Reciprocal kernels and conditional negativity

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

This lesson proves the kernel theorem behind the hot-spots theorem: for every interior point \(p\) of the disc, the kernel \(K(p,s)K(p,t)/N(s,t)\) on the circle is conditionally negative semidefinite. The kernels \(K\) and \(N\) are those of Boundary kernels and the multiplier identity, built from any bounded nonnegative potential satisfying the subcritical inequality (2.1) there. The proof works with matrices that have at most one positive eigenvalue: it regularizes the Green function of the disc, adds integration nodes one at a time by rank-one updates, and passes to the singular kernel and then to the boundary by monotone limits.

All matrices in this lesson are real and symmetric, and the reciprocal of a matrix with nonzero entries is always taken entrywise, never as a matrix inverse.

1. Matrices with at most one positive eigenvalue

A real symmetric matrix has property (P) if it has at most one positive eigenvalue, counted with multiplicity. A subspace \(W\) is positive for \(M\) if \(x^TMx>0\) for all \(0\ne x\in W\).

Lemma 1.1.

  1. \(M\) has (P) if and only if no two-dimensional subspace is positive for \(M\).
  2. Property (P) is preserved by congruence \(M\mapsto S^TMS\) with \(S\) invertible (in particular by positive diagonal congruence), by subtracting a positive semidefinite matrix, by passing to a principal submatrix, and by limits of matrices of a fixed size.
  3. If \(v\) is a vector and \(U\) is positive semidefinite, then \(vv^T-U\) has (P).

Proof. (1) Two orthonormal eigenvectors for positive eigenvalues span a positive plane. Conversely, if \(M\) has at most one positive eigenvalue and \(W\) is a positive plane, then \(W\) meets the orthogonal complement of the eigenvector of that eigenvalue (or the whole space, if there is none) in a nonzero vector \(x\), and \(x^TMx\le0\) because \(x\) is a combination of eigenvectors with eigenvalues \(\le0\).

(2) \(W\) is positive for \(S^TMS\) if and only if \(SW\) is positive for \(M\). A plane positive for \(M-U\) is positive for \(M\). A plane positive for a principal submatrix, extended by zeros, is positive for \(M\). If \(W\) is a positive plane for \(M\), then \(x^TMx\ge c>0\) on the unit circle of \(W\), so \(W\) is positive for every matrix close to \(M\).

(3) Every plane contains \(x\ne0\) with \(v^Tx=0\), and \(x^T(vv^T-U)x=-x^TUx\le0\). \(\square\)

Lemma 1.2 (Schur products). If \(A\) and \(B\) are positive semidefinite (real symmetric, or complex Hermitian), so is the entrywise product \(A\circ B\). Hence entrywise powers \(A^{\circ n}\) and convergent series \(\sum_nc_nA^{\circ n}\) with \(c_n\ge0\) are positive semidefinite, and a real symmetric matrix that is positive semidefinite as a Hermitian matrix is positive semidefinite.

Proof. By the spectral theorem \(A_{ij}=\langle a_i,a_j\rangle\) and \(B_{ij}=\langle b_i,b_j\rangle\) for vectors \(a_i,b_i\). Then \((A\circ B)_{ij}=\langle a_i\otimes b_i,a_j\otimes b_j\rangle\) is a Gram matrix. The last statement holds because \(x^TAx=x^*Ax\) for real \(x\). \(\square\)

2. The theorem

A real symmetric kernel \(D\) on a set, with zero diagonal, is conditionally negative semidefinite if \[ \sum_{i,j}a_ia_jD(s_i,s_j)\le0 \] for every finite family of points \(s_i\) and real \(a_i\) with \(\sum_ia_i=0\).

