A Appendix
In this appendix we collect results for the LS model that we used to establish the results in the previous sections. We start with an estimate between the integrated density of states for finite N and the infinite-volume integrated density of states.
Lemma A.1
For all \(N \in \mathbb {N}\), \(E > 0\), and \(0< {\mathcal {E}} < E^{-1/2}\) we have
$$\begin{aligned} \mathbb {E} \left[ {\mathcal {N}}_N^{\mathrm {I},\omega }(E) \right] \le {\mathcal {N}}_{\infty }^{\mathrm {I}} ([E^{-1/2} - {\mathcal {E}}]^{-2}) \ . \end{aligned}$$
(A.1)
Moreover, for all \(N \in \mathbb {N}\) and Lebesgue-almost all \(E \ge 0\) (i.e., all \(E \ge 0\) except the discontinuity points of \(\mathcal N_{\infty }^{\mathrm {I}}\)) one has
$$\begin{aligned} \mathbb {E} \left[ {\mathcal {N}}_N^{\mathrm {I},\omega }(E) \right] \le {\mathcal {N}}_{\infty }^{\mathrm {I}} (E) \ . \end{aligned}$$
(A.2)
Proof
In this proof, \({\mathcal {N}}_{\Lambda }^{\mathrm {I},\omega }\) denotes the integrated density of states corresponding to the operator \(h^{\Lambda }_{\gamma }(\omega )\) on \(H^1_0(\Lambda )\), defined analogously to (2.4).
In a first step, we realize that, for arbitrary disjoint intervals \(\Lambda ^1, \Lambda ^2 \subset \mathbb {R}\) and for any \(E \ge 0\), the inequality
$$\begin{aligned} |\Lambda ^1|\, {\mathcal {N}}_{\Lambda ^1}^{\mathrm {I}, \omega }(E) + |\Lambda ^2| \, {\mathcal {N}}_{\Lambda ^2}^{\mathrm {I}, \omega }(E) \le |\Lambda ^1 \cup \Lambda ^2| \, {\mathcal {N}}_{\Lambda ^1 \cup \Lambda ^2}^{\mathrm {I}, \omega }(E) \end{aligned}$$
(A.3)
holds for \(\mathbb {P}\)-almost all \(\omega \in \Omega \), see, e.g., [17, 5.39a].
For \(M,N \in \mathbb {N}\) and \(j \in \mathbb {Z}\) we define \(\Lambda ^{j}_N:=\Lambda _N + L_N \cdot j\) and \(\Lambda _{M,N}:=\bigcup _{j \in \mathbb {Z}: |j| \le M} \Lambda _N^j\). Note that \(\{ {\mathcal {N}}_{\Lambda _N^j}^{\mathrm {I},\omega }(E) \}_{j \in \mathbb {Z}}\) is a set of independent and identically distributed random variables for all \(E \ge 0\). Hence, employing the strong law of large numbers, inequality (A.3), and introducing the continuous function
$$\begin{aligned} f_{E, {\mathcal {E}}} : \mathbb {R} \rightarrow \mathbb {R},\ x \mapsto {\left\{ \begin{array}{ll} 0 &{}\quad \text { if }\, x \le - {\mathcal {E}} \\ 1 + {\mathcal {E}}^{-1} x &{}\quad \text { if }\, -{\mathcal {E}}< x< 0 \\ 1 &{}\quad \text { if }\, 0 \le x \le E \\ 1 - \dfrac{x-E}{[E^{-1/2} - {\mathcal {E}}]^{-2} - E} &{}\quad \text { if }\, E< x < [E^{-1/2} - {\mathcal {E}}]^{-2} \\ 0 &{}\quad \text { if }\, x \ge [E^{-1/2} - {\mathcal {E}}]^{-2} \end{array}\right. } \end{aligned}$$
we find that, for all \(E > 0\),
$$\begin{aligned} \mathbb {E} \Big [ {\mathcal {N}}_N^{\mathrm {I},\omega }(E) \Big ]&= \lim \limits _{M \rightarrow \infty } \dfrac{1}{2M+1}\sum \limits _{j \in \mathbb {Z} : |j| \le M} {\mathcal {N}}_{\Lambda _N^j}^{\mathrm {I},\omega }(E)\\&= \lim \limits _{M \rightarrow \infty } \dfrac{1}{|\Lambda _M|} \sum \limits _{j \in \mathbb {Z} : |j| \le M} |\Lambda _N^j| \, \mathcal N_{\Lambda _N^j}^{\mathrm {I},\omega }(E) \\&\le \limsup \limits _{M \rightarrow \infty } \mathcal N_{\Lambda _{M,N}}^{\mathrm {I},\omega }(E) \le \limsup \limits _{M \rightarrow \infty } {\mathcal {N}}_{M}^{\mathrm {I},\omega }(E) \\&\le \lim \limits _{M \rightarrow \infty } \int \limits _{\mathbb {R}} f_{E,{\mathcal {E}}}({{\widetilde{E}}}) \, {\mathcal {N}}_M^{\omega }(\mathrm {d} {{\widetilde{E}}}) = \int \limits _{\mathbb {R}} f_{E,{\mathcal {E}}}(\widetilde{E}) \, {\mathcal {N}}_{\infty }(\mathrm {d} {{\widetilde{E}}}) \\&\le {\mathcal {N}}_{\infty }^{\mathrm {I}}([E^{-1/2} - {\mathcal {E}}]^{-2}) \ . \end{aligned}$$
(A.2) follows from the fact that \({\mathcal {N}}_{\infty }^{\mathrm {I}}\) is monotonically increasing and hence has at most countably many points of discontinuity and by taking the limit \({\mathcal {E}} \searrow 0\). \(\square \)
Using this result we obtain a lower bound for the ground-state energy.
Lemma A.2
For all \(\kappa > 2\) and for \(\mathbb {P}\)-almost all \(\omega \) there exists an \({{\widetilde{N}}} = {{\widetilde{N}}}(\kappa , \omega ) \in \mathbb {N}\) such that for all \(N \ge {{\widetilde{N}}}\) we have
$$\begin{aligned} E_N^{1,\omega } \ge \left( \dfrac{\pi \nu }{\kappa \ln (L_N)} \right) ^{2} \ . \end{aligned}$$
(A.4)
Proof
Let \(\kappa > 2\) be given. We define \({{\widehat{E}}}_N := (\pi \nu / [ \kappa \ln (L_N)])^{2}\) for all \(N \in \mathbb {N}\) with \(L_N> 1\), and pick some \({\mathcal {E}} > 0\).
