Higher-dimensional simple symmetric walks are transient

For fixed $z$, $K_d(z,\cdot )$ is the finite weighted sum of the $2d$ Dirac probability measures at $z\pm e_j$, $1\le j\le d$, with coefficient $1\mathrm{/}(2d)$. By [F1] it is a measure, and its total mass is $2d\mathrm{/}(2d)=1$. For fixed $A\subseteq {\mathbb{Z}}^d$, the function $z\mapsto K_d(z,A)$ is measurable because the source sigma-algebra is the full power set. Thus [F2] makes $K_d$ a probability kernel. The vectors $z\pm e_j$ are pairwise distinct, so its transition matrix is exactly [F5]. [F1, F2, F5, given] 1.2 By [F3], [F4], and [A1], for each fixed $x$ there is a canonical Markov chain law with this matrix and initial state $x$. Also [F8] and [F9] show that its state space ${\mathbb{Z}}^d$ is at most countable, as required by [F18]. The law is unique for each $x$, so no family of laws is selected by choice. [A1, F3, F4, F8, F9, given] 1.3 Repeatedly apply [F7] to the finite-support one-step rows. Induction on the number of steps expands $p^{(n)}(0,0)$ as the sum of the probabilities of all length-$n$ move words that start and end at $0$; each such word has probability equal to the product of its one-step entries [F5]. At every finite time only finitely many words occur, since there are $2d$ choices at each step. In particular, a return after an odd number of steps is impossible: each move changes the parity of the sum of the coordinates. [F5, F6, F7, given] 1.4 Fix $n\ge 1$ and let ${\mathcal{W}}_{n,d}=\{a=(a_1,\dots ,a_d)\in {\mathbb{N}}^d:{\sum }_{j=1}^d{a}_j=n\}\mathrm{.}$ For $a\in {\mathcal{W}}_{n,d}$ define $q_n(a)=d^{-n}\left(\genfrac{}{}{0px}{}{n}{a_1,\dots ,a_d}\right)=\frac{n!}{d^n{\prod }_{j=1}^d{a}_j!},$ where the second equality follows from [F10]. By the real multinomial expansion [F10], evaluating all $d$ variables at $1\mathrm{/}d$ gives ${\sum }_{a\in {\mathcal{W}}_{n,d}}{q}_n(a)=1.$ The index set is finite by [F10], and each $q_n(a)>0$. [F10, given] 2.1 The maximum $M_n:={\max }_{a\in {\mathcal{W}}_{n,d}}{q}_n(a)$ exists by [F10] and [F14]; the set is nonempty, for example $(n,0,\dots ,0)$ belongs to it. If $a_i\ge a_j+2$ for coordinates $1\le i,j\le d$, transferring one unit from coordinate $i$ to coordinate $j$ produces $a^{\mathrm{\prime }}=a-e_i+e_j\in {\mathcal{W}}_{n,d}$ and $\frac{q_n(a^{\mathrm{\prime }})}{q_n(a)}=\frac{a_i}{a_j+1}>1.$ Consequently any maximizing tuple has coordinates differing by at most one. Writing $n=dm+r$ with $0\le r<d$, such a tuple has $r$ coordinates equal to $m+1$ and the others equal to $m$; all such tuples have the same value. [F10, F14, step 1.4, given] 2.2 A $2n$-step word returns to zero exactly when, for each coordinate $1\le j\le d$, it uses $+e_j$ and $-e_j$ equally often. Write their common count as $a_j$; then $a\in {\mathcal{W}}_{n,d}$. For fixed $a$, the number of words is the multinomial coefficient with the $2d$ category counts $(a_1,a_1,\dots ,a_d,a_d)$, so [F10] and [F5] give its return probability as $\frac{(2n)!}{(2d{)}^{2n}{\prod }_{j=1}^d(a_j!{)}^2}\mathrm{.}$ Summing over $a$ and using the definition of $q_n$ yields $p^{(2n)}(0,0)=b_n{\sum }_{a\in {\mathcal{W}}_{n,d}}{q}_n(a{)}^2,b_n=4^{-n}\left(\genfrac{}{}{0px}{}{2n}{n}\right)\mathrm{.}$ Indeed, each summand on the right is $4^{-n}\frac{(2n)!}{(n!{)}^2}\frac{(n!{)}^2}{d^{2n}{\prod }_{j=1}^d(a_j!{)}^2}=\frac{(2n)!