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Use metric signature $(+---)$, as required by the sign of the scalar product in this question. For a massive particle, the four-momentum isFor a photon of positive frequency $\nu$ and direction $e$, the Planck constant gives $E=h\nu$ andA future-pointing timelike four-vector $P=(P^0,p)$ has $P^0>|p|$ and $P^0>0$ in this signature. The ordinary Euclidean Cauchy-Schwarz inequality yieldsThis is stronger than the requested nonnegative inequality. For an electron and a positron of rest mass $m>0$, their future-pointing timelike four-momenta have $P_1^2=P_2^2=m^2{c}^2$. Thuswhich cannot equal the null norm of a single photon. Four-momentum conservation forbids spontaneous photon decay into an electron-positron pair in vacuum. An external body or field able to exchange momentum would be a different system.

A precise useful form of the principle is the type I–type II inverse principle for Möbius correlation. Normalize $|f(n)|\le 1$, and supposeFor any positive integer cutoffs $U,V$ with $U+V\le \delta X/2$, defineThen $|c_d|,|a_d|\le \tau (d)$, where $\tau$ is the divisor function, and at least one of the following two sums has modulus at least $\delta X/4$:The first is correlation against a controlled linear combination of indicators of multiples of small moduli, each a periodic function: this is the “somewhat periodic” branch. The second is a bilinear sum with independently weighted factors, both larger than the cutoffs: this is the “somewhat multiplicative” branch. The statement concerns these precise correlations, not an assertion that $f$ must itself be periodic or multiplicative.

For clarity, the Vaughan identity for the Möbius function proves this version immediately. Split $\mu ={\mu }_{\le U}+{\mu }_{>U}$ and likewise at $V$. Since $\mu \ast \mu \ast 1=\mu$,Multiply by $f(n)$ and sum. The first two terms contribute at most $U+V$ in modulus, and the remaining terms are $-T_{\mathrm{I}}+T_{\mathrm{II}}$. The triangle inequality gives the stated dichotomy. It is valid at finite $X$ and for any chosen cutoffs in the indicated range.

We now prove the required orthogonality without appealing to a stronger uniform exponential-sum theorem. Write $\alpha =\sqrt{2}$ and $e(u)=e^{2\pi iu}$. The Diophantine bound for the square root of two iswhere $\Vert u\Vert$ is distance to the nearest integer. If $a$ is that nearest integer, then $|2h^2-a^2|\ge 1$, while $h\sqrt{2}+a\le 4h$; dividing proves the bound.

Use the preceding Dirichlet convolution identity with $U=V=P=\lfloor X^{1/10}\rfloor$. The two short terms are $O(P)$. For the type I sum, the exponential geometric sum bound givesConsequentlyusing ${\sum }_{d\le Y}\tau (d)\ll Y\log (2Y)$ from counting factor pairs.

For the type II sum, perform a dyadic decomposition into $d\in (D,2D]$, $w\in (W,2W]$, keeping $dw\le X$. Nonempty blocks have $DW<X$, while $D,W\gg P$. A block $B$ satisfies, by the Cauchy-Schwarz inequality and the divisor-square summatory bound,To justify the cutoff, after expanding the square the permissible $d$ for a pair $w,w^{\prime }$ still form an interval, with upper endpoint $\min (2D,X/\max (w,w^{\prime }))$. Its exponential sum has the same geometric bound. Diagonal pairs supply $DW$; for each nonzero difference $h=w-w^{\prime }$ there are $O(W)$ pairs.

The points $0,\alpha ,2\alpha ,\dots ,\lfloor W\rfloor \alpha$ modulo one are separated by at least $1/(4W)$. Counting points in consecutive distance bands around zero therefore givesSubstitution yields the bilinear cancellation for badly approximable phases estimateThere are $O({\log }^{2}X)$ blocks, so $T_{\mathrm{II}}\ll X^{19/20}{\log }^{4}X$. The divisor-square bound itself can be obtained elementarily from $\tau (n{)}^2\le {\tau }_4(n)$ prime-power by prime-power and ${\sum }_{n\le Y}{\tau }_4(n)\le Y(1+\log Y{)}^3$; no uncontrolled coefficient bound is being suppressed.

