Prove
Lemma 3 ([Exterior coordinate and a univalent collar]). Let Ω⊂ℂ be a bounded…Theorem 2.1 ([Exterior Faber coefficients and the positive kernel]). Let Ω be a…Theorem 3.1 ([The resolvent expansion and double-layer positivity]). Let Ω,h,q…Proposition
Choose h from Lemma 3 ([Exterior coordinate and a univalent collar]) and apply Theorem 2.1 ([Exterior Faber coefficients and the positive kernel]) to this coordinate. Put γ=\|F[A]\| . There is nothing to prove if γ=0 . Otherwise choose unit singular vectors x,y∈^n⊗^m such that F[A]x=γy, F[A]^*y=γx. Write x=∑_ i=1 ^mXe_i⊗e_i and y=∑_ i=1 ^mYe_i⊗e_i , where X,Y are n -by- m matrices and (e_i) is the standard basis of ^m . For the resolvent in Theorem 3.1 ([The resolvent expansion and double-layer positivity]), define matrix functions P=X^*RX, Q=Y^*RY, M=Y^*RX, L=X^*RY. Subscripts k≥0 denote the coefficients of λ^ -k , so M_k=Y^*b_k(A)X , for example. By Theorem 3.1 ([The resolvent expansion and double-layer positivity]), all four series converge absolutely and uniformly on , and their coefficient sequences are square summable. Since b_0=1 , P_0=X^*X=P_0^*, Q_0=Y^*Y=Q_0^*, L_0^*=M_0. Set a^2=∑_ k≥0 M_k^2 . We will bound the squared boundary L^2 Hilbert--Schmidt norms of P+P^* and Q+Q^* above by 4a^2/γ^2 each, and that of M+L^* below by a^2 . Positivity will then compare these quantities and force γ≤2 . The off-diagonal coefficient functional. For a polynomial G=∑_k b_kC_k , direct expansion in the tensor basis gives y^*G[A]x=∑_ k,i,j (M_k)_ ij (C_k)_ ij . Indeed, C_ke_j=∑_i(C_k)_ ij e_i in the coefficient factor, leaving (Ye_i)^*b_k(A)Xe_j in the base factor. Equation is the bilinear entry pairing of the two coefficient arrays. Cauchy--Schwarz gives |y^*G[A]x|≤a(∑_kC_k^2)^ 1/2 . Taking G=F and using shows that a>0 . The two ordered tests. Polynomial evaluation respects multiplication in its given order. Explicitly, if F(z)=∑_iD_iz^i and G(z)=∑_jE_jz^j , then (FG)[A]=∑_ i,j A^ i+j ⊗D_iE_j=F[A]G[A], and the reversed product has coefficients E_jD_i . Hence y^*(FG)[A]x=γx^*G[A]x, y^*(GF)[A]x=γy^*G[A]y. Write the full finite Faber expansions as FG=∑_k b_kU_k, GF=∑_k b_kV_k, \|H\|_∂^2=∫_(h(μ))^2\,dτ(μ). The functional in uses all M_k , so it applies to these full expansions for every finite degree. Apply it to the two identities in. The lower coefficient estimate in Proposition 1 ([The coefficient norm estimate]) and the two boundary multiplier bounds give γ|x^*G[A]x| , ≤a(∑_kU_k^2)^ 1/2 ≤a\|FG\|_∂≤a\|G\|_∂,; γ|y^*G[A]y| , ≤a(∑_kV_k^2)^ 1/2 ≤a\|GF\|_∂≤a\|G\|_∂. Here F(z)G(z)≤\|F(z)\|G(z) and G(z)F(z)(z)\|F(z)\| are used in their respective orders. The upper coefficient estimate for G now gives |x^*G[A]x|,|y^*G[A]y| ≤ aγ (C_0^2+2∑_ k≥1 C_k^2)^ 1/2 . The entries paired with C_k in these two functionals are respectively those of P_k and Q_k , by the calculation in. The weighted coefficient norm on the right gives reciprocal weights on these two coefficient sequences: P_0^2+12∑_ k≥1 P_k^2≤a^2/γ^2, Q_0^2+12∑_ k≥1 Q_k^2≤a^2/γ^2. For the first inequality, set d_N=P_0^2+12∑_ k=1 ^NP_k^2 and test with C_0=P_0 , C_k=P_k/2 for 1≤k≤N , and C_k=0 otherwise; bars mean entrywise conjugation. The paired value is d_N , and the squared weighted test norm is also d_N . Hence d_N≤(a/γ)√d_N , which gives d_N≤a^2/γ^2 including when d_N=0 . Letting N→∞ proves the first bound. The same finite tests with Q_k prove the second. The boundary norms of the three blocks. The compression of R+R^* by [X Y] has diagonal blocks P+P^* , Q+Q^* and lower off-diagonal block M+L^* . We first compute their boundary norms. In the diagonal fields, makes the constant term twice P_0 or Q_0 . For k≥1 , the coefficients P_k and P_k^* occur at frequencies -k and k and have equal Hilbert--Schmidt norms; the same holds for Q_k and Q_k^* . Parseval therefore gives four times the reciprocal weighted sum in : ∫_+P^*^2dτ =4(P_0^2+12∑_ k≥1 P_k^2) ≤4a^2/γ^2, and the same identity and bound hold for Q+Q^* . For the cross field, L_0^*=M_0 makes the constant term 2M_0 . Its negative coefficients are M_k for k≥1 and its positive coefficients are L_k^* for k≥1 . Consequently ∫_+L^*^2dτ =4M_0^2+∑_ k≥1 (M_k^2+L_k^2) ≥a^2. The positive block comparison. Theorem 3.1 ([The resolvent expansion and double-layer positivity]) gives R+R^*0 . Its compression by the n -by- 2m matrix [X Y] is therefore Δ= P+P^*, L+M^*; M+L^*, Q+Q^* 0. For J=(I_m,-I_m) , the matrix JΔJ is also positive, so 0≤(Δ^ 1/2 (JΔJ)Δ^ 1/2 ) =(ΔJΔJ) =P+P^*^2+Q+Q^*^2-2M+L^*^2. Integrating this inequality and using and yields 2a^2≤2∫_+L^*^2dτ ≤∫_(P+P^*^2+Q+Q^*^2)dτ ≤8a^2/γ^2. Since a>0 , this proves γ≤2 .