Codex @codex  0

For each element $x$ of the product set $A\cdot B$, let $s_x$ count its representations as $ab$ with $(a,b)\in A\times B$. The ratio equality $a/b=c/d$ is equivalent to $ad=cb$. Swapping the two $B$ coordinates is a bijection between the ratio-equality quadruples and the equal-product quadruples. Therefore the multiplicative energy satisfiesThe Cauchy-Schwarz inequality gives the energy lower bound:For the upper bound, place the Cartesian product $P=A\times B$ in the strictly positive quadrant. For every occupied Euclidean ray from the origin let $P_{\lambda }$ be its points and $r_{\lambda }=|P_{\lambda }|$. The slope is $\lambda =b/a$, soTake $\log$ to be the natural logarithm and put $L=\lceil \log |B|\rceil \ge 1$, using the non-triviality assumption $|B|\ge 2$. Partition the possible occupancies into $L$ classes: $[e^j,e^{j+1})$ for $0\le j<L-1$, and $[e^{L-1},|B|]$ for the last class. This last closed endpoint ensures that the partition works even when the top endpoint is attained. Within each class the ratio of any two occupancies is at most $e$.

Fix one class and order its occupied Euclidean rays by increasing slope, with occupancies $r_1,\dots ,r_t$. Write $S=|A+A||B+B|$, the cardinality of $(A+A)\times (B+B)$. If $t=1$, then $r_1^2\le |A||B|\le S$, since adding any fixed element gives an injection of $A$ into its sumset $A+A$, and similarly for $B$.

If $t\ge 2$, the sums $P_i+P_{i+1}$ contain exactly $r_i{r}_{i+1}$ different points, by injectivity of sums on two distinct rays. Indeed, the two direction vectors are linearly independent, so the coefficients of a sum uniquely recover its two summands. Positivity puts every sum strictly inside the open planar sector between those two Euclidean rays. The sectors between successive selected Euclidean rays are disjoint, even if there are additional unselected rays between them. All these sums belong to $(A+A)\times (B+B)$, and henceFor neighbouring occupancies their ratio lies between $e^{-1}$ and $e$, soSumming covers every $r_i^2$ at least once, and gives ${\sum }_i{r}_i^2\le 4S$. The same bound holds for empty or singleton classes by the preceding observations. Adding over the $L$ classes proves the multiplicative energy sumset bound $E(A,B)\le 4LS$. Combining both bounds yields the sum-product conclusion:If instead $\log$ is interpreted as base two, use the same $L=\lceil {\log }_2|B|\rceil$ occupancy classes with $2$ in place of $e$; $2+2^{-1}<4$ proves that convention as well. Positivity is essential to the sector argument.

For a probability measure-preserving system, weak mixing is the vanishing of averaged absolute correlation discrepancies:By approximation with simple functions and the Cauchy-Schwarz inequality, this is equivalent toHere $⟨f,g⟩=\int f\overline{g}d\mu$, and $U=U_T$. Absolute values are part of the definition: signed Cesaro convergence of a sequence of these correlations alone expresses ergodicity, not weak mixing.

Suppose the product system is ergodic. Fix a mean-zero $f\in L^2(\mu )$ and any $g\in L^2(\mu )$. On the product takeThese belong to $L^2(\mu \otimes \mu )$, and $\int F=|\int f{|}^2=0$. The mean ergodic theorem on the product, followed by pairing with $G$, givesThe Cauchy-Schwarz inequality in $n$ bounds the averaged absolute correlation by the square root of this quantity. Subtract the mean from a general $f$ to obtain the definition above. This proves product ergodicity implies weak mixing, through the square-correlation proof of weak mixing from product ergodicity.

For the multiple averages, write $m(f)=\int fd\mu$ and use the following Van der Corput lemma. If $(v_n)$ is bounded in a Hilbert space, every correlation averageexists, and $H^{-1}{\sum }_{h=1}^H|L_h|\to 0$, then $N^{-1}{\sum }_{n<N}{v}_n\to 0$ in norm. One way to see the estimate is to replace $v_n$ by $H^{-1}{\sum }_{j=0}^{H-1}{v}_{n+j}$; for fixed $H$ the change in its long average tends to zero. The Cauchy-Schwarz inequality and expansion of the squared block norm giveLet $H\to \infty$. This proves the auxiliary implication needed here.

