In the case of small sets, we can present strict polynomial bases as well

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Ann. Funct. Anal. 9 (2018), no. 1, 56–71 https://doi.org/10.1215/20088752-2017-0024 ISSN: 2008-8752 (electronic)

http://projecteuclid.org/afa

BASES IN SOME SPACES OF WHITNEY FUNCTIONS

ALEXANDER GONCHAROV^*and ZEL˙IHA URAL Communicated by R. Mortini

Abstract. We construct topological bases in spaces of Whitney functions on Cantor sets, which were introduced by the first author. By means of suitable individual extensions of basis elements, we construct a linear continuous extension operator, when it exists for the corresponding space. In general, elements of the basis are restrictions of polynomials to certain subsets. In the case of small sets, we can present strict polynomial bases as well.

1. Introduction

This article is supplementary to [5], where the extension problem is discussed for equilibrium Cantor sets K(γ) introduced in [4]. The set K(γ) is defined by means of a sequence of parameters γ = (γs)^∞_s=0 and can be considered as a gen- eralization of the classical quadratic Julia set, but as opposed to Julia sets, it is more flexible with respect to its features.

Following [10] (see also [1] and [2]), we say that a compact set K ⊂ R^d has the extension property (EP) if, for the space E (K) of Whitney jets on K, there exists a linear continuous extension operator W : E (K) → C^∞(R^d). In [5], we present a characterization of EP for K(γ) and, by means of local Newton interpolations, construct an operator W , when it exists. This approach goes back to [8] (see also [9]), so we can say that W is a local version of the Paw lucki–Ple´sniak operator.

Here, we construct topological bases in the spaces E (K(γ)). The construction follows [3]. Besides, for K(γ) with EP, we present an extension operator W by individual extensions of basis elements.

Received Aug. 23, 2016; Accepted Feb. 7, 2017.

First published online Jun. 29, 2017.

*Corresponding author.

2010 Mathematics Subject Classification. Primary 46E10; Secondary 46A35, 28A80.

Keywords. Whitney spaces, extension problem, topological bases.

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The article is organized as follows. Section 2 contains definitions and some auxiliary results about the sets K(γ). In Sections 3 and 4, we show that the method from [3] can be adapted to our case as well. In Section 5, by means of extension of basis elements, we construct an extension operator W for the spaces E(K(γ)), provided K(γ) has EP. In the case when the space has a Faber basis, the operator W coincides with the operator presented in [5].

For a finite set A ⊂ R, let #(A) be the cardinality of A. Given x ∈ R, by dk(x, A) we denote distances from x to the points of A arranged in nondecreasing order, so d_k(x, A) = |x − a_m_k| %. Also, bac is the greatest integer less than or equal to a, and |A| is the diameter of A.

2. Uniform distribution of points

As in [5], we consider γ = (γ_s)^∞_s=1 with 0 < γ_s ≤ 1/32 and P∞

s=1γ_s < ∞.

Let r₀ = 1, let P₂(x) = x(x − 1), and let r_s = γ_sr²_s−1, P₂^s+1 = P₂^s(P₂^s + r_s) where s ∈ N. Then Es := {x ∈ R : P2^s+1(x) ≤ 0} = S2^s

j=1I_j,s, where the sth level basic intervals I_j,s are disjoint. Since E_s+1 ⊂ E_s, we have a Cantor-type set K(γ) :=T∞

s=0E_s.

For the length `_j,s of the interval I_j,s, by Lemma 6 in [4], we have

δ_s < `_j,s< C₀δ_s for 1 ≤ j ≤ 2^s, (2.1) where δ₀ := 1, δ_s := γ₁γ₂· · · γ_s for s ∈ N and C0 = exp(16P∞

k=1γ_k). Clearly, r_k = δ_kδ_k−1δ_k−2² δ_k−3⁴ · · · δ₀²^k−1. (2.2) Each Ij,s contains two adjacent basic subintervals I2j−1,s+1 and I2j,s+1. Let hj,s =

`_j,s− `_2j−1,s+1− `_2j,s+1 be the distance between them. As in [4], Lemma 4, h_j,s≥ 7/8 · `_j,s> 7/8 · δ_s (2.3) for all s and 1 ≤ j ≤ 2^s and

`_2j−1,s+1+ `_2j,s+1 < 4`_j,s. (2.4) We decompose the zeros of P2^s into s groups: X0 = {x1, x2} = {0, 1}, X1 = {x3, x4} = {`1,1, 1 − `2,1}, . . . , Xk= {`1,k, `1,k−1− `2,k, . . . , 1 − `₂^k_,k} for k ≤ s − 1, so X_k contains all zeros of P₂^k+1 that are not zeros of P₂^k. If Y_s =Ss

k=0X_k, then P2^s(x) = Q

xk∈Y_s−1(x − xk). Clearly, #(Xs) = 2^s for s ∈ N and #(Y^s) = 2^s+1 for s ∈ Z+. The elements of X_s are called sth-type points.

We put all points (xk)^∞_k=1fromS∞

k=0Xkin order by means of the rule of increase of type. The order of (xk)⁴_k=1 is given above. To put the points from X2 in order, we increasingly arrange the points from Y1, so Y1 = {x1, x3, x4, x2}. After this we increase the index of each point by 4. This gives the ordering X₂ = {x₅, x₇, x₈, x₆}.

Similarly, indices of increasingly arranged points from Y_k−1 = {x_i₁, x_i₂, . . . , x_i

2k} define the ordering X_k = {x_i₁₊₂^k, x_i₂₊₂^k, . . . , x_i

2k+2^k}. We see that x_j+2^k = x_j ±

`_m,k, where the sign and m are uniquely defined by j.

