Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 9 (Bounded Analytic Functions)

Put $\alpha =-\log |c|>0$. Since $f$ has no zeros and the unit disc is simply connected, it has a holomorphic logarithm. Thus $F=-\log f$ is holomorphic with $\mathrm{Re}F>0$ and $\mathrm{Re}F(0)=\alpha$. The Cayley transform between the half-plane and disk givesThe Schwarz lemma yields $|{\varphi }^{\prime }(0)|\le 1$, so the Carathéodory derivative bound for the right half-plane gives $|F^{\prime }(0)|\le 2\alpha$. Since $f^{\prime }=-F^{\prime }f$, the sharp answer isEquality in the Schwarz lemma holds exactly when $\varphi (z)=\omega z$, $|\omega |=1$. Solving for $F$ and exponentiating gives exactly the extremizersTheir logarithmic real part is $\alpha \mathrm{Re}[(1+\omega z)/(1-\omega z)]>0$, so each is indeed a zero-free map into the unit disc with the prescribed value at zero.

For a fixed $w\in \mathbb{D}$, put $r=|w|$. Apply the Harnack inequality on the unit disk to the positive harmonic function $\mathrm{Re}F$ and then exponentiate its negative. The zero-free Schwarz lemma givesFor $w\ne 0$ both endpoints are attained: take $\omega w=r$ for the lower endpoint and $\omega w=-r$ for the upper endpoint in the displayed extremizers. At zero both are $|c|$. If the question is interpreted as allowing $w$ to vary as well, the sharp infimum and supremum over all admissible pairs $(f,w)$ are respectively $0$ and $1$, approached as $r\uparrow 1$; neither value is attained inside the unit disc. An individual function need not have these global extremal values: the constant function $f=c$ is a simple example.

The Harnack inequality on the unit disk states that a nonnegative harmonic function satisfiesOne proof applies the Poisson integral on the unit disk on a circle of radius $R>|z|$: the Poisson kernel on the circle lies between $(R-|z|)/(R+|z|)$ and its reciprocal. The mean value property for harmonic functions identifies the mean boundary value with $u(0)$; let $R\uparrow 1$.

A harmonic conjugate on the simply connected unit disc makes $H=u+iv$ holomorphic. The Carathéodory derivative bound for the right half-plane gives $|H^{\prime }(0)|\le 2u(0)$. The Cauchy-Riemann equations give $|H^{\prime }|=|\nabla u|$, so $|\nabla u(0)|\le 2u(0)$. For a fixed $z$, use the disk automorphismThe chain rule and conformality multiply the gradient length by $|{\psi }^{\prime }(0)|$. Applying the estimate at zero to $u\circ \psi$ proves the sharp gradient bound for positive harmonic functions:Sharpness follows from $u(\zeta )=\mathrm{Re}[(1+\zeta )/(1-\zeta )]$, which has equality at zero, and then composing with ${\psi }^{-1}$. These extremizers are positive multiples of the Poisson kernel on the circle with its boundary pole fixed at one point.

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 $\log |B|\le 0$.

A nontangential limit at $\xi \in \partial \mathbb{D}$ restricts approach to a fixed Stolz region ${\Gamma }_A(\xi )=\{z\in \mathbb{D}:|z-\xi |<A(1-|z|)\}$, where $A>1$. Radial approach is a special case. The Fatou theorem for bounded holomorphic functions says that every $f$ in H-infinity space has a finite boundary value $f^{\ast }(\xi )$ for almost every $\xi$, with convergence through every fixed Stolz region and $|f^{\ast }|\le \Vert f{\Vert }_{\infty }$ almost everywhere.

Here is a proof outline. Write $f=u+iv$ with bounded real harmonic functions $u,v$. Take a sequence $R_j\uparrow 1$. The bounded functions $u(R_j\xi )$ have a subsequence converging in the weak-star topology of $L^{\infty }(\partial \mathbb{D})$ by the Banach-Alaoglu theorem and weak-star metrizability of the dual ball; its $L^1$ predual is separable. Denote the limit by $U$. For $|z|<R_j$, the Poisson integral on the unit disk givesThe Poisson kernels converge uniformly for this fixed interior point and the boundary functions stay uniformly bounded, so the weak-star topology permits passage to the limit:Repeat for $v$, obtaining $V$. Thus $f$ is the Poisson integral of $F=U+iV\in L^{\infty }$.

