On the exponential integrability of conjugate functions

We relate the exponential integrability of the conjugate function $\tilde{f}$ to the size of the gap in the essential range of $f$. Our main result complements a related theorem of Zygmund.


Introduction
We denote by L p the usual Lebesgue spaces of functions on the unit circle T with norm · p . Given f ∈ L 1 , let u be the Poisson integral of f and denote byũ the harmonic conjugate function of u, normalized so that u(0) = 0. Thenũ(z) has nontangential limitf (θ) almost everywhere on T and we callf the conjugate function of f . Alternatively, the conjugate functionf can be defined as the principal value integral for almost every θ. For further details and references, see Section 2.1 below.
The linear mapping f →f is referred to as the conjugation operator. If f is a trigonometric polynomial N n=−N a n e inθ , thenf is a trigonometric polynomial of the same degreẽ where −i sgn is the Fourier multiplier associated with the conjugation operator.
When 1 < p < ∞, according to a famous theorem of M. Riesz, there is a constant C p such that f p ≤ C p f p Gissy was supported by a University of Reading doctoral grant. Miihkinen was supported by Emil Aaltonen Foundation. Virtanen was supported in part by Engineering and Physical Sciences Research Council grant EP/T008636/1. for all f ∈ L p . In addition, although f ∈ L ∞ does not imply thatf ∈ L ∞ (see, e.g. [4]), the Hilbert transform still has very strong boundedness properties as can be seen in the following theorem, due to Zygmund [19].
Theorem A (Zygmund). For f ∈ L ∞ with f ∞ ≤ π/2 and λ < 1, there is a constant C λ such that For the proof, see Corollary III.2.6 of [4]. It follows that where C(T) stands for the space of continuous functions on T.
Let E be a measurable subset of T and define The Lebesgue measure of E is denoted by |E|. We note that the condition f 1 ∞ < π/2 above is optimal as seen by considering an interval E = [a, b] ⊂ (0, 2π) and showing that (1.6) which is not integrable in any neighborhood of b. More generally, it follows from (1.6) and Theorem III.
Notice that the conditions of (1.4) imply that if the function f is real valued and has jumps, then the size of each jump is strictly less than π, while the size of each jump of π 2 ρ E is exactly π and exp( π 2ρ E ) / ∈ L 1 with E as above. Motivated by the study of the Fredholm properties of Toeplitz operators, Shargorodsky [11] proved that if g is real valued and inf R(g) > π/2, where R(g) stands for the essential range of g, then exp( gρ E ) is not integrable. These observations lead to the question of whether Our main result answers this in the affirmative. Indeed, we give an elementary proof of the following result in Section 3.
Theorem 1. Suppose that f ∈ L ∞ with f ≥ π/2 almost everywhere and 0 < |E| < 2π. Then there is a positive constant C such that for all λ ≥ 0. In particular, Remark 2. In the preceding theorem, the conditions on E and that f ≥ π/2 almost everywhere are optimal-see Remark 5. Notice also that Theorem 1 does not imply Shargorodsky's result mentioned above.
Previously in [9,15], sufficient conditions for exponential integrability of f were obtained in terms of the modulus of continuity of f in L p . In addition to these results and other intrinsic interest [4,7,18], the integrability of the exponential of conjugate functions plays an important role in the spectral theory of Toeplitz and related operators [10,11,12], scalar Riemann-Hilbert problems [13,14,17] and their applications.

Preliminaries
As indicated in the introduction, our approach is elementary and based on classical results of complex analysis which are briefly discussed in this section.

Poisson integrals.
For f ∈ L 1 , denote by P [f ] the Poisson integral of f , that is, where the Poisson kernel P z is defined by Recall that P [f ] is harmonic in D and if the function f is continuous at e iθ , then (see Theorem I.1.3 of [5]). To deal with discontinuities at e iθ , define a cone Γ α by Γ α (e iθ ) = {z ∈ D : |z − e iθ | < α(1 − |z|)} for each α < 1, and recall that a function ϕ : D → C is said to have nontangential limit ϕ * (e iθ ) at e iθ if (2.3) lim for every α < 1. Now, for any f ∈ L 1 , if u = P [f ], then u * = f almost everywhere by Fatou's theorem. Define andũ is the harmonic conjugate function of u normalized so thatũ(0) = 0. By Fatou's theorem and Lemma III.1.1 of [4], for almost every θ. For the integral representation off given in (1.1), see Lemma III.1.2 of [4]. In particular, it follows that the principal value in (1.1) exists almost everywhere. If f ∈ L ∞ , then

