Proof. Write \(\,\mathrm dq=a\,\mathrm dx\), with \(a\) bounded. For \(f\in L^2(q)\), the source \(af\) lies in \(L^2(\,\mathrm dx)\), since \(\int|af|^2\le\lVert a\rVert_\infty\int|f|^2\,\mathrm dq\). Its Dirichlet solution \(\psi=Tf\) satisfies \[\int_\mathbb D|\nabla\psi|^2=\langle \psi,f\rangle_q.\] Combining this with (6) gives \[\lVert \psi\rVert_q^2\le d\lVert \psi\rVert_q\lVert f\rVert_q,\] hence the norm bound. Symmetry follows from the Green kernel, or by testing the two Poisson equations against each other. The Green formula can first be proved for smooth compactly supported sources and then passed to \(L^2(\,\mathrm dx)\) by approximation; \(G(x,\cdot)\in L^2(\,\mathrm dx)\) gives pointwise evaluation. These sections vary continuously in \(L^2\) on compact interior sets, so this is the continuous interior representative. For \(\rho\le q\), its density remains bounded and its weighted Dirichlet inequality follows from (6), proving the same result. ◻

Proof. The elementary estimates \[P_s(y)\le\frac C{|y-s|},\qquad 0\le G(x,y)\le C\log\frac2{|x-y|}\] give uniform bounds for \(P_s\) in \(L^{3/2}(\,\mathrm dx)\) and for \(G(x,\cdot)\) in \(L^3(\,\mathrm dx)\). Hölder and the bounded density imply \[\sup_{s\in\mathbb S^1}\lVert TP_s\rVert_{L^\infty(\mathbb D)}<\infty.\] The section \(P_s\) need not belong to \(L^2(q)\), but one Green integration gives \(F_s=TP_s\in L^\infty(\mathbb D)\subset L^2(q)\). We can now apply the \(L^2(q)\) contraction estimate to the remaining iterates. For \(n\ge2\), \[ |(T^nP_s)(p)| =|\langle G(p,\cdot),T^{n-2}F_s\rangle_q| \le\lVert G(p,\cdot)\rVert_q\,d^{n-2}\lVert F_s\rVert_q \le C d^{n-2}. \tag{16}\] The constants can be chosen uniformly in \(p\) and \(s\). This proves the assertion about \(K\).

For distinct \(s,t\), the product \(P_sP_t\) is integrable: near either boundary pole, the other factor is bounded, and \(|y-s|^{-1}\) is locally integrable in two dimensions. The terms with \(n\ge1\) in (11) are summable by the first smoothing estimate and (16), since \(\int P_s\,\mathrm dq\) is uniformly bounded. Reversing the nonnegative chains proves symmetry. Tonelli’s theorem and \(\int P_s\,\mathrm ds=1\) give \[\iint R(s,t)\,\mathrm ds\,\mathrm dt =\sum_{n\ge0}\langle \mathbf 1,T^n\mathbf 1\rangle_q \le\sum_{n\ge0}d^n\lVert \mathbf 1\rVert_q^2,\] which is (13). Diagonal values of \(R\) play no role in these boundary integrals. Expanding the Neumann series in (9) and using the positive kernels gives (14) and (15); the same bounds justify the exchanges for signed data by absolute domination. ◻

Proof. By 8 and conformal invariance of the scalar Dirichlet integral applied to each Cartesian component, the bulk terms of \(E\) are \[\int_\mathbb D\nabla W_h:\nabla W_k-\langle W_h,W_k\rangle_q.\] The equation for \(W_k\) permits replacing \(W_h\) here by \(Ah\), since their difference has zero trace. Harmonicity of \(Ah\) then replaces \(W_k\) by \(Ak\) in the gradient term. Since \(W_k=(I-T)^{-1}Ak\) in \(L^2(q)\), this gives \[\int_\mathbb D\nabla Ah:\nabla Ak-\langle Ah,(I-T)^{-1}Ak\rangle_q.\] The boundary curvature term changes by the factor \(|\Phi'|\), and (15) identifies the second term.

