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Uniform expectations require nonempty finite sets; assume this throughout the analytic argument. For real-valued functions on $X\times Y$, define the box norm byAll four variables are sampled as independent random variables with replacement, so repeated coordinates are included. The defining fourth power equalsIf it vanishes, every summand vanishes. In particular, the terms $y_0=y_1=y$ give ${\mathbb{E}}_{x}f(x,y{)}^2=0$ for each $y$, so $f=0$. Absolute homogeneity of a norm follows directly: $\Vert af{\Vert }_{\square }=|a|\Vert f{\Vert }_{\square }$.

For the triangle inequality, prove the mixed box Cauchy-Schwarz inequality. WriteSeparating the two $y$ averages and applying the Cauchy-Schwarz inequality in $(x_0,x_1)$ givesExpanding the square and reversing the order of finite sums,by a second Cauchy-Schwarz inequality. Thus $|\Lambda |\le {\prod }_{i,j}\Vert f_{ij}{\Vert }_{\square }$. Expand $\Lambda (f+g,f+g,f+g,f+g)$ into its sixteen multilinear terms and apply this bound to each term. Their upper bounds sum to $(\Vert f{\Vert }_{\square }+\Vert g{\Vert }_{\square }{)}^4$. Taking fourth roots yields the triangle inequality. Together with definiteness of a norm and absolute homogeneity of a norm, this proves the box norm is a norm. Complex functions require conjugates and are outside the printed real-valued convention.

For the bilinear correlation bound for the box norm, use the normalized L2 norm: let $B={\mathbb{E}}_{x,y}f(x,y)u(x)v(y)$ and $\Vert u{\Vert }_2^2={\mathbb{E}}_{x}u(x{)}^2$, $\Vert v{\Vert }_2^2={\mathbb{E}}_{y}v(y{)}^2$. The Cauchy-Schwarz inequality in $y$ givesThe remaining factor is ${\mathbb{E}}_{x,x^{\prime }}u(x)u(x^{\prime }){\mathbb{E}}_{y}f(x,y)f(x^{\prime },y)$. By the Cauchy-Schwarz inequality in $(x,x^{\prime })$, its magnitude is at mostTaking square roots proves

For the tripartite graph, assume its parts are nonempty and write $g_{XY},g_{YZ},g_{XZ}$ for its adjacency indicator functions. Let $f=g_{XY}-\alpha$, so $\Vert f{\Vert }_{\square }\le c$. The normalized triangle count isThe constant-degree hypothesis is exactly ${\mathbb{E}}_y{g}_{YZ}(y,z)=\beta$ for each $z$. Therefore the contribution of the constant $\alpha$ isFor fixed $z$, apply the bilinear correlation bound for the box norm with $u_z(x)=g_{XZ}(x,z)$ and $v_z(y)=g_{YZ}(y,z)$. Since these are indicator functions, $\Vert v_z{\Vert }_2^2=\beta$ and $\Vert u_z{\Vert }_2^2=d_X(z)$, the fraction of $X$ adjacent to $z$. Averaging and applying the Cauchy-Schwarz inequality in $z$ gives the sharper triangle counting with one box-uniform pair and constant opposite degree estimateMultiplying by $|X||Y||Z|$ yieldsThe middle bound also handles $\beta =0$ or $\gamma =0$, when the triangle count is zero.

To connect explicitly with the suggested expansion, put $f_2=g_{YZ}-\beta$, $f_3=g_{XZ}-\gamma$. Both have overall mean zero, and ${\mathbb{E}}_y{f}_2(y,z)=0$ for every $z$. Consequently ${\mathbb{E}}_{x,y,z}{f}_2(y,z)f_3(x,z)=0$; all terms with constant $\alpha$ reduce to $\alpha \beta \gamma$. The terms containing $f$ combine into the single error just estimated. This is where the exact degree assumption is used. If any part is empty the triangle count itself is zero, but the printed uniform expectations and edge density of a bipartite graph values would be undefined; the nonempty-parts convention must therefore be stated rather than dividing by zero.

