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The extreme value theorem ensures that $M$ is finite. Since $f(x)\le |f(x)|\le M$ and $g(x)\ge 0$, multiplication preserves the inequality. Monotonicity of the Riemann integral gives the required boundTo prove the weighted mean value theorem for integrals, put $G={\int }_0^{1}gdx$ and $I={\int }_0^{1}fgdx$. If $G=0$, then $|I|\le {\int }_0^1|f|gdx\le MG=0$, so any $\alpha$ works. This includes $g\equiv 0$.

If $G>0$, the extreme value theorem gives $m_0=\min f$ and $M_0=\max f$. Integrating $m_{0}g\le fg\le M_{0}g$ gives $m_0\le I/G\le M_0$. A continuous function on an interval takes every value between its minimum and maximum by the intermediate value theorem. Choose $\alpha$ with $f(\alpha )=I/G$. Consequently

For $P(x)=x^3+4x-7$, the derivative $P^{\prime }(x)=3x^2+4$ is positive everywhere, so the mean value theorem makes $P$ strictly increasing. Since a polynomial is a continuous function, the intermediate value theorem and the values $P(1)=-2$, $P(2)=9$ locate its unique real root in $(1,2)$. By the rational root theorem, a rational root of a monic integer polynomial must be an integer. There is no integer in $(1,2)$, so the unique real root is an irrational number. The denominator part of that theorem also follows directly: if $a/b$ in lowest terms were a root, $a^3+4a{b}^2-7b^3=0$ would force $b\mid a^3$, hence $b=1$.

Fix $a<x<b$. Injectivity excludes $f(x)=c$ or $d$. If $f(x)<c$, the intermediate value theorem on $[x,b]$ gives a point other than $a$ with value $c$, contradicting injective function. If $f(x)>d$, the same theorem on $[a,x]$ gives a point other than $b$ with value $d$. ThereforeTo prove strict increase, take $x<y$. If $x=a$, the displayed endpoint ordering already gives $f(x)<f(y)$. Otherwise $f(a)<f(y)$, and the same interior-point argument on the subinterval $[a,y]$ yields $f(a)<f(x)<f(y)$. Hence $f$ is a strictly increasing function. This is the interval order theorem for continuous injections; the assumed endpoint order rules out strict decrease.

The intermediate value theorem states that a real-valued continuous function on a closed interval takes every value between its endpoint values. Explicitly, if $v$ lies between $f(s)$ and $f(t)$, there exists $u\in [s,t]$ with $f(u)=v$. No monotonicity is assumed.

With alternative routing, a rejected direct call can try a longer route. The attempt rates now depend on blocking itself, giving an additional feedback absent from fixed routing.

For an explicit example, take three nodes with a unit-resource link between each pair, all of capacity $C=1000$. For every unordered pair, direct calls arrive at rate $v=940$ and have mean holding time one. A call first tries its direct link; if that link is full, it tries the unique two-link route through the third node, and otherwise is lost. Under a symmetric Erlang fixed point approximation, all links have blocking $B$. A given link sees its own direct load $v$. It can also carry overflow from either of the other two pairs; each contributes reduced load $vB(1-B)$, since the direct link must block and the other link of the alternative route must accept. HenceThis is an instance of multiple Erlang fixed points under alternative routing. Put $f(B)=E(1000,940[1+2B(1-B)])-B$. Evaluation using the positive Erlang B formula recurrence $E(0,v)=1$, $E(k,v)=vE(k-1,v)/(k+vE(k-1,v))$ givesThe sign statements can be certified with rational arithmetic at these rational arguments. The intermediate value theorem gives a distinct root in each of $(0,0.01)$, $(0.01,0.10)$ and $(0.10,0.25)$; numerical roots are approximately $0.002748$, $0.071037$ and $0.189785$.

Thus the alternative-routing fixed point is not unique. The mechanism is extra link occupancy from overflow: moderate blocking stimulates two-link calls and can reinforce congestion. This multiplicity belongs to the approximation. The exact finite stochastic model has a unique stationary law on its reachable communicating class; the extra approximate solutions do not create three exact invariant distributions.