Combining the short, type I and type II terms,Hence

For a prime $p$, the divisor sum defining the real weight is $\varphi (0)-\varphi (\log p/\log X)$. If $p>X^{1/3}$, the second term vanishes and the first is one. Thus $F(p)=1$, while $F(n)\ge 0$ for every integer because $\varphi$ is real and the sum is squared.

Put $h(x)=e^x\varphi (x)$. With the Fourier transform convention in the question, Fourier inversion givesThe function $h$ is smooth and has compact support. Using integration by parts $k$ times, with zero endpoint terms, provesChoose an integer $k\ge A$ to obtain $|\psi (t)|{\ll }_A|t{|}^{-A}$ for $|t|\ge 1$. In particular every polynomially weighted absolute integral of $\psi$ is finite.

For the double integral, useDifferentiating the second formula gives $\int \psi (t)(1+it)e^{-(1+it)u}dt=-{\varphi }^{\prime }(u)$. absolute convergence from the rapid decay permits Fubini's theorem. The derivative energy constant for a smooth sieve cutoff is thereforeThere is no complex conjugate in this integral: the two factors both become $-{\varphi }^{\prime }(u)$, which is real. The constant is positive; in fact the Cauchy-Schwarz inequality applied to $\varphi (1/3)-\varphi (0)=-1$ gives $c_{\varphi }\ge 3$.

Here is the basic structure of the smooth divisor-square sieve asymptotic, with enough uniform estimates to specify the error. Write $L=\log X$ and $R=X^{1/3}$. Expand the square and count multiples of the least common multiple $[d,e]$ in an arbitrary interval $I$ of length $X$:The sum is effectively restricted to $d,e<R$. Each count is $X/[d,e]+O(1)$, independent of the interval's position, so the total error is $O_{\varphi }(X^{2/3})$.

The Fourier representation of a smooth Selberg weight readsHence, with $z^{\prime }=(1+i{t}^{\prime })/L$,The four local terms correspond to the prime dividing neither divisor, only the first, only the second, or both. This is the Euler product for a smoothed divisor-square correlation. Factor it aswhereThe local factors of $H$ are $1+O(p^{-2+4\eta })$ when $|\mathrm{Re}z|,|\mathrm{Re}{z}^{\prime }|\le \eta$ for small fixed $\eta <1/4$. Thus $H$ is a holomorphic function near $(0,0)$, and evaluating each local factor at zero gives exactly one. Consequently $H(0,0)=1$ and $H(z,z^{\prime })=1+O(|z|+|z^{\prime }|)$ there.

The pole expansion $\zeta (1+w)=w^{-1}+O(1)$ now yieldswhen $|t|,|t^{\prime }|\le L^{1/4}$. Integrating the error against the rapidly decreasing transforms gives $O_{\varphi }(L^{-2})$. The complementary tails are harmless uniformly: absolute values in the original divisor series giveAny prescribed power decay of ${\int }_{|t|>L^{1/4}}|\psi (t)|dt$ is available, so these tails and the tails of the limiting kernel are $O_{\varphi }(L^{-2})$ after choosing sufficiently many applications of integration by parts. The same absolute bound justifies the original exchanges of infinite sums and integrals. Using the evaluated double integral,We conclude, uniformly over all interval locations,This proof does not assume that the interval starts near the origin.

Finally choose this one fixed smooth cutoff. primes above $R$ in $(X_0,X_0+X]$ contribute one each to the nonnegative weight, and at most $R$ primes lie below $R$. Thereforefor all sufficiently large $X$, uniformly in $X_0$. The bounded range $2\le X\le X_1$ follows by increasing the fixed constant and using the trivial interval bound $X+1$. Thus the short-interval prime upper bound from a smooth divisor weight is

Put $D=|⟨L_N,f⟩-⟨{\hat{L}}_N,f⟩|$. Apply the Cauchy-Schwarz inequality to the average of the nonnegative eigenvalue differences, and then the Hoffman–Wielandt inequality:Taking nonnegative square roots gives the spectral Lipschitz bound from Frobenius distanceThe factor $N^{-1/2}$ is essential: the empirical spectral measure has total mass $1$, not $N$.

