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Write ${\lambda }_n$ for Lebesgue measure on ${\mathbb{R}}^n$. The Prékopa–Leindler inequality says that if $0<\theta <1$ and nonnegative measurable functions $f,g,h$ satisfythen their Lebesgue integrals satisfyIt suffices initially to take $0<\int f,\int g<\infty$. Zero integrals give a trivial bound; infinite integrals can be handled by truncating $f$ and $g$ and applying the monotone convergence theorem. As usual, a zero factor makes the asserted lower bound zero.

Here is a one-dimensional proof using quantile functions. Put $F=\int f$, $G=\int g$, and let $x(u),y(u)$, $0<u<1$, be the quantile functions of the probability density functions $f/F$ and $g/G$. At almost every $u$ the quantile derivative identity givesFor completeness, these reciprocal derivative identities follow by differentiating the corresponding cumulative distribution functions at their Lebesgue points and then differentiating the inverse function relation. Sampling a quantile function at a uniform argument samples the density itself, so the exceptional set, including locations where the density vanishes or is infinite, has probability zero. Gaps in a density's support can give jumps in its quantile function; they do not invalidate the argument below.

The monotone function $z(u)=(1-\theta )x(u)+\theta y(u)$ obeys the monotone substitution inequality:This form uses the ordinary derivative of a monotone function, rather than any jump or singular part: its weighted image measure is dominated by Lebesgue measure. One can see this first for intervals, where ${\int }_{z^{-1}(I)}{z}^{\prime }\le {\lambda }_1(I)$, and then extend to nonnegative measurable functions by simple function approximation. Thus no assumption of strictly positive smooth probability density functions is hidden in the proof.

Using the hypothesis and the weighted arithmetic-geometric mean inequality, we obtainIntegrating over $0<u<1$ proves the one-dimensional Prékopa–Leindler inequality.

For higher dimensions, use mathematical induction and Tonelli theorem. Write $x=(x^{\prime },s)$ and setApply the one-dimensional Prékopa–Leindler inequality to the last-coordinate slices, with $z^{\prime }=(1-\theta )x^{\prime }+\theta y^{\prime }$. It givesApply the induction hypothesis in ${\mathbb{R}}^{n-1}$ and use Tonelli theorem once more. For finite total integrals, infinite slice integrals occur only on null sets and may be replaced by zero there; this only weakens the displayed premise. If the data are only Lebesgue measurable rather than Borel measurable, the exceptional nonmeasurable slices can be handled in the same way for $F_1,G_1$, and by setting $H_1=+\infty$ on its exceptional null set. The total integrals are unchanged, and the slice premise holds everywhere it is needed. The result therefore holds for general nonnegative measurable functions.

Write $J_{n,h}$ for the squared integral norm in question. The scaling property of the Fourier transform gives $\mathcal{F}{K}_h(u)=\mathcal{F}K(hu)$, which vanishes outside $[-1/h,1/h]$. Also $|\mathcal{F}K(v)|\le \Vert K{\Vert }_1$. The condition $\int K=1$ does not imply $\Vert K{\Vert }_1=1$, because $K$ need not be nonnegative.

Part (b), now for observations of $Y$, gives $\mathbb{E}|{\phi }_n^Y(u)-{\phi }^Y(u){|}^2\le 1/n$. Apply the Tonelli theorem to this nonnegative integrand, use the bound on $1/{\phi }_{ϵ}$, and change variables $v=hu$:The characteristic function ${\phi }_{ϵ}$ is continuous and nonzero on the compact interval of integration, so ${\tilde{M}}_h<\infty$ for every $h>0$. The Markov inequality gives, for every $z>0$,Consequently $J_{n,h}^{1/2}=O_{\mathbb{P}}({\tilde{M}}_h/\sqrt{nh})$, including for any deterministic bandwidth sequence $h=h_n>0$. The stochastic order follows directly from this uniform probability bound. Independence of $X$ and $ϵ$ gives ${\phi }^Y={\phi }^X{\phi }_{ϵ}$, explaining why division by ${\phi }_{ϵ}$ is the Fourier deconvolution operation.

Only routes with $n_r>0$ contribute to proportional fairness or to the weighted logarithmic utility; omit inactive routes from expressions with a rate in the denominator. Active routes have positive rates whenever a positive feasible allocation exists. For a feasible competitor $y$, the elementary logarithm inequality $\log u\le u-1$ givesA zero competitor rate on an active route has utility $-\infty$ and also cannot improve the objective. Therefore a proportionally fair allocation maximizes the logarithmic objective. Conversely, the feasible set is a convex set, and differentiating the objective along $x+t(y-x)$ at $t=0+$ shows that an optimizer satisfies the proportional fairness inequality. Strict concavity gives uniqueness on active coordinates; rates assigned to nonexistent flows are immaterial.

