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Choose a directed path ${\eta }_{xy}$ of positive-probability transitions for each ordered pair of states in a finite reversible Markov chain. Write $Q(e)=\pi (u)P(u,v)$ for a directed edge $e=(u,v)$, and let $N_e(\eta )$ count its occurrences. DefineThe Cauchy-Schwarz inequality bounds $(f(x)-f(y){)}^2$ by $|{\eta }_{xy}|$ times the sum of squared differences along the path. Multiply by $\pi (x)\pi (y)/2$, sum and interchange finite sums. The identity ${\mathrm{Var}}_{\pi }(f)=\frac{1}{2}{\sum }_{x,y}\pi (x)\pi (y)(f(x)-f(y){)}^2$ proves the bound. This convention counts directed edges and retains the factor $1/2$ in the Dirichlet form of a Markov chain.

For independent coordinate flips at rate $1/2$, a Walsh character $w_A$ is an eigenfunction with eigenvalue $-|A|$. The generator is self-adjoint and negative semidefinite. Its Dirichlet form of a Markov chain is $-\mathbb{E}[fLf]=\frac{1}{4}{\sum }_i\mathbb{E}(f({\omega }^{(i)})-f(\omega ){)}^2$.

For a transient weighted graph, $H_0$ is the completion of the finitely supported functions in the norm $\Vert f{\Vert }_{H_0}=\mathcal{E}(f,f{)}^{1/2}$. Its elements have finite Dirichlet energy and vanish at infinity in the energy sense.

A Dirichlet form on a real $L^2$ space is a densely defined closed nonnegative symmetric bilinear form with the Markov contraction property. In the real case, truncating $u$ to $(0\vee u)\wedge 1$ keeps it in the form domain and cannot increase $\mathcal{E}(u,u)$. The Dirichlet form of a Markov chain is a discrete energy example; Laplacian energy on a region supplies a continuous example.

Let $P$ be the transition matrix of a finite irreducible Markov chain with at least two states, reversible with stationary distribution $\pi$. Its Dirichlet form of a Markov chain isThe Discrete Poincare inequality states that ${\mathrm{Var}}_{\pi }f\le C\mathcal{E}(f,f)$ for every real $f$. Its optimal constant is $C=(1-{\lambda }_2{)}^{-1}$, where ${\lambda }_2$ is the second largest eigenvalue of $P$. Indeed, detailed balance makes $P$ self-adjoint on $L^2(\pi )$. Expand $f-\pi (f)={\sum }_{j\ge 2}{a}_j{v}_j$ in an orthonormal eigenbasis. Then ${\mathrm{Var}}_{\pi }f=\sum a_j^2$ and $\mathcal{E}(f,f)=\sum (1-{\lambda }_j)a_j^2$, proving the inequality and its sharpness.

The useful constructive version is the canonical paths Poincare bound. Choose transition graph paths ${\gamma }_{xy}$ between each ordered pair, write $Q(u,v)=\pi (u)P(u,v)$, and definewhere $N_{uv}$ counts occurrences of the directed transition. The Cauchy-Schwarz inequality along each path in a graph gives $(f(x)-f(y){)}^2\le |{\gamma }_{xy}|{\sum }_{{\gamma }_{xy}}(f(u)-f(v){)}^2$. Multiply by $\pi (x)\pi (y)/2$ and sum. The identity ${\mathrm{Var}}_{\pi }f=\frac{1}{2}{\sum }_{x,y}\pi (x)\pi (y)(f(x)-f(y){)}^2$ proves ${\mathrm{Var}}_{\pi }f\le \rho \mathcal{E}(f,f)$.

For a lazy Markov chain this also proves the mixing time estimateTo see this, the centred initial density has $L^2(\pi )$ norm at most $\pi (x{)}^{-1/2}$; laziness puts all nonconstant eigenvalues between zero and $1-{\rho }^{-1}$. Their powers contract that norm, and the Cauchy-Schwarz inequality bounds total variation distance by half the density's $L^2(\pi )$ norm.

