This question is related to How to compute the exit probabilities for a random walk? a question concerning a typo on sinai's paper
On Sinai's paper "The Limiting Behavior of a One-Dimensional Random Walk in a Random Medium" (1982) one considers a random walk on $Z^1$ that moves from $x$ to $x+1$ with probability $p(x)$ and moves from $x$ to $x-1$ with probability $q(x)=1-p(x)$.
The author defined (probably with a typo) the following function
$\quad$ The function $w^{(n)}(t)$ is basic for the analysis; it is formed from the transition probabilities $p(x)$ in the following way: $$w^{(n)}(t)=\left\{\begin{array}{lr}\dfrac1{\log n}\sum\limits_{0\leqq x\leqq k}\log\dfrac{q(x)}{p(x)}\text{ for }t=\dfrac k{\log^2n}, & k=1,2,\ldots,\\ \dfrac1{\log n}\sum\limits_{k\leqq x\leqq 0}\log\dfrac{q(x)}{p(x)}\text{ for }t=\dfrac k{\log^2 n}, & \qquad k=-1,-2,\ldots.\end{array}\right.$$
then he states ( the probably wrong) Lemma 1
$$\textbf{2. Exit Probabilities and Their Estimates}$$ $\ \ $ Let $[a,b]$ be a segment with end-points belonging to $Z^1$. We shall denote by $k^+_{[a,b]}(x)$ the $P$-probability of paths starting at $x$ which hit $b$ before $a$; $K^-_{[a,b]}(x)=1-k^+_{[a,b]}(x)$ is the $P$-probability of paths leaving $x$ and hitting $a$ before $b$. Then $$k^+_{[a,b]}(x)=p(x)k^+_{[a,b]}(x+1)+q(x)k^+_{[a,b]}(x-1), \\ k^+_{[a,b]}(a)=0,\qquad k^+_{[a,b]}(b)=1$$ and $$k^-_{[a,b]}(x)=p(x)k^-_{[a,b]}(x+1)+q(x)k^-_{[a,b]}(x-1),\\ k^-_{[a,b]}(a)=1,\qquad k^+_{[a,b]}(b)=0.$$ $\ \ $ The solutions of these equations are easily found. Let us cite the corresponding results in a form suitable for what follows.
$\ \ $ Lemma 1. The following equations hold: $$k^+_{[a,b]}(x)=\left(\sum\limits_{y=a+1}^x\exp\{\log n[w^{(n)}(y\log^{-2}n)-w^{(n)}(a\log^{-2}n)]\}\right) \qquad\qquad\\ \qquad\qquad\cdot\left(\sum_{y=a+1}^b\exp\{\log n[w^{(n)}(y\log^{-2}n)-w^{(n)}(a\log^{-2}n)]\}\right)^{-1},$$
Whose corrected version should be (I believe):
$$k^+_{[a,b]}(x)=\left(\sum\limits_{y=a+1}^x\exp\{\log n[w^{(n)}((y-1)\log^{-2}n)-w^{(n)}(a\log^{-2}n)]\}\right) \qquad\qquad\\ \qquad\qquad\cdot\left(\sum_{x=a+1}^b\exp\{\log n[w^{(n)}((y-1)\log^{-2}n)-w^{(n)}(a\log^{-2}n)]\}\right)^{-1},$$
Most likely, the typo is in the definition of the $w^{(n)}$ where one should have defined
$$w^{(n)}(t)=\left\{\begin{array}{lr}\dfrac1{\log n}\sum\limits_{0\leqq x < k}\log\dfrac{q(x)}{p(x)}\text{ for }t=\dfrac k{\log^2n}, & k=1,2,\ldots,\\ \dfrac1{\log n}\sum\limits_{k < x \leqq 0}\log\dfrac{q(x)}{p(x)}\text{ for }t=\dfrac k{\log^2 n}, & \qquad k=-1,-2,\ldots.\end{array}\right.$$
where the difference lies in the strict inequality in each case.
