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Now $\omega (n)\to -\infty$ makes $\mathbb{E}X=2^{-\omega (n)}\to \infty$. Chebyshev's inequality and the variance bound giveFor every fixed $ϵ>0$, they likewise giveThus the count is concentrated about its growing mean, despite the overlapping runs.

The assumptions implyEventually ${\mu }_n>0$, and $X_n=0$ implies $|X_n-{\mu }_n|\ge {\mu }_n$. Thus Chebyshev's inequality givesNo independence between the different $X_n$ is needed; this is a second moment method.

Chebyshev's inequality states that if $X$ has finite mean $\mu$ and variance ${\sigma }^2$, then for $t>0$,Indeed $(X-\mu {)}^2\ge t^2{1}_{\{|X-\mu |\ge t\}}$ pointwise. Taking expectations proves the result, including the zero-variance case.

Take $k=k_0=\min \{j:j!\ge n\}$. Then $k\to \infty$ and $(k-1)!<n\le k!$. Also $k=O(\log n)$, because $j!\ge 2^{j-1}$, so the needed block lengths are below $n/2$ for large $n$.

If $L\ge k+1$, then $X_{k+1}\ge 1$. Markov's inequality givesFor $j=k-2$, the expected count satisfies $\mu =n/j!>k-1\to \infty$. Part (ii) bounds its variance by $2\mu$, so Chebyshev's inequality givesCombining the tails provesThe blocks are consecutive in the cyclic order throughout; this is not the longest increasing subsequence with arbitrary omitted positions.

This establishes the longest increasing cyclic run of a random permutation concentration bound.

Markov's inequality states that for $Y\ge 0$ and $t>0$, $\mathbb{P}(Y\ge t)\le \mathbb{E}Y/t$. It follows by taking expectations in $Y\ge t{1}_{\{Y\ge t\}}$.

If $X$ has finite mean $\mu$ and variance ${\sigma }^2$, apply Markov to $(X-\mu {)}^2$ to obtain Chebyshev's inequality:For the first moment method, assume ${\mu }_n=\mathbb{E}{X}_n\to 0$. Nonnegative integer values give $\{X_n>0\}=\{X_n\ge 1\}$, soFor a standard second moment method, suppose ${\mu }_n>0$ eventually, the second moments are finite, and $\mathbb{E}{X}_n^2/{\mu }_n^2\to 1$. Equivalently, $\mathrm{Var}(X_n)/{\mu }_n^2\to 0$. The zero event is contained in $\{|X_n-{\mu }_n|\ge {\mu }_n\}$, soHence $\boxed{\mathbb{P}(X_n>0)\to 1}$. The second-moment condition controls the fluctuations relative to the mean; a large mean alone is insufficient.

The zero means and orthogonality give the exact second momentThus $A_N\to 0$ in $L^2$. Chebyshev's inequality gives the requested weak law of large numbers:Independence is unnecessary; the stated second-moment orthogonality is sufficient.

For an explicit almost surely convergent subsequence take $N_j=2^j$. For each positive integer $m$,The First Borel-Cantelli lemma makes only finitely many such exceedances occur. Intersecting the resulting probability-one events over $m$ gives $A_{2^j}\to 0$ almost surely.

The indicator function $I_A$ equals one when the event $A$ occurs and zero otherwise. Its expected value is $\mathbb{E}{I}_A=\mathbb{P}(A)$. By linearity of expectation, no independence assumption is needed to obtainSince $I_i{I}_j=I_{A_i\cap A_j}$, including $p_{ii}=p_i$, the variance isIn particular, correlations between the events cannot simply be discarded.

Write $\mu =\mathbb{E}N$. If $\mu >0$, then $N=0$ implies $|N-\mu |\ge \mu$. Applying Chebyshev's inequality therefore givesEquivalently, the contribution of $N=0$ to $\mathbb{E}[(N-\mu {)}^2]$ is ${\mu }^2\mathbb{P}(N=0)$. A positive mean is necessary for the displayed ratio to be defined. If $\mu =0$, nonnegativity instead gives $N=0$ almost surely, and the printed ratio would be $0/0$.

We use dyadic quadratic variation of Brownian motion. For fixed rational $0\le u<v$, set $\ell =v-u$ and let $Q_n$ be the sum of the squares of the $2^n$ Brownian increments on its equal-length partition. Independence and the normal fourth moment giveFor every $\eta >0$, Chebyshev's inequality bounds $\mathbb{P}(|Q_n-\ell |>\eta )$ by $2{\ell }^2{\eta }^{-2}{2}^{-n}$, a summable sequence. The Borel-Cantelli lemma implies $Q_n\to \ell$ almost surely. Intersect over positive rational $\eta$ and all rational intervals to obtain one probability-one event on which every such interval has this positive quadratic-variation limit.