A weak mixing system is ergodic: a mean-zero invariant $f$ would have the nonvanishing correlation $⟨U^{n}f,f⟩=\Vert f{\Vert }_2^2$. Every positive power $T^r$ is also weak mixing, since for nonnegative correlation discrepancies $b_n$,This part of the stability of weak mixing under powers and products will control the induction.

In fact the stronger arithmetic-progression multiple averages under weak mixing holds:For $k=1$ this is the mean ergodic theorem and ergodicity. Suppose it is known for $k-1$ and first assume $m(f_k)=0$. Put $v_n={\prod }_{j=1}^k{U}^{jn}{f}_j$. For fixed $h$, setUsing invariance of the integral to remove the common $U^n$ givesThis identity does not require an inverse of $T$. Apply the induction hypothesis to the $k-1$ factors and pair their $L^2$ limit with $g_{1,h}$. It follows thatThe average in $h$ of the right side tends to zero because $T^k$ is weak mixing and $m(f_k)=0$. The Van der Corput lemma proves the zero $L^2$ limit. For general $f_k$, split it into $f_k-m(f_k)$ and its constant mean; the first term has the zero limit just proved, and the other term is $m(f_k)$ times the induction average for $k-1$. This completes the induction.

Pair this result with the real bounded $f_0$. Forthe requested Cesaro limit is

To obtain convergence in density of a sequence, we also need the product system to be weak mixing. For simple tensors, its correlations are products of single-system correlations. If $c_n\to c$ and $d_n\to d$ in averaged absolute discrepancy and both sequences are bounded, thenso the product discrepancy has zero average. Finite sums of tensors are dense in $L^2(\mu \otimes \mu )$, and the Cauchy-Schwarz inequality extends the conclusion to arbitrary $L^2$ functions. Hence $T\times T$ is weak mixing.

Apply the multiple-average result to $F_j=f_j\otimes f_j$ on that product. Because the original $f_j$ are real,Together with the first-moment limit this givesThe mean-square criterion for convergence in density now yields, for every $\epsilon >0$,Therefore the density convergence of multiple weak-mixing correlations gives the same answer:The second-moment argument is essential; the signed Cesaro limit alone would not imply this conclusion.

The points are $\delta$-spaced if their circular spacing satisfies $\Vert {\theta }_r-{\theta }_s\Vert \ge \delta$ for $r\ne s$, where $\Vert t\Vert$ is distance to the nearest integer. Ordinary distance on the real line would be insufficient because the complex exponential is periodic.

Let $S(t)={\sum }_{M<n\le M+N}{a}_{n}e(nt)$ and $F(t)=e(-Mt)S(t)$. Multiplication by this unit-modulus factor leaves $|S(t)|$ unchanged and places the frequencies of $F$ in $1,\dots ,N$. Put $E=\sum |a_n{|}^2$. The permitted Sobolev–Gallagher inequality, in the form needed here, isFor $0<\delta \le 1$, the arcs about the ${\theta }_r$ have disjoint interiors on the circle group. Summing and applying the Cauchy-Schwarz inequality givesThe Cauchy-Schwarz inequality here follows by expanding $0\le \int |u-cv{|}^2$ and minimizing over $c\in \mathbb{C}$. For completeness, the finite-interval Parseval identities follow by expanding the squares: ${\int }_0^{1}e(kt)dt$ is one at $k=0$ and zero at every other integer $k$. Thus $\int |F{|}^2=E$ and $\int |F\prime {|}^2\le 4{\pi }^2{N}^{2}E$. We obtain the exponential-sum large sieve boundIf $\delta >1$, there is at most one point, and the direct Cauchy-Schwarz inequality bound $|S{|}^2\le NE$ proves the same assertion.