Here is a precise sufficient density condition for the permanent application: every row and every column has more than $n/2$ ones. Interpret the permanent of a matrix as the number $m_n$ of perfect matchings in its bipartite graph, with $n$ vertices per side. Write $m_k$ for the number of $k$-edge matchings in a graph.

First prove the short augmenting paths in a dense bipartite graph bound. Given a nonperfect matching in a graph, choose unmatched $u,v$ on opposite sides. If either has an unmatched neighbour, augment with one edge. Otherwise the matched partners of $N(u)$ and the neighbours $N(v)$ are subsets of the same $n$-element side whose total size exceeds $n$. Their intersection gives an alternating path in a graph $u,b,a,v$ of length three, with $ab$ a matching edge. Repeating proves $m_n>0$. Choose augmentations by a fixed deterministic rule. A reverse augmentation is specified by at most four vertices, so a next-size matching in a graph has at most $C_0=2n^4$ preimages. Therefore

Use the matching generating function $Z(\lambda )={\sum }_{k=0}^n{m}_k{\lambda }^k$ and the weighted matching Markov chain on all matchings in a graph, with stationary weights ${\pi }_{\lambda }(M)={\lambda }^{|M|}/Z(\lambda )$. It stays put with probability $1/2$; otherwise it chooses a graph edge uniformly and proposes its deletion, its addition if both endpoints are unmatched, or its exchange for the unique incident matching edge if precisely one endpoint is matched. Other proposals stay put. Accept with probability $\min (1,{\lambda }^{|M^{\prime }|-|M|})$. Reverse proposals use the opposite exchanged edge, so detailed balance holds. Deletions connect every state to the empty matching in a graph.

For completeness, the canonical paths for the weighted matching chain give a polynomial bound without assuming a mixing theorem. The symmetric difference $I△F$ consists of alternating graph paths and graph cycles. Order its components deterministically, orient open components from their smaller endpoint, and convert them from $I$ to $F$ by moving an unmatched endpoint with exchanges. If the first edge belongs to $I$, delete it first. For a graph cycle, delete one $I$-edge, move the unmatched endpoint around the opened cycle, then insert the last $F$-edge. Each intermediate state is a matching in a graph, and each entire transition path in a graph has length at most $4n$.

Fix a directed transition with initial state $M$. The complementary edge set $K=I△F△M$ agrees with one of the two matchings in a graph outside the active component and has at most two degree-two vertices on that component. Remove at most two offending edges to obtain a matching in a graph $K^{\prime }$. Store those edges and one bit specifying the initial alternating colour of the active component. This encoding is injective for the fixed transition: it restores $K$ and $I△F=M△K$, identifies the active component from the changed edges, and recovers the two colours using the fixed component order. Completed components have $F$'s colour in $M$, pending components have $I$'s colour, and the stored bit resolves the active component. Common edges are $M\cap K$.

There are at most $2(m+1{)}^2$ auxiliary records per $K^{\prime }$, where $m$ is the number of graph edges. If $d=|K|-|K^{\prime }|\le 2$, thenAlso $Q(M,M^{\prime })\ge {\pi }_{\lambda }(M)/(2m\max (\lambda ,{\lambda }^{-1}))$. Sum over the encodings and use the path length bound to obtainThis is polynomial whenever $\lambda$ and ${\lambda }^{-1}$ are polynomially bounded. Starting at the empty matching in a graph, $\log (1/{\pi }_{\lambda }(\varnothing ))=\log Z(\lambda )\le m\log (1+\lambda )$, so the Discrete Poincare inequality gives polynomial-time sampling to any prescribed total variation distance. The addition/deletion/exchange construction is also described in Section 12.4 of Jerrum and Sinclair's survey.