Then, with Lemma A.1 and Theorem 2.4 we conclude that for all but finitely many \(N \in \mathbb {N}\) one has
$$\begin{aligned} \mathbb {P}\left( \omega : |\Lambda _N| \cdot \mathcal N_N^{\omega }({{\widehat{E}}}_N) \ge 1 \right)&\le L_N \cdot \mathbb {E}\left[ {\mathcal {N}}_N^{\mathrm {I}, \omega }({{\widehat{E}}}_N) \right] \\&\le L_N \cdot {\mathcal {N}}^{\mathrm {I}}_{\infty }([\widehat{E}_N^{-1/2} - {\mathcal {E}}]^{-2}) \\&\le {\widetilde{M}}L_N \cdot \exp \left( - \pi \nu \left[ \widehat{E}_N^{-1/2} - {\mathcal {E}} \right] \right) \\&\le {\widetilde{M}} \mathrm {e}^{\pi \nu {\mathcal {E}}}\cdot L_N^{- \kappa +1} \ . \end{aligned}$$
Hence \(\sum _{N=1}^{\infty } \mathbb {P}\left( \omega : L_N\cdot \mathcal N_N^{\omega }({{\widehat{E}}}_N) \ge 1 \right) < \infty \) and the statement follows from the Borel–Cantelli lemma. \(\square \)
Remark A.3
Lemmas A.1 and A.2 can readily be generalized to Poisson random potentials of the form
$$\begin{aligned} V(\omega , \cdot ) = \sum \limits _{j\in \mathbb {Z}} u(\cdot - x_j(\omega )) \ , \end{aligned}$$
with \(u \in L^{\infty }(\mathbb {R})\) having compact support. What we do not have at our disposal is the \(O(E^{1/2})\)-error bound in the Lifshitz-tail behavior. If we did we could carry over our results to these models.
Note that the critical density \(\rho _c(\beta )\) was defined in (2.14).
Lemma A.4
For all \(\beta > 0\), the critical density \(\rho _c(\beta )\) satisfies
$$\begin{aligned} \rho _c(\beta ) = \int \limits _{(0,\infty )} {\mathcal {B}}(E) \, \mathcal N_{\infty } (\mathrm {d} E) < \infty \ . \end{aligned}$$
Proof
In a first step we show that
$$\begin{aligned} \int \limits _{\mathbb {R}} {\mathcal {B}}(E) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) = \int \limits _{(0,\infty )} {\mathcal {B}}(E) \, \mathcal N_{\infty } (\mathrm {d} E) \end{aligned}$$
(A.5)
is finite. In order to do this, we choose \({{\widetilde{E}}}_1: = {{\widetilde{E}}}\) with \({{\widetilde{E}}} > 0\) as in Theorem 2.4. Moreover, we choose \({{\widetilde{E}}}_2 > {{\widetilde{E}}}_1\) such that for \(E \ge {{\widetilde{E}}}_2\) one has
$$\begin{aligned} {\mathcal {B}}(E) \le \left( 2^{-1/2} \mathrm {e}^{\beta E} \right) ^{-1} \end{aligned}$$
(A.6)
and
$$\begin{aligned} {\mathcal {N}}_{\infty }^{\mathrm {I}}(E) \le c E^{1/2} \end{aligned}$$
(A.7)
with \(c > 0\) as in Theorem 2.4. We write,
$$\begin{aligned} \int \limits _{(0,\infty )} {\mathcal {B}}(E) \, {\mathcal {N}}_{\infty } (\mathrm {d} E)&= \int \limits _{(0,{{\widetilde{E}}}_1]} {\mathcal {B}}(E) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) \end{aligned}$$
(A.8)
$$\begin{aligned}&\quad + \int \limits _{({{\widetilde{E}}}_1,{{\widetilde{E}}}_2]}{\mathcal {B}}(E) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) \end{aligned}$$
(A.9)
$$\begin{aligned}&\quad + \int \limits _{({{\widetilde{E}}}_2,\infty )} {\mathcal {B}}(E) \, \mathcal N_{\infty } (\mathrm {d} E) \end{aligned}$$
(A.10)
The term in line (A.9) is finite since
$$\begin{aligned} \int \limits _{({{\widetilde{E}}}_1,{{\widetilde{E}}}_2]} {\mathcal {B}}(E) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) \le {\mathcal {B}}({{\widetilde{E}}}_1)\, {\mathcal {N}}_{\infty }^{\mathrm {I}} ({{\widetilde{E}}}_2) \le c\,\mathcal B({{\widetilde{E}}}_1)\, {{\widetilde{E}}}_2^{1/2} < \infty \ , \end{aligned}$$
see also (A.7).
In the following, we write \({\mathcal {N}}_{\infty }^{\mathrm {I}}(E+)\) for the right-sided limit of \({\mathcal {N}}_{\infty }^{\mathrm {I}}\) at E.
For the term in line (A.8) we get, using integration by parts, see, e.g., [8, Theorem 21.67], and the fact that \(\mathcal N_{\infty }^{\mathrm {I}}\) is non-decreasing,
$$\begin{aligned} A&:= \lim \limits _{\epsilon _1 \searrow 0} \int \limits _{(\epsilon _1,{{\widetilde{E}}}_1]} {\mathcal {B}}(E) \, \mathcal N_{\infty } (\mathrm {d} E) \\&\,\le \lim \limits _{\epsilon _1 \searrow 0} \left\{ \mathcal B({{\widetilde{E}}}_1) \, {\mathcal {N}}_{\infty }^{\mathrm {I}} (\widetilde{E}_1+) + \int \limits _{\epsilon _1}^{{{\widetilde{E}}}_1} \mathcal N_{\infty }^{\mathrm {I}} (E) \big [{\mathcal {B}}(E)\big ]^2\, \beta \mathrm {e}^{\beta E} \, \mathrm {d} E \right\} \\&\,\le {\mathcal {B}}({{\widetilde{E}}}_1)\, \mathcal N_{\infty }^{\mathrm {I}} (2 {{\widetilde{E}}}_1) + \int \limits _{0}^{{{\widetilde{E}}}_1} {\mathcal {N}}_{\infty }^{\mathrm {I}} (E) \big [{\mathcal {B}}(E)\big ]^2\, \beta \mathrm {e}^{\beta E} \, \mathrm {d} E \end{aligned}$$
and by Theorem 2.4,
