Reverse Carleson measures in Hardy spaces

We give a necessary and sufficient condition for a measure μ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mu $$\end{document} in the closed unit disk to be a reverse Carleson measure for Hardy spaces. This extends a previous result of Lefèvre, Li, Queffélec and Rodríguez-Piazza Lefèvre et al. (Rev Mat Iberoam 28(1):57–76, 2012). We also provide a simple example showing that the analogue for the Paley–Wiener space does not hold. As it turns out the analogue never holds in any model space.


Introduction
For 1 ≤ p < ∞ let H p be the Hardy space on the unit disk D equipped with its usual norm where S I = {z ∈ D: 1 − |I | ≤ |z| ≤ 1, z/|z| ∈ I } is the usual Carleson window. This theorem has been extended to several other spaces, like Bergman, Fock, model spaces etc., and we refer the reader to the huge bibliography on this topic for further information.
Note that H p contains a dense set of continuous functions for which the embedding (1.1) still makes sense when the measure has a part supported on the boundary. Then (1.2) implies that the restriction of the measure μ to the boundary has to be absolutely continuous with respect to Lebesgue measure and with bounded Radon-Nikodym derivative. It is thus possible to consider, more generally, positive, finite Borel measures supported on the closed unit disk: Here, we are interested in reverse Carleson inequalities f p ≤ C f L p (D,μ) , f ∈ C(D) ∩ H p (D), 1 < p < ∞. In [9] Lefèvre et al. proved that when μ is already a Carleson measure these hold if and only it there exists C > 0 such that for all arcs I ⊂ ∂D μ(S I ) ≥ C|I |.
We will show that this result can be deduced from a well-known balayage argument which does not require the Carleson condition. It will be clear from this argument that we have a reproducing kernel thesis for the reverse embedding: if the embedding holds on the reproducing kernels, then it actually holds for every function.
It turns out that the interesting part of the measure has to be supported on the boundary, while the part supported in the disk can be dropped.
Finally, we provide a simple example showing that the analogous reproducing kernel thesis for the reverse embedding in the Paley-Wiener space does not hold. We will actually show that the reproducing kernel thesis for the reverse embedding never holds in any model space. In a previous version of this work, the construction valid in the Paley-Wiener space was generalized to the situation of so-called one-component inner functions. We are very grateful to Anton Baranov who suggested the shorter proof presented below and which gives the result in general model spaces.
We shall use the following standard notation: f g means that there is a constant C independent of the relevant variables such that f ≤ Cg, and f g means that f g and g f .
Our main result reads as follows.
Theorem 2.1 Let 1 < p < ∞ and let μ ∈ M + (D). Then the following assertions are equivalent: The key implication of the above result if of course (2) ⇒ (3) which is based on a balayage argument.
Observe that in this theorem we do not require absolute continuity of the restriction μ| ∂D . Still, if we want to extend (1) to the entire H p -space, then, in order that D | f | p dμ makes sense for every function in H p , we need to impose absolute continuity on μ| ∂D . Note that the integral D | f | p dμ can be infinite for certain f ∈ H p when the Radon-Nikodym derivative of μ| ∂D is not bounded.
(3) ⇒ (4). Take h > 0 so that |I |/ h is a large integer N and consider the modified Carleson window Split I into N subarcs I k such that |I k | = h (and hence S I k ,h = S I k ). Then [11,Theorem 2.18]) we thus have We deduce that the Lebesgue measure on ∂D, denoted by m, is absolutely continuous with respect to the restriction of μ to ∂D and that the corresponding Radon-Nikodym derivative of μ is bounded below by C 3 . In particular one can choose Observe that when p = 2, then |K λ (z)| 2 is nothing but the Poisson kernel, for which the arguments below are very transparent. Let us however do the argument for general p. By hypothesis, integrating over S I,h with respect to area measure d A on D we get so that the previous estimate becomes Indeed, if z / ∈ I , then there are δ, h 0 > 0 such that for every 0 < h < h 0 and for every λ ∈ S I,h , we have |1 − λz| ≥ δ > 0, and the result follows from the estimate Suppose now that z = e iθ 0 ∈ I . Let h ≤ |I |, then setting λ = (1 − t)e iθ for λ ∈ S I,h we have Since 0 ≤ t ≤ h ≤ |I | and z = e it ∈ I , the set {e iθ : |θ − θ 0 | ≤ t, e iθ ∈ I } contains an interval of length at least t/2, we get On the other hand, integrating in polar coordinates, we get Hence ϕ h converges pointwise to a function comparable to χ I , and ϕ h is uniformly bounded in h. Now, from (2.1) and by dominated convergence we finally deduce that Let us see that, on the other hand, the μ-norm of the normalized reproducing kernels is uniformly bounded from below. If λ is such that | Im λ| > 1 then | sin(π(x n − λ))| e π | Im λ| , and hence | Im λ| |x n − λ| 2 1.
It is thus enough to consider points λ ∈ C with | Im λ| ≤ 1. Let x n 0 be the point of S closest to λ; then there is δ > 0, independent of λ, such that It is interesting to point out that μ is a Carleson measure for PW π , since S is in a strip and separated.

