Analytic Hardy Spaces and Canonical Factorisation

Assume countable choice. Let $\mathbb{T}=\mathbb{R}\mathrm{/}\mathbb{Z}$ be the one-dimensional torus, identified with the Euclidean unit circle through $\phi ([t])=e^{2\pi it}$, and let $m$ be its normalized Haar measure, a probability measure on the compact metric space $\mathbb{T}$ (The one-dimensional torus and its normalized Haar integral). Let $\mathbb{D}:=\{z\in \mathbb{C}:\mathrm{\mid }z\mathrm{\mid }<1\}$ be the unit disc of The unit disc, the upper half-plane, and Blaschke factors, and let $f:\mathbb{D}\to \mathbb{C}$ be a function, holomorphic in the sense of Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions when required below. For $0\le r<1$ write $f_r:\mathbb{T}\to \mathbb{C},f_r(\zeta ):=f(r\zeta )\mathrm{.}$ If $f$ is holomorphic, each $f_r$ is continuous, hence measurable with finite modulus; for $0<p<\mathrm{\infty }$ the number $({\int }_{\mathbb{T}}\mathrm{\mid }f(r\zeta ){\mathrm{\mid }}^{p}dm(\zeta ){)}^{1\mathrm{/}p}$ is therefore a well-defined element of $[0,+\mathrm{\infty })$, computed from the continuous representative of $f_r$ in the conventions of Complex Lp classes and Euclidean test-function conventions.

The classes ${\mathbf{H}}^{\mathbf{p}}(\mathbb{D})$. For $0<p<\mathrm{\infty }$ define $H^p(\mathbb{D}):=\{f\text{ holomorphic on }\mathbb{D}:\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}:={\sup }_{0\le r<1}({\int }_{\mathbb{T}}\mathrm{\mid }f(r\zeta ){\mathrm{\mid }}^{p}dm(\zeta ){)}^{1\mathrm{/}p}<+\mathrm{\infty }\},$ and for $p=\mathrm{\infty }$ define $H^{\mathrm{\infty }}(\mathbb{D}):=\{f\text{ holomorphic on }\mathbb{D}:\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}:={\sup }_{z\in \mathbb{D}}\mathrm{\mid }f(z)\mathrm{\mid }<+\mathrm{\infty }\}\mathrm{.}$ Writing $u_r(\zeta ):=u(r\zeta )$ for any function $u$ on $\mathbb{D}$, the definition of $H^{\mathrm{\infty }}$ is equivalently $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_{0\le r<1}{\mathrm{ess\hspace{0.17em}sup}}_{\mathbb{T}}\mathrm{\mid }{f}_r\mathrm{\mid }$: a continuous function on the compact space $\mathbb{T}$ has the same supremum and essential supremum, because a nonempty open subset of $\mathbb{T}$ has positive $m$-measure, and every $z\in \mathbb{D}$ has the form $r\zeta$ with $r=\mathrm{\mid }z\mathrm{\mid }$ and $\zeta \in \mathbb{T}$ (The essential supremum of a measurable function with respect to a measure, The one-dimensional torus and its normalized Haar integral).

The (quasi-)norm assertions. For $1\le p\le \mathrm{\infty }$, $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{H^p}$ is a norm on the complex vector space $H^p(\mathbb{D})$. Homogeneity $\mathrm{\parallel }\lambda f{\mathrm{\parallel }}_{H^p}=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}$ for $\lambda \in \mathbb{C}$ and the triangle inequality $\mathrm{\parallel }f+g{\mathrm{\parallel }}_{H^p}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}+\mathrm{\parallel }g{\mathrm{\parallel }}_{H^p}$ follow by taking suprema over $r$ of the corresponding statements for the $L^p$ classes of the continuous functions $f_r,g_r$ on the probability space $(\mathbb{T},m)$ (Complex Holder, Minkowski, and the quotient norm); the case $p=\mathrm{\infty }$ is the elementary inequality between suprema of moduli, and $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_{r<1}{\sup }_{\zeta }\mathrm{\mid }f(r\zeta )\mathrm{\mid }$ holds because the radii and circle points range over $\mathbb{D}$. All $H^p(\mathbb{D})$ are closed under finite linear combinations: sums and scalar multiples of holomorphic functions are holomorphic (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions), and the (quasi-)norm of a finite linear combination is finite by the inequalities just stated.