In the linear flow network, put $z_0=n_0{x}_0$ and $z_i=n_i{x}_i$ for active routes. A local route $i$ uses only resource $i$. If $n_i>0$ and $z_0+z_i<1$, increasing $x_i$ increases the objective without violating any other constraint. Thus each active local route saturates its resource:Write $L={\sum }_{i=1}^I{n}_i$ and $N=n_0+L$. For $n_0>0$ and $L>0$, substitute $z_i=1-z_0$ into the objective. Apart from constants its dependence on $z_0$ isIts derivative vanishes when $n_0/z_0=L/(1-z_0)$, and its second derivative is negative. Hence the per-flow through-route rate isIf $L=0<n_0$, the optimal through-route aggregate rate is one, so $x_0=1/n_0=1/N$ still holds. If $n_0=0<L$, each active local route has $x_i=1/n_i$, also given by the displayed local formula. When $N=0$ there are no flows and no departures. Inactive coordinates need not be assigned these undefined per-flow formulas.

Independent arrivals and exponential document sizes make this a flow-level network model. A document of type $r$ has residual-size hazard ${\mu }_r$ per unit of data transmitted; when its transmission speed is $x_r(n)$, its completion hazard per unit time is ${\mu }_r{x}_r(n)$. Thus the continuous-time Markov chain of flow counts has transition intensitiesFor $N>0$ these departure intensities are ${\mu }_0{n}_0/N$ for the through route and ${\mu }_{i}L/N$ for an active local route. Total arrival intensity is constant and total departure intensity is bounded by ${\sum }_r{\mu }_r$, ensuring a nonexplosive Markov chain.

Set ${\rho }_r={\nu }_r/{\mu }_r$, the offered traffic, and define the balance function of a flow-level networkThe binomial coefficients satisfyThese identities also cover states with only one kind of active route. They show that the weight $a(n)=\Phi (n){\prod }_{r=0}^I{\rho }_r^{n_r}$ satisfiesThus detailed balance for a continuous-time Markov chain proves that this is a reversible Markov chain and reduces the stationary distribution calculation to normalization.

Sum over $n_0$ first. The negative binomial series gives, for fixed local counts and $L={\sum }_i{n}_i$,All summands are nonnegative, so the Tonelli theorem permits exchanging the sums. The partition sum is thereforeIt is finite exactly when ${\rho }_0+{\rho }_i<1$ for every $i$, for $I\ge 1$. These are the resource-load conditions: resource $i$ receives the through-route offered traffic plus its local offered traffic. The normalized stationary distribution isWith positive arrival rates this is the unique stationary distribution of the irreducible feasible flow-count chain. Zero arrival rates restrict its closed class to the corresponding zero coordinates.

Summing out $n_0$ also shows that the local counts are independent, with geometric distributions on $0,1,\dots$:Hence their mean flow counts areThe binomial balance factor is essential: independent geometric counts for all routes would not give these departure rates or the correct resource-load normalization.

Let $A_2(t)={\int }_0^{t}1\{X(s)=2\}ds$ be the occupation time of a continuous-time Markov chain in the symptoms state. Tonelli theorem permits exchanging the nonnegative integral and expectation. Starting from state 1, put $q=\lambda +\mu >0$. ThensoFor small $t$ the answer is $\lambda t^2/2+O(t^3)$, as expected when onset must first occur. It is between zero and $t$. If $\lambda =\mu =0$, no transition can occur and the answer is zero; the displayed quotient is interpreted separately in that degenerate case.

For a nonnegative continuous event time, the survivor function is $F(t)=\mathbb{P}(T>t)$; here the printed $F$ denotes survival, not the usual cumulative distribution function. A lower median isFor a continuous strictly decreasing survivor function this is the unique solution of $F(m)=1/2$. If the curve has a flat stretch at $1/2$, the distribution has a nonunique median and the infimum specifies the lower endpoint.

The tail integral formula for moments follows from $T={\int }_0^{\infty }1\{T>t\}dt$ and Tonelli theorem. Therefore, when survival vanishes beyond $t^{\ast }$,If $F(t)>0$ at every finite time, the mean is finite precisely when ${\int }_0^{\infty }F(t)dt<\infty$. For a concrete sufficient condition, require $F(t)\le C{t}^{-1-\epsilon }$ for all $t\ge M>0$, with $C<\infty$ and $\epsilon >0$. The integral over $[0,M]$ is at most $M$, and the tail integral is at most $C{M}^{-\epsilon }/\epsilon$. An exponential tail bound also suffices. Merely having $F(t)\to 0$, or even $tF(t)\to 0$, does not suffice: a properly completed decreasing tail proportional to $1/(t\log t)$ for $t\ge e$ has infinite mean.

