Sets of uniqueness, weakly sufficient sets and sampling sets for weighted spaces of holomorphic functions in the unit ball
aa r X i v : . [ m a t h . C V ] A p r SETS OF UNIQUENESS, WEAKLY SUFFICIENT SETSAND SAMPLING SETS FOR WEIGHTED SPACES OFHOLOMORPHIC FUNCTIONS IN THE UNIT BALL
BINGYANG HU , LE HAI KHOI Abstract.
We consider inductive limits of weighted spaces ofholomorphic functions in the unit ball of C n . The relationshipbetween sets of uniqueness, weakly sufficient sets and samplingsets in these spaces is studied. In particular, the equivalence ofthese sets, under general conditions of the weights, is obtained. Introduction
Let B be the unit ball in C n and O ( B ) be the collection of holomor-phic functions on B with the usual compact-open topology. A function ϕ defined on [0 ,
1) is said to be a weight if it takes the values in (1 , + ∞ ).We define the associated weighted space A ϕ as A ϕ := (cid:26) f ∈ O ( B ) : k f k ϕ := sup z ∈ B | f ( z ) | ϕ ( | z | ) < + ∞ (cid:27) . In this paper, we are interested in the case where the function spaceis defined by the inner inductive limit of A ϕ ’s. More precisely, letΦ = ( ϕ p ) ∞ p =1 denote an increasing sequence of weights. For simplicity,we use k f k p instead of k f k ϕ p , and A − pϕ instead of A ϕ p . Then clearly A − pϕ ֒ → A − ( p +1) ϕ (here ֒ → means a continuous embedding), and hence,we let A −∞ Φ := [ p ≥ A − pϕ with the topology induced by the inner inductive limit of A − pϕ , namely( A −∞ Φ , τ ) = lim ind A − pϕ . Now let S be a subset set of B . Define A − p,Sϕ := (cid:26) f ∈ A −∞ Φ : k f k p,S = sup z ∈ S | f ( z ) | ϕ p ( | z | ) < ∞ (cid:27) . Date : April 25, 2019. Supported in part NSF grant DMS 1600458 and NSF grant 1500182. Supported in part by MOE’s AcRF Tier 1 grant M4011724.110 (RG128/16). , LE HAI KHOI Notice that the inclusion relations A − pϕ ֒ → A − p,Sϕ ֒ → A −∞ Φ always hold.Hence, it follows immediately that A −∞ Φ = [ p ≥ A − pϕ ⊆ [ p ≥ A − p,Sϕ ⊆ A −∞ Φ , which implies A −∞ Φ = [ p ≥ A − pϕ = [ p ≥ A − p,Sϕ . Therefore, we can endow A −∞ Φ with another weaker inner inductivelimit topology of seminormed spaces A − p,Sϕ :( A −∞ Φ , τ S ) := lim ind A − p,Sϕ . We note that this type of spaces appeared as duals to Fr´echet-Schwartz (FS) spaces and play an important role in the study of rep-resenting holomorphic functions in series of simpler functions, such asexponential functions, or rational functions, which have many appli-cations in functional equations and approximations of functions (see,e.g., [3, 4, 5, 6, 10, 11, 12] and references therein).We are interested in the following three special cases of S . Definition 1.1.
A set S ⊆ B is said to be weakly sufficient for thespace A −∞ Φ if two topologies τ and τ S are equivalent. Definition 1.2.
A set S ⊆ B is called a set of uniqueness for A −∞ Φ if f ∈ A −∞ Φ and f ( z ) = 0 for all z ∈ S imply that f = 0.Furthermore, let a continuous function ψ : [0 , → (0 , ∞ ) satisfylim r → ψ ( r ) = ∞ . For f ∈ A −∞ Φ , we put T f,ψ = lim sup | z |→ log | f ( z ) | ψ ( | z | ) , and for S ⊂ B , T f,ψ,S = lim sup | z |→ ,z ∈ S log | f ( z ) | ψ ( | z | ) . Definition 1.3.
A set S ⊆ B is called a ψ -sampling set for A −∞ Φ , if T f,ψ = T f,ψ,S for every f ∈ A −∞ Φ .A classical example of Φ and ψ isΦ = (cid:8) ϕ p ( r ) = (1 − r ) − p , p ∈ N (cid:9) , ψ ( r ) = | log(1 − r ) | . Such a choice of Φ and ψ is motivated by the easy fact that dist( z, ∂ B ) =1 − | z | , z ∈ B . Moreover, in this case, the space A −∞ Φ becomes the well-known function space A −∞ , which consists of all the holomorphicfunctions on B with polynomial growth.The property of weakly sufficient sets, sets of uniqueness and sam-pling sets for A −∞ was carefully studied in [9] for the case n = 1 and in[2, 7] for the higher dimension case. More precisely, the the followingresults have been obtained in these papers. Theorem 1.4.
