False. In the grid graph $[9{]}^2$, take the four-row stripIt has $36=6^2$ vertices, and only the nine edges between rows four and five lie in its edge boundary. The corner square $[6{]}^2$ has six boundary edges along each of its two exposed sides, for a total of twelve. Thus a set of size $t^2$ can have smaller edge boundary than $[t{]}^2$.

For $\mathcal{A}$ in the $n$-dimensional Boolean hypercube, the edge-isoperimetric inequality in the discrete cube iswhere $e(\mathcal{A})$ is the number of edges spanned by $\mathcal{A}$. Since the cube is $n$-regular, the equivalent boundary form is

We prove the induced-edge form by induction on $n$. Split the cube according to its last coordinate, and let the two sections have sizes $a\ge b$. At most $b$ edges of $\mathcal{A}$ cross between the sections. The induction hypothesis givesPut $x=b/(a+b)\le 1/2$. The required comparison with $\frac{1}{2}(a+b){\log }_2(a+b)$ is equivalent towhere $H_2$ is the binary entropy function. This follows from concavity because the graph of $H_2$ lies above the chord joining $(0,0)$ to $(1/2,1)$. The induction is complete. Subcubes attain equality when $|\mathcal{A}|$ is a power of two.

Fix a prime number $p>r$. Distinct members of an intersecting $r$-uniform family have intersection size inwhereas every member has size $r\notin L$ modulo $p$. The Frankl-Wilson theorem with $|L|=r-1$ therefore givesFor fixed $r$,This is the asserted asymptotic weakening of the Erdős-Ko-Rado theorem.

No. Let $n=7$, $r=3$, take $\mathcal{A}=\{\{1,2,3\}\}$, and letThe families are nonempty and cross-intersecting. There aremembers, so

Suppose $\mathcal{A},\mathcal{B}\subseteq [n{]}^{(r)}$ are cross-intersecting families. The iterated upper shadow ${\nabla }_{n-r}\mathcal{A}$ is disjoint frombecause $A\subseteq [n]\setminus B$ would mean $A\cap B=\varnothing$. HenceIf $|\mathcal{A}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$, the upper-shadow form of the Kruskal-Katona theorem givesIt follows from Pascal's identity that $|\mathcal{B}|<\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$. Thus the two sizes cannot both exceed that number.

The Erdős-Ko-Rado theorem says that if $n\ge 2r$ and $\mathcal{A}\subseteq [n{]}^{(r)}$ is an intersecting family, thenThe star of all $r$-sets containing one fixed point attains equality.

For the shadow proof, let ${\nabla }_{n-r}\mathcal{A}$ be the iterated upper shadow at level $n-r$, and letThese two families are disjoint: if $A\subseteq [n]\setminus B$, then $A$ and $B$ are disjoint. The upper-shadow form of the Kruskal-Katona theorem says that ifthenBut $|{\mathcal{A}}^c|=|\mathcal{A}|$, so disjointness and Pascal's identity would give more thanmembers at level $n-r$, a contradiction.

For the Katona circle method, place $[n]$ in a cyclic order. At most $r$ of its cyclic intervals of length $r$ can belong to an intersecting family. Indeed, after fixing one selected interval, every selected interval starts at one of the $2r-1$ positions at cyclic distance below $r$ from its start; apart from the fixed interval, these positions form $r-1$ pairs whose corresponding intervals are disjoint. Double-count pairs consisting of $A\in \mathcal{A}$ and a cyclic order in which $A$ is consecutive. There are $(n-1)!$ cyclic orders, at most $r$ selected intervals in each, and each $A$ is consecutive in $r!(n-r)!$ cyclic orders. Thereforewhich rearranges to the required bound.

False. Take $n=6$, $r=3$, andThere are $1+3\cdot 3=10$ members. Its lower shadow consists of the three pairs inside $\{1,2,3\}$ and the nine pairs having one point there and one in $\{4,5,6\}$, so $|\partial \mathcal{A}|=12$. Its upper shadow consists of the twelve four-sets containing at least two points of $\{1,2,3\}$, so $|\nabla \mathcal{A}|=12$. The sum is $24$.

