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Let $f$ be continuous on $(a,b]$, bounded above, and suppose its limit superior at $a$ from the right is at most $L$. If some $c\in (a,b]$ has $f(c)>\max \{L,f(a)\}$, then $f$ attains its maximum on $[a,b]$. A sufficiently short initial interval has all values below $f(c)$. On the remaining closed interval, the extreme value theorem supplies a maximum, which is consequently global. This can prove attainment despite an oscillatory discontinuity at one endpoint.

First, the Cramér model has infinitely many successes almost surely, so every $P_n$ is finite. Indeed, for each fixed $K\ge 3$, independence of the Bernoulli random variables givesThe sum diverges, for example because $1/\log j\ge 1/j$ for $j\ge 3$. Continuity from above of a measure and a countable union bound exclude a final success.

Fix $ϵ>0$ and put $L_k=\lceil (1+ϵ)(\log k{)}^2\rceil$. Let $A_k$ be the event that the whole interval $k+1,\dots ,k+L_k$ consists of failures. By independence and $1-v\le e^{-v}$,Since $L_k=o(k)$, the exponent divided by $\log k$ tends to $-(1+ϵ)$. Thus $\mathbb{P}(A_k)\le k^{-1-ϵ/2}$ for all sufficiently large $k$. The Borel-Cantelli first lemma shows that almost surely every sufficiently large such interval contains a success. Taking $k=P_n$ then gives $P_{n+1}-P_n\le L_{P_n}$ for all sufficiently large $n$.

For each fixed $ϵ$, the limit superior of the normalized gap is therefore at most $1+ϵ$ almost surely. Intersect the probability-one events for $ϵ=1/r$, $r\in \mathbb{N}$, to obtain the Cramér model prime-gap upper bound:

Write ${\mathcal{W}}_n$ for the deterministic set of $n$-step self-avoiding walks from $0$. Each uses $n$ distinct edges, so independence givesFor $n\ge 1$, the function $t\mapsto t^{1/n}$ on $[0,\infty )$ is a concave function. The Jensen inequality gives the root-moment bound for open self-avoiding walksEquivalently, the Lyapunov moment inequality compares exponents $1/n$ and $1$, both strictly positive. Taking the limit superior and the connective constant limit provesThis includes $p=0$ and $n=1$. No independence between the different self-avoiding walk indicators is asserted or needed.

Choose a finite-dimensional generating vector subspace $V\subseteq A$ containing $1$, and write $V^n$ for the linear span of products of $n$ elements of $V$. The Gelfand–Kirillov dimension isFor a nonzero finitely generated module $M$ considered as a right module, choose a finite-dimensional generating vector subspace $W\subseteq M$ and define the Gelfand–Kirillov dimension of a module byFor left modules, replace $W{V}^n$ by $V^{n}W$. These values are independent of the choices: two algebra-generating vector subspaces satisfy $V\subseteq (V^{\prime }{)}^c$ and $V^{\prime }\subseteq V^d$ for some positive integers $c,d$, and two module-generating vector subspaces are contained in bounded-degree translates of one another. The corresponding growth bounds differ only by a constant rescaling and shift of $n$, which do not change the limit superior.

Let $S$ be a nonzero standard graded algebra, generated by $r$ elements of degree one. It is a homogeneous quotient ring of the polynomial ring $k[z_1,\dots ,z_r]$. By the Hilbert-Serre theorem, its Hilbert series is a rational function whose only possible pole is at $t=1$. After canceling, write it as $P(t)/(1-t{)}^d$ with $P(1)\ne 0$, where $0\le d\le r$. The cumulative dimensions are the coefficients of $P(t)/(1-t{)}^{d+1}$ and are eventually a polynomial in $n$ of degree $d$, with positive leading coefficient. With $V=S_0\oplus S_1$, these cumulative dimensions equal $\dim V^n$. ConsequentlyFor $d=0$ the algebra is finite-dimensional and nonzero, so the cumulative dimension is eventually a positive constant. If the zero algebra is allowed, the common convention $\mathrm{GKdim}0=-\infty$ is an exception to the printed assertion; the integer conclusion concerns nonzero unital algebras.