Theorem 2.1 ([OpenAI-HS, Theorem 5.1]). Let \(dq=a\,dx\) on \(\mathbb D\) with \(a\in L^\infty(\mathbb D)\), \(a\ge0\), and suppose that for some \(0<d<1\) \[ \begin{gathered} \int_{\mathbb D}|\psi|^2\,dq\le d\int_{\mathbb D}|\nabla\psi|^2\\ (\psi\in H^1_0(\mathbb D)). \end{gathered} \] For the associated kernels \(K,N\) and every \(p\in\mathbb D\), the kernel \[ \begin{gathered} D_p(s,t)=\frac{K(p,s)K(p,t)}{N(s,t)}\quad(s\ne t),\\ D_p(s,s)=0, \end{gathered} \] is conditionally negative semidefinite on \(\mathbb S^1\).

Remark 2.2 ([OpenAI-HS, Remark 5.2]). For \(q=0\), \(K(p,s)=P_s(p)\) and \(N=N_0\). With the disc automorphism \(\varphi_p(z)=(z-p)/(1-\bar pz)\), \[ \begin{gathered} \varphi_p(s)-\varphi_p(t)=\frac{(s-t)(1-|p|^2)}{(1-\bar ps)(1-\bar pt)},\\ |1-\bar ps|=|s-p|\quad(s\in\mathbb S^1), \end{gathered} \] so \(D_p(s,t)=\pi P_s(p)P_t(p)|s-t|^2\), which equals \(\frac1{4\pi}|\varphi_p(s)-\varphi_p(t)|^2\), a multiple of a squared Euclidean distance. Such kernels are conditionally negative semidefinite: if \(\sum a_i=0\), then \(\sum_{i,j}a_ia_j|x_i-x_j|^2=-2|\sum_ia_ix_i|^2\). Theorem 2.1 says that a squared-distance description survives the addition of a subcritical potential; the last lesson realizes \(D_p\) as a squared distance in a Hilbert space.

The matrix condition behind the proof has a classical meaning. For a positive definite kernel \(k\) whose values never vanish, the condition that every finite matrix \((1/k(x_i,x_j))\) has exactly one positive eigenvalue characterizes the complete Nevanlinna–Pick property; this is the McCullough–Quiggin theorem [AM00, Corollary 1.12]. The arguments below are self-contained.

The fixed point \(p\) links (P) to conditional negativity. Adjoining \(p\) to boundary points gives a bordered reciprocal matrix with zero diagonal, boundary block \(1/N(s_i,s_j)\) and border \(1/K(p,s_i)\) (Lemma 6.1); property (P) of that matrix is exactly what the theorem needs. The bordered matrix is reached through four limits: refinement of a quadrature for a regularized kernel with a compactly supported potential \(\rho\le q\); removal of the regularization; radial passage of evaluation points to the circle; and increase of \(\rho\) to \(q\).

3. A bounded regularization of the Green function

For \(x,y\in\mathbb D\) put \[ \begin{gathered} h(x,y)=\frac{|x-y|^2}{(1-|x|^2)(1-|y|^2)},\\ L(h)=\frac1{\log(1+1/h)}\ (h>0),\\ L(0)=0, \end{gathered} \] and \(f(h)=h+\frac12-L(h)\). By the second formula for \(G\) in Section 1 of the previous lesson, \(G=\frac1{4\pi}\log(1+1/h)\), so \(1/G(x,y)=4\pi L(h(x,y))\), with value \(0\) on the diagonal.

Lemma 3.1 ([OpenAI-HS, Lemma 5.3]). There are bounded continuous strictly positive kernels \(C_\varepsilon\) on \(\mathbb D\times\mathbb D\), \(0<\varepsilon\le1\), such that \(C_\varepsilon\le G\), \(C_\varepsilon\uparrow G\) as \(\varepsilon\downarrow0\) (also on the diagonal), and every finite matrix \((1/C_\varepsilon(x_i,x_j))\) has (P).