If a path were Hölder of exponent $\alpha >1/2$ on a nonempty open interval, choose a compact rational subinterval $[u,v]$ within it and a finite Hölder constant $C$ there. Every partition increment would have absolute value at most $C(\ell 2^{-n}{)}^{\alpha }$, soThis contradicts $Q_n\to \ell >0$. The same pathwise event excludes every exponent greater than $1/2$, without needing an uncountable intersection of events. ThereforeThis is the quadratic variation obstruction to Hölder regularity. It rules out a continuously differentiable trajectory on any interval, since a continuous derivative is bounded on smaller compact intervals and gives a Lipschitz bound. By itself it does not rule out differentiability at an isolated point. The stronger nowhere differentiability of Brownian motion can also be seen directly as follows.

On a mesh of width $h=2^{-n}$ in $[0,M]$, the probability that any specified three consecutive increments all have absolute value at most $Kh$ is at most $C{K}^3{h}^{3/2}$: each increment is $N(0,h)$ and the three are independent. There are at most $M/h$ possible blocks, so the probability of any such block is at most $CM{K}^3{h}^{1/2}$, summable in $n$. By Borel-Cantelli lemma, for every pair of positive integers $M,K$ there are eventually no such blocks. But a finite derivative at some time $t<M$ would imply $|B_s-B_t|\le C_t|s-t|$ near $t$. The three mesh increments beginning at the mesh point immediately to the left of $t$ would then all be at most $6C_{t}h$ for all sufficiently fine meshes. Choosing an integer $K\ge 6C_t$ contradicts the preceding event. Thus finite derivatives do not exist anywhere almost surely, including a finite right derivative at $0$. The Hölder estimates below $1/2$ express continuity with a roughness scale; they do not imply differentiability.

For a transferable utility game on player set $N$, its characteristic function of a coalitional game is $v:2^N\to \mathbb{R}$, where $v(S)$ is the maximum transferable surplus attainable by coalition $S$ and $v(\varnothing )=0$. An imputation is a payoff vector satisfying efficiency ${\sum }_{i\in N}{x}_i=v(N)$ and individual rationality $x_i\ge v(\{i\})$. The core of a cooperative game consists of the efficient payoff vectors satisfyingThese conditions prevent any coalition from blocking the allocation; singleton constraints include individual rationality.

The Shapley value payoff isEquivalently, it is the expected marginal contribution of player $i$ when the players arrive in uniformly random order. The factorial weight counts the orders with exactly $S$ preceding $i$. Summing marginal contributions telescopes along each order, proving Shapley efficiency.

In the four-player market a coalition creates one unit of surplus exactly when it contains the seller and at least one buyer. ThusAll singleton values are zero and $v(N)=1$. Core constraints for the pairs $\{1,j\}$ require $x_1+x_j\ge 1$ for every buyer $j$. Since all payoffs are nonnegative and sum to one, the other two buyers' payoffs must be zero for each such pair. Therefore every buyer receives zero and the seller receives one. This vector satisfies all coalition inequalities, soThe seller's marginal contribution is one unless he arrives before all three buyers, which happens with probability $1/4$. A given buyer contributes one exactly when the seller is first and that buyer is second, with probability $(1/4)(1/3)=1/12$. HenceIn particular the Shapley allocation need not lie in this market's core.

For the market with $k$ sellers and $3k$ buyers, a coalition containing $s$ sellers and $b$ buyers can make precisely $\min (s,b)$ trades, each producing unit surplus. Thus $v(S)=\min (s(S),b(S))$ and $v(N)=k$. Let $s_k$ and $b_k$ denote the common Shapley payoffs within the two roles. Role symmetry and the telescoping efficiency identity giveIt remains to prove $s_k\to 1$, rather than infer it only from competitive intuition.

Generate a uniformly random arrival order by giving every player an independent uniform time in $[0,1]$. Conditional on a specified seller's arrival time $t$, the number $B$ of preceding buyers and number $S$ of preceding other sellers are independent, withThe seller adds a trade precisely when $B>S$. ConsequentlyFor $t>0$, the difference $D=B-S$ has mean $(2k+1)t$ and variance $(4k-1)t(1-t)\le 4kt$. By Chebyshev's inequality,Split the integral at $\delta =k^{-1/2}$. Using the trivial bound one on the first piece givesThis proves the Shapley limit in a buyer-heavy exchange market. Since $b_k=(1-s_k)/3$, we concludeThe proof accounts for sellers arriving very early, where a fixed-time concentration argument alone would not be uniform.

