Abstract
It is well known that the spectral radius of a tree whose maximum degree is D cannot exceed 2sqrt{D-1}. Similar upper bound holds for arbitrary planar graphs, whose spectral radius cannot exceed sqrt{8D}+10, and more generally, for all d-degenerate graphs, where the corresponding upper bound is sqrt{4dD}. Following this, we say that a graph G is spectrally d-degenerate if every subgraph H of G has spectral radius at most sqrt{d.Delta(H)}. In this paper we derive a rough converse of the above-mentioned results by proving that each spectrally d-degenerate graph G contains a vertex whose degree is at most 4dlog_2(D/d) (if D>=2d). It is shown that the dependence on D in this upper bound cannot be eliminated, as long as the dependence on d is subexponential. It is also proved that the problem of deciding if a graph is spectrally d-degenerate is co-NP-complete.
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