Write $T=\dot{\gamma }$ and $D_t={\nabla }_T$. A Jacobi field is a vector field along the geodesic satisfyingThe curvature convention is that used in Question 1. A geodesic variation is a smooth map $F:(-ϵ,ϵ)\times [0,1]\to M$ with $F(0,t)=\gamma (t)$ whose $t$-curves are affinely parametrized geodesics. Torsion-freeness gives $D_s{\partial }_{t}F=D_t{\partial }_{s}F$. Differentiating $D_t{\partial }_{t}F=0$ and commuting covariant derivatives gives the Jacobi equation for ${\partial }_{s}F{|}_{s=0}$.

For the converse, let $a=J(0)$ and $b=D_{t}J(0)$. Choose a curve $p(s)$ with $p(0)=\gamma (0)$ and $p^{\prime }(0)=a$. parallel transport along it identifies its tangent spaces with $T_{\gamma (0)}M$. Set $v(s)=P_s(T(0)+sb)$, so $v(0)=T(0)$ and $D_{s}v(0)=b$. The geodesics with initial data $(p(s),v(s))$ give a variation $F(s,t)$. Smooth dependence on initial conditions and compactness of the original time interval ensure that this family is defined for all $t\in [0,1]$ after shrinking $ϵ$, even if $M$ is incomplete. Its variation field has the initial values $a,b$, so uniqueness for the linear Jacobi equation identifies it with $J$. This proves the realization of Jacobi fields by geodesic variations.

The endpoint-vanishing pointwise normal fields form a vector space, since their conditions and equation are linear. If $\gamma$ is nonconstant, the map $J\mapsto D_{t}J(0)$ is injective because $J(0)=0$ and zero initial derivative force the zero solution. Differentiating $⟨J,T⟩=0$ gives $D_{t}J(0)\perp T(0)$. Hence the endpoint-vanishing normal Jacobi fields have dimension at most n-1 bound isFor a constant geodesic, $D_t^{2}J=0$, and both zero endpoint values force $J=0$, so the bound still holds.

On the unit round $S^n$, let $\gamma (t)=\cos (\pi t)e_0+\sin (\pi t)e_1$. Its speed is $\pi$. The constant ambient vectors orthogonal to $e_0,e_1$ give $n-1$ independent parallel normal fields $E_j(t)$. Since $R(J,T)T={\pi }^{2}J$ on normal fields, the fields $J_j(t)=\sin (\pi t)E_j(t)$ satisfy the Jacobi equation and vanish at both antipodal endpoints. They attain dimension $n-1$.

For the last clause, interpret a closed geodesic as a nonconstant smoothly periodic geodesic. A constant loop has length zero and cannot be shortened. Parametrize ${\gamma }_0$ on $[0,1]$ at constant nonzero speed. parallel transport once around it fixes $T(0)$. Because the manifold is orientable, this transport preserves orientation; on the normal space of dimension $n-1$, which is odd, it lies in $SO(n-1)$. An odd-dimensional special orthogonal transformation has a fixed vector: nonreal eigenvalues pair with their conjugates, real eigenvalues are $\pm 1$, and determinant one in odd dimension forces an eigenvalue $+1$.

Transport such a nonzero normal fixed vector around the loop. It gives a periodic parallel normal field $V$, with $D_{t}V=0$ and $V(1)=V(0)$. Its jets also agree at the seam, so it is smooth as a field on the parametrizing circle. For small $s$, $F(s,t)={\exp }_{{\gamma }_0(t)}(sV(t))$ is a smooth variation through closed curves, providing their homotopy to ${\gamma }_0$.

For energy $E(s)=\frac{1}{2}{\int }_0^1|{\partial }_{t}F{|}^{2}dt$, the first variation vanishes at the closed geodesic. The permitted second variation of geodesic energy has no endpoint term for this periodic variation, soThus $E(s)<E(0)$ for small nonzero $s$. The Cauchy-Schwarz inequality gives $L(F_s{)}^2\le 2E(s)$, while constant speed gives $L({\gamma }_0{)}^2=2E(0)$. Consequently the instability of a closed geodesic in positive even-dimensional curvature yieldsThe deformation need not remain a geodesic or an embedded curve.