The assertion with the same $\epsilon$ in the degree interval and the final colour bound is false as printed. Fix $r=2$, $\epsilon =1/10$, let $G=K_{22m+1}$, and take $D=20m$. Every degree is $22m=(1+\epsilon )D$, while every pair hypergraph codegree is one. For any proposed $\delta >0$, choose $m$ large enough that $1\le \delta D$. Nevertheless a matching in a graph contains at most $11m$ edges, whereas there are $(22m+1)11m$ edges in total. Thus every edge coloring requires at least $22m+1$ colours, exceeding $(1+\epsilon )D=22m$. The obstruction persists with arbitrarily small relative hypergraph codegree.

The meaningful asymptotic target separates the output tolerance from the input regularity error. For fixed rank $r$ and target $\eta >0$, choose a sufficiently small $\delta =\delta (r,\eta )$ and require degrees in $[(1-\delta )D,(1+\delta )D]$ as well as pair hypergraph codegrees at most $\delta D$. The semi-random method then gives ${\chi }^{\prime }(G)\le (1+\eta )D$. Here is the requested proof essay for that corrected conclusion, including its random rounds, tracking requirements and completion step.

A useful way to run all colour classes together is the auxiliary matching construction for hypergraph edge colouring. Use $K=\lceil D\rceil$ initial colours. Create a task vertex for each original edge $e$, and a resource vertex $(v,c)$ for each original vertex and colour. An assignment $(e,c)$ is an $(r+1)$-edge consisting of its task and the resources $(v,c)$ for all $v\in e$. A matching in a hypergraph in this auxiliary system is a proper partial colouring: a task cannot get two colours, and two edges sharing $v$ cannot use the same resource $(v,c)$.

The auxiliary task degrees are $K$, resource degrees are $d_G(v)$, and its pair hypergraph codegrees are at most $\max (1,\delta D)$. It is therefore nearly regular with small hypergraph codegrees. For $r\ge 2$ and a nonempty original hypergraph, the hypergraph codegree condition also forces $D\ge 1/\delta$; taking $\delta$ small absorbs rounding and ensures a large working degree. Rank one is immediate by assigning distinct colours to edges at each vertex.

The extra balance condition is important. Let $S_v$ be the set of tasks corresponding to edges through original vertex $v$. A matching in a hypergraph missing a small fraction of all tasks need not leave few tasks in each $S_v$. The Rödl nibble must leave at most $\beta D$ unmatched tasks in every such star, where one may choose $\beta =\eta /(4r)$. This local requirement, rather than just a global almost-perfect matching in a hypergraph, is what permits a cheap final edge colouring.

Run a Rödl nibble on the auxiliary system, whose rank is $s=r+1$. In a round of working degree $D_j$, sample each surviving assignment independently with probability $\alpha /D_j$, for a small fixed $\alpha$. Accept the sampled assignments that meet no other sampled assignment. Delete every auxiliary vertex touched by any sampled assignment, including conflicted ones, before proceeding. The accepted assignments are a matching in a hypergraph, and later assignments cannot reuse a consumed task or resource. Record each discarded but uncoloured task as waste; it must be counted in the final residual graph.

For an auxiliary vertex of degree $(1\pm \xi )D_j$, the probability of remaining untouched isFor an auxiliary edge through a surviving vertex, the other $s-1$ vertices survive with probability approximately $e^{-(s-1)\alpha }$. Their incident trial families overlap in at most $O_s(\delta D)$ trials by the hypergraph codegree bound, so the overlap error in this product calculation is small. This leads to the degree trajectorywhile the surviving fraction of task vertices is approximately $e^{-\alpha }$ per round. For a sampled assignment, a union bound over at most $s(1+O(\xi ))D_j$ conflicting assignments gives collision probability at most $s\alpha (1+O(\xi ))$. More explicitly, for a fixed task the expected number of ordered sampled conflict pairs having the first assignment through that task is at mostThis bounds its waste probability. Thus productive deletion is first order in $\alpha$, and collision loss is second order.

The technical tracking step makes these expected value calculations locally simultaneous and keeps their errors from accumulating. Its requirements are: residual auxiliary degrees stay near their trajectory after controlled pruning; each original star has the predicted number of surviving tasks; and every star's total collision and pruning waste is small. The standard probabilistic tools enter at distinct points.

Within a fixed round, task-survival indicators are independent: trial families for different tasks are disjoint, even when their assignments share resources. Conditional Chernoff bounds therefore control the number of surviving tasks in each star. Collision loss has short certificates consisting of pairs of selected intersecting assignments. The number selected through one resource has a binomial distribution of bounded mean. Chernoff bounds truncate these numbers at a suitable logarithmic cutoff; after truncation one trial has only a logarithmically bounded effect on the certified collision count. A certifiable function concentration bound, or a corresponding exposure argument, then controls loss in each star. These rare local overload events have polynomially bounded local dependency neighborhoods in the working degree, so the Lovász local lemma can select a simultaneous outcome independent of the total number of vertices.

