Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 27, 2002, 249–255

ASYMPTOTIC MAXIMUM PRINCIPLE

Boris Korenblum

University at Albany, Department of Mathematics and Statistics Albany, NY 12222, U.S.A.

Dedicated to W. K. Hayman on his 75th birthday

Abstract. By definition the asymptotic maximum principle (AMP) is valid for a class 𝒦 of analytic functions in the unit disk D if, whenever f ∈ 𝒦

sup_{0≤θ<2π} lim sup_{r→1} |f(re^{iθ})| = sup_{z∈D} |f(z)|.

Let 𝒦_{t} (t ≥ 0) be the class of all f for which

lim sup_{|z|→1}{(1 − |z|)^{2} log |f(z)|} ≤ t.

Then AMP is valid for 𝒦_{0} and not valid for 𝒦_{t} with t > 0.

1. Historical background and the main result

It is well known (Lusin–Privalov theorem) [3] that there are non-zero analytic functions in D that have zero radial limits on a subset E of ∂D having full Lebesgue measure. (However, this cannot happen if |E| > 0 and at the same time E is of second Baire category.) More specifically, no matter how slowly k(r) tends to +∞ as r → 1, there always is a non-zero analytic f with log |f(z)| ≤ k(|z|) and lim_{r→1} f(re^{iθ}) = 0 almost everywhere on [0, 2π) (see [2] for this and other similar results). Recently [1] a connection was found between such radial null sets E and the notion of k -entropy of E.

The radial null sets represent only one kind of asymptotic behavior, namely convergence to 0. If the radial null sets are replaced by sets of “asymptotic boundedness”, then we are led to the following two general problems whose purpose is to expand the scope of the classical maximum principle.

Problem 1. Given a class 𝒦 , characterize those subsets E of ∂D such that f ∈ 𝒦 together with

sup_{ζ∈E} lim sup_{r→1} |f(rζ)| < ∞

imply f ∈ H^{∞}.

2000 Mathematics Subject Classification: Primary 30H05, 30E25.

Problem 2 (Dual to the above). Given a subset E of ∂D, identify the class 𝒦 , in terms of maximal allowed radial growth, such that the above inequality together with f ∈ 𝒦 imply f ∈ H^{∞}.

Definition. We say that the asymptotic maximum principle (AMP) is valid for the class 𝒦 of analytic functions in the unit disk D if, whenever f ∈ 𝒦 ,

sup_{0≤θ<2π} lim sup_{r→1} |f(re^{iθ})| = sup_{z∈D} |f(z)|.

We give a solution to Problem 2 for the case E = ∂D, i.e. for AMP. More specifically, we identify maximal radial growth of functions that implies validity of the AMP.

Let 𝒦_{t}, t ≥ 0, be the class of all f for which

lim sup_{|z|→1}{(1 − |z|)^{2} log |f(z)|} ≤ t.

(Note that for t = 0 and a non-zero f the above inequality becomes an equality.)

We are now in a position to state our main result.

Theorem. AMP is valid for 𝒦_{0} and not valid for 𝒦_{t} with t > 0.

Remark. A similar result holds for general subharmonic functions instead of log |f(z)|. We emphasize here the analytic case for the sake of comparison with Lusin–Privalov’s theorem and other results.

2. Use of a Baire-category argument

We proceed now to proving the above theorem. If t > 0 then the AMP fails for 𝒦_{t} as the example

f(z) = exp\{ci(\frac{1 + z}{1 − z} )^{2}\}

shows, where c is a suitable positive constant depending on t. We therefore assume t = 0, which means for f ∈ 𝒦_{0} that

lim sup_{|z|→1}{(1 − |z|)^{2} log |f(z)|} = 0,

and proceed to prove that

C = sup_{ζ∈∂D} lim sup_{r→1} |f(rζ)| = sup_{z∈D} |f(z)|.

If the first supremum above is infinite then there is nothing to prove. We therefore suppose that C < ∞.

If now A > C, we define the following sets:

E_{r} = E_{r,f} = \{ζ ∈ ∂D : |f(ϱζ)| ≤ A, r ≤ ϱ < 1\}, \frac{1}{2} ≤ r < 1.

It is easily seen that \{E_{r} : \frac{1}{2} ≤ r < 1\} is a nested family of closed sets whose union equals ∂D. Let G_{r} denote the interior of E_{r} and write G = ⋃_{r≥1/2} G_{r}; ζ ∈ G means that there is some δ = δ(ζ) such that |z − ζ| < δ, z ∈ D imply |f(z)| ≤ A.

