Convex polynomial approximation in the uniform norm: conclusion ∗

Convex polynomial approximation in the uniform norm: conclusion ∗ K. Kopotun†, D. Leviatan‡ and I. A. Shevchuk§

Abstract Estimating the degree of approximation in the uniform norm, of a convex function on a finite interval, by convex algebraic polynomials, has received wide attention over the last twenty years. However, while much progress has been made especially in recent years by, among others, the authors of this article, separately and jointly, there have been left some interesting open questions. In this paper we give final answers to all those open problems. We are able to say, for each rth differentiable convex function, whether or not its degree of convex polynomial approximation in the uniform norm may be estimated by a Jackson-type estimate involving the weighted Ditzian-Totik kth modulus of smoothness, and how the constants in this estimate behave. It turns out that for some pairs (k, r) we have such estimate with constants depending only on these parameters; for other pairs the estimate is valid, but only with constants that depend on the function being approximated; while there are pairs for which the Jackson-type estimate is, in general, invalid.

1

Introduction

Let C[a, b] denote the space of continuous functions f on [a, b], equipped with the uniform norm kf k[a,b] := max |f (x)|. x∈[a,b]

For I := [−1, 1], we omit the interval from this notation and write k · k := k · kI . Also, let Pn be the space of all algebraic polynomials of degree ≤ n − 1, and denote by En (f ) := inf kf − pn k, pn ∈Pn

∗

AMS classification: 41A10, 41A25, 41A29. Keywords and phrases: Convex approximation by polynomials, Degree of approximation. † Department of Mathematics, University of Manitoba, Winnipeg, Manitoba, R3T 2N2, Canada ([email protected]). Supported by NSERC of Canada. ‡ School of Mathematical Sciences, Raymond and Beverley Sackler Faculty of Exact Sciences, Tel Aviv University, Tel Aviv 69978, Israel ([email protected]). § Faculty of Mechanics and Mathematics, National Taras Shevchenko University of Kyiv, 01033 Kyiv, Ukraine ([email protected]). Part of this work was done while the third author was visiting University of Manitoba in 2002.

1

the degree of best uniform polynomial approximation of f . Finally, we denote by ∆2 , the set of convex functions on I, and let En(2) (f ) :=

inf

pn ∈Pn ∩∆2

kf − pn k

denote the degree of best uniform convex polynomial approximation of f ∈ ∆2 ∩C[−1, 1]. Throughout this paper we will have parameters k, l, m, r all of which will denote nonnegative integers, √ with k + r > 0. With ϕ(x) := 1 − x2 , we denote by Br , r ≥ 1, the space of all functions 1] ° fr ∈(r)C[−1, ° ° ° < ∞, with locally absolutely continuous (r−1)st derivative in (−1, 1) such that ϕ f ∞ where for g ∈ L [−1, 1], we denote kgk := ess supx∈[−1,1] |g(x)|.

We use the same notation for the L∞ norm on I, as there can be no confusion. The following estimates of the degree of convex polynomial approximation of functions f ∈ Br ∩ ∆2 were proved by Leviatan [13] (r = 1 and 2) and by Kopotun [5] (r = 3 and r > 5): c(r) En(2) (f ) ≤ r kϕr f (r) k, n ≥ r. (1.1) n Moreover, Kopotun [5] showed that, in general, (1.1) is invalid for r = 4. Namely, for every A > 0 and n ≥ 1, there exists a function f = fn,A ∈ B4 ∩ ∆2 , for which En(2) (f ) > Akϕ4 f (4) k. Nevertheless, Leviatan and Shevchuk [12] have recently proved that, for f ∈ B4 ∩ ∆2 , µ ¶ 1 c 4 (4) (2) En (f ) ≤ 4 kϕ f k + 2 kf k , n ≥ 1, n n with an absolute constant c, which implies (1.1) for n ≥ N (f ) instead of n ≥ r. In fact, Leviatan [13] and Kopotun [7] have obtained estimates refining those in (1.1) and involving, respectively, the Ditzian-Totik (D-T) moduli [2], and the weighted D-T moduli of smoothness (see [15]), defined later in this section. In particular, the following result follows from [13], [6], [7], [5], [9], [12], and this paper (the case α = 4). Theorem 1.1 For f ∈ ∆2 and any α > 0, we have ¢ ¢ ¡ ¡ En (f ) = O n−α , n → ∞ ⇐⇒ En(2) (f ) = O n−α ,

n → ∞.

Let

q p ϕδ (x) := (1 − x − δϕ(x)/2) (1 + x − δϕ(x)/2) = (1 − δϕ(x)/2)2 − x2 .

The weighted D-T modulus of smoothness of a function f ∈ C(−1, 1), is defined by ° ° ϕ ωk,r (f, t) := sup °ϕrkh (·)∆khϕ(·) (f, ·)° , 0 0 is that ϕr f ∈ L∞ (−1, 1). Moreover, if r ≥ 1, then ωk,r (f, t) → 0, as t → 0, if r 0 and only if limx→±1 ϕ (x)f (x) = 0. Therefore, we denote Cϕ := C[−1, 1] and, for r ≥ 1,

Crϕ := {f ∈ Cr (−1, 1) ∩ C[−1, 1] | lim ϕr (x)f (r) (x) = 0}.

(1.7)

x→±1

Clearly Crϕ ⊂ Br . If f ∈ Br , then f ∈ Clϕ for all 0 ≤ l < r, and ° ° ϕ ωr−l,l (f (l) , t) ≤ c tr−l °ϕr f (r) ° ,

(1.8)

t > 0.

Note that for f ∈ Crϕ , and any 0 ≤ l ≤ r and k ≥ 1, the following inequalities hold (see [15]). ϕ ϕ ωk+r−l,l (f (l) , t) ≤ c tr−l ωk,r (f (r) , t),

(1.9)

t > 0,

in particular, if l = 0, then ϕ ϕ ωk+r (f, t) ≤ c tr ωk,r (f (r) , t),

(1.10) Finally, for 0 ≤ l < r/2, (1.11)

t > 0.