Residual-degree counts require additional care because they are not sums of independent indicators. Two survival tests share at most a hypergraph codegree's worth of trials; expanding their joint probabilities bounds their covariance by the corresponding relative overlap. For a residual-degree count $Z$ with $D_j$ summands, overlapping summands and shared survival trials give a variance bound of the formChebyshev's inequality then bounds $\mathrm{Pr}(|Z-\mathbb{E}Z|>\xi D_j)$ by $C_s(D_j^{-1}+\zeta )/{\xi }^2$. This gives a small exceptional fraction when the relative hypergraph codegree is sufficiently small; it does not by itself make every degree good. The tracking argument records and prunes these exceptional configurations, bounding their contribution to each star by codegree-weighted exposure and the same local-loss bookkeeping. Markov's inequality controls exceptional mass, and repeated conditional concentration controls the retained counts. One must not apply an independent Chernoff bound directly to a residual degree or infer a uniform star bound just from a global expected loss. Keeping the pruning loss explicitly is the key distinction between a valid Rödl nibble argument and those shortcuts.

Choose the parameters in this order: first $r,\eta$ and $\beta$; next $\alpha$ so small that $s\alpha \log (4/\beta )\ll \beta$; nextand finally the initial regularity/hypergraph codegree tolerance $\delta$ small enough for all $T$ tracking and pruning errors. This is a bounded number of rounds depending on the tolerances, not on the graph order. After $T$ rounds, the surviving task fraction is at most approximately $\beta /4$, while accumulated collision loss is $O_s(T{\alpha }^2)=O_s(\alpha \log (4/\beta ))$. Reserving the remaining margin for tracking and pruning leaves at most $\beta D$ uncoloured edges at every original vertex. The working degree is still a fixed positive multiple of $D$, namely about $D{e}^{-r\alpha T}$, so the large-degree estimates remain applicable throughout.

Finally use a fresh palette on the residual original hypergraph. Its maximum degree is at most $\beta D$; an edge meets at most $r(\beta D-1)$ others. Greedy colouring therefore needs at most $r\beta D+1$ fresh colours. The total is at mostfor sufficiently large $D$, already ensured by choosing $\delta$ small. This completes the passage from the locally balanced Rödl nibble to the hypergraph chromatic index bound. The degree tolerance must be the small input parameter $\delta$; replacing it by the entire output allowance $\eta$ removes the slack and leads precisely to the complete-graph counterexample above.

The Schumacher compression theorem concerns a memoryless quantum information source with average density matrix $\rho$ on a finite-dimensional Hilbert space. For $n$ uses the average state is ${\rho }_n={\rho }^{\otimes n}$. Compression consists of an encoding quantum channel into a space of dimension $D_n$ and a decoding channel back to the original block space, with asymptotic rate $n^{-1}{\log }_2{D}_n$ qubits per source use. The optimal reliable rate is $S(\rho )$: every larger rate is achievable, and no smaller rate is reliable. This holds for average pure-source fidelity; the direct construction also preserves a purification reference, and the converse for that stronger entanglement fidelity criterion holds with fidelity tending to zero below the threshold. The encoding and decoding are deterministic channels; no free classical transmission of the source label is allowed.

Diagonalize $\rho ={\sum }_j{\lambda }_j|j⟩⟨j|$ and omit zero-probability eigenvalues when sampling. Define the quantum typical subspace projector $P_{n,\delta }$ by retaining product eigenvectors withThe sampled random variable $-{\log }_2{\lambda }_j$ has expectation $S=S(\rho )$. The weak law of large numbers gives $w_n=\mathrm{Tr}({\rho }_n{P}_{n,\delta })\to 1$. Every retained eigenvalue is at least $2^{-n(S+\delta )}$, and the total probability is at most one, soThis proves the two typical subspace theorem estimates used in the direct part.

Write $P=P_{n,\delta }$. Encode its range isometrically into a compressed space and reserve one orthogonal failure flag. Decode the range back isometrically and the failure flag to a fixed pure state $|\phi ⟩$. This typical-subspace compression with a failure flag is a physical channel whose composite action isIndeed its Kraus operators are $P$ and $|\phi ⟩⟨e_a|$, where the $e_a$ form an orthonormal basis of $\ker P$, and their adjoint products sum to $I$. The Kraus formula for entanglement fidelity givesFor any pure-state ensemble of ${\rho }_n$, the average squared quantum fidelity is at least this number: apply convexity of $|z{|}^2$ separately to the amplitudes $⟨{\psi }_x|K_{\alpha }|{\psi }_x⟩$ with their source probabilities. Thus average fidelity also tends to one. The compressed dimension is at most $2^{n(S+\delta )}+1$, and its rate is at most $S+\delta +o(1)$. Choose $\delta <R-S$ for any prescribed $R>S$. One may also take ${\delta }_n=n^{-1/4}$: the finite spectral information variance and Chebyshev's inequality still give $w_n\to 1$, with rate tending to $S$. A pure source, $S=0$, can simply be re-prepared exactly without sending any qubits.

