Bergman and Szegő Kernels: Examples and Counterexamples

By [F1] the family $e_k(z)=\sqrt{(k+1)\mathrm{/}\pi }{z}^k$ is a complete orthonormal system of $A^2(\mathbb{D})$, with $\mathrm{\parallel }{z}^k{\mathrm{\parallel }}_2^2=\pi \mathrm{/}(k+1)$; since $z,w\in \mathbb{D}$ we have $\mathrm{\mid }z\bar{w}\mathrm{\mid }<1$ for the geometric series.

By [F2] applied to this complete orthonormal system, $K_{\mathbb{D}}(z,w)={\sum }_{k\ge 0}{e}_k(z)\overline{e_k(w)}=\frac{1}{\pi }{\sum }_{k\ge 0}(k+1)(z\bar{w}{)}^k$, and by [F3] this sum equals $\frac{1}{\pi (1-z\bar{w}{)}^2}$; this is the displayed series identity.

Fix $k$ and $w$. The holomorphic Fourier sums $T_N:={\sum }_{j\le N}\overline{e_j(w)}{e}_j$ converge in $A^2(\mathbb{D})$ to $K_{\mathbb{D}}(\cdot ,w)$ by [F2]. Continuity of the pairing [F4] gives ${\int }_{\mathbb{D}}{e}_k(z)\overline{K_{\mathbb{D}}(z,w)}d\lambda (z)={\lim }_N⟨e_k,T_N⟩$. For $N\ge k$, conjugate-linearity in the second variable and orthonormality give $⟨e_k,T_N⟩={\sum }_{j\le N}{e}_j(w)⟨e_k,e_j⟩=e_k(w)$. This proves the reproducing integral.

Step 2.1 identifies the displayed series with the disc Bergman kernel of [F3], and step 3.1 verifies the reproducing identity on every monomial. Since the monomial span is dense by [F1] and both sides of the reproducing identity are continuous in the first variable by [F4], the check extends to all of $A^2(\mathbb{D})$ and agrees with the general reproducing theorem.

By [F10] and [F11], the multi-index factorials, lengths and powers in the formulas have their stated meanings. The ball moment in [F1] gives $\mathrm{\parallel }{z}^{\alpha }{\mathrm{\parallel }}_{A^2({\mathbb{B}}^m)}^2=c_{\alpha }$, and [F1] gives completeness of the normalized monomials.

By [F2], $\mathrm{\parallel }{\zeta }^{\alpha }{\mathrm{\parallel }}_{L^2(S,{\sigma }_1)}^2=w_{\alpha }$ and the normalized boundary monomials form a complete orthonormal system of the Hardy space; their evaluations are bounded.

By [F12], $z,w\in {\mathbb{B}}^m$ have Euclidean norms below $1$; Cauchy–Schwarz [F9] gives $\mathrm{\mid }⟨z,w⟩\mathrm{\mid }\le \mathrm{\parallel }z\mathrm{\parallel }\mathrm{\parallel }w\mathrm{\parallel }<1$, so both denominators are nonzero. The model-domain theorem [F3], with Lebesgue measure and normalized polar boundary measure, gives the displayed closed-form kernels.

Fix $w\in {\mathbb{B}}^m$ and let $K_w=K_{{\mathbb{B}}^m}(\cdot ,w)$. By [F4], its Fourier sums over finite $F\subset {\mathbb{N}}^m$ converge in $A^2({\mathbb{B}}^m)$ to $K_w$. For each $\beta$, [F5] gives $⟨e_{\beta },K_w⟩=e_{\beta }(w)$, so conjugate symmetry [F8] makes its Fourier coefficient $⟨K_w,e_{\beta }⟩=\overline{e_{\beta }(w)}$. If $F$ contains $\alpha$, orthonormality gives $⟨e_{\alpha },{\sum }_{\beta \in F}\overline{e_{\beta }(w)}{e}_{\beta }⟩={\sum }_{\beta \in F}{e}_{\beta }(w)⟨e_{\alpha },e_{\beta }⟩=e_{\alpha }(w)\mathrm{.}$ Passing to the norm limit using [F9] proves ${\int }_{{\mathbb{B}}^m}{e}_{\alpha }(z)\stackrel{‾}{K_{{\mathbb{B}}^m}(z,w)}d\lambda (z)=e_{\alpha }(w)$.