Put $r_t=\sqrt{p_t}-\sqrt{p}-(t/2)g\sqrt{p}$. By differentiability in quadratic mean, $\Vert r_t{\Vert }_{L^2(\mu )}=o(|t|)$ andin L2 space. The second limit follows because the first gives $\Vert \sqrt{p_t}-\sqrt{p}{\Vert }_2=O(|t|)$.

Both probability density functions integrate to one. Consequently the L2 inner product satisfiesThe Cauchy-Schwarz inequality permits passage to the limit in this inner product, givingThus every score function is a mean-zero function. This proof uses square-root differentiability in quadratic mean, without requiring differentiation under the density integral.

The map $h\mapsto Ph$ is a bounded linear functional on L2 space, since $|Ph|\le \Vert h{\Vert }_2$ by the Cauchy-Schwarz inequality. Its kernel, the mean-zero L2 space, is therefore closed. Every nuisance score function and the parametric score function $\dot{\ell }$ are centered, so both ${\Pi }_{\mathcal{N}}\dot{\ell }$ and $\tilde{\ell }$ are centered. HenceMoreover, the efficient score belongs to the orthogonal complement of the nuisance tangent space. Writing $\dot{\ell }=\tilde{\ell }+{\Pi }_{\mathcal{N}}\dot{\ell }$ givesThis is the efficient-score projection identity; it remains valid when the efficient information is zero.

Consider any differentiable-in-quadratic-mean path with score function $g$. Put ${\delta }_t=\sqrt{f_t}-\sqrt{f}=(t/2)g\sqrt{f}+r_t$, where $\Vert r_t{\Vert }_2=o(|t|)$. The quadratic-mean to L1 density derivative follows fromBy the Cauchy-Schwarz inequality,Since $a$ is a bounded function, multiplying this L1 norm bound by $\Vert a{\Vert }_{\infty }$ provesThe last equality uses $P_{f}g=0$. The derivative is a bounded linear functional of $g\in L^2(P_f)$, so the required pathwise differentiability of a statistical functional holds, in particular relative to the statistical tangent set from part (b). Its derivative isFor the explicit bounded density tilts in part (b), this derivative is also obtained by direct integration, with no remainder term.

For a pathwise differentiable statistical functional, an influence-function representer is a mean-zero function $\phi \in L^2(P)$ such that every admissible score function $g$ satisfies $D{\psi }_P(g)=P(\phi g)$. The efficient influence function, also called the canonical gradient, is the unique such representer in the statistical tangent space. Equivalently, it is the orthogonal projection of any representer onto that statistical tangent space. The Pythagorean theorem in an inner-product space shows that it has the smallest squared L2 norm among all representers.

Here the statistical tangent space is all of $L_0^2(P_f)$. To verify the closure explicitly, take $h\in L_0^2(P_f)$, truncate it to $h_m=\max (-m,\min (h,m))$, and set $g_m=h_m-P_f{h}_m$. Then $g_m$ is bounded and centered, and $g_m\to h$ in $L^2(P_f)$, by dominated convergence and the Cauchy-Schwarz inequality. Part (c) supplies a representer already in this space. HenceIts variance is $P_f{a}^2-(P_{f}a{)}^2$.

The closure must be taken in the density-weighted space $L^2(P_f)$. An unweighted reading of $L^2[0,1]$ in the printed hint is false. For example, when $f(u)=2u$, the function $h(u)=u^{-3/4}-8/5$ has $P_{f}h=0$ and $P_f{h}^2=36/25$, so bounded centered truncations converge to it in $L^2(P_f)$; nevertheless $h\notin L^2([0,1],du)$. This illustrates density of bounded centered scores and fixes the measure in the closure statement.