Crϕ ⊂ Br ⊂ Cl [−1, 1],

and ϕ ϕ ωk+r−l (f (l) , t) ≤ c tr−2l ωk,r (f (r) , t),

(1.12)

t > 0,

0 ≤ l < r/2.

In this paper, we are interested in determining for which values of parameters k and r, the statement if f ∈ ∆2 ∩ Crϕ , then (1.13)

En(2) (f ) ≤

¢ C ϕ ¡ (r) ωk,r f , 1/n , r n

n≥N,

where C = const > 0 and N = const > 0,

is valid, and for which it is invalid. Here and later in this paper, for clarity of exposition, we denote ω0,r (f, t) := kϕr f k. Hence, in the case k = 0, (1.13) becomes: En(2) (f ) ≤

C r (r) kϕ f k , nr

n≥N,

for f ∈ ∆2 ∩ Br , which is the inequality (1.1). It turns out that the validity of the above statement depends not only on our choice of k and r but also on whether or not we allow constants appearing in (1.13) to depend on the function f . For reader’s convenience we describe our results using an array in Fig. 1 below. There, the symbols “−”, “ª”, and “+”, have the following meaning.

4

• The symbol “−” in the position (k, r) means that there is a function f ∈ ∆2 ∩ Crϕ , such that (2) nr En (f ) lim sup ϕ (r) = ∞. , 1/n) n→∞ ωk,r (f In other words, the estimate (1.13) is invalid even if we allow constants C and N to depend on the function f . • The symbol “+” in the position (k, r) means that (1.13) (or (1.1) if k = 0) is valid with C depending only on k and r, and N = k + r. • The symbol “ª” in the position (k, r) means that (1.13) is valid with C depending only on k and r, and N depending on the function f ; and there are no constants C and N independent of f , such that (1.13) holds for every function f ∈ ∆2 ∩ Crϕ .

r ...

.. .

.. .

.. .

.. .

.. .

.. . ···

6

···

5

···

4

···

3

···

2

···

1 0 0

1

2

3

4

5

6

··· k

¢ ϕ ¡ (r) f , 1/n , n ≥ N Fig. 1. Convex approximation: validity of En(2) (f ) ≤ Cn−r ωk,r

These results are obtained in (or can be derived from) the following papers.

5

1986 1994 1992 1995 2003 —– 1981 1991 1992 1992 2002 —–

Positive results: “+” in position (k, r) (2, 0), hence, {(k, r) | k + r ≤ 2} Leviatan [13] (3, 0), hence, {(k, r) | k + r ≤ 3} Kopotun [6] (see also Hu, Leviatan and Yu [4]) {(0, r) | r ≥ 5} Kopotun [5] {(k, r) | r ≥ 5, k ≥ 1} Kopotun [7] Positive results: “ª” in position (k, r) (4, 0), hence, {(k, r) | k + r = 4} Leviatan and Shevchuk [12] (3, 2), hence (2, 3) and (1, 4) This paper (see Corollary 2.2) Negative results: “+” CANNOT be in position (k, r) (4, 0), hence, {(k, 0) | k ≥ 4} Shvedov [16] (3, 1), hence, {(k, r) | 4 − k ≤ r ≤ 1} Mania (see [15]) (0, 4), hence, {(k, r) | 4 − k ≤ r ≤ 4} Kopotun [5] Negative results: “ª” CANNOT be in position (k, r) (5, 0), hence, {(k, 0) | k ≥ 5} Wu and Zhou [17] (4, 1), hence, {(k, r) | 5 − k ≤ r ≤ 1} Nissim and Yushchenko [14] (2, 4), hence, {(k, r) | 6 − k ≤ r ≤ 4} This paper (see Corollary 2.4)

The structure of our paper is as follows. In Section 2 we state our main results and after collecting some auxiliary results in Section 3, we prove the positive results in Section 4, and the negative results in Sections 5. Section 6 is an appendix which is devoted to a short proof of auxiliary inequalities involving Chebyshev polynomials.

2

Main Results

Theorem 2.1 If f ∈ C2ϕ ∩ ∆2 , then ¡ ¢ ϕ En(2) (f ) ≤ c n−2 ω3,2 (f 00 , 1/n) + n−6 kf 00 k[−1/2,1/2] , (2.1)

where c and N are absolute constants. Hence, ϕ (2.2) En(2) (f ) ≤ cn−2 ω3,2 (f 00 , 1/n),

By virtue of (1.9), an immediate consequence is

n ≥ N,

n ≥ N (f ) .

Corollary 2.2 Let 2 ≤ r ≤ 4, 1 ≤ k ≤ 5 − r. If f ∈ Crϕ ∩ ∆2 , then c ϕ (r) (2.3) (f , 1/n), n ≥ N (f ). En(2) (f ) ≤ r ωk,r n On the other hand, we have the following negative result. Theorem 2.3 There is a function f ∈ C4ϕ ∩ ∆2 , such that (2)

(2.4)

lim sup n→∞

n4 En (f ) = ∞. ϕ ω2,4 (f (4) , 1/n)

Corollary 2.4 Let 0 ≤ r ≤ 4 and k ≥ 6 − r. Then there is a function f ∈ C4ϕ ∩ ∆2 , such that (2) nr En (f ) lim sup ϕ (r) = ∞. , 1/n) n→∞ ωk,r (f 6

3

Auxiliary Results

The following notion of the length of an interval J := [a, b] ⊆ I, relative to its position in I, was introduced in [10]: |J| /J/ := ¡ a+b ¢ , ϕ 2 where |J| := b − a is the length of J. The following was proved in [10] (see inequalities (2.20)-(2.22) there):

(3.1) (3.2) (3.3)

If J1 ⊆ J, then /J1 / ≤ /J/. ωk (f, |J|, J) ≤ ωkϕ (f, /J/) . 1 ϕ ωk,r (f, /J/) , ωk (f, |J|, J) ≤ r w (a, b)

where w (a, b) :=

p (1 + a)(1 − b) .