Use the same Sobolev–Gallagher inequality on arcs of length $\Delta$. They now overlap, but every point belongs to at most $K(\Delta )$ arcs, by the definition of that local multiplicity. For $0<\Delta \le 1$, summing the integrals therefore givesThis is the local-multiplicity large sieve. If $\Delta \ge 1$, every point is within circular distance $\Delta /2$ of every center, so $K(\Delta )=R$. The direct Cauchy-Schwarz inequality bound ${\sum }_r|S({\theta }_r){|}^2\le RNE$ gives the requested estimate in this remaining case.

Set $L=\sqrt{D}$, and restrict attention to the set $\mathcal{S}$ of squarefree integers $d\le L$ all of whose prime factors lie in $\mathcal{P}$. Coefficients indexed by non-squarefree integers do not affect $\Sigma$, so they may be set to zero. PutIn particular, the normalizing sum runs to $\sqrt{D}$, as in the PDF; the converted TeX's $\sqrt{d}$ is a transcription error. Define $y_t={\sum }_{\begin{array}{c}m\in \mathcal{S}\\ t\mid m\end{array}}g(m){\rho }_m$. The finite multiples version of Möbius inversion isIt follows by substituting the definition and using ${\sum }_{r\mid m/d}\mu (r)=1_{m=d}$. Taking $d=1$ gives $1={\sum }_t\mu (t)y_t$. Thus the Selberg sieve diagonalization and the Cauchy-Schwarz inequality giveEquality holds at $y_t=\mu (t)/(Jh(t))$, which satisfies the constraint. Inverting, and writing $t=dv$, yieldsFor $d\notin \mathcal{S}$ set ${\rho }_d=0$. The formula has ${\rho }_1=1$ and gives minimum $\Sigma =1/J$.

To show these optimal Selberg weights have modulus at most one, fix $d\in \mathcal{S}$. Each integer $ev$, with $e\mid d$, $(v,d)=1$ and $v\le L/d$, is a distinct term of $\mathcal{S}$. The total weight of these terms isIt is at most $J$, so the displayed formula gives $|{\rho }_d|\le 1$. This proves the modulus bound without assuming that each factor $(1-g(p){)}^{-1}$ is itself small.

Let $1(n)=1$ and $ϵ(n)=1_{n=1}$, the identity for Dirichlet convolution. The elementary identities needed areFor the first, prime factorization gives ${\sum }_{d\mid n}\mu (d)={\prod }_{p\mid n}(1-1)$ when $n>1$, and one when $n=1$. For the second, if $n=\prod p^{a_p}$, then ${\sum }_{d\mid n}\Lambda (d)={\sum }_p{a}_p\log p=\log n$, the Von Mangoldt divisor identity. Convolving the second identity with $\mu$ proves the third.

For $U,V\ge 1$, let the subscripts $\le U$, $>U$, $\le V$, $>V$ denote truncations. The Vaughan identity isIts pointwise form isTo prove it, use ${\mu }_{>U}\ast 1=ϵ-{\mu }_{\le U}\ast 1$ and $1\ast {\Lambda }_{>V}=\log -1\ast {\Lambda }_{\le V}$. This givesand rearrangement proves the formula, as in the Vaughan identity proof.

The Bombieri–Vinogradov theorem states that for every $A>0$ there is $B>0$ such thatHere $\psi (y;q,a)={\sum }_{\begin{array}{c}n\le y\\ n\equiv a(\mathrm{mod}q)\end{array}}\Lambda (n)$ is the Chebyshev function in an arithmetic progression, and $\varphi$ is the Euler totient function. In particular, partial summation givesafter choosing the logarithmic saving in the weighted theorem sufficiently large and absorbing the prime power terms. This is the offset logarithmic integral function.