2.2.
Harmonic and subharmonic functions. In one of the key steps of the proof of the main theorem, we consider the Dirichlet problem of finding a unique bounded harmonic function on a simply connected domain Ω with prescribed boundary values. If g is a continuous real-valued function on ∂Ω, the Dirichlet problem of finding the bounded harmonic function u : Ω → R such that u = g on ∂Ω can be solved using the Poisson integral and the Riemann mapping theorem, which reduces the problem to the well-known case of the unit disk (see, e.g., [5]).
However, in our case, since the boundary functions are discontinuous (see Lemma 3 below), the following more general result is needed.
Let Ω be a simply connected domain and let g be a piecewise continuous function on ∂Ω with a finite number of discontinuities of the first kind at ξ 1 , . . . ξ k . Then there is at most one bounded harmonic function h on Ω such that h = g on ∂Ω \ {ξ 1 , . . . , ξ k }.
If such a bounded harmonic function h exists, then For the proof of the preceding result, see Theorems 5 and 6 of Section 42 of [8].
Regarding the values of a harmonic function in a domain Ω (a nonempty open connected set), we recall the maximum principle (Theorem 1.8 of [2]): Suppose Ω is a domain, u is real-valued and harmonic on Ω, and u has a maximum or a minimum in Ω. Then u is constant. Lemmas 3 and 4 below are utilized in key steps of the proof of the main result.
be the domain in Figure 1 and put L = ∂G λ ∩ {Im z > λ + 1}. Then there exists a unique bounded harmonic function v λ on G λ with the boundary values Proof. Let τ be a conformal map of G λ onto the unit disk D. Then each straight-line piece of the boundary ∂G λ is mapped to an arc on the circle T (see Theorem II.3.4' of [6]). Now consider the Poisson integral of the function that equals 1 on the arcs to which τ maps L and 0 on the complementary arcs. The composition of the Poisson integral with τ gives the desired harmonic function. Uniqueness and the bounds follow from Theorems B and C.
The following characterization of subharmonic functions is also needed: Theorem D. Let u be a function on D and suppose it satisfies the following conditions: For the proof of the preceding result, see Theorem II.13 of [16].

Lemma 4.
Let λ, f, G λ , and v λ be as in Lemma 3 and E ⊂ T be a measurable set with 0 < |E| < 2π. Define Then H λ is continuous and subharmonic with 0 ≤ H λ < 1.
The local mean value inequality in (2.9) holds for each point of Ω λ because H λ = v λ • X is harmonic in Ω λ , and it holds for each point of D \ Ω λ because H λ ≥ 0 equals zero there. Thus, by Theorem D, the function H λ is subharmonic in D.

Proof of Theorem 1
Suppose that f ∈ L ∞ with f ≥ π/2 almost everywhere and 0 < |E| < 2π. Let where ρ E is defined in (1.5). Then The proof of (1.8) consists of several steps.
Step 1. Since Re X * = f ρ E ≥ π/2 almost everywhere on E and Re X * ≤ −π/2 almost everywhere on T \ E, there is a point w ∈ D such that |Re X(w)| < π/2.
Therefore, using the definition of H λ and the properties of v λ , Now, by (3.2) and Lebesgue's dominated convergence theorem, where the constant is independent of λ.
Step 5. It is difficult to obtain the desired estimate (v λ • X)(w) ≥ Ce −λ directly. Instead we estimate v λ from below by another harmonic function g λ , defined on a vertical strip, which allows us to compute g λ explicitly in the next step. Let g λ be the bounded harmonic function in the strip S = {|Re z| < π/2} with the boundary values g λ (±π/2 + iy) = 1, y > λ + 2 0, y < λ + 2.
Thus, by Theorem B, v 0 (z) − Cg 0 (z) ≥ 0 for all z ∈ S, and so for λ > 0 and Consequently, by (3.1), where the constant C is independent of λ.

