5 Fourier transform of Green’s function

5.1 Fourier transform of the regularized Green’s function

Lemma 5.1

For \(0 \le r {\lt} 1\), the regularized Green’s function \(\mathrm{G}_{r} \colon \mathbb {Z}^d \to \mathbb {R}\) is an element of the Hilbert space \(\ell ^2(\mathbb {Z}^d, \mathbb {C})\) of complex-valued square-summable functions on \(\mathbb {Z}^d\).

Proof ▶

Since \(\ell ^1(\mathbb {Z}^d, \mathbb {C}) \subset \ell ^2(\mathbb {Z}^d, \mathbb {C})\), it suffices to show that \(\mathrm{G}_{r}\) is absolutely summable. This follows from Lemma 4.15.

Definition 5.2

✓

Let \(0 \le r {\lt} 1\). The Fourier transform of the regularized Green’s function \(\mathrm{G}_{r} \, \colon \, \mathbb {Z}^d \to \mathbb {R}\) is the function

\begin{align*} \widehat{\mathrm{G}}_{r} \, \colon \, \mathbb {R}^d \to \mathbb {C}\end{align*}

given by

\begin{align*} \widehat{\mathrm{G}}_{r} (\theta ) = \sum _{x \in \mathbb {Z}^d} e^{\mathrm{i}x \cdot \theta } \, \mathrm{G}_{r}(x) . \end{align*}

5.2 Explicit formula for the Fourier transform

Lemma 5.3

For a time-homogeneous random walk \(X= \big(X(t)\big)_{t \in \mathbb {N}}\) on \(\mathbb {Z}^d\) with step distribution \(p \colon \mathbb {Z}^d \to [0,1]\), the Fourier transform of the Green’s function is

\begin{align*} \widehat{\mathrm{G}}_{r} (\theta ) = \Big(1 - r \sum _{u\in \mathbb {Z}^d} p(u) e^{\mathrm{i}u \cdot \theta }\Big)^{-1} . \end{align*}

Proof ▶

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Lemma 5.4

For the simple random walk \(X= \big(X(t)\big)_{t \in \mathbb {N}}\) on \(\mathbb {Z}^d\), the Fourier transform of the Green’s function is

\begin{align*} \widehat{\mathrm{G}}_{r} (\theta ) = \frac{1}{1-\frac{r}{d} \sum _{j=1}^d \cos (\theta _j)} . \end{align*}

Proof ▶

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5.3 Inversion of the discrete Fourier transform

Lemma 5.5

For any \(x \in \mathbb {Z}^d\) and \(0 \le r {\lt} 1\), we have

\begin{align*} \mathrm{G}_{r} (x) = \frac{1}{(2\pi )^d} \iint _{[-\pi ,\pi ]^d} e^{-\mathrm{i}x \cdot \theta } \, \widehat{\mathrm{G}}_{r} (\theta ) \; \mathrm{d}^d \theta . \end{align*}

Proof ▶

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Lemma 5.6

✓

For any \(x \in \mathbb {Z}^d\) and \(0 \le r {\lt} 1\), we have

\begin{align*} \mathrm{G}_{r} (x) = \frac{1}{(2\pi )^d} \iint _{[-\pi ,\pi ]^d} \Re \mathfrak {e}\Big( e^{-\mathrm{i}x \cdot \theta } \, \widehat{\mathrm{G}}_{r} (\theta ) \Big) \; \mathrm{d}^d \theta . \end{align*}

Proof ▶

The integral of the real part is the real part of the integral so this is obvious from Lemma 5.5 — the left hand side is real to start with, so equal to its own real part.

Recall that we are interested in \(\mathsf{E}[L]\), where \(L\) is the number of visits to the origin by the random walk. Lemma 4.16 states that \(\mathsf{E}[L]\) is the increasing limit of \(\mathrm{G}_{r}(\vec{0})\) as \(r \nearrow 1\), and Lemma 5.6 gives a formula for \(\mathrm{G}_{r}(\vec{0})\) as the integral of the real part of the Fourier transform: \(\mathrm{G}_{r}(\vec{0}) = \frac{1}{(2\pi )^d} I_r\), where

\begin{align} I_r = \iint _{[-\pi ,\pi ]^d} \Re \mathfrak {e}\big( \widehat{\mathrm{G}}_{r} (\theta ) \big) \; \mathrm{d}^d \theta . \end{align}

Corollary 5.7

A random walk \(X= \big(X(t)\big)_{t \in \mathbb {N}}\) on \(\mathbb {Z}^d\) is expectation recurrent if and only if \(\lim _{r \nearrow 1} \, I_r = +\infty \). In other words, \(X\) is expectation transient if and only if \(\lim _{r \nearrow 1} \, I_r \, {\lt} \, +\infty \).

Proof ▶

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