For the coherent converse, we prove the finite-dimensional quantum compression converse. If the compressed space has dimension $D$, the state of reference plus compressed system has a pure-state decomposition with Schmidt rank at most $D$. Applying decoder Kraus operators locally cannot increase the Schmidt rank of those vectors, so the final reference–output state has Schmidt number at most $D$. Let $|{\Psi }_{{\rho }_n}⟩$ purify the source, with squared Schmidt coefficients equal to the decreasing eigenvalues ${\mu }_1\ge {\mu }_2\ge \cdots$ of ${\rho }_n$. For any normalized vector $|v⟩$ of Schmidt rank at most $D$, its reference support lies in a rank-at-most-$D$ projector $Q$. HenceThe last bound follows by maximizing ${\sum }_j{\mu }_j⟨j|Q|j⟩$ subject to $0\le ⟨j|Q|j⟩\le 1$ and their sum at most $D$. Average over a Schmidt-rank-bounded decomposition of the final state to obtain $F_e\le {\sum }_{j=1}^D{\mu }_j$.

At rate $R<S$, choose $0<\delta <S-R$. The total source probability outside the quantum typical subspace tends to zero, while every eigenvalue inside it is at most $2^{-n(S-\delta )}$. Therefore for $D_n\le 2^{nR}$,This argument allows arbitrary encoding and decoding channels, not only subspace projection.

For completeness, average pure-source reliability also forces the same rate, even though it is a weaker criterion than entanglement fidelity. Here is a separate converse for that formulation. Let the emitted block pure states be ${\psi }_x$, with probabilities $p_x$, and let ${\sigma }_x$ be the decoded states. If their mean squared fidelity tends to one, then their mean trace distance ${\bar{\delta }}_n$ tends to zero, since $\frac{1}{2}\Vert {\sigma }_x-{\psi }_x{\Vert }_1\le \sqrt{1-⟨{\psi }_x|{\sigma }_x|{\psi }_x⟩}$. This inequality follows by choosing a purification with that overlap and using contraction of trace distance under a partial trace.

We need only an elementary entropy continuity from Jordan decomposition bound, derivable from Question 4. If $\delta =\frac{1}{2}\Vert \sigma -\tau {\Vert }_1$, write the positive and negative parts of $\sigma -\tau$ as $\delta {\omega }_{+},\delta {\omega }_{-}$. The common density matrixand the two mixture bounds giveHere $g$ is increasing and concave, $g(0)=0$. Apply this in block dimension $d^n$, both to the average states and to individual outputs, using convexity of trace distance and concavity of $g$. Since the original ensemble is pure, its Holevo quantity is $S({\rho }_n)=nS$. The decoded ensemble hasThe Holevo quantity under a quantum channel cannot increase. To specify the entropy ingredient, Strong subadditivity of Von Neumann entropy states $S(XB)+S(BE)\ge S(B)+S(XBE)$. Attach the classical source label $X$, dilate the decoder isometrically to $BE$, and trace out $E$. The displayed inequality says $I(X:B)\le I(X:BE)$; the isometry preserves the input mutual information. Since $I(X:B)$ of a labelled ensemble is its Holevo quantity, this proves the needed decoder monotonicity. Before decoding, the system has dimension $D_n$, so its Holevo quantity is at most ${\log }_2{D}_n$. Thus ${\log }_2{D}_n\ge nS-o(n)$ and the rate cannot be below $S$. Both operational fidelity formulations therefore have the claimed noiseless coding threshold.

If polynomials with integer monomial coefficients converge uniformly to $f$, their values at zero and one are integers converging to $f(0)$ and $f(1)$. A convergent integer sequence is eventually constant: once all its terms lie in an interval of length less than one they must agree. Hence both endpoint values are integers.

Conversely assume those values are integers. The rounded Bernstein polynomial is an integer-coefficient polynomial, because its displayed scalar coefficients are integers and each $x^k(1-x{)}^{n-k}$ expands with integer coefficients. Part b gives its vanishing distance from $B_{n}f$. For completeness, the remaining uniform convergence $B_{n}f\to f$ follows directly from uniform continuity. Let $X$ have the binomial distribution with parameters $n,x$; then $B_n(f,x)=\mathbb{E}f(X/n)$, $\mathbb{E}(X/n)=x$, and $\mathrm{Var}(X/n)=x(1-x)/n\le 1/(4n)$. Splitting into $|X/n-x|\le \delta$ and its complement, and applying Chebyshev's inequality, giveswhere $\omega$ is the modulus of continuity. First choose $\delta$ to make the first term small uniformly, then take $n$ large. The triangle inequality now gives $\Vert B_n^{\ast }f-f{\Vert }_{\infty }\to 0$. We have proved the integer-coefficient polynomial approximation characterization

