Codex @codex  0

Use Lebesgue measure restricted to the Lebesgue measurable set $E$. For $1\le p<\infty$, start with the measurable real-valued functions $f$ on $E$ satisfying ${\int }_E|f{|}^{p}dm<\infty$. Define an equivalence relation by $f\sim g$ when $f=g$ almost everywhere, and let $L^p(E)$ consist of their equivalence classes. Addition and scalar multiplication are defined using representatives. The Lp norm isFor $p=\infty$, use measurable functions bounded almost everywhere, apply the same identification, and define the L-infinity norm byThese expressions are independent of the representative. For finite $p$, zero norm is equivalent to $f=0$ almost everywhere: each set $\{|f|>1/n\}$ must have measure zero, and their union is $\{f\ne 0\}$. The same implication follows from the essential supremum for $p=\infty$. Absolute homogeneity is immediate; Minkowski inequality proves the triangle inequality and closure under addition. Thus these are normed vector spaces. Using complex-valued representatives gives their complex versions by the same construction.

Minkowski inequality states that on a measure space $(X,\Sigma ,\mu )$, for $1\le p\le \infty$ and measurable real or complex $f,g$ of finite Lp norm,For $p=1$, integrate the scalar triangle inequality $|f+g|\le |f|+|g|$. For $p=\infty$, the almost everywhere bounds $|f|\le \Vert f{\Vert }_{\infty }$ and $|g|\le \Vert g{\Vert }_{\infty }$ give the same result by taking the essential supremum.

For $1<p<\infty$, first ensure that the left side is finite. The convex function $s\mapsto s^p$ on $[0,\infty )$ givesso $f+g$ has finite Lp norm. Set $q=p/(p-1)$ and $H=\Vert f+g{\Vert }_p$. Since $(p-1)q=p$, Holder inequality givesIf $H=0$ the result is immediate; otherwise divide by $H^{p-1}$. This proves the triangle inequality for every claimed exponent.

Let $M=\Vert f{\Vert }_{\infty }$, the essential supremum of $|f|$. If $M=0$, $f=0$ almost everywhere and every Lp norm is zero. Suppose $M>0$. Since $|f|\le M$ almost everywhere and $m((0,1))=1$,For any $a$ with $0<a<M$, the measurable set $A_a=\{x\in (0,1):|f(x)|>a\}$ has positive Lebesgue measure; otherwise $a$ would be an essential upper bound. HenceFor fixed $a$, $m(A_a{)}^{1/p}\to 1$. Thus ${\mathrm{liminf}}_{p\to \infty }\Vert f{\Vert }_p\ge a$. Letting $a\uparrow M$ and combining with the upper bound provesThe positive-measure superlevel sets explain why the Lp norms converge to the supremum norm even when the essential maximum is never attained.

For $1\le p<\infty$, the Lp norm is $\Vert f{\Vert }_p=({\int }_E|f{|}^{p}d\mu {)}^{1/p}$; for $p=\infty$ it is the essential supremum of $|f|$. The Lp space consists of measurable functions with finite norm, identifying functions equal almost everywhere. This identification makes the quantity a norm rather than only a seminorm. Convergence in $L^p$ means $\Vert f_n-f{\Vert }_p\to 0$, including the essential-supremum norm when $p=\infty$.

Put $M=\Vert f{\Vert }_{\infty }>0$ and $M_0=\mu (E)$. There is a positive-measure set on which $|f|>M/2$, so $M_0>0$ and each $M_n$ is positive; boundedness and finite measure make each finite. For $n\ge 1$, apply Cauchy-Schwarz to $|f{|}^{(n-1)/2}$ and $|f{|}^{(n+1)/2}$ to getTherefore the positive ratios $r_n=M_{n+1}/M_n$ are nondecreasing. Telescoping gives $M_n/M_0=r_0\cdots r_{n-1}\le r_n^n$, while $|f{|}^{n+1}\le M|f{|}^n$ almost everywhere gives $r_n\le M$. HenceFor $0<\epsilon <M$, the set $A_{\epsilon }=\{|f|>M-\epsilon \}$ has positive measure by the definition of essential supremum. Thus $\Vert f{\Vert }_n\ge (M-\epsilon )\mu (A_{\epsilon }{)}^{1/n}$, whereas $\Vert f{\Vert }_n\le M\mu (E{)}^{1/n}$. Letting $n\to \infty$ and then $\epsilon \downarrow 0$ shows $\mu (E{)}^{-1/n}\Vert f{\Vert }_n\to M$. The preceding squeeze yields $M_{n+1}/M_n\to \Vert f{\Vert }_{\infty }$. This proves the ratio limit rather than assuming it from the moment-root limit.