A useful feature of this order is that for each N , the points Z := (x_k)^N_k=1 are distributed uniformly on K(γ) in the following sense. Suppose that 2ⁿ ≤ N <

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2ⁿ⁺¹. Then the binary representation

N = 2ⁿ+ 2^m+ · · · + 2ⁱ + · · · + 2^w with 0 ≤ w < · · · < i < · · · < m < n (2.5) generates the decomposition Z = Z_n∪ Z_m∪ · · · ∪ Z_w with Z_n = Y_n−1 and Z_m∪

· · · ∪ Z_w ⊂ X_n. Here, #(Z_i) = 2ⁱ for i ∈ N := {w, . . . , m, n}, and each basic interval of ith level contains just one point from Z_i. Also, for each s and i, j ≤ 2^s,

#(Z ∩ I_i,s) − #(Z ∩ I_j,s)

≤ 1. (2.6)

In what follows, we will associate with a number N not only the sets Z and N , but also the product Q

i∈N ri, where ri is defined in (2.2). We combine together all the δk’s that constitute this product and arrange them in nondecreasing order:

Q

i∈Nr_i =QN

m=1ρ_m =Qn

k=0δ_k^s^k^{(N )} with Pn

k=0s_k(N ) = N, ρ_m ≤ ρ_m+1. For example, N = 2ⁿ gives QN

m=1ρ_m = r_n, whereas N = 21 generates N = {0, 2, 4} and Q21

m=1ρ_m = δ₄δ₃δ³₂δ₁⁵δ¹¹₀ . For each N ≥ 1 and k ≥ 0, the corresponding degrees are given by the formula

s_k(N ) = 2^−k−1(N + 2^k).

From here it follows that, for N + 1 = 2^m(2p + 1), the values s_k(N ) and s_k(N + 1) coincide for all k except k = m, where s_m(N +1) = s_m(N )+1. We choose similarly the set Z = (xk,j,s)^N_k=1 on any Ij,s. If 2ⁿ≤ N < 2ⁿ⁺¹, then Z includes 2ⁿ zeros of P₂^s+n on I_j,sand N −2ⁿpoints of the type s+n. Let r_i,s= δ_s+iδ_s+i−1δ_s+i−2² · · · δ²_sⁱ⁻¹ andQ

i∈N r_i,s=QN

m=1ρ_m,s. We estimate the sup-norm of f_N(x) = QN

k=1(x−x_k,j,s) on K(γ) ∩ I_j,s in terms of the last product.

Lemma 2.1. In the above notation, (7/8)^N

N

Y

k=1

ρ_k,s ≤ |f_N|_0,K(γ)∩I_j,s ≤ C₀^N

N

Y

k=1

ρ_k,s.

Proof. For brevity, let us consider the interval Ij,s= [0, 1], since the proof for the general case is the same. Thus, we drop the subscripts j and s. For x ∈ K(γ), we have |f_N(x)| =QN

k=1|x−x_k| =Q

i∈N

Q

xk∈Z_i|x−x_k|. For each i ∈ N , we consider the chain of basic intervals containing x: x ∈ I_j₀_,i ⊂ I_j₁_,i−1 ⊂ · · · ⊂ I_j_i_,0 = [0, 1].

Since the points from Zi are uniformly distributed, the interval Ij0,i contains just one point from Z_i, as well as I_j₁_,i−1\ I_j₀_,i. Also, #(Z ∩ (I_j_k_,i−k\ I_j_k−1_,i−k+1)) = 2^k−1 for 1 ≤ k ≤ i. Therefore, Q

xk∈Zi|x − x_k| ≤ `_j₀_,i`_j₁_,i−1`²_j₂_,i−2· · · `²_jⁱ⁻¹

i,0 < C₀²ⁱr_i, by (2.1) and (2.2). From this, |fN(x)| ≤ C₀^NQ

i∈N ri and

|f_N|_0,K(γ) ≤ C₀^N

N

Y

k=1

ρ_k. (2.7)

The bound (2.7) is sharp with respect to the product QN

k=1ρ_k. Indeed, let us consider |f_N(x_{N +1})|. As above, x_{N +1} ∈ I_j₀_,n ⊂ I_j₁_,n−1 ⊂ · · · ⊂ I_j_n_,0 = [0, 1].

Hence, Q

xk∈Zn|x_{N +1}− x_k| ≥ `_j₀_,nh_j₁_,n−1h²_j

2,n−2· · · h²_j_nⁿ⁻¹_,0 > (7/8)²ⁿr_n, by (2.3).

As for Q

xk∈Z_m|xN +1− xk|, we observe that the point xN +1 must be in some interval I_j,m+1 which is free of points from A := Z_m ∪ · · · ∪ Z_w. Indeed, the set

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{x_{N +1}} ∪ A contains at most 2^m+1 points that are uniformly distributed, so each I_j,m+1 contains at most one point from this set. Thus, x_{N +1} and its closest point from Z_m are located in distinct intervals of the (m + 1)th level. Arguing as above, we see that Q

xk∈Zm|x_{N +1}− x_k| ≥ h_j₀_,mh_j₁_,m−1h²_j

2,m−2· · · h²_j^m−1

m,0 > (7/8)²^mr_m. In a similar fashion, Q

x_k∈Zi|x_{N +1}− x_k| > (7/8)²ⁱr_i for each i ∈ N . Therefore,

|f_N|_0,K(γ)≥

f_N(x_{N +1})

≥ (7/8)^N

N

Y

k=1

ρ_k. (2.8)

Remark. Given x ∈ K(γ), we have |fN(x)| = QN

k=1dk(x, Z). The lengths of basic intervals of the same level may be rather different (we can say only that

`_j,s< C₀`_i,s, by (2.1)). For this reason, as k increases and x, y belong to different parts of K(γ), the values d_k(x, Z) and d_k(y, Z) may increase in quite different fashions. Nevertheless, the product QN

k=1ρ_k is defined by N only, so it does not depend on the choice of x.

In the following technical lemmas, we use the decomposition (2.5). Let 2ⁿ ≤ N < 2ⁿ⁺¹ and a basic interval I = Ij,s be given. Suppose that Z = (xk)^N_k=1 and ˜Z = (xk)^{N +1}_k=1 are chosen on I by the rule of increase of type. Write C1 = 8/7 · (C₀ + 1).