Proof. The kernel \(h+\frac12\) has (P). Multiplying the \(x\)- and \(y\)-entries by \(1-|x|^2\) and \(1-|y|^2\) turns \(h+\frac12\) into \[ \tfrac12(1+|x|^2)(1+|y|^2)-2\,x\cdot y, \] a rank-one positive kernel minus a Gram kernel; Lemma 1.1(3) and (2) apply.

A formula for \(f\). For \(h>0\), \[ \begin{aligned} &\int_0^1h^{1-\alpha}(h+1)^\alpha\,d\alpha\\ &\quad=h\Bigl[\frac{(1+1/h)^\alpha}{\log(1+1/h)}\Bigr]_0^1=L(h). \end{aligned} \] For \(0<\alpha<1\) let \(\mathbb E_\alpha\) be the expectation for the probability measure \(c_\alpha v^{\alpha-1}(1+v)^{-1}\,dv\) on \((0,\infty)\), and put \(r=1/(1+v)\in(0,1)\). Integration by parts gives \[ \mathbb E_\alpha(1-r)=c_\alpha\int_0^\infty\frac{v^\alpha}{(1+v)^2}\,dv=\alpha . \] The substitution \(v=\frac{h+1}hw\) gives \(c_\alpha\int_0^\infty\frac{h(h+1)v^{\alpha-1}}{(h+1)+vh}\,dv=h^{1-\alpha}(h+1)^\alpha\). Since \((h+1)+vh=(h+r)/r\) and \(c_\alpha v^{\alpha-1}dv=(1+v)\cdot c_\alpha v^{\alpha-1}(1+v)^{-1}dv\), \[ \begin{aligned} &h^{1-\alpha}(h+1)^\alpha=\mathbb E_\alpha\frac{h(h+1)}{h+r}\\ &\quad=\mathbb E_\alpha\Bigl[h+(1-r)-\frac{r(1-r)}{h+r}\Bigr], \end{aligned} \] and this also holds at \(h=0\), where both sides vanish. Integrating over \(\alpha\in(0,1)\), \[ \begin{gathered} f(h)=\int_0^1\mathbb E_\alpha\tfrac{r(1-r)}{h+r}\,d\alpha,\\ 0\le f(h)\le f(0)=\tfrac12, \end{gathered} \tag{3.1} \] because \(r(1-r)/(h+r)\le1-r\).

The kernel \(f(h)\) is positive semidefinite. The kernel \[ \begin{aligned} k(x,y)&=\frac1{1+h(x,y)}\\ &=\frac{(1-|x|^2)(1-|y|^2)}{|1-x\bar y|^2} \end{aligned} \] is positive semidefinite: \((1-x\bar y)^{-1}=\sum_n x^n\bar y^n\) is a Gram kernel, and so is its product with its complex conjugate (Lemma 1.2), and with the factors \((1-|x|^2)(1-|y|^2)\). For fixed \(0<r<1\), \[ \frac1{h+r}=\sum_{n\ge0}(1-r)^nk^{n+1} \] with entrywise powers, because \(\sum_n(1-r)^nk^{n+1}=k/(1-(1-r)k)\) and \(0<k\le1\). By Lemma 1.2 this is positive semidefinite, and so is the positive mixture (3.1).

The kernels \(C_\varepsilon\). Define \[ \begin{aligned} \frac1{C_\varepsilon}&=4\pi\Bigl[h+\frac12-(1-\varepsilon)f(h)\Bigr]\\ &=4\pi\bigl[L(h)+\varepsilon f(h)\bigr]. \end{aligned} \] Its finite matrices are a matrix with (P) minus a positive semidefinite one, so they have (P). Since \(L+\varepsilon f=\varepsilon(h+\frac12)+(1-\varepsilon)L\ge\varepsilon/2\), the kernel \(C_\varepsilon\) is bounded by \(1/(2\pi\varepsilon)\); it is continuous and positive. As \(\varepsilon\downarrow0\), \(1/C_\varepsilon\) decreases to \(4\pi L(h)=1/G\), so \(C_\varepsilon\le G\) and \(C_\varepsilon\uparrow G\); on the diagonal \(C_\varepsilon(x,x)=1/(2\pi\varepsilon)\uparrow\infty\). \(\square\)

4. A rank-one update

Lemma 4.1 ([OpenAI-HS, Lemma 5.4]). Let \(M\) be a symmetric matrix with strictly positive entries whose reciprocal has (P). For every index \(z\) and every \(c\ge0\), the matrix \[ \widetilde M_{ij}=M_{ij}+c\,M_{iz}M_{zj} \] also has a reciprocal with (P).