Let S be a subset of the unit disc B (respectively, D ).Consider the following assertions (i) S is classical sampling for A −∞ ( B ) (respectively, D ); (ii) S is weakly sufficient for A −∞ ( B ) (respectively, D ); (iii) S is a set of uniqueness for A −∞ ( B ) (respectively, D ).Then ( i ) = ⇒ ( ii ) = ⇒ ( iii ) . It should be noted that the inverse implications, in general, are nottrue. In [9], two counter-examples are provided to show a failure ofboth inverse implications. Nevertheless, in [7] it was showed, for theclassical ψ , that ( iii ) = ⇒ ( ii ), with an additional assumption.A careful study of proofs of the results above led us to the thoughtthat ( ii ) does not imply ( i ) due to “ too independent” growths of Φand ψ , while ( iii ) does not imply ( ii ) because the assumption of S tobe a set of uniqueness is not strong enough to ensure an inclusion tobecome a continuous embedding.So a question to ask is for what Φ and ψ , we can have( i ) ⇐⇒ ( ii ) ⇐⇒ ( iii ) ?In the recent paper [1], the affirmative answer was given for the case ϕ p ( r ) = e pg ( r ) , < p < ∞ and ψ ( r ) = g ( r ), where g is a so-called everywhere quasi-canonical weight (see, [1, Section 4]).The aim of this paper is to solve the question above for a general Φand ψ , with the “minimal” assumptions imposed on them.The structure of this paper is as follows. In Section 2, we show that( ii ) = ⇒ ( iii ) always. Then we introduce conditions ( C ) − ( C ) for Φ(for which the classical weights are satisfied) and show that with thesame additional assumption as for the classical case, ( iii ) = ⇒ ( ii )(Theorem 2.5). The proof follows the scheme in [7]. Section 3 dealswith the equivalence ( ii ) ⇐⇒ ( i ). We introduce conditions ( C ) − ( C )which “relate” growths of Φ and ψ ” and, together with ( C ) , ( C ), allowus to establish each of implications ( i ) = ⇒ ( ii ) and ( ii ) = ⇒ ( i ).2. Weakly sufficient sets and sets of uniqueness
The following characterization of weakly sufficient sets is needed inthe sequel.
BINGYANG HU , LE HAI KHOI Lemma 2.1. [8, Proposition 2.2]
For the space A −∞ Φ , the followingstatements are equivalent: (a) S is weakly sufficient set. (b) For any p ≥ , there exists m = m ( p ) , such that A − p,Sϕ ֒ → A − mϕ , i.e., for any p ≥ , there exist m = m ( p ) and C = C ( p ) , suchthat k f k m ≤ C k f k p,S , for all f ∈ A − p,Sϕ . (c) For any p ≥ , there exist m = m ( p ) and C = C ( p ) , such that k f k m ≤ C k f k p,S , for all f ∈ A −∞ Φ . The following proposition is an easy modification of [7, Proposition3.1], with replacing the factor (1 − | z | ) p by ϕ p ( | z | ) − , and hence weomit the proof here. Proposition 2.2.
Any weakly sufficient set for A −∞ Φ is a set of unique-ness for this space.Remark . We give an example of a set of uniqueness which is nota weakly sufficient set. Let S = { z ∈ B : | z | ≤ / } , ϕ p ( r ) = (1 − r ) − p , p ∈ N , and the test functions f k ( z ) = 3 − k (1 − z ) − k , k ∈ N , where z = ( z , . . . , z n ) ∈ B . Clearly, S is a set of uniqueness. Moreover,for a fixed p >
0, a direct calculation shows k f k k p ,S is uniformlybounded in k , but for any p > k f k k p = ∞ for k sufficiently large,which implies S fails to be a weakly sufficient set.Now we proceed the other direction, and we are inspired by the ideafrom [7]. For a given Φ = ( ϕ p ) ∞ p =1 , we consider the following conditions( C ) lim r → ϕ p +1 ( r ) ϕ p ( r ) = ∞ , for all p ≥ , and( C ) for all p ≥ , ϕ p +1 ϕ p is bounded on any compact subset of [0 , . Clearly, if all ϕ p are continuous on [0 , C ) is satisfied. Example 2.4.
