Analytic Hardy Spaces and Canonical Factorisation: Examples and Counterexamples

Rationality and holomorphy near the closed disc. Each factor $b_a$ is a quotient of the linear functions $a-z$ and $1-\bar{a}z$ times the constant $\bar{a}\mathrm{/}\mathrm{\mid }a\mathrm{\mid }$ (with $b_0(z)=z$), and its denominator is zero-free on $\bar{\mathbb{D}}$ because $\mathrm{\mid }\bar{a}z\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }<1$. Hence each $b_{a_j}$ is holomorphic on a neighbourhood of $\bar{\mathbb{D}}$, and the finite product $B$ is a rational function holomorphic on such a neighbourhood.

Boundary modulus and value at the origin. By [F1] and [F3], $\mathrm{\mid }B(\zeta )\mathrm{\mid }={\prod }_j\mathrm{\mid }{b}_{a_j}(\zeta )\mathrm{\mid }=1$ for $\zeta \in \mathbb{T}$, and $B(0)={\prod }_j{b}_{a_j}(0)={\prod }_j\mathrm{\mid }{a}_j\mathrm{\mid }$.

Zeros. The zeros of a finite product are the union of the zeros of its factors with multiplicity; by [F1] the zero of $b_{a_j}$ in $\mathbb{D}$ is exactly $a_j$, and $b_{a_j}$ has no other zero in $\mathbb{D}$. Hence the zeros of $B$ in $\mathbb{D}$ are exactly $a_1,\dots ,a_N$ with multiplicity.

Strict decrease of the modulus when $N\ge 1$. Assume $N\ge 1$ and suppose $\mathrm{\mid }B(z_0)\mathrm{\mid }=1$ for some $z_0\in \mathbb{D}$. Since $\mathrm{\mid }B\mathrm{\mid }\le {\prod }_j{\sup }_{\mathbb{D}}\mathrm{\mid }{b}_{a_j}\mathrm{\mid }\le 1$ on $\mathbb{D}$, $\mathrm{\mid }B\mathrm{\mid }$ has at $z_0$ a maximum equal to $1$, so $B$ is constant by [F2]; a constant value of modulus $1$ would give $\mathrm{\mid }B(0)\mathrm{\mid }=1$, but $\mathrm{\mid }B(0)\mathrm{\mid }={\prod }_j\mathrm{\mid }{a}_j\mathrm{\mid }<1$ because $N\ge 1$ and every $\mathrm{\mid }{a}_j\mathrm{\mid }<1$, a contradiction. Hence $\mathrm{\mid }B(z)\mathrm{\mid }<1$ for every $z\in \mathbb{D}$ whenever $N\ge 1$.

The explicit case $a_1=1\mathrm{/}2$, $a_2=-1\mathrm{/}2$. Here $\overline{(1\mathrm{/}2)}\mathrm{/}\mathrm{\mid }1\mathrm{/}2\mathrm{\mid }=1$, so $b_{1\mathrm{/}2}(z)={\phi }_{1\mathrm{/}2}(z)=\frac{1\mathrm{/}2-z}{1-z\mathrm{/}2}$, while $\overline{(-1\mathrm{/}2)}\mathrm{/}\mathrm{\mid }-1\mathrm{/}2\mathrm{\mid }=-1$, so $b_{-1\mathrm{/}2}(z)=-\frac{-1\mathrm{/}2-z}{1+z\mathrm{/}2}=\frac{1\mathrm{/}2+z}{1+z\mathrm{/}2}$. Multiplying and expanding $(1\mathrm{/}2-z)(1\mathrm{/}2+z)=1\mathrm{/}4-z^2$ and $(1-z\mathrm{/}2)(1+z\mathrm{/}2)=1-z^2\mathrm{/}4$ gives $B(z)=\frac{1\mathrm{/}4-z^2}{1-z^2\mathrm{/}4}$, whence $B(0)=1\mathrm{/}4$, $B(1)=\frac{-3\mathrm{/}4}{3\mathrm{/}4}=-1$ and $B(-1)=-1$, in agreement with $\mathrm{\mid }B(1)\mathrm{\mid }=\mathrm{\mid }B(-1)\mathrm{\mid }=1$.

