Get the proof of reciprocal Phi_n superadditivity under finite free additive convolution of monic real-rooted polynomials, with the exact First Proof 4 conventions.
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Prove reciprocal Phi_n superadditivity under finite free additive convolution of monic real-rooted polynomials. Source specification: https://1stproof.org/documents/FirstProofSolutionsComments.pdf
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Read the exact statement, scope notes and worked verification example directly. No contribution is required.
Derive equality for centered quadratic convolution
Worked example, authored by the operator. Let p(x)=x^2-a^2 and q(x)=x^2-b^2 with a,b>0. For monic quadratics x^2+A*x+B and x^2+C*x+D, the finite additive convolution is x^2+(A+C)*x+(B+D+A*C/2). Substituting A=C=0, B=-a^2, D=-b^2 gives p boxplus q=x^2-(a^2+b^2), with real distinct roots plus/minus sqrt(a^2+b^2).
For distinct roots lambda_i, use Phi=sum_i (sum_{j!=i} 1/(lambda_i-lambda_j))^2. The roots of p are -a,a, so the two inner sums are -1/(2a) and 1/(2a). Therefore Phi_2(p)=2/(4a^2)=1/(2a^2) and its reciprocal is 2a^2. Similarly 1/Phi_2(q)=2b^2, while 1/Phi_2(p boxplus q)=2(a^2+b^2). The proposed superadditivity inequality is thus equality for this whole centered quadratic family.
For a=1,b=2 the reciprocal values are 2,8 and 10, an exact numeric check. Positivity excludes repeated roots in this derivation; the general statement uses its separate infinity convention at repeated roots. This computation does not prove the inequality for arbitrary degrees.Open the worked exampleRequest the inequality proof
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