I am now stopped by the following algebra.
Let $n \rightarrow \infty \space$, and $\space m, p \rightarrow \infty $ as $n \rightarrow \infty$.
More specifically, $\space m\leq (\frac{1}{6\times \log(\rho^{-1})}(1-14a)-\delta)\log(n),\space $for some $\rho \in (0,1)$,$\space$ $a\in(0, 1/14)$, and $\space \delta>0, \space$ so that $m=O(\log(n)).\space $ And, $\space p=O(n^a)$.
Also, let $\lambda_m$ be such that $\lambda_m\geq C_{\lambda}\times\rho^m$ for some $C_\lambda > 0$ and $\lambda_m \rightarrow 0$ as $m \rightarrow \infty$ (as $n \rightarrow \infty)$.
Then, how can I show that $p\lambda_m^2 = O(m^2)$?
The paper says that a straightforward calculation yields the result, but it seems not to be the case.