The Lebesgue differentiation theorem makes almost every boundary point a Lebesgue point of $U$ and $V$. The lemma that a Poisson integral converges nontangentially at Lebesgue points then gives $f(z)\to F({\xi }_0)$ in every fixed Stolz region. To see why, put $\delta =1-|z|$ and let $d(\xi ,{\xi }_0)$ be the shorter angular distance. In ${\Gamma }_A({\xi }_0)$ the kernel satisfiesSplit the circle into a central arc of radius comparable to $\delta$ and annuli with radii $2^k\delta$. For a Lebesgue point, averages of $|F-F({\xi }_0)|$ on sufficiently small arcs are arbitrarily small. The annular contributions are bounded by these averages times a summable geometric sequence $C_A{2}^{-k}$. On the remaining fixed distant arc the kernel is $O(\delta )$, so its contribution vanishes. This proves the claimed nontangential limit. The same Lebesgue points work for every integer aperture $A\ge 2$, hence for all finite apertures.

These boundary values recover the interior by the Poisson integral, so they determine $f$ uniquely. They satisfy $\Vert f^{\ast }{\Vert }_{L^{\infty }}=\Vert f{\Vert }_{\infty }$: one inequality follows by taking limits of the bounded interior values, and the other from positivity and unit integral of the Poisson kernel. The dominated convergence theorem also gives $f(r\xi )\to f^{\ast }(\xi )$ in every finite $L^p$ norm. Its negative Fourier coefficients vanish, because the Taylor series of $f$ has only nonnegative powers. This explains the boundary realization of analytic Hardy spaces.

The exceptional set cannot generally be removed. For example, $f(z)=\exp (i\mathrm{Log}(1-z))$ is bounded and holomorphic on the unit disc, using the holomorphic logarithm on the right half-plane: $|f(z)|=e^{-\arg (1-z)}$ lies between $e^{-\pi /2}$ and $e^{\pi /2}$. At the boundary point one its radial values $e^{i\log (1-r)}$ oscillate without a limit. Nor does the theorem assert convergence along arbitrary tangential paths or a continuous boundary extension. Its strength is almost-everywhere control on every fixed approach cone.

First construct a holomorphic function without requiring boundedness. Separate the finitely many zeros at zero into $z^m$, and enumerate the remaining prescribed zeros as $a_n$. Since $|a_n|\to 1$, there is no interior accumulation point and every prescribed point has finite multiplicity. Using the Weierstrass elementary factors, setFor a compact disc $|z|\le r<1$, choose $q$ with $r<q<1$. Eventually $|z/a_n|\le q$. The power series identitygives an absolutely summable uniform bound. The infinite product convergence from logarithmic tails therefore makes the tail holomorphic and nonvanishing on this compact set. The finitely many initial factors have exactly their prescribed zeros, with the prescribed multiplicities. Hence $F$ has exactly the required zero sequence. This is a direct canonical product construction for zeros escaping to the disk boundary.

For a nonzero bounded holomorphic function, Jensen's formula bounds ${\sum }_n\log (1/|a_n|)$ after separating the zero at zero. Thus the Blaschke condition is necessary; the Blaschke product construction makes it sufficient. Use the curvature-minus-one convention for Hyperbolic distance in the Poincare disc:The latter lies between $(1-|z|)/2$ and $1-|z|$. Consequently the precise bounded-function criterion isThis counts multiplicities; the finitely many occurrences at zero cause no problem.

For the requested hyperbolic zero estimate for a Blaschke product, the pseudohyperbolic distance identity isIf $w$ is not a zero, $0<t<1$. The real function $L(t)=\log [(1+t)/(1-t)]-2t$ has $L(0)=0$ and $L^{\prime }(t)=2t^2/(1-t^2)\ge 0$, so $L(t)\ge 0$. Thus $\log |b_a(w)|\le -2e^{-\rho (w,a)}$. Apply this to each factor, including factors at zero, and then pass from finite products to the Blaschke product by locally uniform convergence:At a zero of $B$ the left side is zero, so the inequality is immediate; the logarithmic calculation at $t=1$ is unnecessary. The unimodular constant contributes nothing. Away from the zeros the product is nonzero, so the same inequality also bounds the sum and shows it is finite there.

