Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 29, 2004, 43–58

TRIANGULATIONS OF CLOSED SETS AND BASES IN FUNCTION SPACES

Alf Jonsson

Umeå University, Department of Mathematics SE-901 87 Umeå, Sweden; Alf.Jonsson@math.umu.se

Abstract. Triangulations of closed sets F in R^{n} with certain properties, here called regular sequences of triangulations, appear in a natural way when studing bases in function spaces on F . In this paper a characterization of sets permitting a regular sequence of triangulations is given in the one-dimensional case. Interpolating bases in spaces of functions on such sets are also discussed.

1. Introduction

In the paper [2] bases in different spaces of functions defined on compact subsets F of R^{n} were constructed. It turned out that such bases could be constructed if the set F permits what was called a regular sequence of triangulations. Examples of fractal sets permitting such triangulations were given, but in general it is not easy to see for which sets such triangulations exist. However, it was shown that if a regular sequence of triangulations exists, then the set preserves Markov’s inequality.

In this paper we consider the one-dimensional case only, which of course simplifies things. We show in Section 3 in our main theorem that the existence of a regular sequence of triangulations is then in fact equivalent to the condition that the set preserves Markov’s inequality. We also exhibit, in Section 5, for such sets an interpolating basis in the the Lipschitz spaces Λ_{α}(F), α > 0, α noninteger, thereby extending a result from [2] given for 0 < α < 1, benefitting from working in one dimension only.

2. Regular triangulations and sets preserving Markov’s inequality

Let F be a compact subset of R^{n}. A finite set 𝒯 of n-dimensional closed, non-degenerated, simplices is called a triangulation of F if the following conditions hold. A1. For each pair Δ_{1}, Δ_{2} ∈ 𝒯 , the intersection Δ_{1} ∩ Δ_{2} is empty or a common face of lower dimension. A2. Every vertex of a simplex Δ ∈ 𝒯 is in F. A3. F ⊂ ⋃_{Δ∈𝒯} Δ.

2000 Mathematics Subject Classification: Primary 28A80, 42C15, 46E35.

For a triangulation 𝒯 , let δ = max_{Δ∈𝒯} diam (Δ) be the diameter of the triangulation. When considering a sequence {𝒯_{i}}_{i=0}^{∞} = {𝒯_{i}} of triangulations, we denote by δ_{i} the diameter of the triangulation 𝒯_{i}. In the sequel, we deal with sequences {𝒯_{i}} of triangulations satisfying the following conditions: B1. For each i ≥ 0, 𝒯_{i+1} is a refinement of 𝒯_{i}, i.e., for each Δ ∈ 𝒯_{i+1} there is Δ̃ ∈ 𝒯_{i} such that Δ ⊂ Δ̃. B2. δ_{i} → 0, i → ∞.

It is easy to see that if a triangulation satisfies B1 and B2, then the following condition holds. We denote by 𝒰_{i} the set of vertices of 𝒯_{i}. B3. For i ≥ 0, 𝒰_{i} ⊂ 𝒰_{i+1} .

We now define the class of triangulations which we will use. It is taken from [2] (see also the references given there), but similar classes have been considered elsewhere, for example in connection with the finite element method. Since the definition of this class simplifies in one dimension, we will only give it for this case, but preserve the notation from [2]. In the n-dimensional case, there is also a condition T3 guaranteeing that simplices are not too flat, and the condition T4 is more involved.

Definition 1. Let F ⊂ R, and let {𝒯_{i}} be a sequence of triangulations satisfying B1. Then {𝒯_{i}} is a regular sequence of triangulations if the following conditions hold. T1. There is a constant c_{2} > 0, independent of i, such that, for all Δ_{1}, Δ_{2} ∈ 𝒯_{i},

c_{2}^{−1}diam (Δ_{2}) ≤ diam (Δ_{1}) ≤ c_{2}diam (Δ_{2}).

T2. There are constants 0 < c_{3} < c_{4} < 1 such that, for all i ≥ 0,

c_{3}δ_{i} ≤ δ_{i+1} ≤ c_{4}δ_{i}.

T4. There exist a constant a > 0, independent of i, such that if Δ ∈ 𝒯_{i} and Δ′ ∈ 𝒯_{i} and the distance between these intervals is less than or equal to aδ_{i}, then the intervals intersect.

The following concept was introduced in connection with the study of function spaces on fractals, see e.g. [3], where the concept is discussed in detail. It can be seen as a generalization of the classical Markov inequality, which states that if P is a polynomial in one variable of degree at most m, then max_{x∈[0,1]} |dP/dx| ≤ 2n^{2} max_{x∈[0,1]} |P (x)|.

Definition 2. Denote by P_{m} the set of all polynomials in n variables of total degree less than or equal to m. A closed set F ⊂ R^{n} preserves Markov’s inequality if for every fixed positive integer m there exists a constant c, such that for all polynomials P ∈ P_{m} and all closed balls B = B(x_{0}, r), x_{0} ∈ F, 0 < r ≤ 1, holds

max_{F ∩B} |∇P | ≤ \frac{c}{r} max_{F ∩B} |P |,

where ∇ denotes the gradient.

Examples of sets preserving Markov’s inequality are e.g. selfsimilar fractals not contained in an n − 1-dimensional subspace of R^{n}, and d-sets, as defined in [3], with d > n − 1. There are several different geometric characterizations of such sets. We will use the one given in [3, Section 2.2], stating it in one dimension, only. If F ⊂ R, then a set preserves Markov’s inequality if and only if the following condition holds. (G) There is an ε > 0 such that for any r with 0 < r ≤ 1 and any x_{0} ∈ F, the set F ∩ {x : εr ≤ |x − x_{0}| ≤ r} is nonempty.

The concept of sets preserving Markov’s inequality has appeared also in different contexts, then in general in a geometrical form, see for example [4], where they are called n-thick sets and their invariance properties under different classes of maps are studied.

3. The main theorem

In [2] it is shown that if F ⊂ R^{n} admits a regular sequence of triangulations, then F preserves Markov’s inequality. The converse does not hold, as is seen by the trivial example of the unit ball in R^{n}, n > 1. Here we shall prove that in one dimension the converse holds, and thus we have the following theorem.

Theorem 1. Let F be a compact subset of R. Then there exists a regular triangulation of F if and only if F preserves Markov’s inequality.

As a preparation for the proof of the theorem we give one more charactarization of sets preserving Markov’s inequality, valid in one dimension. For a given compact set F ⊂ R, the complement of F is a numerable union of disjoint open intervals which we denote by O_{ν}, ν = 1, 2, 3, …. Let 0 < c < 1, and let, for a given constant c, and for n ≥ 1,

𝒜_{n} = {O_{ν}; diam O_{ν} ≥ c^{n−1}} and A_{n} = ∪{O_{ν} ∈ 𝒜_{n}}.

