Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 9 / 2

Write a Blaschke product as $B(z)=\lambda z^m{\prod }_n{b}_{a_n}(z)$, where $a_n\ne 0$, $|\lambda |=1$, and the zeros are counted with multiplicity. With normalized angular measure $dm=d\theta /(2\pi )$, Jensen's formula for each Blaschke factor givesIndeed, the mean of $\log |a-r{e}^{i\theta }|$ is $\log \max (r,|a|)$, while $\log |1-\overline{a}r{e}^{i\theta }|$ has mean zero by the mean value property for harmonic functions. Each factor has modulus at most one. Applying the monotone convergence theorem to the negative logarithms of the finite products yields the radial logarithmic mean of a Blaschke product:The Blaschke condition implies ${\sum }_n-\log |a_n|<\infty$: only finitely many zeros have modulus below $1/2$, and $-\log t\le 2(1-t)$ for $1/2\le t<1$. Every summand tends to zero and is bounded in absolute value by $-\log |a_n|$. The dominated convergence theorem for this series, together with $m\log r\to 0$, therefore givesThis also covers finite products and constant unimodular products. Zeros on a circle create integrable logarithmic singularities and do not change the conclusion.

Use the absolute value around the logarithm in the original PDF. Set $v(z)={\log }^{+}|f(z)|=\max (0,\log |f(z)|)$. This is a nonnegative subharmonic function, including at zeros by its continuous extension. Its radial means are nondecreasing: Poisson integral comparison on a larger concentric circle bounds the smaller mean by the larger one. The hypothesis givesConsequently $T(r)=0$ for every $r$, and continuity of $v$ implies $v=0$ throughout the unit disc. Thus $|f|\le 1$ everywhere, which is stronger than the requested conclusion. In particular $f$ has bounded characteristic, because ${\sup }_{r}T(r)<\infty$; it belongs to the Nevanlinna class and to H-infinity space. The hypothesis also excludes the identically zero function, whose absolute logarithmic integral is infinite.

Let $m$ be the multiplicity at zero, and write $f(z)=z^{m}h(z)$ with $h(0)\ne 0$. The proof in solution gives $|f|\le 1$. For radii without boundary zeros, Jensen's formula saysThe absolute logarithmic hypothesis makes the left side tend to zero. The nonnegative zero sum increases to ${\sum }_n\log (1/|a_n|)$ by the monotone convergence theorem, so this series is finite. Hence ${\sum }_n(1-|a_n|)<\infty$, and the Blaschke condition supplies a Blaschke product $B$ with exactly these zeros and their multiplicities.

The quotient $g=f/B$ extends holomorphically across every zero because the multiplicities agree, and the extensions have nonzero values. Thus $f=Bg$ with $g$ zero-free and holomorphic. We will also need the stronger Blaschke factorization of a bounded holomorphic function, namely $|g|\le 1$. Let $B_N$ contain the factor $z^m$ and the first $N$ other zero factors. The quotient $f/B_N$ is holomorphic. On $|z|=R$ enclosing those finitely many zeros, its modulus is at most $1/{\min }_{|z|=R}|B_N(z)|$. The maximum modulus principle transfers this bound to the interior. Since the finite Blaschke product has boundary modulus one continuously, letting $R\uparrow 1$ gives $|f/B_N|\le 1$. Now let $N\to \infty$ using locally uniform convergence of the Blaschke product, first off the zeros and then by continuity at them. This proves $|g|\le 1$.

The zero-free quotient $g$ from solution has a holomorphic logarithm, so $\log |g|$ is harmonic. Its radial mean is the constant $\log |g(0)|$. Subtract the radial logarithmic means of $f$ and $B$. The first tends to zero by the absolute logarithmic hypothesis; the second tends to zero by solution. It follows thatSince $|g|\le 1$ and $|g(0)|=1$, the maximum modulus principle makes $g$ a constant of modulus one. Absorb that constant into the unimodular factor of $B$. Therefore $f$ itself is a Blaschke product. This proves the vanishing absolute logarithmic mean characterization of Blaschke products; conversely a Blaschke product satisfies the absolute version too, since $\log |B|\le 0$.