Return to Article Details Lebesgue constants for Cantor sets. Numerical results

Lebesgue constants for Cantor sets.
Numerical results

Alexander Goncharov, Yİğİt Görgülü and Yaman Paksoy
(Date: November 12, 2024; accepted: May 13, 2025; published online: June 30, 2025.)
Abstract.

We analyze numerically the form of Lebesgue functions and the values of Lebesgue constants in polynomial interpolation for three types of Cantor sets.

Key words and phrases:
Lebesgue constants, Cantor-type sets
2005 Mathematics Subject Classification:
65D05, 65D20, 41A05 and 41A44
The research was partially supported by TÜBİTAK (Scientific and Technological Research Council of Turkey), Project 119F023.
Department of Mathematics, Bilkent University, 06800 Ankara, Turkey, e-mail: goncha@fen.bilkent.edu.tr, yigit.gorgulu@ug.bilkent.edu.tr,
yaman.paksoy@bilkent.edu.tr.

1. Introduction

This article is supplementary to [4], where the problem of boundedness of Lebesgue constants for Cantor-type sets was investigated. Here, we consider the three families of Cantor sets with uniform distribution of 2s interpolating nodes in each corresponding set. The graphs of the corresponding Lagrange fundamental polynomials and the Lebesgue functions are presented. Each family of Cantor sets depends on its own parameter. We analyze the dependence of the Lebesgue constants on these parameters.

First and the second families (Kβ) and K(αs) are geometrically symmetric Cantor-type sets, where, during the Cantor procedure, all intervals of the same level have the same length.

Let (s)s=0 be a sequence such that 0=1 and s13s1 for s. The Cantor set associated with (s)s=0 is K=s=0Es, where E0=I0,1=[0,1],Es is a union of 2s closed intervals Ij,s of length s and Es+1 is obtained by replacing each Ij,s,j=1,2,2s, by two subintervals I2j1,s+1 and I2j,s+1. In what follows, we consider the interpolating set consisting of all 2s endpoints of intervals in Es1, see [4] for details.

A set Kβ with 0<β1/3 is associated with s=βs1 for s, so K1/3 is the classical Cantor ternary set.

Suppose we are given 11/3 and a sequence α=(αs)s=2 such that for s:=s1αs=1α2αs the condition 3ss1 is valid for all s2. The corresponding Cantor-type set is denoted by K(αs). We will use the notation Kα for the case of the constant sequence α

The third family of Cantor sets ([2]) consists of quadratic generalized Julia sets. Given sequence γ=(γs)s=1 with 0<γs<1/4, let r0=1 and rs=γsrs12 for s. Define polynomials P2(x)=x(x1),P2s+1=P2s(P2s+rs) and Es={x:P2s+1(x)0} for s. Then Es=j=12sIj,s and K(γ):=s=0Es.

For a fixed s, let Ys1 be the set of all endpoints (xk)k=12s of intervals from the set Es1. These points determine the polynomial ω2s(x)=k=12s(xxk), the fundamental Lagrange polynomial lk,2s(x)=ω2s(x)(xxk)ω2s(xk), the Lebesgue function λ2s(x)=k=12s|lk,2s(x)| and the Lebesgue constant Λ2s(Ys1,K)=supxKλ2s(x).

2. Lebesgue functions on Kβ

By [4, Theorem 4.4], in the case of small Cantor sets (Kα with α2), the choice of Ys1 as the interpolating set provides a bounded subsequence of the Lebesgue constants. However, for “large” Cantor sets such as Kβ, only one fundamental polynomial at a certain point takes sufficiently large values for large s.

We first consider the classical Cantor ternary set K1/3. Table 1 illustrates the absolute values of fundamental Lagrange polynomials lk,2s evaluated at the first node of the next level s for 1s7, 1k2s. By comparison of these values to the graphs of the corresponding Lebesgue functions (Fig. 1), we observe that for each s, there exist a handful of polynomials that dominate rest of them and determine the behaviour of the Lebesgue constants. For s3, the values marked in red are the largest of their levels and they correspond to the value of the polynomial lm,2s, where m2s is such that xm=n=1s(1)n+1n. Explicitly, if s is odd then m=(2s+1)/3, if s is even then m=(2s+2)/3. Moreover, for these s,m, we have the ratios 5|λ2s(s)lm,2s(s)|6, which exhibit the aforementioned similarity of behaviours.

On the other hand, comparing these results with the lower bounds corresponding to |lk,2s(s)| for k=2s11 which were estimated in a more general setting in [4, Lemma 3.1], we see that the latter are quite rough.

For s, the fundamental polynomials lk,2s,k=1,2,..,2s correspond to interpolating nodes Ys1. In Table 2, we evaluate maximal values of these polynomials over points of Ys and denote this by Mk,s. That is, we have Mk,s=maxxYs|lk,2s(x)|.

