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For $v=\log u_{\epsilon }$, Cauchy-Schwarz inequality and part (a.ii) implyThe John-Nirenberg inequality therefore supplies $p_0=p_0(n,\mu )>0$ such that, with $v_{B_1}$ denoting the average,Since $e^{p_0(v-v_{B_1})}$ and $e^{-p_0(v-v_{B_1})}$ are each bounded by this integrand,

For $s\in {\mathbb{F}}_p^n$, letThe number $T$ of ordered three-term arithmetic progressions in $A$ isBy the Cauchy-Schwarz inequality,The last sum is the additive energy of $A$, namely the number of additive quadruples. If $T\ge \eta |A{|}^2$, then

The Kunita-Watanabe inequality applied to the continuous local martingales $M,N$ gives directlyEquivalently, with the clock and densities from part (i), positivity of the covariance matrix gives $|c|\le \sqrt{ab}$, and the Cauchy-Schwarz inequality gives

Apply the realized absolute covariation theorem to each dyadic partition of an interval. More explicitly, use the continuous increasing clock $C=[M]+[N]$ and the Radon-Nikodym theorem to writeFor each time, let $(U,V)$ have the centered bivariate normal distribution with covariance matrix $\left(\begin{array}{cc}a& c\\ c& b\end{array}\right)$, and defineThis process is continuous and increasing. To prove convergence, localize $M,N$, represent the pair as stochastic integrals against a two-dimensional Brownian motion, and approximate the integrands in $L^2(dC)$ by bounded step previsible processes. For step integrands, the result is the weak law of large numbers applied on each block to independent Gaussian random variables. The Burkholder-Davis-Gundy inequality and the Cauchy-Schwarz inequality make the error uniform on each compact interval in probability. Consequentlyin the sense of uniform convergence on compacts in probability.

PutThe numerator is the linear kernel, while the reciprocal of the denominator is the kernel from part c applied to $\Vert x{\Vert }^2$ and $\Vert y{\Vert }^2$. The product of positive-semidefinite kernels therefore shows that $a$ is a positive-semidefinite kernel. The Cauchy-Schwarz inequality and the arithmetic-geometric mean inequality give $|a(x,y)|\le 1/2$, soEach power is positive semidefinite by the Schur product theorem, and the convergent sum is positive semidefinite.

Moreover $k_2(x,x)=1$. If $\Phi$ is its canonical feature map, thenThe feature-space norm gives symmetry and the triangle inequality. Finally, $d(x,y)=0$ implies $k_2(x,y)=1$, hence $a(x,y)=1/2$ andwhich is equivalent to $\Vert x-y{\Vert }^2=0$. Thus $d$ is a metric rather than merely a pseudometric.

Part b places $\delta =\hat{\beta }-{\beta }^0$ in the Lasso cone condition. Keeping the prediction-error term in the same argument giveswhere the second step is the Cauchy-Schwarz inequality. The restricted eigenvalue condition givesfor nonzero $\delta$ in this cone. Division by $\Vert \delta {\Vert }_2$ provesThe result is immediate when $\delta =0$.

Condition on the independent external sample, and abbreviate ${\delta }_{\pi }={\hat{\pi }}_m-\pi$ and ${\delta }_{\mu }={\hat{\mu }}_m-\mu$. Expanding the plug-in estimator around the true nuisance functions giveswhere $\sqrt{n}{R}_{n,m}=o_p(1)$ because the two nuisance mean squared errors vanish, the evaluation sample is independent of the nuisance fits, $A$ is bounded, and $Y$ has bounded support. The linear term is the empirical average of the influence function, so the central limit theorem givesThe remaining bias obeys the Cauchy-Schwarz inequalityThus the claimed conclusion follows under the standard product-rate conditionequivalently $\sqrt{n}\sqrt{\mathrm{MSE}({\hat{\pi }}_m)operatornameMSE({\hat{\mu }}_m)}\to 0$. Slutsky theorem then yields

As printed, the paper instead assumes only $\sqrt{n}\mathrm{MSE}({\hat{\pi }}_m)\mathrm{MSE}({\hat{\mu }}_m)\to 0$, which is insufficient with its stated definition of MSE. For example, deterministic nuisance errors of size $n^{-3/16}$ have both MSEs equal to $n^{-3/8}$ and satisfy the printed condition, while the scaled product bias is $\sqrt{n}{n}^{-3/8}=n^{1/8}$. The result therefore requires the stronger condition above, or “MSE” in the printed rate must be read as root mean squared error.

