\section{Subadditivity tricks}

\let\lam\lambda
This chapter concerns mainly with one theorem that uses the following
subadditivity property.

We use $\vec{x}$ for a vector of parameters $x_1, x_2, x_3, \dots, x_k$.
The length of the vector may vary and can be deduced from context.

In a subcritical setting ($0 < \lam < \lam_c$), we define a
function 
\[
\xi(\lam) \equiv \lim_{n \rightarrow \infty} (-n^{-1} \log p_n(\lam))
%Equivalently, this can be defined as
=\xi(\lam) = \frac{-1}{n} \log\left(\sum_{k=n}^{\infty} p_k(\lam)\right).
\]

\begin{thm}
Function $\xi$, as defined in the previous paragraph, is a proper function
and moreover it continuously reaches infinity as $\lam$ goes to zero.
\end{thm}

\begin{proof}

First, we will investigate the definition of the functions $p_k(\lam)$.
These can be equivalently expressed as

$$ p_{k+1}(\lam) = (k+1) \lam^k \int_{x_1 \in \mathbb{R}^d} \int_{x_2 \in \mathbb{R}^d} \cdots \int_{x_k \in \mathbb{R}^d} h(0, \vec{x})
\ \exp(-\lam A(0, \vec{x}))\ d\vec{x},$$

where the two new functions are the ``indicator function'' $h$ and an ``area
function'' $A$. Notice the connection between $p_k$ and the Poisson distribution.

More formally, using the Iverson notation:
$$h(0, \vec{x}) \equiv [G(0,\vec{x};1) {\rm\ is\ connected}],$$
$$A(0, \vec{x}) \equiv \bigcup_{i=0}^k {\rm Ball}(x_i, 1).$$

If we multiply $p_{k+1}(\lam)$ with another $p_{j+1}(\lambda)$, we can notice
that the indicator functions merge easily, since the two sets of points have a common
element, which is the origin:
$$h(0,\vec{x}) \cdot h(0,\vec{y}) \le h(0,\vec{x},\vec{y}).$$

Since the area function calculates volume of an area, we have:
$$ A(\vec{x}) + A(\vec{y}) \ge A(\vec{x}, \vec{y}). $$

Substituting the two bounds, we get:
$$ p_{k+1}(\lam) \cdot p_{j+1}(\lam) \le (k+1)(j+1) \lam^{k+j} \int \int \cdots \int
h(0,\vec{x}, \vec{y})\ \exp(-\lam A(\vec{x}, \vec{y}))\ d\vec{x} d\vec{y}.$$

If we define $q_k \equiv - \log \frac{p_k}{k+1}$, the above simplifies to:
$$ q_{m+n} \le q_m + q_n. $$

It remains to show that the subadditivity implies that the limit $\lim_{n \rightarrow \infty} \frac{q_n}{n}$ exists.

We define $\xi(\lam)$ as $-\liminf q_k/(k-1)$. 
That is, for every $\epsi > 0$ exists an $m$ such that $q_{m}/(m-1) < \xi(\lam) + \epsi$.
Then for every $n$, we take nonnegative integers $r<m$ and $k$ such that $n = km + r$ 
and obtain from the subadditivity:
\begin{align*}
q_n &\le q_r + k q_m \\
\frac{q_n}{n-1} &\le \frac{q_r}{n-1} + k \frac{q_m}{k(m-1)} \qquad \text{since $n-1 \ge k(m-1)$} \\
\frac{q_n}{n-1} &\le \frac{\max\{q_1, \dots, q_{m-1}\}}{n-1} + \frac{q_m}{m-1}
\end{align*}
Then, for every $n$ large enough,
\[
\frac{q_n}{n-1} < \xi(\lam) + \epsi.
\]

\end{proof}