We now claim that the open set G is dense in ∂D, and therefore the closed set F = ∂D\G is nowhere dense. In fact, picking an arbitrary monotone sequence \frac{1}{2} ≤ r_{1} < r_{2} < ⋯ → 1 we see that every ζ in G ultimately is in G_{r_{n}} and thus not in E_{r_{n}} \ G_{r_{n}}; on the other hand, every ζ in ∂D is either in some G_{r_{n}} or in some E_{r_{n}} \ G_{r_{n}}. Therefore G ⊃ ∂D \ ⋃_{n}(E_{r_{n}} \ G_{r_{n}}). By the Baire category theorem, the set ⋃_{n}(E_{r_{n}} \ G_{r_{n}}) being a countable union of nowhere dense subsets of ∂D must have a dense complement, which proves our claim.

By the Heine–Borel theorem, in order to prove our theorem it is enough to show that F = ∅. We proceed now to show that the assumption F ≠ ∅ leads to a contradiction.

Recall that F consists of those points ζ ∈ ∂D such that for every δ > 0 there is some z in D with |z − ζ| < δ and |f(z)| > A.

Pick now some Ã, C < Ã < A, and construct the sets Ẽ_{r}, \frac{1}{2} ≤ r < 1, that relate to F essentially as E_{r} relate to ∂D:

Ẽ_{r} : {ζ ∈ F : |f(ϱζ)| ≤ Ã, r ≤ ϱ < 1}.

As before, \{Ẽ_{r} : \frac{1}{2} ≤ r < 1\} is a nested family of closed sets whose union equals F. Since F, as well as all its relatively open subsets, are of second Baire category in themselves, we can use again the Baire category argument to find that there is an open arc I ⊂ ∂D and some r_{0} , \frac{1}{2} ≤ r_{0} < 1, such that ∅ ≠ I ∩ F ⊂ Ẽ_{r_{0}} . Let

K = (Ī\ I) ∪ (I ∩ F).

This set is a closed and nowhere dense subset of Ī such that

|f(ϱζ)| ≤ Ã for all ζ ∈ K and r_{1} ≤ ϱ < 1,

where r_{1} is some number between r_{0} and 1. It then follows that there is a constant B > Ã such that

|f(ϱζ)| ≤ B for all ζ ∈ K, 0 ≤ ϱ < 1.

The critical step now is to extend this estimate to the whole sector S = {z ∈ D : z/|z| ∈ I}. Let Ī \ K = ⋃_{n} J_{n} where J_{n} are disjoint open arcs with end points e^{iα_{n}} and e^{iβ_{n}}. Consider the open sector S_{n} abutting on J_{n}. We need to prove that for all n, |f(z)| ≤ B, z ∈ S_{n}. Recall that J_{n} ⊂ G, which implies that for every closed subarc \overline{γ} ⊂ J_{n} there is some number a = a(\overline{γ}) < 1 so that |f(ϱζ)| ≤ A if ζ ∈ γ and a ≤ ϱ < 1. Construct an increasing sequence of closed arcs J̄_{n1} ⊂ J̄_{n2} ⊂ J̄_{n3} ⊂ ⋯ having a common midpoint with J_{n} and such that ⋃_{k} J̄_{nk} = J_{n}. For the corresponding numerical sequence a_{nk} = a(J̄_{nk}), k = 1, 2, …, we assume that a_{n1} < a_{n2} < ⋯ → 1. This construction yields a piecewise constant function ψ_{n}(θ) equal to a_{nk} if e^{iθ} ∈ J̄_{nk} \ J̄_{n,k−1} . By the construction we have |f(ϱe^{iθ})| ≤ A if ψ_{n}(θ) ≤ ϱ < 1, α_{n} < θ < β_{n}. We would like to replace ψ by a continuous function φ_{n}(θ) ≥ ψ_{n}(θ) having similar properties; this can be achieved by “cutting the corners” of the graph of ψ_{n}, i.e., linearly interpolating between discontinuity points of ψ_{n}. We thus obtain a continuous curve ϱ = φ_{n}(θ), α_{n} ≤ θ ≤ β_{n}, joining the end points e^{iα_{n}} and e^{iβ_{n}} of J̄_{n} and satisfying

|f(ϱe^{iθ})| ≤ A if φ_{n}(θ) ≤ ϱ < 1, α_{n} < θ < β_{n}.

The required estimate

|f(z)| ≤ B, z ∈ S_{n},

is then a consequence of the following lemma that is in fact a sharper form of a classical Phragmén–Lindelöf theorem (see [4], v.2, p. 214).

3. A Phragmén–Lindelöf type lemma

Lemma. Let S be an open sector bounded by an arc J = {z : |z| = 1, −α ≤ Arg z ≤ α} and two radii R^{±} = {z : 0 ≤ |z| < 1, Arg z = ±α}. Let Γ be the subdomain of S bounded by J and a curve γ whose polar equation is ϱ = φ(θ) with φ continuous on [−α, α], 0 < φ(θ) < 1, −α < θ < α, φ(±α) = 1.