Further remarks
In this section we provide remarks and examples related to Theorem 1. We show first that the conditions in the theorem are optimal.
Remark 5. In Theorem 1, (i) the condition on E is optimal, and (ii) the condition that f ≥ π/2 almost everywhere is optimal.
Further, by Corollary III.1.8 of [7], which implies that there is no constant C for which (1.8) holds in this case.
(ii) Suppose that π/4 < f < π/2 on an interval I = (a, b) ⊂ (0, 2π) and Then exp( f ρ E ) is integrable by Zygmund's Theorem A because the gap in the essential range of f ρ E is strictly less than π. Consequently, the condition is optimal for (1.9), and it must also be optimal for (1.8) because it was used to prove (1.9). Remark 6. In addition to the example in the previous proof, there are functions f which are not constant and still satisfy |E λ | e −λ as in (4.1). Indeed, let 0 < |E| < 2π and let f be Hölder with f (θ) = π/2 for all θ ∈ ∂E.
In fact, (0.2) of [18], when ψ = f ρ E =g + π 2ρ E , implies the stronger result Notice that the converse is not true, however, that is, (4.2) does not imply that (see Wolff's counterexample on page 52 of [18]).
Open Problem 1. In the preceding remark, when ψ = 1 2 log |H| with H univalent and zero free, Baernstein [3,18] posed a question of whether (4.2) implies (4.3). This seems to be still open.
Our next example concerns outer functions from the theory of Hardy spaces. Recall that an outer function is a function G on the unit disk which can be written in the form where |α| = 1 and ϕ is a positive measurable function on T such that log ϕ ∈ L 1 . Similarly to (2.4), where u is the Poisson integral of log ϕ and v is the harmonic conjugate function of u so that e iv(0) = α. Let 0 < p < ∞ and let f be analytic in D.
Then the function f is in the Notice that G ∈ H p if and only if ϕ ∈ L p (see, e.g., Section II.4 of [4]). We can now use Theorem 1 to determine when certain outer functions are not in H p as shown in the following example.
Using (4.5), it is easy to see that Φ f and its inverse Φ −1 f are both outer functions. Denote by R(f ) the essential range of f as before. Let be an interval such that I ∩ R(f ) = ∅ and |I| ≥ 2π To see this, notice first that (similarly to (2.5)) . Since Φ f is an outer function, it follows from To illustrate the effect of jumps in relation to exponential integrability (see Example 10), the following lemma will be needed. Its proof is included for completeness because we have not found it in the literature.
cos nx n log n .
Remark 9. In the following example, for small values of |x|, we only need the following consequence of the preceding lemma: ∞ n=2 cos nx n log n ≥ C log log(1/|x|), which can also be obtained using Theorem V.1.5 of [19].
Remark 11. We can use the previous example to construct a function f ∈ L ∞ such that f ∞ = π/2, |f | < π/2 and ef ∈ L 1 . This should be compared with Zygmund's result in (1.2).
Open Problem 2. In Theorem 1, it is assumed that f ∈ L ∞ as in Zygmund's Theorem A. It is natural to ask whether the conclusion of Theorem 1 remains true for unbounded functions.
We finish this section with a connection between the estimate in (1.8) and the distance in BMO to L ∞ (see Section VI.6 of [4]).
Suppose that the conditions of Theorem 1 are satisfied, that is, f ∈ L ∞ with f ≥ π/2 almost everywhere and 0 < |E| < 2π. If g ∞ < π/2, then by Theorems A and 1, which is equivalent to inf g∈L ∞ ϕ − g BMO (see, e.g., page 250 in [4]). By Corollary VI.6.6, there is no ǫ ∈ (0, 1) such that for all λ ≥ 0, where the supremum is taken over all arcs I ⊂ T and the average ϕ I is defined by ϕ I = 1 |I| I ϕ for ϕ ∈ L 1 . The same conclusion also follows directly from (1.8) if f ≥ π/2, which is no surprise because the requirement that f has a gap in its essential range is a stronger assumption than dist( f ρ E , L ∞ ) ≥ π/2. Indeed, assume that (1.8) holds and (4.7) does not hold, so that there exists ǫ ∈ (0, 1) satisfying the estimate (4.7). Denote h = f ρ E so thath 0 = 1 2π 2π 0h , and choose s ≥ 0 so large that λ =h 0 + s ≥ 0. Then there exists a constant C > 0 so that so we have Ce −λ = Ce −(h 0 +s) ≤ e − s ǫ . Therefore Ce −h 0 ≤ e (1− 1 ǫ )s → 0 as s → ∞, which is a contradiction.

Complex-valued functions
While real-valued functions are of particular importance in the study of exponential integrability of their conjugate functions, especially in connection with applications, such as Riemann-Hilbert problems and spectral theory of Toeplitz operators above, it would also be of interest to consider the case of complex-valued functions. Indeed, as in Zygmund's Theorem A, we may consider a complex-valued f ∈ L ∞ and ask under what conditions is exp(| f ρ E |) not integrable. As in the proof of Theorem 1, we can define m(λ) = |{t : exp(| f ρ E (t)|) > λ}| and show that if there is a constant |{t : | f ρ E (t)| > λ}| ≥ Ce −λ for all λ ≥ 0, then exp(| f ρ E |) is not integrable.
However, in the complex case, the function f ρ E no longer has a similar (geometric) meaning as in the real case where it can be related to a gap in the essential range. For this reason, we say that a set A ⊂ C has a gap of size g > 0 if A = B ∪ C for some sets B and C of positive measure with dist(A, B) ≥ g. With this, we can state (1.9) in Theorem 1 as follows: If f ∈ L ∞ is real and R(f ) has a gap of size at least π, then exp(f ) is not integrable.
Open Problem 3. Given a complex-valued function f in L ∞ , find a converse to Zygmund's Theorem A.
It may be useful to try to relate the exponential integrability off to the size of the gap in the essential range of f as in the real case.