Let xj := cos(jπ/n), 0 ≤ j ≤ n, be the Chebyshev knots, and denote Ij := [xj , xj−1 ], ϕ(x ) 1 ≤ j ≤ n. Then, |Ij | ∼ n j + n12 (see (5.6) for exact constants in this equivalence) and (see [10, ineq. (3.1)]) π 2 ≤ /Ij / ≤ , n n for all 1 ≤ j ≤ n. Also, for 1 < j < n we have |Ij | ≤ cw (xj , xj−1 )n−1 . Therefore, for 1 < j < n and 0 ≤ l ≤ r, (3.4)

|Ij |l ωk+r−l (f (l) , |Ij |, Ij ) ≤ c|Ij |r ωk (f (r) , |Ij |, Ij ) ¡ (r) −1 ¢ ¡ (r) ¢ |Ij |r ϕ ϕ ≤c r . f ,n f , /Ij / ≤ cn−r ωk,r ωk,r w (xj , xj−1 )

In the cases j = 1 and j = n, we cannot use the same sequence of estimates since w (xj , xj−1 ) = 0. However, we have ωk (f, t2 ) ≤ 2ωkϕ (f, t),

t > 0,

k ≥ 1,

([10, (2.25)] or [2, Corollary 3.1.3]). Hence, since |I1 |, |In | ∼ n−2 , we conclude by (1.12) that, for all 0 ≤ l < r/2 and j = 1 or n, ¡ (r) ¢ ϕ ϕ (3.5) f , 1/n . |Ij |l ωk+r−l (f (l) , |Ij |, Ij ) ≤ cn−2l ωk+r−l (f (l) , 1/n) ≤ cn−r ωk,r

Let Lm−1 (g; z0 , z1 , . . . , zm−1 ) denote the polynomial of degree ≤ m − 1 which interpolates a function g at the points z0 , z1 , . . . , zm−1 . We remind the reader that [z0 , . . . , zm ; g] stands for the m-th divided difference of a function g at the knots z0 , . . . , zm defined by (3.6)

[z0 , z1 , . . . , zm ; g] :=

g(zm ) − Lm−1 (g; z0 , z1 , . . . , zm−1 )(zm ) . (zm − z0 )(zm − z1 ) . . . (zm − zm−1 ) 7

The following Newton formula for interpolating polynomials is well known: (3.7)

Lm−1 (g; z0 , z2 , . . . , zm−1 )(x) =

m−1 X i=0

(x − z0 ) · · · (x − zi−1 )[z0 , . . . , zi ; g] .

Also, assuming that z0 , z1 , . . . , zm form either a non-increasing or a non-decreasing sequence such that min0≤i≤m−1 |zi+1 − zi | ∼ max0≤i≤m−1 |zi+1 − zi |, and using Whitney’s inequality we have the following estimate: (3.8)|[z0 , z1 , . . . , zm ; g]| ≤ C|zm − z0 |−m ωm (g, |zm − z0 |, [min{z0 , zm }, max{z0 , zm }]) ,

where C depends on m and the ratio min0≤i≤m−1 |zi+1 − zi |/ max0≤i≤m−1 |zi+1 − zi |.

Lemma 3.1 Let f ∈ Crϕ , n ≥ k + 1, and let a polynomial pk+r of degree ≤ k + (i) (r) r − 1 be such that pk+r (x1 ) = f (i) (x1 ) for all i = 0, 1, . . . , r − 1, and pk+r (x) = Lk−1 (f (r) ; x1 , x2 , . . . , xk )(x). Then, c ϕ (r) (3.9) kf − pk+r kI1 ≤ r ωk,r (f , 1/n) . n We remark that a similar estimate holds for the interval In and a polynomial pk+r which is analogously defined on In . Proof. First, we denote Lk−1 := Lk−1 (f (r) ; x1 , x2 , . . . , xk ) and note that it follows by Whitney’s theorem and (3.3), that for any β ∈ [x1 , 1) and the interval J := [xk , β] we have kLk−1 − f (r) kJ ≤ c ωk (f (r) , |J|; J) c c ϕ ϕ (r) ω (f , /J/) ≤ ωk,r (f (r) , 1/n) ≤ k,r r/2 r/2 (1 − β) (1 − β) and, in particular, c (3.10) |Lk−1 (β) − f (r) (β)| ≤ ω ϕ (f (r) , 1/n) . (1 − β)r/2 k,r Now, since for any g ∈ Cr [a, b] and x ∈ [a, b], g(x) =

r−1 (i) X g (a) i=0

i!

1 (x − a) + (r − 1)! i

Z

a

x

(x − u)r−1 g (r) (u) du ,

using (3.10), we conclude that, for any x1 ≤ x < 1, the following holds: ¯Z x ³ ´ ¯¯ ¯ 1 (r) r−1 (r) ¯ (x − u) f (u) − pk+r (u) du¯¯ |f (x) − pk+r (x)| = (r − 1)! ¯ x1 Z x ¯ ¯ ≤ c (x − u)r−1 ¯f (r) (u) − Lk−1 (u)¯ du x1 Z x (x − u)r−1 ϕ (r) ≤ cωk,r (f , 1/n) du r/2 x1 (1 − u) Z 1 ϕ (r) ≤ cωk,r (f , 1/n) (1 − u)r/2−1 du x1 ϕ ωk,r (f (r) , 1/n) (r)

r/2

≤ c(1 − x1 )

ϕ ≤ cn−r ωk,r (f

, 1/n) .

