Suppose $(G_n)$ is a sequence of simple directed graphs with $G_n$ having $n$ edges such that the adjacency matrix $A_n$ of $G_n$ is primitive, and let $(G_n’)$ be a sequence of subgraphs obtained by deleting a single vertex from $G_n$. Suppose also that the degree of any vertex is bounded by some constant $d$ (Not depending on $n$). If we let $\rho(G)$ denote the spectral radius of a graph $G$, does it follow that $\rho(G_n) - \rho(G_n’) \rightarrow 0$ as $n \rightarrow \infty$? Does it follow from some mild assumptions on how "mixing" the graphs are?
When the graphs $G_n$ are undirected, I was able to resolve this question in the positive under the additional assumption that the coordinates of each associated unit eigenvector decayed to zero, i.e., if $x_n$ is the associated unit eigenvector to $\rho(G_n)$, then for all $\epsilon > 0$ there exists some $N$ such that $x_n^i < \epsilon$ for all $n \ge N$ and all $1 \le i \le n$. This follows from Theorem 1 of the following paper: https://www.nas.ewi.tudelft.nl/people/Piet/papers/LAA_2012_MaxEigenvectorComponent.pdf, which gives
$$ \left(1-2(x_n^i)^2\right)\rho(G_n) \le \rho(G_n').$$ However, the proof of this statement uses the inequality $ x^T A_n x \le \rho(G_n)$, which is certainly not true in general when $A_n$ is not symmetric, i.e., when $G_n$ is directed.
It is straightforward that bounds on the degree are necessary: otherwise, we could for example take $G_n$ to have vertices $\{1,\ldots,n\}$ and draw a directed edge from $u$ to $v$ if exactly one of $u,v$ is equal to $1$. I also know that $\rho(G_n') < \rho(G_n)$ by the fact that $A_n$ is primitive. Unfortunately, I know very little about the techniques and strategies used in graph theory (this question came up when I was trying to prove a statement regarding certain family of dynamical systems), so I am sort of stuck.