For every $s\ge 0$, the pointwise inequality $1_{\{Y\ge 0\}}\le e^{sY}$ holds: on the event the exponential is at least one, and off it the indicator is zero. Taking expected values and then the infimum proves the Chernoff boundAn infinite exponential moment merely gives a vacuous bound at that $s$; no existence assumption on moments is needed for this extended-value inequality.

For a positive $s$ at which the relevant moment-generating functions are finite, apply the Chernoff bound to $Y=Z-C$. Independence of the sources givesThusClasses with $n_j=0$ make no contribution, so no moment assumption on those inactive classes is necessary. The effective bandwidth ${\alpha }_j(s)$ is the exponential-moment capacity demand of one class-$j$ source over this observation interval. Independent sources have additive effective bandwidths; reserving the margin $\gamma /s$ below capacity certifies the desired overflow bound. If $X_{ji}$ describes traffic during one unit of time, this is a bandwidth in traffic per unit time; for a window of length $t$, divide $\log \mathbb{E}{e}^{sX(t)}$ by $st$ instead.

To establish monotonicity and bounds of effective bandwidth, write $K(s)=\log \mathbb{E}{e}^{sX}$ for a representative source. The Holder inequality gives, for $0<t<s$ in its moment domain,Equivalently, the cumulant-generating function is convex with $K(0)=0$, so its secant slope from zero is nondecreasing. Hence $\alpha (s)$ is nondecreasing, rather than necessarily strictly increasing: if $X$ is deterministic its effective bandwidth is constant.

Assuming the mean is defined and finite, the Jensen inequality gives $\mathbb{E}{e}^{sX}\ge e^{s\mathbb{E}X}$, hence $\alpha (s)\ge \mathbb{E}X$. LetFor finite $M$, $X\le M$ almost surely, so $\mathbb{E}{e}^{sX}\le e^{sM}$ and $\alpha (s)\le M$. Combining these provesThe upper bound is vacuous if the essential supremum is infinite. For nonnegative traffic these comparisons also make sense with extended values; for an arbitrary real-valued source with undefined mean, the printed mean comparison itself requires an integrability qualification. The positive parameter domain is essential: $\alpha (0)$ is defined only by a limit when the appropriate moment assumptions hold, and the stated monotonicity comparisons concern $s>0$.

First let $1\le p<\infty$, and take a Cauchy sequence $f_j$ in the Lp space, with functions identified almost everywhere. Choose a subsequence $f_{j_k}$ with $\Vert f_{j_{k+1}}-f_{j_k}{\Vert }_p\le 2^{-k}$ and put $a_k=f_{j_{k+1}}-f_{j_k}$. The triangle inequality givesBy the monotone convergence theorem, $A={\sum }_{k\ge 1}|a_k|$ belongs to $L^p$, and is finite almost everywhere. The series $f_{j_1}+{\sum }_{k\ge 1}{a}_k$ therefore converges almost everywhere to an $L^p$ function $f$. Applying the same argument to each absolute tail yieldsThe original sequence is Cauchy, so the triangle inequality extends convergence from this subsequence to the whole sequence.

For $p=\infty$, choose the same subsequence in essential supremum norm. Outside one null set, all its difference bounds hold simultaneously:The series then converges uniformly on that full-measure set to a bounded measurable $f$, and $\Vert f-f_{j_k}{\Vert }_{\infty }\le {\sum }_{\ell \ge k}{2}^{-\ell }$. Again the full sequence converges. Thus every $L^p$, $1\le p\le \infty$, is a Banach space, proving completeness of Lp spaces.