Lemma 2.2. For each x ∈ R with dist(x, K(γ) ∩ Ij,s)) ≤ δ_s+n and z ∈ ˜Z, we have δ_s+nQN

k=2d_k(x, Z) ≤ C₁^NQN +1

k=2 d_k(z, ˜Z).

Proof. As above, for brevity, we take s = 0, j = 1. Let ˜x ∈ K(γ) realize the distance above. Also, let x_p ∈ Z_p ⊂ Z be such that d₁(x, Z) = |x − x_p|. Of course, x_p may coincide with ˜x. Clearly,

δ_n

N

Y

k=2

d_k(x, Z) = Y

i∈N ,i6=p 2ⁱ

Y

k=1

d_k(x, Z_i) · δ_n

2^p

Y

k=2

d_k(x, Z_p) .

For i 6= p, let ˜x ∈ I_j₀_,i ⊂ I_j₁_,i−1 ⊂ · · · ⊂ I_j_i_,0 = I. As in Lemma 2.1, for fixed q with 1 ≤ q ≤ i we consider 2^q−1 points x_kfrom the set I_j_q_,i−q\I_j_q−1_,i−q+1. For each of them we have |x − x_k| ≤ |x − ˜x| + `_j_q_,i−q ≤ (C₀ + 1)δ_i−q. A similar estimation is valid for x_k ∈ I_j₀_,i. Combining these gives Q

xk∈Z_i|x − x_k| ≤ (C₀ + 1)²ⁱr_i. The terms d_k(x, Z_p) for 2 ≤ k ≤ 2^p can be handled in much the same way:

Q2^p

k=2d_k(x, Z_p) ≤ (C₀+1)²^p^{−1 r}_δ^p

p. This yields δ_nQN

k=2d_k(x, Z) ≤ (C₀+1)^NQ

i∈Nr_i, as δ_n ≤ δ_p.

It is sufficient to show that

N +1

Y

k=2

d_k(z, ˜Z) ≥ (7/8)^N Y

i∈N

r_i. (2.9)

The case N + 1 = 2ⁿ⁺¹ follows immediately by the argument of Lemma 2.1.

Suppose that N + 1 < 2ⁿ⁺¹. First consider w = 0 in (2.5), so N = 2ⁿ + 2^m +

· · · + 2^u+ 2^v−1+ 2^v−2+ · · · + 2 + 1 with some 1 ≤ v < u and, correspondingly, N +1 = 2ⁿ+· · ·+2^u+2^v. Fix z ∈ ˜Z and the chain z ∈ I_j₀_,n ⊂ I_j₁_,n−1 ⊂ · · · ⊂ I_j_n_,0.

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Here, z is an endpoint of I_j₀_,n. Suppose that z ∈ Z_n and that another endpoint of I_j₀_,n is z_p ∈ Z_p ⊂ ˜Z. We estimate separately Q

x_k∈Zq|z − x_k| for each q from the binary decomposition of N + 1.

For q = n we have, as in Lemma2.1, d₁(z, Z_n) = `_j₁_,n−1, d₂(z, Z_n) ≥ h_j₂_,n−2, . . . , and Q

xk∈Zn|z − x_k| > (7/8)²ⁿ⁻²r_n/δ_n. If p < q ≤ m, then Q

xk∈Zq|z − x_k| ≥ (7/8)²^qr_q. Indeed, z_p ∈ I_j_n−q−1_,q+1, which may contain at most one point from the set Z_q∪· · ·∪Z_p∪· · ·∪Z_v. Hence, the point x_k∈ Z_q which is closest to z belongs to the adjacent interval of the (q + 1)th level and d₁(z, Z_q) ≥ h_j_n−q_,q. Continuing in this fashion, by (2.3), we get the desired bound for given q.

We now handle the case q = p. Here, d₁(z, Z_p) = |z −z_p| = `_j₀_,n > δ_n. Since z_p ∈ I_j_n−p_,p, this interval cannot contain another point from Z_p. Therefore, d₂(z, Z_p) ≥ h_j_n−p+1_,p−1 ≥ 7/8δ_p−1 and Q

x_k∈Zp|z − x_k| ≥ (7/8)²^pr_pδ_n/δ_p.

To deal with indices q < p, let us take the nearest to z point z_p₁ from the set Z \ (Z˜ _n∪ Z_m∪ · · · ∪ Z_p) = Z_t∪ · · · ∪ Z_p₁∪ · · · ∪ Z_v. If p₁ < q ≤ t, then, as above, Q

xk∈Zq|z − xk| ≥ (7/8)²^qrq. If q = p1, then Q

xk∈Z_p1|z − xk| ≥ (7/8)²^p1rp1δp/δp1. Indeed, the interval I_j_n−p−1_,p+1 contains z_p, so it cannot contain another point from Z_p∪ · · · ∪ Z_p₁ ∪ · · · ∪ Z_v. Therefore, z and z_p₁ must be in different intervals of the (p + 1)th level. Then d₁(z, Z_p₁) ≥ h_j_n−p_,p. Now I_j_n−p1_,p₁ contains z_p₁, so d2(z, Zp1) ≥ hj_n−p1+1,p1−1. The values dk(z, Zp1) for k > 2 we estimate in much the same way as in the case q = p.

We continue in this way and combine all bounds up to pk= v together:

(8/7)^N ·

N +1

Y

k=2

d_k(z, ˜Z) ≥ r_n

δ_nr_m· · · r_pδ_n

δ_pr_t· · · r_p₁ δ_p

δ_p₁ · · · r_p₂δ_p₁

δ_p₂ · · · r_p_kδ_p_k−1 δ_p_k

= r_n· · · r_ur_v/δ_v = Y

i∈N

r_i = r_n· · · r_ur_v−1· · · r₁r₀,

since r_v−1r_v−2· · · r₁r₀ = r_v/δ_v, as is easy to check. This yields (2.9). The cases

z ∈ Z_p ⊂ ˜Z and w > 0 are very similar.