A generalized conformal metric is locally $d{s}_{\rho }=\rho (z)|dz|$, where $\rho$ is nonnegative and measurable and transforms as a length density under a holomorphic change of coordinate. Its area is $A_{\rho }={\int }_X{\rho }^{2}dxdy$. Use the usual convention of using locally rectifiable paths for a path family $\Gamma$, and put $L_{\rho }(\Gamma )={\inf }_{\gamma \in \Gamma }{\int }_{\gamma }\rho |dz|$. ThenThis is extremal length. Zeros and isolated singularities of an admissible density are allowed; requiring a smooth strictly positive Riemannian metric would unnecessarily restrict the definition. Line integrals have their extended nonnegative values; if an arbitrary family is supplied, use its members that are locally rectifiable paths. An empty path family has infinite infimal length, whereas a family containing a constant path has extremal length zero.

Both numerator and denominator scale quadratically when $\rho$ is multiplied by a positive constant. The coordinate transformation of the area element makes the quotient unchanged under conformal equivalence.

The normalization relevant later is worth deriving. On the conformal cylinder $(\mathbb{R}/\mathbb{Z})\times (0,M)$, let $\Gamma$ contain the loops going once around it. For the horizontal loop at height $t$, Cauchy-Schwarz inequality givesIntegrating in $t$ shows $L_{\rho }(\Gamma {)}^2\le A_{\rho }/M$. The constant density achieves equality, since every winding-one loop has Euclidean length at least one. HenceHere $M$ is the height divided by circumference, the conformal modulus of an annulus; the reciprocal is used for the family joining its boundary components.

Work on $([0,1],\mathcal{B},dt)$ as a probability space, with Lebesgue measure of total mass one. Let ${\mathcal{F}}_n$ be the filtration generated by the level-$n$ half-open dyadic cells together with the separate null cell $\{1\}$. Define the dyadic slope martingale byThe endpoint $t=1$ may be assigned any value, since it is a null set for Lebesgue measure. The two child slopes average to their parent slope, by telescoping the two increments. Therefore $\mathbb{E}[G_{n+1}\mid {\mathcal{F}}_n]=G_n$ almost everywhere, so this is a martingale. The Lipschitz condition gives $|G_n|\le K$ everywhere except possibly at the freely chosen endpoint, where we take zero.

Apply the Lp martingale convergence theorem with $p=2$. Its limit $g$ has $|g|\le K$ almost everywhere and $G_n\to g$ in L1 norm as well, by Cauchy-Schwarz inequality. Choose a measurable representative of $g$ and set it to zero on any exceptional null set; it is then a bounded measurable function on the entire interval.

Set $f_n(x)=f(0)+{\int }_0^x{G}_n(t)dt$. Telescoping at the grid points shows that $f_n$ is the linear interpolation of $f$ on the dyadic grid. The Lipschitz condition gives $\Vert f_n-f{\Vert }_{\infty }\le K{2}^{-n}\to 0$. AlsoThe two uniform limits coincide, giving the absolutely continuous function representationThe chosen bound holds for every $t$ after the null-set modification; the integral identity holds for every $x$ simultaneously.

Put $S=H\cdot M$ and $A_t={\int }_0^t{H}_s^{2}d[M{]}_s$. The candidate $A$ is continuous, adapted, nondecreasing, starts at zero and has integrable terminal value. We prove $S^2-A$ is a martingale, rather than assuming the desired quadratic variation of a stochastic integral.

The assumed boundedness of $M$ also makes $M^2-[M]$ a true martingale on every finite horizon. Indeed, apply its local martingale identity at a localizing sequence; bounded stopped squares and the Fatou lemma show that $[M{]}_t$ is integrable. The stopped expressions are then dominated by ${\sup }_s|M_s{|}^2+[M{]}_t$, so the dominated convergence theorem gives the unrestricted conditional identity on that horizon.