Then A_{n}^{c} = R^{n} \ A_{n} consists of a finite union of finite closed intervals which we denote, ordered from left to right, by B_{nm}, m = 1, 2, 3, …, m_{n}. Note that we then have that

(i) the distance between two consecutive intervals B_{nm} is at least c^{n−1} . We shall use the fact that if F preserves Markov’s inequality, then the length

of each B_{nm} is at least c^{n}, where c can be taken as any constant ε admissible in the geometric characterization of sets preserving Markov’s inequality. This property can, in fact, be used to characterize such sets.

Proposition 1. Let F ⊂ R be a compact set. Then F preserves Markov’s inequality if and only if there is a constant c with 0 < c < 1, such that constructing B_{nm} with this constant as the given constant c, one has that (ii) the length of each interval B_{nm}, n ≥ 1, m = 1, 2, …, m_{n}, is at least c^{n}.

Proof. Assume that F preserves Markov’s inequality, and let c be equal to an admissible constant ε in the geometric characterization (G) of such sets. Let [a, b] be one of intervals B_{nm} and denote by l its length. Assume l < c^{n}, and take l′ with l < l′ < c^{n}. Then the set {x; l′ ≤ |x − a| ≤ l′/c} contains no points from F because of the statement (i) above, since l′/c < c^{n−1} . On the other hand, since F preserves Markov’s inequality, the set contains points from F (use the geometric characterization (G) with r = l′/c). Thus, l ≥ c^{n}.

Conversely, let c be a constant such that (ii) holds. Take x_{0} ∈ F and r ≤ 1, and assume that \{x; \frac{1}{2}r < |x − x_{0}| < r\} contains no points from F; otherwise, the condition in (G) holds for these x_{0} and r with ε = \frac{1}{2} . Take n_{0} so that c^{n_{0}−1} < \frac{1}{2}r ≤ c^{n_{0}−2} . Then each one of the two intervals forming the set \{x; \frac{1}{2}r < |x − x_{0}| < r} is a subset of an interval in A_{n_{0}} , which means that, with m such that x_{0} ∈ B_{n_{0}m}, the endpoints of B_{n_{0}m} are in the interval [x_{0} − \frac{1}{2}r, x_{0} + \frac{1}{2}r]. By (ii), since the endpoints are in F, this means that there is an x ∈ F with |x − x_{0}| ≤ \frac{1}{2}r such that |x − x_{0}| ≥ \frac{1}{2}c^{n_{0}} ≥ \frac{1}{4}c^{2}r, so (G) is fulfilled with ε = \frac{1}{4}c^{2} .

Proof of Theorem 1. Assume that F preserves Markov’s inequality, and let c < \frac{1}{2} be a constant as in Proposition 1. Taking c < \frac{1}{2} is allowed, since c can be taken as the constant ε in (G), which, clearly, can be taken arbitrarily small. We prove the theorem inductively in steps, choosing in the n^{th} step a set 𝒫_{n} of points from F, such that the distance between two consecutive points in the same B_{nm} is comparable to c^{n}, and in such a way that 𝒫_{1} ⊂ 𝒫_{2} ⊂ · · ·. A triangulation 𝒯_{n} is then chosen with the points in 𝒫_{n} as vertices.

Assumptions on the set 𝒫_{n−1} . Assume that we have already chosen the set 𝒫_{n−1} of points from F, with 𝒫_{0} meaning the empty set ∅.

Let B_{nm} = [a_{nm}, b_{nm}] be as before. Every point x ∈ F belongs to exactly one interval B_{sm} for each s ≥ 1; we denote the corresponding m by m(s, x) and so the interval is B_{s,m(s,x)} . If x is an endpoint of an interval B_{sm}, then it is, for t > s, an endpoint of some B_{tm}, and we denote by s_{x} the smallest s, if any, such that x is an endpoint of some B_{sm}. If x is not an endpoint of any interval B_{sm} with s ≥ 1, then we put s_{x} = ∞.

For a given n ≥ 1, we now assume the following about the set 𝒫_{n−1} , denoting it the assumption A(n − 1). (If n = 1 the interpretation is that nothing is assumed.) It describes how points z ∈ 𝒫_{n−1} are positioned with respect to the endpoints of the sets B_{sm} with s ≥ n and to each other, in terms of points z′ which will appear naturally in the construction. If z ∈ 𝒫_{n−1} and s_{z} > n (so z is not an endpoint of some B_{nm}), then we assume that to z there is associated a point z′ ∈ F, such that the following conditions hold.

1) z′ − a_{s,m(s,z)} > \frac{1}{2}c^{s} and b_{s,m(s,z)} − z′ > \frac{1}{2}c^{s} for n ≤ s < s_{z}.
2) |z − z′| ≤ \frac{1}{2}c^{s_{z}}.

3) If z_{∗} is a point in 𝒫_{n−1} closest to z satisfying z < z_{∗} < b_{n,m(n,z)} , which in particular means that s_{z_{∗}} > n, then z′_{∗} − z′ > \frac{1}{2}c^{n}.

4) All endpoints of the intervals B_{n−1,m}, 1 ≤ m ≤ m_{n−1} , are in 𝒫_{n−1} .

Note that 1) implies that for z ∈ int B_{nm} we also have z′ ∈ int B_{nm}.

Choice of points in 𝒫_{n}. Let, for some B_{nm}, α ∈ B_{nm} ∩ F and β ∈ B_{nm} ∩ F be points satisfying β − α > \frac{1}{2}c^{n}. We claim that the set (α + \frac{1}{2}c^{n}, α + c^{n−1} + \frac{1}{2}c^{n}] ∩ (α, β] ∩ F is nonempty. Since β ∈ F and β > α + \frac{1}{2}c^{n}, this is of course the case if β ≤ α + c^{n−1} + \frac{1}{2}c^{n}. Otherwise the condition is the same as saying that (α + \frac{1}{2}c^{n}, α + c^{n−1} + \frac{1}{2}c^{n}] ∩ F is nonempty. This follows from the fact that any interval I of length > c^{n−1} and satisfying I ⊂ B_{nm} must intersect F, as otherwise I ⊂ F^{c} and |I| > c^{n−1} implies I ⊂ O_{ν} for some O_{ν} ∈ 𝒜_{n} and thus I ⊂ A_{n} and I ∩ B_{nm} = ∅.

Now we define a point w′′ ∈ B_{nm} ∩ F (depending on α, β, and n) by

(1)  w′′ = max\{x; x ∈ (α + \frac{1}{2}c^{n}, α + c^{n−1} + \frac{1}{2}c^{n}] ∩ (α, β] ∩ F \}.

If for some ν ≥ n the distance from w′′ to the boundary of B_{ν,m(ν,w′′)} is less than or equal to \frac{1}{2}c^{ν}, let ν(w′′) denote the first ν for which this occurs, otherwise put ν(w′′) = ∞. In other words, ν(w′′) is the smallest ν ≥ n, if any, such that the interval [w′′ − \frac{1}{2}c^{ν}, w′′ + \frac{1}{2}c^{ν}] is not “strictly inside” but contains an endpoint of B_{ν,m(ν,w′′)} .

We next select a point w associated to w′′, and consequently it will depend on α, β, and n, too. When doing this we assume more about β. The point β is either the point y′ associated to a point y ∈ 𝒫_{n−1} ∩ B_{nm} with s_{y} > n by the assumption A(n − 1), or the right endpoint b_{nm} of B_{nm}. We make the following choices.