In the next lemma we consider the same N and ˜Z as above, but we now arrange the points in increasing order. Thus, ˜Z = (z_k)^{N +1}_k=1 ⊂ I with z_k %. For q = 2^m− 1 with m < n and 1 ≤ j ≤ N + 1 − q, let J = {z_j, . . . , z_j+q} consist of 2^m consecutive points from ˜Z. Given j, we consider all possible chains of strict inclusions of segments of natural numbers:

[j, j + q] = [a0, b0] ⊂ [a1, b1] ⊂ · · · ⊂ [aN −q, bN −q] = [1, N + 1], (2.10) where a_k = a_k−1, b_k = b_k−1+ 1 or a_k = a_k−1 − 1, b_k = b_k−1 for 1 ≤ k ≤ N − q.

Every chain generates the productQN −q

k=1(z_b_k− z_a_k). For fixed J , let Π(J ) denote the minimum of such products for all possible chains.

Lemma 2.3. For each J ⊂ ˜Z, there is ˜z ∈ J such that QN +1

k=q+2d_k(˜z, ˜Z) ≤ Π(J ).

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Proof. We take I = [0, 1], as the corresponding change of indices gives the proof in the general case. Fix J ⊂ ˜Z. Let Jk = [za_k, zb_k]. Then (2.10), which defines Π(J ), generates inclusions J ⊂ J₁ ⊂ · · · ⊂ J_{N −q} = I with Π(J ) =QN −q

k=1 |J_k|.

Clearly, #( ˜Z \ J ) = N − q and each z ∈ ˜Z \ J appears as an endpoint of some J_k. Let w_k be the endpoint of J_k in its first appearance. This gives an enumeration of ˜Z \ J . We aim to find ˜z ∈ J and a permutation (w_i_k)^{N −q}_k=1 such that for 1 ≤ k ≤ N −q we have d_q+1+k(˜z, ˜Z) ≤ |J_i_k|. Multiplying these inequalities yields the result. Given ˜z, let d_k be shorthand for d_k(˜z, ˜Z).

Recall that 2ⁿ + 1 ≤ #( ˜Z) ≤ 2ⁿ⁺¹, Y_n−1 ⊂ ˜Z ⊂ Y_n, and #(J ) = 2^m. The points from ˜Z are distributed uniformly on I. Hence, for each basic interval of the (n − m + 1)th level, we have

2^m−1 ≤ #( ˜Z ∩ I_j,n−m+1) ≤ 2^m. (2.11) We observe that J may be located on v consecutive intervals of this level with 1 ≤ v ≤ 3. Indeed, if J ⊂ I₁∪ I₂∪ I₃∪ I₄ with J ∩ I_k 6= ∅, then all points from ˜Z in I₂∪ I₃ are included in J and, by (2.11), #(J ) ≥ 2^m+ 2, which is a contradiction.

Let us consider all possible values of v.

(1) Let J ⊂ I₁ := I_j,n−m+1 for some j. Here, J = ˜Z ∩ I₁ and any point z ∈ J may serve as ˜z, since distances d_k for 1 ≤ k ≤ 2^m are implemented on certain points of the set I₁. If q + 2 ≤ k ≤ N + 1, then d_k is |˜z − w_i_k| for some w_i_k ∈ ˜Z \ J and d_k < |J_i_k|, since ˜z ∈ J_i_k and w_i_k is an endpoint of this interval.

(2) Let J ⊂ I₁ ∪ I₂. Suppose first that these intervals are adjacent, that is, I₁ = I_{2j−1,n−m+1} and I₂ = I_2j,n−m+1. Let p := #(J ∩ I₁); that is, z_j, . . . , z_j+p−1 belong to I₁, whereas z_j+p, . . . , z_j+q ∈ I₂. Suppose, for definiteness, that p ≤ 2^m−1; that is, at least a half of J is in the right interval. The right endpoint of I₂ belongs to Y_n−1, so it is z_j+r for some r with r ≥ q. Thus, #( ˜Z ∩ I₂) = r − p + 1, where r − q of these points are from ˜Z \ J . We take ˜z = z_j+p, the left endpoint of I₂. Then d_k = z_j+p+k−1 − ˜z for 1 ≤ k ≤ r − p + 1, since the lengths of basic intervals are smaller than the gaps between them. In particular, d_r−p+1 = z_j+r− ˜z = |I₂|. The next distances will be realized on the points from I₁ : d_r−p+2 = ˜z − z_j+p−1, . . . , d_r+1 = ˜z − z_j. Since #( ˜Z ∩ I₂) ≤ 2^m ≤ r + 1, the value d₂^m is ˜z − z_j+i for some i with j ≤ i ≤ j + p − 1.

Now, for dk with q + 1 ≤ k ≤ r + 1 we take wi_k = zj+k−1 on I2. Then |Ji_k| >

zj+k−1− zj > dk= ˜z − zj+r−k+1, as zj+k−1 > ˜z and zj+r−k+1 ≥ zj. Note that the next values of d_k (for r + 2 ≤ k ≤ N + 1) will be implemented on certain points of the set w_i_k ∈ ˜Z \ J . As in the first case, d_k < |J_i_k|.

The same reasoning applies to the case of nonadjacent intervals. Let I₁ = I_{2j−2,n−m+1} and I₂ = I_{2j−1,n−m+1} ⊂ I_j,n−m. Then we take ˜z as the endpoint of an interval containing at least half of J , let it be again I2. Here, p points from I1∩ J realize some dM +1, . . . , dM +p and we put into correspondence to them the first p points from I_j,n−m\ J. All other d_k’s are realized on some w_i_k with d_k< |J_i_k|.

(3) Let J ⊂ I₁∪ I₂∪ I₃. Here, I₂ is completely filled with points of J . One of the endpoints of I₂ belongs to Y_n−m−1. We take this point as ˜z, which, analogously to the previous cases, satisfies the desired condition.

