Let $P_{n,k}$ be the probability that a simple random walk of length $n$ in $\mathbb{Z}^2$ is such that each vertex of $\mathbb{Z}^2$ is visited at most $k$ times by the walk.
Certainly this probability decays exponentially with $n$, i.e, $P_{n,k} \sim e^{- c(k) n }$, but how fast does the exponent $c(k)$ go to zero with $k$ in the limit of large $k$?
It is reasonable to expect that $c(k) \rightarrow 0$ as $k \rightarrow \infty$, but I am interested in the precise asymptotic behaviour.
This question is related to this other question: Recurrent random walks with bounded local time at each vertex
Here is the brute force union bound based calculation: Let $f_{\ell,m} = P(\sum_{\ell+1 \leq i \leq m} x_i =0) = P(\sum_{\ell+1 \leq i \leq m} y_i =0) = {m-\ell \choose (m-\ell)/2} \frac{1}{2^{m-\ell}}$ where $x_i,y_i$ are the increment of random walk in each co-ordinate with values $+1$ or $-1$ each. I think the probability a state is visited $\geq k+1$ times is $\leq \sum_{n \geq m_{k+1} > ...> m_2 > m_1 \geq 0} \prod_{i = 1}^{k} f^2_{m_i,m_{i+1}}$. If lets say state is visited first time at time $\ell$ and lets say its visited after $\ell$ at time $m$ then $\sum_{\ell+1 \leq i \leq m} x_i = 0$ and $\sum_{\ell+1 \leq i \leq m} y_i = 0$. This is logic behind the upper bound. Hence, probability that each state is visited atmost $k$ times is $\geq 1-\sum_{n \geq m_{k+1} > ...> m_2 > m_1 \geq 0} \prod_{i = 1}^{k} f^2_{m_i,m_{i+1}}$ $\geq 1-\sum_{n \geq m_{k+1} > ...> m_2 > m_1 \geq 0} \frac{1}{2^{2(m_{k+1}-m_1)}}\prod_{i = 1}^{k} {m_{i+1}-m_i \choose (m_{i+1}-m_i)/2} \times {m_{i+1}-m_i \choose (m_{i+1}-m_i)/2}$