Proof. Congruence by the positive diagonal matrix \(\operatorname{diag}(M_{iz})\) turns the reciprocal of \(M\) into \(H_{ij}=M_{iz}M_{zj}/M_{ij}\), which has (P). Its \(z\)-th row and column are constant, equal to \(a=M_{zz}>0\). Order \(z\) last and let \(H_0\) be the remaining principal block. Subtracting the last row from the other rows, and the last column from the other columns, is a congruence: \[ H\ \sim\ \begin{pmatrix}H_0-a\mathbf 1\mathbf 1^T&0\\0&a\end{pmatrix}. \] Since \(a>0\) and \(H\) has (P), the block \(H_0-a\mathbf 1\mathbf 1^T\) has no positive eigenvalue. So \(U=a\mathbf 1\mathbf 1^T-H\), whose \(z\)-th row and column vanish, is positive semidefinite. Since \(H_{ii}>0\), \(0\le U_{ii}<a\), and the Cauchy–Schwarz inequality for \(U\) gives \(|U_{ij}|<a\).

The same congruence turns the reciprocal of \(\widetilde M\) into the matrix with entries \(H_{ij}/(1+cH_{ij})\). With \(H_{ij}=a-U_{ij}\), \[ \begin{aligned} &\frac a{1+ca}-\frac{H_{ij}}{1+cH_{ij}}\\ &\quad=\frac{U_{ij}}{(1+ca)(1+ca-cU_{ij})}\\ &\quad=\sum_{n\ge0}\frac{c^nU_{ij}^{n+1}}{(1+ca)^{n+2}}, \end{aligned} \] the series converging because \(|cU_{ij}|<1+ca\). By Lemma 1.2 the right side is a positive semidefinite matrix, so the transformed reciprocal of \(\widetilde M\) is \(\frac a{1+ca}\mathbf 1\mathbf 1^T\) minus a positive semidefinite matrix, and it has (P) by Lemma 1.1. \(\square\)

5. The interior resolvent kernel

Let \(0\le\rho\le q\) be a measure supported in a compact subset of \(\mathbb D\), let \(T_\rho\) be its Green operator on \(L^2(\rho)\), with \(\|T_\rho\|\le d\) (Lemma 2.2 of the previous lesson), put \(S_\rho=(I-T_\rho)^{-1}\), and define for \(x\ne y\) in \(\mathbb D\) \[ \begin{aligned} &M^\rho(x,y)=G(x,y)\\ &\quad+\bigl\langle G(x,\cdot),S_\rho G(y,\cdot)\bigr\rangle_\rho, \end{aligned} \tag{5.1} \] and \(M^\rho(x,x)=+\infty\). Expanding the Neumann series, \(M^\rho(x,y)\) is the sum over \(n\ge0\) of the nonnegative chain integrals \[ \begin{aligned} &\int\!\cdots\!\int G(x,z_1)G(z_1,z_2)\cdots G(z_n,y)\\ &\qquad d\rho(z_1)\cdots d\rho(z_n), \end{aligned} \] and it is finite for \(x\ne y\): the correction is at most \(\|G(x,\cdot)\|_\rho\|G(y,\cdot)\|_\rho/(1-d)\).

Proposition 5.1 ([OpenAI-HS, Proposition 5.5]). For distinct points \(x_1,\dots,x_m\in\mathbb D\), the matrix \((1/M^\rho(x_i,x_j))\), with zero diagonal, has (P).

