Green-Function Integrability in Dimensions Three and Above

The Green Integrand Near Its Singular Point

For a random walk, the Green function at the origin counts the expected total number of visits there, including the initial visit. Fourier analysis turns that count into an integral. In dimensions three and above, the integrand is unbounded at one point, so finiteness is not immediate: the key is to compare its singularity with the function \( |\theta|^{-2} \) and include the volume factor from radial coordinates.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((X_n)_{n\ge0}\) be the simple symmetric random walk on the state space \(S=\mathbb{Z}^d\), in discrete time \(n\ge0\). From every state \(x\), the walk moves to each of its \(2d\) nearest neighbors \(x\pm e_j\), \(1\le j\le d\), with probability \(1/(2d)\), where \(e_j\) is the \(j\)th coordinate vector. The transition row sums to \(2d\cdot(1/(2d))=1\). Write \(p_{00}^{(n)}=\mathbb{P}_0(X_n=0)\), and define the Green function \(G(0,0)=\sum_{n=0}^{\infty}p_{00}^{(n)}\), allowing the value \(+\infty\).

For \(\theta=(\theta_1,\ldots,\theta_d)\in[-\pi,\pi]^d\), the Fourier symbol of this walk is \(\phi(\theta)=d^{-1}\sum_{j=1}^d\cos\theta_j\). The Fourier Green integrand is \(1/(1-\phi(\theta))\); at \(\theta=0\), where its denominator is zero, regard it as \(+\infty\). The Fourier inversion formula for return probabilities gives \(p_{00}^{(n)}=(2\pi)^{-d}\int_{[-\pi,\pi]^d}\phi(\theta)^n\,d\theta\). We will show directly that the singularity is integrable when \(d\ge3\), and use that inversion formula to justify the resulting Green-function integral.

Proof. Put \(u=|x|/2\), so \(0\le u\le\pi/2\). The function \(\sin u\) is concave on this interval because its second derivative is \(-\sin u\le0\); hence it lies above the line segment joining \((0,0)\) and \((\pi/2,1)\). Thus \(\sin u\ge 2u/\pi=|x|/\pi\). Since \(1-\cos x=2\sin^2(|x|/2)\), it follows that \(1-\cos x\ge2x^2/\pi^2\). Now \(1-\phi(\theta)=d^{-1}\sum_{j=1}^d(1-\cos\theta_j)\). Applying the one-dimensional bound to every coordinate gives \(1-\phi(\theta)\ge 2(d\pi^2)^{-1}\sum_j\theta_j^2=2|\theta|^2/(d\pi^2)\). Taking reciprocals of positive quantities proves the final bound. \(\square\)

Why the Radial Integral Is Finite

Let \(\sigma_{d-1}\) denote the surface area of the unit sphere in \(\mathbb{R}^d\). The polar-coordinate formula gives, for \(0<\varepsilon\le\pi\),

The factor \(r^{d-1}\) is the radial volume element. If \(d\ge3\), then \(d-3\ge0\), and direct integration yields

This calculation controls the part near the origin. Away from the origin, \(1-\phi\) is continuous and positive: it is a sum of nonnegative terms \(1-\cos\theta_j\), and all terms vanish together in the cube only at \(\theta=0\). On the compact set \([-\pi,\pi]^d\cap\{|\theta|\ge\varepsilon\}\), it therefore has a strictly positive minimum. The integrand is bounded there. Together, the local bound and this away-from-zero bound establish integrability over the whole cube.

Proof. The lemma bounds the integrand in the ball \(|\theta|\le\varepsilon\), except at the single point \(\theta=0\), by \(d\pi^2/(2|\theta|^2)\). The radial calculation shows that this bound has finite integral for \(d\ge3\). On the rest of the cube, continuity and strict positivity of \(1-\phi\) give a bounded integrand. Thus the proposed integral is finite.

It remains to verify that it equals the Green function. For \(0<r<1\), the series \(\sum_{n\ge0}r^n\phi(\theta)^n\) converges uniformly, since \(|\phi(\theta)|\le1\). Integrating term by term and using Fourier inversion gives \((2\pi)^{-d}\int(1-r\phi(\theta))^{-1}\,d\theta=\sum_{n\ge0}r^np_{00}^{(n)}\). As \(r\uparrow1\), the right side increases to \(G(0,0)\) by monotone convergence, since each \(p_{00}^{(n)}\ge0\). For the integrands on the left, when \(\phi(\theta)<0\), \(0<(1-r\phi(\theta))^{-1}\le1\). When \(\phi(\theta)\ge0\), \(0<(1-r\phi(\theta))^{-1}\le(1-\phi(\theta))^{-1}\) except at the origin, a set of measure zero. The function equal to \(1\) on \(\{\phi<0\}\) and to \((1-\phi)^{-1}\) on \(\{\phi\ge0\}\) is integrable by the result just proved and the finite volume of the cube. Dominated convergence therefore gives the stated Green-function formula. Its integral is finite, so \(G(0,0)<\infty\). \(\square\)

The dimension threshold comes from the exponent \(d-3\), not from the height of the integrand at the single singular point. In dimension three, the radial power is \(r^0\) and the integral near zero is proportional to \(\varepsilon\). In dimension four it is proportional to \(\varepsilon^2\). In every higher dimension the same calculation gives a finite power of \(\varepsilon\). The value at the origin itself does not affect an integral, but the behavior in every neighborhood of the origin does.

Worked Bounds in Specific Dimensions

A Common Pitfall

A common error is to integrate the bound as though the radial volume element were \(dr\). In dimension three, that incorrect calculation would use \(\int_0^\varepsilon r^{-2}\,dr\), which diverges at zero, and would falsely suggest that the comparison function is not integrable. But in three-dimensional polar coordinates the volume element is \(r^2\,dr\) times surface area. The correct radial integrand is \(r^{-2}r^2=1\), and its integral is \(\int_0^\varepsilon1\,dr=\varepsilon\), which is finite. The Jacobian is essential: the pointwise bound alone does not determine whether the multidimensional integral converges.

The argument proves finiteness by an upper bound; it does not compute the exact value of the Green function. Nor does it require the Fourier integrand to be bounded at zero. The relevant facts are that the singularity is controlled by a multiple of \( |\theta|^{-2}\), that the radial measure contributes \(r^{d-1}\), and that the resulting power \(r^{d-3}\) is integrable at zero when \(d\ge3\).