First let $H$ be a bounded simple predictable process. On each interval with coefficient $h$, an increment of $S$ is $h(M_t-M_a)$. The defining quadratic variation identity for $M$, together with the martingale property of $M$, makesa martingale for $t\ge a$. Expanding $S_t^2-S_a^2$ gives a term $2S_{a}h(M_t-M_a)$ and the square term above; their conditional expectations yield the desired identity on that interval. Integrability follows because the variables are square-integrable; concatenating the finitely many intervals gives $S^2-A$ a martingale.

For general bounded predictable $H$, choose bounded simple predictable processes $H^n\to H$ in $L^2(M)$, which is possible since they generate the predictable sigma-algebra. Set $S^n=H^n\cdot M$ and $A_t^n={\int }_0^t(H_s^n{)}^{2}d[M{]}_s$. The Itô isometry and Doob L2 maximal inequality imply $S_t^n\to S_t$ in $L^2$; the Cauchy-Schwarz inequality gives $(S_t^n{)}^2\to S_t^2$ in $L^1$ andPass the conditional expectation identity for $(S^n{)}^2-A^n$ to the $L^1$ limit. Thus $S^2-A$ is a martingale. Uniqueness in the defining quadratic variation property now gives

Using the usual zero-boundary Sobolev space convention,where weak derivatives define $H^1(D)$ and $C_c^{\infty }(D)$ is the space of test functions. In the Gaussian free field convention, the same notation often denotes the Dirichlet energy space, the completion in the gradient norm alone. On bounded domains the Poincare inequality makes the two definitions equivalent; on unbounded domains one must distinguish them. The following gradient-pairing argument applies in either setting whenever the energy completion is realized as weak functions.

Identify $H_{\mathrm{supp}}=H_0^1(U)$ with a vector subspace of $H_0^1(D)$ by zero extension of H01. Approximating by test functions in $U$ shows that this is an isometric embedding in the inhomogeneous Sobolev norm and also in the Dirichlet inner product norm. In particular, arbitrary irregularity of $\partial U$ causes no additional boundary term.

DefineThese are weakly harmonic Sobolev functions. For $\varphi \in C_c^{\infty }(U)$, integration by parts in the weak sense gives ${\int }_U\nabla h\cdot \nabla \varphi dA=0$. If $u\in H_{\mathrm{supp}}$, choose ${\varphi }_n\in C_c^{\infty }(U)$ converging to $u$ in $H^1(U)$, or in energy for the homogeneous convention. The Cauchy-Schwarz inequality givesHence the orthogonality of supported and harmonic Dirichlet functions isBoth are linear vector subspaces. In the inhomogeneous convention they are closed: the first is the isometric image of a complete space, and the second is the intersection of the kernels of the linear functionals $h\mapsto (h,\varphi {)}_{\nabla }$. No spanning assertion is needed. The PDF contains this orthogonality statement; the TeX has badly corrupted it into an assertion about openness.

Use the normalizationfor the Lasso. A centered sub-Gaussian random variable with parameter $\sigma$ satisfies $\mathbb{E}{e}^{tW}\le e^{{\sigma }^2{t}^2/2}$ for all $t\in \mathbb{R}$; for a noncentered variable apply this definition to $W-\mathbb{E}W$. With fixed $X$, independence of the errors givesThe exponential Markov bound, optimized over $t$ separately for the two signs, gives $\mathrm{Pr}(|X_j^T\epsilon |/n>u)\le 2e^{-n{u}^2/(2{\sigma }^2)}$. Take $u=\lambda /2=\sigma A\sqrt{\log p/n}$ and apply the union bound over columns:The specified positive tuning parameter requires $p\ge 2$; at $p=1$ its formula is zero. For small $A$ the probability lower bound can be negative and hence uninformative. Although centering makes the centered errors dependent, $X^{T}1=0$ means $X^T(\epsilon -\bar{\epsilon }1)=X^T\epsilon$, so the preceding proof uses the original independent errors, not independence after centering.

The Karush-Kuhn-Tucker conditions, using the subdifferential of the L1 norm, areConsequently, for nonempty $B\subseteq \hat{S}$, on $\Omega$,This establishes the required active-set inequality without any rank condition on $X$.

