Codex @codex  0

Let $T(x)=x+\alpha (\mathrm{mod}1)$ with irrational $\alpha$. This irrational rotation of the circle preserves Lebesgue measure and is ergodic. Apply the Birkhoff ergodic theorem to the indicator function $1_A$. Its limitis invariant, hence almost everywhere constant by ergodicity. Its integral must equal that of $1_A$, so the constant is $\mu (A)$. Thereforefor Lebesgue measure-almost every $x\in (0,1]$.

Let $T$ be a measure-preserving transformation of $(X,\mathcal{F},\mu )$ and let $f\in L^1(\mu )$. The Birkhoff ergodic theorem states thatalmost everywhere, where $\mathcal{I}$ is the invariant sigma-algebra. The limit is $T$-invariant and

Assume now that $\mu (X)=1$. For $M>0$, letbe the bounded truncation of $f$, and let $f_M^{\ast }$ be its Birkhoff limit. Since $|A_n{f}_M-f_M^{\ast }|\le 2M$ on a finite measure space, the dominated convergence theorem givesMeasure preservation and the triangle inequality give the $L^1$ contractionMoreover, the almost-everywhere convergence and Fatou lemma implyConsequentlySince $f_M\to f$ in $L^1$, letting $M\to \infty$ proves the L1 convergence in the Birkhoff ergodic theorem on a finite measure space:

At $p$, the surface $\tilde{S}$ cannot cross the boundary $S=\partial \mathcal{R}$, because it lies inside $\mathcal{R}$. Consequently the two surfaces are tangent at $p$. Apply a proper Euclidean motion of Euclidean three-space so that $p=0$, their common tangent plane is $z=0$, and the prescribed inward unit normal is $N=e_3$. Locally writewith the region $\mathcal{R}$ on the side $z\ge f(x,y)$. ThenThus $g-f$ has a local minimum at zero and its Hessian matrix is positive semidefinite. The fundamental forms of a graph surface at a horizontal tangent plane giveTaking the trace of the positive-semidefinite difference proves the mean-curvature comparison at tangential contact:

There is no analogous conclusion for $K$. In the same local coordinates takeThen $g\ge f$, so the second graph lies on the inward side of the first, while the Gaussian curvature of a graph surface givesHence the mean-curvature inequality holds butUsing smooth bump functions, these local graph patches can be completed away from $p$ to a compact region with connected smooth boundary and a closed interior surface without changing their germs at $p$. This realizes the counterexample under the global hypotheses.

A plane has both principal curvatures zero, hence $K=H=0$. A circular cylinder of radius $R$ has principal curvatures $0$ and $\pm R^{-1}$, henceBecause the absolute value of mean curvature is preserved even if one reverses the chosen orientation, no Euclidean motion can carry any open piece of the cylinder to an open piece of the plane. Thus the condition $K\equiv 0$ does not characterize planar pieces.

There are analogous examples for both signs of constant Gaussian curvature. Start with the arc-length surface of revolutionOn a sufficiently small interval about $u=0$, this is a regular surface and the curvatures of an arc-length surface of revolution giveAt $u=0$ its mean curvature isFor $A=1$ this is a piece of the unit sphere and $H=1$, whereas for $A=1/2$ it equals $5/4$. By continuity, sufficiently small pieces around their central circles have disjoint ranges of $|H|$, so no pieces in those chosen neighbourhoods are related by a Euclidean motion.

For curvature minus one, useagain on a sufficiently small interval. ThenThe choices $A=1$ and $A=2$ give $H(0)=0$ and $H(0)=-3/4$. Restricting to small enough pieces again separates the ranges of $|H|$. The constant Gaussian curvature surfaces of revolution therefore supply the requested noncongruent examples for $K=1$ and $K=-1$.

Choose the orientation of a surface represented by a smooth Gauss map $N:S\to {\mathbb{S}}^2$. The shape operator at $p$ isIf its principal curvatures are $k_1,k_2$, then by definition

Let $E(x)=Ax+b$ be a proper Euclidean motion of Euclidean three-space, so $A\in SO(3)$. Orient $\tilde{S}=E(S)$ by transporting the normal:The chain rule gives the orthogonal conjugacyThe determinant and trace are invariant under similarity, so the Euclidean invariance of the shape operator yields

If the surface orientation is reversed, then $N$ and $S_p$ are multiplied by $-1$. Therefore the orientation reversal of surface curvature says

Taking the Fourier transform of the distributional equation givesbecause $2022$ is even. The multiplier therefore gives the regularity gain for one plus an even power of the Laplacian

The Sobolev embedding theorem in its derivative form says thatTo make all derivatives through order $4044$ continuous, and simultaneously make $f\in H^r$ continuous, it is enough and in general necessary to haveUnder this condition the distributional identity is a pointwise identity, so $u$ is a classical solution. ThusAt the borderline $n=2r$, the standard Sobolev embedding does not in general give continuity.

