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Since $a$ is an indicator function of density of a finite subset $\alpha$, the triangle inequality gives $|\hat{a}(r)|\le {\mathbb{E}}_{x}a(x)=\alpha$ for every frequency. Also, character orthogonality and the Parseval identity on a finite group giveFor completeness, the character orthogonality used here iswhich follows by summing a finite geometric series. Expanding the squared Fourier coefficients on a finite abelian group and using this identity proves the displayed Parseval identity on a finite group directly.

Combining the uniform bound with that identity gives the fourth-moment boundIn particular, the Lp norm on the frequency side here is a sum, not a normalized average. This is the fourth Fourier moment bound for an indicator function.

We need independent sets in every sufficiently large remaining induced subgraph, rather than just one in the initial random graph. The useful concentration variable is the maximum edge-disjoint clique packing in the complement graph; changing one edge changes that variable by at most one. This avoids the much larger sensitivity of the ordinary clique count in a binomial random graph.

First work in a binomial random graph $H\sim G(m,1/2)$, take $r=r_0(m)$, and let $X$ count its $r$-cliques. Put $\mu =\mathbb{E}X\ge m^{9/5}$ and let $Z$ be its maximum edge-disjoint clique packing size. Form the clique conflict graph whose vertices are these cliques, adjacent when they share an edge. If $Q$ counts unordered conflicting pairs, the Caro-Wei bound gives, for each realization,To verify the bound, randomly order the vertices of a finite graph and retain each vertex preceding all its graph neighbours. These retained vertices are an independent set, with expected size ${\sum }_v(1+\mathrm{deg}v{)}^{-1}\ge X^2/(X+2Q)$ by the Cauchy-Schwarz inequality. Interpret the ratio as zero when $X=0$. A second application of the Cauchy-Schwarz inequality, now to expectations, givesCounting ordered pairs of $r$-sets with $j$ common vertices, including identical sets when $j=r$, yieldsThe shared edges account for the factor $2^{\left(\genfrac{}{}{0ex}{}{j}{2}\right)}$. We claimHere are the estimates, including the large overlaps. For $2\le j\le \lfloor r/2\rfloor$, the probability that a random $r$-set contains any specified $j$ elements is at most $(r/(m-r){)}^j$. Summing over their choices inside the first $r$-set givesThe logarithm of $T_j$ is a convex function, a quadratic in $j$, so the maximum on this interval is at an endpoint. We have $T_2=O(r^4/m^2)$. As $r=(2+o(1)){\log }_{2}m$, the other endpoint satisfiesand is smaller than $T_2$ for sufficiently large $m$. Summing at most $r$ terms gives $O(r^5/m^2)$.

For the other half of the overlaps, write $j=r-s$, where $0\le s\le \lceil r/2\rceil$. The exact identity and an upper bound areThe quantity in parentheses is $m^{-1/2+o(1)}$, which tends to zero. The geometric series is thus at most $2/\mu$ for sufficiently large $m$. This proves the claim. In fact $r^5/m^2=o(m^{-9/5})$ and $1/\mu \le m^{-9/5}$, so eventuallyNow expose the $N=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ independent edge indicators one at a time. The edge-exposure martingale $M_i=\mathbb{E}[Z\mid \text{first }i\text{ indicators}]$ starts at $\mathbb{E}Z$ and ends at $Z$. Deleting one edge destroys at most one member of any edge-disjoint clique packing. Hence changing one indicator changes $Z$ by at most one; coupling the remaining indicators shows $|M_i-M_{i-1}|\le 1$.

Precisely, the Azuma-Hoeffding inequality says that if $(M_i{)}_{i=0}^N$ is a martingale and $|M_i-M_{i-1}|\le c_i$ almost surely for deterministic $c_i$, then, for $t>0$,If all $c_i=0$, the martingale is constant. Applying the lower-tail inequality with $c_i=1$ and $t=\mathbb{E}Z$ givesfor sufficiently large $m$. This is the only martingale concentration inequality needed.