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Lemma 2.4. Let 2ⁿ ≤ N < 2ⁿ⁺¹ and let Z = (x_k,j,s)^N_k=1 be chosen in I_j,s by the rule of increase of type. Suppose that x, y ∈ I_i,s+n−q ⊂ I_j,sfor some q < n (in general, x, y /∈ K(γ)). Then QN

k=m+1 dk(y,Z)

d_k(x,Z) ≤ exp(2^q), where m = #(I_i,s+n−q∩ Z).

Proof. For brevity, let I_j,s = [0, 1], Z = (x_k)^N_k=1. Fix q < n, any I_i,n−q, and x, y in this interval. Recall that I_i,n−q may contain from 2^q to 2^q+1 points of Z.

The distances d_k(y, Z) and d_k(x, Z) for 1 ≤ k ≤ m are realized on points from Z ∩ I_i,n−q, so we can consider |y − x_p|/|x − x_p| for x_p ∈ Z \ I_i,n−q only. Let I_i,n−q ⊂ I_i₁_,n−q−1 ⊂ · · · ⊂ I_i_n−q_,0. If x_p ∈ I_i₁_,n−q−1\ I_i,n−q, then |y − x_p| ≤ `_i₁_,n−q−1,

|x − x_p| ≥ h_i₁_,n−q−1, and |y − x_p|/|x − x_p| ≤ 8/7, by (2.3). There are at most 2^q+1 such points x_p. They contribute (8/7)²^q+1 into the common product. For the next step, when x_p ∈ I_i₂_,n−q−2 \ I_{i+1,n−q−1}, we have |y − x_p| ≤ |x − x_p| + `_i,n−q with

|x − x_p| ≥ h_i₂_,n−q−2 ≥ 7/8l_i₂_,n−q−2. Here, |y − x_p|/|x − x_p| ≤ 1 + 8/7 · 4⁻², by (2.4). Similarly, we get at most 2^q+k terms in the general product, each of them bounded above by 1 + 8/7 · 4^−k. Here, 2 ≤ k ≤ n − q. Thus, QN

k=m+1 dk(y,Z) dk(x,Z) ≤ (8/7)²^q+1Qn−q

k=2(1 + 8/7 · 4^−k)²^q+k, which does not exceed exp(2^q). As in [5], we use B_k = 2^−k−1 · log_δ¹

k. By Theorem 5.3 in [5], K(γ) has the extension property if and only if Bn+s/Pn+s

k=sBk → 0 as n → ∞ uniformly with respect to s. This condition can be written as

∀M ∃m, k₀: M · B_k ≤ B_k−m+ · · · + B_k for k ≥ k₀. (2.12) Here we will use a stronger one

∀M, Q ∃m, k₀: Q + M · B_k≤ B_k−m+ · · · + B_k for k ≥ k₀, (2.13) which will provide continuity of the extension operator constructed by interpolation of functions on the whole set.

When constructing a Faber basis in the space E (K(γ)), we also use the clause

∀Q ∃m, k₀: Q ≤ B_k−m+ · · · + B_k for k ≥ k₀. (2.14) These conditions have “geometric” forms in terms of (δ_k). For example, (2.13) is

∀M, Q ∃m, k₀: Q²^k · δ_k−m²^m · · · δ²_k−1· δ_k ≤ δ^M_k for k ≥ k₀.

Let us illustrate the difference between conditions (2.12)–(2.14) for the case of a monotone sequence (B_k)^∞_k=1. Since only γ_s ≤ 1/32 are allowed here, we have B_k ≥ log 32 · k2^−k−1. On the other hand, the values of B_k may be as large as we wish for small sets K(γ). Condition (2.12) is valid if B_k &, B_k % B < ∞, or B_k % ∞, but slowly, with subexponential growth (i.e., k⁻¹log B_k → 0), by Theorem 7.1 in [5]. In turn, we have (2.13) for B_k & B > 0, B_k % B < ∞, or B_k % ∞ of subexponential growth, whereas (2.14) is satisfied with B_k & B > 0 and any B_k %. Thus, the sequence of constants B_k satisfies all three conditions.

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3. Faber bases in the space E (K(γ)) We equip the Whitney space E (K(γ)) with the norms kf k_q = |f |_q,K(γ)

+ sup

(R^q_yf )^(k)(x)

· |x − y|^k−q : x, y ∈ K(γ), x 6= y, k = 0, 1, . . . , q for q ∈ Z+, where |f |_q,K(γ) = sup{|f^(k)(x)| : x ∈ K(γ), k ≤ q} and R^q_yf (x) = f (x) − T_y^qf (x) is the Taylor remainder. By the open mapping theorem, for any q there exist r ∈ N, C > 0 such that

k|f k|_q ≤ Ckf k_r (3.1)

for any f ∈ E (K). Here, k|f k|_q = inf |F |_q,[0,1], where the infimum is taken over all possible extensions of f to F ∈ C^∞[0, 1] (see [5] for more details; for more insight into the theory of Whitney spaces, see [12] and [6]).

Here we adjust the construction from [3], where the case of symmetric Cantor sets was considered. Let e₀ ≡ 1 and e_N(x) = QN

1 (x − x_k) for N ∈ N, where the points (x_k)^∞₁ are chosen in K(γ) by the rule of increase of type. The divided differences define linear continuous functionals ξN(f ) = [x1, x2, . . . , xN +1]f on E(K(γ)). By standard properties of divided differences, the system (eN, ξN)^∞_{N =0} is biorthogonal and the functionals (ξ_N)^∞_{N =0}are total on E (K(γ)); that is, whenever ξ_N(f ) = 0 for all N , then f = 0. We show that (e_N)^∞_{N =0} is a topological basis in the space E (K(γ)) provided the set K(γ) is sufficiently small. Thus, for small sets K(γ), the space E (K(γ)) possesses a strict polynomial basis. Recall that a polynomial topological basis (P_n)^∞_n=0 in a functional space is called a Faber (or strict polynomial ) basis if deg P_n = n for all n. Due to a classical result of Faber, the space C[a, b] does not have such a basis.

