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Let $A$ be a nonempty subset of a metric space $X$, and let $f:A\to \mathbb{R}$ satisfy a Lipschitz bound $L$. Thenare real-valued Lipschitz functions on all of $X$, agree with $f$ on $A$, and obey the same bound $L$. Every other extension $H$ satisfying that bound lies between them, $G\le H\le F$. Despite the infimum in its formula, $F$ is the largest such extension: every given sample supplies an upper constraint $H(x)\le f(a)+Ld(x,a)$.

Take three real samples $f(0)=0$, $f(1)=1/2$, $f(2)=0$, and allow the bound $L=1$. Their least Lipschitz constant is only $1/2$, so this is deliberately a looser permitted bound. The two extensions areAt $x=1/2$, the allowable interval is $0=G(1/2)\le H(1/2)\le F(1/2)=1/2$. The interpolant obtained by linear interpolation is a piecewise linear function and has value $1/4$ there. Extending it constantly outside $[0,2]$ obeys the same bound on the entire real line. At $x=5/2$, the allowable interval is $[-1/2,1/2]$. Thus the data and the bound control possible values without selecting a unique curve.

The construction is useful when data are known only on a subset: it extends them without increasing an allowed Lipschitz bound. A short justification explains why the envelopes work. Fix $a_0\in A$. The triangle inequality and the bound on $f$ implyso the infimum is finite below; the term $a=a_0$ makes it finite above. At a sample $x=a_0$, all upper terms are at least $f(a_0)$, while that term gives equality. Finally, $d(x,a)\le d(y,a)+d(x,y)$ gives $F(x)\le F(y)+Ld(x,y)$ after taking an infimum; exchanging $x,y$ proves the bound. The lower envelope follows by applying the same argument to $-f$.

E. J. McShane's paper Extension of range of functions appeared in the Bulletin of the American Mathematical Society, volume 40, in 1934, pp. 837–842. ^{[ref] [ref]}

A minimum of $A$ is a member of $A$ less than or equal to every member of $A$. It equals the infimum when that lower bound is attained. A nonempty finite subset of a totally ordered set has a minimum; a nonempty open real interval does not.

If a convex roof extension is non-increasing on average for every pure input to a quantum instrument, then it is non-increasing on average for mixed inputs too. For an input ensemble $\{p_j,{\pi }_j\}$, let $q_{k|j}$ and ${\sigma }_{jk}$ be its conditional outcome probabilities and states. Linearity of each unnormalized outcome map gives $P_k{\rho }_k={\sum }_j{p}_j{q}_{k|j}{\sigma }_{jk}$. Convexity gives ${\sum }_k{P}_{k}F({\rho }_k)\le {\sum }_{j,k}{p}_j{q}_{k|j}F({\sigma }_{jk})\le {\sum }_j{p}_j\mu ({\pi }_j)$. Taking the infimum over input ensembles proves the result, even when conditional outputs of pure inputs are mixed, provided the averages are well defined.

If $I_1\supseteq I_2\supseteq \cdots$ are nonempty closed intervals with finite endpoints, then at least one real number belongs to every member. If their lengths tend to zero, exactly one real number belongs to every member. The result turns an unlimited sequence of consistent enclosures into an existence guarantee.

Write $I_n=[a_n,b_n]$. Nesting makes the left endpoints nondecreasing and the right endpoints nonincreasing. Their bounds give finite supremum and infimum, withIndeed, every $b_n$ is an upper bound for all the $a_j$, so the least-upper-bound property gives ${\sup }_j{a}_j\le b_n$ for every $n$. This proves the displayed interval is nonempty. If $b_n-a_n\to 0$, its two endpoints must coincide. This is the connection to the completeness of the real numbers discussed in Günther Hörmann and David Langer, Einführung in die Analysis, §§3.15–3.16: nested intervals and order completeness.

The distinction between existence and uniqueness is useful in computation. If an unknown point $x_{\ast }$ lies in every $[a_n,b_n]$, the midpoint $m_n=(a_n+b_n)/2$ satisfiesThus the enclosure itself certifies an absolute error without knowing the answer. The bisection error bound is the special case where each update halves the width.