(a) If β = b_{nm} and the interval [w′′ − \frac{1}{2}c^{n}, w′′ + \frac{1}{2}c^{n}] contains b_{nm}, then we put w = b_{nm}.

(b) If β = y′, and the interval [w′′ − \frac{1}{2}c^{n}, w′′ + \frac{1}{2}c^{n}] contains y′, then we put w = y.

(c) If the interval [w′′ − \frac{1}{2}c^{n}, w′′ + \frac{1}{2}c^{n}] does not contain b_{nm} or y′, respectively, i.e. β, which implies, as we shall see below, that ν(w′′) > n, then, if ν(w′′) = +∞ we let w = w′′, and if n < ν(w′′) < ∞ we let w be an endpoint of B_{ν(w′′),m(ν(w′′),w′′)} which is closest to w′′.

Before continuing the construction of 𝒫_{n}, we make some observations related to this choice of w. In the cases (a) and (b) we obviously have, respectively,

(2)  b_{nm} − w′′ ≤ \frac{1}{2}c^{n} and y′ − w′′ ≤ \frac{1}{2}c^{n},

while in case (c) we have

(3)  |w − w′′| ≤ \frac{1}{2}c^{ν(w′′)} ≤ \frac{1}{2}c^{n+1}.

The inequality ν(w′′) > n in case (c) follows by the following remarks. The point a_{nm} does not belong to [w′′ − \frac{1}{2}c^{n}, w′′ + \frac{1}{2}c^{n}], since, by (1), w′′ > α + \frac{1}{2}c^{n}, and α ≥ a_{nm}. This means that b_{nm} ∈ [w′′ − \frac{1}{2}c^{n}, w′′ + \frac{1}{2}c^{n}] if and only if ν(w′′) = n. If β = y′ as above, then b_{nm} ∉ [w′′− \frac{1}{2}c^{n}, w′′+ \frac{1}{2}c^{n}], since b_{nm}−y′ > \frac{1}{2}c^{n} by 1) in A(n−1), and w′′ ≤ y′. Thus, in case (c), we have that b_{nm} ∉ [w′′− \frac{1}{2}c^{n}, w′′+ \frac{1}{2}c^{n}], so ν(w′′) > n. We also remark that in case (c), if β = y′, then w < y, since, from 2) in A(n−1) and (3), y−w = y−y′+y′−w′′+w′′−w ≥ −\frac{1}{2}c^{n+1}+ \frac{1}{2}c^{n}− \frac{1}{2}c^{n+1} = c^{n}(\frac{1}{2} − c) > 0, since c < \frac{1}{2} .

Now we use the above construction to divide an interval B_{nm} = [a_{nm}, b_{nm}] into subintervals whose endpoints will be the points in B_{nm} ∩ 𝒫_{n}. This shall be done in such a way that the points in B_{nm} ∩ 𝒫_{n−1} are included. We will temporarily denote the points in the new division of B_{nm} by x_{0}, x_{1}, ….

The interval B_{nm} is in a natural way divided by 𝒫_{n−1} ∩ B_{nm} into intervals which are of the form [a_{nm}, y], [y, y_{∗}], [y, b_{nm}], or [a_{nm}, b_{nm}], where y and y_{∗} are consecutive points in 𝒫_{n−1} which are not endpoints of B_{nm}. The last case appears if there are no points from 𝒫_{n−1} strictly inside B_{nm}, which is always the case if n = 1. We start by subdividing an interval of the form [a_{nm}, b_{nm}] or [a_{nm}, y], and take a_{nm} as the first division point, denoting it by x_{0} . Let α = a_{nm} and β = b_{nm} or β = y′, respectively, where y′ is the point associated to y in the assumption A(n − 1). If β = b_{nm}, then by (ii) in Proposition 1, we have β > a_{nm} + \frac{1}{2}c^{n}, and if β = y′, then β > a_{nm} + \frac{1}{2}c^{n} holds by the assumption 1) used with s = n and z′ = y′. Thus we can use the construction above, and associate to α and β points w′′ and w, which we denote by x′′_{1} and x_{1} , respectively. If x_{1} equals b_{nm} or y, i.e., in case (a) or (b), then the subdivision of the interval [a_{nm}, b_{nm}] or [a_{nm}, y], respectively, is completed (without giving new points, except possibly a_{nm} or b_{nm}), otherwise we put α = x′′_{1} but let β be as before. By our construction we have β > x′′_{1} + \frac{1}{2}c^{n}, for if the interval [x′′_{1} − \frac{1}{2}c^{n}, x′′_{1} + \frac{1}{2}c^{n}] had contained β, then x_{1} would have been equal to b_{nm} or y. Thus, we can again use the construction to give us new points w′′ and w, which we denote by x′′_{2} and x_{2} . We continue in this way until we reach b_{nm} or y, say in the k^{th} step, so x_{k} = b_{nm} or x_{k} = y.

Note that in the j^{th} step, if x_{j−1} ≠ b_{nm} or y, we always have x′′_{j} > α+ \frac{1}{2}c^{n} = x′′_{j−1} + \frac{1}{2}c^{n} by (1) and the choice of x′′_{j} := w′′(x′′_{j−1}, β, n). Thus after a finite number of steps the construction ends up with a selection of b_{nm} or y, respectively. We also remark that we have x_{j−1} < x_{j}, j = 1, 2, …, k. We omit the verification of this for the moment; the calculations leading to it are given when estimating z_{2} − z_{1} at the end of this section.

If x_{k} = b_{nm} the division of B_{nm} is completed, otherwise we continue to divide the next subinterval in B_{nm}, which is of the form [x_{k}, b_{nm}] or [x_{k}, y_{∗}]. In these cases we put α = x′_{k} and let β be b_{nm} or the point y′_{∗}, respectively. Here the points x′_{k} and y′_{∗} are the points associated to x_{k} ∈ 𝒫_{n−1} and y_{∗} ∈ 𝒫_{n−1} , in the assumption A(n − 1). Then, by the assumptions 1) and 3), respectively, we again have β > α + \frac{1}{2}c^{n}, and we can use our construction again to give us new points w′′ and w which we denote by x′′_{k+1} and x_{k+1} . Unless we already reached b_{nm} or y_{∗}, we take α = x′′_{k+1} and let β be as before, and continue as above until the present interval is divided and then until all subintervals of B_{nm} have been divided, the last chosen point in B_{nm} being b_{nm}.

Dividing all intervals B_{nm} in this way we get the set 𝒫_{n}, letting it consist of all the chosen points x_{i}. To complete the construction, we state how points z′ are associated to points z ∈ 𝒫_{n}. We do this for points z which are not endpoints of some B_{nm}. If such a z belongs to 𝒫_{n−1} , there is already a point z′ associated to it by our assumption, and we take it as the point associated to z also in 𝒫_{n}. Otherwise, if z ∈ 𝒫_{n} \ 𝒫_{n−1} , we take z′ as the point z′′ (in the notation above w′′) used in the construction of z.