$$\begin{aligned}&\le {\mathcal {B}}({{\widetilde{E}}}_1)\, {\mathcal {N}}_{\infty }^{\mathrm {I}} (2 {{\widetilde{E}}}_1) + \int \limits _{0}^{{{\widetilde{E}}}_1} {{\widetilde{M}}} \mathrm {e}^{- \nu \pi E^{-1/2}} (\beta E)^{-2} \beta \mathrm {e}^{\beta E} \, \mathrm {d} E \\&\le {\mathcal {B}}({{\widetilde{E}}}_1)\, {\mathcal {N}}_{\infty }^{\mathrm {I}} (2 {{\widetilde{E}}}_1) + {{\widetilde{M}}} \beta ^{-1} \mathrm {e}^{\beta {{\widetilde{E}}}_1} \int \limits _{0}^{{{\widetilde{E}}}_1} \dfrac{4! E^2}{(\nu \pi )^4} E^{-2} \, \mathrm {d} E < \infty \ . \end{aligned}$$
As a last step, we show that also the term in line (A.10) is finite. We obtain
$$\begin{aligned} C&:= \lim \limits _{\epsilon _2 \rightarrow \infty } \int \limits _{(\widetilde{E}_2,\epsilon _2]} {\mathcal {B}}(E) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) \\&\, \le \lim \limits _{\epsilon _2 \rightarrow \infty } \left[ \mathcal B(\epsilon _2)\, {\mathcal {N}}_{\infty }^{\mathrm {I}} (\epsilon _2+) + \int \limits _{{{\widetilde{E}}}_2}^{\epsilon _2} \mathcal N_{\infty }^{\mathrm {I}} (E) \big [{\mathcal {B}}(E)\big ]^2\, \beta \mathrm {e}^{\beta E} \, \mathrm {d} E \right] \end{aligned}$$
by using integration by parts. Since \(\mathcal N_{\infty }^{\mathrm {I}}\) is non-decreasing,
$$\begin{aligned} C&\, \le \lim \limits _{\epsilon _2 \rightarrow \infty } \left[ {\mathcal {B}}(\epsilon _2)\, {\mathcal {N}}_{\infty }^{\mathrm {I}} (2\epsilon _2) + \int \limits _{{{\widetilde{E}}}_2}^{\epsilon _2} {\mathcal {N}}_{\infty }^{\mathrm {I}} (E) \big [{\mathcal {B}}(E)\big ]^2\, \beta \mathrm {e}^{\beta E} \, \mathrm {d} E \right] \end{aligned}$$
and using (A.6) and (A.7),
$$\begin{aligned}&\, \le \lim \limits _{\epsilon _2 \rightarrow \infty } \left[ 2 c \, \mathrm {e}^{-\beta \epsilon _2} \epsilon _2^{1/2} + 2 c \beta \, \int \limits _{{{\widetilde{E}}}_2}^{\epsilon _2} E^{1/2} \mathrm {e}^{-2\beta E } \mathrm {e}^{\beta E} \, \mathrm {d} E \right] < \infty \ . \end{aligned}$$
Altogether we have by now proved (A.5).
The final statement about \(\rho _c(\beta )\) then follows, using again dominated convergence, by
$$\begin{aligned} \rho _c(\beta )&= \sup \limits _{\mu \in (-\infty ,0)} \left\{ \int \limits _{\mathbb {R}} {\mathcal {B}}(E - \mu ) \, \mathcal N_{\infty }(\mathrm {d} E) \right\} \\&= \sup \limits _{\mu \in (-\infty ,0)} \left\{ \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu ) \, \mathcal N_{\infty }(\mathrm {d} E) \right\} \\&= \lim \limits _{\mu \rightarrow 0^{-}} \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu ) \, {\mathcal {N}}_{\infty }(\mathrm {d} E) \\&= \int \limits _{(0,\infty )} {\mathcal {B}}(E) \, \mathcal N_{\infty }(\mathrm {d} E) < \infty \ . \end{aligned}$$
The statement in the next lemma is not trivial since we have only vague convergence of the density of states and not weak convergence.
Lemma A.5
For all \(\mu < 0\) we have \(\mathbb {P}\)-almost surely
$$\begin{aligned} \lim \limits _{N \rightarrow \infty } \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu ) \, {\mathcal {N}}_N^{\omega }(\mathrm {d} E) = \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu ) \, \mathcal N_{\infty }(\mathrm {d} E) \ . \end{aligned}$$
(A.11)
Proof
Let \(\mu < 0\) be given. Then, for all \(E_2 > 0\) we get
$$\begin{aligned} \begin{aligned} \limsup \limits _{N \rightarrow \infty } \int \limits _{(0,\infty )}&{\mathcal {B}}(E - \mu ) \, {\mathcal {N}}_N^{\omega }(\mathrm {d} E) \\ \le \,&\limsup \limits _{N \rightarrow \infty } \int \limits _{(0,E_2]} {\mathcal {B}}(E - \mu ) \, {\mathcal {N}}_N^{\omega }(\mathrm {d} E) + \limsup \limits _{N \rightarrow \infty } \int \limits _{[E_2,\infty )} {\mathcal {B}}(E - \mu ) \, {\mathcal {N}}_N^{\omega }(\mathrm {d} E) \ . \end{aligned} \end{aligned}$$
For \(0< {\mathcal {E}} < - \mu \) we define the real function \(g_{{\mathcal {E}}, E_2}\) as
$$\begin{aligned} g_{{\mathcal {E}}, E_2} (E) := {\left\{ \begin{array}{ll} 0 &{} \quad \text { if }\, E \le - {\mathcal {E}} \\ 1 + {\mathcal {E}}^{-1} E &{} \quad \text { if }\, - {\mathcal {E}}< E< 0 \\ 1 &{} \quad \text { if }\, 0 \le E \le E_2 \\ 1 - \dfrac{E - E_2}{{\mathcal {E}}} &{} \quad \text { if }\, E_2< E < E_2 + {\mathcal {E}} \\ 0 &{} \quad \text { if }\, E > E_2 + {\mathcal {E}} \end{array}\right. }\ . \end{aligned}$$
(A.12)
For the first integral we \(\mathbb {P}\)-almost surely obtain, by \(\mathbb {P}\)-almost sure vague convergence,
