A})$ is the Lubell mass.

For equality in the LYM inequality, the integer $|\mathcal{A}\cap \mathcal{C}|\le 1$ must equal one for every maximal chain in a Boolean lattice, since every chain has positive probability. Choose $A\in \mathcal{A}$ of rank $r$. If $0<r<n$, take $x\in A$ and $y\notin A$, and choose a permutation with its first $r$ elements equal to $A$, its $r$th element $x$, and its next element $y$. Interchanging those two adjacent elements changes precisely the rank-$r$ member of the corresponding maximal chain in a Boolean lattice, replacing $A$ by $A^{\prime }=A\setminus \{x\}\cup \{y\}$. All other members are unchanged and none belongs to the antichain. The new chain must still meet $\mathcal{A}$, so $A^{\prime }\in \mathcal{A}$.

Successive exchanges connect all $r$-element subsets: exchange an element outside a desired target for an element missing from it, decreasing the number of differences each time. Thus $\mathcal{A}$ contains the whole rank-$r$ level. Every set of another rank is comparable with some set in that level, so the antichain contains nothing else. If $r=0$ or $r=n$, its sole possible member is already the whole corresponding level. Conversely, every full level meets every maximal chain in a Boolean lattice exactly once. Hence

With the interior ordering correction in part (a), iteration of part (c) givesThe bounded monotone sequence has a pointwise limit $u$. By compact embedding of Hölder spaces, a subsequence converges in $C^{2,\beta }$ for $0<\beta <\alpha$, necessarily to this same limit. All convergent subsequences have that limit, so compactness gives convergence of the full sequence in $C^{2,\beta }$. Passing to the limit in $\Delta u_k=V(u_{k-1})$ gives the equation and boundary values. The uniform Hölder seminorm bounds on second derivatives pass to the limit, so $u\in C^{2,\alpha }(\overline{\Omega })$, and the barriers remain ordered:This proves the intended monotone iteration for a semilinear elliptic equation. With only the boundary ordering printed in the PDF, the counterexample in part (a) shows the requested trapped solution need not exist.

For a sequence trapped between the fixed barriers, let $M$ bound their absolute values and put $A={\sup }_{|s|\le M}|V^{\prime }(s)|$. The mean value theorem givesThe global Schauder estimate, including the fixed boundary data $\psi$, and $\Vert u_k{\Vert }_{\infty }\le M$ implyApply the supplied Hölder interpolation inequality with $KA\epsilon \le 1/2$. Its remaining supremum-norm term is bounded by $M$, yieldingIf $A=0$, the stronger bound without a previous-iterate term holds directly. The constant depends on the fixed domain, exponent, barriers, boundary data and $V$, independently of $k$.

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.

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.

Let $u=v+\psi$, so $v$ has zero Dirichlet boundary condition. Using the supplied form of the minimal surface equation for a graph, setFor $\Vert \psi {\Vert }_{C^{2,\alpha }}\le \beta \le 1$ and $\Vert v\Vert ,\Vert w\Vert \le 1$, the difference bound impliesandThe shifted arguments stay in the permitted radius-two ball, and $\Vert f_{\psi }{\Vert }_{C^{0,\alpha }}\le C{\beta }^2+C_n\beta$. The Laplacian with zero Dirichlet boundary condition has trivial null space by the maximum principle for harmonic functions. Choose $\beta >0$ small enough that $2C\beta \le \delta$ and $C{\beta }^2+C_n\beta \le {\epsilon }_0$, for the constants in part (c) with $L=\Delta$. Its small-data existence for a nonlinear elliptic Dirichlet problem supplies $v$, and $\boxed{u=v+\psi }$ solves the minimal surface equation for a graph with the required boundary values.

The Fredholm alternative for an elliptic Dirichlet problem and part (b) give an inverse $S:Y\to X$ of norm at most $K\ge 1$. ChooseOn the closed radius-$\epsilon$ ball $\mathcal{B}\subset X$, set $T(v)=S(\mathcal{Q}(v)+f)$. The stated inequalities giveandThus $T$ is a contraction mapping of the complete metric space $\mathcal{B}$ into itself. The Banach fixed-point theorem gives $u=T(u)$, solving the required nonlinear elliptic boundary value problem. This small-data existence for a nonlinear elliptic Dirichlet problem also gives uniqueness within this small ball; it does not assert global uniqueness.

Failure of the bound would give normalized $v_j$ with $\Vert v_j{\Vert }_{\infty }=1$, zero Dirichlet boundary condition, and $\Vert L{v}_j{\Vert }_{C^{0,\alpha }}\to 0$. The global Schauder estimate bounds them in $C^{2,\alpha }$. By compact embedding of Hölder spaces, a subsequence converges in $C^{2,\beta }$ to $v$ with $Lv=0$, zero boundary data and $\Vert v{\Vert }_{\infty }=1$. The uniform Hölder seminorm bounds pass to the limit, so $v\in C^{2,\alpha }$. This contradicts the trivial null space, provingThe global Schauder estimate then gives $\Vert u{\Vert }_{C^{2,\alpha }}\le K\Vert f{\Vert }_{C^{0,\alpha }}$, the bounded-inverse estimate used next.

Interpret strict ellipticity as uniform ellipticity up to the boundary, as required for the global elliptic boundary value problem. Mere pointwise positive definiteness in the interior is insufficient: $L=x^2{D}^2$ on $(0,1)$ has trivial homogeneous Dirichlet kernel but $Lu=1$ has no $C^2$ solution up to $x=0$.