For the harmonic energy, \[ \int_\mathbb D\nabla Ah:\nabla Ak =\frac12\iint N_0(s,t)(h(s)-h(t))\cdot(k(s)-k(t))\,\mathrm ds\,\mathrm dt. \tag{19}\] Indeed both sides diagonalize in Fourier modes. The Hermitian energy of \(s^m\) is \(2\pi|m|\), while \[\iint\frac{|s^m-t^m|^2}{|s-t|^2}\,\mathrm ds\,\mathrm dt=(2\pi)^2|m|\] by a geometric sum, verifying the normalization. Smooth convolution approximation, with uniformly bounded Lipschitz constants and convergence in \(H^{1/2}\), extends the identity to the stated data. This proves (18). ◻

Proof. Since \(E\) is positive semidefinite and \(E(g)=0\), its bilinear form annihilates \(g\): this follows by considering \(E(g+th)\) for real \(t\). In particular, \(E(g,b^2g)=0\). Subtract (18) for this pair from the formula for \(E(bg,bg)\). The curvature terms cancel. The harmonic term gives \[\frac12N_0(s,t)(b(s)-b(t))^2g(s)\cdot g(t)\] in the double integral. Symmetry of \(R\) gives the same expression with \(R\) in place of \(N_0\). Their sum is (22). All terms are integrable: the squared Lipschitz difference cancels the singularity of \(N_0\), and \(R\) is integrable. ◻

Proof. Put \[h(x,y)=\frac{|x-y|^2}{(1-|x|^2)(1-|y|^2)},\qquad L(h)=\frac1{\log(1+1/h)}\ (h>0),\quad L(0)=0,\] and \(f(h)=h+\tfrac12-L(h)\). After multiplying the \(x\)- and \(y\)-entries by \(1-|x|^2\) and \(1-|y|^2\), respectively, the kernel \(h+\tfrac12\) becomes \[ \tfrac12(1+|x|^2)(1+|y|^2)-2x\cdot y. \tag{24}\] This is a positive rank-one kernel minus a positive semidefinite kernel, so \(h+\tfrac12\) has (P) on finite sets.

Since \(1/G=4\pi L(h)\), the identity \(L=(h+\tfrac12)-f\) will be useful once we prove that \(f(h(x,y))\) is positive semidefinite. For \(h>0\), \[ L(h)=\int_0^1h^{1-\alpha}(h+1)^\alpha\,\mathrm d\alpha. \tag{25}\] For \(0<\alpha<1\), let \(c_\alpha\) normalize \(v^{\alpha-1}(1+v)^{-1}\,\mathrm dv\) to a probability measure on \((0,\infty)\), and set \(r=(1+v)^{-1}\). Integration by parts gives \[\mathbb E_\alpha(1-r) =c_\alpha\int_0^\infty\frac{v^\alpha}{(1+v)^2}\,\mathrm dv =\alpha.\] For \(h>0\), scaling \(v\) in the normalized integral yields \[\begin{align*} h^{1-\alpha}(h+1)^\alpha &=c_\alpha\int_0^\infty \frac{h(h+1)}{(h+1)+vh}v^{\alpha-1}\,\mathrm dv\\ &=\mathbb E_\alpha\frac{h(h+1)}{h+r} =\mathbb E_\alpha\left[h+(1-r)-\frac{r(1-r)}{h+r}\right]. \tag{26}\end{align*}\] The last equality remains true at \(h=0\). Integrating in \(\alpha\) therefore gives \[ f(h)=\int_0^1\mathbb E_\alpha\frac{r(1-r)}{h+r}\,\mathrm d\alpha, \qquad 0\le f(h)\le f(0)=\tfrac12. \tag{27}\]

Now \[k(x,y)=\frac1{1+h(x,y)} =\frac{(1-|x|^2)(1-|y|^2)}{|1-x\overline y|^2}\] is positive semidefinite. Indeed \((1-x\overline y)^{-1}\) is a Gram kernel by its power series; multiplying it by its complex conjugate and the positive diagonal factors preserves positive semidefiniteness. For fixed \(0<r<1\), \[ \frac1{h+r}=\sum_{n\ge0}(1-r)^n k^{n+1}, \tag{28}\] where powers are entrywise. This series converges even when \(h=0\). Each power is positive semidefinite, as follows by taking tensor products of Gram vectors. Thus \(1/(h+r)\) is positive semidefinite, and the positive mixture (27) proves the claim. The mixture has finite entries by the displayed bound, including on the diagonal.

Define \[ \frac1{C_\epsilon(x,y)} =4\pi\bigl[h+\tfrac12-(1-\epsilon)f(h)\bigr] =4\pi\bigl[L(h)+\epsilon f(h)\bigr]. \tag{29}\] Its reciprocal matrix has (P) by (24) and the positive semidefiniteness just proved. Its denominator is at least \(2\pi\epsilon\), because \[L+\epsilon f=\epsilon(h+\tfrac12)+(1-\epsilon)L\ge\epsilon/2.\] It is positive and continuous, so \(C_\epsilon\) is bounded, continuous, and strictly positive. Since \(f\ge0\), these kernels increase to \(G\) by (7), and \(0<C_\epsilon\le G\). On the diagonal \(C_\epsilon(x,x)=1/(2\pi\epsilon)\), as required. ◻