The sequence is Blaschke. Since $\mathrm{\mid }{a}_n\mathrm{\mid }=1-1\mathrm{/}{n}^2$, one has $1-\mathrm{\mid }{a}_n\mathrm{\mid }=1\mathrm{/}{n}^2$ and ${\sum }_{n\ge 2}1\mathrm{/}{n}^2<\mathrm{\infty }$ by [F2]; by [F1] the product $B$ converges normally, is holomorphic with $\mathrm{\mid }B\mathrm{\mid }\le 1$, has exactly the simple zeros $1-1\mathrm{/}{n}^2$, and has $\mathrm{\mid }{B}^{\ast }\mathrm{\mid }=1$ $m$-almost everywhere.

Value at the origin. By [F1], $B(0)={\prod }_{n\ge 2}\mathrm{\mid }{a}_n\mathrm{\mid }={\prod }_{n\ge 2}(1-\frac{1}{n^2})$, and the finite products telescope: using $1-\frac{1}{n^2}=\frac{(n-1)(n+1)}{n^2}$, ${\prod }_{n=2}^N(1-\frac{1}{n^2})=\frac{(N-1)!\cdot (N+1)!\mathrm{/}2}{(N!{)}^2}=\frac{N+1}{2N}⟶\frac{1}{2}\mathrm{.}$

No continuous extension. The zeros $a_n=1-1\mathrm{/}{n}^2$ converge to $1\in \mathbb{T}$. If $B$ had a continuous extension to $\bar{\mathbb{D}}$, then along $a_n\to 1$ one would get $\mathrm{\mid }B(1)\mathrm{\mid }={\lim }_n\mathrm{\mid }B(a_n)\mathrm{\mid }=0$, while on the other hand the boundary values of the extension agree $m$-almost everywhere with $B^{\ast }$, so the continuous function $\mathrm{\mid }B\mathrm{\mid }$ restricted to $\mathbb{T}$ equals $1$ on a set of full measure, hence equals $1$ everywhere by continuity; at $\zeta =1$ this gives $\mathrm{\mid }B(1)\mathrm{\mid }=1$, a contradiction. Thus $B$ has no continuous extension to $\bar{\mathbb{D}}$; in particular $B$ is not a finite Blaschke product, since a finite product would extend continuously.

The $H^2$ norm. Since $\mathrm{\mid }B\mathrm{\mid }\le 1$ and $B(r\zeta )\to B^{\ast }(\zeta )$ for $m$-almost every $\zeta$ (the nontangential limits of [F1] in particular give radial limits almost everywhere), [F3] gives ${\int }_{\mathbb{T}}\mathrm{\mid }B(r\zeta ){\mathrm{\mid }}^{2}dm(\zeta )\to {\int }_{\mathbb{T}}\mathrm{\mid }{B}^{\ast }{\mathrm{\mid }}^{2}dm=1$ as $r\uparrow 1$. The radial means are nondecreasing in $r$ by [F3], so their supremum is the limit, that is, $\mathrm{\parallel }B{\mathrm{\parallel }}_{H^2}^2=1$ and hence $\mathrm{\parallel }B{\mathrm{\parallel }}_{H^2}=1$.

The measure and the kernel value. By [F1], ${\delta }_1$ is a finite positive measure singular with respect to $m$; evaluating the kernel at $\zeta =1$ gives $K(z,1)=\frac{1+z}{1-z}$, so $S(z)=\exp (-K(z,1))=\exp (-\frac{1+z}{1-z})$.

Modulus, value at the origin, zero-freeness. By [F2], $\mathrm{Re}K(z,1)=P(z,1)=\frac{1-\mathrm{\mid }z{\mathrm{\mid }}^2}{\mathrm{\mid }1-z{\mathrm{\mid }}^2}$, hence $\mathrm{\mid }S(z)\mathrm{\mid }=e^{-P(z,1)}\le 1$ and $S(0)=e^{-P(0,1)}=e^{-1}>0$; $S$ is an exponential, hence zero-free.