We remark that we have, with z_{1} and z_{2} being consecutive points with z_{1} < z_{2} from 𝒫_{n} in B_{nm}, with z_{2} ∈ 𝒫_{n−1} or z_{2} = b_{nm}, replacing z′_{1} by a_{nm} if z_{1} = a_{nm}, that

(4)  z′_{2} − z′_{1} > \frac{1}{2}c^{n}, z_{2} ≠ b_{nm}  and  z_{2} − z′_{1} > \frac{1}{2}c^{n}, z_{2} = b_{nm}.

These estimates can be found above in the description of the subdivision of B_{nm}, taking into account that z′_{1} = z′′_{1} if z_{1} ∈ 𝒫_{n} \ 𝒫_{n−1} , z_{1} ≠ a_{nm}.

The set 𝒫_{n} satisfies A(n). To complete the induction step we must check that this subdivision fulfills the condition A(n). Let us make some observations first. Assume that w ∈ 𝒫_{n} \ 𝒫_{n−1} , w ≠ a_{nm}, b_{nm}. Then w is chosen in step n, and emanates, by means of case (c) in the construction, from a point w′′ with ν(w′′) > n, in such a way that w is an endpoint of the interval B_{sm} with s = ν(w′′) which contains w′′ if n < ν(w′′) < ∞, and w = w′′ if ν(w′′) = ∞. Because of 4) in A(n − 1), w is not an endpoint of some B_{sm} with s < n, and since by our construction the intervals [w′′ − \frac{1}{2}c^{s}, w′′ + \frac{1}{2}c^{s}] do not contain an endpoint of any B_{sm} for all s with n ≤ s < ν(w′′) but contain w, it follows that s = ν(w′′) is the first s such that w is an endpoint of some B_{sm}, i.e. s_{w} = ν(w′′). Note also that, if ν(w′′) < ∞, w and w′′ are both in the same B_{ν(w′′),m}, so, since the intervals are nesting, they are also in the same interval B_{sm} if s < ν(w′′), which means that for s ≤ ν(w′′) we have e.g. a_{s,m(s,w′′)} = a_{s,m(s,w)} .

To check A(n), let z ∈ 𝒫_{n} with s_{z} > n+1. If z ∈ 𝒫_{n−1} , then the conditions 1), and 2), in A(n) hold by assumption and the choice of z′. If instead w ∈ 𝒫_{n}\𝒫_{n−1}, s_{w} > n+1, then as we saw above the interval [w′′− \frac{1}{2}c^{s}, w′′+ \frac{1}{2}c^{s}] does not contain a_{s,m(s,w′′)} = a_{s,m(s,w)} or b_{s,m(s,w′′)} = b_{s,m(s,w)} for n ≤ s < ν(w′′), but does for s = ν(w′′) = s_{w}. Thus, recalling that w′ = w′′ if w ∈ 𝒫_{n} \ 𝒫_{n−1} , s_{w} > n, we see that we have have 1) even for n ≤ s < s_{w}, and 2) (with z in 1) and 2) equal to w). Let z and z_{∗} be as in 3) in A(n). Then the point α in the construction of z_{∗} is taken as z′, and if z_{∗} ∈ 𝒫_{n} \ 𝒫_{n−1} , so z′_{∗} = z′′_{∗}, then (1) gives z′_{∗} − z′ > \frac{1}{2}c^{n} > \frac{1}{2}c^{n+1} , and if z_{∗} ∈ 𝒫_{n−1} , then we have the same estimate by (4). Condition 4) is immediate from the construction.

A regular triangulation from the constructed points. Having defined the point sets 𝒫_{1} , 𝒫_{2} , 𝒫_{3}, …, we define the triangulation 𝒯_{nm} as consisting of the intervals connecting consecutive points of 𝒫_{n} ∩ B_{nm}, and then the triangulation 𝒯_{n} as the union of 𝒯_{nm} over 1 ≤ m ≤ m_{n}. We must check that the triangulation obtained in this way is regular; it will turn out that to assure that the condition T2 is fulfilled, one has to choose a subsequence of 𝒯_{n}.

Recall that if z ∈ 𝒫_{n−1} ∩ B_{nm}, z ≠ a_{nm}, or z = b_{nm}, then z′′ is the point which in the construction step above (there denoted w′′) leads to the incorporation of z in 𝒫_{n}, and if z ∈ (𝒫_{n} \ 𝒫_{n−1}) ∩ B_{nm}, s_{z} > n, then z′′ is the point w′′ which leads to a new point z and then z′′ = z′. Note also that if z ∈ 𝒫_{n} ∩ B_{nm}, s_{z} > n, then

(5)  |z − z′| ≤ c^{s_{z}} ≤ \frac{1}{2}c^{n+1}

holds by 2) in A(n − 1) if z ∈ 𝒫_{n−1} and by (3) otherwise.

Suppose z_{1} and z_{2} are consecutive points with z_{1} < z_{2} from 𝒫_{n} in the same B_{nm} = [a_{nm}, b_{nm}]. Then z_{2} was constructed with the aid of the point z′′_{2} chosen as the point w′′ obtained from (1), with α = z′_{1} in (1) if z_{1} ≠ a_{nm} and α = a_{nm} if z_{1} = a_{nm}. Thus we have, respectively,

(6) \frac{1}{2}c^{n} < z′′_{2} − z′_{1} ≤ \frac{1}{2}c^{n} + c^{n−1} and \frac{1}{2}c^{n} < z′′_{2} − a_{nm} ≤ \frac{1}{2}c^{n} + c^{n−1}.

Assume now, furthermore, that z_{1} and z_{2} are not endpoints of B_{nm}. Then we have the following estimates.

(i) If z_{2} ∈ 𝒫_{n−1} , then, from (5) and (4),

\begin{aligned}z_{2} − z_{1} = z_{2} − z′_{2} + z′_{2} − z′_{1} + z′_{1} − z_{1} & ≥ −\frac{1}{2}c^{n+1} + \frac{1}{2}c^{n} − \frac{1}{2}c^{n+1} \\ & = \frac{1}{2}c^{n} − c^{n+1}. \end{aligned}

(ii) If z_{2} ∈ 𝒫_{n} \ 𝒫_{n−1} , so z′_{2} = z′′_{2} , then, from (5) and (6), we have,

\begin{aligned}z_{2} − z_{1} & = z_{2} − z′_{2} + z′_{2} − z′_{1} + z′_{1} − z_{1} \\ & ≥ −\frac{1}{2}c^{n+1} + \frac{1}{2}c^{n} − \frac{1}{2}c^{n+1} = \frac{1}{2}c^{n} − c^{n+1}. \end{aligned}

(iii) If z_{2} ∈ 𝒫_{n−1} , then, from (5), (2), and (6),

z_{2} − z_{1} = z_{2} − z′_{2} + z′_{2} − z′′_{2} + z′′_{2} − z′_{1} + z′_{1} − z_{1}
≤ \frac{1}{2}c^{n+1} + \frac{1}{2}c^{n} + \frac{1}{2}c^{n} + c^{n−1} + \frac{1}{2}c^{n+1}
= c^{n−1} + c^{n} + c^{n+1}.