Lemma 3.1. Let p = 2^u < N/2. Then ke_Nk_p ≤ C · C₀^NN^p·QN

k=2p+1ρ_k, where C does not depend on N and QN

k=1ρ_k is the product generated by N . Proof. By Lemma2.1, for Ij,s= [0, 1], we have |eN|q,K(γ) ≤ C₀^NQN

k=1ρk for q = 0.

Our first goal is to generalize it to q < N . Let us show that

|e_N|_q,K(γ) ≤ C₀^{N −q}N^q

N

Y

k=q+1

ρ_k. (3.2)

Fix x. Then |e_N(x)| =QN

k=1d_k(x, Z) and the qth derivative of e_N at x is the sum of _{(N −q)!}^{N !} products, where each product contains N − q terms of the form (x − x_k).

Hence, |e^(q)_N (x)| ≤ N^qQN

k=q+1d_k(x, Z). Here, for each k, we take the smallest m = m(k) with d_k(x, Z) ≤ `_j_m_,i−m < C₀δ_i−m. By the remark after Lemma 2.1, δ_i−m ≤ ρ_k. The last inequality may be strict if we take x in a part of the set with a high density of points x_k, for example, near the origin. To deal with ke_Nk_p, let us fix i ≤ p and x 6= y in K(γ). For brevity, let R := (R^p_xe_N)⁽ⁱ⁾(y). We consider two cases: (a) x, y belong to the same interval or (b) two different intervals of the level n − u.

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In the first case, let x, y ∈ I_j,n−u. By the Lagrange form for the Taylor remainder, we have |R| · |x − y|^i−p ≤ |e^(p)_N (θ)| + |e^(p)_N (x)| for some θ ∈ I_j,n−u. Let m := #(I_j,n−u∩Z). Since the points from Z are distributed uniformly on K(γ), we have p ≤ m ≤ 2p. Hence, |e^(p)_N (θ)| ≤ N^p·QN

k=p+1d_k(θ, Z) ≤ N^p·QN

k=m+1d_k(θ, Z), as all distances here do not exceed 1. By Lemma 2.4 and the argument in the proof of (3.2), |e^(p)_N (θ)| ≤ e^pN^p · QN

k=m+1d_k(x, Z) ≤ e^pN^pC₀^{N −m} · QN

k=m+1ρ_k. On the other hand, |e^(p)_N (x)| ≤ N^pC₀^{N −p} · QN

k=p+1ρ_k. Thus, in the first case,

|R| · |x − y|^i−p ≤ 2e^pN^pC₀^N ·QN

k=2p+1ρ_k, since the last product dominates the products of ρ_k involved in the estimation of both terms.

In the second case, let y /∈ I_j,n−u, so |x − y| ≥ h_j₁_,n−u−1, where x ∈ I_j,n−u ⊂ I_j₁_,n−u−1. By (2.3), |x − y| > 7/8 · δ_n−u−1. Now,

|R| · |x − y|^i−p ≤

e⁽ⁱ⁾_N(y)

· |x − y|^i−p+

p

X

k=i

e^(k)_N (x)

· |x − y|^k−p (k − i)! . By (3.2), |e^(k)_N (x)| · |x − y|^k−p ≤ C₀^{N −k}N^kQN

m=k+1ρ_m· (8/7)^p−kδ_n−u−1^k−p . Recall that I_j,n−u contains at least p points from Z. Therefore, ρ_k+1· · · ρ_p ≤ δ^p−k_n−u−1 and

|e^(k)_N (x)| · |x − y|^k−p ≤ (8/7)^pC₀^NN^kQN

m=p+1ρ_m. Clearly, Nⁱ+Pp k=i

N^k

(k−i)! < N^pe.

Combining these yields |R| · |x − y|^i−p ≤ (8/7)^pC₀^NN^pQN

m=p+1ρ_m. The result follows from a comparison of the estimates in both cases. We proceed to estimate |ξ_N(f )| for f ∈ E (K(γ)), ξ_N(f ) = [z₁, . . . , z_{N +1}]f . As in Lemma2.3, ˜Z = (z_k)^{N +1}_k=1 is the set (x_k)^{N +1}_k=1 arranged in increasing order. Here, N generates the product QN

k=1ρ_k = Qn

k=0δ^s_k^k^{(N )}, whereas for N + 1 we have QN +1

k=1 ρ˜_k = Q

k=0δ_k^s^k^{(N +1)} with s_k(N ) = s_k(N + 1) for all k except one value, which is not larger than n + 1. Therefore,

N +1

Y

k=1

˜ ρ_k ≥

N

Y

k=1

ρ_k· δ_n+1. (3.3)

Lemma 3.2. For each N and q = 2^m− 1 < N , we have ξ_N(f )

≤ (16/7)^Nk|f k|_q

N

Y

k=q+1

ρ⁻¹_k . Proof. As in (17) from [5],

ξ_N(f )

≤ 2^{N −q}k|f k|_q Π(J₀)−1

, (3.4)

where Π(J0) = min1≤j≤N +1−qΠ(J ) for Π(J ) defined in Lemma2.3, so it is enough to estimate from below QN +1

k=q+2d_k(z, ˜Z) uniformly for z ∈ ˜Z.

Arguing as in the proof of (2.8), we see that for each x ∈ K(γ) e_{N +1}(x)

=

N +1

Y

k=1

d_k(x, ˜Z) ≥ d₁(x, ˜Z) · (7/8)^N

N +1

Y

k=2

˜ ρ_k.

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Similarly, QN +1

k=q+2d_k(x, ˜Z) ≥ (7/8)^{N −q−1}QN +1

k=q+2ρ˜_k, due to the correspondence between dk(x, ˜Z) and ˜ρk for 1 ≤ k ≤ N + 1. Removing q + 1 smallest terms from both parts of (3.3) gives QN +1

k=q+2ρ˜_k ≥QN

k=q+1ρ_k, and the lemma follows. The following is Theorem 1 from [3] adapted to our case.

Theorem 3.3. Suppose that (Bk)^∞_k=1 satisfies (2.14). Then the sequence (eN)^∞_{N =0} is a Schauder basis in the space E (K(γ)).