At each finite stopping time ${\sigma }_a^k$, the Strong Markov property says that $B_{{\sigma }_a^k+t}-B_{{\sigma }_a^k}$ is a standard Brownian motion independent of the past sigma-algebra. Its exit time from $(-a,a)$ is exactly ${\sigma }_a^{k+1}-{\sigma }_a^k$. By induction the stopping times are finite almost surely, and these durations are independent with common law ${\sigma }_a$, not merely identically distributed. This is the Brownian exit-time skeleton.

Put $v=\mathrm{Var}({\sigma }_1)<\infty$, which follows from the moment estimate. Brownian scaling gives $\mathrm{Var}({\sigma }_a)=a^{4}v$. Forthere are $2^{2n}$ independent durations of mean $C{2}^{-2n}$ and variance $v{2}^{-4n}$. ConsequentlyFor each $\epsilon >0$, Chebyshev's inequality givesApply the Borel-Cantelli first lemma for a countable sequence of positive errors tending to zero. We obtain the Brownian exit-skeleton clock convergence:No independence between the clocks for different values of $n$ is required.

For fixed $t$, let $B$ have the binomial distribution with parameters $m,t$. The Bernstein polynomial is $f_m(t)=\mathbb{E}[f(B/m)]$. Continuity on the compact interval gives uniform continuity and a finite bound $M=\Vert f{\Vert }_{\infty }$.

Given $ϵ>0$, choose $\delta >0$ so $|f(s)-f(t)|<ϵ$ whenever $|s-t|<\delta$. Since $\mathbb{E}[B/m]=t$ and $\mathrm{Var}(B/m)=t(1-t)/m\le 1/(4m)$, Chebyshev's inequality givesThe bound is independent of $t$, including both endpoints. Taking the supremum and then letting $m$ tend to infinity proves $f_m\to f$ uniformly on $[0,1]$. This provides the required polynomial-approximation theorem with its proof.

Here is a tripartite Schmidt projection entanglement concentration protocol. For $0<p<1$, set $q=1-p$. Regrouping the $n$ local qubits of each party giveswhere $|x|$ counts ones. Alice performs a projective measurement of that count, without learning the individual string. The orthogonal projector for outcome $k$ is ${\Pi }_k={\sum }_{|x|=k}|x⟩⟨x|$. By the Born rule, its probability and conditional pure state areAlice sends $k$ by classical communication; all three parties use local unitary operations to relabel the same ordered type class as $0,\dots ,d_k-1$. There are now $d_k$ equal common-label amplitudes. Measuring the individual qubits instead would have destroyed precisely this coherence.

To obtain an integer number $f$ of target GHZ states, choose $m=2^f\le d_k$. The following exact LOCC step handles type classes whose sizes are not powers of two. For every $m$-element subset $S\subseteq \{0,\dots ,d_k-1\}$, Alice uses the Kraus operatorEvery label occurs in $\left(\genfrac{}{}{0ex}{}{d_k-1}{m-1}\right)$ such subsets, proving ${\sum }_S{K}_S^{†}{K}_S=I$ on the occupied subspace. Thus this is a complete measurement in quantum mechanics. Each outcome has probability $m/[d_k(\genfrac{}{}{0ex}{}{d_k-1}{m-1})]=1/\left(\genfrac{}{}{0ex}{}{d_k}{m}\right)$, and leaves the normalized pure state $m^{-1/2}{\sum }_{r\in S}|r,r,r⟩$. After Alice communicates $S$, all parties relabel its elements in the same order. Binary expansion of the labels then factors the state into exactly $f$ independent copies of the target GHZ state. Local extensions of these operations handle unused basis vectors and do not affect the protocol.

For example, taking $f(k)=\lfloor {\log }_2{d}_k\rfloor$ gives an outcome-dependent yield without any failure branch. Since $K$ has a binomial distribution with parameters $(n,q)$, the weak law of large numbers gives $K/n\to q$ in probability. Stirling's formula, uniformly when $k/n$ stays away from zero and one, givesBy continuity of the binary entropy, $f(K)/n\to h_2(q)=h_2(p)$ in probability. The integer rounding and logarithmic correction have no first-order effect.