We give some examples to show that these conditionsare independent each of other. (1)
The classical weights
Φ = ((1 − r ) − p ) ∞ p =1 satisfy both ( C ) and ( C ) . (2) For any a > , the weights Φ = (cid:16) a + r − p (cid:17) ∞ p =1 satisfy ( C ) but not ( C ) . (3) For each p ∈ N , let ϕ p ( r ) = ((cid:0) − r (cid:1) − p , r ∈ (cid:2) , (cid:1) ,p p , r ∈ (cid:2) , (cid:1) . Then the weights
Φ = ( ϕ p ) ∞ p =1 satisfy ( C ) but not ( C ) . (4) For p ∈ N , consider ϕ p ( r ) = ((cid:0) − r (cid:1) − p , r ∈ (cid:2) , (cid:1) , , r ∈ (cid:2) , (cid:1) . Then the weights
Φ = ( ϕ p ) ∞ p =1 do not satisfy neither ( C ) nor ( C ) . The following is the main result in this section.
Theorem 2.5.
Suppose Φ satisfies conditions ( C ) and ( C ) . Then S ⊂ B is a weakly sufficient set for A −∞ Φ if and only if (a) S is a set of uniqueness for A −∞ Φ . (b) For any p ≥ , there exists an m = m ( p ) , such that A − p,Sϕ ⊂ A − mϕ . This result shows that under conditions ( C ) − ( C ), a set-inclusion(statement (b)) implies a continuous embedding. Proof.
Since the necessity is obvious, it suffices to show the sufficiency.For any p ≥ m = m ( p ), such that A − p,Sϕ ⊂ A − mϕ . Without loss of generality, we may assume that m ≥ p .Now we take and fix an integer q > m + 2 and let us denote U p,S theunit ball in A − p,Sϕ and U q the unit ball in A − qϕ . Sincesup f ∈ U p,S k f k q ≤ sup f ∈ U p,S \ U q k f k q + sup f ∈ U q k f k q , it suffices for us to show that U := U p,S \ U q is bounded in the space A − qϕ .At this point we pause and prove the following result for the norm k f k q . Lemma 2.6.
Let ℓ > m be some integer. Then there exists some s = s ( ℓ ) ≥ , s ∈ N , such that the k f k ℓ is attained on the compact set K s := (cid:8) z ∈ B : | z | ≤ ss +1 (cid:9) for all f ∈ U . BINGYANG HU , LE HAI KHOI Proof.
We prove the result by contradiction. Namely, for any s ∈ N ,there exists a f s ∈ U , such thatsup K s | f s ( z ) | ϕ ℓ ( | z | ) < k f s k ℓ , from which it follows that(2.1) sup B \ K s | f s ( z ) | ϕ ℓ ( | z | ) = k f s k ℓ . The proof will complete if we can construct a function u ∈ A − p,Sϕ but u / ∈ A − ( ℓ − ϕ , which will contradict to our early assumption A − p,Sϕ ⊂ A − mϕ ⊂ A − ( ℓ − ϕ (as m ≤ ℓ − u ( z ) = ∞ X k =1 c k u k ( z )where u k ∈ U := n f k f k ℓ ; f ∈ U o will be defined inductively as follows.Take an arbitrary u ∈ U . If we have u , u , . . . , u t − ∈ U ( t ≥ u t is determined in the following way: Step I:
By condition ( C ), we can choose s t ∈ N large enough sothat(2.2) ϕ ℓ ( | z | ) ϕ ℓ − ( | z | ) ≥ · t +1 t − X k =1 k u k k ℓ − , for all z ∈ B \ K s t . Note that here the quantity k u k k ℓ − is well-defined. Indeed, for any f ∈ U = U p,S \ U ℓ , we have f ∈ A − p,Sϕ ⊂ A − mϕ ⊂ A − ( ℓ − ϕ ⊂ A − ℓϕ , whichimplies the desired claim. Step II:
For the K s t chosen above, there is an f t ∈ U such that (2.1)holds, and we define u t ( z ) = f t ( z ) k f t k ℓ ∈ U .Thus from the above two steps, a sequence { u k } ⊂ U is defined.Taking c k = 4 − k , k = 1 , , . . . , we have u ( z ) = ∞ X k =1 − k u k ( z ) , which implies k u k ℓ ≤ ∞ X k =1 k u k k ℓ k = ∞ X k =1 k < ∞ . This shows that u ∈ A − ℓϕ . Claim I: u ∈ A − p,Sϕ . Note that for any f ∈ U = U p,S \ U ℓ , we have k f k p,S ≤ k f k ℓ > g ∈ U , it always holds that k g k p,S <