Proof. By the Dynin–Mityagin criterion (see [7, Theorem 9]), it is enough to show that for each p there is r such that the sequence (ke_Nk_p· |ξ_N|−r)^∞_{N =0} is bounded.

Here, | · |−r is the dual norm: for ξ ∈ E⁰(K(γ)), let |ξ|−r = sup{|ξ(f )|, kf k_r ≤ 1}.

We consider only p of the form p = 2^u. In order to apply Lemma2.3, we have to take q of the type q = 2^k− 1. For this reason, given arbitrary u, let q = 2^v+1− 1, where v = v(u) will be specified later. Then r = r(q) is defined by (3.1).

Fix N with 2ⁿ≤ N < 2ⁿ⁺¹. We take N so large that Lemmas 3.1 and 3.2 can be applied. Then |ξ_N|_−r≤ C(16/7)^NQN

k=q+1ρ⁻¹_k for C defined by (3.1), and ke_Nk_p· |ξ_N|−r≤ ˜C(3C₀)^NN^p

q

Y

k=2p+1

ρ_k ≤ ˜C(3C₀)^NN^p

2^v

Y

k=2p+1

ρ_k, (3.5) where ˜C does not depend on N . We can decrease the upper index of the product above as all ρ_k’s do not exceed 1.

The productQ2^v

k=2p+1ρktakes its maximal value in the case of minimal density of points of Z, when each basic interval of the level n − u contains p points from Z. At worst, ρ₁, . . . , ρ_p do not exceed δ_n−u, whereas ρ_p+1, . . . , ρ_2p = δ_n−u−1, ρ_2p+1, . . . , ρ_4p= δ_n−u−2, and so on. On the other hand, values N close to 2ⁿ⁺¹ will give maximal density of Z on K(γ) with ρ₁, . . . , ρ_p ≤ δ_n−u+1, ρ_p+1, . . . , ρ_2p= δ_n−u, and so on. Thus, Q2^v

k=2p+1ρ_k ≤ δ_n−u−2^2p δ_n−u−3^4p · · · δ_n−v²^v−1 = exp[−2ⁿ(B_n−u−2+ · · · + Bn−v)]. We claim that the right-hand side of (3.5) is bounded for a suitable v.

It suffices to prove that N log(3C0) + p log N ≤ 2ⁿ(Bn−u−2+ · · · + Bn−v). Since N < 2ⁿ⁺¹, it is reduced to 2 log(3C₀) + (n + 1)2⁻ⁿp log 2 ≤ B_n−u−2+ · · · + B_n−v, which is valid for large n if we take v = m + u + 2, where m is chosen in (2.14)

for Q = 2 log(3C₀) + 1.

Remarks.

1. Under the stronger assumption (2.13), the set K(γ) has, in addition, the extension property. The second part of the proof of Theorem 5.3 from [5]

(for j = 1, s = 0) actually shows that the sequence (k˜e_Nk_p· |ξ_N|−r)^∞_{N =0} is bounded. Here, ˜e_N is a suitable extension of e_N. Since for each extension f of f ∈ E (K(γ)) we have kf k˜ _p ≤ 3| ˜f |_p (by means of the Lagrange form for the Taylor remainder), this proof implies also that (e_N)^∞_{N =0} is a basis provided (2.13).

2. We conjecture that using analytic properties of P2^s, one can replace C₀^N in Lemma 3.1 with N^Q for some Q. It would be interesting to analyze whether one can replace the exponential growth of the constant in (3.4) by a polynomial growth.

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4. Local polynomial bases

In general, the system (e_N, ξ_N)^∞_{N =0} is not a basis. Following [3], we use local interpolations to construct bases for any considered case. Suppose we are given a nondecreasing sequence of natural numbers (ns)^∞_s=0. Let Ns = 2ⁿ^s, Ms^(l) = Ns−1/2 + 1, Ms^(r) = Ns−1/2 for s ≥ 1 and M0 = 0. Here, (l) and (r) mean left and right, respectively. We choose, as above, N_s points (x_k,j,s)^N_k=1^s in each sth level basic interval I_j,s. Set e_N,1,0(x) = QN

k=1(x − x_k,1,0) =QN

k=1(x − x_k) for x ∈ K(γ) and N = 0, 1, . . . , N₀. Given s ≥ 1 and j with 1 ≤ j ≤ 2^s, for Ms^(a) ≤ N ≤ N_s we take eN,j,s(x) = QN

k=1(x − xk,j,s) if x ∈ K(γ) ∩ Ij,s and eN,j,s = 0 on K(γ) otherwise. Here, the superscript a in Ms is l for odd j and a = r if j is even.

Thus, we interpolate a function f on the interval I_j,s up to degree N_s, where- upon we continue this process on subintervals, preserving the previous nodes of interpolation. All x_k,j,s’s are taken from the sequence (x_k)^∞₁ .

As above, Z = (x_k,j,s)^N_k=1 and (z_k,j,s)^N_k=1 is the same set arranged in increasing order. The functionals ξ_N,j,s(f ) = [z_1,j,s, . . . , z_{N +1,j,s}]f are biorthogonal to e_k,i,sfor N, k ∈ [Ms^(a), N_s] and i, j ∈ [1, 2^s]. But, in general, the ξ_N,j,s are not biorthogonal to ek,i,q. For example, ξN,1,s+1(eNs,1,s) 6= 0 for M_s+1^(l) ≤ N ≤ Ns. For this reason, as in [3], we consider the functionals

η_N,j,s(f ) = ξ_N,j,s(f ) −

Ns−1

X

k=N

ξ_N,j,s(e_k,i,s−1)ξ_k,i,s−1(f ) (4.1)

for I_j,s ⊂ I_i,s−1 and N = Ms^(a), Ms^(a) + 1, . . . , N_s. We see that the subtrahend in ηN,j,s is a kind of biorthogonal projection of ξN,j,s in the dual space on the subspace spanned by (ξ_k,i,s−1)^N_k=N^s−1. Also, let η_N,1,0 = ξ_N,1,0 for 0 ≤ N ≤ N₀. By Lemma 2 in [3], the system (e, η) := (e_N,j,s, η_N,j,s)^∞,_s=0,²_j=1,^s^, ^N_{N =M}^s

s is biorthogonal.