The initial lex segment of length ten is the star of all triples containing $1$; its lower and upper shadows have sizes $15$ and $10$. The initial colex segment consists of all triples in $[5]$; its lower and upper shadows have sizes $10$ and $15$. Both sums are $25$, so neither canonical segment minimizes the sum.

Let $c(A)=[n]\setminus A$ and reverse the order of the ground set by $j\mapsto n+1-j$. The resulting bijection sends lexicographic order on $[n{]}^{(r)}$ to colexicographic order on $[n{]}^{(n-r)}$. It also sends the iterated upper shadow of an $r$-uniform family at level $r+1$ to the lower shadow of the corresponding $(n-r)$-uniform family, up to the same harmless reversal of the ground set.

The Kruskal-Katona theorem says that this lower shadow is smallest for an initial colex segment. Undoing the complement and reversal therefore says that the upper shadow of a family of fixed size in $[n{]}^{(r)}$ is smallest for the initial lex segment.

Every positive integer $m$ has a unique binomial representationDefineThe Kruskal-Katona theorem says that every $r$-uniform family $\mathcal{A}$ of size $m$ hasand the initial segment of length $m$ in colexicographic order attains equality.

We prove the inequality by simultaneous mathematical induction on the ground-set size and on $r$. We use the elementary binomial-shadow arithmetic lemmaTo verify the lemma, greedily remove the largest term $\left(\genfrac{}{}{0ex}{}{q}{r}\right)$ from the binomial representation of $a+b$. If $a$ reaches that term first, apply the induction hypothesis to the remainders; otherwise transfer the excess to $b$. In the crossing case Pascal's identitygives exactly the two terms on the right. This induction also proves uniqueness of the greedy representation.

Choose the largest ground-set element $x$ and splitThe members of the lower shadow that avoid $x$ form $\partial {\mathcal{A}}_0\cup {\mathcal{A}}_1$, while those containing $x$ are the sets $B\cup \{x\}$ with $B\in \partial {\mathcal{A}}_1$. Hence, writing $a=|{\mathcal{A}}_0|$ and $b=|{\mathcal{A}}_1|$,This completes the induction.

Finally, the initial colex segment first contains all $r$-subsets of $[a_r]$, followed recursively by sets containing $a_r+1$ whose remaining elements form the appropriate initial segment at level $r-1$. Its shadow has the same recursive decomposition, and therefore has exactly $f_r(m)$ members. This proves sharpness.

For every ${\Omega }^{\prime }⋐\Omega$, the interior Schauder estimate isThe constant depends on $n$, $\alpha$, the ellipticity constants, the coefficient $C^{0,\alpha }$ norms, $\Omega$, ${\Omega }^{\prime }$, and in particular the distance from ${\Omega }^{\prime }$ to the boundary. It is independent of $u$ and $f$.

The Hölder interpolation inequality says that for $u\in C^{l,\alpha }(B_R(x_0))$ and every $\epsilon >0$,More generally, each lower derivative norm can be bounded by an arbitrarily small multiple of the top $C^{l,\alpha }$ seminorm plus a constant multiple of the $C^0$ norm. The powers of $R$ make the inequality invariant under scaling of the ball.

The Leray-Schauder fixed point theorem says that a continuous compact map $T$ on a Banach space has a fixed point if the homotopy setis bounded. If $u=tT(u)$ with $t>0$, multiplying the equation for $T(u)$ by $t$ shows that $u$ solvesThe case $t=0$ gives $u=0$. Therefore it is enough to prove one uniform $C^{1,\beta }$ estimate for all solutions of this family and all $t\in [0,1]$. The theorem then yields a fixed point $u=T(u)$, which solves the original quasilinear problem.

