Consider a non-constant harmonic function $f$ on $\mathbb{Z}^d$ (meaning this that $f(x)$ if the average of the $2d$ values $f(y)$ such that the distance between $x$ and $y$ is one). Let $M\_n$ denote the maximum of the absolute values $|f(x)|$ for all $x$ such that $||x||\_{\mathbf{L}^1} \leq n$ (where $n$ is a positive whole number). I am guessing that there is a constant $k>0$ such that, no matter what the whole positive number $n$ might be, $M\_n \geq kn$. However, I am not sure how to tackle this. Any suggestions?



On lower bounds for harmonic functions on $\mathbb{Z}^d$
The way your statement is formulated, it is wrong. The simplest counterexample is the function $f(x,y)=xy$ on $\mathbb Z^2$ which vanishes on the ball of radius 1. In Heilbronn's pioneeriing 1949 paper it is, in particular, proved that any polynomial growth harmonic function for the simple random walk on $\mathbb Z^d$ is polynomial. This result is actually valid for any group of polynomial growth and any finitely supported averaging probability measure, see Alexopoulos <https://projecteuclid.org/download/pdf_1/euclid.aop/1023481007>.


However, the answer to the corrected version of your question, whether 
$$
\liminf \frac{M\_n}n >0
$$ 
for any non-constant harmonic function on $\mathbb Z^d$, is "yes". Indeed, if the limit above is 0, then by the maximum principle applied to the $\ell^1$ balls in $\mathbb Z^d$, the growth of $f$ is at most linear, so that by the above theorem of Heilbronn, $f$ is actually linear, which yields a contradiction. 