$$\begin{aligned} \limsup \limits _{N \rightarrow \infty } \int \limits _{(0,E_2]}&{\mathcal {B}}(E - \mu ) \, {\mathcal {N}}_N^{\omega }(\mathrm {d} E) \\ \le \,&\limsup \limits _{N \rightarrow \infty } \int \limits _{\mathbb {R}} g_{{\mathcal {E}}, E_2} (E) {\mathcal {B}}(E - \mu ) \, \mathcal N_N^{\omega }(\mathrm {d} E) \\ \le \,&{\mathcal {B}}(E_2 - \mu ) {\mathcal {N}}_{\infty }^{\mathrm {I}}(E_2 + {\mathcal {E}})+ \int \limits _{(0,E_2]} {\mathcal {B}}(E - \mu ) \, {\mathcal {N}}_{\infty }(\mathrm {d} E). \end{aligned}$$
Furthermore,
$$\begin{aligned} \limsup \limits _{N \rightarrow \infty }&\int \limits _{[E_2,\infty )}{\mathcal {B}}(E - \mu ) \, \mathcal N_N^{\omega }(\mathrm {d} E) \le \limsup \limits _{N \rightarrow \infty } \lim \limits _{\epsilon _2 \rightarrow \infty } \int \limits _{[E_2,\epsilon _2]} {\mathcal {B}}(E - \mu ) \, {\mathcal {N}}_N^{\omega }(\mathrm {d} E)\ , \end{aligned}$$
and integrating by parts gives
$$\begin{aligned} \le \,&\limsup \limits _{N \rightarrow \infty } \lim \limits _{\epsilon _2 \rightarrow \infty } \left[ {\mathcal {B}}(\epsilon _2- \mu ) \mathcal N_N^{\mathrm {I},\omega }(\epsilon _2) + \beta \int \limits _{E_2}^{\epsilon _2} \mathcal N_N^{\mathrm {I},\omega }(E) \big [{\mathcal {B}}(E - \mu )\big ]^2\, \mathrm {e}^{\beta (E - \mu )} \, \mathrm {d} E\right] \ . \end{aligned}$$
If we denote with \({\mathcal {N}}_N^{\mathrm {I},(0)}\) the integrated density of states of the free Hamiltonian \(-\mathrm {d}^2/\mathrm {d}x^2\) on \(H^1_0(\Lambda _N)\) then we can further bound this by
$$\begin{aligned} \le \,&\limsup \limits _{N \rightarrow \infty } \lim \limits _{\epsilon _2 \rightarrow \infty } \left[ {\mathcal {B}}(\epsilon _2 - \mu ) \mathcal N_N^{\mathrm {I},(0)}(\epsilon _2) + \beta \int \limits _{E_2}^{\epsilon _2} {\mathcal {N}}_N^{\mathrm {I},(0)}(E) \big [{\mathcal {B}}(E - \mu )\big ]^2\, \mathrm {e}^{\beta (E - \mu )} \, \mathrm {d} E\right] \ . \end{aligned}$$
Since \({\mathcal {N}}_N^{\mathrm {I},(0)}(E) \le \pi ^{-1} E^{1/2}\) for all \(E \ge 0\) and all \(N \in \mathbb {N}\) we get
$$\begin{aligned} = \,&\lim \limits _{\epsilon _2 \rightarrow \infty } \left[ \pi ^{-1} {\mathcal {B}}(\epsilon _2 - \mu ) \epsilon _2^{1/2} + \beta \pi ^{-1}\int \limits _{E_2}^{\epsilon _2} E^{1/2} \big [\mathcal B(E - \mu )\big ]^2\, \mathrm {e}^{\beta (E - \mu )} \, \mathrm {d} E \right] \\ = \,&\beta \pi ^{-1}\int \limits _{E_2}^{\infty } E^{1/2} \big [{\mathcal {B}}(E - \mu )\big ]^2\, \mathrm {e}^{\beta (E - \mu )} \, \mathrm {d} E \ . \end{aligned}$$
Hence, in total we obtain \(\mathbb {P}\)-almost surely
$$\begin{aligned} \limsup \limits _{N \rightarrow \infty }&\int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu ) \, {\mathcal {N}}_N^{\omega }(\mathrm {d} E) \\ \le \,&\lim \limits _{E_2 \rightarrow \infty } {\mathcal {B}}(E_2 - \mu ) {\mathcal {N}}_{\infty }^{\mathrm {I}}(E_2 + {\mathcal {E}})+\lim \limits _{E_2 \rightarrow \infty } \int \limits _{(0,E_2]} {\mathcal {B}}(E - \mu ) \, \mathcal N_{\infty }(\mathrm {d} E)\\&+ \, \beta \pi ^{-1} \lim \limits _{E_2 \rightarrow \infty } \int \limits _{E_2}^{\infty } E^{1/2} \big [{\mathcal {B}}(E - \mu )\big ]^2\, \mathrm {e}^{\beta (E - \mu )} \, \mathrm {d} E \\ = \,&\int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu ) \, \mathcal N_{\infty }(\mathrm {d} E), \end{aligned}$$
where we also used Theorem 2.4.
On the other hand, for all \(E_2 > 0\)\(\mathbb {P}\)-almost surely,
$$\begin{aligned} \liminf \limits _{N \rightarrow \infty } \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu ) \, {\mathcal {N}}_N^{\omega }(\mathrm {d} E)&\ge \liminf \limits _{N \rightarrow \infty } \int \limits _{\mathbb {R}} g_{{\mathcal {E}}, E_2}(E) {\mathcal {B}}(E - \mu ) \, {\mathcal {N}}_N^{\omega }(\mathrm {d} E) \\&= \int \limits _{\mathbb {R}} g_{{\mathcal {E}}, E_2}(E) {\mathcal {B}}(E - \mu ) \, {\mathcal {N}}_{\infty }(\mathrm {d} E) \\&\ge \int \limits _{(0,E_2]} {\mathcal {B}}(E - \mu ) \, \mathcal N_{\infty }(\mathrm {d} E) \ . \end{aligned}$$
Thus, \(\mathbb {P}\)-almost surely,
$$\begin{aligned} \liminf \limits _{N \rightarrow \infty } \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu ) \, {\mathcal {N}}_N^{\omega }(\mathrm {d} E)&\ge \lim \limits _{E_2 \rightarrow \infty } \int \limits _{(0,E_2]} {\mathcal {B}}(E - \mu ) \, \mathcal N_{\infty }(\mathrm {d} E) \\&=\, \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu )\, {\mathcal {N}}_{\infty }(\mathrm {d} E) \ . \end{aligned}$$
\(\square \)
The next lemma is taken from [12]. We follow in parts their proof.
Lemma A.6
If \(\rho < \rho _c(\beta )\), then \(\mu ^{\omega }_{N}\) converges \(\mathbb {P}\)-almost surely to a non-random limit point \({{\widehat{\mu }}} < 0\). On the other hand, if \(\rho \ge \rho _c(\beta )\), then \(\mu ^{\omega }_{N}\) converges \(\mathbb {P}\)-almost surely to 0.
