Inner-outer factorisation of a Hardy-space function

Put $F:=[\mathrm{\mid }{f}^{\ast }\mathrm{\mid }]$. By [L1] and [L2], it is holomorphic, zero-free and outer, with $F(0)>0$, $\mathrm{\mid }{F}^{\ast }\mathrm{\mid }=\mathrm{\mid }{f}^{\ast }\mathrm{\mid }$ almost everywhere, $F\in H^p$ and $\mathrm{\parallel }F{\mathrm{\parallel }}_{H^p}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}$. For finite $p$, [L3] gives $f=Bg$ with $g$ zero-free and $\mathrm{\parallel }g{\mathrm{\parallel }}_{H^p}=\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}$. For $p=\mathrm{\infty }$, apply [L3] with exponent $1$ (bounded $f$ belongs to $H^1$) to obtain the same holomorphic zero-free quotient $g=f\mathrm{/}B$.

For $p=\mathrm{\infty }$, fix $N$ and $0<\epsilon <1$. The holomorphic quotient $g_N=f\mathrm{/}{B}_N$ on every radius-$R$ circle sufficiently near $1$ satisfies $\mathrm{\mid }{g}_N\mathrm{\mid }\le (1-\epsilon {)}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, since $B_N$ is continuous on the closed disc with unit boundary modulus. The boundary maximum principle [L6] gives the same bound inside that circle. Let $\epsilon \downarrow 0$, then $N\to \mathrm{\infty }$ off the zeros of $B$, where $g_N\to g$; continuity extends the bound to those zeros. Thus $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, and $\mathrm{\mid }B\mathrm{\mid }\le 1$ gives equality. In every case [L1] now supplies $g^{\ast }$, so $f^{\ast }=B^{\ast }{g}^{\ast }$ and $\mathrm{\mid }{B}^{\ast }\mathrm{\mid }=1$ yield $\mathrm{\mid }{g}^{\ast }\mathrm{\mid }=\mathrm{\mid }{f}^{\ast }\mathrm{\mid }$ almost everywhere.

The quotient is a zero-free inner function. Let $S:=g\mathrm{/}F$, holomorphic and zero-free because both factors are. By [L4] applied to $g$, $\log \mathrm{\mid }g\mathrm{\mid }\le P[\log \mathrm{\mid }{g}^{\ast }\mathrm{\mid }]=\log \mathrm{\mid }F\mathrm{\mid }$ on $\mathbb{D}$, since $\log \mathrm{\mid }F\mathrm{\mid }=P[\log \mathrm{\mid }{F}^{\ast }\mathrm{\mid }]=P[\log \mathrm{\mid }{g}^{\ast }\mathrm{\mid }]$; hence $\mathrm{\mid }S\mathrm{\mid }\le 1$. On the boundary, $\mathrm{\mid }{S}^{\ast }\mathrm{\mid }=\mathrm{\mid }{g}^{\ast }\mathrm{\mid }\mathrm{/}\mathrm{\mid }{F}^{\ast }\mathrm{\mid }=1$ a.e. By [L5] there are ${\lambda }^{\mathrm{\prime }}\in \mathbb{T}$ and a unique finite positive $\mu \perp m$ with $S={\lambda }^{\mathrm{\prime }}{S}_{\mu }$ and $S_{\mu }(0)>0$.

The factorization and its uniqueness. Substituting gives $f=Bg=BFS={\lambda }^{\mathrm{\prime }}B{S}_{\mu }F$, which is the asserted factorization. Uniqueness: $B$ is determined by the zeros of $f$ by [L3]; $\mathrm{\mid }{F}^{\ast }\mathrm{\mid }=\mathrm{\mid }{f}^{\ast }\mathrm{\mid }$ a.e., $F$ outer and $F(0)>0$ determine $F$ by [L2]; then $S=f\mathrm{/}(BF)$ is determined, and its representation ${\lambda }^{\mathrm{\prime }}{S}_{\mu }$ with $S_{\mu }(0)>0$ is unique by [L5]; hence $\lambda ={\lambda }^{\mathrm{\prime }}$ is determined, and writing $\lambda$ for ${\lambda }^{\mathrm{\prime }}$ gives the stated normalized factorization.

Assembly. Steps 1.1, 3.1 and 4.1 produce the factorization $f=\lambda B{S}_{\mu }F$ with the asserted bounds and prove that the normalized factors $B,F,S_{\mu }$ and the constant $\lambda$ are uniquely determined.