1, which implies k u k p,S ≤ ∞ X k =1 k k u k k p,S ≤ ∞ X k =1 k < ∞ , which implies the first claim. Claim II: u / ∈ A − ( ℓ − ϕ . Take and fix any t ∈ N . Then from the Step II above, there existsan z t ∈ B \ K s t such that(2.3) | f t ( z t ) | ϕ ℓ ( | z t | ) ≥ k f t k ℓ . Thus | u ( z t ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X k =1 u k ( z t )4 k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ (cid:12)(cid:12)(cid:12)(cid:12) u t ( z t )4 t (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X k = t u k ( z t )4 k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ (cid:12)(cid:12)(cid:12)(cid:12) u t ( z t )4 t (cid:12)(cid:12)(cid:12)(cid:12) − X k = t | u k ( z t ) | k = (cid:12)(cid:12)(cid:12)(cid:12) u t ( z t )4 t (cid:12)(cid:12)(cid:12)(cid:12) − t − X k =1 | u k ( z t ) | k − ∞ X k = t +1 | u k ( z t ) | k = J − J − J . Estimate of J . Applying (2.3), we have J ≥ t · ϕ ℓ ( | z t | )2 . Estimate of J . For each k ∈ N , we have | u k ( z t ) | = | u k ( z t ) | ϕ ℓ − ( | z t | ) · ϕ ℓ − ( | z t | ) ≤ k u k k ℓ − ϕ ℓ − ( | z t | ) , and hence J ≤ ϕ ℓ − ( | z t | ) · t − X k =1 k u k k ℓ − k . Estimate of J . Similarly, for each k ∈ N , we have | u k ( z t ) | = | u k ( z t ) | ϕ ℓ ( | z t | ) · ϕ ℓ ( | z t | ) ≤ ϕ ℓ ( | z t | ) , BINGYANG HU , LE HAI KHOI which gives J ≤ ϕ ℓ ( | z t | ) · ∞ X k = t +1 k = ϕ ℓ ( | z t | )3 · t . Combining the three estimates above yields | u ( z t ) | ≥ (cid:18) · t − · t (cid:19) · ϕ ℓ ( | z t | ) − ϕ ℓ − ( | z t | ) · t − X k =1 k u k k ℓ − k = ϕ ℓ − ( | z t | )6 · t · ϕ ℓ ( | z t | ) ϕ ℓ − ( | z t | ) − ϕ ℓ − ( | z t | ) · t − X k =1 k u k k ℓ − k ≥ ϕ ℓ − ( | z t | )6 · t · · t +1 t − X k =1 k u k k ℓ − − ϕ ℓ − ( | z t | ) · t − X k =1 k u k k ℓ − k (by (2.2)) ≥ ϕ ℓ − ( | z t | ) · t − X k =1 k u k k ℓ − . Noting the easy fact that k u k k ℓ − ≥ k u k k ℓ = 1 , for all k ≥
1, we havefor any t ≥ k u k ℓ − = sup z ∈ B | u ( z ) | ϕ ℓ − ( | z | ) ≥ | u ( z t ) | ϕ ℓ − ( | z t | ) ≥ t − X k =1 k u k k ℓ − > t − , which implies k u k ℓ − = ∞ and therefore u / ∈ A − ( ℓ − ϕ .Thus, there exists some s , such that the k f k ℓ is attained on K s uniformly for all f ∈ U . (cid:3) Now return back to the proof of Theorem 2.5. By Lemma 2.6, wetake and fix some s ∈ N so that k f k q − is attained on K s for all f ∈ U (note that q > m + 2, and hence q − > m ).By condition ( C ), there exists M = M ( q, s ) >
0, such that ϕ q ( | z | ) ϕ q − ( | z | ) ≤ M, z ∈ K s and hence for any f ∈ U , k f k q − = sup z ∈ B | f ( z ) | ϕ q − ( | z | ) = sup z ∈ K s | f ( z ) | ϕ q − ( | z | ) = sup z ∈ K s | f ( z ) | ϕ q ( | z | ) · ϕ q ( | z | ) ϕ q − ( | z | ) ≤ M sup z ∈ K s | f ( z ) | ϕ q ( | z | ) ≤ M k f k q , which shows that the set (cid:26) f k f k q (cid:27) f ∈ U is bounded in A − ( q − ϕ . An easy application of Montel’s theorem andcondition ( C ) imply that the identity map i : A − ( q − ϕ → A − qϕ is com-pact, and hence the set n f k f k q o f ∈ U is relatively compact in A − qϕ .Recall that we have to show U is bounded in A − qϕ , namely(2.4) sup f ∈ U k f k q < ∞ . We prove it by contradiction. Assume (2.4) does not hold. Then wecan take a sequence { f k } ⊂ U such that(2.5) lim k →∞ k f k k q = ∞ . By the remark above, we have the set n f k k f k k q o k is sequentially compact,namely, there is a sequence(2.6) g ℓ := f k l k f k l k q → g as ℓ → ∞ in the sense of A − qϕ . Clearly, g ∈ A − qϕ with k g k q = 1. Thisimplies k g k q,S = C >