Fix $w\in {\mathbb{B}}^m$ and let $S_w$ be the Riesz representer of the bounded Hardy evaluation at $w$. By [F4], its finite-subset Fourier sums in the complete system $e_{\beta }^{\mathrm{\partial }}$ converge to $S_w$. By [F6], $⟨e_{\beta }^{\mathrm{\partial }},S_w⟩=e_{\beta }^{\mathrm{\partial }}(w)$, so conjugate symmetry [F8] gives Fourier coefficient $⟨S_w,e_{\beta }^{\mathrm{\partial }}⟩=\overline{e_{\beta }^{\mathrm{\partial }}(w)}$. For a finite $F$ containing $\alpha$, orthonormality gives $⟨e_{\alpha }^{\mathrm{\partial }},{\sum }_{\beta \in F}\overline{e_{\beta }^{\mathrm{\partial }}(w)}{e}_{\beta }^{\mathrm{\partial }}⟩=e_{\alpha }^{\mathrm{\partial }}(w)\mathrm{.}$ Passing to the norm limit using [F9] proves $⟨e_{\alpha }^{\mathrm{\partial }},S_w⟩=e_{\alpha }^{\mathrm{\partial }}(w)$, the Hardy reproducing identity for that basis vector. This uses the boundary Hilbert-space representer $S_w$; the interior kernel is not evaluated at a boundary point.

The finite linear spans of the complete systems in [F1] and [F2] are dense in their respective Hilbert spaces. Bounded evaluation from [F5] and [F2], and continuity of inner products from [F9], extend the identities of steps 1.4 and 1.5 from monomials to the full Bergman and Hardy spaces. At $\alpha =0$, $c_0={\pi }^m\mathrm{/}m!$ and $w_0=1$; setting $z=0$ in the kernels gives $m!\mathrm{/}{\pi }^m$ and $1$. When $m=1$, ${\mathbb{B}}^1=\mathbb{D}$ and the formulas specialize to the disc kernels.

The map $\phi$ has the Möbius form $\frac{1z-i}{1z+i}$ with $ad-bc=1\cdot i-(-i)\cdot 1=2i\mathrm{\ne }0$, so by [F1] it is holomorphic away from its pole $z=-i$ and is a bijection of the sphere with Möbius inverse. Its inverse is $\psi (u)=\frac{i(1+u)}{1-u}$: indeed $\psi (u)+i=\frac{2i}{1-u}$ and $\psi (u)-i=\frac{2iu}{1-u}$, so $\phi (\psi (u))=u$, and $1+\phi (z)=\frac{2z}{z+i}$, $1-\phi (z)=\frac{2i}{z+i}$ give $\psi (\phi (z))=z$.

For $z\mathrm{\ne }-i$, [F3] gives $\mathrm{\mid }\phi (z)\mathrm{\mid }<1⟺\mathrm{\mid }z-i{\mathrm{\mid }}^2<\mathrm{\mid }z+i{\mathrm{\mid }}^2⟺-2\mathrm{Im}z<2\mathrm{Im}z⟺\mathrm{Im}z>0$; thus $\phi (\mathbb{H})\subseteq \mathbb{D}$. Likewise, for $\mathrm{\mid }u\mathrm{\mid }<1$ the real part $(1-\mathrm{\mid }u{\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }1-u{\mathrm{\mid }}^2$ of $\frac{1+u}{1-u}$ is positive, so $\mathrm{Im}\psi (u)=\frac{1-\mathrm{\mid }u{\mathrm{\mid }}^2}{\mathrm{\mid }1-u{\mathrm{\mid }}^2}>0$ and $\psi (\mathbb{D})\subseteq \mathbb{H}$. Since the two maps are inverse bijections by step 1.1 and both are holomorphic on these domains, [F6] makes $\phi :\mathbb{H}\to \mathbb{D}$ a biholomorphism with $J_{\phi }(z)={\phi }^{\mathrm{\prime }}(z)=\frac{2i}{(z+i{)}^2}$.