The Quantum de Finetti theorem says that the fixed-$k$ reduced density matrix of an exchangeable $N$-particle state approachesas $N\to \infty$. Finite versions bound the trace norm error by a constant of order $d_{\mathrm{loc}}^{2}k/N$. On a bipartite lattice, the corresponding two-sublattice form is a mixtureThis is the mean-field ansatz from the quantum de Finetti theorem.

The bond energy is a linear functional of ${\rho }_{AB}$. A convex combination cannot have energy below its lowest product component, so it remains only to minimizeThe Cauchy-Schwarz inequality gives ${\mathbf{r}}_A\cdot {\mathbf{r}}_B\ge -|{\mathbf{r}}_A||{\mathbf{r}}_B|\ge -1$, with equality for pure antiparallel vectors. This reproduces the Néel state and $-J/4$ per bond found in part (a).

Yes. The limiting state saturates the de Finetti mean-field lower bound: the minimizing product state belongs to the allowed de Finetti mixture, and the finite-de-Finetti error tends to zero as $z\to \infty$. At finite $z$ the theorem gives only an approximation; entanglement and correlated fluctuations can lower the energy below the product-state value by corrections that vanish in the infinite-coordination limit.

Testing the weak formulation with the constant function $1$ proves the necessary compatibility conditionAssume first that $U$ is connected and this condition holds. On the mean-zero Sobolev spaceuse the norm $\Vert v{\Vert }_{†}=\Vert Dv{\Vert }_{L^2(U)}$. To prove the needed Poincare-Wirtinger inequality, suppose it failed. There would be $v_k\in H_{†}^1(U)$ with $\Vert v_k{\Vert }_2=1$ and $\Vert D{v}_k{\Vert }_2\to 0$. The Rellich-Kondrachov compactness theorem gives a subsequence converging strongly in $L^2$ and weakly in $H^1$ to a function $v$. Its weak derivative vanishes, so connectedness makes $v$ constant; its zero mean makes it zero. This contradicts $\Vert v{\Vert }_2=1$. Hence $\Vert Dv{\Vert }_2$ is an equivalent Hilbert space norm on $H_{†}^1(U)$.

DefineBoundedness of $a^{ij}$ makes $B$ a bounded bilinear form, while uniform ellipticity givesso it is a coercive bilinear form. The Cauchy-Schwarz inequality and the Poincare-Wirtinger inequality make $\ell$ a bounded linear functional. The Lax-Milgram theorem supplies a unique $u\in H_{†}^1(U)$ satisfying $B(u,v)=\ell (v)$ for every mean-zero $v$. For arbitrary $v\in H^1(U)$, subtract its mean; the omitted constant contributes zero on both sides because $D1=0$ and ${\int }_{U}f=0$. Thus $u$ solves the original problem.

If two solutions exist, their difference $w$ satisfies $B(w,w)=0$, so uniform ellipticity gives $Dw=0$. It is therefore constant on $U$. The solution is unique up to an additive constant, and its mean-zero representative is unique. If $U$ is disconnected, the precise condition is ${\int }_{U_j}f=0$ on every connected component $U_j$, and one independent additive constant remains on each component.

Let $T=AB$ with $A,B$ Hilbert--Schmidt, and write $T=U|T|$. Since $|T|=U^{\ast }AB$, part iii and the Cauchy-Schwarz inequality for series giveThus $T$ is trace class with the required bound.

Factor $T=AB$ as in part b(v). Then Parseval identity for a Hilbertian basis and Cauchy-Schwarz inequality giveThe defining series for the operator trace is therefore absolutely convergent and satisfies $|\mathrm{tr}T|\le \Vert T{\Vert }_1$.

For unit vectors $x,y$, the adjoint operator of $R=x\otimes y$ is $R^{\ast }=y\otimes x$, andHence $|R|=P_{\mathbb{C}y}$ and part b(iii), followed by Parseval, givesScaling proves $\Vert x\otimes y{\Vert }_1=\Vert x\Vert \Vert y\Vert$. A second use of Parseval yieldswhich proves the asserted rank-one operator formulas.