Cones at $1$. Let $A>1$ and $z\in {\mathrm{\Gamma }}_A(1)$. By [F4], $\mathrm{\mid }1-z\mathrm{\mid }\le A(1-\mathrm{\mid }z\mathrm{\mid })$, so $P(z,1)=\frac{1-\mathrm{\mid }z{\mathrm{\mid }}^2}{\mathrm{\mid }1-z{\mathrm{\mid }}^2}\ge \frac{(1-\mathrm{\mid }z\mathrm{\mid })(1+\mathrm{\mid }z\mathrm{\mid })}{A^2(1-\mathrm{\mid }z\mathrm{\mid }{)}^2}=\frac{1+\mathrm{\mid }z\mathrm{\mid }}{A^2(1-\mathrm{\mid }z\mathrm{\mid })}⟶+\mathrm{\infty }$ as $\mathrm{\mid }z\mathrm{\mid }\to 1$; therefore $\mathrm{\mid }S(z)\mathrm{\mid }=e^{-P(z,1)}\to 0$ and hence $S(z)\to 0$ along every cone at $1$. Combined with the a.e. boundary values of [F3] (which give $\mathrm{\mid }{S}^{\ast }\mathrm{\mid }=1$ a.e. because ${\delta }_1\perp m$) and the fact that $S^{\ast }(\zeta )=1$-modulus for every $\zeta \mathrm{\ne }1$ (where the exponent $-\frac{1+z}{1-z}$ has a finite limit), the boundary modulus of $S$ fails to be $1$ exactly on the null set $\{1\}$.

The logarithm $\log \mathrm{\mid }1-\zeta \mathrm{\mid }$ is integrable with zero mean. By [F2] and [F5], ${\int }_{\mathbb{T}}\log \mathrm{\mid }1-\zeta \mathrm{\mid }dm={\lim }_{r\uparrow 1}{\int }_{\mathbb{T}}\log \mathrm{\mid }1-r\zeta \mathrm{\mid }dm=0$. Moreover $\log \mathrm{\mid }1-\zeta \mathrm{\mid }\in L^1$: its positive part is bounded by $\log 2$, and its negative part is integrable by the bound of [F5].

Fourier coefficients and the holomorphic kernel. For $0<r<1$, the power series in [F1] shows that $\log \mathrm{\mid }1-r\zeta \mathrm{\mid }$ has Fourier coefficients $-r^n\mathrm{/}(2n)$ at $\pm n$ and zero mean. By the $L^1$ limit in [F5], the function $u(\zeta )=\log \mathrm{\mid }1-\zeta \mathrm{\mid }$ has $\widehat{u}(0)=0$ and $\widehat{u}(\pm n)=-1\mathrm{/}(2n)$. The holomorphic kernel is $K(z,\zeta )=1+2{\sum }_{n\ge 1}{z}^n{\zeta }^{-n}$, uniformly convergent in $\zeta$ for fixed $\mathrm{\mid }z\mathrm{\mid }<1$. Multiplication by $u\in L^1$ and termwise integration are legitimate under the uniform convergence, so ${\int }_{\mathbb{T}}K(z,\zeta )u(\zeta )dm(\zeta )=\widehat{u}(0)+2{\sum }_{n\ge 1}{z}^n\widehat{u}(n)=-{\sum }_{n\ge 1}{z}^n\mathrm{/}n=\log (1-z)$, the holomorphic branch with value zero at the origin. Its real part is $\log \mathrm{\mid }1-z\mathrm{\mid }$.

The outer function is $(1-z{)}^{\alpha }$. Since $\log h=\alpha \log \mathrm{\mid }1-\zeta \mathrm{\mid }\in L^1$ by step 1.1, the outer function is well defined and, by step 2.1, $[h](z)=\exp ({\int }_{\mathbb{T}}K(z,\zeta )\log h(\zeta )dm(\zeta ))=\exp (\alpha \log (1-z))=(1-z{)}^{\alpha },$ the principal branch, with $[h](0)=1$.

