Sharp Lp Bound for Holomorphic Functions on the Unit Disc

For any 1 < p < ∞ and any X,Y∈R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${X, Y\in \mathbb{R}}$$\end{document} satisfying |X|≤Y\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${|X|\leq Y}$$\end{document} , we determine the optimal constant Cp(X,Y) such that the following holds. If F is a holomorphic function on the unit disc satisfying ReF(0) = X and ||ReF||Lp(T)=Y\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${||{\rm Re}F||_{L^{p}(\mathbb{T})}=Y}$$\end{document} , then ||F||Lp(T)≥Cp(X,Y).\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$||F||_{L^p(\mathbb{T})}\geq C_p(X,Y).$$\end{document}This can be regarded as a reverse version of the classical estimates of Riesz and Essén. The proof rests on the exploitation of certain families of special subharmonic functions on the plane.


Introduction
Let u, v be two harmonic functions on the unit disc D, satisfying Cauchy-Riemann equations and normalized so that v(0) = 0. A classical problem, which interested many mathematicians at the beginning of the previous century, is the following: How is the size of v controlled by the size of u? Here, the size of a function is measured, for instance, by the L p norm on the unit circle T equipped with the normalized Haar measure m. In other words, for which 1 ≤ p ≤ ∞ is there a finite constant C p , depending only on p, such that for all u, v as above? This question was answered by Riesz [17,18]: the above estimate holds with some C p < ∞ if and only if 1 < p < ∞. This result is of fundamental importance to harmonic and complex analysis and has been modified and extended in numerous directions (cf. [2,[4][5][6]9,11,23] to name just a few). Moreover, the methods presented in the works of Riesz have led to the development of many areas of research (e.g., interpolation 128 A. Osȩkowski MJOM theory, functional analysis) and have had a profound influence on the shape of contemporary mathematics. The question about the best value of C p has gained some interest in the literature. It was answered partially by Gohberg and Krupnik [8]: by the use of a clever inductive argument, it was shown there that C p = cot(π/(2p)) if p = 2, 4, 8, . . .. The identification of C p in the full range p ∈ (1, ∞) is due to Pichorides [16] and Cole (unpublished: see Gamelin [7]): we have C p = cot(π/(2p * )), where p * = max{p, p/(p − 1)}. There are several papers which treat related sharp estimates for conjugate harmonic functions on the disc; see e.g., Aarão and O'Neill [1], Davis [5], Janakiraman [10], Nazarov and Treil [14], Osȩkowski [15], and consult references therein.
One can look at the estimate (1.1) from a slightly different perspective. Obviously, u can be regarded as the real part of the holomorphic function u+iv on the unit disc. Consequently, by the triangle inequality, the inequality (1.1) is equivalent to the following statement: if F is a holomorphic function on D satisfying the normalization condition Im F (0) = 0, then for some finite E p depending only on p. Actually, as Essén [6] proved, the choice E p = sin −1 (π/(2p * )) is optimal. See also Tomaszewski [21] for the sharp weak-type counterpart of this estimate. The purpose of this paper is to study a certain reverse version of (1.2). Clearly, if F is a holomorphic function on the unit disc (with no additional assumptions on Im F (0)), then we have Of course, this bound is sharp: equality holds for constant real functions. However, one can study the following more sophisticated version of this problem: namely, find the sharp analog of (1.3) subject to the restrictions Clearly, the answer to this question provides us with more detailed information on the behavior of the operator F → Re F . Such a type of problems appears in many places in the literature, in the study of other classical operators and objects in harmonic analysis. See e.g., Melas [12], Melas et al. [13] and Slavin et al. [19] and equality can hold, for instance, if we take F (z) = z √ 2Y 2 − 2X 2 + X. What about other values of p? This question is answered in Theorem 1.1 to formulate which we will need some auxiliary notation. Let X, Y be two real numbers satisfying |X| ≤ Y . For 1 < p ≤ 2, define The existence and uniqueness of φ p follow from the fact that the right-hand side of (1.5) is a continuous and strictly decreasing function of φ, which takes value 1 at φ = 0 and converges to 0 as φ → π/(2p).
We are ready to state our main result.
For each p, X and Y , the number C p (X, Y ) cannot be replaced by a smaller number.
The proof of this statement rests on the existence of certain families of subharmonic functions on the plane. In the next section, we study the case 1 < p ≤ 2 of the above theorem, and Sect. 2 of the paper is devoted to the case 2 < p < ∞. In the final part of the paper, we sketch some ideas which lead to the discovery of special functions used in Sects. 2 and 3.