0, since S is a set of uniqueness. Moreover,by (2.6), one can also see that lim ℓ →∞ k g ℓ − g k q,S = 0, which implies forsufficient large ℓ , we have k g ℓ k q,S ≥ k g k q,S C . On the other hand, k g ℓ k q,S = (cid:13)(cid:13)(cid:13)(cid:13) f k l k f k l k q (cid:13)(cid:13)(cid:13)(cid:13) q,S = k f k l k q,S k f k l k q ≤ k f k l k q , where in the last inequality, we use the fact that k f k l k q,S ≤ k f k l k p,S ≤ k f k l k q ≤ C , which contradicts to the assumption (2.5). The proof is complete. (cid:3) , LE HAI KHOI Sampling sets and weakly sufficient sets
In this section, we study the relation between sampling sets andweakly sufficient sets, in particular we investigate under what condi-tions weakly sufficient sets are sampling, and vice versa.As said in Introduction, weakly sufficient sets fail to be sampling setsbecause the growths of Φ and ψ are too “independent”. Motivated bythis, we introduce some conditions which “relate” growths of the pair(Φ , ψ ).( C ) for each p ∈ N , there exists lim r → log ϕ p ( r ) ψ ( r ) := α p , ( C ) ( α p ) is strictly increasing , ( C ) ( α p ) is bounded . In case ( C ) − ( C ) are satisfied, we denote α := lim p →∞ α p , and for each ξ ∈ B , t ∈ [0 , α ), we put R ( ψ, ξ, t ) := (cid:8) g ∈ O ( B ) : | g ( ξ ) | = e tψ ( | ξ | ) (cid:9) . Finally, we introduce the following condition for (Φ , ψ ):( C ) there exists M >
0, such that for each ξ ∈ B and t ∈ [0 , α ),there is a f ξ,t ∈ R ( ψ, ξ, t ) for which | f ξ,t ( z ) | ≤ M e tψ ( | z | ) , for all z ∈ B .Now we can state the main result in this section. Theorem 3.1. (1)
Let Φ and ψ satisfy conditions ( C ) − ( C ) .Then every ψ -sampling set is a weakly sufficient set. (2) Let Φ and ψ satisfy conditions ( C ) − ( C ) . Let further, ϕ is bounded on any compact subset of [0 , . Then every weaklysufficient set is a ψ -sampling set.Remark . Let us consider the classical case, that is ϕ p ( r ) = (1 − r ) − p , p ∈ N and ψ ( r ) = | log(1 − r ) | .First, it is easy to check that conditions ( C ) − ( C ) are satisfied. SoTheorem 3.1 (1) contains the classical result of the implication “sam-pling sets = ⇒ weakly sufficient sets” as a particular case.On the other hand, the condition ( C ) is not satisfied, as α ( p ) = p ,is unbounded. That is, the classical pair (Φ , ψ ) does not satisfy theassumption in Theorem 3.1 (2). Remark . Concerning condition ( C ), we give some motivations forits appearance.First, a typical example that makes condition ( C ) non-trivial is(3.1) ψ ( r ) = − log(1 − r ) , < r < . Actually, we can prove a slightly stronger assertion for such a choice of ψ : for any ξ ∈ B and t >