Proof
In a first step, we show that the sequence \((\mu ^{\omega }_{N})_{N=1}^{\infty }\) has \(\mathbb {P}\)-almost surely at least one accumulation point in both cases: Note that
$$\begin{aligned} \limsup \limits _{N \rightarrow \infty } \int \limits _{(0,\infty )} \mathrm {e}^{-\beta E} {\mathcal {N}}_N^{\omega }(\mathrm {d} E) \le \limsup \limits _{N \rightarrow \infty }\frac{1}{L_N} \sum \limits _{j =1}^{\infty } \mathrm {e}^{-\beta (j\pi / L_N)^2} = (4\pi \beta )^{-1/2} < \infty \end{aligned}$$
(A.13)
for \(\mathbb {P}\)-almost all \(\omega \in \Omega \). We define
$$\begin{aligned} \phi _N^{\omega }(\beta ) := \dfrac{1}{L_N} \sum \limits _{j=1}^{\infty } \mathrm {e}^{-\beta E_N^{j,\omega }} = \int \limits _{(0,\infty )} \mathrm {e}^{-\beta E} \, {\mathcal {N}}_N^{\omega }(\mathrm {d} E)\ . \end{aligned}$$
The relation between \(\phi _N^{\omega }(\beta )\) and \(\rho \) is simply
$$\begin{aligned} \rho&= \dfrac{1}{L_N} \sum \limits _{j = 1}^{\infty } \left( \mathrm {e}^{\beta ( E_N^{j,{\omega }} - \mu _N^{\omega })} -1 \right) ^{-1} = \dfrac{1}{L_N} \sum \limits _{j= 1}^{\infty } \mathrm {e}^{-\beta E_N^{j,\omega }} \dfrac{1}{ \mathrm {e}^{- \beta \mu _N^{\omega }} - \mathrm {e}^{-\beta E_N^{j,\omega }} } \\&\le \left( \dfrac{1}{L_N} \sum \limits _{j = 1}^{\infty } \mathrm {e}^{- \beta E_N^{j,{\omega }}} \right) \dfrac{\mathrm {e}^{\beta \mu _N^{\omega }}}{1 - \mathrm {e}^{-\beta (E_N^{1,{\omega }} - \mu _N^{\omega })}} = \phi _N^{\omega }(\beta ) \dfrac{\mathrm {e}^{\beta \mu _N^{\omega }}}{1 - \mathrm {e}^{-\beta (E_N^{1,{\omega }} - \mu _N^{\omega })}} \ . \end{aligned}$$
Consequently, we conclude that
$$\begin{aligned} \rho - \rho \mathrm {e}^{-\beta E_N^{1,\omega }} \mathrm {e}^{\beta \mu _N^{\omega }} \le \phi _N^{\omega }(\beta ) \mathrm {e}^{\beta \mu _N^{\omega }} \end{aligned}$$
and
$$\begin{aligned} \beta ^{-1} \ln \left( \dfrac{\rho }{\phi _N^{\omega }(\beta ) + \rho \mathrm {e}^{-\beta E_N^{1,\omega }}} \right)&\le \mu _N^{\omega } \ . \end{aligned}$$
(A.14)
Due to (A.13) we have \(\limsup _{N \rightarrow \infty } \phi ^{\omega }_N(\beta ) < \infty \). Moreover, since \(\mu _N^{\omega }< E_N^{1,\omega }\) for all \(N \in \mathbb {N}\) and since \(E_N^{1,\omega }\) converges \(\mathbb {P}\)-almost surely to 0, we obtain \(\mathbb {P}\)-almost surely
$$\begin{aligned} \beta ^{-1} \ln \left( \dfrac{\rho }{\limsup \limits _{N \rightarrow \infty } \phi _N^{\omega }(\beta ) + \rho } \right) \le \liminf \limits _{N \rightarrow \infty } \mu _N^{\omega }\le \limsup \limits _{N \rightarrow \infty } \mu _N^{\omega }\le 0 \ . \end{aligned}$$
This shows that there exists a set \({{\widetilde{\Omega }}} \subset \Omega \) with measure \(\mathbb {P}({{\widetilde{\Omega }}}) = 1\) such that for all \(\omega \in {{\widetilde{\Omega }}}\) equation (A.11) from Lemma A.5 holds and the sequence \((\mu ^{\omega }_{N})_{N=1}^{\infty }\) has at least one accumulation point \(\mu _{\infty }^{\omega } \in \mathbb {R}\).
Now, for \(\omega \in {{\widetilde{\Omega }}}\) consider the case where \(\rho < \rho _c(\beta )\): Since
$$\begin{aligned} (-\infty ,0) \rightarrow \mathbb {R}, \quad \mu \mapsto \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu ) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) \end{aligned}$$
(A.15)
is a strictly increasing function, there is a unique, non-random solution \({{\widehat{\mu }}} < 0\) to
$$\begin{aligned} \rho = \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu ) \, \mathcal N_{\infty } (\mathrm {d} E) \ . \end{aligned}$$
(A.16)
Suppose to the contrary that the accumulation point \(\mu _{\infty }^{\omega }\) is zero. Then there exists a subsequence \((\mu _{N_j}^{\omega })_{j=1}^{\infty }\) which converges to 0. It follows that
$$\begin{aligned} \dfrac{{{\widehat{\mu }}} }{2} < \mu _{N_j}^{\omega } \end{aligned}$$
(A.17)
for all but finitely many \(j \in \mathbb {N}\). Because
$$\begin{aligned} (-\infty ,0) \rightarrow \mathbb {R}, \quad \mu \mapsto \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu ) \, \mathcal N_N^{\omega }(\mathrm {d} E) \end{aligned}$$
(A.18)
is strictly increasing,
$$\begin{aligned} \int \limits _{(0,\infty )} {\mathcal {B}}(E-{\widehat{\mu }}/2)\, \mathcal N_{N_j}^{\omega }(\mathrm {d} E) < \int \limits _{(0,\infty )} \mathcal B(E-\mu _{N_j}^{\omega }) \, {\mathcal {N}}_{N_j}^{\omega }(\mathrm {d} E) = \rho \end{aligned}$$
(A.19)
for all but finitely many \(j \in \mathbb {N}\). However, due to (A.16) and employing Lemma A.5
$$\begin{aligned} \begin{aligned} \rho&= \int \limits _{(0,\infty )} {\mathcal {B}}(E - {{\widehat{\mu }}}) \, {\mathcal {N}}_{\infty }(\mathrm {d} E) \\&< \int \limits _{(0,\infty )} {\mathcal {B}}(E-{\widehat{\mu }}/2) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) \\&= \lim \limits _{N \rightarrow \infty } \int \limits _{(0,\infty )} \mathcal B(E-{\widehat{\mu }}/2)\, {\mathcal {N}}_{N}^{\omega } \, (\mathrm {d} E) \ , \end{aligned} \end{aligned}$$
(A.20)
which is a contradiction to (A.19). As a consequence, any accumulation point \(\mu _{\infty }^{\omega }\) is strictly smaller than 0. In addition, using Lemma A.5, \(m \in \mathbb {N}\),
$$\begin{aligned} \limsup \limits _{j \rightarrow \infty } \int \limits _{(0,\infty )} \mathcal B(E-\mu _{N_j}^\omega ) {\mathcal {N}}_{N_j}^{\omega }(\mathrm {d}E)&\le \lim \limits _{j \rightarrow \infty } \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu _{\infty }^{\omega } + m^{-1} \mu _{\infty }^{\omega }/2) \, \mathcal N_{N_j}^{\omega }(\mathrm {d}E) \\&= \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu _{\infty }^{\omega } + m^{-1}\mu _{\infty }^{\omega }/2) \, {\mathcal {N}}_{\infty }(\mathrm {d}E) \end{aligned}$$
and
$$\begin{aligned} \liminf \limits _{j \rightarrow \infty } \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu _{N_j}^{\omega }) \, {\mathcal {N}}_{N_j}^{\omega }(\mathrm {d}E)&\ge \lim \limits _{j \rightarrow \infty } \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu _{\infty }^{\omega } - m^{-1} \mu _{\infty }^{\omega }/2) \, {\mathcal {N}}_{N_j}^{\omega }(\mathrm {d}E) \\&= \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu _{\infty }^{\omega } - m^{-1} \mu _{\infty }^{\omega }/2)\, {\mathcal {N}}_{\infty }(\mathrm {d}E)\ . \end{aligned}$$
Hence, since \(m \in \mathbb {N}\) was arbitrary,
$$\begin{aligned} \lim \limits _{j \rightarrow \infty } \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu _{N_j}^{\omega }) \, {\mathcal {N}}_{N_j}^{\omega }(\mathrm {d}E)&= \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu _{\infty }^{\omega }) \, {\mathcal {N}}_{\infty }(\mathrm {d}E) \ . \end{aligned}$$
We conclude that
$$\begin{aligned} \begin{aligned} \int \limits _{(0,\infty )} {\mathcal {B}}(E - {{\widehat{\mu }}}) \, \mathcal N_{\infty }(\mathrm {d} E)&= \rho = \lim \limits _{j \rightarrow \infty } \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu _{N_j}^{\omega }) \, {\mathcal {N}}_{N_j}^{\omega }(\mathrm {d} E) \\&= \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu _{\infty }^{\omega }) \, {\mathcal {N}}_{\infty }(\mathrm {d} E)\ , \end{aligned} \end{aligned}$$
(A.21)
holds \(\mathbb {P}\)-almost surely for all convergent subsequences of \((\mu ^{\omega }_{N})_{N=1}^{\infty }\) with corresponding limit point \(\mu _{\infty }^{\omega }\). Hence, due to the strict monotonicity of the function (A.15) and due to (A.21), any accumulation point \(\mu _{\infty }^{\omega }\) is equal to \(\widehat{\mu }\). In other words, the sequence \((\mu ^{\omega }_{N})_{N=1}^{\infty }\) converges to the non-random limit \({{\widehat{\mu }}} < 0\).