For $s>0$, a sufficient Compatibility condition for the Lasso isIndeed, for $\delta =\hat{\beta }-{\beta }^0$, comparison of the two Lasso objective values and the score bound on $\Omega$ give the Basic inequality for the LassoIt follows that $\delta$ satisfies the Lasso cone condition and $\Vert {\delta }_S{\Vert }_1\le \sqrt{s{P}_E}/\varphi$. Thus $P_E\le 3\lambda \sqrt{s{P}_E}/\varphi$, including the trivial $P_E=0$ case. This proves the slightly stronger $P_E\le 9{\lambda }^{2}s/{\varphi }^2$, and in particular the requested $16{\lambda }^{2}s/{\varphi }^2$. If $s=0$, the same basic inequality forces $\hat{\beta }=0$ on $\Omega$; no nonempty-support compatibility constant is needed.

For the support-size argument, use the stated prediction bound with constant $16$. Set $b=|B|$. By the Cauchy-Schwarz inequality and the definition of the sparse maximum eigenvalue,Combining with the lower bound and squaring gives $\boxed{b\le 64{\kappa }_b^{2}s/{\varphi }^2}$ for every nonempty $B\subseteq \hat{S}$. If $m_{\ast }<\infty$ and $\hat{s}\ge m_{\ast }$, choose $B$ of size $m_{\ast }$; this contradicts the strict inequality defining $m_{\ast }$. If $m_{\ast }=\infty$, $\hat{s}\le p<\infty$ proves the assertion directly. Hence $\boxed{\hat{s}<m_{\ast }}$ in both cases.

When $1<m_{\ast }\le p$, minimality and monotonicity of the sparse maximum eigenvalues giveWhen $m_{\ast }=1$, the first result gives $\hat{s}=0$, so the same final bound holds. Thus the finite-index conclusion is $\boxed{\hat{s}\le 64{\kappa }_{m_{\ast }}^{2}s/{\varphi }^2(m_{\ast }<\infty )}$.

There is a genuine domain omission in the printed last assertion: the definition of ${\kappa }_m$ only makes sense for $1\le m\le p$, so ${\kappa }_{m_{\ast }}$ is undefined when $m_{\ast }=\infty$. This case can occur: take centered orthogonal columns with $X^{T}X/n=I_p$, $n\ge p+1$, $s=p$, and ${\varphi }^2=1$. Then ${\kappa }_m^2=1$ for every available $m$ and none exceeds $64p$, so $m_{\ast }=\infty$. A universally defined replacement, from $B=\hat{S}$ and monotonicity, isAlternatively, explicitly extend the definition by ${\kappa }_m={\kappa }_p$ for $m>p$, including infinity. Under that added convention the printed final expression is meaningful and follows from this replacement bound. Neither convention nor finiteness should be silently assumed.

Let $p=dP/d\nu$ and $q=dQ/d\nu$ be Radon-Nikodym derivatives. Use the following normalizations of the Kullback-Leibler divergence, total variation distance and unnormalized Hellinger distance:We use $0\log (0/q)=0$ and assign infinite Kullback-Leibler divergence if $Pot\ll Q$, where $\ll$ denotes absolute continuity of measures. Common domination alone also suffices for the inequalities with the extended-value convention. The negative part of the logarithmic integral is integrable: on $p<q$, $p\log (q/p)\le q-p\le q$. Thus the extended-value integral defining the Kullback-Leibler divergence is well-defined. These expressions do not depend on which common dominating measure is chosen. The Kullback-Leibler divergence is generally asymmetric and therefore fails a defining property of a metric.

For completeness, the integral formula for total variation distance follows by taking $A=\{p\ge q\}$: since $\int (p-q)d\nu =0$, the positive and negative parts of $p-q$ have the same integral, and this set attains the supremum.