For $0<p<1$, $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{H^p}$ is a quasi-norm, not a norm. For complex numbers $a,b$ one has $\mathrm{\mid }a+b{\mathrm{\mid }}^p\le \mathrm{\mid }a{\mathrm{\mid }}^p+\mathrm{\mid }b{\mathrm{\mid }}^p$, because the power function is subadditive on $[0,\mathrm{\infty })$ when $0<p\le 1$: for $x,y\ge 0$ it suffices to prove $(x+y{)}^p\le x^p+y^p$, which is trivial when $y=0$, and for $y>0$ and $t=x\mathrm{/}y\ge 0$ is the claim $(1+t{)}^p\le 1+t^p$; the function $h(t):=1+t^p-(1+t{)}^p$ is continuous on $[0,\mathrm{\infty })$ with $h(0)=0$ and, for $t>0$, $h^{\mathrm{\prime }}(t)=p(t^{p-1}-(1+t{)}^{p-1})\ge 0$ because $p-1\le 0$ and $t\le 1+t$, so $h$ is nondecreasing and $h\ge 0$. Integrating at each radius gives ${\int }_{\mathbb{T}}\mathrm{\mid }f(r\zeta )+g(r\zeta ){\mathrm{\mid }}^{p}dm(\zeta )\le {\int }_{\mathbb{T}}\mathrm{\mid }f(r\zeta ){\mathrm{\mid }}^{p}dm(\zeta )+{\int }_{\mathbb{T}}\mathrm{\mid }g(r\zeta ){\mathrm{\mid }}^{p}dm(\zeta ),$ and taking suprema over $r$ yields $\mathrm{\parallel }f+g{\mathrm{\parallel }}_{H^p}^p\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}^p+\mathrm{\parallel }g{\mathrm{\parallel }}_{H^p}^p$. Writing $A:=\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}$, $B:=\mathrm{\parallel }g{\mathrm{\parallel }}_{H^p}$ and $x:=A\mathrm{/}(A+B)$ when $A+B>0$, the bound $x^p+(1-x{)}^p\le 2^{1-p}$ for $0\le x\le 1$ (the maximum of the concave left side is at $x=1\mathrm{/}2$) gives $A^p+B^p\le 2^{1-p}(A+B{)}^p$, hence $\mathrm{\parallel }f+g{\mathrm{\parallel }}_{H^p}\le 2^{\frac{1}{p}-1}(\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}+\mathrm{\parallel }g{\mathrm{\parallel }}_{H^p}),$ the quasi-norm statement; homogeneity and the case $A+B=0$ are immediate. The quasi-norm inequality for $p<1$ is the only place where the constant $2^{1\mathrm{/}p-1}$ enters, and no further structure (completeness, separability, duality) of these spaces is asserted here or used later.

The ordinary triangle inequality does fail for each $0<p<1$. Put $f_n(z)=(1+z{)}^n$, $g_n(z)=(1-z{)}^n$ for positive integers $n$, holomorphic polynomials (Complex polynomials are entire with the power-rule derivative, and rational functions are holomorphic wherever their denominator is nonzero). Any polynomial $h$ is bounded on the closed disc, hence lies in $H^p$; the radial-means certifier Radial p-means of a holomorphic function are nondecreasing ↗ and uniform boundary continuity give $\mathrm{\parallel }h{\mathrm{\parallel }}_{H^p}^p={\int }_{\mathbb{T}}\mathrm{\mid }h(\zeta ){\mathrm{\mid }}^{p}dm$. Write $A_n=\int \mathrm{\mid }{f}_n{\mathrm{\mid }}^{p}dm=\int \mathrm{\mid }{g}_n{\mathrm{\mid }}^{p}dm$, using Haar invariance under $\zeta \mapsto -\zeta$, and $D_n=\int \min (\mathrm{\mid }{f}_n{\mathrm{\mid }}^p,\mathrm{\mid }{g}_n{\mathrm{\mid }}^p)dm$. The already proved subadditivity gives $\mathrm{\mid }u+v{\mathrm{\mid }}^p\ge \max (\mathrm{\mid }u{\mathrm{\mid }}^p,\mathrm{\mid }v{\mathrm{\mid }}^p)-\min (\mathrm{\mid }u{\mathrm{\mid }}^p,\mathrm{\mid }v{\mathrm{\mid }}^p)$, hence $\int \mathrm{\mid }{f}_n+g_n{\mathrm{\mid }}^{p}dm\ge 2A_n-2D_n$. Since $\mathrm{\mid }1+\zeta {\mathrm{\mid }}^2+\mathrm{\mid }1-\zeta {\mathrm{\mid }}^2=4$, $D_n\le 2^{np\mathrm{/}2}$. The open set $U=\{\zeta \in \mathbb{T}:\mathrm{\mid }1+\zeta \mathrm{\mid }>\sqrt{3}\}$ contains $1$, so $c=m(U)>0$ and $A_n\ge c{3}^{np\mathrm{/}2}$ (The one-dimensional torus and its normalized Haar integral). Thus $D_n\mathrm{/}{A}_n\le c^{-1}((2\mathrm{/}3{)}^{p\mathrm{/}2}{)}^n\to 0$ (For $\mathrm{\mid }r\mathrm{\mid }<1$ the sequence $r^k$ is null, and for $\mathrm{\mid }r\mathrm{\mid }>1$ the sequence $\mathrm{\mid }r{\mathrm{\mid }}^k$ diverges to $+\mathrm{\infty }$). For sufficiently large $n$, $D_n\mathrm{/}{A}_n<1-2^{p-1}$, so $\mathrm{\parallel }{f}_n+g_n{\mathrm{\parallel }}_{H^p}^p>2^p{A}_n$ and $\mathrm{\parallel }{f}_n+g_n{\mathrm{\parallel }}_{H^p}>2A_n^{1\mathrm{/}p}=\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_{H^p}+\mathrm{\parallel }{g}_n{\mathrm{\parallel }}_{H^p}$. This proves the claimed failure without any later factorization or outer-function existence result.

