Annales Academiæ Scientiarum Fennicæ Series A. I. Mathematica Volumen 18, 1993, 395–401

ON THE MULTIPLICATIVE BEHAVIOR OF BLOCH FUNCTIONS

Nikolaos Danikas and Peter Lappan

Aristotle University of Thessaloniki, Department of Mathematics, 54006 Thessaloniki, Greece Michigan State University, Department of Mathematics, East Lansing, MI 48824, U.S.A.

Abstract. We prove that for each unbounded Bloch function f in the open unit disk D there exists a singular inner function S and a bounded outer function Q such that neither product f · S nor f · Q is a normal function.

A function f meromorphic in D = {z : |z| < 1} is called a normal function if

sup_{z∈D} (1 − |z|^{2})\frac{|f′(z)|}{1 + |f(z)|^{2}} < ∞.

A well known subfamily of the normal functions is the so-called Bloch space ℬ, which consists of those functions f analytic in D for which

∥f∥_{ℬ} = sup_{z∈D} (1 − |z|^{2})|f′(z)| < ∞

(see, for example, [1]). It is well known that the Bloch functions form a linear space, but the collection of all normal functions do not, and, moreover, neither class is closed under multiplication (see [5], [6]). It has been shown that for each unbounded normal analytic function in D, there exists a Blaschke product B such that the product f · B is not a normal function (see [2], [6]), and that, in fact, the Blaschke product B can be replaced by a function g of the disk algebra, where g has zeros in D (see [3]).

In this paper, we show the following result.

Theorem. If f is an unbounded Bloch function in D, then there exists both a singular inner function S and a bounded outer function Q such that neither the product f · S nor the product f · Q is a normal function.

Since neither S nor Q has zeros in D, this result is not contained in any of the results mentioned above.

We recall that a singular inner function is a function of the form

S(z) = exp \{ − ∫_{0}^{2π} \frac{e^{it} + z}{e^{it} − z} dμ(t)\}

1991 Mathematics Subject Classification: Primary 30D45; Secondary 30D40, 30D55.

where μ is a finite positive Borel measure on ∂D which is singular with respect to Lebesgue measure. Also, a bounded outer function is a function of the form

Q(z) = e^{iγ} exp \{ − ∫_{0}^{2π} \frac{e^{it} + z}{e^{it} − z} log ψ(e^{it}) dt\}

where γ is a real number, ψ ∈ L^{∞}(∂D), and log ψ(e^{it}) ∈ L^{1}(∂D) (see [4], [8]).

Proof of the Theorem. We first construct the appropriate singular inner function S.

We note that an unbounded Bloch function f must satisfy the growth condition

|f(z)| = O( log \frac{1}{1 − |z|}) as |z| → 1,

(see [1, p. 13]), so it follows that there exists a sequence {z_{n}} in D for which

(1)  |f(z_{n})| → ∞  and  (1 − |z_{n}|) log |f(z_{n})| → 0 as n → ∞.

Without loss of generality, we may assume that z_{n} → 1, that 0 < arg z_{n+1} < arg z_{n} < \frac{1}{2}π, and that |f(z_{n+1})| > |f(z_{n})| > 1 for each n.

Now consider the angular domain

A = \{z ∈ D : Im z > 0, | arg(1 − z)| < \frac{3π}{8} \}

and an arc Γ ⊂ E = {z ∈ D \ A : Im z > 0} where Γ has an endpoint at z = 1. If all but a finite number of the points z_{n} lie in A, then, by a result of J. McMillan (see [1, p. 33] or [7, p. 269]), |f(z)| cannot be bounded on Γ, which means that we may assume that the sequence {z_{n}} lies on the arc Γ, in addition to the other properties we have assumed above. Now an elementary calculation shows that 1 − |z| < |e^{i arg z} − 1| for each z ∈ E. From this, it is easy to see that we may assume that the sequence {z_{n}} is such that it also has all of the following properties:

(2)  1 − |z_{n}| < |e^{i arg z_{n}} − z_{k}|  for each k ≠ n,

(3)  (1 − |z_{n}|) log |f(z_{n})| < (1 − |z_{n−1}|)\frac{log 2}{2^{n+1}} ,  n ≥ 1,

and

(4)  (1 − |z_{n}|^{2}) ∑_{k=1}^{n−1} \frac{log |f(z_{k})|}{1 − |z_{k}|} < \frac{log 2}{2},  n ≥ 2.