The Case 1 < p ≤ 2
As we have mentioned above, the proof will rest on the existence of certain special subharmonic functions. Introduce U p : [0, ∞) 2 → R by where R ≥ 0 and θ ∈ [0, π/2], stand for the polar coordinates. Let us extend U p to the whole plane R 2 by the requirement for all x, y ∈ R. One easily checks that the function U p is continuous; further properties of U p are gathered in a lemma below.
Proof of (1.6). We are ready to establish the lower bound'. Let us fix a function F as in the statement and gather some information. First, since Re F belongs to L p (T), the restriction F | T also has this property, by Riesz' theorem. But, clearly, we have |U p (x, y)| ≤ c p R p for all x, y and some c p depending only on p, so the restriction of U p • F to the unit circle is integrable. Finally, observe that by the third part of the above lemma, the composition U p • F is a subharmonic function on the unit disc. Therefore, using the first two parts of the lemma and the mean-value property, we obtain This is equivalent to which is precisely the desired lower bound.
Sharpness. Let X be an arbitrary real number and fix ε > 0. Pick ϕ 0 ∈ (0, π/2 − π/(2p)) and and This will yield the claim: if we take F to be the conformal map sending D onto A and 0 onto (X, 0), then we will have Re F (0) = X and, by (2.2) and (2.3), so the sharpness will hold due to the fact that ε is arbitrary. By symmetry and continuity, we may and do assume that X > 0. Let us start with (2.2). The left-hand side does not change if we translate the angle A and the point (X, 0) by the vector (0, −M ). For the translated angle, the analysis of the harmonic measure is simpler: the function is harmonic on A + (0, −M ) and equals (x, y) → |x| p on the boundary of this set. Hence where ψ is the angle corresponding to the point (X, −M ) (that is, ψ = arctan(−M/X)). So, we can rewrite the above identity in the equivalent form Now, suppose that ϕ 0 is close to π/2 − π/(2p). The largest allowed value of M is −X tan ϕ 0 : then the point (X, 0) lies at the boundary of A, so μ (X,0) A = δ (X,0) and hence ∂A |x| p dμ (X,0) A = X p . On the other hand, if we let M → −∞, then ψ → π/2 and ∂A |x| p dμ (X,0) A → ∞. Finally, if we fix X and ϕ 0 , then the right-hand side of (2.4) is a strictly increasing function of M . Hence there is a unique number M = M (ϕ 0 ) such that (2.2) holds. The crucial observation is that M (ϕ) is bounded and ψ → π/2 − π/(2p) as ϕ 0 → π/2 − π/(2p). Indeed, when ϕ 0 approaches π/2 − π/(2p), then we have

The Case 2 ≤ p < ∞
Recall the number φ p defined in (1.5) and introduce the parameters .
Consider the function U p : [0, ∞) × R → R, given by where, as previously, we have used the polar coordinates. Let us extend U p to the whole R 2 , setting U p (x, y) = U p (−x, −y) for all x, y ∈ R. As in the previous section, first we study some elementary properties of this special function.

134
A. Osȩkowski MJOM (ii) For all x, y ∈ R we have the majorization Proof. (i) It suffices to show the inequality ∂ ∂y U p (x, y) ≥ 0 for x, y > 0. This is evident if θ ≥ φ p , so let us assume that θ ∈ (0, φ p ). A direct differentiation gives as needed. (ii) By symmetry, we may assume that x, y > 0. Clearly, it suffices to verify the majorization for θ ∈ (0, φ p ). We rewrite the bound in the equivalent form Both sides are equal when θ = φ p , so it is enough to prove that G is nondecreasing. We derive that G (θ) equals Since p − 1 ≥ 1 and (p − 1)θ ≤ (p − 1)φ p ≤ π/2, we will be done if we prove that for any fixed 0 < s < t, the function ξ(α) = sin αs sin αt is nondecreasing on (0, π/(2y)). We compute that and note that the function ζ(u) = u cot u is decreasing on (0, π/2): ζ (u) = (2 sin 2 u) −1 (sin 2u − 2u) ≤ 0. This implies that ξ ≤ 0 and the majorization follows. (iii) The function U p is of class C 1 on the plane, and it is harmonic on the set {|θ| < φ p }. Consequently, it suffices to check that the Laplacian of U p is nonnegative on {|θ| > φ p }. We derive that on this set, we have Vol. 13 (2016) Holomorphic Functions 135 Note that lim r→0 p p − 1 − sin(pr) cos r sin((p − 1)r) = 0.
Equipped with the above lemma, we turn our attention to Theorem 1.1.
Proof. Proof of (1.6) The reasoning is the same as in the case 1 < p ≤ 2: we obtain or, equivalently, ||F || p p ≥ c p Y p + α p |X| p . Now, by the definition of c p , α p and the identity (1.5), the latter estimate is equivalent to which is the claim.

