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.