Initially the construction gives $u\in C^{2,\alpha \beta }$. This makes $u$ and $Du$ Lipschitz, so composing the original $C^{0,\alpha }$ coefficients with the jet of $u$ produces $C^{0,\alpha }$ coefficients. A second application of the global Schauder estimate gives $u\in C^{2,\alpha }(\overline{\Omega })$. The fixed-point theorem is an existence result and supplies no uniqueness.

Suppose $v_k\to v$ in $C^{1,\beta }$ and write $u_k=T(v_k)$. The preceding uniform Schauder estimate bounds $(u_k)$ in $C^{2,\alpha \beta }$. Every subsequence therefore has a further subsequence converging in $C^{1,\beta }$ by the compact embedding of Hölder spaces. The composed coefficients converge uniformly, and lower-exponent Hölder compactness provides enough convergence of the second derivatives to pass to the linear equation. Every subsequential limit solves the problem defining $T(v)$.

Uniqueness of that linear Dirichlet problem forces every such limit to equal $T(v)$. Since every subsequence has a further subsequence with this same limit, the whole sequence converges to $T(v)$ in $C^{1,\beta }$. Thus $T$ is continuous.

Let $v$ range over a bounded subset of $C^{1,\beta }$. All jets $(x,v,Dv)$ then lie in one compact set, so the composed coefficients have uniform $C^{0,\alpha \beta }$ bounds and one ellipticity constant. The weak maximum principle for elliptic operators bounds $T(v)$ in $C^0$ by the boundary data and the bounded forcing term. The global Schauder estimate consequently bounds $T(v)$ in $C^{2,\alpha \beta }$.

The compact embedding of Hölder spacesthen makes the image relatively compact. Hence $T$ is a compact operator on $C^{1,\beta }(\overline{\Omega })$.

Fix $v\in C^{1,\beta }(\overline{\Omega })$. Its jet map $x\mapsto (x,v(x),Dv(x))$ is $C^{0,\beta }$, while the barred coefficients are $C^{0,\alpha }$ in all their variables. Composition therefore makes$C^{0,\alpha \beta }$. The image of the jet map is compact, so strict ellipticity on the coefficient domain has a positive uniform lower bound on this image. The assumed linear Dirichlet problem theory now gives a uniquesolving ${\bar{a}}^{ij}(x,v,Dv)D_{ij}u+\bar{b}(x,v,Dv)=0$ with boundary value $\phi$. Thus $T:C^{1,\beta }\to C^{1,\beta }$ is well defined.

The global Schauder estimate isThe constant depends on the dimension, $\alpha$, the domain and its boundary regularity, the ellipticity constants, and the $C^{0,\alpha }$ norms of the coefficients, but not on $u$, $f$, or $\phi$.

Forthe weak maximum principle for elliptic operators states that $Lu\ge 0$ impliesIn particular, a solution of $Lu=0$ cannot have a positive interior maximum exceeding its boundary maximum.

Because the coefficient matrix is positive definite and the closure of the smooth bounded domain is compact, strict ellipticity supplies a uniform lower bound after restricting to $\overline{\Omega }$. Rotate and translate coordinates so that $\Omega$ is bounded in the $x_1$ direction, and set $h=e^{\gamma x_1}$. For sufficiently large $\gamma$,If $u+\epsilon h$ had a positive interior maximum, its gradient would vanish and its Hessian matrix would be negative semidefinite there, giving $L(u+\epsilon h)\le 0$. This contradicts $L(u+\epsilon h)=\epsilon Lh>0$. Comparing on the boundary and sending $\epsilon \downarrow 0$ proves the assertion.

Seek a radial function $u=u(r)$ on the unit ball. The equation and regularity at the origin giveHence $u^{\prime }(r)=(r/n{)}^{1/(p-1)}$, and the zero Dirichlet boundary condition givesThe power $|x{|}^q$ is twice differentiable at the origin exactly when $q\ge 2$. Here $q=p/(p-1)$, so $u\in C^2$ at the origin exactly when $1<p\le 2$. Under the assumption $2<p<n$, this weak solution is never $C^2$ at the origin, exhibiting the regularity loss caused by degenerate ellipticity.