Proof. Positive diagonal congruence transforms the reciprocal of \(M\) into \[H_{ij}=\frac{M_{iz}M_{zj}}{M_{ij}}.\] Writing \(a=M_{zz}>0\), its \(z\)-row and column are identically \(a\). Put \(z\) last, and let \(H_0\) denote the remaining principal block. Subtracting the last row from every other row, and doing the same to the columns, gives the congruence \[H\ \sim\ \begin{pmatrix}H_0-a\mathbf 1\mathbf 1^T&0\\0&a\end{pmatrix}.\] Because \(a>0\) and \(H\) has (P), the first block is nonpositive. Hence \[U=a\mathbf 1\mathbf 1^T-H\succeq 0.\] Since \(H_{ii}>0\), one has \(0\le U_{ii}<a\), and positive-semidefinite Cauchy–Schwarz gives \(|U_{ij}|<a\).

Using the same diagonal normalization, the new reciprocal has entries \(H_{ij}/(1+cH_{ij})\). The difference from the constant rank-one kernel is \[\begin{align*} \frac a{1+ca}-\frac{H_{ij}}{1+cH_{ij}} &=\frac{U_{ij}}{(1+ca)(1+ca-cU_{ij})}\\ &=\sum_{n\ge0}\frac{c^nU_{ij}^{n+1}}{(1+ca)^{n+2}}. \tag{30}\end{align*}\] The series converges absolutely because \(|cU_{ij}|<1+ca\). Its coefficients are nonnegative and every entrywise power of \(U\) is positive semidefinite, even if \(U\) has negative entries. Thus the normalized reciprocal of \(\widetilde M\) is a positive rank-one kernel minus a positive semidefinite matrix, proving (P). For \(c=0\) the identity reduces directly to \(U\). ◻

Proof. First fix \(\epsilon>0\) and replace \(G\) throughout (31) by \(C=C_\epsilon\). Its integral operator \(T_C\) on \(L^2(\rho)\) satisfies \[|T_Cf|\le T_\rho|f|,\qquad \lVert T_C\rVert\le d<1.\] The absolute pointwise domination proves the norm bound without requiring a separate positivity assertion about its quadratic form.

Partition a compact integration set into finitely many measurable cells with uniformly vanishing diameters. For each cell of positive \(\rho\)-mass, choose a representative, and let \(\pi\) send that cell to the representative. Use its mass as the quadrature weight. Uniform continuity gives uniform convergence of \[C(\pi(\cdot),\pi(\cdot))\longrightarrow C\] on the integration set up to null sets. The associated operators \(T_\pi\) on \(L^2(\rho)\) converge to \(T_C\) in operator norm; for example their norm difference is at most \(\rho(\mathbb D)\) times the uniform kernel error. Likewise \(C(x,\pi(\cdot))\to C(x,\cdot)\) in \(L^2(\rho)\) for each fixed evaluation point \(x\). In particular, sufficiently fine quadratures have \(\lVert T_\pi\rVert<1\), and \[ C(x,y)+\langle C(x,\pi(\cdot)),(I-T_\pi)^{-1}C(y,\pi(\cdot))\rangle_\rho \tag{32}\] converges to the regularized continuum resolvent evaluation. This follows from convergence of the source vectors and of the inverses in operator norm.

For clarity, if the nodes and weights are \(z_\alpha,w_\alpha\), set \[\mathsf T_{\alpha\beta} =\sqrt{w_\alpha}\,C(z_\alpha,z_\beta)\sqrt{w_\beta}, \qquad (\xi_x)_\alpha=\sqrt{w_\alpha}\,C(x,z_\alpha).\] The normalized cell indicators form an orthonormal basis for the step functions. On this subspace \(T_\pi\) is represented by \(\mathsf T\), and it vanishes on the orthogonal complement. Thus \(\lVert \mathsf T\rVert<1\), and the correction in (32) is \(\xi_x^T(I-\mathsf T)^{-1}\xi_y\). Evaluation points enter only these finite vectors, including when they coincide with nodes.

We next prove (P) for each sufficiently fine quadrature. On the union of the evaluation points and quadrature nodes, start with \(M_{ij}=C(x_i,x_j)\). By 15, its reciprocal has (P). Expanding the finite resolvent sums all chains, multiplying kernel entries along each chain and the quadrature weight at every intermediate occurrence. These nonnegative sums are finite by the preceding matrix formula. Restricting the allowed intermediate nodes gives a subseries and hence remains finite.