0, there exists a f ξ,t ∈ R ( ψ, ξ, t ), such that | f ξ,t ( z ) | ≤ e tψ ( | z | ) . Indeed, this follows from(3.2) f ξ,t ( z ) = (1 − | ξ | ) t (1 − h z, ξ i ) t and an easy fact1 − | Φ ξ ( z ) | = (1 − | ξ | )(1 − | z | ) | − h ξ, z i| , ξ, z ∈ B (see, [13, Lemma 1.2]), where Φ ξ is the involutive automorphism in B associated to ξ .The example above works for all t >
0, that is why for ψ chosenin (3.1) we get a slightly stronger result, which is independent of thechoice of α in ( C ). So is the collection Φ.This example leads us to the following.1. There are many “ ψ ” that satisfy ( C ). One can take, say ψ ( r ) = − − r ) or ψ ( r ) = − log 2(1 − r ), 0 < r < C ).3. There is an alternative way to think about ( C ). More precisely,for each t >
0, we define the Banach space A t,ψ := (cid:26) f ∈ O ( B ) : sup z ∈ B | f ( z ) | e − tψ ( | z | ) < ∞ (cid:27) . Then ( C ) can be restated as follows: there exists M > ξ ∈ B and t ∈ [0 , α ), there is f ξ,t ∈ B t,ψ ( M ),satisfying | f ξ,t ( ξ ) | = e tψ ( | ξ | ) , where B t,ψ ( M ) is the ball of radius M in A t,ψ .Note also that the space A t,ψ can be thought as of the “growthspace” of certain space of holomorphic functions on the unitball. For instance, in our above example, A t,ξ is the Bergman-type spaces A − t , which is the “growth space” of the classicalBergman space A n +1 − t . We refer the reader to [13] for detailedinformation about these spaces. , LE HAI KHOI Proof of Theorem 3.1 .3.1.
Proof of (1).
Let S ⊂ B be a ψ -sampling set for A −∞ Φ . Sinceconditions ( C ) − ( C ) are satisfied, we prove that two conditions inTheorem 2.5 are true. (a). S is a set of uniqueness .Suppose f ∈ A −∞ Φ and f ( z ) = 0 for all z ∈ S . Then since S is a ψ -sampling set, we have T f,ψ = T f,ψ,S = lim sup | z |→ ,z ∈ S log | f ( z ) | ψ ( | z | ) = −∞ . In particular, T f,ψ = lim sup | z |→ log | f ( z ) | ψ ( | z | ) < − δ >
0, such that log | f ( z ) | ≤ − ψ ( | z | ) for all z with 1 − δ < | z | < w ∈ B . By the Maximum modulus principle, wehave | f ( w ) | ≤ sup | z | = r | f ( z ) | ≤ e − ψ ( r ) , for all r ∈ (max { − δ, | w |} , . Letting r → − , since lim r → | ψ ( r ) | = ∞ , we get f ( w ) = 0. (b). For any p ∈ N , there exists an m = m ( p ) , such that A − p,Sϕ ⊂ A − mϕ .Let p ∈ N . For every f ∈ A − p,Sϕ , sincesup z ∈ S | f ( z ) | ϕ p ( | z | ) < ∞ , there exists C > | f ( z ) | < Cϕ p ( | z | ) , for all z ∈ S, which gives log | f ( z ) | ψ ( | z | ) < log Cψ ( | z | ) + log ϕ p ( | z | ) ψ ( | z | ) , for all z ∈ S. Since S is a ψ -sampling set, by condition ( C ), we have T f,ψ = T f,ψ,S = lim sup | z |→ ,z ∈ S log | f ( z ) | ψ ( | z | ) ≤ lim sup | z |→ ,z ∈ S (cid:18) log Cψ ( | z | ) + log ϕ p ( | z | ) ψ ( | z | ) (cid:19) ≤ lim sup | z |→ (cid:18) log Cψ ( | z | ) + log ϕ p ( | z | ) ψ ( | z | ) (cid:19) = lim sup | z |→ log ϕ p ( | z | ) ψ ( | z | ) = α p . Furthermore, by condition ( C ), since α p < α p +1 , there exists some δ ∈ (0 , | f ( z ) | ψ ( | z | ) < α p + α p +1 < log ϕ p +1 ( | z | ) ψ ( | z | ) , for all δ < | z | < . From this it follows that | f ( z ) | ϕ p +1 ( | z | ) < , for all δ < | z | < . On the other hand, since ϕ p +1 ( w ) ∈ (1 , ∞ ), we also have | f ( z ) | ϕ p +1 ( | z | ) ≤ sup | z |≤ δ | f ( z ) | < ∞ . So we get sup z ∈ B | f ( z ) | ϕ p +1 ( | z | ) < ∞ , that is f ∈ A − mϕ , with m = p + 1.3.2. Proof of (2).