Return to the original binomial random graph $G\sim G(n,1/2)$ and set $m=\lceil n/({\log }_{2}n{)}^2\rceil$. For each fixed $m$-element vertex set $U$, the complement graph of $G[U]$ has distribution $G(m,1/2)$. The union bound therefore givessince $m^{8/5}/n\to \infty$. Thus with high probability every vertex set of size at least $m$ contains an independent set of sizeApply greedy colouring by removing independent sets: remove such an independent set, give it a fresh colour, and repeat while at least $m$ vertices remain. Give each of the fewer than $m$ final vertices its own colour. This uses at mostcolours, since $m=o(n/{\log }_{2}n)$. Together with the lower bound, this gives the chromatic number of the half-density binomial random graph:

A feasible per-flow allocation satisfies $x_r>0$ for each active route and ${\sum }_r{A}_{jr}{n}_r{x}_r\le C_j$. It has proportional fairness if every feasible competitor obeysBy the first-order inequality for the strictly concave logarithm, this is equivalent to maximizing weighted logarithmic utility ${\sum }_{r:n_r>0}{n}_r\log x_r$ over the feasible allocations. Inactive-route rates are immaterial; set them to zero as a convention.

Write $a=n_{\{1\}}$, $b=n_{\{2\}}$, $c=n_{\{1,2\}}$, and $N=a+b+c$. The capacity constraints are $a{x}_1+c{x}_0\le 1$ and $b{x}_2+c{x}_0\le 1$. For $c>0$, put $q=c{x}_0$. Each active local route uses its remaining capacity $1-q$. Omitting constants, the utility isIts maximum has $q=c/N$; when $a+b=0$, the maximum is the endpoint $q=1$. Hence for active routesThese formulas also give $x_1=1/a$ and $x_2=1/b$ when $c=0$. When $N=0$ there is no allocated service. Equivalently, the aggregate route rates for $N>0$ areThis is the proportionally fair allocation on a two-resource linear network for a two-resource linear flow network.

For independent Poisson processes of documents and independent document sizes with exponential distribution of rate ${\mu }_r$, the memorylessness of the exponential distribution makes the population a continuous-time Markov chain. Its transition intensities areThe departure intensity sums the $n_r$ independent completion hazards ${\mu }_r{x}_r$. There is no departure on an inactive route. Set ${\rho }_r={\nu }_r/{\mu }_r$, with ${\mu }_r>0$.

To prove the proposed stationary distribution, define the balance function of a flow-level networkFor an active local route, $\Psi (n-e_r)/\Psi (n)=(a+b)/N$, while for the through route this ratio is $c/N$. Thus ${\Lambda }_r(n)=\Psi (n-e_r)/\Psi (n)$. The weightssatisfy detailed balance on every arrival-departure pair:Each total jump rate is bounded by ${\sum }_r({\nu }_r+{\mu }_r)$, so the chain is a nonexplosive Markov chain; normalized detailed-balance weights therefore give its stationary distribution, the stationary law of a two-resource linear flow network.

Use the negative binomial series to sum first over $c$:The remaining two geometric series giveAll summands are nonnegative. Consequently the sum is finite exactly under the stability conditions for a two-resource linear flow networkThese say that the offered load on each unit-capacity resource is strictly below one. Equality is insufficient: at least one geometric series then diverges. If all arrival rates are positive, the chain is irreducible and the normalizable law is its unique equilibrium law.

Finally put $q_i={\rho }_i/(1-{\rho }_0)$, $i=1,2$. Summing over $c$ yieldsThus the two local-route counts are independent, each with geometric distribution on the nonnegative integers. The conditional through-route count has negative binomial distributionFor positive traffic rates this depends on the local counts, so all three route counts are not independent. More quantitatively, $\mathrm{Cov}(a,c)={\rho }_0\mathrm{Var}(a)/(1-{\rho }_0)>0$, and likewise for $b$. If zero arrival rates are allowed, the exceptional independent cases are ${\rho }_0=0$, or ${\rho }_1={\rho }_2=0$, when the relevant counts are deterministic zero. This distinction is local-count independence with through-flow dependence.