In the next step we assume that \(\rho \ge \rho _c(\beta )\): Suppose to the contrary that the sequence \((\mu ^{\omega }_{N})_{N=1}^{\infty }\) has an accumulation point \(\mu _{\infty }^{\omega } < 0\) with the subsequence \((\mu _{N_j}^{\omega })_{j=1}^{\infty }\) converging to it. As in (A.21) we get
$$\begin{aligned} \rho = \lim \limits _{j \rightarrow \infty } \int \limits _{(0,\infty )} \mathcal B(E - \mu _{N_j}^\omega ) \, {\mathcal {N}}_{N_j}^{\omega }(\mathrm {d} E) = \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu _{\infty }^{\omega }) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) \ge \rho _c(\beta ) \ . \end{aligned}$$
However, one also has
$$\begin{aligned} \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu _{\infty }^{\omega }) \, {\mathcal {N}}_{\infty }(\mathrm {d} E) < \sup \limits _{\mu \in (-\infty , 0)} \left\{ \int \limits _{(0,\infty )} {\mathcal {B}}(E - \mu ) \, \mathcal N_{\infty }(\mathrm {d} E) \right\} = \rho _c(\beta ) \ , \end{aligned}$$
yielding a contradiction. Since this holds for any subsequence, we conclude the statement. \(\square \)
The next lemma is essential in the proof of generalized BEC in the supercritical region \(\rho > \rho _c(\beta )\). We do not know whether the limit of \(\int \limits _{(\epsilon ,\infty )} \mathcal B(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E)\) as \(N\rightarrow \infty \) exists \(\mathbb {P}\)-almost surely. Therefore we state bounds on the \(\limsup \) and \(\liminf \), which, most importantly, coincide in the limit \(\epsilon \searrow 0\).
Lemma A.7
If \(\rho \ge \rho _c(\beta )\) and \(\epsilon > 0\), then \(\mathbb {P}\)-almost surely,
$$\begin{aligned} \limsup \limits _{N \rightarrow \infty } \int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E)&\le \int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E) \, \mathcal N_{\infty } (\mathrm {d} E) + \dfrac{2}{\beta \epsilon } \mathcal N_{\infty }^{\mathrm {I}}(\epsilon ) \ , \nonumber \\ \liminf \limits _{N \rightarrow \infty } \int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E)&\ge \int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E) \, \mathcal N_{\infty } (\mathrm {d} E) - \dfrac{4}{\beta \epsilon } \mathcal N_{\infty }^{\mathrm {I}}(2\epsilon ) \ , \end{aligned}$$
(A.22)
and
$$\begin{aligned} \begin{aligned}&\lim \limits _{\epsilon \searrow 0} \limsup \limits _{N \rightarrow \infty } \int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) \\&\quad = \, \lim \limits _{\epsilon \searrow 0} \liminf \limits _{N \rightarrow \infty } \int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) = \rho _c(\beta ) \ . \end{aligned} \end{aligned}$$
(A.23)
Proof
In a first step we note that \(\mathbb {P}\)-almost surely and for all \(E_2> \epsilon > 0\)
$$\begin{aligned} \lim \limits _{N \rightarrow \infty } \int \limits _{\mathbb {R}} g_{\epsilon }^{(E_2)}(E) {\mathcal {B}}(E-\mu _N^{\omega }) \, \mathcal N_N^{\omega } (\mathrm {d} E) = \int \limits _{\mathbb {R}} g_{\epsilon }^{(E_2)}(E) {\mathcal {B}}(E) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) \end{aligned}$$
(A.24)
where
$$\begin{aligned} g_{\epsilon }^{(E_2)}(E):= {\left\{ \begin{array}{ll} 0 \quad &{} \text { if }\, E< \epsilon /2 \\ \dfrac{E - \epsilon /2}{\epsilon /2} &{} \text { if }\, \epsilon /2 \le E \le \epsilon \\ 1 \quad &{} \text { if }\, \epsilon< E \le E_2 \\ 1 - \dfrac{E - E_2}{E_2} \quad &{} \text { if }\, E_2< E < 2 E_2 \\ 0 \quad &{} \text { if } E \ge 2 E_2 \end{array}\right. } \ . \end{aligned}$$
(A.25)
This can be shown as follows: One has
$$\begin{aligned}&\left| \, \int \limits _{\mathbb {R}} g_{\epsilon }^{(E_2)}(E) \mathcal B(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) - \int \limits _{\mathbb {R}} g_{\epsilon }^{(E_2)}(E) {\mathcal {B}}(E) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) \, \right| \end{aligned}$$
(A.26)
$$\begin{aligned}&\quad \le \, \left| \, \int \limits _{\mathbb {R}} g_{\epsilon }^{(E_2)}(E) \left( {\mathcal {B}}(E-\mu _N^{\omega }) - {\mathcal {B}}(E) \right) \, {\mathcal {N}}_N^{\omega }(\mathrm {d} E) \, \right| \end{aligned}$$
(A.27)
$$\begin{aligned}&\qquad + \, \left| \, \int \limits _{\mathbb {R}} g_{\epsilon }^{(E_2)}(E) {\mathcal {B}}(E) \mathbb \, \big [ \mathcal N_N^{\omega }(\mathrm {d} E) - {\mathcal {N}}_{\infty }(\mathrm {d} E) \big ] \, \right| \ . \end{aligned}$$
(A.28)
For the term in line (A.27) we get \(\mathbb {P}\)-almost surely, using that \(\mathbb {P}\)-almost surely \(\mu _N^{\omega }\) converges to 0 and (2.8),
$$\begin{aligned}&\lim \limits _{N \rightarrow \infty } \left| \, \int \limits _{\mathbb {R}} g_{\epsilon }^{(E_2)}(E)\dfrac{\mathrm {e}^{\beta E } - \mathrm {e}^{\beta ( E - \mu _N^{\omega } ) }}{\left( \mathrm {e}^{\beta ( E - \mu _N^{\omega } ) } - 1 \right) \cdot \Big ( \mathrm {e}^{\beta E } - 1 \Big )} \, {\mathcal {N}}_N^{\omega }(\mathrm {d} E) \, \right| = 0 \ . \end{aligned}$$
The term in line (A.28) \(\mathbb {P}\)-almost surely converges to zero for \(N \rightarrow \infty \) by vague convergence.