For a kernel for density estimation $K$ with integral one and a smoothing bandwidth $h>0$, the univariate kernel density estimator isIts expected value isThis is the convolution identity. It holds wherever the integral is defined, in particular everywhere for the bounded kernel used below. For a nonnegative kernel for density estimation, Tonelli theorem also gives the identity as an extended nonnegative integral. A signed kernel for density estimation instead requires absolute integrability for this interchange.

For the centred unit-width uniform kernel, substitute $u=(x-y)/h$:Put $M={\sup }_z|f^{\prime \prime }(z)|$. The Taylor theorem with Lagrange remainder givesThe odd function term integrates to zero, while ${\int }_{-1/2}^{1/2}{u}^{2}du=1/12$. Hence the bias of a kernel density estimator obeysThis is a second-order pointwise bias bound for kernel density estimation. Existence and boundedness of the second derivative suffice for the stated remainder; continuity of that derivative is not an extra assumption.

For concentration, writeThe $B_i$ are independent and identically distributed random variables with a Bernoulli distribution, and ${\hat{f}}_h(x)=(nh{)}^{-1}{\sum }_i{B}_i$. The required concentration bound follows from the Hoeffding lemma. We prove its Bernoulli case directly. Withwe have $\psi (0)={\psi }^{\prime }(0)=0$ and ${\psi }^{\prime \prime }(s)=p_s(1-p_s)\le 1/4$, where $p_s=p{e}^s/(1-p+p{e}^s)$ is the exponentially tilted success probability. Therefore $\psi (s)\le s^2/8$ for every real $s$. If $p=0$ or $p=1$, the centred variable is identically zero and the same bound holds directly.

Independence and the exponential form of the Markov inequality now give, for $s>0$,Choose $s=4ht$. Applying the same argument to the lower tail and adding the probabilities proves the Hoeffding inequality

To obtain the requested variance constant, use the additional bound that a probability is at most one. Set $Z={\hat{f}}_h(x)-\mathbb{E}{\hat{f}}_h(x)$ and $a=2n{h}^2$. The tail integral formula for moments, justified by Tonelli theorem, yieldsThusThe clipping at one is essential for this integration argument to give that constant. It is an instance of a second moment bound from a clipped Gaussian tail. The exact Bernoulli distribution calculation also gives $\mathrm{Var}({\hat{f}}_h(x))=p(1-p)/(n{h}^2)\le 1/(4n{h}^2)$, an even stronger bound.

Group the random resistance cost by unoriented edges. For $e=\{x,y\}$, its resistance contributes once whenever the walk traverses it in either direction. Pathwise,The graph is finite and the commute has finite expected value. Linearity of expectation, or Tonelli theorem for these nonnegative terms, therefore permits taking expectations of the sum. By the directed edge occupation in a random-walk commute, both directed traversal counts have expected value $R_{\mathrm{eff}}(a,z)$. ThusApply Foster's theorem from part (a) to obtainThe factor two counts the two orientations of each edge. It would be lost by treating the expected directed count in part (b) as an expected count for both directions together. The same distinct-terminal and simple-random-walk conventions from part (b) remain in force.

Put $L={\sup }_x\Vert \nabla f(x){\Vert }_{\infty }$, the uniform bound on the supremum norm of the gradient. The fundamental theorem of calculus on a line segment, together with the inner product inequality $|⟨a,b⟩|\le \Vert a{\Vert }_{\infty }\Vert b{\Vert }_1$, givesThus $f$ is Lipschitz continuous with respect to the L1 norm. The Gaussian random variables comprising $X$ and $Y$ have finite exponential moments of their absolute values. More explicitly, for $c\ge 0$, Hölder's inequality gives $\mathbb{E}{e}^{c\Vert X{\Vert }_1}\le {\prod }_{j=1}^n(\mathbb{E}{e}^{nc|X_j|}{)}^{1/n}<\infty$. Consequently $\mathbb{E}f(X)$ and every exponential expression below are finite, including when the covariance matrix is singular.

Use the convex function $\Psi (v)=e^{\lambda |v|}$. Since $Y$ is an independent copy of $X$, Jensen inequality for the conditional expectation givesFor $0\le \theta \le \pi /2$, define the Gaussian rotation of independent copiesWriting $\Sigma =\mathrm{Cov}(X)$, both rotated covariance matrices are $\Sigma$ and their cross-covariance is zero. The pair has a centered multivariate normal distribution, so its components are independent and $(U_{\theta },V_{\theta })$ has the same probability law as $(X,Y)$. This calculation uses no inverse of $\Sigma$ and therefore also proves the assertion for degenerate multivariate normal distributions.