If f is analytic in S, continuous on R^{+} ∪ R^{−} ∪ S, and satisfies

(∗)  lim sup_{|z|→1, z∈S}{(1 − |z|)^{2} log |f(z)|} = 0,

then

sup{|f(z)| : z ∈ R^{+} ∪ R^{−} ∪ Γ} = sup{|f(z)| : z ∈ S}.

Proof. It is convenient to change the setting from the sector S to that of the half-strip

Ω = {x + iy : −1 < x < 1, y > 0},

which is done by an elementary conformal mapping. We have

∂Ω = [−1, 1] ∪ l^{−} ∪ l^{+} ∪ {∞},

where l^{±} = {±1 + iy : y > 0}. We are going to prove a somewhat stronger version of the lemma (cf. Remark at the end of Section 1), in which log |f(z)| is replaced by an arbitrary subharmonic v: Ω → [−∞, ∞) which extends to be continuous on l^{−} ∪ l^{+} ∪ Ω ∪ {∞}. (Note that continuity here is equivalent to the usual continuity of e^{v}.) Our assumptions are:

(i) there is a region Γ between [−1, 1] and a continuous curve γ : y = φ(x), with φ(−1) = φ(1) = 0, φ(x) > 0, −1 < x < 1, such that

sup{v(z) : z ∈ l^{−} ∪ l^{+} ∪ Γ} = M < ∞;

(ii)

(∗∗)  lim sup_{y→0+, −1<x<1}{y^{2}v(x + iy)} = 0.

Our aim is to show that

sup{v(z) : z ∈ Ω} = M.

First we modify v by subtracting a harmonic function that is positive in Ω. Let

Ψ(x, y) = −1/2 Im z^{−2} = \frac{xy}{(x^{2} + y^{2})^{2}} .

Clearly, Ψ is harmonic and positive for x > 0, y > 0, Ψ = 0 if x = 0 or y = 0, and Ψ(x, mx) = m/(1 + m^{2})^{2}x^{2} .

Consider now

v_{a}(z) = v(z) − a(Ψ(1 + x, y) + Ψ(1 − x, y)), a > 0.

Let Ω_{1} = {z = x + iy ∈ Ω : y > min(1 − x, 1 + x)}. By (∗∗) v_{a}(z) is bounded above on ∂Ω_{1} . By applying the classical Phragmén–Lindelöf theorem referred to earlier, we obtain that v_{a}(z) is bounded above in Ω_{1} ; it is also continuous on \overline{Ω}_{1} except, perhaps, at ±1. To make it continuous on \overline{Ω}_{1} we adjust it as follows with ε_{1} > 0:

v_{a,ε_{1}}(z) = v_{a}(z) + log |(z + 1)/(z + 1 + ε_{1})| + log |(1 − z)/(1 + ε_{1} − z)|.

We then have v_{a,ε_{1}}(z) < v_{a}(z) < v(z), z ∈ Ω_{1} ; also, v_{a,ε_{1}} is continuous on \overline{Ω}_{1} , with v_{a,ε_{1}}(±1) = −∞.

The next step requires the curve γ : y = φ(x), −1 ≤ x ≤ 1, to satisfy the following additional condition: φ is differentiable, with |φ′(x)| ≤ \frac{1}{2} ; this does not result in any loss of generality, as φ(x) can be replaced by any smaller positive function. With this assumption made, we see that Γ ∩ Ω_{1} = ∅; also, every straight line with slope m, |m| > 1, intersects the boundary of the region Ω_{2} lying between Ω_{1} and Γ at most at two points.

Consider now:

v_{a,ε_{1},ε_{2}}(z) = v_{a,ε_{1}}(z) − a(Ψ(x + 1 − ε_{2}, y) + Ψ(1 − ε_{2} − x, y)), 0 < ε_{2} < 1.

This function is subharmonic in the region Ω^{(ε_{2})} ⊂ Ω lying between the two lines l_{ε_{2}}^{±} = {z = x + iy : x = ±(1 − ε_{2})}; also,

v_{a,ε_{1},ε_{2}}(z) < v_{a,ε_{1}}(z) < v_{a}(z) < v(z), z ∈ Ω^{(ε_{2})}.