Comparing these values to the graphs of the Lebesgue functions of corresponding degree, it is evident that the fundamental polynomials that have comparable maxima with respect to the corresponding Lebesgue constant, attain their maximal values in either the first or the last interval of their level. These correspond to the polynomial lm,2s (marked by red) and the ones corresponding to the adjacent nodes of xm. In order to see this explicitly, one can also compare the values from Table 1 and Table 2 directly and notice for the aforementioned polynomials the agreement of values Mk,s and the values at the first node of the next level.

s k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 0,667 0,333
2 0,494 0,741 0,296 0,062
3 0,41 1,107 0,885 0,43 0,203 0,221 0,096 0,016
4 0,363 1,421 1,679 1,231 2,305 \cellcolorpink4,244 3,229 0,968 0,475 1,326 1,439 0,632 0,139 0,113 0,037 0,005
s k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
5 0,335 1,672 2,532 2,387 9,753 23,5 23,51 9,308 72,26 283,5 \cellcolorpink435,2 272,8 183,2 221,7 107,8 20,04
6 0,317 1,863 3,318 3,683 24,7 70,3 83,17 38,99 1441 6755 12407 9315 10812 15741 9218 2065
7 0,306 2,001 3,97 4,91 45,57 144,6 190,8 99,75 9925 52003 1E+05 89595 1E+05 2E+05 2E+05 38918
s k 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
5 9,957 50,63 98,26 76,5 94,49 141,1 85,81 20,37 1,126 2,48 2,126 0,74 0,082 0,054 0,014 0,001
6 3E+05 2E+06 4E+06 4E+06 1E+07 \cellcolorpink2E+07 1E+07 4E+06 2E+06 6E+06 7E+06 3E+06 8E+05 7E+05 2E+05 30040
7 1E+08 8E+08 2E+09 2E+09 8E+09 2E+10 2E+10 5E+09 8E+09 3E+10 3E+10 2E+10 6E+09 6E+09 2E+09 3E+08
s k 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48
6 14989 1E+05 3E+05 4E+05 1E+06 3E+06 3E+06 9E+05 1E+06 5E+06 6E+06 3E+06 1E+06 1E+06 5E+05 73047
7 1E+13 9E+13 3E+14 4E+14 2E+15 6E+15 6E+15 3E+15 2E+16 6E+16 \cellcolorpink9E+16 6E+16 3E+16 4E+16 2E+16 3E+15
s k 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64
6 255,3 1092 1784 1170 868,5 1096 563,8 113,3 1,446 2,717 1,989 0,591 0,041 0,023 0,005 4E-04
7 1E+15 5E+15 1E+16 8E+15 1E+16 1E+16 9E+15 2E+15 1E+14 3E+14 3E+14 1E+14 1E+13 8E+12 2E+12 2E+11
s k 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
7 1E+11 1E+12 4E+12 6E+12 5E+13 1E+14 2E+14 7E+13 1E+15 4E+15 7E+15 4E+15 3E+15 4E+15 2E+15 4E+14
81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96
7 1E+15 7E+15 1E+16 1E+16 2E+16 3E+16 2E+16 5E+15 8E+14 2E+15 2E+15 7E+14 1E+14 9E+13 3E+13 3E+12
97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112
7 4E+07 3E+08 7E+08 7E+08 2E+09 4E+09 3E+09 9E+08 5E+08 1E+09 2E+09 8E+08 2E+08 2E+08 6E+07 9E+06
113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128
7 1479 5671 8305 4885 2617 2963 1368 246,7 1,201 2,03 1,336 0,357 0,018 0,009 0,002 1E-04
Table 1. Values |lk,2s(s)| for 1s7 and k2s.
s k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 1 1
2 1 1,037 1,037 1
3 1 1,274 1 1 1 1,274 1
4 1 1,445 1,679 1,231 2,305 \cellcolorpink4,244 3,229 1 3,229 4,244 2,305 1,231 1,679 1,445 1
s k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
5 1 1,672 2,532 2,387 9,753 23,5 23,51 9,308 72,26 283,5 \cellcolorpink435,2 272,8 183,2 221,7 107,8 20,04
6 1 1,863 3,318 3,683 24,7 70,3 83,17 38,99 1441 6755 12407 9315 10812 15741 9218 2065
7 1 2,001 3,97 4,91 45,57 144,6 190,8 99,75 9925 52003 1E+05 89595 1E+05 2E+05 2E+05 38918
s k 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
5 20,04 107,8 221,7 183,2 272,8 435,2 283,5 72,26 9,308 23,51 23,5 9,753 2,387 2,532 1,672 1
6 3E+05 2E+06 4E+06 4E+06 1E+07 \cellcolorpink2E+07 1E+07 4E+06 2E+06 6E+06 7E+06 3E+06 8E+05 7E+05 2E+05 30040
7 1E+08 8E+08 2E+09 2E+09 8E+09 2E+10 2E+10 5E+09 8E+09 3E+10 3E+10 2E+10 6E+09 6E+09 2E+09 3E+08
Table 2. Values Mk,s for 1s7 and k2s.
s k 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48
6 30040 2E+05 7E+05 8E+05 3E+06 7E+06 6E+06 2E+06 4E+06 1E+07 2E+07 1E+07 4E+06 4E+06 2E+06 3E+05
7 1E+13 9E+13 3E+14 4E+14 2E+15 6E+15 6E+15 3E+15 2E+16 6E+16 \cellcolorpink9E+16 6E+16 3E+16 4E+16 2E+16 3E+15
s k 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64
6 2065 9218 15741 10812 9315 12407 6755 1441 38,99 83,17 70,3 24,7 3,683 3,318 1,863 1
7 1E+15 5E+15 1E+16 8E+15 1E+16 1E+16 9E+15 2E+15 1E+14 3E+14 3E+14 1E+14 1E+13 8E+12 2E+12 2E+11
s k 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
7 2E+11 2E+12 8E+12 1E+13 1E+14 3E+14 3E+14 1E+14 2E+15 9E+15 1E+16 1E+16 8E+15 1E+16 5E+15 1E+15
81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96
7 3E+15 2E+16 4E+16 3E+16 6E+16 9E+16 6E+16 2E+16 3E+15 6E+15 6E+15 2E+15 4E+14 3E+14 9E+13 1E+13
97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112
7 3E+08 2E+09 6E+09 6E+09 2E+10 3E+10 3E+10 8E+09 5E+09 2E+10 2E+10 8E+09 2E+09 2E+09 8E+08 1E+08
113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128
7 38918 2E+05 2E+05 1E+05 89595 1E+05 52003 9925 99,75 190,8 144,6 45,57 4,91 3,97 2,001 1
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Figure 1. β=1/3, λ22 (left), λ23 (right)
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Figure 2. β=1/3, λ24 (left), λ25 (right)
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Figure 3. β=1/3, λ26 (left), λ27 (right)

