Extend the spectral-radius ordering theorem to lower values of alpha

Determine whether a new method can establish that, for two strongly connected digraphs G_1 and G_2 with n≥4 vertices and m arcs, the condition Δ⁺(G_1)≥2α(1−α)(m−n+1)+2α together with Δ⁺(G_1)>Δ⁺(G_2) implies λ_α(G_1)>λ_α(G_2) for α∈[1/2,1/√2).

Background

Theorem 1.1 proves an ordering result for the A_α spectral radii of two strongly connected digraphs with the same number of vertices and arcs. Under a lower bound on the maximum outdegree of G_1 and the strict inequality Δ⁺(G_1)>Δ⁺(G_2), the theorem establishes λα(G_1)>λα(G_2), but only for α∈[1/√2,1).

The authors explain that their proof does not extend directly to α∈[1/2,1/√2), because one estimate in Case 1 yields the lower bound α−1/(2α), which is negative in that interval. They therefore leave unresolved whether another method can prove the same spectral-radius ordering conclusion throughout the remaining parameter range.

References

For \alpha\in [\frac{1}{2},\frac{1}{\sqrt{2})$, the method used in the proof of Theorem 1.1 may not be true. Because in Case 1, we have $\lambda_{\alpha}(G_1)-\lambda_{\alpha}(G_2)\ge \alpha-\frac{1}{2\alpha}$ and $\alpha-\frac{1}{2\alpha}<0$ under this condition. Thus, we may not be able to get $\lambda_{\alpha}(G_1)>\lambda_{\alpha}(G_2)$. Therefore, we pose a question: whether is there a new method that can lead to the conclusion holding at $\alpha\in [\frac{1}{2},\frac{1}{\sqrt{2})$?

— Ordering digraphs with maximum outdegrees by their $A_α$ spectral radius  (2501.12412 - Xu et al., 18 Jan 2025) in Remark following the proof of Theorem 1.1, Section 3