Closedness and a finite initial enclosure matter. The open intervals $(0,1/n)$ have no common point, while the closed but unbounded intervals $[n,\infty )$ also have empty intersection. A picture of a few intervals cannot certify the intersection of an infinite family; the hypotheses of the nested interval theorem do that work.

Take $f_2(x)=x$. It is continuous and bounded on $(0,1)$, with infimum zero and supremum one, neither attained on this open interval.

Let $f$ be continuous on a closed bounded interval $[a,b]$. First prove that $f$ is a bounded function. If it failed, we could choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_j}\to x_{\ast }$. Because the interval is closed, $x_{\ast }\in [a,b]$. Continuity gives $f(x_{n_j})\to f(x_{\ast })$, contradicting $|f(x_{n_j})|>n_j\to \infty$.

Now let $M=\sup \{f(x):x\in [a,b]\}$, which exists by completeness of the real numbers and the bound just proved. Choose $y_n\in [a,b]$ with $f(y_n)>M-1/n$. A convergent subsequence has limit of a sequence $y_{\ast }\in [a,b]$, and continuity gives $f(y_{\ast })=M$. Applying the same argument to $-f$ gives attainment of the infimum. This proves the extreme value theorem: the maximum and minimum both exist and are attained. The argument also covers a one-point interval, where the conclusions are immediate.

For any trigonometric polynomial $t_n$ of degree at most $n$, reproduction from (a) and the operator norm bound from (b) implyTaking the infimum over $t_n$ givesThis is the polynomial reproduction error bound with $\Vert v_n\Vert \le 3$; attaining the best uniform approximation is not needed for this inequality.

Put $\alpha =-\log |c|>0$. Since $f$ has no zeros and the unit disc is simply connected, it has a holomorphic logarithm. Thus $F=-\log f$ is holomorphic with $\mathrm{Re}F>0$ and $\mathrm{Re}F(0)=\alpha$. The Cayley transform between the half-plane and disk givesThe Schwarz lemma yields $|{\varphi }^{\prime }(0)|\le 1$, so the Carathéodory derivative bound for the right half-plane gives $|F^{\prime }(0)|\le 2\alpha$. Since $f^{\prime }=-F^{\prime }f$, the sharp answer isEquality in the Schwarz lemma holds exactly when $\varphi (z)=\omega z$, $|\omega |=1$. Solving for $F$ and exponentiating gives exactly the extremizersTheir logarithmic real part is $\alpha \mathrm{Re}[(1+\omega z)/(1-\omega z)]>0$, so each is indeed a zero-free map into the unit disc with the prescribed value at zero.

For a fixed $w\in \mathbb{D}$, put $r=|w|$. Apply the Harnack inequality on the unit disk to the positive harmonic function $\mathrm{Re}F$ and then exponentiate its negative. The zero-free Schwarz lemma givesFor $w\ne 0$ both endpoints are attained: take $\omega w=r$ for the lower endpoint and $\omega w=-r$ for the upper endpoint in the displayed extremizers. At zero both are $|c|$. If the question is interpreted as allowing $w$ to vary as well, the sharp infimum and supremum over all admissible pairs $(f,w)$ are respectively $0$ and $1$, approached as $r\uparrow 1$; neither value is attained inside the unit disc. An individual function need not have these global extremal values: the constant function $f=c$ is a simple example.

For a nonempty set $S\subseteq \mathbb{R}$, a real number $s$ is its supremum if every $x\in S$ satisfies $x\le s$ and every upper bound $u$ of $S$ satisfies $s\le u$. Equivalently, $s$ is an upper bound and, for every $\epsilon >0$, some $x\in S$ has $s-\epsilon <x\le s$. The least upper bound axiom states that every nonempty set of real numbers bounded above has a real supremum.