A fixed yield with success probability tending to one is also explicit. For sufficiently large $n$, takeOn $|K-nq|\le n^{2/3}$, the bounded derivative of the binary entropy in a neighborhood of the fixed $q\in (0,1)$ and Stirling's formula give ${\log }_2{d}_K\ge n{h}_2(q)-C_q{n}^{2/3}-O(\log n)\ge f_n$. Chebyshev's inequality bounds the complementary probability by $q(1-q)n^{-1/3}$. Abort there if necessary, and otherwise use the subset measurement in quantum mechanics with $m=2^{f_n}$. ThusThis matches the first-order rate in part (i), with exact outputs on every successful forward branch. If $p=1/2$, simply retain the original $n$ target copies; if $p$ is an endpoint, output zero target copies.

Identify the vertices of the Boolean hypercube $Q_n$ with subsets of $[n]$, adjacent when their symmetric difference has size one. Write $N(\mathcal{A})$ for the closed graph neighbourhood, including $\mathcal{A}$ itself. In simplicial order on the discrete cube, smaller sets come first, and equal-sized sets are ordered by lexicographic order: the least element of the symmetric difference belongs to the earlier member. Let $I_m$ denote the first $m$ vertices. The vertex-isoperimetric inequality in the discrete cube statesSubtracting $|\mathcal{A}|$ gives the equivalent assertion for the external vertex boundary.

We first check that $N(I_m)$ is also an initial segment in simplicial order on the discrete cube. Apart from the empty and full cases, $I_m$ consists of complete lower layers and a lexicographic initial segment $\mathcal{F}$ in its top rank $r$. Its closed graph neighbourhood contains all ranks at most $r$ and the upper shadow $\nabla \mathcal{F}$ in rank $r+1$. This upper shadow is lexicographic initial: an $(r+1)$-subset $C$ belongs to it precisely when its first $r$-subset in lexicographic order, obtained by deleting its largest element, belongs to $\mathcal{F}$. Taking the first $r$ elements preserves weak lexicographic order, proving the assertion. This also handles $r=0$, where $N(\{\varnothing \})$ consists of ranks zero and one.

We prove the vertex-isoperimetric inequality in the discrete cube by mathematical induction on $n$, with $n=0,1$ immediate. Fix a coordinate $i$ and write ${\mathcal{A}}_0,{\mathcal{A}}_1\subseteq Q_{n-1}$ for the sections without and with $i$. The two sections of its closed graph neighbourhood areA simplicial section compression replaces ${\mathcal{A}}_j$ by the initial segment $I_{a_j}$ of size $a_j=|{\mathcal{A}}_j|$ in the other coordinates. By the preceding observation, both $N(I_{a_j})$ and $I_{a_{1-j}}$ are initial segments, so their union has size $\max \{|N(I_{a_j})|,a_{1-j}\}$. The mathematical induction hypothesis givesAdding the two section bounds proves that simplicial section compression cannot increase the closed graph neighbourhood.

Apply changing simplicial section compressions until none remains. The sum of the global simplicial order on the discrete cube positions strictly decreases at each step: within a section the order is exactly the restriction of the global order. Thus the procedure terminates at a set family $\mathcal{B}$ every coordinate section of which is initial. We now classify the possible terminal families for simplicial section compression, rather than assuming this terminal set family must be globally initial.

If $X<Y$, $X\notin \mathcal{B}$ and $Y\in \mathcal{B}$, then $X,Y$ cannot agree in any coordinate, since that coordinate section would have an omitted earlier member and an included later one. Therefore $Y=X^c$. Let $X$ be the first omitted vertex and $Y$ the last included vertex. If $\mathcal{B}$ is noninitial, $X<Y$ and $Y=X^c$. Any vertex $Z$ strictly between them would have to be included, since an omitted $Z$ paired with $Y$ would have to equal $Y^c=X$. But it would also have to be omitted, since an included $Z$ paired with $X$ would have to equal $X^c=Y$. Thus there is no such $Z$: $X,Y$ are consecutive complementary vertices. Everything before $X$ is included and everything after $Y$ omitted. Hence $\mathcal{B}$ is obtained from the initial segment of its size by replacing $X$ with its immediate successor $X^c$.

There are only two forms of this exception. If $n=2r+1$, consecutive complementary vertices must straddle ranks $r,r+1$; otherwise an intermediate rank or a further vertex in one of their ranks separates them. They are consequentlywith $X$ the last $r$-subset and $Y$ the first $(r+1)$-subset. The corresponding initial segment is all ranks at most $r$. For $r\ge 1$, the omitted $X$ remains in $N(\mathcal{B})$ through a lower neighbour, and every $(r+1)$-subset has at least two $r$-subsets, only one of which was omitted. Thus $N(\mathcal{B})$ contains every rank at most $r+1$, namely $N(I_{|\mathcal{B}|})$. For $r=0$ both singleton closed graph neighbourhoods are the whole $Q_1$.

