Given an infinite connected graph $G$ of bounded degree with vertex set $X$, let $P_x^n$ the time $n$ distribution of the simple random walk started at the vertex $x$ (so $P^n_x(y)$ is the probability that a simple random walk started at $x$ ends at $y$ after $n$ steps).
Let further
$$
H_n :=\displaystyle \sup_{x \in X} h(P_x^n),
\qquad
\eta_n := \displaystyle \inf_{x \in X} h(P_x^n)
\quad \text{ and } \quad
r_n := \displaystyle \sup_{x,y \in X} P_x^n(y).
$$
Here $h(\mu) = \sum_y -\mu(y) \ln \mu(y)$ is the entropy of $\mu$.
Given that for any measure $\mu$ one has $\displaystyle \sup_{y \in X} \mu(y) \geq e^{-h(\mu)}$, one gets the bound $r_n \geq e^{-H_n}$. One question is:
Question 1: what are upper bounds on $r_n$ in terms of $H_n$ and $h_n$?
Note that in the case of [Cayley graphs of] groups, there is such a bound (the sharpest version known to me is by Peres & Zheng). My question is for "generic" [infinite connected] graphs [of bounded degree].
There is a natural counterpart to the question (which is hopefully easier to get):
Question 2: what are [family of] examples where upper bounds of $r_n$ in terms of $H_n$ and $\eta_n$ are "bad"?
Of course "bad" is not well-defined, but in Cayley graphs it may happen that $r_n$ behaves roughly like $C_1e^{-C_2\sqrt{H_n}}$. I would expect much worse for a graph.
PS: if "generic" needed to be made explicit, then I would say something like: "no finitely generated subgroup of the subgroup of self quasi-isometries of the graph acts co-compactly".
PPS: any result with "lazy random walk" in place of "simple random walk" is also OK.
Best Answer
This is a very partial answer. For simple random walks on Cayley graphs, linear growth of $H_n$ implies that for any $\epsilon>0$, the inequality $r_n \le \exp((\epsilon-1/2)n)$ must hold for infinitely many $n$, see Theorem 1.1 in [1].
For infinite connected graphs of bounded degree, it is possible for $H_n$ to grow linearly while $r_n \asymp cn^{-3/2}$ for even $n$. A simple example is a graph $G$ obtained by adjoining an infinite path to the root of an infinite binary tree. I suspect that slower decay of $r_n$ is not compatible with linear growth of entropy.
(Edit: @tmh Indicated in a comment below that one could get close to decay rate of $1/n$ for $r_n$. So perhaps $1/n$ is a barrier? recall that without any entropy assumptions, the slowest possible decay rate for $r_n$ on an infinite graph of bounded degree is $\;$ const.$/\sqrt{n}$.)
[1] Peres, Yuval, and Tianyi Zheng. "On groups, slow heat kernel decay yields liouville property and sharp entropy bounds." International Mathematics Research Notices 2020, no. 3 (2020): 722-750. https://arxiv.org/abs/1609.05174