8

Note that since f − pk+r is continuous, the above inequality is also valid for x = 1. The proof of the lemma is now complete. ¤ Lemma 3.2 Let α, β ∈ R be such that α + β > 1, and let 1 ≤ ν ≤ b n2 c. Then, n

b2c X

(3.11)

i=ν

n

|Ii |−α ϕ−β (xi ) ∼ n2α+β

b2c X i=ν

i−α−β ≤ C

n2α+β , ν α+β−1

where C and the constants in the equivalence relation depend only on α and β, and are independent of ν and n. Furthermore, n−1 X (3.12) |Ii |−α ϕ−β (xi ) ≤ Cn2α+β . i=1

Proof. Taking into account that, for any 1 ≤ i ≤ b n2 c, ϕ(xi ) |Ii | ∼ n

and

we have n

n

b2c X i=ν

q i ϕ(xi ) = 1 − x2i = sin(iπ/n) ∼ , n

|Ii |−α ϕ−β (xi ) ∼

b2c X i=ν

n

nα ϕ−α−β (xi ) ∼ n2α+β

b2c X i=ν

i−α−β ≤ C

n2α+β . ν α+β−1

Inequality (3.12) immediately follows from (3.11) with ν = 1 taking into account that ϕ(xi ) = ϕ(xn−i ) and |In−i | = |Ii+1 | ∼ |Ii |. ¤ Lemma 3.3 Let k ≥ 2, r ∈ N, 1 ≤ m ≤ k − 1, n ≥ 3m, and let j ∈ N be such that [xj+m , xj ] ⊂ [−1/2, 1/2] (which is equivalent to n/3 ≤ j ≤ 2n/3 − m). Then, for any f ∈ C(−1, 1), ¢ ¡ ϕ (3.13) |[xj , xj+1 , . . . , xj+m ; f ]| ≤ c nk ωk,r (f, 1/n) + kf k[−1/2,1/2] , where c depends only on k and r.

Note that Lemma 3.3 is valid if m = 0 and m = k as well. However, since (3.13) becomes quite weak in these cases, they are excluded from the statement. Proof. First, recall the Marchaud inequality for a function f ∈ C[a, b] and 1 ≤ m < k: ½Z b−a ¾ ωk (f, s, [a, b]) ds m −m ωm (f, t, [a, b]) ≤ ct + (b − a) kf k[a,b] . sm+1 t

9

Now, taking [a, b] = [−1/2, 1/2] in the above estimate, using (3.8), and taking into account that |xj − xj+m | ∼ 1/n for all j such that xj ∈ [−1/2, 1/2], we have |[xj , xj+1 , . . . , xj+m ; f ]| ≤ c(xj − xj+m )−m ωm (f, xj − xj+m , [xj+m , xj ]) ≤ cnm ωm (f, 1/n, [−1/2, 1/2]) Z 1 ωk (f, s, [−1/2, 1/2]) ds ≤ c + ckf k[−1/2,1/2] sm+1 1/n Z 1 ωk (f, s, [−1/2, 1/2]) k−1−m ·s ds + ckf k[−1/2,1/2] = c sk 1/n Z 1 k sk−1−m ds + ckf k[−1/2,1/2] ≤ cn ωk (f, 1/n, [−1/2, 1/2]) 1/n

k

≤ cn ωk (f, 1/n, [−1/2, 1/2]) + ckf k[−1/2,1/2] ϕ ≤ cnk ωk,r (f, 1/n) + ckf k[−1/2,1/2] ,

where the last inequality follows from (1.5).

¤

The following lemma generalizes Lemma 3.3 to the cases when xj can be close to the endpoints of [−1, 1]. Note that the condition r > k − 2m in its statement is essential. ϕ In fact, this is the main reason why (2.2) is no longer valid with ωk,2 , k ≥ 4, instead of ϕ ω3,2 . Lemma 3.4 Let k ≥ 2, n ≥ max{3k, 6}, 1 ≤ m ≤ k − 1, and let r ∈ N0 , r > k − 2m. Then, for f ∈ C(−1, 1) and for every 1 ≤ j ≤ n − m − 1, (3.14)

|[xj , xj+1 , . . . , xj+m ; f ]| µ ¶r+2m−k n ϕ k ≤ cn ωk,r (f, 1/n) + ckf k[−1/2,1/2] . min{j, n − j}

Proof. It is sufficient to prove this lemma for 1 ≤ j ≤ b n2 c, the other case being symmetric. For convenience, everywhere in the proof below, we write [xj , . . . , xj+µ ] ϕ instead of [xj , . . . , xj+µ ; f ], and w instead of ωk,r (f, 1/n). Also, recall that, for all 1 ≤ i ≤ b n2 c, ϕ(xi ) ∼ i/n and |Ii | ∼ i/n2 . Now, for all 1 ≤ j ≤ n − k − 1, the inequalities (3.8) and (3.3) imply |[xj , xj+1 , . . . , xj+k ]| ≤ c|Ij |−k ωk (f, xj − xj+k , [xj+k , xj ]) ϕ ≤ c|Ij |−k w −r (xj+k , xj )ωk,r (f, /[xj+k , xj ]/)

≤ c|Ij |−k ϕ−r (xj )w . © ª Therefore, if 1 ≤ j ≤ min n − k − 1, b n2 c , then

µ ¶k+r n w, |[xj , xj+1 , . . . , xj+k ]| ≤ cn j k

10

which is stronger than (3.14) for m = k. We now use induction in m (the case m = k being its base). Suppose that k and r are fixed, that m is such that 1 ≤ m ≤ k − 1 and m > (k − r)/2, and that (3.14) is valid with m replaced by m + 1. We will now show that it has to be valid for m as well which will complete the proof of the lemma. Let 1 ≤ µ ≤ b n2 c be an index such that [xµ+m , xµ ] ⊂ [−1/2, 1/2]. Lemma 3.3 implies that ¢ ¡ |[xµ , . . . , xµ+m ]| ≤ c nk w + kf k[−1/2,1/2] . Also,

−[xj , xj+1 , . . . , xj+m ] = ([xµ , . . . , xµ+m ] − [xj , . . . , xj+m ]) − [xµ , . . . , xµ+m ] µ−1 X (xi+m+1 − xi )[xi , xi+1 , . . . , xi+m+1 ] − [xµ , . . . , xµ+m ] . = i=j

Therefore, n

|[xj , xj+1 , . . . , xj+m ]| ≤ c

b2c X i=j

|Ii | |[xi , xi+1 , . . . , xi+m+1 ]| + cnk w + ckf k[−1/2,1/2] ,

and using our induction hypothesis we have |[xj , xj+1 , . . . , xj+m ]| Ã µ bn c 2 X ≤ c |Ii | nk i=j

n min{i, n − i}

¶r+2(m+1)−k

w + kf k[−1/2,1/2]

!