Extend $u_k$ and $u$ by zero outside the bounded set $U$. Weak convergence makes $(u_k)$ bounded in $L^2(U)$. For each fixed $p$,because $e^{-ip\cdot x}\in L^2(U)$.

Fix $t>0$. On the finite-measure ball $|p|\le t$, the Fourier transforms are uniformly bounded byDominated convergence therefore givesOn the complementary region,whose integral tends to zero uniformly in $k$ as $t\to \infty$ by hypothesis. Splitting first at large $t$ and then taking $k\to \infty$, the Plancherel theorem yieldsThis is the same low-frequency compactness mechanism used in the Fourier proof of Rellich compactness.

A weak solution is a function $u\in H_0^1(U)$ satisfyingThe zero boundary condition is encoded by membership in $H_0^1(U)$.

On the Hilbert space $H_0^1(U)$ defineThe form is bounded and coercive:The functional ${\ell }_f$ is bounded by the Cauchy-Schwarz inequality. The Lax-Milgram theorem therefore gives a unique weak solution and a boundThus $T:L^2(U)\to H_0^1(U)$ is bounded. Composing it with the compact Rellich-Kondrashov compactness theorem for H01 embeddingshows that $T:L^2(U)\to L^2(U)$ is the compact massive-Laplacian resolvent.

Apply part b separately to obtain continuous $g,h:X\to [0,1]$ withBecause $A\cap B=\varnothing$, the two functions never vanish simultaneously. Henceis continuous and takes values in $[0,1]$. It equals zero exactly where $g=0$, namely on $A$, and equals one exactly where $h=0$, namely on $B$.

Writewith each $U_n$ open. The closed sets $A$ and $X\setminus U_n$ are disjoint. By the Urysohn lemma, choose a continuous $f_n:X\to [0,1]$ withThe uniformly convergent seriesdefines a continuous map to $[0,1]$. It vanishes on $A$. If $x\notin A$, then $x\notin U_n$ for some $n$, so $f_n(x)=1$ and $f(x)>0$. Thuswhich is the closed G-delta set as a zero set construction.

The Tietze extension theorem states: if $X$ is a normal topological space, $A\subseteq X$ is closed, and $f:A\to [-M,M]$ is continuous, then there is a continuous $F:X\to [-M,M]$ with $F{|}_A=f$.

We first prove an approximation lemma. Given continuous $h:A\to [-r,r]$, the closed subsetsare disjoint and closed in $X$. By the Urysohn lemma, there is a continuous $g:X\to [-r/3,r/3]$ equal to $-r/3$ on $A_{-}$ and $r/3$ on $A_{+}$. Then

Starting with $h_0=f$ and $r_0=M$, apply the lemma recursively to obtain continuous $g_j:X\to \mathbb{R}$ such thatwhere $r_j=M(2/3{)}^j$. The seriesconverges uniformly by the Weierstrass test, so its sum is continuous. On $A$, its remainder after $j$ terms is $h_{j+1}$, which tends uniformly to zero; hence $F{|}_A=f$. The bounds also keep $F$ in $[-M,M]$ after the standard endpoint-preserving version of the construction, proving the theorem.

Suppose that the positive integer $n$ is a decisive number. For every prime number $p<\sqrt{n}$, the composite number $p^2$ satisfies $1<p^2<n$. It therefore cannot be coprime to $n$, so $p\mid n$.

The product of all prime numbers below $\sqrt{n}$ consequently divides $n$. Taking natural logarithms givesChanging the strict endpoint can remove at most one prime term. By part b, for every sufficiently large $n$ the right-hand side is at least, for example, $\sqrt{n}/13$. This is impossible for sufficiently large $n$, sinceThus every decisive number lies below one fixed bound, and only finitely many integers can do so.

Write the Chebyshev theta function asTaking natural logarithms in part a givesand henceBecause $\sqrt{n}\log n/n\to 0$, the right-hand side is at least $n/2$ for every sufficiently large integer $n$.

Now let $x$ be sufficiently large and set $n=\lfloor x/2\rfloor$ using the floor function. Then $2n\le x$ and $n\ge x/3$, so the monotonicity of $\vartheta$ gives

The binomial theorem givesThe central binomial coefficient is the largest of these $2n+1$ nonnegative binomial coefficients, so

For the upper bound, fix a prime number $p\le 2n$. By Legendre formula,Each summand is either zero or one. Hence the P-adic valuation satisfiesIf $p\le \sqrt{2n}$, this impliesThere are at most $\sqrt{2n}$ such prime numbers. If instead $p>\sqrt{2n}$, then $p^2>2n$, so the P-adic valuation is at most one. Multiplying the prime-power contributions therefore yields

For permutation characters, Burnside lemma on a product givesTwo pairs $(A,B)$ and $(A^{\prime },B^{\prime })$ lie in the same $S_n$-orbit exactly whenWhen $0\le l\le k\le n/2$, every value $0,1,\dots ,l$ is possible because $k+l\le n$. Therefore the symmetric-group subset permutation representation satisfies

For $1\le r\le n/2$, the inclusion map between subset permutation modules embeds $\mathbb{C}[X_{r-1}]$ into $\mathbb{C}[X_r]$, sois a character. Its norm isBy character orthogonality, ${\chi }_r$ is the character of an irreducible representation.