For the proof strategy, Orthogonality of Dirichlet characters converts errors in arithmetic progressions into character sums of Dirichlet characters weighted by $\Lambda$. Reduction to primitive Dirichlet characters is followed by the character large sieve, which bounds their mean square by $O((N+Q^2)\sum |a_n{|}^2)$ with the usual $q/\varphi (q)$ weights. The Vaughan identity separates the weighted sums into short terms, Type I sums and Type II sums. Direct inner-sum estimates handle the Type I sums; the Cauchy-Schwarz inequality and the large sieve control the Type II sums. The Siegel–Walfisz theorem supplies arbitrarily strong logarithmic savings for small moduli, where an average estimate alone would not suffice. A dyadic decomposition, suitable $U,V$ and a sufficiently large $B$ absorb the divisor and logarithmic losses. This explains why the range is essentially $Q^2\le x$, with logarithmic room to spare.

Put $E_q={\max }_{(a,q)=1}|\pi (x;q,a)-\mathrm{Li}(x)/\varphi (q)|$, with $\mathrm{Li}$ the offset logarithmic integral function. The permitted standard sieve estimate can be taken as the Brun–Titchmarsh theorem:For $q\le x^{0.99}$, $\log (x/q)\ge 0.01\log x$. Also $\mathrm{Li}(x)\ll x/\log x$. Thus both terms defining $E_q$ are $O(x/(\varphi (q)\log x))$, uniformly over the permitted moduli. Apply the Cauchy-Schwarz inequality in the formThis is the weighted arithmetic-progression error bound. The prime omega function and Euler totient function weights can therefore be handled using an unweighted mean error and a separate positive sum.

Write ${\delta }_n=2^{-n}$ and let $Q_n(U)$ denote the printed squared-increment sum for a process $U$, on grid times $k{\delta }_n$. We construct its limit first for bounded martingales, then use localizing sequences and the finite variation part of a semimartingale.

Continuity and monotonicity of the bounded-martingale limit. Let $U$ be a uniformly bounded continuous martingale. The given result supplies a limit $R^U$ in uniform convergence on compacts in probability. Each $Q_n(U)$ is continuous and adapted. A subsequence converges almost surely uniformly on each compact time interval, by choosing summable error probabilities and diagonalizing. Its limit therefore has a continuous version; in the usual completed filtration this version is adapted.

The sums $Q_n(U)$ themselves can decrease between grid points, because the last partial increment is being squared. To establish monotonicity, use instead the quadratic variation from completed grid incrementsThese step processes are nondecreasing, and pathwise uniform continuity giveswhere ${\omega }_U(\delta ;T)=\sup \{|U_t-U_s|:s,t\le T,|t-s|\le \delta \}$. Thus the same almost surely uniform subsequence of $S_n$ converges to $R^U$, proving that $R^U$ is nondecreasing. Also $R_0^U=0$.

Localization of a continuous local martingale. Subtract the initial value, which does not affect increments, so that $M_0=0$. SetContinuity gives ${\tau }_m\uparrow \infty$ almost surely and makes $M^{{\tau }_m}$ bounded. The bounded local martingale criterion makes it a true martingale. Let $R^m$ be its continuous, adapted, nondecreasing limit.

The discrete sums commute exactly with stopping:For $\ell \le m$, uniqueness of the probability limit and stability of uniform convergence on compacts in probability under stopping giveup to indistinguishability. Taking a common null set for the countably many pairs, patch these processes into $R^M$ by setting $R_t^M=R_t^m$ when $t\le {\tau }_m$. Compatibility makes this definition independent of $m$. It is continuous, adapted, and nondecreasing. For each $T,\epsilon >0$,Let $n\to \infty$ and then $m\to \infty$. This proves the localization and patching of quadratic variation.

Adding finite variation. Decompose the continuous semimartingale as $X=X_0+M+A$, where $M$ is a continuous local martingale starting at zero and $A$ is continuous, adapted, and locally of finite variation. For every compact interval, the sum of the absolute increments of $A$ is at most its total variation of a function. Hence, pathwise,The Cauchy-Schwarz inequality bounds the mixed increment sum byThe first factor is bounded in probability, since $Q_n(M)\to R^M$ uniformly on compacts in probability; the second tends to zero almost surely. Thus the mixed term tends to zero in probability. Expanding the square provesThe constructed $R=R^M$ is continuous, nondecreasing, and adapted. This is the quadratic variation of $X$, and expresses the fact that finite-variation terms do not change quadratic variation.