Proof. Step 1: regularization. Fix \(\varepsilon\) and write \(C=C_\varepsilon\). Let \(M_C\) be defined by (5.1) with \(G\) replaced by \(C\) throughout, including in the operator: \(T_Cf(x)=\int C(x,z)f(z)\,d\rho(z)\). Since \(0<C\le G\), \(|T_Cf|\le T_\rho|f|\) pointwise, so \(\|T_C\|\le d\), and \(M_C(x,y)\) is the sum of the chain integrals of \(C\), finite for all \(x,y\).

Step 2: quadrature. Let \(Q\) be the support of \(\rho\). Partition it into finitely many Borel cells of diameter less than \(\eta\) and positive \(\rho\)-mass (cells of mass zero are discarded); choose a node \(z_\beta\) in each cell \(Q_\beta\), put \(w_\beta=\rho(Q_\beta)\), and let \(\pi(y)=z_\beta\) for \(y\in Q_\beta\). By uniform continuity of \(C\) on \(Q\times Q\), as \(\eta\to0\) the operators \(T_\pi\) with kernel \(C(\pi(\cdot),\pi(\cdot))\) converge to \(T_C\) in norm (the norm difference is at most \(\rho(\mathbb D)\sup|C(\pi(y),\pi(z))-C(y,z)|\)), and \(C(x,\pi(\cdot))\to C(x,\cdot)\) in \(L^2(\rho)\) for each \(x\). For fine partitions \(\|T_\pi\|<1\); with \(S_\pi=(I-T_\pi)^{-1}\) and \(c_x=C(x,\pi(\cdot))\in L^2(\rho)\), \[ \begin{aligned} &M_\pi(x,y)\\ &\quad=C(x,y)+\langle c_x,S_\pi c_y\rangle_\rho\\ &\quad\longrightarrow M_C(x,y). \end{aligned} \tag{5.2} \] The functions \(e_\beta=\mathbf 1_{Q_\beta}/\sqrt{w_\beta}\) are orthonormal, \(T_\pi\) maps \(L^2(\rho)\) into their span and vanishes on its orthogonal complement, and its matrix there is \(\mathsf T_{\beta\gamma}=\sqrt{w_\beta}\,C(z_\beta,z_\gamma)\sqrt{w_\gamma}\). With \((\xi_x)_\beta=\sqrt{w_\beta}\,C(x,z_\beta)\), the correction in (5.2) equals \(\xi_x^T(I-\mathsf T)^{-1}\xi_y\), and \(\|\mathsf T\|<1\).

Step 3: adding nodes one at a time. Let \(S\) be the finite set consisting of the points \(x_i\) and the nodes \(z_\beta\). For a set \(A\) of nodes let \(M^A\) be the matrix on \(S\times S\) whose \((x,y)\) entry is the sum over all chains \(x,z_{\beta_1},\dots,z_{\beta_n},y\) (\(n\ge0\)) with internal nodes in \(A\) of \[ C(x,z_{\beta_1})w_{\beta_1}C(z_{\beta_1},z_{\beta_2})\cdots w_{\beta_n}C(z_{\beta_n},y). \] For \(A\) the set of all nodes this is \(C(x,y)+\sum_{n\ge0}\xi_x^T\mathsf T^n\xi_y=M_\pi(x,y)\), which is finite; for smaller \(A\) the sum is a subseries, hence finite. For \(A=\emptyset\), \(M^A\) is the restriction of \(C\) to \(S\), whose reciprocal has (P) by Lemma 3.1. Add a node \(z\notin A\) of weight \(w\). The chains with exactly \(k\ge1\) internal occurrences of \(z\) contribute \(w^kM^A_{xz}(M^A_{zz})^{k-1}M^A_{zy}\); all factors are positive, and finiteness forces \(wM^A_{zz}<1\). Summing over \(k\), \[ M^{A\cup\{z\}}_{xy}=M^A_{xy}+\frac w{1-wM^A_{zz}}\,M^A_{xz}M^A_{zy}, \] so Lemma 4.1 propagates (P) of the reciprocal from \(M^A\) to \(M^{A\cup\{z\}}\). After all nodes are added, the reciprocal of \(M_\pi\) on \(S\) has (P), and so does its principal submatrix on the points \(x_i\).

