In the paper Large Deviations for Random Walks in a Random Environment (2003)
One reads on page 1226:
$h_c$ is obtained as the following limit:
$$h_c(y) = \lim_n \frac{1}{n} -\log \sup_m \big(\pi^m(\omega, 0,n y) e^{-c|m-n|}\big) $$
where $\pi^n(\omega, 0 , ny)$ is the probability of the random walk in the environment $\omega$ leave the origin and reach the site ny in n steps.
To obtain that for $y \leq y'$ we have $h_c(y)\leq h_c(y')$ we compute $$h^n_c(y) = \frac{1}{n} -\log \sup_m \big(\pi^m(\omega, 0,n y) e^{-c|m-n|}\big) \\ = \frac{1}{n} -\log \big(\pi^{m(n,y)}(\omega, 0,n y) e^{-c|m(n,y)-n|}\big)$$
and we compare $$h^n_c(y') - h^n_c(y) = \frac{1}{n}\bigg[c|m(n,y') - n| - c | m(n,y)-n| + \log \frac{\pi^{m(n,y)}(\omega, 0,n y)}{\pi^{m(n,y')}(\omega, 0, n y')}\bigg]$$
Heuristically one thinks that if $y'> y$ then $m(n,y')\geq m(n,y)$ since it takes longer to leave 0 and reach $n y'$ therefore to conclude the comparison, we would need to decide what is the behavior of $$\log \frac{\pi^{m(n,y)}(\omega, 0,n y)}{\pi^{m(n,y')}(\omega, 0, n y')} $$
Maybe we could find estimates on $\pi^{m(n,y)}(\omega, 0,n y)$ that would allow us to decide the sign of $h^n_c(y') - h^n_c(y)$.
Any ideas?