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Extremal sequences for a problem related to the Hardy operator

We prove a characterization of the extremal sequences of the extremal problem for the Hardy operator, related to the Bellman function of the dyadic maximal operator. In fact we prove that they behave approximately like eigenfunctions of the Hardy operator for a specific eigenvalue.


Introduction
The dyadic maximal operator on R n is defined by M d φ(x) = sup 1 |Q| Q |φ(y)|dy : x ∈ Q, Q ⊆ R n in a dyadic cube (1.1) for every φ ∈ L 1 loc (R n ), where the dyadic cubes are those formed by the grids 2 −N Z n , for N = 0, 1, 2, . . . .
It is well known that it satisfies the following weak type (1,1) inequality: for every φ ∈ L 1 (R n ) and every λ > 0.
(1.2) as it is easily seen implies the following L p -inequality It is also easy to see that the weak type inequality (1.2) is best possible while (1.3) is also sharp.
Further study of the dyadic maximal operator leads us to introduce the following function of two variables, defined by (1.4) where Q is a fixed dyadic cube and 0 < f p ≤ F . The function (1.4), which is called the Bellman function of two variables of the dyadic maximal operator, is in fact independent of the cube Q and it's value has been given in [3]. More precisely it is proved there that where ω p : [0, 1] → 1, p p − 1 denotes the inverse function H −1 p of H p which is defined by H p (z) = −(p − 1)z p + pz p−1 , for z ∈ 1, p p − 1 .
As a matter of fact this evaluation has been done in a much more general setting where the dyadic sets are given now as elements of a tree T on a non-atomic probability space (X, µ). Then the associated dyadic maximal operator is defined by: |φ|dµ : x ∈ I ∈ T , (1.5) Additionally the inequalities (1.1) and (1.3) remain true and sharp even in this setting. Moreover, if we define for 0 < f p ≤ F , then B ′ p,T (f, F ) = B p (f, F ). In particular the Bellman of the dyadic maximal operator is independent of the structure of the tree T .
Another approach for finding the value of B p (f, F ) is given in [4] where the following function of two variables has been introduced: The first step, as it can be seen in [4], is to prove that S p (f, F ) = B p (f, F ). This can be viewed as a symmetrization principle of the dyadic maximal operator with respect to the Hardy operator. The second step is to prove that S p (f, F ) has the value that was mentioned above. Now the proof of the fact that S p = B p can be given in an alternative way as can be seen in [7]. In fact it is proved there the following result.
It is obvious that Theorem A implies the equation S p = B p , and gives an immediate connection of the dyadic maximal operator with the Hardy operator.
An interesting question that arises now is the behaviour of the extremal sequences of functions for the quantities (1.6) and (1.7). The problem concerning (1.6) has been solved in [5] where it is proved the following: Now it is interesting to search for the opposite problem concerning (1.7). In fact we will prove the following: Theorem 1. Let g n : (0, 1] → R + be a sequence of non-increasing functions continuous such that The proof is based on the proofs of Theorem A and B. Concerning now the problem (1.6) it can be easily seen that extract functions do not exist (when the tree T differentiates L 1 (X, µ)). That is for every φ ∈ L p (X, µ) with φ ≥ 0 and This is because a self-similar property that is mentioned in [6] and states that for every extremal sequence φ n for (1.6) the following is true: φ p dµ = F and if the tree T differentiates L 1 (X, µ) then we must have that µ-a.e the following equalities hold φ(x) = f and φ p (x) = F , that is f p = F which is the trivial case. It turns out that the above doesn't hold for the extremal problem (1.7). That is there exist extremal functions for (1.7). We state it as: Theorem 2. There exist g : (0, 1] → R + non-increasing and continuous with 1 0 g(u)du = f and As it is expected due to Theorem 1 it satisfies the following equality for every t ∈ (0, 1] which gives immediately gives (1.9).
After proving Theorem 2 we will be able to prove the following In this way we complete the discussion about the characterization of the extremal functions for this problem related to the Hardy operator.
The paper is organized as follows: We prove Theorem 1 and 2 and 3 in Sections 3 and 4 and 5 respectively while in Section 2 we give some preliminary definitions and results.