Step 4: the quadrature limit. By (5.2) and Lemma 1.1(2), the reciprocal matrix of \(M_C\) at the points \(x_i\) has (P).

Step 5: removing the regularization. As \(\varepsilon\downarrow0\), each chain integral of \(C_\varepsilon\) increases to the corresponding chain integral of \(G\) by monotone convergence, and so do their sums. Off the diagonal the limit is \(M^\rho(x_i,x_j)<\infty\). On the diagonal \(M_{C_\varepsilon}(x_i,x_i)\ge C_\varepsilon(x_i,x_i)\), which is \(1/(2\pi\varepsilon)\to\infty\), so its reciprocal tends to \(0\). Lemma 1.1(2) gives the assertion. \(\square\)

6. The boundary limit and the proof of the theorem

For a compactly supported \(0\le\rho\le q\), let \(K^\rho\), \(R^\rho\), \(N^\rho=N_0+R^\rho\) be the kernels of Section 4 of the previous lesson built from \(\rho\).

Lemma 6.1 ([OpenAI-HS, Lemma 5.6]). For distinct \(s_1,\dots,s_m\in\mathbb S^1\) and \(p\in\mathbb D\), the bordered matrix \[ \begin{gathered} \begin{pmatrix}B&\ell\\\ell^T&0\end{pmatrix},\qquad\ell_i=\frac1{K(p,s_i)},\\ B_{ij}=\begin{cases}1/N(s_i,s_j),&i\ne j,\\0,&i=j,\end{cases} \end{gathered} \] has (P).

Proof. Fix a compactly supported \(0\le\rho\le q\). For \(|p|<r<1\), apply Proposition 5.1 to the distinct points \(rs_1,\dots,rs_m,p\). From the second formula for \(G\), \[ \begin{gathered} \frac{G(rs,x)}{1-r}\longrightarrow P_s(x),\\ \frac{G(rs,rt)}{(1-r)^2}\longrightarrow N_0(s,t)\quad(s\ne t), \end{gathered} \] as \(r\uparrow1\), the first uniformly for \(x\) in compact subsets of \(\mathbb D\); for the second use \(G(rs,rt)=\frac1{4\pi}\log\bigl(1+\frac{(1-r^2)^2}{r^2|s-t|^2}\bigr)\). On the support of \(\rho\) these convergences are uniform and \(P_s\) is bounded. Expanding \((I-T_\rho)^{-1}\) in (5.1), and using \(R^\rho(s,t)=\sum_n\langle P_s,T_\rho^nP_t\rangle_\rho\) and \(\langle G(p,\cdot),T_\rho^nP_s\rangle_\rho=(T_\rho^{n+1}P_s)(p)\), we get \[ \begin{gathered} \frac{M^\rho(rs_i,rs_j)}{(1-r)^2}\to N^\rho(s_i,s_j)\quad(i\ne j),\\ \frac{M^\rho(p,rs_i)}{1-r}\to K^\rho(p,s_i). \end{gathered} \] Congruence by the diagonal matrix with entries \(1-r\) at the moving points and \(1\) at \(p\) turns the reciprocal matrix of Proposition 5.1 into a matrix converging to the bordered matrix built from \(K^\rho\), \(N^\rho\); the zero diagonal stays zero. By Lemma 1.1(2), that bordered matrix has (P).