For $p=\mathrm{\infty }$, $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}=0$ immediately gives $f\equiv 0$. For $0<p<\mathrm{\infty }$, if $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}=0$, then ${\int }_{\mathbb{T}}\mathrm{\mid }f(r\zeta ){\mathrm{\mid }}^{p}dm(\zeta )=0$ for every $0\le r<1$, so $\mathrm{\mid }{f}_r{\mathrm{\mid }}^p=0$ $m$-almost everywhere (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere), hence $f_r=0$ everywhere since $f_r$ is continuous; taking $r=\mathrm{\mid }z\mathrm{\mid }$ for $z\mathrm{\ne }0$ and $r=0$ gives $f\equiv 0$. Thus $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{H^p}$ is a genuine norm on $H^p(\mathbb{D})$ for $1\le p\le \mathrm{\infty }$ and a genuine quasi-norm for $0<p<1$.

This item defines the classes and their (quasi-)norms only. The monotonicity of the radial $p$-means in $r$, and with it the containments $H^q(\mathbb{D})\subseteq H^p(\mathbb{D})$ for $q>p$ together with $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^q}$, are proved in Radial p-means of a holomorphic function are nondecreasing ↗, which is the item that certifies this definition. No choice principle beyond countable choice is used.

Let $f:\mathbb{D}\to \mathbb{C}$ be holomorphic and let $0<p<\mathrm{\infty }$. For all $0<r\le R<1$, ${\int }_{\mathbb{T}}\mathrm{\mid }f(r\zeta ){\mathrm{\mid }}^{p}dm(\zeta )\le {\int }_{\mathbb{T}}\mathrm{\mid }f(R\zeta ){\mathrm{\mid }}^{p}dm(\zeta )\mathrm{.}$ Consequently, for $f\in H^p(\mathbb{D})$ the nondecreasing radial means satisfy $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}^p={\lim }_{r\uparrow 1}{\int }_{\mathbb{T}}\mathrm{\mid }f(r\zeta ){\mathrm{\mid }}^{p}dm(\zeta ),$ so that $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}={\lim }_{r\uparrow 1}\mathrm{\parallel }{f}_r{\mathrm{\parallel }}_{L^p}$. Moreover, for $0<p<q\le \mathrm{\infty }$ one has $H^q(\mathbb{D})\subseteq H^p(\mathbb{D})$ and $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^q}$ for every $f\in H^q(\mathbb{D})$.

