The disc trace space is the Hardy boundary space and the Szegő family reproduces $H^2$

For every $f\in H^2(\mathbb{D})$, [F3] gives $f^{\ast }\in L^2$ and $\mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_{H^2}$; boundary limits are linear, so $f\mapsto f^{\ast }$ is a linear isometry. If $f^{\ast }=0$, then $f\in H^1$ by [F2] and [F4] gives $f(w)=0$ for every $w\in \mathbb{D}$, proving injectivity.

Fix $f\in H^2(\mathbb{D})$ and $0<r<1$. The dilate $g_r(z):=f(rz)$ is holomorphic on $\{\mathrm{\mid }z\mathrm{\mid }<1\mathrm{/}r\}$: for $z$ there, its difference quotient tends to $r{f}^{\mathrm{\prime }}(rz)$ by [F8]. It is continuous on a neighbourhood of $\bar{\mathbb{D}}$ by [F9], and its boundary trace is $f(r\zeta )$. By [F3], these traces converge to $f^{\ast }$ in $L^2$ as $r\uparrow 1$, so the neighbourhood-holomorphic traces are dense in the boundary-value range.

Let $G\in \mathcal{O}(\mathbb{D})\cap C(\bar{\mathbb{D}})$, the generating class in [F6]. By [F11] and [F12], $\mathrm{\mid }G\mathrm{\mid }$ is bounded on $\bar{\mathbb{D}}$, hence every radial $L^2$ mean of $G$ is bounded and $G\in H^2(\mathbb{D})$ by [F2]. Continuity on $\bar{\mathbb{D}}$ makes its radial boundary limit equal to $G{\mathrm{\mid }}_{\mathbb{T}}$ at every point, so [F3] identifies its trace with $G^{\ast }$ in $L^2$. Thus every generating trace in [F6] belongs to the boundary-value range.

Let $h_n=f_n^{\ast }$ be a sequence in the boundary-value range converging in $L^2$ to $h$. The preimage $f_n$ is unique by step 1.1, so this sequence is well defined; [F3] applied to differences shows $(f_n)$ is Cauchy in $H^2$. For fixed $z\in \mathbb{D}$, [F4] applies to $f_n-f_k\in H^1$. With ${\eta }_z(\zeta ):=1\mathrm{/}(1-z\bar{\zeta })$, [F1] and [F10] give $\mathrm{\mid }{\eta }_z(\zeta )\mathrm{\mid }\le (1-\mathrm{\mid }z\mathrm{\mid }{)}^{-1}$ and $\mathrm{\parallel }{\eta }_z{\mathrm{\parallel }}_2\le (1-\mathrm{\mid }z\mathrm{\mid }{)}^{-1}$. Cauchy–Schwarz in [F5] therefore yields $\mathrm{\mid }{f}_n(z)-f_k(z)\mathrm{\mid }\le \frac{\mathrm{\parallel }{f}_n^{\ast }-f_k^{\ast }{\mathrm{\parallel }}_2}{1-\mathrm{\mid }z\mathrm{\mid }}\mathrm{.}$

By [F14], $f_n(z)$ has a limit $F(z)$ for each $z\in \mathbb{D}$. Given $z_0\in \mathbb{D}$, choose $\rho$ with $\mathrm{\mid }{z}_0\mathrm{\mid }<\rho <1$; the estimate of step 2.1, uniformly for $\mathrm{\mid }z\mathrm{\mid }<\rho$, makes $f_n$ uniformly Cauchy on that neighbourhood. Hence $f_n\to F$ locally uniformly, and [F15] makes $F$ holomorphic.

The Cauchy sequence $(f_n)$ is bounded in $H^2$. For every $0\le r<1$, pointwise convergence on $r\mathbb{T}$ and [F16] give ${\int }_{\mathbb{T}}\mathrm{\mid }F(r\zeta ){\mathrm{\mid }}^{2}dm(\zeta )\le {\mathrm{lim\hspace{0.17em}inf}}_{n\to \mathrm{\infty }}{\int }_{\mathbb{T}}\mathrm{\mid }{f}_n(r\zeta ){\mathrm{\mid }}^{2}dm(\zeta )\le {\sup }_n\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_{H^2}^2,$ so $F\in H^2$. Given $\epsilon >0$, choose $N$ with $\mathrm{\parallel }{f}_n-f_k{\mathrm{\parallel }}_{H^2}<\epsilon$ for $n,k\ge N$. For fixed $n\ge N$ and each $r$, another application of [F16] as $k\to \mathrm{\infty }$ gives ${\int }_{\mathbb{T}}\mathrm{\mid }{f}_n(r\zeta )-F(r\zeta ){\mathrm{\mid }}^{2}dm(\zeta )\le {\epsilon }^2$. Taking the supremum over $r$ proves $f_n\to F$ in $H^2$.