By [F4] applied to $\phi$, $K_{\mathbb{H}}(z,w)=K_{\mathbb{D}}(\phi (z),\phi (w)){\phi }^{\mathrm{\prime }}(z)\overline{{\phi }^{\mathrm{\prime }}(w)}$. By [F5] and [F3], $1-\phi (z)\overline{\phi (w)}=1-\frac{(z-i)(\bar{w}+i)}{(z+i)(\bar{w}-i)}=\frac{(z+i)(\bar{w}-i)-(z-i)(\bar{w}+i)}{(z+i)(\bar{w}-i)}=\frac{-2i(z-\bar{w})}{(z+i)(\bar{w}-i)}\mathrm{.}$ Substituting this and ${\phi }^{\mathrm{\prime }}(z)=\frac{2i}{(z+i{)}^2}$, $\overline{{\phi }^{\mathrm{\prime }}(w)}=\frac{-2i}{(\bar{w}-i{)}^2}$ into the transformation law gives $K_{\mathbb{H}}(z,w)=\frac{(z+i{)}^2(\bar{w}-i{)}^2}{\pi (-2i(z-\bar{w}){)}^2}\cdot \frac{2i}{(z+i{)}^2}\cdot \frac{-2i}{(\bar{w}-i{)}^2}=\frac{-1}{\pi (z-\bar{w}{)}^2},$ since $(2i)(-2i)=4$ and $(-2i{)}^2=-4$.

Setting $w=z$ in step 3.1 and using $z-\bar{z}=2i\mathrm{Im}z$ gives $(z-\bar{z}{)}^2=-4(\mathrm{Im}z{)}^2$, hence $K_{\mathbb{H}}(z,z)=\frac{1}{4\pi (\mathrm{Im}z{)}^2}>0$ for $z\in \mathbb{H}$. This is transport of the disc's diagonal: $K_{\mathbb{H}}(z,z)=\mathrm{\mid }{\phi }^{\mathrm{\prime }}(z){\mathrm{\mid }}^2{K}_{\mathbb{D}}(\phi (z),\phi (z))$ by [F4], and by [F4] the pullback along the biholomorphism $\phi$ carries the disc reproducing property to the reproducing property of $K_{\mathbb{H}}$ on $A^2(\mathbb{H})$.

Set $w=z$ in the two formulas of [F2]. Since $1-z\bar{z}=1-\mathrm{\mid }z{\mathrm{\mid }}^2>0$ on $\mathbb{D}$, the diagonal values are $K_{\mathbb{D}}(z,z)=\frac{1}{\pi (1-\mathrm{\mid }z{\mathrm{\mid }}^2{)}^2}$ and $S_{\mathbb{D}}(z,z)=\frac{1}{1-\mathrm{\mid }z{\mathrm{\mid }}^2}$; at $z=0$ these are $1\mathrm{/}\pi$ and $1$.

By [F1] the first kernel reproduces point evaluations in $A^2(\mathbb{D})$ for the area pairing, while the second reproduces the extended evaluations of the Szegő-regular pair $(\mathbb{D},{\mu }_{\mathbb{T}})$, whose Hardy space is its boundary-trace closure; [F3] records the classical radial-mean form of the disc Hardy class against the same normalized Haar boundary measure.

As $\mathrm{\mid }z\mathrm{\mid }\to 1$, the formulas of step 1.1 show $K_{\mathbb{D}}(z,z)=\frac{1}{\pi }(1-\mathrm{\mid }z{\mathrm{\mid }}^2{)}^{-2}$ and $S_{\mathbb{D}}(z,z)=(1-\mathrm{\mid }z{\mathrm{\mid }}^2{)}^{-1}$, so the boundary blow-up exponents are $2$ and $1$. Their ratio $K_{\mathbb{D}}(z,z)\mathrm{/}{S}_{\mathbb{D}}(z,z)=\frac{1}{\pi (1-\mathrm{\mid }z{\mathrm{\mid }}^2)}$ is unbounded on $\mathbb{D}$; in particular the two diagonal kernels are not equal, so neither normalization may be read off the other without declaring which measure is used.

The conclusion $A^2({\mathbb{C}}^m)=\{0\}$ and $K_{{\mathbb{C}}^m}\equiv 0$ is already proved in The Bergman space $A^2(\mathrm{\Omega })$ and the Bergman kernel, so this example uses that earlier result without repeating its large-polydisc argument.