+cnk w + ckf k[−1/2,1/2] n

≤ cnk w

b2c X i=j

c bn 2

|Ii |

³ n ´r+2(m+1)−k i

+ cnk w + ckf k[−1/2,1/2]

X nr+2m−k + cnk w + ckf k[−1/2,1/2] r+2m−k+1 i i=j µ ¶r+2m−k n + ckf k[−1/2,1/2] , ≤ cnk w j

≤ cnk w

and the proof is now complete.

¤

We need the following special case of Lemma 3.4 for j = 1, k = 3, r = 2, and m = 1 or 2. Corollary 3.5 Let n ≥ 9, m = 1 or m = 2, and f ∈ C(−1, 1). Then, (3.15)

ϕ (f, 1/n) + ckf k[−1/2,1/2] . |[x1 , x2 , . . . , xm+1 ; f ]| ≤ cn2m+2 ω3,2

11

4

Proofs of positive results

Let Σk,n be the collection of all continuous piecewise polynomials of degree k − 1, on the Chebyshev partition {xj }nj=0 = {cos(jπ/n)}nj=0 . The following lemma is a corollary of (more general) Theorem 3 in [11]. Lemma 4.1 For every k ∈ N there are constants c = c(k) and c∗ = c∗ (k), such that if n ∈ N and S ∈ Σk,n ∩ ∆2 , then there is a polynomial Pn ∈ ∆2 of degree ≤ c∗ n, satisfying kS − Pn k ≤ cωkϕ (S, 1/n).

(4.1)

Hence, in order to obtain direct estimates for polynomial approximation, we only need to construct suitable piecewise polynomials S ∈ Σk,n ∩ ∆2 . In order to construct such piecewise polynomials, we use the following result which was proved in [12] (see Corollary 2.4 there). Lemma 4.2 Let k ∈ N, and let f ∈ C2 [a, a + h], h > 0, be convex. Then there exists a convex polynomial P of degree ≤ k + 1 satisfying P (a) = f (a), P (a + h) = f (a + h), P 0 (a) ≥ f 0 (a), and P 0 (a + h) ≤ f 0 (a + h), and such that kf − P k[a,a+h] ≤ c(k)h2 ωk (f 00 , h, [a, a + h]) . Lemma 4.3 If f ∈ C2ϕ ∩ ∆2 , then for n ≥ 9, there is a continuous piecewise quartic polynomial sn ∈ Σ5,n ∩ ∆2 , such that (4.2)

kf − sn k[xn−1 ,x1 ] ≤

c ϕ 00 ω (f , 1/n) n2 3,2

and, for j = 1 and j = n, (4.3)

ϕ kf − sn kIj ≤ cn−2 ω3,2 (f 00 , 1/n) + cn−6 kf 00 k[−1/2,1/2] .

Proof. First, we apply Lemma 4.2 on each interval Ij , j = 2, . . . , n − 1 (i.e., set a := xj and h := |Ij | = xj−1 − xj ), with k = 3, and define sn |Ij := P . On the intervals I1 and In , we define sn as follows sn(i) (x1 +) = f (i) (x1 ) , i = 0, 1 ,

and s00n (x) = f 00 (x1 ) ,

x ∈ I1 ,

and sn(i) (xn−1 −) = f (i) (xn−1 ) , i = 0, 1 ,

and s00n (x) = f 00 (xn−1 ) ,

x ∈ In .

Then, sn is a continuous piecewise quartic polynomial on [−1, 1] which is convex (since s0n is non-decreasing for all x ∈ [−1, 1]) and such that, for every j = 2, . . . , n − 1, kf − sn kIj ≤ c|Ij |2 ω3 (f 00 , |Ij |, Ij ) . Now, we use the estimate (3.4) with l = r = 2 and k = 3 to conclude (4.4)

kf − sn kIj ≤

c ϕ 00 ω (f , 1/n) n2 3,2 12

for j = 2, . . . , n − 1. Hence, (4.2) is proved, and it remains to estimate kf − sn kI1 and kf − sn kIn . We only estimate the former since the latter can be dealt with analogously. First, Lemma 3.1 with k = 3 and r = 2 implies that kf − l1 kI1 ≤

(4.5)

c ϕ 00 ω (f , 1/n) , n2 3,2 (i)

where the polynomial l1 of degree ≤ 4 is such that l1 (x1 ) = f (i) (x1 ), i = 0, 1, and l100 (x) := L2 (f 00 ; x1 , x2 , x3 )(x) = f 00 (x1 ) + (x − x1 )[x1 , x2 ; f 00 ] + (x − x1 )(x − x2 )[x1 , x2 , x3 ; f 00 ] . Now, using Corollary 3.5, we have for every x ∈ I1 |l100 (x) − s00n (x)| = = ≤ ≤

|l100 (x) − f 00 (x1 )| |(x − x1 )[x1 , x2 ; f 00 ] + (x − x1 )(x − x2 )[x1 , x2 , x3 ; f 00 ]| n−2 |[x1 , x2 ; f 00 ]| + n−4 |[x1 , x2 , x3 ; f 00 ]| ¢ ¡ ϕ cn−2 n4 ω3,2 (f 00 , 1/n) + kf 00 k[−1/2,1/2] ¡ ¢ ϕ +cn−4 n6 ω3,2 (f 00 , 1/n) + kf 00 k[−1/2,1/2]

ϕ ≤ cn2 ω3,2 (f 00 , 1/n) + cn−2 kf 00 k[−1/2,1/2] ,

and, therefore, ¯ ¯Z x ¯ ¯ 00 00 ¯ |l1 (x) − sn (x)| = ¯ (x − u) (l1 (x) − sn (x)) du¯¯ x1

≤ cn−4 kl100 − s00n (x)kI1 ϕ ≤ cn−2 ω3,2 (f 00 , 1/n) + cn−6 kf 00 k[−1/2,1/2] .