Use the convention required by the printed expansion:where $u_i\in Y$ and $v_i\in X$ form the positive-singular-value parts of the singular system of a compact operator. Consequently $A^{\ast }A{v}_i={\sigma }_i^2{v}_i$. Apply part (b)'s polynomial to each such eigenvector, using noisy data:The geometric series then gives the full inverse coefficient, or Landweber full inverse filter,In particular, for the unrelaxed iteration $\tau =1$, $g_{\alpha }(\sigma )=[1-(1-{\sigma }^2{)}^{1/\alpha }]/\sigma$. The dimensionless Landweber spectral filter is instead $\sigma g_{\alpha }(\sigma )$; the printed expression uses the full coefficient and therefore needs the denominator $\sigma$.

For each fixed finite $n$, the filter extends continuously at zero by $g_{1/n}(0)=0$, since $g_{1/n}(\sigma )=n\tau \sigma +O({\sigma }^3)$ as $\sigma \to 0$. Small singular values are suppressed rather than immediately divided into noisy data. A component of $y^{(\delta )}$ in $\ker A^{\ast }$ contributes nothing. The parameter here is discrete: $1/\alpha$ is an integer. This matters when $1-\tau {\sigma }^2<0$; arbitrary real powers are not intended. A family for every $\alpha >0$ can use $n=\lfloor 1/\alpha \rfloor$ instead.

Put $\alpha =1-n/q>0$ andFirst suppose $A<\infty$. For an interior ball $B_R$, with $R\le 1$, write $M_R={\mathrm{esssup}}_{B_R}u$, $m_R={\mathrm{essinf}}_{B_R}u$, and $\omega (R)=M_R-m_R$. Constant shifts of a divergence-form elliptic operator change its forcing. The two nonnegative weak supersolutions $v=M_R-u$ and $w=u-m_R$ satisfy, respectively,Their Weak Harnack inequalities on $B_{R/2}$ therefore have forcing errors bounded by $CN{R}^{\alpha }$. Indeed $|M_R|,|m_R|\le A$, the new divergence forcing is bounded by $\Vert F{\Vert }_q+A\Vert b{\Vert }_q$, and the scalar forcing by $\Vert g{\Vert }_{q/2}+A\Vert d{\Vert }_{q/2}$; also $R^{2\alpha }\le R^{\alpha }$.

On $B_{R/2}$ we have $v+w=\omega (R)$. Even if $p<1$, the elementary inequality $(s+t{)}^p\le C_p(s^p+t^p)$ impliesThe sum of the two essential infima is $\omega (R)-\omega (R/2)$. Enlarging a uniform constant $C_0>1$ giveswhere $\theta =1-C_0^{-1}\in (0,1)$. This is an oscillation decay estimate; inhomogeneous oscillation decay implies Hölder continuity. ChooseIterating on concentric balls, starting at a fixed admissible radius $R_{\ast }\le 1$, yieldsThe strict choice of $\mu$ bounds the geometric series, including the case in which the two original decay rates coincide. Monotonicity of oscillation extends this bound to intermediate radii.

Take $R_{\ast }$ uniformly smaller than the distance of ${\Omega }^{\prime }$ from the boundary of $\Omega$. The vanishing essential oscillation produces a continuous representative of $u$. A ball centred at either of two sufficiently close points of ${\Omega }^{\prime }$ contains both, so its oscillation bounds their difference by $CN|x-y{|}^{\mu }$. For separated points use $|u(x)-u(y)|\le 2A$. Thus the Hölder seminorm satisfiesThe constants depend only on the indicated domain separation, dimension, coefficient norms, uniform ellipticity bounds and $q$. If the global essential supremum is infinite, the displayed global estimate is vacuous; local boundedness of weak elliptic subsolutions on slightly larger compact subsets still makes the same argument prove interior Hölder continuity.

