For a (symmetric) random walks on countable groups generated by $\mu$, there is a "brute-force computation" argument of Avez (1974) that shows that if the entropy $h\_\mu$ is trivial then there are no non-constant $\mu$-harmonic functions on the group. [Definitions are below.]


Given it is fairly easy to show that zero speed implies zero entropy (the converse is also true but requires the Gaussian estimates from Varopoulos/Carne)...


$\textbf{Question:}$ is there a simpler "brute-force" way of seeing that zero speed implies the absence of non-constant $\mu$-harmonic functions?


Definitions:


$\cdot$ A random walk on group $G$ is generated by $\mu$ symmetric (i.e. $\mu(g)= \mu(g^{-1})$) and finitely supported if the kernel of the Markov chain is $K(g,h) = \mu(g^{-1}h)$. In other words, the law of the $n^{\text{th}}$-step distribution is $\mu^{\*n} = \mu\*\mu\*\cdots\*\mu$ (convolution with $n$ appearances of $\mu$).


$\cdot$ $f$ is said $\mu$-harmonic if $f\*\mu = f$. 


$\cdot$ The entropy $h\_\mu$ is defined as $\lim\_{n \to \infty} \frac{1}{n} H\_{n,\mu}$ and $H\_{n,\mu} = - \sum\_{g \in G} \mu^{\*n}(g) \ln \mu^{\*n}(g)$. Subadditivity of $n \mapsto H\_{n,\mu}$ ensures the existence of the limit.


$\cdot$ The speed $l\_\mu$ is defined as $\lim\_{n \to \infty} \frac{1}{n} L\_{n,\mu}$ and $L\_{n,\mu} = \sum\_{g \in G} \mu^{\*n}(g) |g|$ (where $|g|$ is the word length for some fixed generating set). Subadditivity ensures the existence of the limit again.



Speed and absence of non-constant bounded harmonic functions
Here is an alternative simple proof (without any entropy) of non-Liouville $\Rightarrow$ positive speed using some of the theory of the Martin boundary/positive harmonic functions. It's quite high-level so not really what you're looking for, but is too long for a comment. 


Let $G$ be a transient graph with associated simple random walk transition probabilities $p(x,y)$ and let $\rho$ be a fixed root vertex. The **Martin kernel** of a vertex $v$ is defined to be
\begin{equation} M\_v(u) = \frac{\mathbb{E}\_u[\text{visits to }v]}{\mathbb{E}\_\rho[\text{visits to }v]}.\end{equation}
Some standard results from the Martin boundary theory (see e.g. Woess): If there are non-constant bounded harmonic functions on $G$, the Martin kernels $M\_{X\_n}$along the random walk converge pointiwse to some non-constant positive harmonic function $h$, and the law of the random walk conditioned on $h$ is the Markov chain with transition probabilities
\begin{equation} p^h(x,y) = \frac{h(y)}{h(x)}p(x,y). \end{equation}
This is the **$h$-transform** of the simple random walk on $G$.


Now, using the fact that we are on a Cayley graph, the sequence $h(X\_{n+1})/h(X\_n)$ is stationary and ergodic and, using the above expression for the law of $X$ conditional on $h$, we see that $\mathbb{E}[\log h(X\_1)]>0$, and so the Ergodic Theorem implies that 
\begin{equation} \lim \frac{1}{n}\log h(X\_n) = \lim \frac{1}{n}\sum \log \frac{h(X\_n)}{h(X\_{n-1})} = \mathbb{E}[\log h(X\_1)]>0 \end{equation}
so that $h$ grows exponentially along the random walk path almost surely. Now, again because we are in a Cayley graph and in particular have bounded degree, and $h$ is positive harmonic, $h(v)/h(u) \leq \deg(u)$ (we implicitly used this to get integrability in our application of the Ergodic Theorem also) and so $\log h$ is Lipschitz in the graph distance, so that the walk must have positive speed amost surely.