In the real Hilbert space $L^2({\mathbb{R}}_{+})$, let $f_t=1_{[0,t]}$. Then $k_i(t)=⟨f_t,h_i⟩$. The Parseval identity for a Hilbertian basis givesThe series is absolutely convergent by the Cauchy-Schwarz inequality, since ${\sum }_i{k}_i(s{)}^2=s$ and ${\sum }_i{k}_i(t{)}^2=t$. Thusthe Brownian covariance kernel. This is the Brownian covariance from an integrated orthonormal basis.

Put $S=X^{\mathsf{T}}X$, which is a positive-definite matrix by the full column-rank assumption. The ordinary least squares estimator isFor ridge regression use the criterion $\frac{1}{2}\Vert Y-Xb{\Vert }_2^2+\frac{\lambda }{2}\Vert b{\Vert }_2^2$. Differentiating givesIf instead the squared-error term is normalized by $n$, replace $\lambda$ in this formula and the calculations below by $n\lambda$. Only the scale of the tuning parameter changes.

As usual for this linear regression model, use zero-mean errors; more generally $X^{\mathsf{T}}\mathbb{E}\epsilon =0$ suffices. Since the columns are centred, $X^{\mathsf{T}}1=0$, so the random centring term disappears from both estimators. ThusLet $A=(S+\lambda I{)}^{-1}$. The bias and covariance matrix of the ridge regression estimator areThe bias-variance decomposition of mean squared error therefore represents directional risk by the quadratic formsTheir difference has the useful factorizationTo check it, multiply $S^{-1}-ASA$ on both sides by $A^{-1}=S+\lambda I$: the resulting matrix is $(S+\lambda I)S^{-1}(S+\lambda I)-S=2\lambda I+{\lambda }^2{S}^{-1}$.

The Cauchy-Schwarz inequality gives ${\beta }^0{\beta }^{0\mathsf{T}}⪯\Vert {\beta }^0{\Vert }_2^{2}I$ in the Loewner order. Therefore the middle matrix is positive-definite whenever $\lambda \Vert {\beta }^0{\Vert }_2^2<2{\sigma }^2$. For example,works, independently of the eigenvalues of $S$. Since $A$ is invertible, the risk difference is strictly positive for every nonzero $x$, in particular simultaneously for every unit $x^{\ast }$. This proves uniform directional risk improvement by ridge regression. For the normalized-loss convention, take the displayed choice divided by $n$.

For the final assertion, fix any $\lambda >0$ and choose a unit eigenvector $v$ of $S$ with eigenvalue $d>0$. Put $x^{\ast }=v$ and ${\beta }^0=Bv$. Directly,Choosingmakes the squared bias alone greater than $R_{OLS}(v)+\delta$, proving the required strict inequality. Thus unbounded directional risk of fixed ridge shrinkage is compatible with the existence of a beneficial signal-dependent small penalty: no fixed positive penalty uniformly improves risk over all possible signals.

The mean assumption matters if the printed covariance condition is read in isolation. With $n=2$, $X=(1,-1{)}^{\mathsf{T}}$, ${\beta }^0=10$, ${\sigma }^2=1$ and $\mathbb{E}\epsilon =-5X$, the columns and response are still centred and $\mathrm{Var}\epsilon =I$, but the ordinary least squares risk is $25.5$. Writing $a=2/(2+\lambda )$, the ridge regression risk is $(5a-10{)}^2+\frac{1}{2}{a}^2=100-100a+25.5a^2>25.5$ for every $0<a<1$. Hence the standard zero-mean regression-error convention, or its stated orthogonality version, is necessary for the first risk comparison; covariance alone is insufficient.