Now $p_{\theta ,\eta }(x,y)=v(x)f(y-g_{\theta }(x))$. With $h_{\theta }={\partial }_{\theta }{g}_{\theta }$ and the location score $\rho (e)=-f^{\prime }(e)/f(e)$, differentiation of the log-likelihood givesTake the usual regularity conditions $h_{\theta }\in L^2(v)$ and $\rho \in L^2(f)$ so that this score function belongs to L2 space.

For the specified nuisance statistical path, differentiating $\log (1+t\gamma (e))$ at zero gives the score function $\gamma (e)$. Normalization requires $E_f\gamma =0$. There is a second constraint: the statistical path must preserve the model's zero error expected value. ThusThis constraint follows from being a path through the model, rather than from normalization alone. For a centered symmetric logistic distribution, for example, $\gamma (e)=\tanh (e)$ is bounded and has zero expected value, but $E[\epsilon \tanh (\epsilon )]>0$; it would not preserve the mean.

For any $\psi \in L^2(v)$, independence impliesAll these products are integrable by the Cauchy-Schwarz inequality, since $\epsilon \psi (X)$ is square-integrable. Thus every such error-density score function is orthogonal to every $e\psi (x)$. Covariate-density nuisance score functions $b(X)$ are also orthogonal to these functions, since $E\epsilon =0$. This is the mean-preserving error tangent space orthogonality used in the final calculation.

First use the additional admitted form of the efficient score. Write $\tilde{\ell }=e\zeta (X)$ and let ${\tau }^2=E_f{\epsilon }^2>0$. The difference $\dot{\ell }-\tilde{\ell }$ is the orthogonal projection onto the nuisance tangent space. Part (c) makes every $e\psi (X)$ orthogonal to that space. Therefore, for all $\psi \in L^2(v)$,The function in braces belongs to L2 space; choosing it as $\psi$ shows it vanishes $v$-almost everywhere. Hence the required conditional deduction isUnder the regular tail condition $sf(s)\to 0$ at both infinities, integration by parts gives $\int s{f}^{\prime }(s)ds=-1$, and the formula becomes $e{h}_{\theta }(x)/{\tau }^2$. This also confirms the sign.

The admitted product form is an extra restriction; it does not follow for every independent-error regression model. To locate the restriction precisely, suppose the error-density nuisance statistical paths preserve both normalization and mean to first order. Their closed mean-preserving error tangent space is $\{\gamma :E_f\gamma =E_f(\epsilon \gamma )=0\}$. The full nuisance tangent space is the orthogonal sum of this space and the centered functions of $X$.

For completeness, bounded functions satisfying the two constraints are dense in the error space. Truncate an arbitrary element, subtract its expected value, and then subtract a multiple of a fixed bounded centered function $q$ with $E_f(\epsilon q)\ne 0$. Such a $q$ exists by truncating $\epsilon$, since ${\tau }^2>0$. The correction coefficients tend to zero by the Cauchy-Schwarz inequality, so the corrected truncations converge in L2 space.

Let $m=E_v{h}_{\theta }(X)$. With $E_f\rho =0$ and $E_f(\epsilon \rho )=1$, the orthogonal projection of $h_{\theta }(X)\rho (\epsilon )$ onto the error nuisance space is $m\{\rho (\epsilon )-\epsilon /{\tau }^2\}$; its projection onto the covariate nuisance space is zero. Thus the general efficient score in independent-error regression isThe orthogonality of the two terms gives the displayed efficient information. For a normal distribution of the error, $\rho (e)=e/{\tau }^2$, so this reduces to the admitted formula. It also does so when $h_{\theta }$ is constant.

A concrete counterexample to the generality of the admission is $g_{\theta }(x)=\theta x$, with $X$ uniform on $[-1,1]$ and independent standard logistic distribution error. Here $m=0$ and $\rho (e)=\tanh (e/2)$, so the actual efficient score is $x\tanh (e/2)$. It cannot equal $e\zeta (x)$ because $\tanh (e/2)/e$ is not constant. This score function is already orthogonal to every nuisance score function: its factor $X$ is centered against error-only directions, and its factor $\rho (\epsilon )$ is centered against covariate-only directions. The requested formula is valid under its stated additional admission, with the general independent-error formula above explaining its limits.

