Let $G$ be the Green function of the simple random walk on $\mathbb{Z}^d,\:d\geq 3$; i.e. $$G(x) = E \sum_{i=0}^{+\infty} 1_{X_i=x},$$ where $X$ is the simple random walk starting from $0$. The behaviour of $G(x)$ is well know when $x$ is large. But do we know how the function behaves when $x$ is small?

I tried to look in book of Spitzer (Principles of RW) and Lawler, Limic (Random Walk: Modern Introduction). I only found "inverse Fourier transform" formula which I do not have idea how to use (besides some numeric calculations).

May be some of you know other references?

My question stems from esimation of $$S_d:=\sum_{x\in \mathbb{Z}^d\setminus \lbrace 0 \rbrace} a_x \left(\sqrt{1+\frac{G(x)^2}{G(0)}}-1\right),$$ where $a_x$ are some coefficients. In the simplest case $a_x =1$ (or $a_x$ = 1 if $\sum_{i=1}^d x_i$ is odd and -1 otherwise; "chessboard of -1, 1").

In the case of $a_x=1$ one can easily prove that $S=\infty$ if $d=3,4$ and $S<\infty$ if $d\geq5$ (just using the fact that $G(x) \sim |x|^{2-d}$ when $x$ is large.)

But is it possible to say anything more about $S_d$ when $d>5$? E.g. I expect that $S_d \rightarrow 0$ as $d\rightarrow +\infty$.

I made a mistake in the first version the sum was over $\mathbb{Z}^d$ and it should be over $\mathbb{Z}^d\setminus \lbrace0\rbrace$.

Update 24.05.2011 Thanks to observations made in Didier Piau proof I was able to prove that indeed $S_d \rightarrow 0$ (in the case when $a_x=1$). I put the proof here.


1 Answer 1


In fact $S_d$ does not converge to zero when $d\to\infty$, at least if $a_x=1$ for every $x$. Here is a proof.

For every $x$, $G_d(x)=P_0^{(d)}(\mathtt{hit}\ x)G_d(0)\le G_d(0)$ hence $G_d(x)^2/G_d(0)\le G_d(0)$. For every $z$ in $[0,G_d(0)]$, $\sqrt{1+z}-1\ge c_dz$ with $$ c_d=\frac1{G_d(0)}[\sqrt{1+G_d(0)}-1]=\frac1{\sqrt{1+G_d(0)}+1}. $$ Hence, for every $(a_x)$, $$ S_d\ge c_dT_d,\quad\mbox{with}\ T_d=\sum_xa_xG_d(x)^2/G_d(0). $$ In the special case where $a_x=1$ for every $x$, by reversibility, $$ \sum_xG_d(x)^2=\sum_{i,j}\sum_xp_i^{(d)}(0,x)p_j^{(d)}(x,0)=\sum_{i,j}p_{i+j}^{(d)}(0,0)=\sum_{i}(i+1)p_i^{(d)}(0,0), $$ hence $$ \sum_xG_d(x)^2>\sum_{i}p_i^{(d)}(0,0)=G_d(0). $$ In particular, $T_d>1$. Now $G_d(0)$ is a nonincreasing function of $d$ hence $c_d\ge c_5$ for every $d\ge5$ and $S_d>c_5$ for every $d\ge5$.

Edit Much simpler: if $a_x\ge0$ for every $x$, then $S_d>a_0(\sqrt{1+G_d(0)}-1)$ and $G_d(0)>1$ hence $S_d>a_0(\sqrt2-1)$.

  • $\begingroup$ What is the error? $\endgroup$
    – Did
    May 22, 2011 at 19:35
  • $\begingroup$ Sorry, no error. I corrected my previous comment. $\endgroup$ May 24, 2011 at 9:16
  • $\begingroup$ As you observed the question in the first version was somehow trivial. What I meant was a sum over Zd∖{0} :), sorry for that. I am not sure at the momement if your proof can be improved to handle this situations as well $\endgroup$ May 24, 2011 at 9:17

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