Set $X=\{u\in C^{2,\alpha }(\overline{\Omega }):u{|}_{\partial \Omega }=0\}$ and $Y=C^{0,\alpha }(\overline{\Omega })$. The Fredholm alternative for an elliptic Dirichlet problem says $L:X\to Y$ is a Fredholm operator of index zero. Its null space $N$ is finite dimensional, its range is closed, and the range codimension equals $\dim N$. Hence $\boxed{N=\{0\}⟺L:X\to Y\text{ is invertible}}$; in this case every forcing has a unique solution.

Otherwise there are nonzero homogeneous solutions and finitely many compatibility conditions: $Lu=f$ is solvable exactly when $\ell (f)=0$ for every continuous linear functional on $Y$ vanishing on $L(X)$. When solvable, its solutions are $u_0+N$. This Fredholm solvability condition can be expressed with formal-adjoint null solutions for smoother coefficients. The functional formulation avoids differentiating the merely $C^{0,\alpha }$ coefficients.

Take ${\Omega }^{\prime }$ nonempty. If the bound failed, normalization would give nonnegative $v_j$ with supremum one, zero boundary values, ${\inf }_{{\Omega }^{\prime }}{v}_j\to 0$, and $\Vert L{v}_j{\Vert }_{C^{0,\alpha }}\to 0$. The global Schauder estimate bounds them in the Hölder space $C^{2,\alpha }$. By compact embedding of Hölder spaces, a subsequence converges in $C^{2,\beta }$, $0<\beta <\alpha$, to $v\ge 0$ with $Lv=0$, zero boundary data and supremum one.

Points approaching the infima have a subsequence converging in $\overline{{\Omega }^{\prime }}\subset \Omega$. Uniform convergence gives an interior zero of $v$, and part (c), the strong minimum principle for elliptic operators after replacing $c$ by $\min (c,0)$, gives $v=0$. This contradicts its supremum one, proving

Nonnegativity lets us dispense with a sign restriction on $c$. Set $c_0=\min (c,0)$ and $L_0=a^{ij}{D}_{ij}+b^i{D}_i+c_0$. ThenThe strong minimum principle for elliptic operators forces $u$ to vanish identically if it has an interior zero. Otherwise it is strictly positive:Connectedness is part of the usual domain hypothesis; on a disconnected open set this conclusion applies separately to each component.

For zero Dirichlet boundary condition, the global Schauder estimate isThe constant depends on $n,\alpha$, domain regularity, the ellipticity lower bound, and coefficient $C^{0,\alpha }$ norms. No sign restriction on $c$ is needed. Nonzero boundary data contribute the $C^{2,\alpha }$ norm of an extension.

Assume $c\le 0$, bounded coefficients and uniform ellipticity near the boundary point, an interior tangent ball there, and enough regularity for the boundary normal derivative. The Hopf boundary point lemma says that a nonnegative maximum at $x_0$ with strictly smaller values inside the tangent ball satisfiesfor the outward unit normal. If $c=0$, the maximum can have either sign. Without existence of a classical derivative, one states a strict one-sided derivative bound.

The PDF also asks to deduce the strong maximum principle for elliptic operators; that request is missing from the TeX transcription. Suppose $v$ attains a nonnegative global maximum $M$ inside a connected domain but is not constant. A ball in $\{v<M\}$ can be chosen relatively compact in the domain and tangent to $\{v=M\}$ at an interior point $z$: use the distance from a nearby point to the closed maximum set. The Hopf boundary point lemma on that ball gives a nonzero outward derivative at $z$, contradicting $Dv(z)=0$ at an interior maximum. Therefore $v$ is constant. When $c=0$, the same proof works for a maximum of either sign.

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.

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:

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 $c=0$.

The conclusion holds in dimension two. Shift and change sign to obtain a nonnegative harmonic function $v$. Its circular mean satisfiesHere the angular second derivative in the polar-coordinate Laplacian integrates to zero. Thus $m(r)=a\log r+b$; nonnegativity for every $r>0$ forces $a=0$. For $|x|=r$, the mean value property for harmonic functions and nonnegativity giveThe removable singularity for a bounded harmonic function fills in the origin, and the harmonic Liouville theorem makes the extension constant.

The conclusion fails for $n\ge 3$: $\boxed{u(x)=|x{|}^{2-n}}$ is positive and nonconstant, and its radial Laplacian $u^{\prime \prime }+(n-1)u^{\prime }/r$ vanishes away from zero. For $n=1$, $\boxed{u(x)=|x|}$ is a nonconstant function bounded below whose second derivative vanishes on both components of the punctured line.

Shift $u$ by a constant and, if necessary, change sign to obtain an entire nonnegative harmonic function $v$. The Harnack inequality for harmonic functions on arbitrarily large balls centred at $x_0$ gives $v(x)\le 3^{n}v(x_0)$ for every $x$. Thus $v$ is globally bounded. The interior derivative estimate for a harmonic function now gives $|Dv(x)|\le C\Vert v{\Vert }_{\infty }/R$ on balls of arbitrarily large radius $R$. Letting $R\to \infty$ proves $Dv=0$, hence $u$ is constant. This proves the one-sided harmonic Liouville theorem in every dimension.

For $x,z\in B_{\rho }(y)$, the open balls satisfy $B_{\rho }(x)\subset B_{3\rho }(z)\subset B_{4\rho }(y)$. Nonnegativity and the mean value property for harmonic functions giveTaking the supremum over $x$ and the infimum over $z$ proves the Harnack inequality for harmonic functions:This includes a zero infimum, which forces $u=0$ on the smaller ball.