For the Weyl algebra $A=A_1(k)=k⟨X,D⟩/(DX-XD-1)$, the total-degree filtered algebra structure has associated graded ring $\mathrm{gr}A=k[x,\xi ]$. Its ordered monomials $X^i{D}^j$ form a basis. Every finitely generated module has a good filtration of a module, and its associated graded module is finitely generated over $k[x,\xi ]$. The Hilbert-Serre theorem therefore shows that its Gelfand–Kirillov dimension is an integer in $\{0,1,2\}$.

A nonzero module of Gelfand–Kirillov dimension zero here would have eventually constant cumulative dimension, hence be finite-dimensional. This is impossible in characteristic zero: taking the trace of the endomorphisms representing $DX-XD=1$ gives $0={\dim }_{k}M$. For right modules the right-action operators reverse composition and give the negative identity instead, with the same contradiction. ThusBoth occur. The regular module $A$ has $\dim V^n=\left(\genfrac{}{}{0ex}{}{n+2}{2}\right)$, hence Gelfand–Kirillov dimension two. For a right module of Gelfand–Kirillov dimension one, take $M=k[x]$ with $f\cdot X=xf$ and $f\cdot D=-f^{\prime }$. These actions satisfy the defining relation because $(f\cdot D)\cdot X-(f\cdot X)\cdot D=f$. The module is a cyclic module generated by $1$, and $\dim (1\cdot V^n)=n+1$. This also gives the lower bound for this particular Weyl algebra without invoking a general inequality for higher Weyl algebras.

Write $S_n={\sum }_{j=1}^n{Y}_j$ for the partial sum. Fix an integer $M\ge 1$. The tail-sum formula for expectation, applied to the nonnegative random variable $|Y_1|$, givesbecause $\mathbb{E}|Y_1|=\infty$. The events $\{|Y_n|>Mn\}$ are independent, so the second Borel--Cantelli lemma shows that $|Y_n|/n>M$ infinitely often with probability one. Taking the countable intersection over $M\in \mathbb{N}$ yieldsalmost surely.

Now $Y_n=S_n-S_{n-1}$, and the triangle inequality givesIf the limit superior of $|S_n|/n$ were finite, the right-hand side would have finite limit superior, contradicting the preceding conclusion. Thereforealmost surely.

The odd $X_1$ copy number decreases by two until it reaches one, so $⟨x_1(t)⟩\to 1$. The exact $X_2$ first-moment equation isBecause copy numbers are nonnegative integers, $x_2^2\ge x_2$, and the last mixed moment is nonnegative. HenceThe integrating factor $e^{{\alpha }_{3}t}$, together with $⟨x_1(t)⟩\to 1$, gives the comparison bound

Subtracting the exact $X_4$ first-moment equation from the $X_3$ equation cancels reaction 5:The assumed inequality is preciselyChoose a positive $\epsilon$ smaller than this gap divided by ${\alpha }_6$. The limit superior bound implies that, for all sufficiently large $t$,Thus $⟨x_3⟩-⟨x_4⟩$ grows at least linearly. Since $⟨x_4⟩\ge 0$,so the required species index is $\boxed{i=3}$.

Assume ${\sup }_n\mathbb{E}|M_n|<\infty$. For every pair of rationals $a<b$, Doob upcrossing inequality givesby monotone convergence theorem. Thus every rational interval is upcrossed only finitely often almost surely. If a real sequence has distinct limit inferior and limit superior, it completes infinitely many upcrossings of some rational interval between them. Hence $M_n$ converges in the extended real line almost surely.

Fatou lemma givesso the limit is finite almost surely and integrable. This is the $L^1$-bounded form of the Martingale convergence theorem.

In independent bond percolation on a graph, let $c_n$ be the number of $n$-step self-avoiding walks from a fixed graph vertex and ${\kappa }_n$ the number that are open. Then $\mathbb{E}{\kappa }_n=c_n{p}^n$. The Jensen inequality for the concave function $t^{1/n}$ gives $\mathbb{E}({\kappa }_n^{1/n})\le p{c}_n^{1/n}$. No independence between different self-avoiding walk indicators is needed. When $c_n^{1/n}\to \mu$, the limit superior is at most $p\mu$.