On the multiplicative behavior of Bloch functions 397

Now, for each positive integer n, let

t_{n} = arg z_{n},
c_{n} = \frac{1 − |z_{n}|}{1 + |z_{n}|} log |f(z_{n})|,

and

g(z) = ∑_{n=1}^{∞} c_{n} \frac{exp(it_{n}) + z}{exp(it_{n}) − z} .

It follows from (3) that ∑_{n=1}^{∞} c_{n} < log 2, and so the sum defining the function g(z) is convergent in D. Finally, we set

S(z) = exp ( − g(z)),

and we note that S(z) is a singular inner function.

We first show that \frac{1}{2} < |f(z_{n})S(z_{n})| < 1. Note that

Re (g(z_{n})) = ∑_{j=1}^{∞} c_{n} \frac{(1 − |z_{n}|^{2})}{| exp(it_{j}) − z_{n}|^{2}}

= c_{n} \frac{(1 − |z_{n}|^{2})}{| exp(it_{n}) − z_{n}|^{2}} + ∑_{j≠n} c_{j} \frac{(1 − |z_{n}|^{2})}{| exp(it_{j}) − z_{n}|^{2}}

< c_{n} \frac{(1 − |z_{n}|^{2})}{| exp(it_{n}) − z_{n}|^{2}} + ∑_{j=1}^{n−1} \frac{c_{j}(1 − |z_{n}|^{2})}{(1 − |z_{j}|)^{2}} + ∑_{j=n+1}^{∞} \frac{c_{j}}{1 − |z_{n}|}.

From (4), we have

∑_{j=1}^{n−1} \frac{c_{j}(1 − |z_{n}|^{2})}{(1 − |z_{j}|)^{2}} < \frac{log 2}{2}

and from (3),

∑_{j=n+1}^{∞} \frac{c_{j}}{1 − |z_{n}|} = ∑_{j=n+1}^{∞} \frac{(1 − |z_{j}|) log |f(z_{j})|}{(1 + |z_{j}|)(1 − |z_{n}|)}

< ∑_{j=n+1}^{∞} \frac{1 − |z_{j−1}|}{1 − |z_{n}|} \frac{log 2}{2^{j}} < \frac{log 2}{2}.

Finally,

c_{n} \frac{(1 − |z_{n}|^{2})}{| exp(it_{n}) − z_{n}|^{2}} = \frac{(1 − |z_{n}|)^{2} log |f(z_{n})|}{| exp(it_{n}) − z_{n}|^{2}} = log |f(z_{n})|,

so we may combine these inequalities to obtain that

log |f(z_{n})| < Re (g(z_{n})) < log |f(z_{n})| + log 2.

It follows that

\frac{1}{2|f(z_{n})|} < |S(z_{n})| < \frac{1}{|f(z_{n})|},

which is the desired conclusion. We note also that |S(z_{n})| < 1.

Also, if h(z) = f(z)S(z) then \frac{1}{2} < |h(z_{n})| < 1, and so

\begin{aligned}(1 − |z_{n}|^{2})\frac{|h′(z_{n})|}{1 + |h(z_{n})|^{2}} & ≥ \frac{1}{2}(1 − |z_{n}|^{2})(|S′(z_{n})f(z_{n})| − |S(z_{n})f′(z_{n})|) \\ & > \frac{1}{4}(1 − |z_{n}|^{2})|g′(z_{n})| − \frac{1}{2} ∥f∥_{ℬ} \end{aligned}

and we will be finished if we can show that

(1 − |z_{n}|^{2})|g′(z_{n})| → ∞.

But

(1 − |z_{n}|^{2})|g′(z_{n})| = (1 − |z_{n}|^{2})| ∑_{j=1}^{∞} 2c_{j} \frac{exp(it_{j})}{( exp(it_{j}) − z_{n})^{2}} |

> 2c_{n} \frac{1 + |z_{n}|}{1 − |z_{n}|} − 2(1 − |z_{n}|^{2}) ∑_{j=1}^{n−1} \frac{(1 − |z_{j}|) log |f(z_{j})|}{| exp(it_{j}) − z_{n}|^{2}}

− 2(1 − |z_{n}|^{2}) ∑_{j=n+1}^{∞} \frac{(1 − |z_{j}|) log |f(z_{j})|}{| exp(it_{j}) − z_{n}|^{2}}

> 2 log |f(z_{n})| − 2 log 2 → ∞,

again making use of (2), (3), and (4). This completes the proof of the existence of the appropriate singular inner function.

