Let $G\_0(x)=G(x,0)$ be the Green's function of the simple symmetric random walk on $\mathbb Z^d$, $d\geq 3$. The question is whether $G\_0$ must always vary locally, i.e., whether


$$
\sum\_{\substack{y\in\mathbb Z^d:\\ |y-x|=1}} |G\_0(x)-G\_0(y)| >0
$$


holds for all $x\in\mathbb Z^d$.


The claim seems intuitive in view of the random walk interpretation of $G\_0$, i.e., $G\_0(x)=\mathbb P(\exists n: S\_n=x) G\_0(0)$.


For large $x$, the claim can be verified from the asymptotics of $G\_0$.



Does the Green's function of the simple random walk on $\mathbb Z^d$ always vary locally?
We will use the elementary fact that for $m \ge k \ge m/2$, the binomial coefficients satisfy
$${m\choose k+1} <{m \choose k}. \quad (\#)$$


The case $x=0$ is obvious so we may assume $x$ has some nonzero coordinate.
By symmetry, we may assume that $x\_1>0$. Then it suffices to show that for every $y \in \mathbb Z^d$ that satisfies $y\_1 \ge 0$, the point $z=z(y)$ that agrees with $y$ in all coordinates except the first, where $z\_1=y\_1+2$, we have the strict inequality $$P(S\_n=z)<P(S\_n=y)$$ for all $n$ such that $P(S\_n=z)>0$. Let $A\_n$ be the (random) set of steps among the first $n$ when the random walk moved in the first coordinate, and let $w^\*$ denote the projection of a node $w \in\mathbb Z^d$ to coordinates $2,3 ,\ldots, d$. Fix $y$ with $y\_1 \ge 0$ and let $z=z(y)$ as above, so that $z^\*=y^\*$. If $A\_n$ satisfies $$P(S\_n=z \,|A\_n)>0 \,,$$ then the cardinality $|A\_n|$ and $z\_1$ must have the same parity, and by $(\#)$,
$$P(S\_n=z\, | \, A\_n)=P(S\_n^\*=z^\* \,|A\_n) \cdot {|A\_n| \choose \frac{|A\_n|+z\_1}{2}}2^{-|A\_n|} \; < $$ $$ \: P(S\_n^\*=y^\* \,|A\_n) \cdot {|A\_n| \choose \frac{|A\_n|+y\_1}{2}}2^{-|A\_n|}= P(S\_n=y\, | \, A\_n) \,.$$
Taking expectations (i.e., averaging over $A\_n$) gives $P(S\_n=z)<P(S\_n=y)$.