Let S be a weakly sufficient set for A −∞ Φ . We showthat T f,ψ,S = T f,ψ for every f ∈ A −∞ Φ .Assume in contrary that there exists a function f ∈ A −∞ Φ , such that T f,ψ,S < T f,ψ . Then we can take d ∈ (0 ,
1) sufficiently small so that T f,ψ,s < T f,ψ − d .Furthermore, by ( C ) − ( C ), there is some p ∈ N big enough, suchthat(3.3) α p > α − d and f ∈ A − pϕ . Since S is weakly sufficient for A −∞ Φ , by Lemma 2.1, there existssome m = m ( p ) ≥ p and C p , such that(3.4) k g k m ≤ C p k g k p,S , for all g ∈ A −∞ Φ . We need the following result.
Lemma 3.4. T f,ψ ≤ α p .Proof. Since f ∈ A − pϕ , there exists some C = C ( p ) >
1, such that forany z ∈ B , | f ( z ) | ≤ C ϕ p ( | z | ) , which implies that log | f ( z ) | log ϕ p ( | z | ) ≤ log C log ϕ p ( | z | ) + 1 . , LE HAI KHOI Furthermore, note that by ( C ) and the assumption lim r → ψ ( r ) = ∞ , wehave lim r → log ϕ p ( r ) = ∞ . Then T f,ψ = lim sup | z |→ log | f ( z ) | ψ ( | z | ) = lim sup | z |→ (cid:26) log | f ( z ) | log ϕ p ( | z | ) · log ϕ p ( | z | ) ψ ( | z | ) (cid:27) ≤ α p . The lemma is proved. (cid:3)
Note that Lemma 3.4 gives T f,ψ ≤ α m < α, because m ≥ p and ( α p ) is strictly increasing to α .Next we can choose x, y > α m + 2 y < x + T f,ψ < α − y. Indeed, first since α m < α , there is y > α m + 3 y < α ( ⇐⇒ α m + 2 y < α − y ) . Then since 0 ≤ α m − T f,ψ < α m + 2 y − T f,ψ < α − y − T f,ψ , any choiceof x ∈ ( α m + 2 y − T f,ψ , α − y − T f,ψ ) gives (3.5).Now by the definition of T f,ψ,S and T f,ψ , there exists some A > | f ( z ) | ≤ Ae ( T f,ψ + y ) ψ ( | z | ) , for all z ∈ B , and(3.7) | f ( z ) | ≤ Ae ( T f,ψ − d ) ψ ( | z | ) , for all z ∈ S. Also there exists a sequence { z k } ⊂ B with lim k →∞ | z k | = 1, such that(3.8) | f ( z k ) | ≥ e ( T f,ψ − y ) ψ ( | z k | ) , for every k = 1 , , . . . By condition ( C ) with t = x and ξ ∈ B , we can have a function g ξ,x ∈ R ( ψ, ξ, x ), satisfying(3.9) | g ξ,x ( z ) | ≤ M e xψ ( | z | ) , for all z ∈ B . Consider the function h ξ,x = f · g ξ,x ∈ O ( B ). We prove the following. Lemma 3.5. h ξ,x ∈ A −∞ Φ .Proof. From (3.6) and (3.9), it follows that | h ξ,x ( z ) | ≤ AM e ( T f,ψ + x + y ) ψ ( | z | ) , for all z ∈ B . By (3.5), we have T f,ψ + x + y < α , and hence by ( C ), there is e p largeenough, such that T f,ψ + x + y < α e p − < α e p < α, which gives(3.10) | h ξ,x ( z ) | ≤ AM e α e p − ψ ( | z | ) , for all z ∈ B . Furthermore, by ( C ), for 0 < ε ′ <
13 ( α e p − α e p − ), there exists δ ( e p ) ∈ (0 , δ ( e p ) < | z | < α e p − − ε ′ < log ϕ e p − ( | z | ) ψ ( | z | ) < α e p − + ε ′ and α e p − ε ′ < log ϕ e p ( | z | ) ψ ( | z | ) < α e p + ε ′ , or equivalently, ϕ e p − ( | z | ) e − ε ′ ψ ( | z | ) < e α e p − ψ ( | z | ) < ϕ e p − ( | z | ) e ε ′ ψ ( | z | ) and ϕ e p ( | z | ) e − ε ′ ψ ( | z | ) < e α e p ψ ( | z | ) < ϕ e p ( | z | ) e ε ′ ψ ( | z | ) . Hence, when δ ( e p ) < | z | <
1, we have e α e p − ψ ( | z | ) < ϕ e p − ( | z | ) e ε ′ ψ ( | z | ) < e ( α e p − +2 ε ′ ) ψ ( | z | ) < e ( α e p − ε ′ ) ψ ( | z | ) < ϕ e p ( | z | ) . Thus for δ ( e p ) < | z | < | h ξ,x ( z ) | ≤ AM e α e p − ψ ( | z | ) ≤ AM ϕ e p ( | z | ) . On the other hand, since ψ is continuous on [0 , δ ( e p )], there exists apositive constant e C (depending only on e p and α e p − ), such thatsup | z |≤ δ ( e p ) e α e p − ψ ( | z | ) ≤ e C. Combining the last two inequalities, we conclude that for some e A >