Let $f$ be holomorphic on $\mathbb{D}$ with $f\equiv ̸0$, and suppose that ${\mathrm{lim\hspace{0.17em}inf}}_{r\uparrow 1}{\int }_{\mathbb{T}}\log \mathrm{\mid }f(r\zeta )\mathrm{\mid }dm(\zeta )<+\mathrm{\infty },$ where $\log \mathrm{\mid }f\mathrm{\mid }=-\mathrm{\infty }$ at the zeros of $f$. This hypothesis holds in particular for every $f\in H^p(\mathbb{D})$, $0<p\le \mathrm{\infty }$: the logarithmic Jensen inequality and Radial p-means of a holomorphic function are nondecreasing give ${\int }_{\mathbb{T}}\log \mathrm{\mid }{f}_r\mathrm{\mid }dm\le \log \mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}$ for $0<p<\mathrm{\infty }$ and ${\int }_{\mathbb{T}}\log \mathrm{\mid }{f}_r\mathrm{\mid }dm\le \log \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ for $p=\mathrm{\infty }$. Let $(a_n{)}_{n\ge 1}$ be the zeros of $f$ in $\mathbb{D}$ repeated according to multiplicity. Then $f$ has finite vanishing order $m\ge 0$ at the origin (with $m=0$ when $f(0)\mathrm{\ne }0$), and the nonzero zeros satisfy ${\sum }_{n:a_n\mathrm{\ne }0}\log \frac{1}{\mathrm{\mid }{a}_n\mathrm{\mid }}<+\mathrm{\infty },\text{hence}{\sum }_{n\ge 1}(1-\mathrm{\mid }{a}_n\mathrm{\mid })<+\mathrm{\infty }\mathrm{.}$ (The second inequality uses $\log (1\mathrm{/}x)\ge 1-x$ for $0<x\le 1$. The finite vanishing order at the origin contributes finitely many terms equal to $1$ to the second sum and does not affect convergence of the nonzero part.)

For $a\in \mathbb{D}$, let ${\phi }_a(z)=\frac{a-z}{1-\bar{a}z}$ be the published Blaschke factor of The unit disc, the upper half-plane, and Blaschke factors, a biholomorphic self-map of $\mathbb{D}$ by Blaschke factors are automorphisms of the disc. Define the normalized Blaschke factor $b_a(z):=\frac{\bar{a}}{\mathrm{\mid }a\mathrm{\mid }}{\phi }_a(z)(a\mathrm{\ne }0),b_0(z):=z\mathrm{.}$ Since $\mathrm{\mid }\bar{a}\mathrm{/}\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }=1$, each $b_a$ is holomorphic on an open neighbourhood of the closed disc (the denominator $1-\bar{a}z$ does not vanish there), $b_a$ is zero-free on $\mathbb{D}$ except for the simple zero at $a$ (the zero of ${\phi }_a$), and $b_a(0)=\frac{\bar{a}}{\mathrm{\mid }a\mathrm{\mid }}a=\mathrm{\mid }a\mathrm{\mid }\ge 0,\mathrm{\mid }{b}_a(z)\mathrm{\mid }\le 1(z\in \mathbb{D}),$ the last inequality because ${\phi }_a$ maps $\mathbb{D}$ into itself. On the unit circle one has $\mathrm{\mid }{b}_a(\zeta )\mathrm{\mid }=1$ for every $\zeta \in \mathbb{T}$ (identified with the unit circle): for $\mathrm{\mid }\zeta \mathrm{\mid }=1$, $\mathrm{\mid }1-\bar{a}\zeta \mathrm{\mid }=\mathrm{\mid }\zeta \mathrm{\mid }\cdot \mathrm{\mid }\bar{\zeta }-\bar{a}\mathrm{\mid }=\mathrm{\mid }a-\zeta \mathrm{\mid }$ by Conjugation is an involutive real-field automorphism, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, and modulus is definite, multiplicative, and subadditive and Real and imaginary parts, complex conjugation, and modulus, so $\mathrm{\mid }{\phi }_a(\zeta )\mathrm{\mid }=1$.