In this way it makes more sense to define explicitly $w^{(n)}(0) =0$ since the previous definition does not includes (by "continuity") the case $k = 0$. More precisely, If the $<$ was indeed a $\leq$ one could have defined $w^{(n)}(0) = \frac{1}{\log n} \log\frac{q(0)}{p(0)}$.
Moving on to the corollary (and having in mind the corrected version of the function $w^{(n)}$) one reads
$\ \ $ We shall state two consequences of these relations. Let us introduce the notation $T_{r,s}=[r\log^{-2}n,s\log^{-2}n]$.
$\ \ $ Corollary 1. Let the variable $x$ be such that $$w^{(n)}(x\log^{-2}n)=\max_{t\in T_{a,b}}w^{(n)}(t),\quad w^{(n)}(a\log^{-2}n)=\min_{t\in T_{a,x}}w^{(n)}(t), \quad w^{(n)}(b\log^{-2}n)=\min_{t\in T_{x,b}}w^{(n)}(t).$$ Then $k^+_{[a,b]}(x)\geqq(b-a)^{-1}$, $k^-_{[a,b]}(x)\geqq(b-a)^{-1}$.
$\ \ $ In fact, $$\max_{y\in[a,b]}[w^{(n)}(y\log^{-2}n)-w^{(n)}(a\log^{-2}n)]=w^{(n)}(x\log^{-2}n)-w^{(n)}(a\log^{-2}n).$$ Therefore, $$\begin{align}k^+_{[a,b]}(x) & \geqq \exp\{\log n[w^{(n)}(x\log^{-2}n)-w^{(n)}(a\log^{-2}n)]\} \\ & \color{white}{\geqq}\cdot[(b-a)\max_{y\in[a,b]}\exp\{\log n[w^{(n)}(y\log^{-2}n)-w^{(n)}(a\log^{-2} n)]\}]^{-1} \\ & =(b-a)^{-1}.\end{align}$$ $\ \ $ The estimate for $k^-_{[a,b]}(x)$ is obtained similarly.
In this case, with the new definitions the proof of $k^+_{[a,b]}(x) \geq \frac{1}{b-a}$ follows. However it seems that we don't need the assumptions $$w^{(n)}\left(\frac{x}{\log^2 n}\right) = \max_{t \in T_{a,b}} w^{(n)}(t)\\ w^{(n)}\left(\frac{a}{\log^2 n}\right) = \max_{t \in T_{a,b}} w^{(n)}(t)\\ w^{(n)}\left(\frac{b-1}{\log^2 n}\right) = \max_{t \in T_{x,b}} w^{(n)}(t)$$
but only $$w^{(n)}\left(\frac{x}{\log^2 n}\right) = \max_{t \in T_{a,b-1}} w^{(n)}(t)\\ $$
However, the result seems to fail for $k^-_{[a,b]}(x)$ indeed, consider
case $p(0) = \epsilon $ $p(1) = 1-\epsilon$ $p(2) = 1/2$ with $\epsilon$ small
then $$w^{(n)}(0) = 0\\ w^{(n)}(1/\log^2 n) = \frac{ \log\left(\frac{1-\epsilon}{\epsilon}\right)}{\log n}\\ w^{(n)}(2/\log^2 n) = 0 \\ w^{(n)}(3/\log^2 n) = 0$$
In this case $w^{(n)}(1\log^{-2}n)$ satisfies all the assumptions of corollary 1 and $$k^+_{[0,3]}(1) = \frac{1}{1 +\frac{1-\epsilon}{\epsilon} + 1}< 1/3$$ while $$k^-_{[0,3]}(1) = 1 - k^+_{[0,3]}(1) = \frac{\frac{1-\epsilon}{\epsilon} +1}{1 +\frac{1-\epsilon}{\epsilon} + 1} > 1/3 $$ therefore, the function $w^{(n)}$ satisfies the condition of the corollary but for $\epsilon <1/2$ the result fails.
Is it the case? Or am I missing something? Can we correct the proof of this result and go on with the main result of the paper?