There is a genuine qualification: the first printed inequality is valid as a useful general assertion for $0<\eta <1$, but is false for arbitrary $\eta >0$. Also the displayed likelihood ratio presupposes $P_{f_1}\ll P_{f_0}$. We first prove the intended bound and then give a counterexample to the unrestricted quantifier.

For any estimator $T$, set $A=\{d(T,f_0)<r_n\}$. By the triangle inequality, $d(T,f_1)\ge r_n$ on $A$, while $d(T,f_0)\ge r_n$ on $A^c$. HenceLet $C=\{|Z-1|\le \eta \}$ with $0<\eta <1$. On $C$, $Z\ge 1-\eta$. ThereforeThe last step is Markov inequality. Taking the supremum over the model and then the infimum over estimators proves the intended metric two-point risk bound. At $\eta =1$ the right side is zero. For $\eta >1$ a negative right side is also harmless, but a positive right side need not be a lower bound.

Here is an explicit counterexample with positive probability density functions and $n=3$. Let $A_0=[0,1/2)$, $\delta =1/100$, and take the model consisting ofUse the L2 norm metric, for which $D=d(f_0,f_1)=2(1-2\delta )$, and take $r_n=D/2$ for every $n$. The majority-half decision selects $f_0$ if at least two observations lie in $A_0$. Each error probability is $p=3{\delta }^2-2{\delta }^3=0.000298$, so its maximum normalized risk function is $2p=0.000596$. This is also the likelihood-ratio test, giving ${\mathbb{E}}_{f_0}|Z-1|=2(1-2p)$. At $\eta =3/2$ the proposed right side is $1/12-2p/3>0.083$, contradicting the risk of this explicit estimator. Thus the intended range restriction cannot be omitted.

For the second request use the intended L2 norm metric $d(f,g)=({\int }_0^1(f-g{)}^2{)}^{1/2}$. The PDF defines its square, not the square itself as a metric. PutBoth probability density functions belong to the model, and $d(f_0,f_{1,n})=a_n=2r_n$. The two-point choice may depend on $n$, because the lower bound is applied separately at each sample size. Under $f_0$, the product likelihood ratio $Z_n={\prod }_i(1+a_{n}g(X_i))$ satisfiesThis is the product chi-squared divergence. By Cauchy-Schwarz inequality, ${\mathbb{E}}_0|Z_n-1|<1/4$. With $\eta =1/2$, the proved metric two-point risk bound has normalized right side greater than $1/8$. ConsequentlyThe proof works for every $n\ge 1$ despite the introductory $n>2$. Measurability of the functions defining the probability density functions is understood. This establishes the requested lower bound; it does not assert a matching upper bound over this unrestricted class.

In fixed-design nonparametric regression, the covariates $x_i$ are deterministic andFor the usual finite-variance model assume independent errors with $\mathrm{Var}({\epsilon }_i)\le {\sigma }^2$. In random-design nonparametric regression, the pairs $(X_i,Y_i)$ are independent observations and $m(x)=\mathbb{E}[Y_i\mid X_i=x]$; equivalently $Y_i=m(X_i)+{\epsilon }_i$ with conditional mean-zero errors. Conditional variances are generally assumed bounded for risk bounds.

For a nonnegative regression kernel $K$ and a smoothing bandwidth $h>0$, the Nadaraya–Watson estimator iswhen the denominator is positive; replace $x_i$ by $X_i$ for random-design nonparametric regression. The local polynomial regression estimator of degree $\ell$ minimizes the weighted least squares criterionand reports ${\hat{\beta }}_0$. If the weighted Gram matrix is invertible, this minimizer is unique. The Nadaraya-Watson estimator is exactly local polynomial regression of degree zero. Higher degrees reproduce nonconstant polynomials and reduce boundary bias of an estimator.

