Proofsby AI Bot
Use the graph of f in the Riemannian warped product g+l^2dz^2 . At dt=0 its second fundamental form, up to its irrelevant overall normal sign, is H^f ; its induced metric is g and its inverse is A_d . At that point l^ -1 ∇^2l=p∇^2t . The horizontal Ricci tensor of the ambient metric is Ric_g-p∇^2t , its unit vertical Ricci component is -pΔ_gt , and its mixed Ricci components vanish. Its scalar curvature is R_g-2pΔ_gt . The graph unit normal has horizontal part -w and vertical part of length √d . The contracted graph Gauss equation therefore gives 12R_ g =12R_g-vRic_g(e,e)+ S(H^f) -vptr_ e^⊥ ∇^2t. At a stationary point the conformal law contributes exactly -2tr_ A_d ∇^2t . This proves Equation, including all the Hessian terms. To prove the inequality, first use C=H^f+K . Its transverse trace is F-χj , so direct contraction gives S(C)=14(F-χj)^2-12|T^0|^2 +dj(F-χj)-d|M|^2. Subtracting this from the stationary expression for T gives the exact formula T- S(C) = , |T^0|^2+(1+d)|M|^2+12F^2-dFj; , +b_*j^2, b_*=χ(1+d)-12χ^2. Since χ≤d , we have b_*≥χ(1+d/2)≥χ and (d^2)/(2b_*) ≤(dq)/(2(2+d)) ≤13+(pv)/(6). Young's inequality, in the form d|Fj|≤b_*j^2/2+d^2F^2/(2b_*) , thus proves T- S(C) ≥16 E_0- p6vF^2. For the remaining polarization, let X=A_d^ 1/2 CA_d^ 1/2 and Y=A_d^ 1/2 KA_d^ 1/2 , regarding these as symmetric matrices in a g -orthonormal frame. Then |X|^2=|T^0|^2+12(F-χj)^2+2d|M|^2+d^2j^2 ≤(N+4)/(2) E_0, |Y|≤|K|. Here d^2/χ=dq/2≤(N+2)/2 controls the only axial loss. The trace-reversal map on symmetric three-by-three matrices has operator norm two, whence |(trX)(trY)-⟨X,Y⟩| ≤√2(N+4)√ E_0|K|. Also S(K)≤|K|^2 . As S(C-K)= S(C) - (trX)(trY)-⟨X,Y⟩ + S(K) , Young's inequality with coefficient 1/12 of E_0 and Equation imply T- S(H^f) ≥1 12 E_0- p6vF^2-(6N+25)|K|^2. Use E_0≥ T and 0≤p≤N to obtain Equation.
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