The polydisc Bergman product and the different distinguished-torus Hardy kernel

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.