The path $\theta \mapsto U_{\theta }$ runs from $Y$ to $X$. The chain rule and the fundamental theorem of calculus give the pathwise identityApply Jensen inequality to the uniform probability measure on $[0,\pi /2]$. Then use Tonelli theorem and the Gaussian rotation of independent copies to obtainThis proves the Gaussian rotation interpolation inequality in the required form:The stated almost-sure domination now gives an exponential moment of an absolute standard normal variable bound. If $a=\lambda \pi /2$, completing the square in the standard normal density yieldswhere $\Phi$ is the standard normal distribution function. Hence Markov inequality gives, for every $\lambda >0$,The exponent is minimized at $\lambda =4u/{\pi }^2>0$. Therefore the resulting Gaussian tail bound isNo independence of $Z$ from $X$ or $Y$ is needed: only the given almost-sure domination is used.

Put $Q=Q_r(x)$ and $v_Q=|Q{|}^{-1}{\int }_{Q}v$. The fundamental theorem of calculus along a line segment, followed by averaging, givesFor fixed $t$, use the change of variables formula $w=y+t(z-y)$. Its Jacobian determinant is $t^n$. The transformed domain $Q_t=y+t(Q-y)$ lies in $Q$, because a cube is a convex set. Moreover, $w\in Q_t$ implies $|w-y|\le \sqrt{n}rt$. Thus Tonelli theorem givesThe diagonal $w=y$ is a null set. The Holder inequality with $p^{\prime }=p/(p-1)$ now applies: $p>n$ implies $(n-1)p^{\prime }<n$, and integration in spherical coordinates bounds the kernel byThis proves the Morrey inequality on a cube, uniformly even when $y$ approaches its boundary:Apply this at both $y$ and the center $x$ and use the triangle inequality to obtain

Use the probability-system convention $\mu (X)=1$. The Birkhoff ergodic theorem, also called the pointwise ergodic theorem, states that for a measure-preserving system and $f\in L^1(\mu )$,where $\mathcal{I}$ is the invariant sigma-algebra. The limit is integrable and has the same integral as $f$. On a probability space the convergence also holds in $L^1$, as in the allowed mean ergodic theorem. If $T$ is an ergodic transformation, $\mathcal{I}$ is trivial modulo null sets, giving

The integer multiplication map on the circle preserves Lebesgue measure: for any integrable $g$ on $[0,1)$,To prove the ergodicity of integer multiplication on the circle, suppose $u\in L^2(m)$ satisfies $u\circ T_K=u$. Let $c_j$ be its Fourier coefficients in the Fourier basis $e_j(x)=e^{2\pi ijx}$. Since $e_j\circ T_K=e_{Kj}$, the Fourier coefficients of $u\circ T_K$ at index $j$ are zero if $K$ does not divide $j$, and are $c_{j/K}$ otherwise. This identity holds for all $L^2$ functions by approximation with trigonometric polynomials and the isometry $u\mapsto u\circ T_K$. Invariance givesEvery nonzero integer $j$ can be divided by $K$ only finitely often. Thus $c_j=0$ for every $j\ne 0$, and completeness of the Fourier basis makes $u$ constant almost everywhere. Applying this to the indicator function of an invariant set gives measure zero or one, so

A normal number in base $K$ has every word $w=(w_1,\dots ,w_r)$ of $r$ base-$K$ digits occurring with limiting overlapping frequency $K^{-r}$. Use the expansion that is not eventually equal to $K-1$ when there are two expansions. The word $w$ corresponds to the half-open intervalA word starting at position $n+1$ occurs exactly when $T_K^{n}x\in I_w$. The Birkhoff ergodic theorem, applied to $1_{I_w}$, gives frequency $m(I_w)=K^{-r}$ almost everywhere. There are countably many pairs $(K,w)$, so their full-measure sets have a full-measure intersection. In particular,These are absolutely normal numbers, so existence follows as well. This interval description also proves normality and equidistribution under integer multiplication: the base-$K$ intervals form arbitrarily fine grids, so their frequencies imply the correct frequency for every interval by approximation from inside and outside.