Now let \(\rho_n=\mathbf 1_{\{|x|\le r_n\}}q\) with \(r_n\uparrow1\). Every term of \(K^{\rho_n}\) and \(R^{\rho_n}\) is a chain integral of nonnegative functions, so by monotone convergence \(K^{\rho_n}(p,s)\uparrow K(p,s)\) and \(N^{\rho_n}(s,t)\uparrow N(s,t)\) for \(s\ne t\). The limits are finite (Lemma 4.1 of the previous lesson), so the bordered matrices converge, and Lemma 1.1(2) concludes. No boundary value of a Poisson kernel in \(L^2(q)\) has been used: the boundary passage is made for the compactly supported \(\rho\). \(\square\)

Proof of Theorem 2.1. Let \(\mathcal B\) be the bordered matrix of Lemma 6.1. First, \(x^TBx\le0\) whenever \(\ell^Tx=0\). Otherwise, choose \(y\) with \(\ell^Ty\ne0\) (possible since \(\ell\ne0\)) and replace it by \(y-\frac{x^TBy}{x^TBx}x\); then \(x^TBy=0\) and still \(\ell^Ty\ne0\). The vectors \((x,0)\) and \((y,c)\) are \(\mathcal B\)-orthogonal, and for a suitable \(c\ne0\) \[ (y,c)\,\mathcal B\,(y,c)^T=y^TBy+2c\,\ell^Ty>0 . \] They span a positive plane for \(\mathcal B\), contradicting (P).

Now let \(a_1,\dots,a_m\) be real with \(\sum a_i=0\), for distinct \(s_i\), and put \(x_i=K(p,s_i)a_i\). Then \(\ell^Tx=\sum_ia_i=0\), and \[ \sum_{i,j}a_ia_jD_p(s_i,s_j)=x^TBx\le0 . \] If some points coincide, combine their coefficients. \(\square\)

7. Exercises

7.1. Show that \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\) has (P) and the identity matrix of size two does not. Show that a sum of two matrices with (P) need not have (P).

7.2. Check numerically or by hand that \(L(1)=1/\log2\) agrees with \(\int_0^12^\alpha\,d\alpha\).

7.3. Verify the formula for \(\varphi_p(s)-\varphi_p(t)\) in Remark 2.2.

7.4. Let \(M=\begin{pmatrix}1&1/2\\1/2&1\end{pmatrix}\). Check that its reciprocal has (P), compute \(\widetilde M\) of Lemma 4.1 for \(z=2\) and \(c=1\), and check that its reciprocal has (P).

7.5. Show that a kernel \(D\) with zero diagonal on a set with at least two points, all of whose finite matrices are positive semidefinite, is identically zero; so conditional negativity is a genuinely weaker condition than negative semidefiniteness of \(-D\).

8. Solutions

7.1. The eigenvalues of \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\) are \(\pm1\); the identity has two eigenvalues \(1\). The identity is the sum of \(\operatorname{diag}(1,0)\) and \(\operatorname{diag}(0,1)\), each with (P).

7.2. \(\int_0^12^\alpha\,d\alpha=1/\log2\), and \(L(1)=1/\log(1+1)\).

7.3. The difference \((s-p)(1-\bar pt)-(t-p)(1-\bar ps)\) equals \((s-t)(1-|p|^2)\) after cancellation of \(-\bar pst+\bar pst\). For \(|s|=1\), \(|1-\bar ps|=|\bar s-\bar p|=|s-p|\).

7.4. The reciprocal \(\begin{pmatrix}1&2\\2&1\end{pmatrix}\) has eigenvalues \(3\) and \(-1\). \(\widetilde M=\begin{pmatrix}5/4&1\\1&2\end{pmatrix}\), with reciprocal \(\begin{pmatrix}4/5&1\\1&1/2\end{pmatrix}\), whose determinant \(2/5-1<0\) shows one positive and one negative eigenvalue.

7.5. For two points \(s,t\), positive semidefiniteness of \(\begin{pmatrix}0&D(s,t)\\D(s,t)&0\end{pmatrix}\) forces \(D(s,t)=0\).

References