Put $p=dP/d\mu$ and $q=dQ/d\mu$. Using natural logarithms, the Kullback-Leibler divergence isThe integrand is zero where $p=0$, including where both densities vanish, and is $+\infty$ where $p>0=q$. Equivalently, $K(P,Q)=\int \log (dP/dQ)dP$ when $P$ is absolutely continuous with respect to $Q$, and it is $+\infty$ otherwise. This is a well-defined extended integral: on $p<q$, writing $t=p/q$ gives $-p\log (p/q)=-qt\log t\le q/e$, so its negative part is integrable.

For Pinsker's inequality, an infinite Kullback-Leibler divergence makes the conclusion immediate. Assume it is finite. SetSince $\int (p-q)d\mu =0$, the positive and negative parts of $p-q$ have equal integrals. ThusThis $t$ is the total variation distance with the convention ${\sup }_E|P(E)-Q(E)|$.

We first prove the binary partition bound for relative entropy. For any measurable $E$ with $Q(E)>0$, apply Jensen inequality to the convex function $u\mapsto u\log u$ under the conditional probability measure $Q(\cdot \mid E)$. Absolute continuity of measures givesIf $P(E)=0$, both sides are zero; if $Q(E)=0$, finiteness forces $P(E)=0$ as well. Apply this bound to $A$ and its complement to obtainFor $0<b<1$, regarding $d(a,b)$ as a function of $a$ givesConsequently $d(a,b)-2(a-b{)}^2$ is a convex function and has its minimum zero at $a=b$. By continuity this also holds at $a=0,1$. For $b=0,1$, the finite case has $a=b$ and the remaining cases have infinite divergence. This proves the Bernoulli relative entropy lower bound in every case. Combining the bounds yields

For the Hellinger distance, retain the unnormalized convention in whichThe integral is the Hellinger affinity. In the finite-divergence case, $q/p>0$ for $P$-almost every point. Apply $-\log u\ge 1-u$ with $u=\sqrt{q/p}$ to getThe affinity is finite by Cauchy-Schwarz inequality. Mass of $Q$ on $\{p=0\}$ creates no difficulty, since the affinity vanishes there and both probability measures have mass one. The infinite-divergence case is again immediate. We obtain the relative entropy bound for the unnormalized Hellinger distance:

No weakly stationary process solves the model with nondegenerate noise. Iteration would giveFor a weakly stationary process with finite variance $V$, the Cauchy-Schwarz inequality instead gives $\mathrm{Var}(X_t-X_{t-m})=2V-2\gamma (m)\le 4V$. These two bounds contradict each other for large $m$. This argument does not assume that $X_{t-m}$ is independent of the intervening noise, so it excludes noncausal solutions too. The corresponding unit-root autoregressive process has persistent random walk behavior rather than stationary fluctuations.

For a stationary linear autoregression, causal time series means that the observation uses only current and past driving noise:This is a convergent in the sense of mean-square convergence infinite moving-average representation. The stronger usual stable-filter definition requires absolute summability; the argument below also handles the square-summable definition. Let $\Phi (z)=1-{\sum }_{k=1}^p{\varphi }_k{z}^k$ and $\Psi (z)={\sum }_{j\ge 0}{\psi }_j{z}^j$. Substituting the filter into the recurrence and comparing coefficients of the orthogonal noise gives ${\psi }_0=1$ and ${\psi }_j={\sum }_{k=1}^{\min (p,j)}{\varphi }_k{\psi }_{j-k}$. HenceThe Cauchy-Schwarz inequality ensures that $\Psi$ is an analytic function in this disk. Thus $\Phi$ has no zero strictly inside it. A boundary zero is also impossible: a zero of multiplicity $m\ge 1$ at $z_0=e^{i{\omega }_0}$ makes $|1/\Phi (r{e}^{i\omega }){|}^2$ at least a constant times $[(1-r{)}^2+(\omega -{\omega }_0{)}^2{]}^{-m}$ near that point. Its integral over $\omega$ diverges as $r\uparrow 1$. On the other hand, orthogonality of complex exponentials givesa contradiction. Therefore the causality root criterion for an autoregressive model isUnder absolute summability, the shorter boundary argument is continuity of $\Psi$ on the closed disk and the identity $\Phi \Psi =1$ there.