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For nearest-neighbour bond percolation with $0<p<1$ on the strip $T_k=\mathbb{Z}\times \{-k,\dots ,k\}$, put $q_k(n)={\mathbb{P}}_p((0,0)\leftrightarrow (n,0)\text{ in }{T}_k)$. The Harris-FKG inequality and translation invariance imply $q_k(n+m)\ge q_k(n)q_k(m)$. Thus $-\log q_k(n)$ is a subadditive sequence, and the Fekete lemma givesThe direct horizontal graph path gives $0\le f_k(p)\le -\log p$. Enlarging the strip increases every percolation two-point connection probability, so $f_k(p)$ is nonincreasing in $k$ and has a nonnegative limit of a sequence.

For a real subadditive sequence $(x_n{)}_{n\ge 1}$, the normalized sequence has the extended-real limit of a sequenceNo lower-bound hypothesis is needed for this conclusion. The limit of a sequence is finite precisely when the ratios are bounded below. The example $x_n=-n^2$ is a subadditive sequence with normalized limit of a sequence $-\infty$.

To prove the result, fix $k\ge 1$ and, for $n\ge k$, write $n=qk+r$ with $q\ge 1$ and $0\le r<k$. Iterating subadditivity gives $x_n\le q{x}_k+C_k$, where $C_k=\max (0,x_1,\dots ,x_{k-1})$ and $C_1=0$. Hence ${\mathrm{limsup}}_n{x}_n/n\le x_k/k$ for every $k$. Every ratio is at least their infimum. If that infimum is finite these bounds identify the limit of a sequence; if it is $-\infty$, choosing $k$ with an arbitrarily negative ratio gives convergence to $-\infty$.

A real sequence $(x_n)$ has limit of a sequence $L\in \mathbb{R}$ when for every $\epsilon >0$ there is $N$ such that $|x_n-L|<\epsilon$ for all $n\ge N$. Its limit of a sequence is unique. Extended-real convergence to $-\infty$ means that every real upper bound eventually exceeds all the terms; convergence to $+\infty$ is the corresponding lower-bound condition. The extended-real Fekete lemma is an example where the limit of a sequence need not be finite.

A real sequence $(x_n)$ has limit of a sequence $x$ whenA series ${\sum }_{n\ge 1}{x}_n$ is a convergent series with sum $s$ when its partial sums $S_N={\sum }_{n=1}^N{x}_n$ have limit of a sequence $s$.

For the first comparison test for series, either of the two allowed bounds implies $0<x_n\le a_n+b_n$. HenceThe partial sums form a monotone bounded sequence, so they converge. The series $\sum x_n$ is convergent.

For the exponent $2$, use $1/n^2\le 1/[n(n-1)]=1/(n-1)-1/n$ for $n\ge 2$. The resulting telescoping series bounds every partial sum by $2$, proving that the series converges by the monotone bounded sequence property. To prove divergence of the harmonic series, group the terms with $2^j<n\le 2^{j+1}$: their sum is at least $2^j/2^{j+1}=1/2$. Thus its partial sums are unbounded. If $\alpha \le 1$, then $n^{-\alpha }\ge n^{-1}$, so the comparison test for series proves

For the weighted square assumption, the finite-dimensional Cauchy-Schwarz inequality gives, for every $N$,Again the positive partial sums form a monotone bounded sequence. This is the weighted square roots of a summable sequence argument, with $a_n=n^2{x}_n^2$. The weighted square condition implies $\sum x_n<\infty$.

The converse is false. Take $x_n=n^{-3/2}$. Each block $2^j\le n<2^{j+1}$ contributes at most $2^j{2}^{-3j/2}=2^{-j/2}$, and the resulting geometric series converges. But $n^2{x}_n^2=1/n$, whose harmonic series diverges. This proves the counterexample without assuming the full P-series criterion.

Use integer graph vertices in the boxes: ${\Lambda }_n=[-n,n{]}^2\cap {\mathbb{Z}}^2$, and set ${\Lambda }_{-1}=\varnothing$. Thus $\partial {\Lambda }_0=\{0\}$, $g_0=1$, and $|\partial {\Lambda }_m|=8m$ for $m\ge 1$.

If $0\leftrightarrow \partial {\Lambda }_{m+n}$, erase loops from an open graph path to obtain a self-avoiding walk. Let $x$ be its first graph vertex on $\partial {\Lambda }_m$. Split at $x$. The prefix certifies $A_x=\{0\leftrightarrow x\}$, while a segment of the suffix certifies $B_x=\{x\leftrightarrow x+\partial {\Lambda }_n\}$: the final graph vertex is at supremum norm distance at least $n$ from $x$, so the suffix first hits that translated boundary. The two certificates use disjoint edges. Therefore the BK boundary-splitting estimate, the union bound, the van den Berg-Kesten inequality, and translation invariance giveThe cases $m=0$ or $n=0$ follow directly from $g_0=1$, so this proves the bound including the endpoints.

Here is an explicit way to remove the polynomial boundary factor. For $p>0$ put $b_n=32n^2{g}_n$, $n\ge 1$. By symmetry in $m,n$, assume $m\le n$. ThenConsequently $a_n=\log b_n$ is a subadditive sequence. The Fekete lemma states that any real subadditive sequence satisfies $\lim a_n/n={\inf }_{n\ge 1}{a}_n/n$, possibly $-\infty$. In this case the direct horizontal open graph path gives $g_n\ge p^n$, so this limit of a sequence is bounded below by $\log p$. Since $g_n\le 1$, its upper bound is zero. Finally,has the same limit of a sequence. Thus the percolation one-arm decay rate exists andFor $p=0$, every $g_n$ with $n\ge 1$ is zero and $\gamma =0$. No exponential-decay theorem or assumption that $p$ is below the percolation critical probability is needed.