Let $a=(z,w)\in \mathbb{D}\times \mathbb{C}$ and set $\rho =(1-\mathrm{\mid }z\mathrm{\mid })\mathrm{/}2>0$. For every $R>0$, the closed polydisc ${\bar{\mathrm{\Delta }}}_{(\rho ,R)}(a)$ lies in $\mathbb{D}\times \mathbb{C}$, since $\mathrm{\mid }z\mathrm{\mid }+\rho =(1+\mathrm{\mid }z\mathrm{\mid })\mathrm{/}2<1$ and the second coordinate is unrestricted. These polydiscs show that the product set is open. Let $f$ be the unique holomorphic representative of any class in $A^2(\mathbb{D}\times \mathbb{C})$, as in [F6]. The mean bound [F3] and integral monotonicity [F4] give $\mathrm{\mid }f(a){\mathrm{\mid }}^2\le ({\pi }^2{\rho }^2{R}^2{)}^{-1}{\int }_{{\mathrm{\Delta }}_{(\rho ,R)}(a)}\mathrm{\mid }f{\mathrm{\mid }}^{2}d{\lambda }_4\le \mathrm{\parallel }[f]{\mathrm{\parallel }}_{L^2(\mathbb{D}\times \mathbb{C})}^2\mathrm{/}({\pi }^2{\rho }^2{R}^2)$. Letting $R\to \mathrm{\infty }$ gives $f(a)=0$. Since $a$ was arbitrary, every holomorphic representative is identically zero, hence $A^2(\mathbb{D}\times \mathbb{C})=\{0\}$.

By [F7], each point evaluation on the zero space $A^2(\mathbb{D}\times \mathbb{C})$ has the unique representer $k_w=0$, so $K_{\mathbb{D}\times \mathbb{C}}(z,w)=0$ for all $z,w$. Its diagonal is zero, making the logarithm in the kernel-derived Bergman metric undefined.

By [F5], the constant function $1$ is in $A^2(\mathbb{D})$. Its evaluation at $0$ is $1$, so its Riesz representer $k_0$ is nonzero by [F7]. Reproduction with $f=k_0$ gives $K_{\mathbb{D}}(0,0)=k_0(0)=⟨k_0,k_0⟩=\mathrm{\parallel }{k}_0{\mathrm{\parallel }}^2>0$, proving the claimed contrast with the two unbounded examples.

By [F1] and [F2], the complete monomial expansion is $K_{{\mathbb{D}}^m}(z,w)={\pi }^{-m}{\sum }_{\alpha \in {\mathbb{N}}^m}{\prod }_{j<m}({\alpha }_j+1)(z_j\overline{w_j}{)}^{{\alpha }_j}$. The model theorem [F3] identifies this sum as ${\pi }^{-m}{\prod }_{j<m}(1-z_j\overline{w_j}{)}^{-2}$, and its disc formula identifies each factor as $K_{\mathbb{D}}(z_j,w_j)$. This is the stated Bergman product.

The monomial classes are orthonormal by [F5]. Let $H_{\mathrm{\partial }}^2$ be their closed linear span in the Hilbert space [F12]; they form a complete orthonormal system there by construction. For $z\in {\mathbb{D}}^m$, [F7] gives ${\sum }_{\alpha }\mathrm{\mid }{z}^{\alpha }{\mathrm{\mid }}^2={\prod }_{j<m}(1-\mathrm{\mid }{z}_j{\mathrm{\mid }}^2{)}^{-1}<\mathrm{\infty }$. More strongly, ${\sum }_{\alpha }\mathrm{\mid }{z}^{\alpha }\mathrm{\mid }={\prod }_{j<m}(1-\mathrm{\mid }{z}_j\mathrm{\mid }{)}^{-1}<\mathrm{\infty }$: box partial sums of the nonnegative family factor as products of finite geometric sums, and every finite index set lies in a box. Thus $\mathrm{\Sigma }(z,\zeta ):={\sum }_{\alpha }{z}^{\alpha }\overline{{\zeta }^{\alpha }}$ converges absolutely and uniformly for $\zeta \in {\mathbb{T}}^m$. For complex $\mathrm{\mid }t\mathrm{\mid }<1$, absolute convergence follows from ${\sum }_{k\ge 0}\mathrm{\mid }t{\mathrm{\mid }}^k<\mathrm{\infty }$, so [F14] supplies the complex sum. The finite telescoping identity $(1-t){\sum }_{k=0}^N{t}^k=1-t^{N+1}$ gives ${\sum }_{k\ge 0}{t}^k=(1-t{)}^{-1}$, since the real geometric convergence in [F7] makes $\mathrm{\mid }t{\mathrm{\mid }}^{N+1}\to 0$. Applying this identity coordinatewise and taking limits of finite box products gives $\mathrm{\Sigma }(z,\zeta )={\prod }_{j<m}(1-z_j\overline{{\zeta }_j}{)}^{-1}$. Haar measure has mass one, so uniform convergence implies $L^2$ convergence; conjugating the polynomial sums shows $k_z:=\overline{\mathrm{\Sigma }(z,\cdot )}\in H_{\mathrm{\partial }}^2$, with $\mathrm{\parallel }{k}_z{\mathrm{\parallel }}_2^2={\sum }_{\alpha }\mathrm{\mid }{z}^{\alpha }{\mathrm{\mid }}^2$.