Blaschke sequences and products. A sequence $(a_n{)}_{n\ge 1}$ in $\mathbb{D}$ is a Blaschke sequence if ${\sum }_n(1-\mathrm{\mid }{a}_n\mathrm{\mid })<+\mathrm{\infty }$; its Blaschke product is the holomorphic product $B(z):={\prod }_{n\ge 1}{b}_{a_n}(z),$ defined as the locally uniform limit of the partial products $B_N:={\prod }_{n\le N}{b}_{a_n}$; for the empty sequence set $B:=1$. This definition is meaningful because the product converges normally on $\mathbb{D}$ in the sense of Normal convergence of holomorphic products: a Blaschke sequence satisfies $\mathrm{\mid }{a}_n\mathrm{\mid }\to 1$, so for every compact $K\subseteq \mathbb{D}$ only finitely many $a_n$ lie in $K$, and for every $a\in \mathbb{D}\setminus \{0\}$ and $\mathrm{\mid }z\mathrm{\mid }\le r<1$ the identity $1-b_a(z)=(1-\mathrm{\mid }a\mathrm{\mid })\frac{1+\frac{\bar{a}}{\mathrm{\mid }a\mathrm{\mid }}z}{1-\bar{a}z},\mathrm{\mid }1-b_a(z)\mathrm{\mid }\le (1-\mathrm{\mid }a\mathrm{\mid })\frac{1+r}{1-r}$ holds: the identity is a direct computation from $b_a=\frac{\bar{a}}{\mathrm{\mid }a\mathrm{\mid }}\frac{a-z}{1-\bar{a}z}$ using $\mathrm{\mid }a{\mathrm{\mid }}^2=a\bar{a}$, and the estimate uses $\mathrm{\mid }1+\frac{\bar{a}}{\mathrm{\mid }a\mathrm{\mid }}z\mathrm{\mid }\le 1+r$ and $\mathrm{\mid }1-\bar{a}z\mathrm{\mid }\ge 1-r$. For $a=0$, $b_0(z)=z$ and $\mathrm{\mid }1-b_0(z)\mathrm{\mid }\le 1+r=(1-\mathrm{\mid }0\mathrm{\mid })(1+r)\le (1-\mathrm{\mid }0\mathrm{\mid })(1+r)\mathrm{/}(1-r)$, so the same estimate holds without dividing by $\mathrm{\mid }a\mathrm{\mid }$. Hence ${\sum }_n{\sup }_{\mathrm{\mid }z\mathrm{\mid }\le r}\mathrm{\mid }1-b_{a_n}(z)\mathrm{\mid }\le \frac{1+r}{1-r}{\sum }_n(1-\mathrm{\mid }{a}_n\mathrm{\mid })<+\mathrm{\infty }\mathrm{.}$ The published normal-convergence theorem therefore applies: the partial products converge locally uniformly to a holomorphic $B$ that has exactly the zeros $a_n$ with the multiplicity with which they occur, and satisfies $\mathrm{\mid }B(z)\mathrm{\mid }\le 1$ on $\mathbb{D}$ (Normally convergent products define holomorphic functions with the expected zeros). The limit does not depend on the enumeration of the sequence: the normal-convergence criterion is a condition on the set of factors, and for two enumerations and finite initial segments $A,B$ containing a common block $\{1,\dots ,N\}$ the quotient ${\prod }_{n\in A}{b}_{a_n}\mathrm{/}{\prod }_{n\in B}{b}_{a_n}$ deviates from $1$ by at most a constant times the tail sum ${\sum }_{n>N}{\sup }_{\mathrm{\mid }z\mathrm{\mid }\le r}\mathrm{\mid }1-b_{a_n}(z)\mathrm{\mid }$, which tends to $0$; hence the two partial-product sequences have the same locally uniform limit. The Blaschke sequence is reproduced by The zero set of a Hardy function satisfies the Blaschke condition from the zero set of a Hardy function. The boundary-modulus property $\mathrm{\mid }{B}^{\ast }\mathrm{\mid }=1$ almost everywhere is not part of this definition; it is proved in Boundary values and zeros of a Blaschke product ↗.

Let $(a_n{)}_{n\ge 1}$ be a Blaschke sequence with Blaschke product $B={\prod }_n{b}_{a_n}$ and partial products $B_N={\prod }_{n\le N}{b}_{a_n}$. Then:

(i) $\mathrm{\mid }B(z)\mathrm{\mid }\le 1$ for every $z\in \mathbb{D}$, and the zeros of $B$ are exactly the points $a_n$, with multiplicity;

(ii) for every $N$, $B\mathrm{/}{B}_N$ is the Blaschke product of the tail $(a_n{)}_{n>N}$; it is holomorphic on $\mathbb{D}$, satisfies $\mathrm{\mid }(B\mathrm{/}{B}_N)(z)\mathrm{\mid }\le 1$, and $(B\mathrm{/}{B}_N)(0)={\prod }_{n>N}\mathrm{\mid }{a}_n\mathrm{\mid }$, a product that is eventually positive and tends to $1$ as $N\to \mathrm{\infty }$;

(iii) $B$ has finite nontangential limits $B^{\ast }(\zeta )$ for $m$-almost every $\zeta \in \mathbb{T}$, and $\mathrm{\mid }{B}^{\ast }(\zeta )\mathrm{\mid }=1$ for $m$-almost every $\zeta$. In particular $B\equiv ̸0$.