(iv) If z_{2} ∈ 𝒫_{n} \ 𝒫_{n−1} , then, from (5) and (6),

z_{2} − z_{1} = z_{2} − z′′_{2} + z′′_{2} − z′_{1} + z′_{1} − z_{1}
≤ \frac{1}{2}c^{n+1} + \frac{1}{2}c^{n} + c^{n−1} + \frac{1}{2}c^{n+1}
= c^{n−1} + \frac{1}{2}c^{n} + c^{n+1}.

If instead z_{2} = b_{nm}, then we make the same estimates as in (i) and (iii), except that we now do not add and subtract z′_{2} , and if z_{1} = a_{nm} we do not add and subtract z′_{1} in the various cases. In any case this clearly leads to slightly better estimates. Thus, in general we have

c^{n}(\frac{1}{2} − c) = \frac{1}{2}c^{n} − c^{n+1} ≤ z_{2} − z_{1} ≤ c^{n−1} + c^{n} + c^{n+1} = c^{n}(\frac{1}{c} + 1 + c),

from which T1, and also the left inequality in T2, clearly follows since c < \frac{1}{2} . Condition T4 is easily checked. Concerning the right inequality in T2, the construction gives δ_{i+1} ≤ δ_{i}, only. The inequality δ_{i+1} ≤ c_{4}δ_{i} is obtaind by considering a subsequence of the triangulations {𝒯_{n}}, considering the sequence {𝒯′_{n}} where 𝒯′_{n} = 𝒯_{nk}, n = 1, 2, …, for a positive integer k big enough. This concludes the proof of the theorem.

4. The spaces C^{k}(F) and Λ_{α}(F)

The spaces C^{k}(F) of k times differentiable functions on F and the Lipschitz spaces Λ_{α}(F) can be defined on arbitrary closed subsets of R^{n}, but the definitions simplify if F is a set in one dimension preserving Markov’s inequality, and we define them here on such sets only, referring to e.g. [3] for the general case.

Let F be a compact subset of R preserving Markov’s inequality and f a function defined on F. Then, since F is perfect, derivatives can be defined in the usual way by Df(x_{0}) = lim_{x→x_{0},x∈F} (f(x)−f(x_{0}))/(x−x_{0}), x_{0} ∈ F. For a given k ≥ 0, assuming that derivatives of orders ≤ k exist, denote by R_{j} the Taylor remainders given by

D^{j}f(x) = ∑_{l=0}^{k−j}(x − y)^{l}D^{j+l}f(y)/l! + R_{j}(x, y), 0 ≤ j ≤ k.

A function f belongs to the space C^{k}(F) if for every ε > 0 there is a δ > 0 such that |R_{j}(x, y)| < ε|x − y|^{k−j} for 0 ≤ j ≤ k and |x − y| < δ. The Whitney extension theorem given in [5] gives that there is a linear extension operator E from C^{k}(F) to C^{k}(R). It should be noted that in the present setting derivatives are uniquely determined by f, which means that elements in C^{k}(F) are functions rather than families of functions as in the general Whitney extension theorem.

For k < α < k+1, a function f is in the Lipschitz space Λ_{α}(F) if |D^{j}f(x)| ≤ M and |R_{j}(x, y)| ≤ M|x − y|^{α−j} for x, y ∈ F and 0 ≤ j ≤ k. The norm of f in Λ_{α}(F) is the infimum of the possible constants M. If F = R, we will in general work with the equivalent norm given by the infimum of the constants M such that |D^{j}f(x)| ≤ M, 0 ≤ j ≤ k and |D^{k}f(x) − D^{k}f(y)| ≤ M|x − y|^{α−k}, |x − y| ≤ 1. For this case the Whitney extension theorem gives that there is a bounded linear operator E from Λ_{α}(F) to Λ_{α}(R), see e.g. [3].

If instead α = k > 0, we say, following [3, p. 62], that f ∈ Λ_{α}(F) if there is a sequence {f_{n}}_{n=1}^{∞} of functions in Λ_{γ}(F) where k < γ < k + 1, such that, for n ≥ 1, ∥D^{j}(f − f_{n})∥_{∞,F} ≤ M2^{−n(k−j)} for 0 ≤ j < k, ∥D^{j}(f_{n+1} − f_{n})∥_{∞,F} ≤ M for j = k, and ∥f_{n}∥_{Λ_{γ}(F)} ≤ M2^{−n(k−γ)} (here ∥ · ∥_{∞,F} denotes the maximum norm on F). The norm in Λ_{α}(F) is again the infimum of the possible constants M. The version of the Whitney extension theorem given by Theorem 2 in Chapter 3 in [3] gives the same conclusion as in the noninteger case.

We will need the following lemma on Hermite interpolation. We give it for functions defined on R, but because of the Whitney extension theorem we will be able to apply it for functions defined on sets preserving Markov’s inequality as well. For f ∈ C(R), let ω(f, t) = sup{|f(x) − f(y)|; |x − y| ≤ t}.

Lemma 1. Let k ≥ 0, f ∈ C^{k}(R), a ∈ R, h > 0, and let P be the polynomial of degree at most 2k + 1 with D^{ν}P (a) = D^{ν}f(a) and D^{ν}P (a + h) = D^{ν}f(a + h) for ν = 0, 1, …, k. Then

|D^{ν}P (x) − D^{ν}f(x)| ≤ ch^{k−ν}ω(D^{k}f, h), x ∈ [a, a + h], ν = 0, 1, …, k,

where c is a constant depending on k, only.

Proof. Assume first that a = 0 and h = 1, let 0 ≤ x ≤ 1, and let T denote the Taylor polynomial of f around 0 of degree k. Then, for ν < k,

\begin{aligned}D^{ν}f(x) − D^{ν}T (x) & = D^{ν}f(x) − ∑_{l=0}^{k−ν−1} x^{l}D^{ν+l}f(0)/l! − x^{k−ν}D^{k}f(0)/(k − ν)! \\ & = x^{k−ν}(D^{k}f(ξ) − D^{k}f(0))/(k − ν)! \end{aligned}

for some ξ between 0 and x, so we get |D^{ν}f(x)−D^{ν}T (x)| ≤ 1/(k −ν)!ω(D^{k}f, 1). For ν = k, the same estimate is immediate since D^{k}T (x) = D^{k}f(0).