Combining this with (4.5) and using the triangle inequality we get (4.3).

¤

Proof of Theorem 2.1. By virtue of Lemma 4.1, the inequality ϕ kf − sn k ≤ cn−2 ω3,2 (f 00 , 1/n) + cn−6 kf 00 k[−1/2,1/2] ,

which follows from Lemma 4.3, and the estimate ϕ ω5ϕ (sn , 1/n) ≤ ckf − sn k + cω5ϕ (f, 1/n) ≤ ckf − sn k + cn−2 ω3,2 (f 00 , 1/n)

(see (1.10)), we conclude that there exists a polynomial Pn ∈ ∆2 of degree ≤ cn such that kf − Pn k ≤ kf − sn k + ksn − Pn k ≤ kf − sn k + cω5ϕ (sn , 1/n) ϕ ≤ cn−2 ω3,2 (f 00 , 1/n) + cn−6 kf 00 k[−1/2,1/2] . This completes the proof of the estimate (2.1). 13

In order to prove (2.2) note that (1.6) implies that ϕ n3 ω3,2 (f 00 , 1/n) ≥ C(f ) ,

for all n ∈ N .

Hence, for n ≥ kf 00 k[−1/2,1/2] /C(f ) =: N (f ), 1 00 ϕ kf k[−1/2,1/2] ≤ C(f ) ≤ n3 ω3,2 (f 00 , 1/n) . n Therefore,

ϕ kf − Pn k ≤ cn−2 ω3,2 (f 00 , 1/n) ,

which completes the proof of the theorem.

5

n ≥ N (f ) , ¤

Proofs of negative results

In order to prove Theorem 2.3, let b ∈ (0, 1) and, for x ∈ [−1, 1], set Z x b gb (x) := ln and Gb (x) := (x − u)gb (u) du . 1+x+b −1 Clearly, Gb is in C∞ [−1, 1] (note also that Gb is concave on [−1, 1]). First, we prove Lemma 5.1 The following estimates hold: (4)

ϕ ω2,4 (Gb , t) ≤ c

(5.1) and (5.2)

b ln

3e2 1 |gb (x)| ≤ (1 + x) ln . b 1+x (4)

Proof. First, taking into account that Gb = gb00 (x) = (1 + x + b)−2 we have (4)

ϕ ϕ (gb00 , t) ≤ ckϕ4 gb00 k ≤ c . ω2,4 (Gb , t) = ω2,4

To prove (5.2) we have to check the inequality b ln

3e2 1 1+x+b ln ≤ (1 + x) ln . b b 1+x

Indeed, this inequality holds for x = −1. At the same time, for the derivatives of both sides we have µ ¶ µ ¶ d 1 1+x+b b ln 1/b 3e2 3e 3e d b ln ln = (1 + x) ln , ≤ ln ≤ ln = dx b b 1+x+b 1+x+b 1+x dx 1+x which completes the proof.

¤ 14

Denote by P∗n the set of polynomials pn of degree ≤ n − 1, such that p00n (−1) ≥ 0. Clearly, every polynomial pn from Pn ∩ ∆2 is also in P∗n . In the proof of Lemma 5.3 below we need the following Dzyadyk type inequality (see [15, Lemma 14.1 and (14.9)]) which is a generalization of the classical Dzyadyk inequality [3]. Lemma 5.2 (Dzyadyk type inequality) Let m ∈ N0 , n ∈ N, and y be any point in [−1, 1]. Then, for any ν ∈ N0 , and any polynomial Pn of degree ≤ n − 1, ° ° ° ° P (·) n ° (5.3) |Pn(ν) (y)| ≤ C(m, ν)ρnm−ν (y) ° m ° [| · −y| + ρn (y)] ° , where ρn (x) :=

√

1−x2 n

+

1 . n2

Since the references [3] and [15] may not be readily accessible, for the sake of completeness, we give a short proof of Lemma 5.2 here. Proof. Everywhere in this proof, C denotes constants that may depend only on m and ν. Now, we ¢ recall that xj = cos(jπ/n), 0 ≤ j ≤ n, Ij = [xj , xj−1 ], and let ¡ x˜j := cos j − 21 πn , 1 ≤ j ≤ n, be the zeros of the Chebyshev polynomial Tn (x) := cos(n arccos x) of degree n. For any 1 ≤ j ≤ n and x ∈ Ij , (5.4)

4 |x − x˜j | ≤ |Tn (x)||Ij | ≤ 4|x − x˜j | , 3

(5.5)

|Ij | ≤ 4 min {|xj−1 − x˜j |, ||xj − x˜j |} ,

and

π2 ρn (x) . 2 We note that all constants in (5.4)-(5.6) are exact. Estimates (5.5)-(5.6) are simple trigonometric inequalities and so we omit their proofs. Inequalities (5.4) can also be verified using standard Calculus techniques, and are certainly known (perhaps not with the exact constant). For reader’s convenience, we give a short proof in the last section “Appendix”. The right inequality in (5.4), (5.5), and the observation kTn k = 1 imply that, for any x ∈ [−1, 1] and 1 ≤ j ≤ n, (5.6)

(5.7)

ρn (x) ≤ |Ij | ≤

|Tn (x)|(|Ij | + |x − x˜j |) ≤ 5|x − x˜j | .

Now, let y ∈ [−1, 1] be fixed and denote by µ the index such that y ∈ Iµ , 1 ≤ µ ≤ n. Without loss of generality, we assume that y 6= x˜µ and ° ° ° ° Pn (·) −m ° ° ° [| · −y| + ρn (y)]m ° = ρn (y) . 15

Hence, we need to show that, for every ν ∈ N0 , |Pn(ν) (y)| ≤ Cρ−ν n (y) .

