Spectral density and return probability Consider a random graph $G$, and I define the Laplacian matrix $L$ of this graph.
After choosing an origin point 'O' from where the walker starts at time '0', one calculates the probability $P_{0}(t)$ of the returning to the origin as
\begin{equation}
\label{return-probability1}
P_{0}(t)= \int_{0}^{\infty}  e^{-\lambda t}\rho(\lambda) \mathrm{d} \lambda,
\end{equation}
where, $\lambda$ and $\rho(\lambda)$ are the eigenvalues and eigenvalues density of $L$.
There is a statement which says: at very large times 't', one can notice that the spectral dimension has an exponential falloff determined by the lowest eigenvalues.
Or equivalently, the behavior of $P_{0}(t)$ is dominated by the behavior of eigenvalues density $\rho(\lambda)$ when $\lambda$ goes to zero.
How can you explain this? Under which condition is this true? 
 A: Using our friend the dominated convergence theorem, we can make a nice statement about $P_0(t)$ for large t. Asymptotically, we can say that for large t
$$P_0(t) = a_0 + \frac{a_1}{t} + \frac{a_2}{t^2} + \cdots$$
Now let's find the first two terms of the expansion. Assuming $\rho(\lambda)\leq M$ for some constant:
$$a_0 = \lim_{t\to\infty} P_0(t) = \lim_{t\to\infty}\int_0^\infty e^{-\lambda t}\rho(\lambda) d\lambda = 0$$
$$a_1 = \lim_{t\to\infty}t(P_0(t)-P_0(\infty)) = \lim_{t\to\infty}\int_0^{\infty}e^{-\lambda t}\rho(\lambda)td\lambda$$ $$= \lim_{t\to\infty}\int_0^{\infty}e^{-\gamma }\rho(\frac{\gamma}{t})d\gamma = \int_0^\infty e^{-\gamma}\rho(0)d\gamma = \rho(0)$$
Thus we have for $t\gg1$
$$P_0(t) = \frac{\rho(0)}{t} + O\left(\frac{1}{t^2}\right)$$
So for large t, the behavior of $P_0(t)$ is dominated by the density at $0$.
A: Setup: As $\lambda$ are the eigenvalues of a Laplacian matrix generated from a random graph, they are real valued and non-negative. Hence, $p(\lambda)$ has a support in $[0, \infty)$ or $(0, \infty).$ Furthermore, let $p(\lambda)$ be such that $P_0(t)$ exists.
Return to origin: We assume that we can separate, for simplicity, the (possibly mixed) density $p(\lambda)$ into $N_C$ number of mass points at $l_1 < l_2 < \dots < l_{N_C}$ with probabilities $\mu_1, \mu_2, \dots, \mu_{N_C}$ and a continuous part $p_c(\lambda)$ with compact support $(a,b).$ By approximating the integral of the continuous part by Riemann–Stieltjes sum with bin centres denoted by $a < \lambda_1 < \lambda_2 < \dots < \lambda_N < b$ we have:
\begin{align}
P_0(t)  &=  \sum_{m = 1}^{N_C} \exp(-l_m t) \mu_m + \int_{a}^{b} \exp(-\lambda t) p_c(\lambda) \text{d}\lambda \\\\
&\approx \sum_{m = 1}^{N_C} \exp(-l_m t) \mu_m + \sum_{i = 1}^{N} \exp(-\lambda_i t) p_c(\lambda_i) \Delta_\lambda.
\end{align}
Some notes and conditions:


*

*If the graph is trivial with no edges, then $\boldsymbol{L} = \boldsymbol{0}$ and $p(\lambda)$ is a mass point with probability one at $l_1 = 0.$ Hence, $\lim_{t \to \infty} P_0(t) = 1.$

*A return to origin is only possible for $t \to \infty$ with non-zero probability if a mass point exists at $l_1 = 0.$ Then $\lim_{t \to \infty} P_0(t) = \mu_1.$

*For large enough $t,$ $P_0(t)$ is dominated by $\exp(-l_1 t) \mu_1$ or $\exp(-\lambda_1 t) p_c(\lambda_1),$ i.e., typically the density around the smallest eigenvalue. 

*If the density $p(\lambda)$ is bounded away from zero, then $P_0(t)$ is not dependent on $p(\lambda)$ for $\lambda \to 0.$

Spectral dimension: when a mass point is present at $\lambda = 0$ it is clear that the spectral dimension is zero. In the remainder, for simplicity, we assume that the density $p(\lambda)$ is continuous with no mass points. Furthermore, we assume that $p(\lambda)$ has support and is analytic around $\lambda = 0.$
Substituting the Taylor series for $p(\lambda)$ around zero in the expression for $P_0(t)$ we have
\begin{align}
P_0(t) &= \int_{0}^{\infty} \exp(-\lambda t) \sum_{n = 0}^{\infty} \frac{p^{(n)}(0)}{n!} \lambda^n \text{d}\lambda \\\\
&= \sum_{n = 0}^{\infty} \frac{p^{(n)}(0)}{n!} \int_{0}^{\infty} \exp(-\lambda t) \lambda^n \text{d}\lambda,
\end{align}
where $p^{(n)}(0)$ denotes the $n$-th derivative evaluated at $\lambda = 0,$ and $p^{(0)}(0) = p(0)$ by convention.
It can be easily verified that
\begin{align}
\int_{0}^{\infty} \exp(-\lambda t) \lambda^n \text{d}\lambda = O\left(\frac{1}{t^{(1+n)}}\right),
\tag{1}
\label{eqn:O1}
\end{align}
hence
\begin{align}
P_0(t) = O\left(\frac{1}{t^{(1 + n_0)}}\right),
\end{align}
where $n_0$ is the smallest $n$ such that $p^{(n)}(0)$ is non-zero. Therefore, it follows that
\begin{align}
-\lim_{t \to \infty} \frac{\log P_0(t)}{\log t} = 1 + \lim_{\lambda \to 0} \frac{\log p(\lambda)}{\log \lambda} = 1 + n_0,
\end{align}
obtaining the result shown in the comments above. For non-analytic functions, an analogous result based on Laurent or other series definitions can be shown. Notably, (\ref{eqn:O1}) also holds for any real-valued, non-negative $n.$