Fig. 1Fig. 3 give the graphs of λ2s, 2s7 for K1/3. We observe fast growth of Λ2s(Ys1,K1/3).

Numerical results demonstrate the exponential growth of these values, which correspond to in [4, Theorem 3.2]. Fig. 4Fig. 9 contain the graphs of λ2s,2s8 for Kβ with β=1/5 and β=1/10.

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Figure 4. β=1/5, λ22 (left), λ23 (right)
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Figure 5. β=1/5, λ24 (left), λ25 (right)
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Figure 6. β=1/5, λ26 (left), λ27 (right)
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Figure 7. β=1/10, λ24 (left), λ25 (right)
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Figure 8. β=1/10, λ26 (left), λ27 (right)
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Figure 9. β=1/10, λ27 (left), λ28 (right)

From these figures, we observe that the local maximum values λ2s become smaller when β decreases. However, even for small β the figures illustrate a fast growth of Λ2s(Ys1,Kβ). Of course this does not support from [5, Theorem 6.2] (see [4, Section 3] for a more detailed discussion of this contradiction).

3. Lebesgue functions on Kα

In the Cantor process corresponding to Kα, the length of intervals converge to zero exponentially, whilist for Kβ it was geometrically. So in this sense the sets Kα are smaller than Kβ. Based upon the previous results, we have the intuition that smaller sets correspond to smaller Lebesgue constants. The results from the numerical experiments for Kα support this intuition and illustrate numerically in [4, Corollary 4.5]: the sequence (Λ2s(Ys1,Kα))s=1 is bounded if and only if α2.

Fig. 10Fig. 13 contain the graphs of λ2s for K2 with 1=1/3 and 1=1/10. We see that the local maxima of the Lebesgue functions decrease fast to one.

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Figure 10. α=2, 1=1/3, λ22 (left), λ23 (right)
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Figure 11. α=2, 1=1/3, λ24 (left), λ25 (right)
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Figure 12. α=2, 1=1/10, λ22 (left), λ23 (right)
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Figure 13. α=2, 1=1/10, λ24 (left), λ25 (right)

In general, we have a function of two variables Fs(1,α):=Λ2s(Ys1,Kα(1)), where α>1, 11/3 with

(1) 31α11.

By [4, Theorem 4.4], the fixed values α<2 and 1(13)1α1 gives Fs(1,α) as s. Following our intuition, in order to get fast growth of Fs we have to take values of α close to 1 and not very small values of 1. However, by the restriction (1), this is impossible. We guess that for this reason and since our computational abilities are limited to values of s6, the results from Fig. 14Fig. 16 do not show the growth of Fs.