Let $\mathcal{A}$ be the closure of the polynomials in $C(K)$ for the uniform norm. It is a closed algebra, since products of uniformly convergent sequences converge uniformly on the compact set $K$.

Choose a point $W$ with $|W|>{\sup }_{z\in K}|z|$. The uniformly convergent geometric seriesshows that this reciprocal belongs to $\mathcal{A}$. Join $w$ to $W$ by a continuous path in $\mathbb{C}\setminus K$. Its image is compact and disjoint from $K$, so their distance $d$ is positive. Partition the path into finitely many steps shorter than $d/2$.

Suppose $1/(b-z)\in \mathcal{A}$ and $|a-b|<d/2$, with $b$ on the path. Thenconverges uniformly on $K$, since $|b-z|\ge d$. Every partial sum belongs to $\mathcal{A}$, by its algebra property, and so does the limit. Working backwards along the partition transfers the pole from $W$ to $w$. Thus $1/(w-z)\in \mathcal{A}$: a polynomial $P$ exists withThis pole-moving polynomial approximation is the elementary mechanism behind the polynomial Runge theorem.

The reciprocal lattice is ${\Lambda }^{\ast }=\{G:G\cdot \ell \in 2\pi \mathbb{Z}\text{ for every }\ell \in \Lambda \}$. With oriented primitive-cell volume $v_c=a_1\cdot (a_2\times a_3)$, its basis is $b_1=2\pi (a_2\times a_3)/v_c$ and the cyclic variants; thus $b_i\cdot a_j=2\pi {\delta }_{ij}$.

The total potential is $V(x)={\sum }_{\ell }U(x-\ell )$. Using the Fourier transform convention $\tilde{U}(q)=\int e^{-iq\cdot x}U(x)d^{3}x$, translation gives $\tilde{V}(q)=\tilde{U}(q)\Delta (q)$, withThe given Born approximation therefore supplies the amplitude $\boxed{-m\Delta (q)\tilde{U}(q)/(2\pi {\hslash }^2)}$. A finite geometric series gives, when the denominators are nonzero,At reciprocal-lattice vectors each phase is one, so $|\Delta |=N_1{N}_2{N}_3$, the number of atoms. For large finite crystals these narrow peaks represent coherent Bragg scattering, with peak intensity proportional to the squared number of atoms; their finite width reflects the finite crystal size.

For the specified face-centered cubic lattice, $v_c=a^3/4$ andThe reciprocal lattice have coordinates $(2\pi /a)(h,k,l)$ with $h,k,l$ all even or all odd. The two shortest nonzero length families are $G_1=2\pi \sqrt{3}/a$ for $(111)$ and $G_2=4\pi /a$ for $(200)$. Elastic scattering satisfies $|q|=2|k|\sin (\theta /2)$. The assumed incident magnitude exceeds $G_2/2$, so both families are kinematically accessible, andThis is the angular prediction for suitable crystal orientations, or for a powder containing those orientations. A fixed single-crystal orientation must additionally satisfy the vector Elastic Bragg scattering condition $2k\cdot G+|G{|}^2=0$; the magnitude bound alone cannot guarantee both peaks for every incident direction.

For $q\in \mathbb{Q}\setminus \{0\}$ write $q=p^{k}u/v$, where $k\in \mathbb{Z}$ and neither integer $u$ nor integer $v$ is divisible by $p$. Define the P-adic valuation $v_p(q)=k$ and the p-adic absolute value $|q{|}_p=p^{-k}$, with $|0{|}_p=0$. The p-adic metric isFactoring out the smaller power of $p$ from two rational numbers shows $v_p(q+r)\ge \min \{v_p(q),v_p(r)\}$, with the usual convention $v_p(0)=+\infty$. Consequently the ultrametric inequality givesTogether with symmetry, positivity and separation of points, this verifies that $d_p$ is a metric.

