An outer function with a prescribed power of a vanishing modulus

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.