Suppose $(a_n)$ is nondecreasing and bounded above. Apply the least upper bound axiom to $S=\{a_n:n\ge 1\}$, obtaining $s=\sup S$. For any $\epsilon >0$, $s-\epsilon$ cannot be an upper bound, so there is $N$ with $a_N>s-\epsilon$. For every $n\ge N$, monotonicity givesso $|a_n-s|<\epsilon$. This is precisely the definition of a convergent sequence. If $(a_n)$ is nonincreasing and bounded below, apply the proved case to $(-a_n)$; its limit is the negative of the infimum of the original terms. Thus the bounded monotone sequence theorem follows directly from the axiom:

Put $s_i={\sup }_{C_i}f$ and omit empty sets. Since the sets cover the space, boundedness givesFor nonnegative numbers $u_{n,i}$, their sum lies between their maximum and m times their maximum. Consequently its scaled logarithmic upper limit is at most ${\max }_i{\mathrm{limsup}}_n{n}^{-1}\log u_{n,i}$; the added $n^{-1}\log m$ vanishes. Each $C_i$ is closed, so the upper bound in the large deviation principle givesThe convention $\log 0=-\infty$ also handles sets of zero probability or infinite infimum rate.

Take $m=\lceil (b-a)/\epsilon \rceil$. The ceiling, as printed in the PDF, makes the intervals cover the bounded range of f. If $a=b$, use the single set $\mathcal{X}$ instead, since f is constant. By continuity their preimages are closed. On every nonempty preimage $C_i$, the oscillation of f is at most epsilon. Hence for every $y\in C_i$,Taking the supremum over y in the second expression gives the bound for ${\sup }_{C_i}f-{\inf }_{C_i}I$; if that infimum is infinite, the bound is automatic. Apply the finite-cover estimate to obtainThis is the exponential-integral upper estimate in the bounded form of Varadhan lemma.

Write $a_n=n^{2(1-H)}$ and $b_n=n^{2(1-G)}$. Then $a_n/b_n=n^{2(G-H)}\to \infty$. The large deviation principle on the whole space forces $\inf I=0$; compact sublevel sets and lower semicontinuity make that infimum attained. The unique possible zero is therefore mu, and $I(\mu )=0$.

If a closed set C contains mu, its new infimum rate is zero and the claimed upper bound follows from probabilities being at most one. If it excludes mu, then $\eta ={\inf }_{C}I>0$, possibly infinite: otherwise a sequence of rates tending to zero would have a convergent subsequence in a compact sublevel set, with limit in C and zero rate, contradicting uniqueness. This is the fact that a good rate function is separated from zero away from its minimizer.

Choose any $0<c<\eta$. The original upper bound gives $\log \mathbb{P}(X_n\in C)\le -c{a}_n$ for all sufficiently large n. ThusTogether these cases prove the closed-set upper bound at the slower speed, with rate zero only at mu. Its finite sublevels are the compact singleton $\{\mu \}$, so it is a good rate function.

Let $I_X,I_Y$ be the original rates and ${\mu }_X,{\mu }_Y$ their unique zero points. Suppose first that $H<G$. At speed $b_n=n^{2(1-G)}$, the preceding argument gives X the rate zero at ${\mu }_X$ and infinity elsewhere, while Y retains $I_Y$. Independence allows the product large-deviation principle, and addition is continuous, so the contraction principle for large deviations gives the sum rateThis is good, has unique zero ${\mu }_X+{\mu }_Y$, and inherits a strictly positive finite value by translation from $I_Y$. Thus the sum has Hurstiness G. The case $G<H$ is identical with the roles exchanged.

If $G=H$, the same product and contraction results apply to the original rates at their common speed, giving their infimal convolutionIt is good. Every finite fiber infimum is attained: a minimizing sequence lies in a product of compact sublevel sets and has a subsequence converging in the closed addition fiber. Consequently its only zero is ${\mu }_X+{\mu }_Y$. Choose $\hat{x}$ with $0<I_X(\hat{x})<\infty$ and set $\hat{z}=\hat{x}+{\mu }_Y$. Then $\hat{z}$ differs from the zero point and $0<K(\hat{z})\le I_X(\hat{x})<\infty$. The nondegeneracy requirement is therefore satisfied here too. In all cases,The general results used are the independent product large-deviation theorem and the continuous contraction principle. Checking the zero set and a positive finite value is necessary in addition to obtaining an LDP.

Use the stated continuity of the infinite-buffer and finite-buffer queue workload maps on the input path space. The contraction principle for large deviations gives, at the same speed as the input sequence,Empty fibers have infimum infinity. Both rates are good because they are continuous contractions of the good rate function I. In particular, at speed L the infinite-buffer queue workload satisfies, for every Borel set E,and the identical bounds with bars give the finite-buffer queue workload LDP. This states the entire principles, not just an informal tail approximation.