Preliminaries
Let (X, µ) be a non-atomic probability measure space. A set T of measurable subsets of X will be called a tree if it satisfies conditions of the following Definition 2.1.
i) X ∈ T and for every I ∈ T we have that µ(I) > 0. ii) For every I ∈ T there corresponds a finite or countable subset C(I) ⊆ T containing at least two elements such that (a) the elements of C(I) are pairwise disjoint subsets of I Examples of trees are given in [3]. The most known is the one given by the family of all dyadic subcubes of [0, 1] n .
The following has been proved in [3]. We will also need the following fact obtained in [7].
Now given a tree on (X, µ) we define the associated dyadic maximal operator as follows We will also the following from [7]. Lemma 2.3. Let k ∈ (0, 1] and K measurable subset of X with µ(K) = k. Then the following inequality holds

Proof of Theorem 1
We will prove Theorem 1 by repeating the proof of Theorem A and by using Theorem B.
We begin with a sequence (g n ) n of non-increasing continuous functions g n : (0, 1] → ω p (f p /F ) and we suppose that (g n ) n is extremal for (1.7), that is Our aim is to prove that For this reason it is enough to prove that We consider the quantity in (3.1) and similarly we work in the second. Set A n = t ∈ (0, 1] : 1 t t 0 g n > cg n (t) so we need to prove that Since (x − y) p < x p − y p , for x > y > 0 and p > 1 it is enough to prove that For each A n , which is an open set of (0, 1] we consider it's connected components I n,i , It is clear that such choice of i n exists. Then |II n − II 1,n | < 2ε where We need to find a n 0 ∈ N such that II 1,n < ε, ∀ n ≥ n 0 . Fix now a g n =: g. We prove the following Lemma 3.1. There exists a family φ a : (X, µ) → R + of rearrangements of g (φ * a = g for each a ∈ (0, 1)) such that for each γ ∈ (0, 1] there exists a family of measurable subsets of X, S (γ) a satisfying the following: for each a γ < γ ′ ≤ 1 and a ∈ (0, 1).
Proof. We follow [7]. Let a ∈ (0, 1). By using Lemma 2.1 we choose for every I ∈ T a family F(I) ⊆ T of disjoint subsets of I such that We define S = S a to be the smallest subset of T such that X ∈ S and for every I ∈ S, F(I) ⊆ S. We write for I ∈ S, A I = I We define also for I ∈ S a , rank(I) = r(I) to be the unique integer m such that I ∈ S a,(m) .
Additionally, we define for every I ∈ S a with r(I) = m We also set for I ∈ Using now Lemma 2.2 we see that there exists a rearrangement of τ a |S m S m+1 = τ Define now φ a : Let now I ∈ S a,(m) . Then Now for x ∈ S m S m+1 , there exists I ∈ S a,(m) such that x ∈ I so

4)
Since µ(S m ) = (1 − a) m , for every m ≥ 0 we easily see from the above that we have where ∪S a,(m,a) denotes the union of the elements of S a,(ma) . This is S ma = I∈S a,(ma)
Since (1 − a) ma → γ and the right hand side of (3.6) expresses a Riemann sum of the Then by (3.5) we have equality on (3.7).
We thus constructed the family (φ a ) a∈(0,1) we need for which we easily see that if for each ℓ ∈ (0, 1). We now return to the proof of Theorem 1. We remind that (a n,i 1 b n,i ), which is a disjoint union. Thus Now, for every n ∈ N we consider the corresponding to g n , family (φ a,n ) a∈(0,1) and the respective subsets of X S (a n,i ) a,n , S (b n,i ) a,n , a ∈ (0, 1), i = 1, 2, . . . , n i for which µ S (a n,i ) a,n → a n,i and µ S We can also suppose that a n,i < b n,i ≤ a n,i+1 < b n,i+1 , i = 1, 2, . . . , i n − 1.
Then we also have that and similarly for the other endpoint b n,i of I n,i . Therefore, by (3.8) there exists for every n ∈ N an a 0,n ∈ (0, 1) such that 0 < a < a 0,n ⇒ |III 1,n Additionally, we can suppose because of the relation , for each n ∈ N and since g n is extremal for the problem (1.7) that a 0,n can be chosen such that for every a ∈ (0, a 0,n ) Choose a ′ n ∈ (0, a n ) and from the sequence φ a ′ n ,n =: φ n . Then, because of (3.9) and since φ * n = g n we have that φ n is extremal for (1.6). Because of the Remark, after the proof of Lemma 3.1 we have now for every ℓ ∈ N, each n ∈ N and a ∈ (0, 1), that φ a,n S a,(ℓ) * n ) = |F n |, for every n ∈ N we can suppose that a 0,n satisfies the following we must have additionally, since φ a ′ n,n = φ n , that and that µ(Λ n ) − |F n | < 1 n , for every n ∈ N. It is also easy to see because of the above relations and the Remark 3.1 and the form of Λ n (passing to a subsequence if necessary) that We now take advantage of Theorem B.
Since φ n is extremal for (1.6) we must have that X |M T φ n − cφ n | p dµ → 0, as n → ∞ where c = ω p (f p /F ) p . This implies: we must have, because of the definition of Λ ′ n that: In the same way we prove that: Because now of (3.10) we have that and from the choice of F n we see that we must have that II n < 2ε, for n ≥ n 0 , for a suitable n 0 ∈ N. And this is what we wanted to prove.