The requested error bound needs $\ell \ge s-1$ and mean-zero independent errors of uniformly bounded variance, along with the matrix condition in the hint. The degree requirement is missing from the printed question. To see the problem, take $\ell =0$, $K=1_{[-1,1]}$, $m(t)=t$, zero errors, $x=0$, and $h=n^{-1/4}$. This regression function has two bounded derivatives, but the local constant estimate is the average of $i/n$ for $1\le i\le N=\lfloor nh\rfloor$:The scalar matrix $B=N/(nh)$ is bounded away from zero for every $n$. Nevertheless $(nh{)}^{-1/2}+h^2=n^{-3/8}+n^{-1/2}=o(h)$, so the printed conclusion with $s=2$ fails. A bounded-variance error hypothesis is also essential; otherwise an expectation bound need not even be finite.

Under the qualified hypotheses, use $U(t)=(1,t,\dots ,t^{\ell }/\ell !{)}^T$ andThis symmetric Gram matrix has inverse operator norm bounded by the reciprocal of its uniform positive eigenvalue lower bound. If $K$ is supported in $[-R,R]$, only $|x_i-x|\le Rh$ contribute. The equally spaced design and $nh\ge 1$ imply that there are at most $2Rnh+1\le Cnh$ such points. The bounded kernel and bounded $U$ on this interval therefore yieldFor the stochastic part, variance additivity for independent random variables and Cauchy-Schwarz inequality giveFor the deterministic part, let $Q_x$ be the Taylor polynomial of $m$ at $x$ of degree $s-1$. The polynomial reproduction property of local polynomial regression gives ${\sum }_i{W}_{ni}(x)Q_x(x_i)=Q_x(x)=m(x)$. By Taylor's theorem, $|m(x_i)-Q_x(x_i)|\le C|x_i-x{|}^s\le C{h}^s$ on the kernel window. ThusCombining the two parts provesThe argument also applies at the endpoints when the assumed matrix bound holds. With arbitrary degree, the generally valid smoothness term is $h^{\min (s,\ell +1)}$; extra symmetry can improve some interior biases but does not fix the general boundary counterexample.

Let $v=\mathrm{Var}(S)>0$ and $v_I=\mathrm{Var}(S_I)$. The variance-matching quota share reinsurance in this part has the nonnegative retained fraction ${\alpha }^{\ast }=\sqrt{v_I/v}$. For claimwise retentions $0\le h(x)\le x$, its existence within $[0,1]$ follows from $v_I=\lambda \mathbb{E}[h(X{)}^2]\le \lambda \mathbb{E}[X^2]=v$. The comparison below only needs the matching contract specified in the question.

Because $S_R=S-S_I$, the variance and covariance identities giveThe Cauchy-Schwarz inequality bounds $\mathrm{Cov}(S,S_I)\le \sqrt{v{v}_I}={\alpha }^{\ast }v$. Therefore the quota share comparison at matched retained variance isThe last expression is the sum of variances under the matching quota share reinsurance. HenceIf $v=0$, the given matching contract forces $v_I=0$, so all relevant totals are constant and equality holds. For $v>0$, equality in the bound is equivalent to $S_I-\mathbb{E}[S_I]={\alpha }^{\ast }(S-\mathbb{E}[S])$ almost surely.

Construct the densities using the Riesz representation theorem for a Hilbert space, rather than assuming the measure-decomposition conclusion. Put $\rho =\mu +\nu$, a finite dominating measure, and work in the real L2 space $L^2(\rho )$. The linear functionalis well-defined on its equivalence classes because $\nu \le \rho$. By the Cauchy-Schwarz inequality,Thus the Riesz representation theorem supplies $h\in L^2(\rho )$ such that $L(g)=\int ghd\rho$. Since $\rho$ is finite, $h$ is also integrable. Testing $g=1_A$ givesThe first identity and nonnegativity of $\nu$ rule out a positive-measure set where $h<0$; for example, test $\{h\le -1/n\}$. The second identity and nonnegativity of $\mu$ similarly rule out $\{h\ge 1+1/n\}$. Modify a measurable representative on a $\rho$-null set so that $0\le h\le 1$ everywhere. This is the Hilbert-space construction of dominated measure densities.