Put $g={\Lambda }^{\ast }$ and $\gamma ={\inf }_{v>C}g(v)/(v-C)$, allowing gamma to be infinity if no rate above C has finite cost. We prove the linear workload rate for a local convex action, which in particular proves monotonicity on the physical domain $[0,\infty )$.

For an absolutely continuous finite-cost path and any interval of length t with positive net work $z(t)$, its average arrival rate is $C+z(t)/t$. Nonnegativity of g and Jensen inequality giveIf $q(a)=x>0$, choose intervals with $z(t)$ approaching its supremum x, obtaining $I(a)\ge \gamma x$. If gamma is infinite, the existence of any positive net interval instead forces infinite cost, giving the same extended-value conclusion.

Conversely, choose any $v>C$ of finite cost, use arrival rate v during the last $t=x/(v-C)$ time units and the typical rate mu before that. This constant-rate burst path has queue workload exactly x: the lookback net input increases to x over the burst and then decreases with slope $\mu -C<0$. Its action is $tg(v)=xg(v)/(v-C)$. Taking the infimum over v proves the matching upper bound. The typical-rate path has zero queue workload and zero cost. ThereforeStrict convexity and $g(\mu )=0$ ensure $\gamma >0$ whenever it is finite. Indeed, choose a finite-cost $v>C$ and some $u\in (\mu ,C)$; then $g(u)>0$, and convexity gives $g(v)/(v-C)\ge g(u)/(u-\mu )$ for every finite-cost $v>C$. Thus J is increasing on nonnegative queue workloads, and the preceding fiber bound yields $\boxed{\bar{J}(x)\ge J(x)\text{ for }x\ge 0}$. Negative values are impossible for both queue workloads and have infinite rate. No monotonicity is asserted across that impossible negative domain.

Take $0<x\le B$ and a finite-cost burst rate $v>C$. For the constant-rate burst path just constructed, let $t_0=x/(v-C)$. On $0\le s\le t_0$ its net input is $z(s)=(v-C)s$, so the supremum is x and the infimum is zero. The finite-buffer formula evaluated at $t=t_0$ is therefore $\min (x,B)=x$. Since $\bar{q}\le q=x$, we conclude $\bar{q}=x$ exactly: the burst never needs more space than the permitted capacity.

That path's action bounds the contracted finite-buffer rate, and optimizing givesAt zero use the typical-rate path to obtain $\bar{J}(0)=J(0)=0$. If the positive infinite-buffer rate is infinite, the asserted upper inequality is automatic. Approximating burst rates suffice, so the argument does not rely on an unproved existence of an optimal duration. Consequently $\boxed{\bar{J}(x)\le J(x)\text{ for }0\le x\le B}$.

Take $c=1$ and normalize the Big Bang to $t=0$. The cosmological perfect-fluid continuity equation for pressureless matter makes $C=8\pi G{\rho }_m{a}^3/3$ a positive constant. The flat Friedmann equation consequently readsPutting $u=a^{3/2}$ gives ${\dot{u}}^2=\frac{9}{4}(C+\Lambda u^2/3)$, which is directly integrable. The expanding flat dust universe with a cosmological constant hasFor negative cosmological constant, the last expression describes expansion and contraction on $0<t<2\pi /\sqrt{3|\Lambda |}$. Its maximum is $a_m=(3C/|\Lambda |{)}^{1/3}$ at $t_m=\pi /\sqrt{3|\Lambda |}$; it reaches zero again at $2t_m$. Positive or zero cosmological constant gives unbounded expansion and no finite maximum. All three have infimum zero at their initial singularity; negative cosmological constant also gives zero at the final singularity, with no positive minimum.

For positive cosmological constant, expanding the hyperbolic sine shows $a\sim (9C/4{)}^{1/3}{t}^{2/3}$ at early times: pressureless matter dominates. At late times,Thus the expansion changes from matter-dominated deceleration to exponential acceleration. Indeed $\ddot{a}/a=-C/(2a^3)+\Lambda /3$ changes sign at $a^3=3C/(2\Lambda )$.