(5.8) Now,

qn (x) :=

y − x˜µ Tn (x) x − x˜µ Tn (y)

is a polynomial of degree n − 1 such that qn (y) = 1, and inequalities (5.4) and (5.7) imply ¯ ¯ ¯ ¯ ¯ y − x˜µ ¯ ¯ Tn (x) ¯ 3|Iµ | 5 15 |Iµ | ¯ ¯≤ ¯ ¯ ·¯ · = · |qn (x)| = ¯ ¯ ¯ Tn (y) x − x˜µ 4 |x − x˜µ | + |Iµ | 4 |x − x˜µ | + |Iµ | for all x ∈ [−1, 1]. Using (5.6) we get

|x − x˜µ | |˜ xµ − y| π 2 |x − x˜µ | π 2 |x − y| + ρn (y) ≤ + +1≤ + +1 ρn (y) ρn (y) ρn (y) 2 |Iµ | 2 |x − x˜µ | + |Iµ | ≤ 6· |Iµ | and, therefore, |qn (x)| ≤

23ρn (y) , |x − y| + ρn (y)

x ∈ [−1, 1] .

Hence, Qn (x) := (qn (x))m is a polynomial of degree m(n − 1), and the following inequalities are satisfied: ° ° ° ° Pn (·) m m ° °=C kPn Qn k ≤ 23 ρn (y) ° [| · −y| + ρn (y)]m °

and

kQn k ≤ C .

We now use the well known Markov-Bernstein inequality kρνn p(ν) n k ≤ Ckpn k which is satisfied for every polynomial pn of degree ≤ n − 1, to conclude that −ν |[Pn (y)Qn (y)](ν) | ≤ Cρ−ν n+m(n−1) (y) ≤ Cρn (y)

and −ν −ν |Q(ν) n (y)| ≤ Cρm(n−1)+1 (y) ≤ Cρn (y)

for ν ∈ N0 . We will now use strong induction in ν to prove (5.8). If ν = 0, (5.8) is obvious. Suppose now that (5.8) is proved for all 0 ≤ l ≤ µ − 1. Using Leibniz Identity for derivatives ν µ ¶ X ν (l) (ν) f (x)g (ν−l) (x) , [f (x)g(x)] = l l=0 16

the fact that Qn (y) = 1, and the induction hypothesis we have |Pn(ν) (y)|

ν−1 µ ¶ ¯ ¯ X ν (ν) = ≤ ¯[Pn (y)Qn (y)] ¯ + |Pn(l) (y)| |Qn(ν−l) (y)| l l=0 µ ¶ ν−1 X ν −ν+l ≤ Cρ−ν ρ−l (y) = Cρ−ν n (y) + C n (y)ρn n (y) , l l=0

|Pn(ν) (y)Qn (y)|

and so (5.8) is verified for l = ν as well. This completes the proof of the lemma. Lemma 5.3 For each b ∈ (0, n−2 ), and every polynomial pn ∈ P∗n , we have kGb − pn k ≥

1 c 1 ln 2 − 4 . 4 n nb n

Proof. Put gb∗ (x) := − ln (n2 (1 + x + b)),

l(x) := gb (x) − gb∗ (x) = ln n2 b,

so that l(x) is a constant. Let Z x Z x 1 ∗ ∗ (x − u)l(u) du = (x + 1)2 ln n2 b . (x − u)gb (u) du and L(x) := Gb (x) := 2 −1 −1 Then we have G∗b (x) + L(x) = Gb (x). Also, for every pn ∈ P∗n , p00n (−1) − L00 (−1) ≥ −l(−1) = ln 1/n2 b.

(5.9)

Straightforward computations yield Z x |gb∗ (u)| du ≤ c/n2 + cn2 (1 + x)2 , −1

−1 ≤ x ≤ 1,

whence |G∗b (x)|

≤ (x + 1)

Z

x

−1

|gb∗ (u)| du ≤ c(1 + x)/n2 + cn2 (1 + x)3 ≤

c (1 + n2 (1 + x))3 . 4 n

Hence, |pn (x) − L(x)| ≤ kpn − Gb k +

|G∗b (x)|

≤ cn

6

µ

1 kpn − Gb k + 4 n

¶

(1/n2 + (1 + x))3 .

We now apply (5.3) with y = −1, k = 2, and m = 3: ° ° ° ° 1 00 00 −2 ° pn (x) − L(x) ° ≤ cn4 (kpn − Gb k + 4 ) . |pn (−1) − L (−1)| ≤ cn ° ° −2 3 (x + 1 + n ) n 17

¤

This combined with (5.9), in turn completes the proof of the lemma.

¤

We are now ready to prove Theorem 2.3 by constructing a counterexample. Proof of Theorem 2.3. Let bn ∈ (0, 1/e), n ≥ 2, be such that bn ln

1 1 = 2, bn n

and set

1 Gb (x), n2 n where c < 1 (which is independent of n) is taken so small that (5.12) and (5.13) below are fulfilled. We summarize the properties of fn as follows from Lemma 5.1. Namely, for every n ≥ 2, fn ∈ C∞ [−1, 1] fn (x) := c

3e2 . 1+x fn (−1) = fn0 (−1) = fn00 (−1) = 0, |fn00 (x)| ≤ (1 + x) ln

(5.10) (5.11) (5.12)

kfn(j) k < 1,

j = 0, 1, 2,

and (5.13)

and kϕ2j−4 fn(j) k < 1,

j = 3, 4,

ϕ ω2,4 (fn(4) , 1/n) ≤ n−2 .