Sharpness.
Here the reasoning is a bit simpler than in the case 1 < p ≤ 2. Consider the angle A = {(x, y) : x > 0, |θ| ≤ φ p } and let μ be the harmonic measure on ∂A with respect to the point (X, 0). The restriction of the function U p to the set A is harmonic and U p (x, y) = x p (cos −p φ p − c p ) for (x, y) ∈ ∂A. Consequently, by the mean-value property, we see that Consequently,

136
A. Osȩkowski MJOM Therefore, if F is the univalent mapping which sends D onto A and 0 onto (X, 0), then Re F (0) = X, || Re F || L p (T) = Y and So, the lower bound (1.6) is attained and the proof is complete.

On the Search of the Special Functions
In this section, we will explain briefly some informal argumentation which leads to the discovery of the special functions U p used above. We will focus on the case 2 ≤ p < ∞, for 1 < p < 2 the reasoning is similar. For the sake of clarity, let us start with the general idea behind the proof of Theorem 1.1. Given 2 ≤ p < ∞ and a constant β > 0, one searches for the optimal (i.e., the largest) constant γ(p, β) such that for all holomorphic functions F on the unit disc. This clearly gives some initial insight into (1.6): having analyzed (4.1), we see that Is this bound optimal? To answer this question, suppose that for each p and β, there is an extremizer: a nonzero function F = F p,β for which both sides are equal. Clearly, for any p and β such an object is not unique: the inequality (4.1) is homogeneous of order p, so if we multiply an extremizer by a constant, we again obtain an extremizer. It is evident how to proceed: we take the number β for which the supremum in (4.2) is attained, consider the extremizer of (4.1), scaled so that Re F p,β (0) = X, and verify that it satisfies || Re F || p = Y . So, we see that the problem boils down to a thorough analysis of the inequality (4.1). Next, the reasoning presented in the papers [2,6,16] links the validity of the estimate (4.1) with the existence of special functions on the plane. Namely, given p and β, we search for a largest subharmonic function U p,β on R 2 , satisfying the majorization Then, as we have already seen in the previous sections, the mean-value property yields: ). Hence, one is forced to take γ(p, β) = inf x,y U p,β (x, y)/|x| p .
How to find U p,β ? Such a function, if it exists, must satisfy the symmetry condition U p,β (x, y) = U p,β (x, −y) = U p,β (−x, y) for all x, y ∈ R; indeed, if this did not hold, we could replace U p,β with a larger subharmonic function for some function g p,β to be found. Now, we assume that U p,β is of class C 2 ; despite the fact that the function we obtain at the end does not have this regularity, it will facilitate our further considerations. A closer look at the papers [2,6,16] suggests that there is a number κ(p, β) > 0 such that either (i) U p,β is harmonic on the set {(x, y) : |y| < κ(p, β)|x|}, and U p,β (x, y) So, we have two possibilities to check. Suppose that (i) holds true and takes φ(p, β) ∈ (0, π/2) such that κ(p, β) = tan φ(p, β). A direct calculation shows that ΔU p,β = 0 if and only if g (θ) + p 2 g(θ) = 0, so we see that for some unknown constants a 1 , a 2 . Since U p,β (x, y) = U p,β (x, −y), we conclude that a 2 = 0. To derive a 1 , we use the fact that U p,β is of class C 1 and see what happens on the common boundary of the sets {(x, y) : |y| > κ(p, β)|x|} and {(x, y) : |y| ≤ κ(p, β)|x|}. By the continuity of U p,β , we get that a 1 cos(pφ(p, β)) = 1 − β cos p φ(p, β), while the comparison of the partial derivatives yields −a 1 sin(pφ(p, β)) = β cos p−1 φ(p, β) sin φ(p, β).
It remains to check that the right-hand side of (4.2) is precisely the constant C p (X, Y ); when we take φ(p, β) = φ p (defined in (1.5)), then U p,β is precisely the function U p used above.

138
A. Osȩkowski MJOM Finally, let us say a few words about the search for the appropriate extremizers F in (4.1). A look at the above proof immediately gives three conditions on F . First, we must have equality in the majorization (4.3), i.e., U p,β (F (e iθ )) = |F (e iθ )| p − β| Re F (e iθ )| p for almost all θ ∈ [−π, π). That is, F must send the unit circle into the set {(x, y) : |x| ≥ κ(p, β)|y|}. The second condition is that the mean value property, applied to the subharmonic function U p,β • F , must return equality: this suggests that F must send the open unit disc into {(x, y) : x > 0, |y| < κ(p, β)|x|}, inside which U p,β is harmonic. Finally, we must have F (0) = X; this will guarantee the equality U p,β (F (0)) = γ(p, β)|X| p . Let us combine the three observations: we see that a natural choice for F is a conformal mapping of D onto the angle A = {(x, y) : x > 0, |y| ≤ κ(p, β)x}, sending 0 ∈ D onto (X, 0) ∈ A. One can check that this is indeed the right choice.