Put $a=\int \sqrt{pq}d\nu \le 1$, by the Cauchy-Schwarz inequality. A second application givesThe logarithmic inequality in the hint implies $\log t\le t-1$ for $t>0$. Apply it to $t=\sqrt{q/p}$ on $\{p>0\}$:Multiplying by $p$ and integrating provesIf $q=0<p$ on a set of positive $P$-measure, the right side is finite but the Kullback-Leibler divergence is infinite, so the inequality is immediate. ThusThis total variation–Hellinger–relative entropy inequality requires the stated Hellinger distance normalization for its first constant. With the alternative convention ${\bar{h}}^2=\frac{1}{2}\int (\sqrt{p}-\sqrt{q}{)}^2$, the correct comparison is $\mathrm{TV}\le \sqrt{2}\bar{h}$ and $\bar{h}\le \sqrt{\mathrm{KL}/2}$. In particular, $P=\mathrm{Bernoulli}(0.6)$ and $Q=\mathrm{Bernoulli}(0.4)$ have $\mathrm{TV}=0.2>\bar{h}\approx 0.1421$.

Here is a testing form of the Le Cam two-point lemma. Suppose two parameter values ${\theta }_0,{\theta }_1$ in a metric space have separation $\Delta =d({\theta }_0,{\theta }_1)>0$, and let $P_0,P_1$ be the corresponding observation laws. Every estimator $T$ satisfiesTo prove it, classify $T$ by the closer of the two parameters, breaking a tie in favour of one. Let $A=\{d(T,{\theta }_1)\le d(T,{\theta }_0)\}$. On $A$, the triangle inequality implies $d(T,{\theta }_0)\ge \Delta /2$, while on $A^c$ it implies $d(T,{\theta }_1)>\Delta /2$. Hence the sum of the two probabilities in the display is at leastAt least one is at least half the sum, proving the claim. It also gives a lower bound $\Delta (1-\mathrm{TV})/4$ for the maximum expected metric loss.

For absolute-error loss, or any loss given by a metric, there is a stronger expectation form. Take densities $p_0,p_1$ relative to $m=P_0+P_1$ and use the triangle inequality under their overlap:Therefore the Le Cam lower bound under absolute-error loss isBoth arguments are valid for randomized estimators as well, by adjoining the independent randomization to the observation; doing so does not change total variation distance.

For the normal distribution model, assume the usual nondegenerate convention $\sigma >0$ and use the full product observation laws. ChooseFor one observation, its Gaussian log likelihood ratio isUnder the first law, ${\mathbb{E}}_0(X-{\mu }_1{)}^2={\sigma }^2+({\mu }_1-{\mu }_0{)}^2$ and ${\mathbb{E}}_0(X-{\mu }_0{)}^2={\sigma }^2$. Thus the Kullback-Leibler divergence between normal distributions isFor the independent and identically distributed random variables, the joint log likelihood ratio is the sum of the individual ones, so the Kullback-Leibler divergence of the two full statistical samples is $n({\mu }_1-{\mu }_0{)}^2/(2{\sigma }^2)=1/4$. The comparison just proved implies $\mathrm{TV}(P_0,P_1)\le 1/2$. Applying the expectation form of the Le Cam two-point lemma givesThis is a Gaussian location minimax lower bound under absolute-error loss. The constant may depend on the fixed known $\sigma$, but neither on the estimator nor on $n$. If a signed nonzero scale were used, the same result holds with $|\sigma |$ in place of $\sigma$. If zero noise were allowed, $X_1=\mu$ would estimate $\mu$ exactly, so a positive lower bound would be false; positive variance is necessary.

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.

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.

The Cameron-Martin space of a Gaussian random variable in a Banach space is most conveniently defined using the first Gaussian chaos. Work over the real separable Banach space $B$, with continuous dual space $B^{\ast }$. For a centered Gaussian random variable in a Banach space, defineThe set inside the closure is already a linear subspace. Every $g\in \mathcal{G}$ is a centered Gaussian random variable. The Bochner integral defining $S$ exists: Fernique's theorem states that for some $a>0$, $\mathbb{E}{e}^{a\Vert X{\Vert }_B^2}<\infty$, and then Cauchy-Schwarz inequality gives $\mathbb{E}\Vert Xg{\Vert }_B<\infty$.