Next, we obtain, for all \(E_2 > \epsilon \) and all \(N \in \mathbb {N}\),
$$\begin{aligned} \int \limits _{(\epsilon ,\infty )}&{\mathcal {B}}(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) \\ = \,&\int \limits _{(\epsilon ,E_2]} {\mathcal {B}}(E-\mu _N^{\omega })\, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) + \int \limits _{(E_2,\infty )} {\mathcal {B}}(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) \ . \end{aligned}$$
On the one hand, by (A.24) we have \(\mathbb {P}\)-almost surely,
$$\begin{aligned} \limsup \limits _{N \rightarrow \infty } \int \limits _{(\epsilon ,E_2]} \mathcal B(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E)&\le \lim \limits _{N \rightarrow \infty } \int \limits _{\mathbb {R}} g_{\epsilon }^{(E_2)}(E) {\mathcal {B}}(E-\mu _N^{\omega }) \, \mathcal N_N^{\omega } (\mathrm {d} E) \\&= \int \limits _{\mathbb {R}} g_{\epsilon }^{(E_2)}(E) {\mathcal {B}}(E) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) \ . \end{aligned}$$
As for the second integral, we obtain \(\mathbb {P}\)-almost surely,
$$\begin{aligned} \begin{aligned} \limsup \limits _{N \rightarrow \infty } \int \limits _{(E_2,\infty )} \mathcal B(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E)&\le \limsup \limits _{N \rightarrow \infty } \int \limits _{[E_2,\infty )} \mathcal B(E-\epsilon /2) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) \\&\le \beta \pi ^{-1} \int \limits _{E_2}^{\infty } E^{1/2} \big [\mathcal B(E-\epsilon /2)\big ]^{2} \mathrm {e}^{\beta (E - \epsilon /2)} \, \mathrm {d} E \end{aligned} \end{aligned}$$
where the last step is as in the proof of Lemma A.5 (with \(\mu = \epsilon /2\)).
We conclude that
$$\begin{aligned} \limsup \limits _{N \rightarrow \infty }&\int \limits _{(\epsilon ,\infty )}{\mathcal {B}}(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) \\&\le \int \limits _{\mathbb {R}} g_{\epsilon }^{(E_2)}(E) {\mathcal {B}}(E) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) + \beta \pi ^{-1/2}\int \limits _{E_2}^{\infty } E^{1/2} \big [\mathcal B(E-\epsilon /2)\big ]^2 \mathrm {e}^{\beta (E - \epsilon /2)} \, \mathrm {d} E \end{aligned}$$
for all \(E_2 > \epsilon \) and hence
$$\begin{aligned} \limsup \limits _{N \rightarrow \infty }&\int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) \\ \le \,&\lim \limits _{E_2 \rightarrow \infty } \int \limits _{\mathbb {R}} g_{\epsilon }^{(E_2)}(E) {\mathcal {B}}(E) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) \\&+ \, \lim \limits _{E_2 \rightarrow \infty } \beta \pi ^{-1/2}\int \limits _{E_2}^{\infty } E^{1/2} \big [\mathcal B(E-\epsilon /2)\big ]^2 \mathrm {e}^{\beta (E - \epsilon /2)} \, \mathrm {d} E \\ \le \,&\lim \limits _{E_2 \rightarrow \infty } \int \limits _{[\epsilon /2,2E_2]} {\mathcal {B}}(E) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) \\ \le \,&\int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E) \, \mathcal N_{\infty } (\mathrm {d} E) + \dfrac{2}{\beta \epsilon } \mathcal N_{\infty }^{\mathrm {I}}(\epsilon ) \ . \end{aligned}$$
On the other hand, for all \(E_2 > \epsilon \) and all \(N \in \mathbb {N}\),
$$\begin{aligned} \int \limits _{(\epsilon ,\infty )}&{\mathcal {B}}(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) \\ \ge \,&\int \limits _{\mathbb {R}} g_{\epsilon }^{(E_2)}(E) \mathcal B(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) - \int \limits _{[\epsilon /2,\epsilon ]} g_{\epsilon }^{(E_2)}(E) \mathcal B(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) \ . \end{aligned}$$
For the first integral, by (A.24), \(\mathbb {P}\)-almost surely
$$\begin{aligned} \lim \limits _{N \rightarrow \infty } \int \limits _{\mathbb {R}} g_{\epsilon }^{(E_2)}(E) {\mathcal {B}}(E-\mu _N^{\omega }) \, \mathcal N_N^{\omega } (\mathrm {d} E)&= \int \limits _{\mathbb {R}} g_{\epsilon }^{(E_2)}(E) {\mathcal {B}}(E) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) \\&\ge \int \limits _{(\epsilon ,E_2]} {\mathcal {B}}(E) \, \mathcal N_{\infty } (\mathrm {d} E) \ . \end{aligned}$$
Since \(\mu _N^{\omega }\) converges \(\mathbb {P}\)-almost surely to zero and \(\, {\mathcal {N}}_N^{\omega }\) converges \(\mathbb {P}\)-almost surely vaguely to \({\mathcal {N}}_{\infty }\), the second integral \(\mathbb {P}\)-almost surely converges to
$$\begin{aligned} \limsup \limits _{N \rightarrow \infty }&\int \limits _{[\epsilon /2,\epsilon ]} g_{\epsilon }^{(E_2)}(E) {\mathcal {B}}(E-\mu _N^{\omega }) \, \mathcal N_N^{\omega } (\mathrm {d} E) \\ \le \,&\limsup \limits _{N \rightarrow \infty } \int \limits _{[\epsilon /2,\epsilon ]} g_{\epsilon }^{(E_2)}(E) \mathcal B(E-\epsilon /4) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) \\ \le \,&{\mathcal {B}}(\epsilon /4) \int \limits _{\mathbb {R}} g_{\epsilon }^{(\epsilon )}(E) \, {\mathcal {N}}_{\infty } \, (\mathrm {d} E) \\ \le \,&\dfrac{4}{\beta \epsilon } \mathcal N_{\infty }^{\mathrm {I}}(2 \epsilon ) \ . \end{aligned}$$