It remains to turn sampling into a relative count; merely sampling perfect matchings would not do that. For $0<\epsilon \le 1/2$, choose ${\lambda }_0=\epsilon /(8m)$ and ${\lambda }_{\ast }=8C_0/\epsilon$. The elementary bounds above giveChoose a schedule ${\lambda }_0<{\lambda }_1<\cdots <{\lambda }_J={\lambda }_{\ast }$ whose successive ratios are at most $1+1/n$. It has $J=O(n\log (n/\epsilon ))$ stages. The annealing ratios for a matching generating function satisfyEach observable lies in $[1,e]$, so the Hoeffding inequality gives relative accuracy $\eta =\epsilon /(10J)$ with $O({\eta }^{-2}\log (J/\delta ))$ independent samples per stage. Run independent copies of the chain from the empty matching in a graph; make the total variation distance per sample at most $\delta /(3JK)$, where $K$ is that sample count. The coupling characterization of total variation distance and the union bound make the entire sample collection coincide with exact stationary samples except with probability at most $\delta /3$. Choose the sample-count constant so all stationary sample means are accurate except with probability at most $\delta /3$.

Let $\hat{R}$ be the product of the sample means, and output $\hat{p}=\hat{R}/{\lambda }_{\ast }^n$. On the successful event its multiplicative error relative to $Z({\lambda }_{\ast })/Z({\lambda }_0)$ lies between $e^{-\epsilon /5}$ and $e^{\epsilon /5}$. Combining the two endpoint bounds provesActivities, sample counts, simulation lengths and rational bit lengths are polynomial in $n,{\epsilon }^{-1},\log ({\delta }^{-1})$. Thus this is a fully polynomial randomized approximation scheme. Small dimensions can be evaluated directly.

Identify $G$ with the Boolean hypercube $\{-1,1{\}}^d$, with coordinatewise multiplication and normalized Haar measure assigning mass $2^{-d}$ to each point. Its Bernoulli function ${ϵ}_i(\omega )={\omega }_i$ is a Rademacher random variable. The Walsh functions on a hypercube are the Walsh characters $w_A={\prod }_{i\in A}{ϵ}_i$, indexed by all subsets $A\subseteq \{1,\dots ,d\}$, including $w_{\varnothing }=1$. Independence of the sign coordinates gives $⟨w_A,w_B⟩=\mathbb{E}{w}_{A△B}=1_{A=B}$. There are $2^d$ characters, equal to the dimension of $L^2(G)$, so they form an orthonormal basis.

Flipping coordinate $i$ negates $w_A$ exactly when $i\in A$. Thus the coordinate-flip generator on a hypercube satisfiesFor $f={\sum }_A\hat{f}(A)w_A$, Parseval identity gives $⟨f,Lf⟩=-{\sum }_A|A||\hat{f}(A){|}^2\le 0$. Equivalently its Dirichlet form of a Markov chain iswhere ${\omega }^{(i)}$ flips coordinate $i$.

Set $S(\omega )={\sum }_i{a}_i{ϵ}_i(\omega )$ and $F(\omega )=\Vert S(\omega )\Vert$. Since $F(-\omega )=F(\omega )$, its Walsh expansion contains only sets $A$ of even size. Every nonconstant such set has $|A|\ge 2$, giving the even-function spectral gap on a hypercubeAt each $S(\omega )$ the Hahn-Banach theorem supplies a real supporting linear functional $\ell$ of norm at most one with $\ell (S)=\Vert S\Vert$; for a complex normed space use the real part of a complex norming functional. At $S=0$ take $\ell =0$. Convexity of the norm gives $F({\omega }^{(i)})-F(\omega )\ge \ell (S({\omega }^{(i)})-S(\omega ))$. Since $LS=-S$, sum these inequalities to get $LF\ge -F$. Therefore $\mathcal{E}(F)=-\mathbb{E}[FLF]\le \mathbb{E}{F}^2$. Combining the two bounds yields the sharp Rademacher second-moment inequalityNo smoothness of the norm is required. The constant $2$ is sharp: two equal nonzero real coefficients give modulus $0$ or $2|a|$ with equal probabilities. This is the second-versus-first moment case of the Kahane-Khintchine inequality for arbitrary normed vector spaces.