}{(2d{)}^{2n}{\prod }_{j=1}^d(a_j!{)}^2},$ the word probability just computed. [F5, F7, F10, step 1.3, step 1.4, given, algebra] 3.1 The limit in [F12], and positivity of its terms for $m\ge 1$, imply constants $c,C>0$ such that for every integer $m\ge 1$, $c\sqrt{m}(m\mathrm{/}e{)}^m\le m!\le C\sqrt{m}(m\mathrm{/}e{)}^m\mathrm{.}$ For fixed $d$ and $n\ge 2d$, a balanced tuple has every coordinate $a_j\ge n\mathrm{/}(2d)>0$. Put $r_j=a_j\mathrm{/}n$ for $1\le j\le d$. These positive weights sum to one; applying weighted AM–GM [F11] to $u_j=1\mathrm{/}(d{r}_j)$ gives ${\prod }_{j=1}^d{\left(\frac{1}{d{r}_j}\right)}^{r_j}\le {\sum }_{j=1}^d\frac{r_j}{d{r}_j}=1,$ and raising to the $n$th power yields $\frac{n^n}{d^n{\prod }_{j=1}^d{a}_j^{a_j}}={\prod }_{j=1}^d{\left(\frac{1}{d{r}_j}\right)}^{a_j}\le 1.$ Use the upper Stirling bound for $n!$ and the lower one for each $a_j!$ in the factorial formula for $q_n$. Since ${\sum }_{j=1}^d{a}_j=n$, the exponential factors cancel, and the last display gives $M_n\le \frac{C}{c^d}\frac{\sqrt{n}}{{\prod }_{j=1}^d\sqrt{a_j}}\le \frac{C}{c^d}(2d{)}^{d\mathrm{/}2}{n}^{-(d-1)\mathrm{/}2}\mathrm{.}$ For $1\le n<2d$, [F10] gives $M_n\le 1$; increasing the constant therefore produces $C_d>0$ with $M_n\le C_d{n}^{-(d-1)\mathrm{/}2}(n\ge 1)\mathrm{.}$ Here $C_d$ may depend on the fixed dimension $d$. [F10, F11, F12, step 2.1, step 1.4, given, algebra] 4.1 Since $0\le q_n(a)\le M_n$, normalization [step 1.4] gives ${\sum }_{a\in {\mathcal{W}}_{n,d}}{q}_n(a{)}^2\le M_n{\sum }_{a\in {\mathcal{W}}_{n,d}}{q}_n(a)=M_n\mathrm{.}$ By [F13] there is a constant $C_0$ with $b_n\le C_0{n}^{-1\mathrm{/}2}$ for all $n\ge 1$. Combining the return factorization, the bound from step 3.1, and the square-sum estimate just proved, there is a constant $D_d>0$ such that $0\le p^{(2n)}(0,0)\le D_d{n}^{-d\mathrm{/}2}(n\ge 1)\mathrm{.}$ The odd-time return probabilities vanish by step 1.3. [F13, step 1.3, step 1.4, step 3.1, step 2.2, algebra] 5.1 For fixed integer $d\ge 3$, the rational exponent $d\mathrm{/}2$ exceeds one, so [F15] gives convergence of ${\sum }_{n\ge 1}{n}^{-d\mathrm{/}2}$. The comparison theorem [F16] and step 4.1 show that ${\sum }_{n=0}^{\mathrm{\infty }}{p}^{(n)}(0,0)=1+{\sum }_{n=1}^{\mathrm{\infty }}{p}^{(2n)}(0,0)<\mathrm{\infty }\mathrm{.}$ The initial term is one by [F6]. [F6, F15, F16, step 1.3, step 4.1, algebra] 6.1 The Green-series criterion [F18] implies that $0$ is not recurrent; [F17] gives the exhaustive recurrent/transient alternatives, so $0$ is transient. For every $x\in {\mathbb{Z}}^d$, translation by $x$ is a probability-preserving bijection from the finite move words from $0$ back to $0$ to the words from $x$ back to $x$. Therefore $p^{(n)}(x,x)=p^{(n)}(0,0)$ for every $n$, and the same finite Green-series criterion makes each $x$ transient. [F5, F6, F17, F18, step 1.3, step 5.1, given] 7.1 The time-zero return contributes exactly one; all odd positive returns have probability zero; and the estimates apply at the threshold $d=3$, where the bounding exponent is $3\mathrm{/}2$. The constant $D_d$ is allowed to depend on fixed $d$, so no uniform-in-d claim is made. AC [A1] is used for the canonical chain law and recurrence criterion; the path counting, finite maximization, and translation argument require no choice. The assertion is one-way, not an iff statement.