By [F3] applied to $f_n-F$, its boundary traces converge in $L^2$ to $F^{\ast }$. Since $f_n^{\ast }\to h$ as well and $L^2$ limits are unique, $h=F^{\ast }$. This proves that the boundary-value range is closed.

The neighbourhood-holomorphic traces in step 1.2 lie in the generating trace space of [F6], and step 1.3 shows every generator lies in the boundary-value range. Step 1.2 gives density of the smaller trace space in that range, and step 5.1 proves the range is closed. Therefore the closure defining $H^2(\mathrm{\partial }\mathbb{D},m)$ equals the boundary-value range; step 1.1 makes $f\mapsto f^{\ast }$ an isometric isomorphism onto it.

Fix $w\in \mathbb{D}$. If $w=0$, $s_w(z)=1$; if $w\mathrm{\ne }0$, choose $R=(1+1\mathrm{/}\mathrm{\mid }w\mathrm{\mid })\mathrm{/}2$, so $1<R<1\mathrm{/}\mathrm{\mid }w\mathrm{\mid }$ and $\mathrm{\mid }\bar{w}z\mathrm{\mid }<\mathrm{\mid }w\mathrm{\mid }R=(1+\mathrm{\mid }w\mathrm{\mid })\mathrm{/}2<1$ for $\mathrm{\mid }z\mathrm{\mid }<R$, hence $1-\bar{w}z\mathrm{\ne }0$ by [F10]. Thus $s_w(z):=1\mathrm{/}(1-\bar{w}z)$ is holomorphic on a neighbourhood of $\bar{\mathbb{D}}$ by [F13]. Step 1.3 identifies its trace $S_w(\zeta )=s_w(\zeta )$ with $s_w^{\ast }$ in the boundary-value range, and step 6.1 puts it in the boundary space. For $\zeta \in \mathbb{T}$, [F1] and [F10] give $\mathrm{\mid }1-\bar{w}\zeta \mathrm{\mid }\ge 1-\mathrm{\mid }w\mathrm{\mid }$, so $\mathrm{\mid }{S}_w(\zeta )\mathrm{\mid }\le (1-\mathrm{\mid }w\mathrm{\mid }{)}^{-1}$, hence $\mathrm{\parallel }{S}_w{\mathrm{\parallel }}_2\le (1-\mathrm{\mid }w\mathrm{\mid }{)}^{-1}$. For $f\in H^2\subseteq H^1$, [F4] gives $f(w)=⟨f^{\ast },S_w⟩$; [F5] and [F3] then give the stated bounded-evaluation estimate and reproducing identity.

For every generating trace $\mathrm{tr}G$ from [F6], step 1.3 gives $G\in H^2$ and $G^{\ast }=G{\mathrm{\mid }}_{\mathbb{T}}$, so step 7.1 shows $G(w)=⟨\mathrm{tr}G,S_w⟩$. Thus trace evaluation is well defined and bounded on the generating space, and $E_w(h):=⟨h,S_w⟩$ is its continuous extension to the closure, as required by [F17]. By step 6.1 every $h$ in that closure is $f^{\ast }$ for a unique $f\in H^2$; step 7.1 gives $E_z(h)=f(z)$, which is holomorphic in $z$. In particular $S_w=s_w^{\ast }$, so $E_z(S_w)=s_w(z)$. The boundary range is closed by step 5.1 in the Hilbert space [F5], hence it is a Hilbert space; [F7] and [F18] identify $S_w$ as the unique representer used in the kernel definition. Therefore $S(z,w)=E_z(S_w)=s_w(z)=\frac{1}{1-z\bar{w}},$ proving Szegő regularity and the claimed normalization.