Let $f\in H_{\mathrm{\partial }}^2$. By [F6] its Fourier sums $p_N(\zeta )={\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le N}{c}_{\alpha }{\zeta }^{\alpha }$, with $c_{\alpha }=⟨f,{\zeta }^{\alpha }⟩$, converge in $L^2$ to $f$; degree cutoffs contain every finite index set eventually. Orthogonality gives ${\sum }_{\alpha }\mathrm{\mid }{c}_{\alpha }{\mathrm{\mid }}^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$ by passing to the limit in $\mathrm{\parallel }{p}_N{\mathrm{\parallel }}_2^2$. Cauchy–Schwarz and step 1.2 show that $\tilde{f}(z):={\sum }_{\alpha }{c}_{\alpha }{z}^{\alpha }$ is absolutely convergent. On $\mathrm{\mid }{z}_j\mathrm{\mid }\le r_j<1$, the degree tail is bounded by $({\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }>N}\mathrm{\mid }{c}_{\alpha }{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}{\prod }_{j<m}(1-r_j^2{)}^{-1\mathrm{/}2}$, which tends to zero uniformly. The polynomial extensions therefore converge locally uniformly, and [F13] makes $\tilde{f}$ holomorphic. Pairing the finite sums with $k_z$ and passing to the norm limit gives $⟨f,k_z⟩=\tilde{f}(z)$ and $\mathrm{\mid }\tilde{f}(z)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_2{\prod }_{j<m}(1-\mathrm{\mid }{z}_j{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}$. The extension of $k_w$ has coefficients $\overline{w^{\alpha }}$, so its value at $z$ is $\mathrm{\Sigma }(z,w)$, the displayed reproducing kernel.

The Bergman product in step 1.1 has reciprocal powers $2$, whereas the torus kernel in steps 1.2–2.1 has powers $1$. For $m\ge 2$, [F10] shows that the distinguished torus is a proper subset of the nonsmooth topological boundary, so the surface-measure Szegő definition does not apply to ${\mathbb{D}}^m$. For $m=1$, the torus is $\mathrm{\partial }\mathbb{D}$ and [F3] gives the same normalized-circle Szegő kernel $(1-z\bar{w}{)}^{-1}$. These are exactly the asserted comparisons.

Let $p=(1,\dots ,1)$. For each $j<m$, the circle curve ${\gamma }_j(t)=(1,\dots ,e^{it},\dots ,1)$ lies in $\mathrm{\partial }{\mathbb{D}}^m$ and has velocity $i{e}_j$ at $t=0$ by [F1, F4]. For each $j<m$, the one-sided radial curve ${\delta }_j(t)=(1,\dots ,1-t,\dots ,1)$, $0\le t<1$, also lies in $\mathrm{\partial }{\mathbb{D}}^m$: for $t>0$ its $j$th coordinate has modulus below one, while another coordinate remains of modulus one since $m\ge 2$. Its right velocity at $0$ is $-e_j$. The $2m$ velocities $i{e}_j,-e_j$ form a real basis by [F3].