Recurrence of an irreducible random walk (Chung - Fuchs Theorem)

Question

I am studying the Theorem by Chung and Fuchs (1951) from the book Probability Theory, 3rd version, by A. Klenke and I have a couple of steps for which I need some clarification.

The theorem states that an irreducible random walk $X$ on $\mathcal{Z}^D$ with characteristic function $\phi$ is recurrent if and only if, for every $\epsilon>0$,

$$ \lim_{\lambda \uparrow 1}\int_{(-\epsilon,\epsilon)^D} Re\bigg(\frac{1}{1-\lambda\phi(t)}\bigg)dt=\infty $$

In the book the proof begins with:

$$ G(0,0) = \lim_{\lambda \uparrow 1} \sum_{n=0}^\infty \lambda^np^n(0,0) \\ =\lim_{\lambda \uparrow 1}\int_{[-\pi,\pi]^D} (2\pi)^{-D}\frac{1}{1-\lambda\phi(t)}dt \\ =\lim_{\lambda \uparrow 1}\int_{[-\pi,\pi]^D} (2\pi)^{-D} Re\bigg(\frac{1}{1-\lambda\phi(t)}\bigg)dt $$

1)Why does the last equality holds?

Then, to obtain the statement of the theorem, the author argues as follow:

If we had $\phi(t)=1$ for some $t\in(-2\pi,2\pi)^D\setminus\{0\}$, then we would have $\phi^n(t)=1$ for every $n\in\mathcal{N}$ and hence $P_0[\langle X_n,t/(2\pi)\rangle\in \mathcal{Z}]=1$. Thus $X$ would not be irreducible contradicting the assumption.

2) Why this is a contradiction?

He concludes is this way:

Due to the continuity of $\phi$ for all $\epsilon>0$, we thus have $$ \inf\{|1-\phi(t)|: t\in[-\pi,\pi)^D \setminus (-\epsilon,\epsilon)^D\}>0 $$

3) How is the author using the continuity of the characteristic function?

4) How is this last result linked with the condition stated in the theorem?

Please, let me know if more context is needed. Thank you.

Answer 1

1. The equality is an equality between real numbers (clearly G(0,0) is real valued) so if we split the integral between it's real and imaginary part, the imaginary part is null and follows that the integral of the imaginary part of the function under the integral must be null. (Because $\int f(z)dz = \int \Re f(z)dz + i\int \Im f(z)dz$ by definition of the integral of a complex valued function).

2. The theorem makes the assumption that the random-walk is irreducible. But here we show that if $\phi(t)=1$ then only the points $z\in \mathbb{Z}^d$ that verify $\langle z | \frac{t}{2\pi}\rangle \in \mathbb{Z}$ can be reached by the random walk (and $t\not\in \mathbb Z^d$ because all the coordinates are smaller than 1 and we exclude the case $t=0$).

3. By continuity, the inf is actually a min (you may want to take the inf over $[-\pi, \pi]^d\setminus (-\epsilon, \epsilon)^d$ to make the compacity evident. If the inf was 0 then there would be $t\in[-\pi, \pi]^d\setminus (-\epsilon, \epsilon)^d$ such that $\phi(t) = 1$ which contradicts the previous point.

4. The result gives you that $\frac{1}{|1-\phi(t)|}$ is bounded outside $(-\epsilon, \epsilon)^d$ and thus outside this cube, the integral stays bounded. To see if the integral diverges or not (and thus $G(0,0)=\infty$ or not ie if the walk is recurrent or not) one can focus on the behavior of the integral over a neighborhood of 0.