We apply now the classical maximum principle to v_{a,ε_{1},ε_{2}}(z) in the region Ω_{3}^{(ε_{2})} ⊂ Ω^{(ε_{2})} \ Γ whose boundary is made up of the following parts:

B_{1} : the segment of γ between the lines λε_{2}^{+} : y = −2(x − 1 + ε_{2}) and λε_{2}^{−} : y =

2(x + 1 − ε_{2}); B_{2} : the two segments of the above lines lying in Ω_{2} ; B_{3} : the two segments of ∂Ω_{1} lying between lε_{2}^{−} and λε_{2}^{−}, and between λε_{2}^{+} and

lε_{2}^{+} , respectively; B_{4} : parts of l_{ε_{2}}^{±} lying in Ω_{1} .

We thus obtain for z_{0} ∈ Ω_{3}^{(ε_{2})} that v_{a,ε_{1},ε_{2}}(z_{0}) is less than or equal to

max(max{v(z) : z ∈ B_{1}}, max{v_{a,ε_{1},ε_{2}}(z) : z ∈ B_{2} ∪ B_{3}}, max{v_{a,ε_{1}}(z) : z ∈ B_{4}}).

Now, we let ε_{2} → 0, holding ε_{1} and a fixed. The second maximum in parentheses then tends to 0 due to (∗∗), and the third maximum tends to max{v_{a,ε_{1}}(z) : z ∈ l^{+}∪l^{−}\} due to continuity of v_{a,ε_{1}} on \overline{Ω}_{1} ; at the same time v_{a,ε_{1},ε_{2}}(z_{0}) → v_{2a,ε_{1}}(z_{0}) and Ω_{3}^{(ε_{2})} eventually includes all z_{0} ∈ Ω_{3} = Ω \ Γ. Thus for all z_{0} ∈ Ω \ Γ

v_{2a,ε_{1}}(z_{0}) ≤ max(sup{v(z) : z ∈ γ}, sup{v(z) : z ∈ l^{+} ∪ l^{−}}).

The right-hand side depends neither on a nor on ε_{1} . Letting a → 0, ε_{1} → 0 we get

sup{v(z_{0}) : z_{0} ∈ Ω \ Γ} ≤ max(sup{v(z) : z ∈ γ}, sup{v(z) : z ∈ l^{+} ∪ l^{−}}),

which obviously implies

sup{v(z) : z ∈ Ω} = sup{v(z) : z ∈ l^{−} ∪ l^{+} ∪ Γ}.

4. Completion of the proof of the theorem

Applying the above lemma to each S_{n} we obtain

sup{|f(z)| : z ∈ S_{n}} ≤ B.

Since the closure of ⋃_{n} S_{n} is the whole sector \overline{S} we deduce that |f(z)| is bounded by B in S. We also have lim sup_{ϱ→1} |f(ϱζ)| ≤ C < Ã on I. On the two boundary radii R_{1} and R_{2} |f(ϱζ)| ≤ Ã < A, r_{1} ≤ ϱ < 1, ζ ∈ Ī \ I. This implies that log |f(z)| is dominated in S by a harmonic function U(z) whose boundary values on R_{1} and R_{2} are log B, 0 ≤ |z| < r_{1} , log Ã, r_{1} ≤ |z| < 1, and on I, U(ζ) = log C. To make U(z) continuous we increase its piecewise constant boundary values by linearly interpolating them between the discontinuity points. The resulting harmonic function U_{1}(z) ≥ U(z) is continuous in the closed sector \overline{S} , and on I U_{1}(ζ) = log Ã. This implies that there is some r_{2} , r_{1} < r_{2} < 1, such that

|f(z)| ≤ A, z ∈ S, r_{2} ≤ |z| < 1,

which means that I ∩ F = ∅, contrary to the assumption. This contradiction proves our theorem.

Acknowledgements. 1. The author thanks Charles Horowitz, Dmitry Khavinson, and Pascal Thomas for helpful criticism of the first draft of this paper. The author also thanks the referee for a careful reading of the manuscript and numerous helpful suggestions. 2. Part of this paper was done during the author’s visit to the Gelbart Research Insitute for the Mathematical Sciences (Bar-Ilan University, Israel). The author thanks this institution and its director, Professor B. Pinchuk, for generous hospitality and support.

References

[1] Bénéteau, C., and B. Korenblum: Jensen type inequalities and radial null sets. - Analysis 21, 2001, 99–105.

[2] Kahane, J. P., and Y. Katznelson: Sur le comportement radial des fonctions analytiques. - C. R. Acad. Sci. Paris Sér. A 272, 1971, 718–719.

[3] Lusin, N. N., and I. T. Privalov: Sur l’unicité et la multiplicité des fonctions analytiques. - Ann. Sci. École Norm. Sup. (3) 42, 1925, 143–191.

[4] Markushevich, A. I.: Theory of Functions of a Complex Variable. - Chelsea Publishing Co., New York, 1977.

Received 21 May 2001