We conclude that, \(\mathbb {P}\)-almost surely,
$$\begin{aligned} \liminf \limits _{N \rightarrow \infty } \int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) \ge \int \limits _{(\epsilon ,E_2]} {\mathcal {B}}(E) \, \mathcal N_{\infty } (\mathrm {d} E) - \dfrac{4}{\beta \epsilon } \mathcal N_{\infty }^{\mathrm {I}}(2 \epsilon ) \end{aligned}$$
for all \(E_2 > \epsilon \) and hence
$$\begin{aligned} \liminf \limits _{N \rightarrow \infty } \int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E-\mu _N^{\omega })\, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) \ge \int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E) \, \mathcal N_{\infty } (\mathrm {d} E) - \dfrac{4}{\beta \epsilon } \mathcal N_{\infty }^{\mathrm {I}}(2 \epsilon ) \ . \end{aligned}$$
Finally, with Theorem 2.4, we obtain
$$\begin{aligned} \lim \limits _{\epsilon \searrow 0} \dfrac{4}{\beta \epsilon } \mathcal N_{\infty }^{\mathrm {I}}(2 \epsilon ) \le \lim \limits _{\epsilon \searrow 0} \dfrac{4}{\beta \epsilon } {{\widetilde{M}}} \dfrac{4! (2 \epsilon )^2}{(\pi \nu )^4} = 0 \ , \end{aligned}$$
which, taking Lemma A.4 into account, shows that \(\mathbb {P}\)-almost surely
$$\begin{aligned} \begin{aligned} \lim \limits _{\epsilon \searrow 0} \limsup \limits _{N \rightarrow \infty } \int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E)&= \lim \limits _{\epsilon \searrow 0} \liminf \limits _{N \rightarrow \infty } \int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E-\mu _N^{\omega }) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) \\&= \lim \limits _{\epsilon \searrow 0} \int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E) \, {\mathcal {N}}_{\infty } (\mathrm {d} E) \\&= \rho _c(\beta ) \ . \end{aligned} \end{aligned}$$
\(\square \)
Finally, we present a proof of generalized BEC and we follow in parts the proof in [12].
Proof of Theorem 2.7
Assume firstly that \(\rho \ge \rho _c(\beta )\): According to Lemma A.6, the sequence \((\mu ^{\omega }_{N})_{N=1}^{\infty }\) converges to 0 \(\mathbb {P}\)-almost surely. Also, for all \(\epsilon > 0\), all \(N \in \mathbb {N}\) and \(\mathbb {P}\)-almost all \(\omega \in \Omega \) one has
$$\begin{aligned} \rho = \int \limits _{(0,\epsilon ]} {\mathcal {B}}(E-\mu _N^{\omega }) \,{\mathcal {N}}_N^{\omega } (\mathrm {d} E) + \int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E-\mu _N^{\omega }) \,{\mathcal {N}}_N^{\omega } (\mathrm {d} E) \end{aligned}$$
and hence
$$\begin{aligned} \begin{aligned} \rho _0 (\beta )&= \lim \limits _{\epsilon \searrow 0} \liminf \limits _{N \rightarrow \infty } \int \limits _{(0,\epsilon ]} \mathcal B(E-\mu _N^{\omega }) \,{\mathcal {N}}_N^{\omega } (\mathrm {d} E) \\&= \rho - \lim \limits _{\epsilon \searrow 0} \limsup \limits _{N \rightarrow \infty } \int \limits _{(\epsilon ,\infty )} {\mathcal {B}}(E-\mu _N^{\omega }) \,{\mathcal {N}}_N^{\omega } (\mathrm {d} E) \ . \end{aligned} \end{aligned}$$
(A.29)
Consequently, by (A.23),
$$\begin{aligned} \rho _0(\beta ) = \rho - \rho _c(\beta ) \end{aligned}$$
holds \(\mathbb {P}\)-almost surely, implying the statement.
In a next step assume that \(\rho < \rho _c(\beta )\): According to Lemma A.6, the sequence \((\mu ^{\omega }_{N})_{N=1}^{\infty }\)\(\mathbb {P}\)-almost surely converges to a limit \({{\widehat{\mu }}} < 0\). Consequently, \(\mathbb {P}\)-almost surely there exists a \(\delta > 0\) such that \(- \mu _N^{\omega } > \delta \) for all but finitely many \(N \in \mathbb {N}\). Hence, \(\mathbb {P}\)-almost surely, with \(g_{\epsilon , \epsilon }(E)\) defined as in (A.12),
$$\begin{aligned} \lim \limits _{\epsilon \searrow 0} \lim \limits _{N \rightarrow \infty } \dfrac{1}{N} \sum \limits _{j : E_N^{j,\omega } \le \epsilon } n_N^{j,\omega }&= \rho ^{-1} \lim \limits _{\epsilon \searrow 0} \lim \limits _{N \rightarrow \infty } \int \limits _{(0,\epsilon ]} \mathcal B(E-\mu _N^{\omega }) \,{\mathcal {N}}_N^{\omega } (\mathrm {d} E) \\&\le \rho ^{-1} \lim \limits _{\epsilon \searrow 0} \lim \limits _{N \rightarrow \infty } \int \limits _{(0,\epsilon ]} {\mathcal {B}}(E+\delta ) \,\mathcal N_N^{\omega } (\mathrm {d} E) \\&\le \rho ^{-1} {\mathcal {B}}(\delta ) \lim \limits _{\epsilon \searrow 0} \lim \limits _{N \rightarrow \infty } \int \limits _{\mathbb {R}} g_{\epsilon ,\epsilon }(E) \, {\mathcal {N}}_N^{\omega } (\mathrm {d} E) \\&= \rho ^{-1} {\mathcal {B}}(\delta )\lim \limits _{\epsilon \searrow 0} \int \limits _{\mathbb {R}} g_{\epsilon ,\epsilon }(E) \, \mathcal N_{\infty } (\mathrm {d} E) \\&\le \rho ^{-1} {\mathcal {B}}(\delta ) \lim \limits _{\epsilon \searrow 0} {\mathcal {N}}_{\infty }^{\mathrm {I}} (2 \epsilon ) \\&= 0 \end{aligned}$$
employing \(\mathbb {P}\)-almost sure vague convergence of \(\mathcal N_N^{\omega }\) to \({\mathcal {N}}_{\infty }\) and Theorem 2.4. \(\square \)