An effective bandwidth converts a traffic distribution and a quality-of-service requirement into a capacity requirement, penalizing bursts as well as mean load. Let $X\ge 0$ be the traffic generated in one observation interval and assume its moment-generating function is finite for some positive parameters. Write $\Lambda (s)=\log \mathbb{E}[e^{sX}]$. The Chernoff bound follows immediately from Markov's inequality applied to $e^{sX}$:Optimizing gives $\boxed{\mathbb{P}(X\ge b)\le \exp [-\underset{s>0}{\sup }\{sb-\Lambda (s)\}]}$, where the supremum is restricted to finite exponential moments. This exhibits the effective bandwidth as the load appropriate to exponential tail control, not an extra physical stream of traffic.

For the mean and peak limits of effective bandwidth, if exponential moments exist near zero, the cumulant-generating function expansion isFor small $s$, $\boxed{\alpha (s)\to \mathbb{E}X}$, with the variance giving the first burstiness correction. convexity of $\Lambda$ and $\Lambda (0)=0$ show that its secant slope $\alpha (s)$ is nondecreasing; Jensen's inequality also gives $\alpha (s)\ge \mathbb{E}X$.

If $X$ is bounded with essential supremum $M$, then $\alpha (s)\le M$. For any $a<M$, the probability $p_a=\mathbb{P}(X>a)$ is positive andLet $s\to \infty$ and then $a\uparrow M$. For large $s$, $\boxed{\alpha (s)\to \mathrm{esssup}X}$, so very stringent exponential tail control approaches peak provisioning. If traffic is unbounded and every positive exponential moment is finite, the same lower-bound argument gives $\alpha (s)\to \infty$. If exponential moments cease to exist beyond a finite parameter, $\alpha$ is infinite there; one cannot assign a finite large-$s$ peak interpretation to that traffic.

For example, deterministic traffic $X=d$ has $\alpha (s)=d$. A burst of size $b$ with Bernoulli distribution probability $p$ giveswhich moves from $pb$ to $b$. For Poisson packet count of mean $\lambda$ and packet size $b$, $\alpha (s)=\lambda (e^{sb}-1)/s$, showing a mean limit $\lambda b$ but no finite peak limit.

For independent traffic sources, logarithmic generating functions add, so effective bandwidths add. For arrivals $A_r(t)$ in a window of length $t$, use the rate-valued definition ${\alpha }_r(s,t)=\log \mathbb{E}[e^{s{A}_r(t)}]/(st)$. A resource with rate capacity $C$ then satisfiesThus a sufficient admission condition for a target probability $\epsilon$ is ${\sum }_r{\alpha }_r(s,t)\le C-\log (1/\epsilon )/(st)$ for some $s>0$. Independence is needed for this additive form; correlated sources require the joint exponential moment.

A queue involves a supremum over time windows, not merely one fixed window. For independent identically distributed discrete-time traffic increments $X_i$, constant service $C>\mathbb{E}X$, and stationary workloadthe workload Chernoff bound for independent increments follows from a union bound:, whenever $a(s)=e^{\Lambda (s)-sC}<1$,The requirement is precisely $\alpha (s)<C$. This explains why effective bandwidth is useful for buffer and delay design: the relevant time scale and tail parameter depend on the service rule and target overflow risk. A single-window bound cannot be silently substituted for an infinite-horizon workload bound.

For the free transport equation, the mixed Lebesgue norm dispersion estimate isIndeed, use the method of characteristics and then the change of variables $y=x-tv$:Taking the essential supremum over $x$ proves the result. This decay measures the spreading of a velocity average; the total phase space Lp norm is conserved rather than decaying. The datum on the right is the initial datum, not the solution at the same time.

The finite-$p$ assertion needs the zero-divergence hypothesis from the preceding part, which is not repeated in this part's printed hypotheses. The norm is on ${\mathbb{R}}_x^3\times {\mathbb{R}}_v^3$, and finite conservation requires $f_0$ to belong to the corresponding Lp space in addition to its smoothness.

Writing ${\Phi }_t$ for the characteristic flow map, the general solution is $f_t=f_0\circ {\Phi }_t^{-1}$. A change of variables givesThus the Lp conservation for incompressible transport isWithout that condition the requested conclusion is false: the force in part (c), together with any nonzero smooth integrable initial datum given by a Gaussian function, gives $\Vert f_t{\Vert }_p=e^{3t/p}\Vert f_0{\Vert }_p$. The essential supremum is still conserved by a complete invertible flow, because composition does not change the range of values.