Uniqueness of extremal functions
In this section we will prove that there exists unique g 0 : (0, 1] → R + with This is the statement of Theorem 2. Proof of Theorem 2. By Theorem 1 it is obvious that if such a function g 0 exists, it must satisfies 1 t  So, in order that g 0 satisfies (4.1) we need to set g 0 (t) = kt −1+ 1 c , t ∈ (0, 1], and search for a constant k such that (by solving the respective first order linear differential equation) Se, we ask if g 0 for this k satisfies the second equation. This is Because now of the form of g 0 : (0, 1] → R + we have that So g 0 is the only extremal function in (0, 1].

Uniqueness of extremal sequences
We are now able to prove Theorem 3. The direction ii)⇒i) is obvious from the conditions that g satisfies. We now proceed to ii)⇒i) We suppose that we are given g n : (0, 1] → R + non-increasing, continuous, such Using Theorem 2 we conclude that Thus there exists a subsequence (g kn ) n such that if F n (t) = 1 t t 0 g n − cg n (t), t ∈ (0, 1], n ∈ N, then F kn → 0 almost everywhere (with respect to Lesbesgue measure). By a well known theorem in measure theory we have because of the finiteness of the measure space that F kn → 0 uniformly almost every where on (0, 1]. This means that there exists a g kn , (5.1) where f = 1 0 g kn . Now by Holder's inequality we have that Thus II ≤ δf a + 1 a δ 1− 1 p F . We consequently have that for a given ε > 0 there exists δ = δ a,ε > 0 for which the following implication holds t, t ′ ∈ [a, 1] H kn , |t − t ′ | < δ ⇒ |g kn (t) − g kn (t ′ ) < ε, for every n ∈ N. (5.2) Thus (g kn ) n has a property of type of equicontinuity on a certain set that depends on a. We consider now an enumeration of the rationals in (0, 1], let {q 1 , q 2 , . . . , q k 1 , . . .} = Q ∩ (0, 1].
For every q ∈ Q∩(0, 1] we have that (g kn (q)) n is a bounded sequence of real numbers, because g kn is a sequence of non-negative, non-increasing functions on (0, 1] satisfying sup n 1 0 g kn < +∞.
By a diagonal argument we produce a subsequence which we denote again by g kn such that g kn (q) → λ q , n → ∞ where λ q ∈ R + , q ∈ Q ∩ (0, 1].
H kn , which is a set of Lesbesgue measure zero, and suppose that x ∈ (a, 1) H. Then x > a, and there exist a n 0 ∈ N such that x / ∈ H kn 0 , so that x / ∈ H kn , ∀ n ≥ n 0 . Additionally, choose a sequence (p k ) k of rationals on (a, 1) H n 0 such that p k → x. This is possible because the set (q, 1) H n 0 is an open set. Thus, we have that p k > a and p k / ∈ H kn , n ≥ n 0 , k ∈ N. Let now k 0 ∈ N : |p k − x| < δ, ∀ k ≥ k 0 , where δ is one given in (5.2). We then have that |g kn (x) − g km (p k 0 )| < ε, for every n ∈ N. Thus, for every such x, and every n, m ∈ N we have that |g kn (x) − g km (x)| ≤ |g kn (x) − g kn (p k 0 )| + |g kn (p k 0 ) − g km (p k 0 )| + |g km (p k 0 ) − g km (x)| < 2ε + |g kn (p k 0 ) − g km (p k 0 )|.
But (g kn (p k 0 )) n is convergent sequence, thus Cauchy. Then (g kn (x)) n is a Cauchy sequence for every x ∈ (a, 1) H for every a ∈ (0, 1]. Thus (g kn (x)) n is a Cauchy sequence in all (0, 1] H. As a consequence there exists g ′ 0 : (0, 1] → R + such that g kn → g ′ 0 a.e. on (0, 1] ⇒ g kn → g ′ 0 uniformly a.e. on (0, 1]. (5.3) But this easily implies since F kn (t) → 0, a.e. that This gives us that g ′ 0 is the function constructed in Theorem 2. Additionally we obtain The validity of (5.4) can be concluded from the following Remark 5.2. In [3] it can be seen that the following is true Then the supremum is (5.6) is bounded above B p (ℓ, F, δ) for every δ ∈ (0, 1]. Thus, if we work on every subsequence of (g n ) n which is again extremal we produce a subsequence of it for which (5.5) is satisfied. Therefore the proof of Theorem 3 is completed.