Apply the Cauchy-Schwarz inequality successively in the three shift variables to the correlation with the quadratic phase $h$. At the third step the third additive derivative of its phase vanishes, leavingThus $|⟨f,h⟩|\ge \delta$ implies

Repeated Cauchy-Schwarz inequality shows that the $U^3$ norm dominates the absolute mean of a function:Taking $f=1_A$ and using $\mathbb{E}{1}_A=\alpha$ gives

Only the $k$th block can have nonzero subgradient coordinates. If ${\beta }^{(k)}\ne 0$, the Cauchy-Schwarz inequality shows that the unique supporting vector is ${\beta }^{(k)}/\Vert {\beta }^{(k)}{\Vert }_2$. At zero, the defining inequality is $\Vert h{\Vert }_2\ge u^{T}h$ for every block vector $h$, which is equivalent to $\Vert u{\Vert }_2\le 1$. Thus

Subtract a constant so that $f(0)=0$, and write $m=\mathbb{E}f(Y)$, $V=\mathrm{Var}f(Y)$, and $A=\mathbb{E}{f}^{\prime }(Y{)}^2$. The supplied identity applied to $f$ and $f^2$ givesBy Cauchy-Schwarz inequality, $m^2\le A$ andThus $V+m^2\le 2\sqrt{(V+m^2)A}$, so $V+m^2\le 4A$ and in particular $V\le 4A$. Therefore the standard Laplace distribution has $C_P(Y)\le 4$.

A kernel for density estimation is an integrable function $K$ with $\int K=1$, usually also bounded and nonnegative, and its scaled version is $K_h(u)=h^{-1}K(u/h)$. For suitable $g_1,g_2$, their convolution is

Because the observations have length-biased density $g(u)=uf(u)/\mu$,Thus the exact bias is $(K_h\ast f)(x)-f(x)$, exactly the same as for the ordinary kernel density estimator based directly on observations from $f$.

Write $R(K)=\int K(u{)}^{2}du$. The estimator is the average of $Y_i(x)=\mu X_i^{-1}{K}_h(x-X_i)$, soIts mean is $K_h\ast f$. Integrating the pointwise variance therefore givesFinally, Cauchy-Schwarz inequality under density $f$ givesEquality would require $X$ to be constant almost surely, impossible for a density, so $\mu \bar{\mu }>1$. The ordinary KDE has the same negative term but leading integrated variance $R(K)/(nh)$; length-biased sampling strictly inflates it.

With densities $p,q$,By Cauchy-Schwarz inequality,Also $(\sqrt{p}-\sqrt{q}{)}^2\le |p-q|$, so $H^2\le 2\mathrm{TV}$.

The Hellinger affinity is $\rho (P,Q)=\int \sqrt{pq}d\mu =1-H^2/2$. Product densities and Fubini's theorem give $\rho (P^n,Q^n)=\rho (P,Q{)}^n$, hence

Le Cam two-point lemma states, for squared-error estimation at parameter points ${\theta }_0,{\theta }_1$, thatTake ${\theta }_0=0$, ${\theta }_1=\delta =1/(4n)$. The one-observation uniform densities overlap on length $1-\delta$, so $H^2(P_0,P_1)=2\delta$. ThereforeThe first distance inequality gives $\mathrm{TV}(P_0^n,P_1^n)\le 1/\sqrt{2}$. Le Cam's lemma now yieldsThis proves the claim with the displayed universal positive constant.

Thomson principle states that the effective resistance is the minimum energy of a unit flow from $a$ to $z$:An edge cutset separating $a$ and $z$ is a set of edges whose removal disconnects them. Every unit flow has net flux one across each such cutset ${\Pi }_k$. By Cauchy-Schwarz inequality,For disjoint cutsets, summing these energy lower bounds and applying Thomson's principle gives the Nash-Williams inequalityThe commute time identity is

The limiting conditional error of one-nearest-neighbour classification is $1-{\sum }_k{p}_k(x{)}^2$. Put $M={\max }_k{p}_k(x)$ and $r=1-M$. By the Cauchy-Schwarz inequality, the other $K-1$ probabilities satisfyConsequentlyNow $\mathbb{E}r=R_{\mathrm{Bayes}}$, and Jensen inequality gives $\mathbb{E}{r}^2\ge (\mathbb{E}r{)}^2$. Taking expectations proves

Take the $\tilde{H}$ inner product of the finite-dimensional equation with $u_m$. Part a gives exact cancellation of the nonlinear term:By the Cauchy-Schwarz inequality and the Poincare inequality,Thus every solution satisfiesThe bound is independent of $m$.