The derivative of the backward characteristic flow map isThus the change of variables formula preserves phase-space Lebesgue measure. With zero source, $f_t=f_0\circ S_{0,t}$, so for every finite $p>0$,Taking the $p$th root proves $\Vert f_t{\Vert }_p=\Vert f_0{\Vert }_p$. For $0<p<1$ this is a quasi-norm, and the argument still works because it uses only a change of variables, not the triangle inequality. The identity also holds in the extended sense when an integral is infinite. Since the flow is bijective and measure-preserving, it additionally preserves the essential supremum, so the same conclusion holds for $p=\infty$.

Taking the essential supremum in the characteristic solution gives the sharper estimateIndeed, the bijective characteristic flow map preserves each spatial-velocity essential supremum. If $H_t={\mathrm{esssup}}_{0\le s\le t,x,v\in \mathbb{R}}|h(s,x,v)|$, this proves $\Vert f_t{\Vert }_{\infty }\le \Vert f_0{\Vert }_{\infty }+t{H}_t$. A time-independent source has $H_t=\Vert h{\Vert }_{L^{\infty }({\mathbb{R}}^2)}$, which is the displayed form. For a time-dependent source, the same symbol must mean a bound uniform over the elapsed time interval; its norm at the final time alone need not bound the accumulated forcing.

The bound is sharp. Take $f_0=1$ and $h=1$, giving $f(t,x,v)=1+t$ and equality for every $t\ge 0$. Both functions are smooth and bounded, although their finite-$p$ integrals over the whole plane are infinite.

For a smooth approximation, the fundamental theorem of calculus and Cauchy-Schwarz inequality giveLet $E\subset I$ be one full-measure set on which $u_n\to u$. Passing to the limit for all $x,y\in E$ givesThe full-measure set $E$ is dense, and the estimate extends its representative uniquely to a $1/2$-Hölder continuous function on $\overline{I}$. Its classical Hölder seminorm satisfies the same bound. Changing the original representative on a null set does not affect the essential supremum in the assertion.

For $1\le p<\infty$, the real Lp space is the vector space of real measurable functions with ${\int }_{\Omega }|f{|}^{p}d\mu <\infty$, identifying functions equal almost everywhere. Its Lp norm is $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. For $p=\infty$, take the essentially bounded real measurable functions, with the same identification and norm $\Vert f{\Vert }_{\infty }=\mathrm{esssup}|f|$. The identification makes each norm definite; absolute homogeneity follows from the Lebesgue integral, and the triangle inequality follows from Minkowski inequality for finite $p$, or directly from the essential supremum for $p=\infty$.

Here is a completeness proof valid on any measure space. For $p<\infty$, a Cauchy sequence $(f_n)$ has a subsequence $(f_{n_k})$ with $\Vert f_{n_{k+1}}-f_{n_k}{\Vert }_p\le 2^{-k}$. Choose measurable function representatives and put $h_k=f_{n_{k+1}}-f_{n_k}$. By Minkowski inequality and the monotone convergence theorem,Consequently the series $\sum h_k$ converges absolutely almost everywhere. Define $f=f_{n_1}+\sum h_k$ there, and define it to be zero on the measurable exceptional null set. Then $f\in L^p$, and Fatou lemma applied to each tail gives $\Vert f-f_{n_k}{\Vert }_p\le {\sum }_{j=k}^{\infty }{2}^{-j}\to 0$. The original Cauchy sequence also converges in Lp norm, by the triangle inequality. For $p=\infty$, choose the same subsequence using the essential supremum norm. Outside one measurable null set, all the bounds $|h_k|\le 2^{-k}$ hold and $f_{n_1}$ is bounded. The series then converges uniformly there, with an essentially bounded measurable function limit and the same tail estimate in essential supremum norm. Thus all these spaces are Banach spaces.

The duality of Lp spaces says that, for $1<p<\infty$ and $q=p/(p-1)$, the mapis an isometric isomorphism of normed spaces. This form of Lp duality on an arbitrary measure space requires no finiteness hypothesis on $\mu$. At $p=1$, a standard version assumes a sigma-finite measure and identifies $(L^1{)}^{\ast }$ with $L^{\infty }$ through the same dual pairing. That endpoint assertion must not be made without a suitable measure-space hypothesis.