Consider R = P − T. Then R is a polynomial of degree at most 2k + 1 such that D^{ν}R(0) = 0 and |D^{ν}R(1)| ≤ 1/(k − ν)!ω(D^{k}f, 1) for ν ≤ k. Let P^{s} be the polynomial of degree ≤ 2k + 1 defined in the beginning of Section 5. Then we have R(x) = ∑_{ν=0}^{k} D^{ν}R(0)(−1)^{ν}P^{ν}(1 − x) + D^{ν}R(1)P^{ν}(x). With c_{1} = max_{x∈[0,1],0≤s≤2k+1} |P^{s}(x)| this gives |R(x)| ≤ c_{1} ∑_{ν=0}^{k}(|D^{ν}R(0)| + |D^{ν}R(1)|) ≤ c_{1}(k + 1)ω(D^{k}f, 1) for 0 ≤ x ≤ 1. By using the classical Markov inequality (see Section 2) repeatedly one obtains |D^{ν}R(x)| ≤ 2^{ν}(2k + 1)^{2ν} max_{x∈[0,1]} |R(x)| and so

|D^{ν}P (x) − D^{ν}f(x)| ≤ |D^{ν}P (x) − D^{ν}T (x)| + |D^{ν}T (x) − D^{ν}f(x)|
≤ c_{1}2^{ν}(2k + 1)^{2ν}(k + 1)ω(D^{k}f, 1) + 1/(k − ν)!ω(D^{k}f, 1)
= cω(D^{k}f, 1).

If instead f is defined on the interval [a, a + h], then a linear change of variables gives, with g(x) = hf(hx + a), that ω(D^{ν}g, 1) = h^{ν}ω(D^{ν}f, h), from which the lemma follows.

5. Interpolation bases in C^{k}(F) and the characterization of Lipschitz spaces

Now we define the functions which will give a basis in C^{k}(F). Let a sequence {𝒯_{i}}_{i=0}^{∞} of regular triangulations of F be given, and let as before 𝒰_{i} be the set of vertices of 𝒯_{i}. Let 𝒱_{i} = 𝒰_{i} \ 𝒰_{i−1} , i > 0, 𝒱_{0} = 𝒰_{0} , and 𝒱 = ⋃_{i=0}^{∞} 𝒱_{i}, and let ⪯ be a linear order on 𝒱 satisfying the following condition: if ξ ∈ 𝒱_{i} and η ∈ 𝒱_{j} with i < j, then ξ ⪯ η.

Let k ≥ 0, and let, for s = 0, 1, …, k, P^{s} be the polynomial of degree 2k + 1 which satisfies D^{ν}P^{s}(0) = 0 for ν = 0, 1, …, k, D^{ν}P^{s}(1) = 0 for ν = 0, 1, …, k, ν ≠ s, and D^{ν}P^{s}(1) = 1 if ν = s. Let ξ ∈ 𝒱_{i}. Then ξ is the endpoint of one or two intervals in 𝒯_{i}. If ξ is the right endpoint of Δ ∈ 𝒯_{i} of length l, define ϕ_{i,ξ}^{s} on Δ by ϕ_{i,ξ}^{s}(x) = l^{s}P^{s}((x−ξ+l)/l), and if ξ is the left endpoint of Δ′ ∈ 𝒯_{i} of length l′, define ϕ_{i,ξ}^{s} on Δ′ by ϕ_{i,ξ}^{s}(x) = l′^{s}(−1)^{s}P^{s}((ξ −x+l′)/l′). If ξ is an endpoint of both Δ and Δ′, this means that ϕ_{i,ξ}^{s} is a spline function defined on Δ ∪ Δ′ which is k times differentiable, equals a polynomial of degree at most 2k + 1 on each of the intervals Δ and Δ′, and whose derivatives of orders less than or equal to k are equal to zero at the endpoints of these intervals, except that D^{s}ϕ_{i,ξ}^{s}(ξ) = 1. Note that for x ∈ Δ we have D^{ν}ϕ_{i,ξ}^{s}(x) = l^{s−ν}(D^{ν}P^{s})((x − ξ + l)/l) and consequently for x ∈ Δ we have, with a constant c depending on ν,

(7)  |D^{ν}ϕ_{i,ξ}^{s}(x)| ≤ cl^{s−ν},  ν ≥ 0,

an estimate which clearly also holds if x ∈ Δ′, with l replaced by l′.

For ξ ∈ 𝒱_{i}, let ψ_{ξ}^{s} = ϕ_{i,ξ}^{s} on the one or two intervals in 𝒯_{i} which have ξ as an endpoint, and put ψ_{ξ}^{s} = 0 elsewhere on {∪Δ; Δ ∈ 𝒯_{i}}. Note that ψ_{ξ}^{s} is defined with respect to the triangulation 𝒯_{i} in which ξ first appears as a vertex. We let C^{k}(F) be normed by max_{0≤ν≤k} ∥D^{ν}f∥_{∞,F} , although C^{k}(F) is, in general, not complete in this norm, see e.g. [1]. Then the functions ψ_{ξ}^{s}|_{F} form a basis in C^{k}(F) which interpolates to f and its derivatives of degree at most k as explained by Proposition 2 below. Here f|_{F} denotes the restriction of f to F.

Given f ∈ C^{k}(F), let S_{i}(f) denote the spline function defined on {∪Δ; Δ ∈ 𝒯_{i}} which coincides with a polynomial of degree at most 2k + 1 on each interval in 𝒯_{i}, and interpolates to f and all its derivatives of orders less than or equal to k at each point in 𝒰_{i}.

Proposition 2. Let F ⊂ R be a compact set with a regular sequence {𝒯_{i}}_{i=0}^{∞} of triangulations of F, and let k be a nonnegative integer. Then the system of functions {ψ_{ξ}^{s}|_{F} , ξ ∈ 𝒱 , s = 0, 1, …, k}, ordered by ⪯, is an interpolating Schauder basis in C^{k}(F).

More precisely, every f ∈ C^{k}(F) has a unique representation

(8)  f = ∑_{i=0}^{∞} ∑_{ξ∈𝒱_{i}} ∑_{s=0}^{k} c_{ξ}^{s}ψ_{ξ}^{s}|_{F}

in C^{k}(F), where c_{ξ}^{s} = D^{s}f(ξ) − D^{s}S_{i−1}(f)(ξ) for ξ ∈ 𝒱_{i}, i > 0, and c_{ξ}^{s} = D^{s}f(ξ) for ξ ∈ 𝒱_{0} = 𝒰_{0} . In addition, for N ≥ 0,

D^{ν}f(η) = ∑_{i=0}^{N} ∑_{ξ∈𝒱_{i}} ∑_{s=0}^{k} c_{ξ}^{s}D^{ν}ψ_{ξ}^{s}(η)

for 0 ≤ ν ≤ k and η ∈ 𝒰_{N}.