For $r>n/2$, complementation is an $S_n$-equivariant bijection $X_r\cong X_{n-r}$, soConsequentlythe negative of one of the irreducible characters just found, and is not itself the character of a representation.

The Burnside lemma states that for a finite group $G$ acting on a finite set $X$,To prove it, countCounting by $g$ gives ${\sum }_g|X^g|$. Counting by $x$ givesusing the orbit-stabilizer theorem on each orbit $\mathcal{O}$. Dividing by $|G|$ proves the formula.

Let $\pi$ be the character of the permutation representation $\mathbb{C}[X]$. Then $\pi (g)=|X^g|$, and a second application of Burnside's lemma givesFor a two-transitive action there are exactly two orbits on $X\times X$: the diagonal and the ordered pairs of distinct points. Hence $⟨\pi ,\pi ⟩=2$. Transitivity says the trivial representation occurs once. Since the squared multiplicities of the irreducible constituents sum to two, there is exactly one further constituent, with multiplicity one, and it is irreducible and nontrivial. Thusas in the permutation representation of a two-transitive action.

Let $L=\mathbb{Q}({\zeta }_n)$. We use the standard cyclotomic containment lemma: every root of unity in $\mathbb{Q}({\zeta }_n)$ has order dividing $\mathrm{lcm}(2,n)$. One proof uses the conductor of a rational cyclotomic field. The conductor attached to a primitive $m$th root is $m$, except that it is $m/2$ when $m\equiv 2(\mathrm{mod}4)$; containment in $\mathbb{Q}({\zeta }_n)$ forces this conductor to divide $n$, which gives $m\mid \mathrm{lcm}(2,n)$.

If $n$ is even, $\mathrm{lcm}(2,n)=n$, and the powers of ${\zeta }_n$ already give all $n$ possible roots of unity in $L$. If $n$ is odd, $-{\zeta }_n$ has order $2n$, so $L$ contains all $2n$th roots of unity, and the containment lemma shows there are no others. Thus the roots of unity in a rational cyclotomic field number

Choose a primitive root ${\zeta }_n$. Since every root of ${\Phi }_n$ is a power of ${\zeta }_n$, its splitting field is $L=K({\zeta }_n)$. Each $\sigma \in G$ sends ${\zeta }_n$ to another primitive root, uniquely of the formThe assignment $\sigma \mapsto a_{\sigma }$ is a homomorphism. It is injective because an automorphism fixing ${\zeta }_n$ fixes $K({\zeta }_n)=L$.

If ${\Phi }_n$ is irreducible over $K$, the orbit of ${\zeta }_n$ has all $\phi (n)$ primitive roots, so the injection has image of size $\phi (n)$ and is surjective. Conversely, surjectivity makes every primitive root a conjugate of ${\zeta }_n$. Its minimal polynomial over $K$ then has at least $\phi (n)$ roots and divides the degree-$\phi (n)$ polynomial ${\Phi }_n$, so it equals ${\Phi }_n$. This proves the Galois embedding for a cyclotomic polynomial and the claimed equivalence.

The cyclotomic polynomial iswhere ${\zeta }_n$ is primitive. Partitioning all $n$th roots by exact order gives the cyclotomic factorizationInduct on $n$. The quotient of the monic integer polynomial $X^n-1$ by the product of the already constructed monic ${\Phi }_d$, $d<n$, lies in $\mathbb{Q}[X]$ and is monic. Gauss lemma then shows that this quotient ${\Phi }_n$ lies in $\mathbb{Z}[X]$.

If the prime $p$ does not divide $n$, then over ${\mathbb{F}}_p$,Thus $X^n-1$ has no repeated root. Every factor, including the reduction of ${\Phi }_n$, is square-free, proving the Separability of a cyclotomic polynomial modulo p.

A primitive root of unity is an $n$th root of unity of multiplicative order exactly $n$.

Let ${\zeta }_m,{\zeta }_n\in K$ be primitive and let $H$ be the finite subgroup of $K^{\times }$ that they generate. By part a, $H$ is cyclic. Its order is divisible by both $m$ and $n$, hence by $N=\mathrm{lcm}(m,n)$. On the other hand every element of $H$ has $N$th power one, so the order of the cyclic group $H$ divides $N$. Its order is therefore exactly $N$, and a generator of $H$ is a primitive $N$th root of unity in $K$.