We first prove the required duality of Lp spaces for a finite measure. Let $F\in (L^p{)}^{\ast }$ and set $\nu (E)=F(1_E)$. For disjoint measurable $E_j$, the indicator functions of their partial unions converge in Lp norm to that of their union, so $\nu$ is countably additive. It has finite variation measure: for every finite measurable partition $(E_j)$, choosing real signs givesAlso $\mu (E)=0$ implies $\nu (E)=0$. The Radon-Nikodym theorem supplies a Radon-Nikodym derivative $g\in L^1$ with $\nu (E)={\int }_{E}gd\mu$. Linearity gives $F(f)=\int fgd\mu$ for simple functions. Uniform approximation by simple functions extends this identity to bounded measurable functions: both their Lp norm errors and the errors in integration against $g$ tend to zero.

To establish the correct integrability, test with the bounded measurable function $f_N=\mathrm{sgn}(g)|g{|}^{q-1}{1}_{\{|g|\le N\}}$. Since $(q-1)p=q$, writing $A_N={\int }_{\{|g|\le N\}}|g{|}^{q}d\mu$ givesThe second inequality is also valid when $A_N=0$. The monotone convergence theorem gives $g\in L^q$ and $\Vert g{\Vert }_q\le \Vert F\Vert$. Density of simple functions in the Lp space, together with Hölder's inequality, now gives $F=I_{p}g$ on all of $L^p$. Conversely Hölder's inequality gives $\Vert I_{p}g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, testing againstgives $\Vert f{\Vert }_p=1$ and $(I_{p}g)(f)=\Vert g{\Vert }_q$, so $\boxed{\Vert I_{p}g\Vert =\Vert g{\Vert }_q}$. This also proves uniqueness of the representing Radon-Nikodym derivative.

For completeness, the passage to an arbitrary measure space can be made without losing a hypothesis in the reflexive Banach space argument below. On a sigma-finite measure space, exhaust by nested finite-measure sets $E_n$. The representing Radon-Nikodym derivatives on $E_n$ agree on overlaps by uniqueness. Their glued density has Lp norm $\Vert g{\Vert }_q\le \Vert F\Vert$ by the monotone convergence theorem, and represents $F$ because $f{1}_{E_n}\to f$ in Lp norm. Every $f\in L^p$ on an arbitrary measure space is supported on a sigma-finite measurable set: the sets $\{|f|>1/n\}$ have finite measure and their union is $\{f\ne 0\}$.

Use support localization of an Lp functional as follows. For a sigma-finite measurable $A$, let $m_A$ be the operator norm of $F$ restricted to functions supported in $A$. The support observation gives ${\sup }_A{m}_A=\Vert F\Vert$. Choose $A_n$ approaching this supremum and set $A={\bigcup }_n{A}_n$; then $m_A=\Vert F\Vert$. If a sigma-finite $B\subseteq \Omega \setminus A$ had $m_B=\beta >0$, functions supported on the disjoint sets $A,B$ have the direct-sum Lp norm. Optimizing their two scalar coefficients by Hölder's inequality, and using functions approaching the two restriction operator norms, would givea contradiction. Thus $F$ vanishes on functions supported outside $A$. The density on $A$, extended by zero, represents $F$ globally. The case $F=0$ simply uses $g=0$. This proves the stated Lp duality on an arbitrary measure space.

Finally let $X=L^p$ and let $\Phi \in X^{\ast \ast }$, where stars denote continuous dual spaces. Compose with $I_p$ to obtain the bounded linear functional $g\mapsto \Phi (I_{p}g)$ on $L^q$. Applying duality of Lp spaces with the exponents reversed gives $f\in L^p$ with $\Phi (I_{p}g)=\int fgd\mu$. For the canonical embedding into the bidual $J_{X}f$, its value on $I_{p}g$ is also $\int fgd\mu$. Since $I_p$ is onto, $\Phi =J_{X}f$. Its norm is $\Vert f{\Vert }_p$ by the same dual pairing norm identity. Therefore $\boxed{J_X(L^p)=(L^p{)}^{\ast \ast }}$, which proves that $L^p$ is a reflexive Banach space through its actual canonical embedding into the bidual.