Let ${\Delta }_n={\sup }_{\theta \in \Theta }|Q_n(\theta )-Q(\theta )|$. For any $\epsilon >0$, the setis compact. If it is empty there is nothing to prove. Otherwise continuity and the unique minimum give a strictly positive separation gapTake any measurable attained minimizer of $Q_n$. Its defining inequality givesConsequentlyThis proves argmin consistency under uniform convergence in probability. No continuity of $Q_n$ is needed once the minimizer exists; compactness and continuity concern the deterministic separation gap. The uniform convergence in probability assumption controls all candidate parameters, including the random minimizer.

For the estimating equation, fix $\epsilon >0$ and put $a={\theta }_0-\epsilon$, $b={\theta }_0+\epsilon$. The prescribed signs make $\eta =\frac{1}{2}\min \{-S(a),S(b)\}>0$. Pointwise convergence in probability at just these two points impliesand similarly $\mathrm{Pr}(S_n(b)\le 0)\to 0$. With probability tending to one, $S_n(a)<0<S_n(b)$. The intermediate value theorem then gives a zero inside $(a,b)$, and uniqueness identifies it with ${\hat{\theta }}_n$. HenceThis is consistency of a uniquely bracketed zero. It needs no monotonicity, no continuity of the limit $S$, and no uniform convergence of the $S_n$. The bracket interval must lie in the domain of $S_n$: read literally, the printed sign condition for every positive $\epsilon$ puts every real point into $\Theta$, so this requirement is satisfied. More generally it is enough to have an interval about ${\theta }_0$ and sign brackets arbitrarily close to it.

Consider the secant slopeThe moment-generating function is continuous on the interior of its finite domain and has right derivative $M^{\prime }(0)=\mu$. Thus $H(r)\to \mu$ as $r\downarrow 0$. For every fixed positive claim amount $x$, the function $(e^{rx}-1)/r$ is strictly increasing in $r>0$: its derivative has numerator $e^{rx}(rx-1)+1>0$, since this numerator starts at zero and has derivative $rx{e}^{rx}$ as a function of $rx$. Taking expected values preserves the strict inequality. Therefore $H$ is continuous and strictly increasing. Equivalently, the strictly convex transform $M$ has strictly increasing secant slopes from the origin.

If $r_{\infty }<\infty$, the assumed blow-up of $M$ gives $H(r)\to \infty$. If $r_{\infty }=\infty$, choose $a>0$ with $d=\mathbb{P}(X_1\ge a)>0$. Such an $a$ exists because the claims are positive. Then $M(r)\ge d{e}^{ar}$, so again $H(r)\to \infty$. This exponential lower bound is needed at an infinite endpoint: mere divergence of $M$ would not, by itself, establish divergence of $M(r)/r$.

The target $(1+\theta )\mu$ strictly exceeds the limiting slope $\mu$. The intermediate value theorem and strict monotonicity therefore giveMultiplying by $R$ gives the defining adjustment coefficient equation. The zero root of the undivided equation is excluded. This is the secant-slope existence criterion for an adjustment coefficient.

For a chord of length $1/2$, consider the continuous function $h(x)=f(x+1/2)-f(x)$ on $[0,1/2]$. Since $f(0)=f(1)$,An endpoint zero already supplies the chord. Otherwise the endpoint values have opposite signs, so the intermediate value theorem supplies $h(\alpha )=0$ at some $\alpha \in (0,1/2)$. Thus a horizontal chord of length $1/2$ exists.

For each integer $n>1$, define $h_n(x)=f(x+1/n)-f(x)$ on $[0,1-1/n]$. The sampled differences telescope:If any sampled value is zero, it gives the required chord. Otherwise there must be both a positive and a negative sampled value. The intermediate value theorem, applied between their sample points, then gives $h_n(\alpha )=0$. Hence a horizontal chord of length $1/n$ exists for every positive integer $n$; for $n=1$, the endpoint chord itself suffices. This proves the horizontal chord of reciprocal-integer length result, without restricting $n$ to powers of two.