Local Integrability at the Singular Point

Why the Origin Controls the Integral

For a random walk on a lattice, the Fourier representation of the Green function turns expected visit counts into an integral. For the simple symmetric walk, the integrand is singular at the origin: its denominator vanishes there. The key local question is how quickly that denominator approaches zero, and whether the surrounding volume is large enough to make the resulting singularity integrable. A quadratic expansion answers the first question; radial integration answers the second.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, and fix an integer dimension $d\ge1$. Consider the discrete-time walk $(X_n)_{n\ge0}$ on the state space $S=\mathbb{Z}^d$, started at $X_0=0$, whose independent increments are each uniformly distributed over the $2d$ vectors $\pm e_1,\ldots,\pm e_d$. Here $e_j$ is the $j$-th coordinate vector. Its Fourier symbol, the characteristic function of one increment, is

\phi(\theta)=\mathbb{E}\!\left[e^{i\theta\cdot(X_1-X_0)}\right] =\frac{1}{d}\sum_{j=1}^{d}\cos\theta_j, \qquad \theta\in[-\pi,\pi]^d.

The Fourier Green-function formula from the earlier material on Fourier integrals uses the integrand $1/(1-\phi(\theta))$, with normalized measure $(2\pi)^{-d}d\theta$. At $\theta=0$, both numerator and denominator in this expression are not the issue: the denominator itself is zero. We therefore study the integral on a neighborhood of the origin, excluding the point $0$. A value assigned to an integrand at a single point does not change its Lebesgue integral.

Definition: The integrand $1/(1-\phi(\theta))$ is locally integrable at the origin if its integral over some neighborhood of $0$, with the origin omitted, is finite. For this walk, $1-\phi(\theta)\ge0$, so this is an ordinary nonnegative integral.

The Quadratic Expansion and a Useful Bound

For each coordinate, the Taylor expansion of cosine begins with $1-\theta_j^2/2$. Summing those terms gives the leading behavior of $1-\phi$. The following bound makes the approximation precise and will also let us compare the integrand with a simple radial function.

Proposition: If $|\theta|\le1$, where $|\theta|=(\sum_{j=1}^d\theta_j^2)^{1/2}$, then $$ 1-\phi(\theta)=\frac{|\theta|^2}{2d}+R(\theta), \qquad |R(\theta)|\le\frac{|\theta|^4}{24d}, $$ and, in particular, $$ \frac{11}{24d}|\theta|^2 \le 1-\phi(\theta) \le\frac{1}{2d}|\theta|^2. $$

Proof. For a real number $x$, Taylor's theorem with a fourth-order remainder gives $\left|1-\cos x-x^2/2\right|\le |x|^4/24$. Because $|\theta|\le1$, every coordinate satisfies $|\theta_j|\le1$, and thus $1-\cos\theta_j$ lies between $\theta_j^2/2-\theta_j^4/24$ and $\theta_j^2/2$. Now $1-\phi(\theta)=d^{-1}\sum_{j=1}^d(1-\cos\theta_j)$. The quadratic terms sum to $|\theta|^2/(2d)$, and the absolute remainder is at most $(24d)^{-1}\sum_j\theta_j^4$. Since $\sum_j\theta_j^4\le(\sum_j\theta_j^2)^2=|\theta|^4$, this proves the remainder bound. Also, $\theta_j^4\le\theta_j^2$, so $\theta_j^2/2-\theta_j^4/24\ge(11/24)\theta_j^2$. Summing this lower bound and the upper bound $\theta_j^2/2$, then dividing by $d$, proves both inequalities. $\square$

In particular, the denominator is comparable to $|\theta|^2$ near zero. Taking reciprocals of the positive bounds gives

\frac{2d}{|\theta|^2} \le\frac{1}{1-\phi(\theta)} \le\frac{24d}{11|\theta|^2}, \qquad 0<|\theta|\le1.

The leading approximation is therefore $1/(1-\phi(\theta))\sim 2d/|\theta|^2$, where the ratio of the two sides tends to $1$ as $\theta$ tends to $0$. The two-sided bound is stronger for deciding integrability: it shows that the actual integrand has the same local integrability as the reciprocal-square function.

Worked Example: The One-Dimensional Singularity

For $d=1$, the symbol is $\phi(\theta)=\cos\theta$. The expansion gives $1-\phi(\theta)\sim\theta^2/2$, so the leading approximation to the integrand is $2/\theta^2$. To see its local behavior explicitly, integrate this approximation over the punctured interval $\epsilon<|\theta|<\delta$, where $0<\epsilon<\delta\le1$. The normalized integral is

\frac{1}{2\pi}\int_{\epsilon<|\theta|<\delta}\frac{2}{\theta^2}\,d\theta =\frac{1}{2\pi}\left(4\int_\epsilon^\delta\frac{1}{\theta^2}\,d\theta\right) =\frac{2}{\pi}\left(\frac{1}{\epsilon}-\frac{1}{\delta}\right).

Here the factor $4$ comes from the integrand's factor $2$ and the two equal intervals, one on each side of zero. Since $\int_\epsilon^\delta\theta^{-2}d\theta=1/\epsilon-1/\delta$, the displayed value diverges as $\epsilon$ decreases to zero. The comparison bounds in the proposition show that the true integrand also has infinite integral in every neighborhood of the origin.

