In a random graph $G(n,p)$, with Laplacian matrix $L = A - D$, for adjacency matrix $A$ and degree matrix $D$.
Is there a way to lower-bound the value of the biggest eigenvalue for the Laplacian matrix? Something that looks like the following: $$ Prob(\lambda_\max(L) > \lambda_0) \geq \delta $$
Note: I know of "Convergence in Probability" results. For example, Ding and Jian, 2010 (Theorem 1) have: $$ \frac{ \lambda_\max(L) }{\sqrt{n \log n}} \overset{P} \rightarrow \sqrt{2} $$ which I don't think it gives what I want.
Why wouldn't it give you this though, I concluded the opposite:
That $\left[ \frac{\lambda_{max}(L_n)}{\sqrt{n \log n}} \right]$ converges in probability to $\sqrt{2}$ (as $n$ goes to infinity) implies that for any $\epsilon, \tau > 0$ there exists an $n$ such that
$\mathbb{P}\left[ \frac{\lambda_{max}(L_n)}{\sqrt{n \log n}} \right] \geq \sqrt{2} - \epsilon]$ $\geq 1-\tau$ and
$\mathbb{P}\left[ \frac{\lambda_{max}(L_n)}{\sqrt{n \log n}} \right] \geq \sqrt{2} + \epsilon]$ $\leq \tau$.
So for any fixed $\epsilon >0$ and $\lambda_0 \leq \sqrt{(2-\epsilon) n \log n}$ there is a $\delta >0$ such that the inequality
$\mathbb{P}\left[{\lambda_{max}(L_n)} \geq {\lambda_0} \right]$ $\geq \delta$
holds; in fact $\delta$ can be close to 1. And for $\lambda_0 \geq \sqrt{(2+\epsilon) n \log n}$ there exists no such $\delta >0$ such that the above ineqaulity holds.