0, depending only on A , M , e p and and α e p − ,(3.11) | h ξ,x ( z ) | ≤ e Cϕ e p ( | z | ) , for all z ∈ B . This shows that h ξ,x ∈ A − e pϕ , and hence h ξ,x ∈ A −∞ Φ . The lemma isproved. (cid:3) Now we choose and fix ε ∈ (0 , d ). In this case, since T f,ψ − d + x < α − y − d < α − d < α p − ε (here p is given in (3.3)), by a similar way as the proof of (3.11), due toconditions (3.7) and (3.9), there exists B > z ∈ S ,(3.12) | h ξ,x ( z ) | ≤ AM e ( T f,ψ − d + x ) ψ ( | z | ) ≤ AM e ( α p − ε ) ψ ( | z | ) ≤ Bϕ p ( | z | ) . , LE HAI KHOI This shows that f · g ξ,x ∈ A − p,Sϕ . Here, the constant B depends on A , M , p and ε . In particular, it is independent of the choice of ξ ∈ B .Similarly, applying (3.4) to h ξ,x , by (3.12) and Lemma 3.5, we seethat there exists D >
0, such that(3.13) | h ξ,x ( z ) | ≤ C p Bϕ m ( | z | ) ≤ De ( α m + δ ) ψ ( | z | ) , for all z ∈ B . Here, δ > α m + 2 y + δ < x + T f,ψ , while the constant D only depends on C p , B , δ , boundedness of ϕ oncompact subsets of [0 , C ) − ( C ).In particular,it is independent of the choice of ξ ∈ B .For each k ≥
1, we let z = ξ = z k , where ( z k ) are taken from (3.8).Then since g z k ,x ∈ R ( ψ, z k , x ), we have | g z k ,x ( z k ) | = e xψ ( | z k | ) . Putting g z k ,x ( z k ) into (3.11) and taking into account (3.14), we have | f ( z k ) | = (cid:12)(cid:12)(cid:12)(cid:12) h z k ,x ( z k ) g z k ,x ( z k ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ De ( α m − x + ε ) ψ ( | z k | ) ≤ De ( T f,ψ − y ) ψ ( | z k | ) , which contradicts to (3.8) if k is sufficiently large.The theorem is proved completely.As a consequence of Theorem 2.5 and Theorem 3.1, we have thefollowing result. Corollary 3.6.
Let S be a subset of B . Let further, Φ and ψ satisfy theconditions ( C ) − ( C ) . If in addition, the weight ϕ is bounded on anycompact subset of [0 , , then the following assertions are equivalent: (i) S is ψ -sampling for A −∞ Φ relative to ψ . (ii) S is weakly sufficient for A −∞ Φ . (iii) S is a set of uniqueness for A −∞ Φ .Proof. It suffices to recall that the condition ( C ) is guaranteed if Φconsists of continuous weights. (cid:3) Example 3.7.
To complete our exposition, we provide some examplesof weights ( ϕ p ) and ψ that satisfy all conditions ( C ) − ( C ) in themain results of our paper. Clearly, there is a lot of such weights. (1) Φ = (cid:16) (1 − r ) − p (cid:17) ∞ p =1 and ψ ( r ) = − log(1 − r ) . (2) Φ = (cid:16) (1 − e r − ) − p (cid:17) ∞ p =1 and ψ ( r ) = − log(1 − r )10 . (3) Φ = (cid:16) sin (cid:0) − r (cid:1) − p ) (cid:17) ∞ p =1 and ψ ( r ) = − − r ) . References [1] Abanin, A.V., Effective and sampling sets for Hormander spaces,
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Bingyang Hu: Department of Mathematics, University of Wisconsin,Madison, WI 53706-1388, USA.
E-mail address : [email protected] Le Hai Khoi: Division of Mathematical Sciences, School of Phys-ical and Mathematical Sciences, Nanyang Technological University(NTU), 637371 Singapore, Singapore
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