Suppose some nodes have been allowed and their chain sum is \(M\). Admit one more node \(z\) of weight \(w>0\). Chains with exactly \(k\ge1\) internal occurrences of \(z\) have total weight \[w^kM_{iz}M_{zz}^{k-1}M_{zj}.\] All factors are positive. Finiteness of the full series forces \(wM_{zz}<1\), and summing over \(k\) updates the matrix to \[ M_{ij}+\frac{w}{1-wM_{zz}}M_{iz}M_{zj}. \tag{33}\] 16 therefore propagates (P) one node at a time. This counting remains valid when an endpoint is a quadrature node, since only internal occurrences receive weights. Coincident node locations may be merged by adding their weights. Thus the discrete kernel (32) has the required reciprocal property. The quadrature limit gives it for the regularized continuum kernel. If \(\rho=0\), this conclusion follows directly from 15.

Finally let \(\epsilon\downarrow0\). Every chain integral increases to its counterpart with \(G\) by monotone convergence, as does the sum of these nonnegative integrals. Off the diagonal, its limit is (31), and the correction is finite because \[\big|\langle G(x,\cdot),(I-T_\rho)^{-1}G(y,\cdot)\rangle_\rho\big| \le\frac{\lVert G(x,\cdot)\rVert_\rho\lVert G(y,\cdot)\rVert_\rho}{1-d}.\] The direct diagonal term tends to infinity, so its reciprocal tends to zero. Finite entrywise limits preserve (P), completing the proof. ◻

Proof. Fix compactly supported \(0\le\rho\le q\). Apply 17 to the points \(r s_1,\ldots,r s_m,p\), with \(r\uparrow1\) and \(r>|p|\). The explicit Green formula gives \[\begin{align*} \frac{G(rs,x)}{1-r}&\longrightarrow P_s(x),\tag{35}\\ \frac{G(rs,rt)}{(1-r)^2}&\longrightarrow N_0(s,t)\qquad(s\ne t). \tag{36}\end{align*}\] The first convergence is uniform on compact interior sets. For the second, one may use \[G(rs,rt)=\frac1{4\pi}\log\left(1+\frac{(1-r^2)^2}{r^2|s-t|^2}\right).\] The bounded resolvent in (31) and the uniform convergence on \(\operatorname{supp}\rho\) therefore imply \[\frac{M^\rho(rs_i,rs_j)}{(1-r)^2}\longrightarrow N^\rho(s_i,s_j) \quad(i\ne j),\qquad \frac{M^\rho(p,rs_i)}{1-r}\longrightarrow K^\rho(p,s_i).\] Here the first resolvent correction tends to \(\langle P_{s_i},(I-T_\rho)^{-1}P_{s_j}\rangle_\rho=R^\rho(s_i,s_j)\). The second tends to \(P_{s_i}(p)+\langle G(p,\cdot),(I-T_\rho)^{-1}P_{s_i}\rangle_\rho=K^\rho(p,s_i)\). These pairings are legitimate on the compact support, where the Poisson kernels are bounded.

Scale the reciprocal matrix by positive diagonal congruence with factors \(1-r\) at the moving points and factor \(1\) at \(p\). Its limit is (34) with \(K^\rho,N^\rho\), so it has (P). The zero diagonal stays zero throughout.

Now choose increasing disk cutoffs \(\,\mathrm d\rho_n=\mathbf 1_{\{|x|\le r_n\}}\,\mathrm dq\), with \(r_n\uparrow1\). Every term of \(K^{\rho_n}\) and \(R^{\rho_n}\) is a nonnegative chain integral. Monotone convergence in each finite product integral and then in the series gives \[K^{\rho_n}(p,s)\uparrow K(p,s),\qquad N^{\rho_n}(s,t)\uparrow N(s,t)\quad(s\ne t).\] These limits are finite by 10. Taking the reciprocals and the finite matrix limit proves (34). No \(L^2(q)\) boundary limit of an individual Poisson kernel has been used. ◻

Proof of 13. Use the matrix of 18. We first show that \(x^TBx\le0\) whenever \(\ell^Tx=0\). If instead \(x^TBx>0\), choose \(y\) with \(\ell^Ty\ne0\) and replace it by \[y-\frac{x^TBy}{x^TBx}x.\] Then \(x^TBy=0\) and still \(\ell^Ty\ne0\). In the bordered form, \((x,0)\) and \((y,c)\) are orthogonal, and \(c\) can be chosen so that \[(y,c)^T\begin{pmatrix}B&\ell\\\ell^T&0\end{pmatrix}(y,c) =y^TBy+2c\ell^Ty>0.\] They would span a two-dimensional positive subspace, contradicting (P).

Finally, for coefficients with \(\sum_i a_i=0\), put \(x_i=K(p,s_i)a_i\). Then \(\ell^Tx=0\) and \[\sum_{i,j}a_i a_jD_p(s_i,s_j)=x^TBx\le0.\] This proves the assertion on distinct points; repeated points are handled by combining their coefficients. ◻