For the given geometric partial sums, use the finite geometric series identity. Since $p$ does not divide $p-1$,Thus the limit is the rational number $\boxed{1/(1-p)}$, even though the partial sums diverge in the ordinary real metric.

If $|a{|}_p>|b{|}_p$, the ultrametric inequality gives $|a+b{|}_p\le |a{|}_p$. Applying it to $a=(a+b)-b$ gives $|a{|}_p\le \max \{|a+b{|}_p,|b{|}_p\}$. Since $|b{|}_p<|a{|}_p$, this forces $|a+b{|}_p=|a{|}_p$. Interchanging $a,b$ handles the other unequal case, provingFinally, let $b$ be outside the open ball $B(a,\delta )$, so $d_p(a,b)\ge \delta$. If $d_p(b,c)<\delta$, the preceding unequal-norm identity gives $d_p(a,c)=d_p(a,b)\ge \delta$. Therefore $B(b,\delta )$ also lies outside $B(a,\delta )$. Its complement is open, so every p-adic open ball is also closed, that is, a clopen set.

Let $T_0=0$ and let $T_{r+1}$ be the first visit to $i$ strictly after $T_r$, with $T_{r+1}=\infty$ if there is none. Put $q={\mathbb{P}}_i(T_1<\infty )$. The Strong Markov property at each finite first return time gives, by induction,The number of visits, including time zero, is $N={\sum }_{r\ge 0}{1}_{\{T_r<\infty \}}={\sum }_{n\ge 0}{1}_{\{X_n=i\}}$. By Tonelli theorem for nonnegative terms,If $q<1$, this geometric series is $1/(1-q)$, and $1-q$ is the probability of never returning. If $q=1$, every term of the final series is one and both sides are infinite. Thuswith the prescribed infinite-value convention. This covers both transient states and recurrent states.

Rewrite the ratio of hyperbolic sines asFor $\mathrm{Re}p>0$, expand the denominator by the geometric series. Part (a) inverts each delayed exponential:At each finite time only finitely many terms are nonzero, so this reflected-step solution of the wave equation is locally a finite sum. The first front travels from $x=0$; the second is its sign-reversed reflection from the fixed endpoint, with repeated round trips. At $x=L$ the delays coincide and cancel. At $x=0$ the sum telescopes to $a$ for $t>0$.

An equivalent standing-wave Fourier series isIts static ramp has exactly the sine coefficients $2a/(n\pi )$, so cancellation gives the zero initial displacement in the interior. Both representations solve the equation away from the fronts and in the distributional sense across them; the convention $H(0)=1/2$ agrees with Fourier midpoint values.

Put $p_j=2{\kappa }^j$, where $\kappa =n/(n-2)>1$. Since $p_j^2/(p_j-1)\le 2p_j$, part (ii), followed by a $p_j$-th root, gives the Moser iteration stepAfter $N$ steps,The geometric series and its differentiated form giveIn particular the exact convergent product isThe printed hint's equality to a single power of $2\kappa$ is not correct in general; its claimed finiteness is correct and the displayed expression supplies the correction.

On a finite-measure domain, Lp norms converge to the supremum norm for a bounded continuous function. Indeed, the upper bound is $\Vert u{\Vert }_p\le |\Omega {|}^{1/p}\Vert u{\Vert }_{\infty }$; for every $a<\Vert u{\Vert }_{\infty }$, the set $\{|u|>a\}$ has positive measure, giving $\Vert u{\Vert }_p\ge a|\{|u|>a\}{|}^{1/p}$. Let $N\to \infty$ in the iteration and use part (i), which gives $\lambda >0$. The Dirichlet eigenfunction supremum estimate isThe eigenvalue exponent is exactly $n/4$. The constant depends only on $n$: the supplied zero-boundary Sobolev inequality has a dimension-only constant, and every product factor above depends only on $\kappa$.