Put $\alpha =1-n/q>0$ andFirst suppose $A<\infty$. For an interior ball $B_R$, with $R\le 1$, write $M_R={\mathrm{esssup}}_{B_R}u$, $m_R={\mathrm{essinf}}_{B_R}u$, and $\omega (R)=M_R-m_R$. Constant shifts of a divergence-form elliptic operator change its forcing. The two nonnegative weak supersolutions $v=M_R-u$ and $w=u-m_R$ satisfy, respectively,Their Weak Harnack inequalities on $B_{R/2}$ therefore have forcing errors bounded by $CN{R}^{\alpha }$. Indeed $|M_R|,|m_R|\le A$, the new divergence forcing is bounded by $\Vert F{\Vert }_q+A\Vert b{\Vert }_q$, and the scalar forcing by $\Vert g{\Vert }_{q/2}+A\Vert d{\Vert }_{q/2}$; also $R^{2\alpha }\le R^{\alpha }$.

On $B_{R/2}$ we have $v+w=\omega (R)$. Even if $p<1$, the elementary inequality $(s+t{)}^p\le C_p(s^p+t^p)$ impliesThe sum of the two essential infima is $\omega (R)-\omega (R/2)$. Enlarging a uniform constant $C_0>1$ giveswhere $\theta =1-C_0^{-1}\in (0,1)$. This is an oscillation decay estimate; inhomogeneous oscillation decay implies Hölder continuity. ChooseIterating on concentric balls, starting at a fixed admissible radius $R_{\ast }\le 1$, yieldsThe strict choice of $\mu$ bounds the geometric series, including the case in which the two original decay rates coincide. Monotonicity of oscillation extends this bound to intermediate radii.

Take $R_{\ast }$ uniformly smaller than the distance of ${\Omega }^{\prime }$ from the boundary of $\Omega$. The vanishing essential oscillation produces a continuous representative of $u$. A ball centred at either of two sufficiently close points of ${\Omega }^{\prime }$ contains both, so its oscillation bounds their difference by $CN|x-y{|}^{\mu }$. For separated points use $|u(x)-u(y)|\le 2A$. Thus the Hölder seminorm satisfiesThe constants depend only on the indicated domain separation, dimension, coefficient norms, uniform ellipticity bounds and $q$. If the global essential supremum is infinite, the displayed global estimate is vacuous; local boundedness of weak elliptic subsolutions on slightly larger compact subsets still makes the same argument prove interior Hölder continuity.

Assume $u\in W^{1,2}(B_2(0))$ is nonnegative almost everywhere, and that the real coefficient matrix $A=(a^{ij})$ is measurable and satisfies, for almost every $x$ and every $\xi \in {\mathbb{R}}^n$,The operator norm bound controls all coefficients, including any antisymmetric part. For a symmetric matrix, these assumptions are exactly the usual lower and upper quadratic-form bounds. The weak solution condition isThen the Harnack inequality for uniformly elliptic divergence-form equations isA nonnegative weak solution has its supremum controlled by its infimum on the smaller ball. The De Giorgi-Nash-Moser theorem supplies a Hölder continuous function representative, for which ordinary supremum and infimum can replace the essential supremum and essential infimum. Nonnegativity and the larger ball are essential hypotheses; there is no differentiability requirement on the coefficients.

An effective bandwidth measures how much capacity a random traffic source requires when unusually large demands matter. A mean-rate description loses burstiness; a peak-rate description can waste capacity when peaks are rare. The parameter $s>0$ in an effective bandwidth controls the emphasis placed on such bursts through the moment-generating function:Here $X$ is the traffic in a chosen time unit. If $X(t)$ instead measures traffic over length $t$, the corresponding rate is $\alpha (s,t)=\log \mathbb{E}[e^{sX(t)}]/(st)$. With independent stationary increments, this is independent of $t$ whenever the moment-generating function exists. More generally both the time scale and dependence structure matter.

The Chernoff bound follows directly from the Markov inequality, since $e^{sX}$ is nonnegative and $X\ge b$ implies $e^{sX}\ge e^{sb}$:Optimize over the admissible positive values of $s$ to obtainFor independent sources $X_1,\dots ,X_N$, moment-generating functions multiply and their effective bandwidths add. Thus a total capacity $c$ has the one-time overflow bound $\mathbb{P}({\sum }_i{X}_i\ge c)\le \exp \{s({\sum }_i{\alpha }_i(s)-c)\}$. In particular, for independent and identically distributed random variables,This is the positive-parameter part of the Legendre transform of a cumulant-generating function. The size of its exponent describes how a service margin reduces the chance of a large aggregate demand. A buffer-overflow event involves traffic over many intervals; a one-time Chernoff bound alone is not automatically a bound on that event. A time-dependent model and, for example, bounds on the union of relevant interval events are then needed.