Failure in general model spaces
The previous construction can be generalized to model spaces in the disk. The model space associated to an inner function is K = H 2 H 2 , and the reproducing kernel corresponding to λ ∈ D is given by If is a finite Blaschke product of degree strictly bigger than one, picking for instance μ = δ 0 we immediatly get the reverse inequality on reproducing kernels [see (3.1)]. Clearly μ is Carleson, and since the degree of is not one, we can combine two linearly independent functions of K vanishing at 0.
In the general case we need to construct a measure supported on T.

Theorem 3.1 Let be an inner function which is not a finite Blaschke product.
Then there exists a measure μ on T such that K ⊂ L 2 (μ), the measure μ satisfies the reverse estimate on reproducing kernels k λ ,

1)
but the reverse Carleson embedding for the space K does not hold.
Proof Let us first assume that vanishes at z 0 = 0, and let = z 0 . Denote by μ the Clark measure for 0 , that is μ is defined by Clark introduced these measures in [6]. Observe first that K = C⊕z K 0 which implies that K ⊂ L 2 (μ). Continuous functions are dense in K (see [3] or [5, p. 187]) and functions in K are μ-measurable (see [10]). In particular, if we had reverse embedding on continuous functions then we would have it on the whole space K . It is thus sufficient to find a non-zero function f ∈ K with zero L 2 (μ)-norm. To this end, pick f (z) = 1 − 0 (z) which belongs to K . Clearly f = 0 μ-a.e, so that | f | 2 dμ = 0. Let us show that (3.1) is satisfied. Any reproducing kernel k λ has representation as the (orthogonal) sum In particular, Also, since μ is a Clark measure for K 0 , we have Thus, we clearly have Assume that there exists a sequence λ n such that k λ n 2 ≤ 2(1 + μ(T) 1/2 ) and (3.1) does not hold for λ n with any positive C. Since the norms of the kernels are supposed bounded on λ n , this implies that Passing if necessary to a subsequence we may assume that λ n → λ 0 , and it follows from the fact that the norms k λ n 2 are uniformly bounded that even in the case when λ 0 ∈ T we still have that the kernel k λ 0 is correctly defined and belongs to K . Then, by the Fatou lemma, we have that k λ 0 L 2 (μ) = 0. Hence, 1 − (λ 0 ) (z) = 0 μ-a.e. But 0 (z) = 1 μ-a.e. Thus, 1 − (λ 0 )z = 0 μ-a.e. z, which is impossible if the support of μ contains at least two points (but it does, since 0 is not a single Blaschke factor). Let us now discuss the situation when does not vanish at 0. For a = (0), the Frostman shift has a zero at 0. By the above discussions, there is a measure μ a such that K a ⊂ L 2 (μ a ), the reverse estimate holds on the kernels: and there is a non zero function f 0 ∈ K a with f 0 L 2 (μ a ) = 0. Recall that the Crofoot transform U a : K −→ K a , defined by is isometric onto K a . An easy computation gives k a λ = U a (c a,λ k λ ), λ ∈ D, where c a,λ = 1 − |a| 2 1 − a (λ) .
Using (3.2) and the isometry property of the Crofoot transform, we get C k λ 2 = C U a k λ 2 ≤ U a k λ L 2 (μ a ) = k λ L 2 (μ) , and so μ satisfies (3.1). We also have the Carleson measure condition for this measure: for every f ∈ K Finally, since there is 0 = f 0 ∈ K a with f 0 L 2 (μ a ) = 0, take the unique 0 = g 0 ∈ K with U a g 0 = f 0 , then g 0 L 2 (μ) = U a g 0 L 2 (μ a ) = f 0 L 2 (μ a ) = 0.
Note that the above proof actually works for finite Blaschke product with degree at least 3.