Increasing N by 1 means an inclusion of one more point into the interpolation set, so, if η_N,j,s(f ) = 0 for all functionals, then f (x_k) = 0 for all k. Since the set under consideration is perfect, the functionals η are total on E (K(γ)). Provided a suitable choice of the sequence (n_s)^∞₀ , the system (e, η) has the basis property.

As in [5], let n₀ = n₁ = 2 and n_s = blog₂log_δ¹

sc for s ≥ 2. Then n_s≤ n_s+1 and 1

2log 1

δ_s < N_s ≤ log 1

δ_s for s ≥ 2. (4.2)

In the next lemma, we consider any s, I_j,s ⊂ I_i,s−1 and N, k, as in (4.1), that is, Ms^(a)≤ N ≤ k ≤ N_s−1. Also,QN

m=1ρ_m,scorresponds (in the sense of the proof of (2.8)) to the product QN

m=1d_m(x_{N +1,j,s}, Z), whereas Qk

m=1ρ_m,s−1 does so for Qk

m=1d_m(x_k+1,i,s−1, W ), where the points W = (x_m,i,s−1)^k_m=1 are chosen by the same rule as Z, but in the interval I_i,s−1. Let q < N .

Lemma 4.1. In the above notation, δ^k−N_s−1 QN

m=q+1ρm,s≤Qk

m=q+1ρm,s−1.

Proof. We have N_s−1/2 ≤ M_s≤ N ≤ N_s−1, so 2ⁿ≤ N ≤ 2ⁿ⁺¹with n := n_s−1− 1.

We can omit the case N = 2ⁿ⁺¹ with QN

m=1ρ_m,s = δ_n+s+1δ_n+sδ_n+s−1² · · · δ_s²ⁿ,

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since then k = N and Qk

m=1ρ_m,s−1 = δ_n+sδ_n+s−1δ_n+s−2² · · · δ_s−1²ⁿ , so the desired inequality is evident. Therefore, 2ⁿ ≤ N < 2ⁿ⁺¹, ρm,s ∈ {δn+s, . . . , δs}, whereas ρ_m,s−1∈ {δ_n+s, . . . , δ_s−1} for all considered m.

Let U = W ∩ I_2i−1,s and V = W ∩ I_2i,s. Let us show that #(U ) ≤ #(Z) and #(V ) ≤ #(Z). By that, the densities of points from W on each sth level subinterval of I_i,s−1 do not exceed that of Z and ρ_m,s ≤ ρ_m,s−1 for 1 ≤ m ≤ N . Since the number of points of W in the interval adjacent to I_j,s is not smaller than k − N , the result follows.

Suppose that I_j,s is the right subinterval of I_i,s−1; that is j = 2i. If k = 2q + 1, then #(Z) = N ≥ q + 1. Here, #(U ) = q + 1 and #(V ) = q. If k = 2q, then N ≥ q = #(U ) = #(V ). Similarly, for j = 2i − 1 and k = 2q + 1 we have N ≥ q + 1 = #(U ) with #(V ) = q, whereas k = 2q gives #(U ) = #(V ) = q ≤ N . Since 2ⁿ ≤ N < 2ⁿ⁺¹, the set Z contains all points of the (n + s − 1)th level on Ij,s and at least one point of the (n + s)th level is not included in the set.

Therefore, ρ1,s = δn+s, ρ2,s = δn+s−1 and ρ3,s ∈ {δn+s−1, δn+s−2} depending on N . As in Section2, we haveQN

m=1ρ_m,s=Qs+n

r=s δr^s^r^{(N )}. But #(W ∩ I) ≤ #(Z), where I is the subinterval of the sth level containing x_k+1,i,s−1. As was mentioned in Section2, the distribution of points from W ∩ I is uniform or bilateral symmetric to uniform. This means that ρ_m,s−1≥ ρ_m,sfor 1 ≤ m ≤ N and the degrees s_r(k) in the representation QN

m=1ρm,s−1 =Qs+n

r=s−1δ^sr^r^(k) do not exceed the corresponding s_r(N ), except for the value r = s − 1 if #(W ∩ I_j,s) < #(Z). In addition, we have ρ_{N +1,s−1} = · · · = ρ_k,s−1= δ_s−1. This completes the proof.

We are able to give an analogue of Theorem 2 from [3].

Theorem 4.2. Let (n_s)^∞_s=0 be chosen as above. Then the system (eN,j,s, ηN,j,s)^∞,_s=0,²_j=1,^s^, ^N_{N =M}^s _s

is a Schauder basis in the space E (K(γ)).

Proof. Since in the proofs of Lemmas 3.1 and 3.2 we use only (2.1), (2.3), and the properties of uniformly distributed points, the same reasoning applies to the local case. For 2ⁿ≤ N < 2ⁿ⁺¹ with N > 2p we have

keN,j,skp ≤ C · C₀^NN^p ·

N

Y

m=2p+1

ρm,s,

where the values ρ_m,sbelong to the set {δ_n+s, . . . , δ_s} and correspond to the points Z = (x_m,j,s)^N_m=1 ⊂ I_j,s. In the proof, we use the notation C for any constant that does not depend on N , s, and j.

As in Lemma3.2, |ξ_N,j,s(f )| ≤ (16/7)^Nk|f k|_qQN

m=q+1ρ⁻¹_m,s for N > q + 1 with q of the form 2^m− 1. In order to estimate the subtrahend in (4.1), we use (8) from [3]:

ξ_N,j,s(e_k,i,s−1)

= |e^{(N )}_k,i,s−1(θ)|

N ! ≤ k N

`^k−N_i,s−1 ≤ (2C₀)^kδ_s−1^k−N,