Radial Integration Gives the Threshold

In dimension $d$, the volume of a thin radial shell at distance $r$ from the origin is proportional to $r^{d-1}dr$. Multiplying this shell factor by an integrand proportional to $r^{-2}$ gives the power $r^{d-3}dr$. This is the source of the dimension-dependent threshold.

Theorem: For the simple symmetric nearest-neighbor walk on $\mathbb{Z}^d$, the Green-function integrand $1/(1-\phi(\theta))$ is locally integrable at the origin if and only if $d\ge3$. Equivalently, its local radial comparison is $r^{d-3}dr$.

Proof. Choose a radius $0<\delta\le1$. By the proposition, throughout the punctured ball $0<|\theta|<\delta$, the integrand is bounded below by $2d/|\theta|^2$ and above by $24d/(11|\theta|^2)$. It is therefore locally integrable exactly when $|\theta|^{-2}$ is integrable on this ball. In polar coordinates, the latter integral is a positive constant (the surface area of the unit sphere) times $\int_0^\delta r^{d-3}\,dr$. If $d=1$, this is $\int_0^\delta r^{-2}\,dr$, which diverges. If $d=2$, it is $\int_0^\delta r^{-1}\,dr$, which also diverges. If $d\ge3$, its exponent $d-3$ is nonnegative, and direct integration gives $\delta^{d-2}/(d-2)<\infty$. The upper comparison then proves local integrability for $d\ge3$; the lower comparison proves failure of local integrability for $d=1,2$. This establishes both directions. $\square$

The result concerns the singular point, but for this walk it also identifies whether any other point of the Fourier domain causes a singularity. Indeed, each $\cos\theta_j\le1$, so $\phi(\theta)=1$ requires every $\cos\theta_j=1$. On the torus represented by $[-\pi,\pi]^d$, this happens only at the origin. Away from any neighborhood of the origin, $1-\phi$ is continuous and strictly positive on a compact set, so its reciprocal is bounded there. Thus no additional singularity changes the local diagnosis.

Worked Example: The Two-Dimensional Logarithmic Comparison

For $d=2$, $\phi(\theta)=\tfrac12(\cos\theta_1+\cos\theta_2)$, and the quadratic term is $(\theta_1^2+\theta_2^2)/4$. Writing $r=|\theta|$, the leading approximation to the integrand is $4/r^2$. On the annulus $\epsilon<r<\delta$, the normalized integral of this approximation is

\frac{1}{(2\pi)^2}\int_0^{2\pi}\int_\epsilon^\delta \frac{4}{r^2}\,r\,dr\,d\alpha =\frac{1}{4\pi^2}\cdot 8\pi\int_\epsilon^\delta\frac{1}{r}\,dr =\frac{2}{\pi}\log\!\left(\frac{\delta}{\epsilon}\right).

The angular integral contributes $2\pi$; the factor $4$ in the integrand therefore gives $8\pi$, and the radial integral is $\log(\delta/\epsilon)$. This comparison diverges as $\epsilon$ decreases to zero. The theorem's two-sided bounds ensure the true integrand has the same integrability outcome, even though this calculation integrates the leading approximation.

Worked Example: The Three-Dimensional Radial Comparison

For $d=3$, the quadratic term in $1-\phi(\theta)$ is $r^2/6$, so the leading integrand is $6/r^2$. In three dimensions the area of the unit sphere is $4\pi$. Integrating this approximation over $\epsilon<r<\delta$ gives

\frac{1}{(2\pi)^3}\int_\epsilon^\delta \frac{6}{r^2}\,4\pi r^2\,dr =\frac{24\pi}{8\pi^3}(\delta-\epsilon) =\frac{3}{\pi^2}(\delta-\epsilon).

The factors $6$, $4\pi$, and $r^2$ come, respectively, from the reciprocal quadratic term, the sphere area, and the radial volume element. The resulting expression remains finite as $\epsilon$ decreases to zero. The theorem's upper bound on the true integrand confirms its local integral is finite as well.

A Common Miscount and Why the Criterion Matters

Common mistake. One might reason that a ball of radius $r$ has volume proportional to $r^d$, multiply that by the pointwise size $r^{-2}$, and conclude that integrability requires $d>1$. This omits the fact that an integral accumulates contributions from shells across a range of radii. For a shell of thickness $dr$, the volume is proportional to $r^{d-1}dr$, not $r^d$. The correct product is $r^{-2}r^{d-1}dr=r^{d-3}dr$, which is nonintegrable in dimensions one and two and integrable from dimension three onward.

This distinction matters when using Fourier integrals to study expected visit counts. The quadratic expansion alone describes how the integrand blows up; the radial shell factor determines whether the blow-up has finite total mass. Local integrability provides that part of the Green-function calculation. It does not, by itself, supply return-probability asymptotics or a recurrence classification; those require additional arguments.

Check Your Understanding

Use the quadratic comparison and radial shell factor to answer the questions.

For dimension $d=4$, what power of $r$ appears in the radial integral near zero, and is that integral finite?
Explain why the value assigned to $1/(1-\phi)$ at $\theta=0$ does not determine local integrability.
For $d=2$, write the leading quadratic term of $1-\phi(\theta)$ in terms of $\theta_1,\theta_2$.
What incorrect threshold results from using $r^d$ instead of $r^{d-1}dr$ as the shell-volume factor?
Why does the bound comparing $1-\phi(\theta)$ to $|\theta|^2$ establish the same integrability criterion for the actual integrand as for $1/|\theta|^2$?

Local Integrability at the Singular Point

What You'll Learn