A continuous map $F$ of a compact metric space is uniquely ergodic if it has exactly one invariant measure that is a Borel probability measure. The uniqueness requirement ranges over all invariant Borel probability measures.

For an irrational rotation of the circle, let $R_{\alpha }x=x+\alpha$ on $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, and let $\nu$ be any invariant Borel probability measure. For each integer $r$, set ${\chi }_r(x)=e^{2\pi irx}$. Invariance givesFor $r\ne 0$, irrationality forces $e^{2\pi ir\alpha }\ne 1$, so the corresponding Fourier coefficient is zero. For $r=0$ it is one. These are exactly the Fourier coefficients of normalized Lebesgue measure $m$. By the Stone-Weierstrass theorem, trigonometric polynomials are uniformly dense in the continuous functions on the circle group; hence $\nu$ and $m$ integrate every continuous function equally and are the same Borel probability measure. Since $m$ is invariant, the irrational rotation of the circle is uniquely ergodic, with unique measure $m$.

For the irrational skew shift on the two-dimensional torus, writeThe map is invertible, with $T^{-1}(x,y)=(x-\alpha ,y-x+\alpha )$ modulo one, and it preserves $m_2$ as allowed in the question. We first prove the ergodic transformation property by the invariant-function characterization of ergodicity.

Let $F\in L^2(m_2)$ satisfy $F\circ T=F$. Its expansion in the Fourier basis is $F={\sum }_{r,s}{c}_{r,s}{e}_{r,s}$ in $L^2$. Direct calculation of the Koopman operator givesUniqueness of the Fourier coefficients therefore impliesFor $s\ne 0$, the magnitudes of the Fourier coefficients along all distinct indices $(r+js,s)$, $j\in \mathbb{Z}$, are equal. By the Bessel inequality they are square summable, so every such coefficient must be zero. When $s=0$, the relation becomes $c_{r,0}=e^{2\pi ir\alpha }{c}_{r,0}$, which forces $c_{r,0}=0$ for $r\ne 0$. Only $c_{0,0}$ remains. Thus every invariant $L^2$ function is constant, and

To prove unique ergodicity, we will establish uniform averages for every continuous function directly. No theorem on unique ergodicity of skew products is needed. Induction on $n$, using the old first coordinate in the second coordinate of $T$, yieldsIn particular, for a Fourier basis element,If $s=0$ and $r\ne 0$, this is a geometric series in $n$, anduniformly in $(x,y)$.

For $s\ne 0$, we give the finite Van der Corput inequality for finite scalar sequences and its proof. For $|u_n|\le 1$, extend $u_n$ by zero outside $0\le n<N$. Fix an integer $1\le H\le N$ and set $C_h={\sum }_{n=0}^{N-h-1}{u}_{n+h}\overline{u_n}$. Every original summand appears $H$ times in the identityThe Cauchy-Schwarz inequality, followed by expansion of the squared window sums, givesConsequently the Van der Corput inequality for finite scalar sequences isThe finite prefactor is important; the order of limits will be $N\to \infty$ with $H$ fixed, followed by $H\to \infty$.

For the irrational skew shift character sequence above, differencing cancels the quadratic term:For every fixed $h\ge 1$, $s\alpha h$ is irrational, so another geometric series estimate givesuniformly in $(x,y)$. With fixed $H$, the Van der Corput inequality for finite scalar sequences therefore impliesLetting $H\to \infty$ proves that the averages of every nonconstant Fourier basis element converge uniformly to zero. The constant character has average one. This establishes uniform equidistribution of an irrational skew shift on all trigonometric polynomials.

The Stone-Weierstrass theorem makes these trigonometric polynomials uniformly dense in $C({\mathbb{T}}^2)$. If $p$ approximates a continuous $g$ with $\Vert g-p{\Vert }_{\infty }<\epsilon$, thenTaking $N\to \infty$ and then $\epsilon \downarrow 0$ proves uniform convergence to $\int gd{m}_2$ for every continuous $g$.