For the growth assertion, put $g=|f|$. For every $\epsilon >0$, the Tonelli theorem gives the useful summability boundSince $T$ is a measure-preserving transformation, $\mu (g\circ T^n>\epsilon n)=\mu (g>\epsilon n)$. The first Borel-Cantelli lemma shows that $g(T^{n}x)>\epsilon n$ occurs only finitely often almost everywhere. Intersecting the resulting full-measure sets for $\epsilon =1,1/2,1/3,\dots$ proves the linear growth bound for integrable observables, $f(T^{n}x)/n\to 0$. Multiplication by $n^{1-a}\le 1$ then gives

The threshold is sharp. For $0<a<1$, choose $p$ with $1<p<1/a$, and take the Bernoulli shift on $X=(0,1{)}^{\mathbb{Z}}$ with the product measure of independent uniform coordinates. The left shift preserves that measure because it preserves the probability of every finite-coordinate event. Define $f(x)=x_0^{-1/p}$; it is integrable becauseThe variables $f(T^{n}x)=x_n^{-1/p}$ are independent. For any fixed $M\ge 1$,The probability sum diverges, so the second Borel-Cantelli lemma makes these events occur infinitely often almost surely. Intersecting over positive integer $M$ even yields ${\mathrm{limsup}}_{n}f(T^{n}x)/n^a=\infty$. For $a\le 0$, the constant observable $f=1$ already fails to give limit zero. Thus the sharpness of the linear growth bound for integrable observables gives

Pointwise, an integer-valued nonnegative random variable has the representationwhen $T=k$, exactly the terms $n=0,\dots ,k-1$ equal one. Applying the Tonelli theorem, or the monotone convergence theorem to the finite sums of indicator random variables, gives the tail-sum formula for expectationBoth sides may be infinite; no integrability assumption is necessary.

Write $\Delta X_{n+1}=X_{n+1}-X_n$ and $Y={\sum }_{n\ge 0}|\Delta X_{n+1}|1_{\{T>n\}}$. Since $T$ is a stopping time, $\{T>n\}\in {\mathcal{F}}_n$. The Tonelli theorem, conditional expectation, and (c)'s tail-sum formula for expectation yieldThus $Y$ is an integrable random variable. Since $X_0=0$, telescoping gives $|X_T|\le Y$ and $|X_n|1_{\{T>n\}}\le Y{1}_{\{T>n\}}$. The dominated convergence theorem gives $\mathbb{E}[|X_n|1_{\{T>n\}}]\to 0$, so (b) gives uniform integrability of the stopped martingale.

The optional stopping theorem at the bounded stopping time $n\wedge T$ gives $\mathbb{E}{X}_{n\wedge T}=\mathbb{E}{X}_0=0$. Passing to the limit using the convergence in L1 proved in (b),One can also pass directly by the dominated convergence theorem, since $|X_{n\wedge T}|\le Y$.

Write $Z(u)$ for the continuous-time Markov chain and ${\mathbb{E}}_r$ for expected value conditional on $Z(0)=r$. The occupation time of a continuous-time Markov chain in state $s$ is the integral of its indicator random variable. By Tonelli theorem,The interchange is valid because the integrand is nonnegative; indeed the random integral is bounded by $t$.

Integrate the expected Markov-chain entry count density and use the occupation time of a continuous-time Markov chain formula:Each term is a transition intensity multiplied by the expected exposure time to that transition. This counts every entry, including returns, rather than only the first visit. For countably many states the same nonnegative interchange follows from Tonelli theorem, provided the resulting expected count is finite. The sum excludes $i=s$.

Put $v_p(u)=(1,u,\dots ,u^p{)}^T$ and $w_i(x)=K((x_i-x)/h)$, where $p$ is a nonnegative integer and $h$ is the smoothing bandwidth. The local polynomial estimator is the intercept of the weighted least squares fitWhen the local polynomial Gram matrix $G_p(x)={\sum }_i{w}_i(x)v_p((x_i-x)/h)v_p((x_i-x)/h{)}^T$ is invertible, the normal equations giveA singular local polynomial Gram matrix requires a selection convention and may leave the intercept unidentified; an arbitrary nonnegative regression kernel alone does not guarantee uniqueness. At $p=0$, provided ${\sum }_i{w}_i(x)>0$, the Nadaraya–Watson estimator isMultiplying every weight by $1/h$ changes neither fit.

For the uniform kernel, let $I_x=\{i:|i/n-x|\le h\}$ and $N_x=|I_x|$. Then ${\hat{m}}_h(x)=N_x^{-1}{\sum }_{i\in I_x}{Y}_i$. For $0<h<1/2$ and $x\in [h,1-h]$, the interval $[x-h,x+h]$ is inside $[0,1]$, and the regular design hasThe same lower bound holds when the interval starts at zero, despite the absence of the design point zero. In particular the Nadaraya–Watson estimator is defined throughout the integration interval.