If $n=2r$, the two complementary vertices must both have rank $r$. The complement map reverses the lexicographic order in that rank, so a consecutive complementary pair must be its central pair. Exactly half the $r$-subsets contain $1$, and all of those precede those missing $1$. Therefore the pair isHere $I_{|\mathcal{B}|}$ comprises all ranks below $r$ and all $r$-subsets containing $1$. For $r\ge 2$, $N(\mathcal{B})$ contains every rank at most $r$, because all lower ranks were retained. Every $(r+1)$-subset containing $1$ has $r\ge 2$ different $r$-subsets containing $1$, so deletion of the one member $X$ removes no vertex of the upper shadow. Consequently $N(\mathcal{B})\supseteq N(I_{|\mathcal{B}|})$. For $r=1$, the two closed graph neighbourhoods both have size three in $Q_2$. These cases exhaust the exceptions. Since the successive simplicial section compressions never increased the closed graph neighbourhood, this proves the vertex-isoperimetric inequality in the discrete cube.

The same argument gives all integer radii. Write $N^t$ for $t$ repetitions of the closed graph neighbourhood. Its application to an initial segment is again initial; the sizes of these initial segments are monotone in the original size. Inducting on $t$ therefore gives

For the concentration question, use uniform probability measure on a finite connected graph and normalize its graph distance by its graph diameter. A sequence is a Lévy family of graphs if, for every fixed $\epsilon >0$ and $\delta >0$, the radius-$\epsilon$ neighbourhood of every set of measure at least $\delta$ has measure tending to one, uniformly over those sets. One may equivalently require this only for sets of measure at least one half. To see the reverse implication, first the radius-$\epsilon /2$ neighbourhood of every positive-density set must eventually have measure greater than one half. Otherwise its complement has measure at least one half, while the radius-$\epsilon /4$ neighbourhood of that complement misses the original positive-density set, contradicting the half-measure condition. Apply the half-measure condition once more to the radius-$\epsilon /2$ neighbourhood, enlarging it by another $\epsilon /2$, to obtain the full claim. The metric normalization is essential here.

For $Q_n$, the graph diameter is $n$. A uniformly chosen vertex has size $Z_n$ with binomial distribution $\mathrm{Bin}(n,1/2)$, expectation $n/2$ and variance $n/4$. Fix $\delta >0$ and choose $C>0$ so that $1/(4C^2)<\delta$. By Chebyshev's inequality,For sufficiently large $n$, put $k_n=\lfloor n/2-C\sqrt{n}\rfloor \ge 0$. Every initial segment of measure at least $\delta$ contains the complete ranks through $k_n$. Its radius-$t$ graph neighbourhood therefore contains every rank at most $k_n+t$. Taking $t=\lfloor \epsilon n\rfloor$, the all-radius vertex-isoperimetric inequality in the discrete cube gives, uniformly for $|\mathcal{A}|\ge \delta 2^n$,The last Chebyshev's inequality is used only when its denominator has positive square root. These are precisely the radius-$\epsilon$ neighbourhoods in the normalized graph distance. The discrete cubes form a Lévy family. In fact, taking $t=\lfloor n^{3/4}\rfloor$ in the same bound makes the omitted measure $O(n^{-1/2})$ for each fixed $\delta$: a radius that is $o(n)$ already suffices.

For fixed $d\ge 1$, the nearest-neighbour grid $[n{]}^d$ has graph distance ${\sum }_i|x_i-y_i|$ and graph diameter $d(n-1)$. Take the slab ${\mathcal{A}}_n=\{x:x_1\le \lceil n/2\rceil \}$, whose uniform measure is at least one half. Its integer radius-$t$ graph neighbourhood is exactly $\{x:x_1\le \min (n,\lceil n/2\rceil +t)\}$: changing the first coordinate alone attains the minimum distance to the slab. For $0<\epsilon <1/(2d)$ and $t=\lfloor \epsilon d(n-1)\rfloor$, the omitted proportion tends toThus there is failure of concentration in fixed-dimensional grids, and the fixed-dimensional grids do not form a Lévy family with normalized graph distance.

On the real line, Prohorov's theorem states that a sequence of probability measures is uniformly tight if and only if every subsequence has a further subsequence converging weakly to a probability measure. Here uniformly tight means that, for every $\epsilon >0$, some compact interval contains mass at least $1-\epsilon$ for every measure in the sequence. This is relative sequential compactness for weak convergence of probability measures, not merely the existence of one convergent subsequence.

We first prove the characteristic-function tail bound. Put $c_0=1-\sin 1>0$. For $y\ne 0$,At $y=0$ the expression is zero by continuity. If $|y|\ge \lambda$, the supplied elementary bound, with $t=|y|/\lambda$, makes it at least $c_0$. The integrand is nonnegative, so Tonelli's theorem yieldsThus one valid universal constant isThe estimate controls spatial tails using only the characteristic function near zero.