Modulus and $H^{\mathrm{\infty }}$ membership. Since $0\le h\le 2^{\alpha }$, the function $h$ lies in $L^{\mathrm{\infty }}(\mathbb{T},m)\subseteq L^p$ for every $p$, and [F4] gives $[h]\in H^{\mathrm{\infty }}$ with $\mathrm{\mid }[h]\mathrm{\mid }\le 2^{\alpha }$ and $\mathrm{\mid }[h{]}^{\ast }\mathrm{\mid }=h$ a.e.; explicitly $\mathrm{\mid }(1-z{)}^{\alpha }\mathrm{\mid }=\mathrm{\mid }1-z{\mathrm{\mid }}^{\alpha }\le 2^{\alpha }$ on $\mathbb{D}$ with equality along $z\to -1$, so $\mathrm{\parallel }[h]{\mathrm{\parallel }}_{\mathrm{\infty }}=2^{\alpha }$. On the boundary, for every $\zeta \mathrm{\ne }1$ the principal branch is continuous and $\mathrm{\mid }(1-\zeta {)}^{\alpha }\mathrm{\mid }=\mathrm{\mid }1-\zeta {\mathrm{\mid }}^{\alpha }=h(\zeta )$; combined with the a.e. identity this gives $\mathrm{\mid }[h{]}^{\ast }\mathrm{\mid }=h$ off the single point $1$.

Non-rationality for non-integer $\alpha$. If $\alpha \mathrm{\notin }\mathbb{N}$ and $(1-z{)}^{\alpha }$ agreed on $\mathbb{D}$ with a rational function $R$, then $R$ has a finite integer order $m$ at $1$, obtained by factoring its numerator and denominator into their powers of $z-1$. Because $R(x)=(1-x{)}^{\alpha }\to 0$ along real $x\uparrow 1$, this order is positive, so $R$ is holomorphic near $1$. But near $z=1$ the principal branch behaves as $(1-z{)}^{\alpha }$, which is not of the form $(1-z{)}^{m}g(z)$ with $m\in \mathbb{Z}$ and $g$ holomorphic and nonzero at $1$ unless $\alpha \in \mathbb{Z}$ (compare the growth of $(1-z{)}^{\alpha -m}$ along real $z\to 1^{-}$: it tends to $0$ for $\alpha >m$ and to $+\mathrm{\infty }$ for $\alpha <m$); a rational function has such a finite-order behaviour at each of its singularities, so $\alpha \in \mathbb{N}$, a contradiction.

Part (a). The function $f(z)=1-z$ is a polynomial, hence holomorphic, with $\mathrm{\mid }f\mathrm{\mid }\le 2$ on $\mathbb{D}$, so $f\in H^{\mathrm{\infty }}(\mathbb{D})$; it is not identically zero. Its radial limits are $f^{\ast }(\zeta )=1-\zeta$, continuous on $\mathbb{T}$ and vanishing exactly at $\zeta =1$ by [F5], a set of measure zero.

Part (b). Let $f\in H^p$ with $\{f^{\ast }=0\}$ of positive measure. If $f\equiv ̸0$, then [F1] gives $\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }\in L^1$, so $f^{\ast }\mathrm{\ne }0$ $m$-almost everywhere, contradicting positive measure of the zero set; hence $f\equiv 0$.

Part (c). Let $f,g\in H^p$ with $f^{\ast }=g^{\ast }$ a.e. By [F2], the difference $f-g$ lies in $H^{p^{\mathrm{\prime }}}$ for some $0<p^{\mathrm{\prime }}\le \mathrm{\infty }$ (take $p^{\mathrm{\prime }}=p$); its boundary function $(f-g{)}^{\ast }=f^{\ast }-g^{\ast }$ vanishes a.e., a set of full measure. If $f-g\equiv ̸0$, then by step 1.2 applied to $f-g$ its zero set would have to be null, contradicting that it has full measure; hence $f=g$.

Part (d): $1-z$ is outer. By [F3] and [F4], $1-z=[\mathrm{\mid }1-\zeta \mathrm{\mid }]$; since $[h](0)=1>0$ and $e^{i\gamma }$ is normalized by the value at the origin, the unimodular constant is $1$. Hence $1-z$ is outer, and its canonical factorization $1-z=\lambda B{S}_{\mu }F$ has $B=1$ (no zeros in $\mathbb{D}$), $S_{\mu }=1$ (no singular factor for an outer function) and $F=1-z$ with $\lambda =1$.

Assembly. Step 1.1 proves (a), step 1.2 proves (b), step 2.1 proves the uniqueness statement (c), and step 2.2 identifies $1-z$ as the outer function $[\mathrm{\mid }1-\zeta \mathrm{\mid }]$ with the stated trivial canonical factors, proving (d).