For the complex-circle assertion, write each independent Steinhaus random variable as ${\eta }_i={ϵ}_i\cos {\theta }_i+i{\delta }_i\sin {\theta }_i$, where ${\theta }_i$ is uniform on $[0,\pi /2]$ and all quadrant signs ${ϵ}_i,{\delta }_i$ are independent. Conditional on the angles, ${\sum }_i{a}_i{\eta }_i$ is a Rademacher sum with $2d$ complex coefficients $a_i\cos {\theta }_i$ and $i{a}_i\sin {\theta }_i$. Its conditional second moment is ${\sum }_i|a_i{|}^2$, independent of the angles. The preceding inequality gives its conditional first moment at least $\sqrt{\frac{1}{2}{\sum }_i|a_i{|}^2}$. Average over the angles and square to obtain the Steinhaus first-moment lower bound

On a finite state space with at least two states, as in the surrounding questions, for a real function write $\pi (f)={\sum }_x\pi (x)f(x)$ and ${\mathrm{Var}}_{\pi }(f)={\sum }_x\pi (x)(f(x)-\pi (f){)}^2$. The Poincare inequality for a reversible Markov chain with constant $C$ iswhere the Dirichlet form of a Markov chain isThe associated spectral gap is the ordinary gap $\gamma =1-{\lambda }_2$, with variational formulaEquivalently minimize $\mathcal{E}(f,f)$ subject to $\pi (f)=0$ and $⟨f,f{⟩}_{\pi }=1$. Thus the Poincare inequality for a reversible Markov chain holds precisely for $C\ge 1/\gamma$, and the optimal constant is $1/\gamma$. It controls the ordinary rather than absolute spectral gap, so negative eigenvalues do not invalidate it. If the unspecified state space in this part is infinite, the Poincare inequality for a reversible Markov chain and the variational infimum still make sense on nonconstant functions in $L^2(\pi )$, using the self-adjoint operator on that Hilbert space; the resulting spectral gap need not be represented by an actual second eigenvalue. The finite-state formula above is the intended specialization.

For a finite irreducible Markov chain satisfying detailed balance, define the directed-edge capacity $Q(u,v)=\pi (u)P(u,v)$. For every ordered pair $(x,y)$ choose a path ${\eta }_{xy}$ from $x$ to $y$ using only positive-capacity edges; take the empty path for $x=y$. Let $N_e(\eta )$ count occurrences of the directed edge $e$. The canonical paths Poincare bound says thatis a valid Poincare inequality for a reversible Markov chain constant. Simple paths give $N_e\in \{0,1\}$; allowing repetitions requires the multiplicities shown.

To prove the theorem, expand the variance using two independent samples from $\pi$:Telescoping along ${\eta }_{xy}$ and applying the Cauchy-Schwarz inequality givesInterchange the finite sums, then use the definition of $\rho$:Consequently $\boxed{{\mathrm{Var}}_{\pi }(f)\le \rho \mathcal{E}(f,f),\gamma \ge 1/\rho }$. All edges here are directed, so the $1/2$ in the Dirichlet form of a Markov chain is retained consistently.

Use $⟨f,g{⟩}_{\pi }={\sum }_x\pi (x)f(x)g(x)$. The Dirichlet form of a Markov chain isThe equality follows by expanding the square and using row sums one and the stationary distribution identity. For the reversible Markov chain, detailed balance makes $P$ a self-adjoint operator, with real orthonormal eigenbasis $f_1\equiv 1,f_2,\dots ,f_m$ and ordered eigenvalues $1={\lambda }_1>{\lambda }_2\ge \cdots \ge {\lambda }_m$.

Adding a constant changes neither the variance nor the Dirichlet form of a Markov chain. Expand the centered function as $f-\pi (f)={\sum }_{j\ge 2}{a}_j{f}_j$. ThenFor every nonconstant $f$ the quotient is at least $1-{\lambda }_2$, and equality is attained at $f=f_2$. This provesEquivalently minimize the Dirichlet form of a Markov chain over centered unit-norm functions. This is the variational characterization underlying the Poincare inequality for a reversible Markov chain, and does not require the transition matrix to have nonnegative eigenvalues. For a one-state chain there is no nonconstant test function; the displayed minimization concerns $m\ge 2$.