We prove necessity of the geometric condition from the embedding condition. Use normalized angular measure $m$ and the H1 space on the unit disk norm $\Vert f{\Vert }_1={\sup }_{r<1}\int |f(r\xi )|dm(\xi )$. Testing $f=1$ first gives $\mu (\mathbb{D})\le N(\mu )$.

For a short arc $I$ of normalized length $\ell =m(I)$ centered at ${\xi }_0$, put $a=(1-\ell ){\xi }_0$ and use the normalized squared Cauchy kernelIts boundary modulus is the Poisson kernel on the circle, so $\Vert f_a{\Vert }_1=1$. The geodesic-cap and Carleson-box comparison gives $|z-{\xi }_0|\le C\ell$ for $z\in Q(I)$. Therefore $|1-\overline{a}z|\le C^{\prime }\ell$, whereas $1-|a{|}^2=2\ell -{\ell }^2\ge \ell$. Thus $|f_a(z)|\ge c/\ell$ throughout the hyperbolic geodesic cap. The embedding bound givesFor arcs longer than a fixed small threshold, the total-mass estimate gives the same form after changing the absolute constant. HenceThis is precisely the Carleson measure condition, in its geodesic-cap version.

We prove sufficiency by a non-tangential maximal function and the layer cake representation. Write $K=C(\mu )$. First $\mu (\mathbb{D})\le CK$: three hyperbolic geodesic caps corresponding to arcs of normalized length $2/3$, with equally spaced centers, cover the unit disc. This handles interior points and large arcs, not only a thin boundary strip.

Fix aperture two and set $Nf(\xi )={\sup }_{z\in {\Gamma }_2(\xi )}|f(z)|$. The crucial H1 non-tangential maximal inequality on the disk isHere is its endpoint proof. For $0<R<1$ let $f_R(z)=f(Rz)$, which is holomorphic across the closed unit disc. Since modulus powers of holomorphic functions are subharmonic, Poisson comparison givesThe estimate that a Poisson maximal function is dominated by the Hardy-Littlewood maximal function yields $(N{f}_R{)}^{1/2}\le CM{g}_R$. The strong $L^2$ Hardy-Littlewood maximal inequality now givesFor each interior point $f_R(z)\to f(z)$, hence $Nf\le {\mathrm{liminf}}_{R\uparrow 1}N{f}_R$. The Fatou lemma proves the maximal inequality. This uses a genuine $L^2$ maximal bound; the Hardy-Littlewood maximal function is not strongly bounded on $L^1$.

For $t>0$, the set $O_t=\{\xi :Nf(\xi )>t\}$ is open, since each interior point witnessing this strict inequality has an open boundary shadow. Write its disjoint arc components as $I_j$. If $|f(z)|>t$ and $\delta =1-|z|$, the arc centered at $z/|z|$ with angular half-width $\delta$ lies in $O_t$, because $|z-\xi |<2\delta$ there. Thus it lies in one component $I_j$, and $z$ has angular projection in $I_j$ with $\delta \le \pi m(I_j)$. The geodesic-cap and Carleson-box comparison puts all such points in $Q(\kappa I_j)$ for a fixed enlargement factor $\kappa$ whenever $I_j$ is small. For large $I_j$, use $\mu (\mathbb{D})\le CK\le C^{\prime }Km(I_j)$. If $z=0$ or $O_t$ is the entire circle, that same total-mass estimate applies. Summing, without needing the enlarged caps to be disjoint, givesThe layer cake representation and the proved H1 non-tangential maximal inequality on the disk finish the argument:Thus the two conditions are equivalent, with their least admissible constants comparable: $C(\mu )\le C_{1}N(\mu )$ and $N(\mu )\le C_{2}C(\mu )$. This is the Carleson embedding theorem on the disk, proved at the analytic $H^1$ endpoint.