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Figure 14. α=1.5, 1=1/3, λ22 (left), λ23 (right)
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Figure 15. α=1.5, 1=1/3, λ24 (left), λ25 (right)
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Figure 16. α=1.5, 1=1/10, λ26

Let us note that Kα is polar if and only if α2, see, e.g., in [6, Chapter V, §6, Theorem 3] or [1, Chapter IV, Theorem 3]. It gives immediately.

Proposition 1.

The sequence (Λ2s(Ys1,Kα))s=1 is bounded if and only if the set Kα is polar.

In the next section we will show that there are both polar and non-polar sets from the third family with a bounded sequence (Λ2s(Ys1,K(γ)))s=1. Now we present an example of a polar set K(αs) for which the corresponding subsequence of Lebesgue constants is not bounded. It should be mentioned that Privalov constructed in [7] a polar set K (in fact countable) such that for any array of interpolating nodes from K the corresponding sequence of Lebesgue constants is not bounded, so K is outside 𝒞 in notations from [4].

Example 2.

Let αk=22δk and πs:=k=2s(1δk). If πs0 as s with the divergent series s=2πs, then the set K(αs) is polar and Λ2s(Ys1,K(αs)) as s.

Indeed, by [1], the set K(αs) is polar if and only if the series s=2α2α3αs2s diverges. Hence, by the condition, K(αs) is polar. On the other hand, by [4, Lemma 3.1],

|lk(s)|s(1s11+s1)2s1,

where k=2s11 with xk=1s1. Here, s3, so 1(1s)>s1, which implies 1s11+s1>111. Therefore,

Λ2s(Ys1,K(αs))|lk(s)|1α2α3αs(111)2s1=(1πs11)2s1.

For large enough s, by the condition, 1πs>11. Hence, Λ2s(Ys1,K(αs)) as s.

4. Lebesgue functions on K(γ)

Finally, we turn our attention to the family of weakly-equilibrium sets K(γ). Here each set is the intersection of the inverse polynomial images (2rsP2s+1)1([1,1]).

By (3.1) in [3], the set K(γ) is non-polar if and only if

(2) k=12klog1γk<.

[4, Theorem 6.3] gives boundedness of (Λ2s(Ys1,K(γ)))s=1 provided γs1/32 for s and s=1γs<. It is easy to find sequences γ satisfying these conditions with (2), as well as without it.

Here, in Fig. 17Fig. 20, we observe that even for the large values of the parameters (γs=0.24 for all s), the magnitudes of Λ2s(Ys1,K(γ)) are not large for s8.

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Figure 17. γ0.24, λ22 (left), λ23 (right)
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Figure 18. γ0.24, λ25 (left), λ26 (right)
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Figure 19. γ0.24, λ26 (left), λ27 (right)
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Figure 20. γ0.24, λ28

On the other hand, in the limit case, when γs=1/4 for all s, we have K(γ)=[0,1], see [2, Example 1 from Section 4]. In this case, Ys1 consists of zeros of the Chebyshev polynomial T2s,[0,1], corresponding to the interval [0,1] with the logarithmic growth of the Lebesgue constants. The numerical evidence from Fig. 17Fig. 20 is quite consistent with the hypothesis about the logarithmic growth of Λ2s(Ys1,K(γ)) for such values of the parameters. We guess that the condition s=1γs< is more important for the boundedness of (Λ2s(Ys1,K(γ)))s=1 than the second restriction. Indeed, even in the case of arbitrary small γ0, the condition γs=γ0 for all s gives a uniformly perfect set K(γ) ([2, Theorem 3]), which is more close in its nature to Kβ than to Kα. Recall that the sequence (Λ2s(Ys1,Kβ))s=1 is not bounded for any β. For the definition of uniformly perfect sets, see, e.g., [2].

5. Conclusions

Based on numerical experiments as well as theoretical results from [4], we conclude that for Cantor-type sets K, the values Λ2s(Ys1,K)

1. are smaller for smaller sets,

2. may be bounded for sufficiently small K,

3. increase fast to infinity for symmetric uniformly perfect Cantor sets,

4. may be bounded even for non-polar sets, if K is a “good” generalized Julia set constructed by means of a proper sequence of polynomials.

At the same time, the sequence (Λ2s1(Z,K))s=1 is not bounded ([4, Theorems 4.6 and 6.4]) for any uniformly distributed set of nodes Z. What is more, by Theorem 6.4, even in the case of a “good” generalized Julia set, this sequence has a linear or faster growth.

Comparison of Corollary 4.5 and Theorem 4.6 allows us to assume that, at least for small Cantor-type sets, 5. uniform distribution of nodes is preferable.

Moreover, by [4, Theorem 5.1] we know that for α>2 and for any distribution of nodes the Lebesgue Constants on Kα are unbounded, which gives Kα𝒞.

References