To show the existence of an appropriate outer function, we again begin by selecting a sequence {z_{n}} in D such that, without loss of generality, z_{n} → 1, |f(z_{n})| → ∞, 0 < |z_{n}| < |z_{n+1}|, and 0 < arg z_{n+1} < arg z_{n} < \frac{1}{2}π for each positive integer n. As in the proof of the first part, we can assume that the sequence {z_{n}} also satisfies the following conditions:

(5)  (1 − |z_{n}|)( log |f(z_{n})|)^{2} < \frac{1}{16}  for each n,

(6)  1 − |z_{j}| < | exp(it_{j}) − z_{n}|  for j ≠ n and t = arg(z_{j}),

(7)  2t_{n+1} < t_{n}  for each n,

(8)  (1 − |z_{n}|^{2}) ∑_{j=1}^{n−1} \frac{log |f(z_{j})|}{1 − |z_{j}|} < \frac{1}{16} log 2,  for n ≥ 2,

and

(9)  (1 − |z_{n}|) log |f(z_{n})| < (1 − |z_{n−1}|)\frac{log 2}{2^{n+1}} ,  for n ≥ 2.

Now set c_{n} = 4(1 − |z_{n}|) log |f(z_{n})| for each n, let

χ_{n}(t) = \{ \begin{matrix}1, & for t ∈ [t_{n}, t_{n} + c_{n}^{2}], \\ 0, & otherwise \end{matrix}

let ψ(e^{it}) = exp ( ∑_{j=1}^{∞} χ_{n}(t)/c_{n}), and define

φ(z) = ∫_{0}^{2π} \frac{e^{it} + z}{e^{it} − z} log ψ(e^{it}) dt

and Q(z) = exp { − φ(z)}.

First, we show that the intervals [t_{j}, t_{j} + c_{j}^{2}] are mutually disjoint. If we fix j > 1, the definition of c_{j} and (5) yield

t_{j} + c_{j}^{2} = t_{j} + 16(1 − |z_{j}|)^{2}( log |f(z_{j})|)^{2} < t_{j} + 1 − |z_{j}|,

and from (6), we get t_{j} + 1 − |z_{j}| < t_{j} + | exp(it_{j}) − z_{n}| for each n ≠ j. But z_{n} → 1, and 0 < t_{j} = arg(z_{j}) < \frac{1}{2}π so we conclude that

t_{j} + 1 − |z_{j}| < t_{j} + | exp(it_{j}) − 1| < 2t_{j}.

Now, (7) says that 2t_{j} < t_{j−1} , so by combining the inequalities in this paragraph we conclude that t_{j} + c_{j}^{2} < t_{j−1} , which means that the intervals [t_{j}, t_{j} + c_{j}^{2}] are mutually disjoint.

From the disjointness of the intervals and (9), we have that

∫_{0}^{2π} log ψ(e^{it}) dt = ∑_{j=1}^{∞} (\frac{1}{c_{j}} )c_{j}^{2} = ∑_{j=1}^{∞} c_{j} < \frac{log 2}{2}.

Noting that Re (φ(z)) > 0, we conclude that Q(z) is a bounded outer function.

Further, for t ∈ [t_{n}, t_{n} + c_{n}^{2}], we have

\begin{aligned}|e^{it} − z_{n}|^{2} & = (1 − |z_{n}|)^{2} + 4|z_{n}| sin^{2} ((t − t_{n})/2) \\ & < (1 − |z_{n}|)^{2} + (t − t_{n})^{2} ≤ (1 − |z_{n}|)^{2} + c_{n}^{4}. \end{aligned}

Now, making use of (5), we obtain

(10)

∫_{t_{n}}^{t_{n}+c_{n}^{2}} \frac{1 − |z_{n}|^{2}}{|e^{it} − z_{n}|^{2}} \frac{1}{c_{n}} dt > \frac{c_{n}(1 − |z_{n}|)}{(1 − |z_{n}|)^{2} + c_{n}^{4}}

= \frac{4 log |f(z_{n})|}{1 + (4 log |f(z_{n})|)^{4}(1 − |z_{n}|)^{2}} > log |f(z_{n})|

for each n. Also

∑_{j≠n} ∫_{t_{j}}^{t_{j}+c_{j}^{2}} \frac{1 − |z_{n}|^{2}}{|e^{it} − z_{n}|^{2}} \frac{1}{c_{j}} dt ≤ 2(1 − |z_{n}|) ∑_{j=1}^{n−1} \frac{c_{j}}{(1 − |z_{j}|)^{2}} + ∑_{j=n+1}^{∞} \frac{2c_{j}}{1 − |z_{n}|} < log 2

as consequences of (6), (8), and (9). It follows that

log |f(z_{n})| < Re φ(z_{n}) < log |f(z_{n})| + log 2,

which means that

\frac{1}{2} < |f(z_{n})Q(z_{n})| < 1.