Proof. Assume that f ∈ C^{k}(F). Consider the sum ∑_{i=0}^{N} ∑_{ξ∈𝒱_{i}} ∑_{s=0}^{k} c_{ξ}^{s}ψ_{ξ}^{s}. Using induction over N it is easy to see that this sum coincides with the function S_{N}(f), which, in particular, gives the interpolation property. Denote by Ef the Whitney extension of f belonging to C^{k}(R) satisfying D^{ν}f(x) = D^{ν}(Ef)(x), x ∈ F, ν ≤ k. We take, as we may, E so that Ef has compact support. Note that S_{N}(f) = S_{N}(Ef) on {∪Δ; Δ ∈ 𝒯_{N}}. By Lemma 1 we then have the estimate |D^{ν}Ef(x)−D^{ν}S_{N}(Ef)(x)| ≤ c(diam Δ)^{k−ν}ω(D^{k}(Ef), diam Δ) for x ∈ Δ ∈ 𝒯_{N}, so

(9) |D^{ν}f(x) − D^{ν}S_{N}(f)(x)| ≤ cδ_{N}^{k−ν}ω(D^{k}(Ef), δ_{N}), x ∈ F,

which shows that the partial sums of the sum in (8) corresponding to summation over all terms with i ≤ N converge to f in C^{k}(F) as N → ∞. The estimate (9) also gives |c_{ξ}^{s}| ≤ cδ_{N}^{k−s}ω(D^{k}(Ef), δ_{N}) for ξ ∈ 𝒱_{N+1} , and from (7) it follows, using T1, that |D^{ν}ψ_{ξ}^{s}(x)| ≤ cδ_{N+1}^{s−ν}, ξ ∈ 𝒱_{N+1} . Thus we get with the aid of T2 that |c_{ξ}^{s}D^{ν}ψ_{ξ}^{s}| ≤ cδ_{N}^{k−ν}ω(D^{k}(Ef), δ_{N}), ξ ∈ 𝒱_{N+1} , from which it follows that we have convergence to f in C^{k}(F) for the sequence of all partial sums. Uniqueness follows from the fact that if we have a representation of f in C^{k}(F) as in (8), then it is easy to realize that the coefficients must be the ones given in the proposition.

Remark. If k = 0, then the theorem holds under the much weaker condition that the triangulations {T_{i}} satisfy the conditions B1 and B2, only. This is not true for k > 0, even if F = [0, 1]. To see this, let for i > 0 𝒯_{i} denote a subdivision of [0,1] into equal intervals of length δ_{i} = 2^{−2^{i}}. Define f_{i} on I_{i} = [δ_{i}, 2δ_{i}] by f_{i}(x) = δ_{i}^{−2}(x − δ_{i})^{2}(x − 2δ_{i})^{2} and let f_{i} = 0 elsewere. Put f = ∑_{i=1}^{∞} f_{i}. Then f ∈ C^{1}[0, 1]. Let ξ = \frac{3}{2}δ_{i}. Then ξ ∈ 𝒱_{i+1} , and it is easy to see that the derivative of c_{ξ}^{0}ψ_{ξ}^{0} is at least \frac{1}{16} at some point, from which it follows that f can not have a representation in C^{1}(F) as in the proposition.

Theorem 2. Let F ⊂ R be a compact set with a regular sequence of triangulations {𝒯_{i}}_{i=0}^{∞} of F, and let k < α < k + 1, k ≥ 0. Assume that f has the representation in C^{k}(F)

(10)  f = ∑_{i=0}^{∞} ∑_{ξ∈𝒱_{i}} ∑_{s=0}^{k} a_{ξ}^{s}ψ_{ξ}^{s}|_{F} .

Then f belongs to Λ_{α}(F) if and only if the coefficients a_{ξ}^{s} satisfy

(11)  |a_{ξ}^{s}| ≤ cδ_{i}^{α−s},  i ≥ 0, ξ ∈ 𝒱_{i}, 0 ≤ s ≤ k,

and the norm of f in Λ_{α}(F) is equivalent to the infimimum of the possible constants c in (11).

In addition, convergence of (10) in C^{k}(F) follows from (11) assuming pointwise convergence of the sum in (10), only.

Proof. If f ∈ Λ_{α}(F) with the representation (10) in C^{k}(F), then a_{ξ}^{s} = c_{ξ}^{s} by the uniqueness of the representation in C^{k}(F), and (11) follows immediately from the definition of the coefficients c_{ξ}^{s} and the estimate (9), with E in (9) being a bounded operator from Λ_{α}(F) to Λ_{α}(R) as in the Whitney extension theorem. In fact, for ξ ∈ 𝒱_{i}, i > 0, (9) gives |c_{ξ}^{s}| ≤ cδ_{i−1}^{k−s}ω(D^{k}(Ef), δ_{i−1}) which gives (11), using the boundedness of E, the definition of Λ_{α}(R), and T2.

To prove the converse, assume that (11) holds. We first extend the functions ψ_{ξ}^{s} to functions defined on R. If ξ is the endpoint of just one interval Δ ∈ 𝒯_{i}, say the right endpoint, denote by Δ_{1} the interval of length \frac{1}{4}aδ_{i} such that ξ is the left endpoint of the interval, where a is the constant in T4. Define ψ_{ξ}^{s} on Δ_{1} in the same way as it was previously defined on Δ′ when ξ was the left endpoint of Δ′. After this, whether ξ is the endpoint of one or two intervals in 𝒯_{i}, put ψ_{ξ}^{s} = 0 everywhere where ψ_{ξ}^{s} is not already defined. We consider the sum in (10) with these extended functions ψ_{ξ}^{s} in the place of ψ_{ξ}^{s}|_{F} , denoting it again by f, and show that it is in Λ_{α}(R) using the classical definition of Λ_{α}(R) as in Section 4, and that the convergence of the sum is in C^{k}(R). Since the extension does not change the the sum on F we clearly then have the desired result. We remark that this can be used to extend a function f ∈ Λ_{α}(F) to a function in Λ_{α}(R).

By (7) and T1 we have, for ξ ∈ 𝒱_{i}, |D^{ν}ψ_{ξ}^{s}| ≤ cδ_{i}^{s−ν}, and thus
(12)  |a_{ξ}^{s}D^{ν}ψ_{ξ}^{s}| ≤ cδ_{i}^{α−ν}.

For any point x, ψ_{ξ}^{s}(x) is nonzero for at most two ξ ∈ 𝒱_{i}, so for 0 ≤ ν ≤ k we have ∑_{i=0}^{∞} ∑_{ξ∈𝒱_{i}} ∑_{s=0}^{k} |a_{ξ}^{s}D^{ν}ψ_{ξ}^{s}| ≤ c ∑_{i=0}^{∞} δ_{i}^{α−ν} ≤ c ∑_{i=0}^{∞}(c_{4}^{i}δ_{0})^{α−ν} ≤ c, where c_{4} is the constant in T2. This shows that f ∈ C^{k}(R) and that convergence to f is in C^{k}(R).

Consider next

D^{k}f(x) − D^{k}f(y) = (∑_{i=0}^{i_{0}−1} + ∑_{i=i_{0}}^{∞}) ∑_{ξ∈𝒱_{i}} ∑_{s=0}^{k} a_{ξ}^{s}(D^{k}ψ_{ξ}^{s}(x) − D^{k}ψ_{ξ}^{s}(y)) = I+II,

and choose i_{0} so that δ_{i_{0}+1} < |x−y| ≤ δ_{i_{0}} (assume |x−y| ≤ δ_{0} , and interpret I as vanishing if i_{0} = 0). The sum taken over i ≥ i_{0} is estimated in an analoguous way as in the estimate above, using |D^{k}ψ_{ξ}^{s}(x) − D^{k}ψ_{ξ}^{s}(y)| ≤ |D^{k}ψ_{ξ}^{s}(x)| + |D^{k}ψ_{ξ}^{s}(y)|, and gives the estimate

\begin{aligned}II & ≤ ∑_{i=i_{0}}^{∞} cδ_{i}^{α−k} ≤ ∑_{i=i_{0}}^{∞}(c_{4}^{i−i_{0}}δ_{i_{0}})^{α−k} \\ & = cδ_{i_{0}}^{α−k} ≤ cδ_{i_{0}+1}^{α−k} ≤ c|x − y|^{α−k}. \end{aligned}

Let next i < i_{0} and let x ∈ Δ ∈ 𝒯_{i} and y ∈ Δ′ ∈ 𝒯_{i}; the case when x or y is in an interval of the type Δ_{1} is treated in the same way. If Δ = Δ′ then, by the mean value theorem, (7), and T1, for ξ ∈ 𝒱_{i},

(13)  |D^{k}ψ_{ξ}^{s}(x) − D^{k}ψ_{ξ}^{s}(y)| ≤ cδ_{i}^{s−k−1}|x − y|.