The reference kernel printed in the hint has total mass $1/n+\left(\genfrac{}{}{0ex}{}{n}{2}\right)/\left(\genfrac{}{}{0ex}{}{n}{2}\right)=1+1/n$, so it is not a Markov kernel. We explicitly replace it by the normalized random transposition shuffle: choose two positions independently and uniformly and swap them. Its probabilities areThese sum to one, and its ordinary spectral gap is $\tilde{\gamma }=2/n$, the value intended in the hint. This value can also be certified by the Aldous spectral gap theorem: the single-label generator has rate $2/n^2$ between distinct positions, so on centered functions it acts as multiplication by $-2/n$. Both shuffles have the same uniform distribution on a finite set as stationary distribution.

We use the Canonical paths comparison theorem in the following precise form. For two finite reversible Markov chains with the same positive stationary distribution $\pi$, route each directed transition $(x,y)$ of $\tilde{P}$ along a positive-capacity path ${\eta }_{xy}$ of $P$. Ifthen $\tilde{\mathcal{E}}(f,f)\le A\mathcal{E}(f,f)$ and $\gamma \ge \tilde{\gamma }/A$. Indeed, telescope each difference along its path, apply the Cauchy-Schwarz inequality, and sum with weights $\pi (x)\tilde{P}(x,y)/2$. The load definition gives the Dirichlet form of a Markov chain inequality; the common variance denominator then gives the spectral gap inequality. Identity transitions use empty paths.

Put $s_k=(k,k+1)$. A transposition $(i,j)$ is routed using the wordof length $L_{ij}=2(j-i)-1\le 2n$. Each generator occurs at most twice; write its occurrence count as $N_k(i,j)\le 2$. For a fixed directed adjacent edge $(\sigma ,\sigma s_k)$ and each occurrence of $s_k$ in a word, there is exactly one starting permutation whose translated word crosses this edge at that occurrence. This follows because right multiplication by the prefix is a bijection of the symmetric group. Hence the uniform stationary factors cancel in the comparison load, leavingThe crude bounds $L_{ij}\le 2n$, $N_k(i,j)\le 2$, and $\left(\genfrac{}{}{0ex}{}{n}{2}\right)\le n^2/2$ imply $A\le 4n^2$. ThereforeOnly pairs with $i\le k<j$ contribute, and there are $k(n-k)\le n^2/4$ of these. Keeping this restriction gives the stronger $A\le 2n^2$ and $\gamma \ge 1/n^3$. Neither comparison bound needs the exact adjacent-shuffle spectral gap; the normalization repair of the reference kernel is essential.

For a path $\Gamma$,by the Cauchy-Schwarz inequality. Average over ${\nu }_{xy}$, multiply by $\tilde{Q}(x,y)$, and sum. Reversing the order of summation in the definition of the two Dirichlet forms givesLet $M={\max }_x\pi (x)/\tilde{\pi }(x)$. The variational formula for variance givesTake a nonconstant eigenfunction attaining the Rayleigh quotient $\gamma$ for $P$. Then

By the stationary distribution property, $X_1$ and $X_2$ have the same marginal distribution $\pi$. Expanding the square givesThis is the probabilistic representation of the Dirichlet form of a Markov chain.

The Dirichlet form of a Markov chain isWhen $P$ is reversible, this equals $⟨f,(I-P)g{⟩}_{\pi }$.

The Dirichlet energy space with zero boundary at infinity iswhere $C_c(G)$ is the vector space of finitely supported real functions on the vertices. Transience makes the Dirichlet form of a Markov chain positive definite on this completion.

One Markov kernel dominates another in Peskun ordering when it assigns at least as much transition probability to every measurable set not containing the current state. For reversible kernels this increases Dirichlet energy and cannot decrease the spectral gap.

For a finite irreducible Markov chain with at least two states satisfying detailed balance, its Dirichlet form of a Markov chain obeysExpand $f-\pi (f)$ in an orthonormal eigenbasis. The denominator is ${\sum }_{j\ge 2}{a}_j^2$, and the numerator is ${\sum }_{j\ge 2}(1-{\lambda }_j)a_j^2$. Their ratio is at least $1-{\lambda }_2$ and equality holds for the second eigenfunction. Thus the optimal Poincare inequality for a reversible Markov chain constant is $1/\gamma$, for the ordinary rather than absolute spectral gap.