The union bound gives $g_k\le {\sum }_{x\in \partial {\Lambda }_k}{\mathbb{P}}_p(0\leftrightarrow x)$. Hence some boundary graph vertex has percolation two-point connection probability at least $g_k/(8k)$. The rotations and reflections of the square lattice let us choose such a graph vertex as $x=(k,j)$, $-k\le j\le k$. Reflection in the vertical line through $x$ sends $0$ to $e_{2k}=(2k,0)$ and fixes $x$. ThereforeThe Harris-FKG inequality applied to these two increasing events proves the reflection lower bound for two-point percolation:Every graph path from $0$ to $e_{2k}$ reaches $\partial {\Lambda }_{2k}$, so $h_{2k}\le g_{2k}$. Taking roots givesBoth outside expressions have limit of a sequence $\gamma$, proving the even case. For $p>0$, the Harris-FKG inequality with the last horizontal edge gives $h_{2k+1}\ge p{h}_{2k}$, and also $h_{2k+1}\le g_{2k+1}$. The lower bound has rootThe upper bound has the same limit of a sequence. At $p=0$ all positive-distance connection probabilities vanish. Thus ${\lim }_{n\to \infty }{h}_n^{1/n}=\gamma$ for every $p\in [0,1]$.

Let $\alpha ={\inf }_{k\ge 1}{x}_k/k$. The relevant version of the Fekete lemma is the extended-real Fekete lemma: $\alpha$ may be $-\infty$, but cannot be $+\infty$ because $x_1$ is real. We show that this is exactly the limit of a sequence.

Fix $k\ge 1$ and, for $n\ge k$, write $n=qk+r$, where $q\ge 1$ and $0\le r<k$. Iterating the defining inequality gives $x_{qk}\le q{x}_k$. If $r>0$, it also gives $x_n\le q{x}_k+x_r$; if $r=0$, there is no remainder term. Thus, with $C_k=\max (0,x_1,\dots ,x_{k-1})$ and $C_1=0$,As $n\to \infty$, $q/n\to 1/k$, so ${\mathrm{limsup}}_n{x}_n/n\le x_k/k$. This holds for every positive $k$. On the other hand every $x_n/n\ge \alpha$. If $\alpha$ is finite, these two bounds give the desired equality. If $\alpha =-\infty$, for every real $M$ choose $k$ with $x_k/k<M-1$; the same upper bound makes $x_n/n<M$ for all sufficiently large $n$. ThereforeThe index $k=0$ is excluded because its ratio is undefined, and $x_0$ is irrelevant since the defining inequality was required only at positive indices. If the printed word “limit of a sequence” is interpreted as a finite real limit of a sequence, an additional lower bound is necessary: $x_n=-n^2$ defines a subadditive sequence but $x_n/n=-n\to -\infty$. The extended-real statement is the correct unrestricted conclusion.

For each integer $j$, consider the cut of horizontal edges joining column $j$ to column $j+1$ inside the strip. It contains exactly $2k+1$ edges. Its being completely closed has probabilityDifferent cuts use disjoint edge sets, so these closed-cut events are independent. Any graph path from column zero to column $n>0$, even one that wanders to other columns first, must cross each cut with $j=0,\dots ,n-1$. None of those $n$ cuts may therefore be completely closed. This gives the closed-cut bound for percolation in a strip:Taking negative logarithms, dividing by $n$ and taking the established limit of a sequence yieldsThe strict inequality uses both finite width and $p<1$. It is consistent with $f_k(p)\le -\log p$, since $1-(1-p{)}^{2k+1}\ge p$. At $p=1$ all strip connections occur and the decay rate is zero, so that excluded endpoint would invalidate strict positivity. The mechanism is a positive probability of an impermeable finite cut, rather than any assumption that the unrestricted lattice is subcritical.

The strips are nested. Every connection permitted inside $T_k$ is also permitted inside $T_{k+1}$, so $q_k(n)\le q_{k+1}(n)$ for every fixed separation. ConsequentlyThe preceding nonnegative bound makes this a decreasing sequence bounded below. By monotone convergence of real sequences,There is no assertion that this limit of a sequence is strictly positive: positivity at each fixed width need not survive an increasing-width limit of a sequence.

One can also identify the limit of a sequence. Let $q(n)$ be the percolation two-point connection probability in the whole square lattice. Every finite connecting graph path has a bounded vertical extent, so $q_k(n)\uparrow q(n)$. Commuting infima gives the strip approximation to the planar connection decay rate:The same positive-association argument identifies the final infimum with the whole-plane normalized logarithmic limit of a sequence. This is an interchange of infima justified by monotonicity at fixed $n$. For example, when $p>1/2$, the threshold proved in question 2 and uniqueness give an infinite percolation cluster with root probability $\theta (p)>0$. The Harris-FKG inequality makes the probability that both endpoints belong to it at least $\theta (p{)}^2$, so $q(n)\ge \theta (p{)}^2$. The whole-plane rate, and hence ${\lim }_k{f}_k(p)$, is then zero, despite the strict positivity of every finite-strip rate.