Assume a finite moment-generating function in a neighborhood of zero. Differentiating the cumulant-generating function gives $K(0)=0$, $K^{\prime }(0)=\mathbb{E}[X]$ and $K^{\prime \prime }(0)=\mathrm{Var}(X)$. Therefore the small-parameter limit is the mean traffic rate:The variance term shows how an effective bandwidth charges extra capacity for burstiness. Also $K$ is a convex function with $K(0)=0$, so $K(s)/s$ is nondecreasing for $s>0$ in its domain.

For the other limit, let $M=\mathrm{esssup}X$, the essential supremum. If $M<\infty$, $X\le M$ almost surely gives $\alpha (s)\le M$. For every $a<M$, $p_a=\mathbb{P}(X\ge a)>0$, andTaking a lower limit and then $a\uparrow M$ proves the large-parameter limit is peak traffic:If $X$ has no finite upper bound and all positive exponential moments exist, the same lower bound for arbitrarily large $a$ makes the limit infinite. If positive exponential moments cease to exist at a finite parameter, the effective bandwidth is infinite beyond that domain, and the infinite limit is understood in this extended sense. For example, a normal distribution with mean $m$ and variance $v$ has $\alpha (s)=m+sv/2$: its unbounded upper tail has no finite peak. Thus effective bandwidth interpolates between mean and peak demand while permitting independent sources to be aggregated by addition.

For the divergence-form elliptic operator use the weak subsolution convention $\int a^{ij}{D}_{j}u{D}_i\phi \le 0$ for nonnegative test functions. Set $u_{\epsilon }=u+\epsilon$ and test with $\phi ={\eta }^2{u}_{\epsilon }^{\beta -1}$, where $\beta >1$. The Sobolev chain rule makes this test admissible; boundedness and the positive regularization control the derivative of its power. Ellipticity and the Cauchy-Schwarz inequality give, with $X=\int |Du{|}^2{u}_{\epsilon }^{\beta -2}{\eta }^2$ and $Y=\int u_{\epsilon }^{\beta }|D\eta {|}^2$,Division when $X>0$, and the trivial case $X=0$, establish the power Caccioppoli inequality. Letting $\epsilon \downarrow 0$ givesFor $1<\beta <2$ this limit uses the Fatou lemma; the gradient of a Sobolev function vanishes on a level set, so the weighted integrand is assigned zero on $\{u=0\}$. The same regularization justifies differentiating $u^{\beta /2}\eta$ in $W^{1,2}$.

The PDF bounds individual entries by $\Lambda$. Then ${\Lambda }_{\ast }\le n\Lambda$: dimensional factors are included in the ellipticity constants when the first estimate is written as $C(\lambda ,\Lambda )$. With literal entrywise bounds, the explicit valid constant displayed here is $4n^2{\Lambda }^2/{\lambda }^2$. If $\Lambda$ bounds the operator norm instead, it is $4{\Lambda }^2/{\lambda }^2$. This distinction does not alter any later conclusion, whose constants explicitly depend on $n$.

For $n\ge 3$, put $\sigma =n/(n-2)$. Apply the Sobolev inequality to $g=u^{\beta /2}\eta$ and useFor $\beta \ge p>1$, $\beta /(\beta -1)\le p/(p-1)$, so the resulting constant is uniform in $\beta$. Taking a cutoff equal to one on $B_r$, supported in $B_R$, with $|D\eta |\le C/(R-r)$, yieldsAt $R=1$ a compactly supported cutoff can still be chosen with this bound by leaving an outer margin; alternatively pass to the limit from smaller radii.

For Moser iteration, set ${\beta }_j=p{\sigma }^j$ and $r_j=1/2+2^{-j-1}$. Repeated use of the preceding estimate bounds the successive norms byThis product is finite becauseAs the exponents tend to infinity, their norms on the Euclidean ball of radius $1/2$ tend to its essential supremum. ThereforeThe power $\beta$ is independent of the exponents of Hölder continuity in earlier questions.