Further,

φ′(z_{n}) = ∑_{j=1}^{∞} ∫_{t_{j}}^{t_{j}+c_{j}^{2}} \frac{2e^{it}}{(e^{it} − z_{n})^{2}} \frac{1}{c_{j}} dt,

and, for large n, the argument of the integrand of the term corresponding to j = n is always approximately −t_{n}, so that the integral is at least 1/2 of the what its value would be if the integrand were the constant exp(−it_{n})(1 − |z_{n}|)^{−2} . Now, applying (6), we obtain

|φ′(z_{n})| ≥ \frac{c_{n}}{(1 − |z_{n}|)^{2}} − ∑_{j=1}^{n−1} \frac{2c_{j}}{(1 − |z_{j}|)^{2}} − ∑_{j=n+1}^{∞} \frac{2c_{j}}{(1 − |z_{n}|)^{2}} .

Applying (8), we get

(1 − |z_{n}|^{2}) ∑_{j=1}^{n−1} \frac{2c_{j}}{(1 − |z_{j}|)^{2}} = (1 − |z_{n}|^{2}) ∑_{j=1}^{n−1} \frac{8(1 − |z_{j}|) log |f(z_{j})|}{(1 − |z_{j}|)^{2}}

= (1 − |z_{n}|^{2}) ∑_{j=1}^{n−1} \frac{8 log |f(z_{j})|}{1 − |z_{j}|} < \frac{log 2}{2}.

On the multiplicative behavior of Bloch functions 401

Also, using (9), we get

(1 − |z_{n}|^{2}) ∑_{j=n+1}^{∞} \frac{2c_{j}}{(1 − |z_{j}|)^{2}} = (1 − |z_{n}|^{2}) ∑_{j=n+1}^{∞} \frac{8(1 − |z_{j}|) log |f(z_{j}))}{(1 − |z_{j}|)^{2}}

< 16 ∑_{j=n+1}^{∞} \frac{1 − |z_{j−1}|}{1 − |z_{n}|} \frac{log 2}{2^{j+1}} < log 2 for n ≥ 3.

Finally,

\frac{c_{n}}{(1 − |z_{n}|)^{2}} = \frac{4 log |f(z_{n})|}{1 − |z_{n}|}

and so

(1 − |z_{n}|^{2})|φ′(z_{n})| ≥ \frac{4 log |f(z_{n})|}{1 − |z_{n}|} − \frac{3}{2} log 2 → ∞

as n → ∞. Now, if k(z) = f(z)Q(z), a computation similar to the one performed in the first part of the proof shows that

(1 − |z_{n}|)\frac{|k′(z_{n})|}{1 + |k(z_{n})|^{2}} → ∞,

and this completes the proof.

References

[1] Anderson, J.M., J. Clunie, and Ch. Pommerenke: On Bloch functions and normal functions. - J. Reine Angew. Math. 270, 1974, 12–37.

[2] Campbell, D.M.: Nonnormal sums and products of unbounded normal functions II. - Proc. Amer. Math. Soc. 74, 1979, 202–203.

[3] Danikas, N.: On the multiplicative behaviour of normal functions. - Analysis 10, 1990, 187–192.

[4] Duren, P.L.: Theory of H^{p} spaces. - Academic Press, New York, 1970.

[5] Hayman, W.K., and D.A. Storvick: On normal functions. - Bull. London Math. Soc. 3, 1971, 193–194.

[6] Lappan, P.: Non-normal sums and products of unbounded normal functions. - Michigan Math. J. 8, 1961, 187–192.

[7] Pommerenke, Ch.: Univalent functions. - Vandenhoeck & Ruprecht, Göttingen, 1975.

[8] Rudin, W.: Real and complex analysis. - McGraw-Hill, New York, 1974.

Received 5 August 1992