If Δ ≠ Δ′ and Δ and Δ′ intersect at the point η ∈ 𝒱_{i}, we get the same estimate by inserting ±D^{k}ψ_{ξ}^{s}(η). If Δ ≠ Δ′ and Δ and Δ′ do not intersect, then |x−y| ≥ aδ_{i}, so by (7) we have |D^{k}ψ_{ξ}^{s}(x) − D^{k}ψ_{ξ}^{s}(y)| ≤ cδ_{i}^{s−k} = cδ_{i}^{s−k}a ≤ cδ_{i}^{s−k}|x − y|/δ_{i} = cδ_{i}^{s−k−1}|x − y|, so we again have (13). In case one of the points x or y, say y, is not in some Δ ∈ 𝒯_{i}, or in some Δ_{1} , then we use, if D^{k}ψ_{ξ}^{s}(x) ≠ 0, that D^{k}ψ_{ξ}^{s}(x) − D^{k}ψ_{ξ}^{s}(y) = D^{k}ψ_{ξ}^{s}(x) = D^{k}ψ_{ξ}^{s}(x) − D^{k}ψ_{ξ}^{s}(y∗) where y∗ is the endpoint of the interval Δ, or Δ_{1} , containing x not equal to ξ.

Note that D^{ν}ψ_{ξ}^{s}(x)−D^{ν}ψ_{ξ}^{s}(y) is nonzero for at most four vertices ξ ∈ 𝒱_{i} and

that, by T2, δ_{i} ≥ c_{4}^{−(i_{0}−i)}δ_{i_{0}} if i < i_{0} . Using this, (11), and (13), we get

I ≤ c ∑_{i=0}^{i_{0}−1} ∑_{s=0}^{k} δ_{i}^{s−k−1}δ_{i}^{α−s}|x − y|

≤ c ∑_{i=0}^{i_{0}−1} c_{4}^{−(i_{0}−i)(α−k−1)}δ_{i_{0}}^{α−k−1}|x − y|

≤ c|x − y|^{α−k}.

Thus we have |D^{k}f(x)−D^{k}f(y)| ≤ c|x−y|^{α−k}, |x−y| ≤ δ_{0} , and since the desired norm estimates are implicit in the proof, this concludes the proof of the theorem.

The construction above can be used to give a characterization of Λ_{α}(F) also if α = k, where k is a positive integer, although the representation is not unique and somewhat less straightforward. We suggest a way to find a kind of atomic decomposition which in spirit is similar to the non-integer case.

Given a sequence δ_{i} → 0 satisfying T2, a definition of Λ_{k}(F) equivalent to the one given in Section 4 is given by f ∈ Λ_{k}(F) if and only if there is a sequence {f_{i}}_{i=0}^{∞} of functions in Λ_{γ}(F) where k < γ < k + 1, such that, for i ≥ 0, ∥D^{j}(f − f_{i})∥_{∞,F} ≤ Mδ_{i}^{(k−j)} for 0 ≤ j < k, ∥D^{j}(f_{i+1} − f_{i})∥_{∞,F} ≤ M for j = k, and ∥f_{i}∥_{Λ_{γ}(F)} ≤ Mδ_{i}^{(k−γ)} , with the norm in Λ_{k}(F) equal to the infimum of the possible constants M. We omit the straightforward verification.

Let f ∈ Λ_{k}(F), let 𝒯_{i} be a regular triangulation of F with diameters δ_{i}, and let {f_{i}}_{i=0}^{∞} be a sequence associated to f and {δ_{i}} as in the above characterization of Λ_{k}(F). Define, inductively for i ≥ 0, g_{i} as the spline function which coincides with a polynomial of degree at most 2k+1 on each interval in 𝒯_{i}, and interpolates to f_{i}−S_{i−1}(f_{i−1}) (to f_{i} if i = 0) and all its derivatives of orders less than or equal to k at each point in 𝒰_{i}. Then we will have ∑_{i=0}^{n} g_{i} = S_{n}(f_{n}) on {∪Δ, Δ ∈ 𝒯_{n}}. The function g_{i} has a representation

g_{i} = ∑_{ξ∈𝒰_{i}} ∑_{s=0}^{k} c_{i,ξ}^{s}ϕ_{i,ξ}^{s},

where c_{i,ξ}^{s} = D^{s}f_{i}(ξ) − D^{s}S_{i−1}(f_{i−1})(ξ) for ξ ∈ 𝒰_{i}. Writing f − S_{n}(f_{n}) as f − f_{n} + f_{n} − S_{n}(f_{n}) and estimating with the aid of (9), one obtains that on F the function f has the representation, in C^{k−1}(F),

(14)  f = ∑_{i=0}^{∞} ∑_{ξ∈𝒰_{i}} ∑_{s=0}^{k} c_{i,ξ}^{s}ϕ_{i,ξ}^{s}.

Proposition 3. Let F ⊂ R be a compact set with a regular sequence of triangulations {𝒯_{i}}_{i=0}^{∞} of F, and let k > 0. Then, if f ∈ Λ_{k}(F), in the representation (14) we have

(15)  |c_{i,ξ}^{s}| ≤ cδ_{i}^{k−s}.

Conversely, if f has a representation (14) with some coefficients c_{i,ξ}^{s} satisfying (15), then f ∈ Λ_{k}(F).

The proof is similar to, but more involved than, the proof of Theorem 2.

58 Alf Jonsson

References

[1] Glaeser, G.: Étude de quelques algèbres tayloriennes. - J. Anal. Math. 6, 1958, 1–124.

[2] Jonsson, A., and A. Kamont: Piecewise linear bases and Besov spaces on fractal sets. - Anal. Math. 27, 2001, 77–117.

[3] Jonsson, A., and H. Wallin: Function Spaces on Subsets of R^{n}. - Harwood Acad. Publ., 1984.

[4] Väisälä, J., M. Vuorinen, and H. Wallin: Thick sets and quasisymmetric maps. - Nagoya Math. J. 135, 1994, 121–148.

[5] Whitney, H.: Analytic extensions of differentiable functions defined in closed sets. - Trans. Amer. Math. Soc. 36, 1934, 63–89.

Received 22 January 2002