Consider the graph $\mathbb{Z}^d$, where two vertices $x,y \in \mathbb{Z}^d$ are said to be neighbours if their $\mathrm{L}^1$ distance is $1$ (they differ in exactly one coordinate by one unit); I'll denote neighbours as $x \sim y$. Consider a function $f:\mathbb{Z}^d \to \mathbb{R}$ and the average $\displaystyle \bar f(x) = \dfrac{1}{2d} \sum_{y \sim x} f(y)$: say $f$ is harmonic if $f(x) = \bar f(x).$ It is fairly well known that bounded harmonic functions on $\mathbb{Z}^d$ are constant (one can prove this in a rather fascinating manner considering a coupling argument, as shown here -theorem 4.4-: http://www.math.cornell.edu/~durrett/CPSS2009/peres6.pdf)
It is not hard to show that when $d = 1$ there is a recurrence (one can assume that the harmonic function is zero at zero, as linear combination of harmonic functions gives rise to a harmonic function and because constants themselves are harmonic functions) of the form $f(n) = n f(1)$ for any $n \in \mathbb{Z}$, making $f$ a "linear kind" function.
Is this still true in $\mathbb{Z}^d$ in the sense that if $f:\mathbb{Z}^d \to \mathbb{R}$ is harmonic, then there exists a vector $A \in \mathbb{R}^d$ and a number $b \in \mathbb{R}$ for which $f(x) = (A,x) + b$ for all $x \in \mathbb{Z}^d$ (I denote the inner product of two vectors $A$ and $x$ as $(A,x)$)?
If this is not true, would the following weaker statement be true:
if $f:\mathbb{Z}^d \to \mathbb{R}$ is harmonic and $\displaystyle \max_{||x||_{\mathrm{L}^1} < r} |f(x)| = o(r)$ as $r \to \infty$, then $f$ is constant (where the relation "$u(r) = o(r)$ as $r \to a$" means that $\displaystyle \lim_{r \to a} \dfrac{u(r)}{r} = 0$)?
Take $d=2$ and $f(x,y)=x^2-y^2$.