The definition of bn yields ln ln n ≤ ln ln n2 ≤ ln ln 1/bn = ln

1 n2 b

. n

Hence, by virtue of Lemma 5.3, there exists n1 ≥ 2 such that, for all n ≥ n1 and pn ∈ P∗n , (5.14)

kfn − pn k ≥ c∗

ln ln n , n6

for some suitable constant c∗ > 0, where we used the fact that qn := cn2 pn ∈ P∗n . Now, we put D0 := 1 and Dσ :=

1 1 Dσ−1 = 6 ··· 6 , 6 nσ n1 nσ

where nσ is defined by induction as follows. Suppose that n1 , . . . , nσ−1 , σ > 1, have been selected. We write σ−1 X Fσ−1 (x) := Dj−1 fnj (x) j=1

and select nσ > nσ−1 to be so large that the following inequalities are satisfied: (5.15) (5.16)

(6)

max{σ, kFσ−1 k} < Dσ−1 ln ln ln nσ , (9)

kFσ−1 k < Dσ−1 nσ . 18

Note that (5.16) implies 00 Enσ −2 (Fσ−1 )≤

(5.17)

c Dσ−1 nσ (9) kFσ−1 k < c = cDσ . 7 nσ n7σ

Now, let Φσ (x) :=

∞ X

Dj−1 fnj (x),

j=σ

where the uniform convergence of the series as well as its four times term-by-term differentiation for x ∈ (−1, 1), is justified by (5.12). In fact, since (5.18)

∞ X

Dj−1 = Dσ−1

j=σ

∞ X j=σ

¶j−σ ∞ µ X 1 Dσ−1 n6σ 1 < D < 2Dσ−1 , = σ−1 n6σ · · · n6j−1 n6σ n6σ − 1 j=σ

the inequalities (5.12) imply that (5.19)

kΦσ k < 2Dσ−1

and kϕ4 Φ(4) σ k < 2Dσ−1 .

Setting f1 (x) := Φ1 (x), ϕ and using properties of the ω2,4 modulus, we have by (5.13), (5.15) and (5.19):

(5.20)

(4)

ϕ ω2,4 (f1 ,

1 1 1 1 (4) (4) ϕ ϕ ϕ ) ≤ ω2,4 (Fσ−1 , ) + ω2,4 (Dσ−1 fn(4) (Φσ+1 , ) , ) + ω2,4 σ nσ nσ nσ nσ 1 D σ−1 (6) ≤ kFσ−1 k + 2 + cDσ 2 nσ nσ Dσ−1 Dσ−1 (ln ln ln n + c) ≤ 2 ln ln ln nσ ≤ σ n2σ n2σ

for sufficiently large σ. On the other hand, by virtue of (5.11) and (5.17) there exists a polynomial qnσ of degree nσ − 3, such that 00 qnσ (−1) = Fσ−1 (−1) = 0, and 00 kFσ−1 − qnσ k ≤ 2cDσ .

Hence, for Qnσ (x) :=

Z

x

−1

we have

(x − u)qnσ (u) du,

Q00nσ (−1) = qnσ (−1) = 0, and (5.21)

kFσ−1 − Qnσ k ≤ 4cDσ , 19

where we have used the relations 0 Fσ−1 (−1) = Fσ−1 (−1) = 0.

Now, if pnσ ∈ Pn∗σ , then f1 − pnσ = (Fσ−1 − Qnσ ) + Dσ−1 (fnσ − Rnσ ) + Φσ+1 , where we observe that Rnσ := (5.19) and (5.21),

1 (pnσ Dσ−1

− Qnσ ) ∈ Pn∗σ . Therefore, by virtue of (5.14),

(5.22) kf1 − pnσ k ≥ Dσ−1 kfnσ − Rnσ k − kFσ−1 − Qnσ k − kΦσ+1 k c∗ ≥ Dσ−1 6 ln ln nσ − (4c + 2)Dσ = Dσ (c∗ ln ln nσ − (4c + 2)) nσ 1 ≥ Dσ c∗ ln ln nσ , 2 where the last inequality holds for all sufficiently large σ. The inequalities (5.10) and (5.18) imply that |f001 (x)| ≤ 2(1 + x) ln

3e2 , 1+x

f002 (x) := 2(1 + x) ln

3e2 1+x

so that if we put

and f2 (x) :=

Z

x

−1

and if we denote

(x − u)f002 (u)du,

f(x) := f1 (x) + f2 (x) , then f is a convex function on [−1, 1]. Note also that f2 is twice continuously differentiable in [−1, 1], and it is in B6 . We will show that there exists a polynomial Qn ∈ Pn such that Q00n (−1) = 0 and (5.23)

kf2 − Qn k ≤

c . n6

It follows from [8, Theorem 1] that there exists a polynomial Rn such that kf2 − Rn k ≤ cω6ϕ (f2 , 1/n) and

kf002 − Rn00 k ≤ cω4ϕ (f002 , 1/n) .

Therefore, since (see (1.8)) (6)

ω6ϕ (f2 , 1/n) ≤ cn−6 kϕ6 f2 k ≤ cn−6 20

and (see (1.9)) (4)

(4)

ϕ ω4ϕ (f002 , 1/n) ≤ cn−2 ω2,2 (f2 , 1/n) ≤ cn−2 kϕ2 f2 k ≤ cn−2 ,

we conclude that Rn satisfies the inequalities kf2 − Rn k ≤ cn−6 and kf002 − Rn00 k ≤ cn−2 . In particular, since f002 (−1) = 0, the estimate |Rn00 (−1)| ≤ cn−2 holds. Now, it follows from [5, Lemma 8] (or see [7, Lemma 14c]) that there exists a polynomial Mn ∈ Pn such that kMn k ≤ cn−4 and Mn00 (−1) ≥ 2−10 . We now define Qn as follows: Qn (x) := Rn (x) −

Rn00 (−1) Mn (x) . Mn00 (−1)

Clearly, Q00n (−1) = 0, and ¯ 00 ¯ ¯ Rn (−1) ¯ ¯ ¯ kMn k ≤ kf2 − Rn k + cn−2 kMn k ≤ cn−6 , kf2 − Qn k ≤ kf2 − Rn k + ¯ 00 Mn (−1) ¯

which proves (5.23). Now, (1.8) implies that

ϕ ω2,4

µ

(4) f2 ,

1 n

¶

(6)

≤ cn−2 kϕ6 f2 k ≤ cn−2 ,

and hence, using (5.20) and (5.15), we have ϕ ω2,4 (f(4) ,

1 (4) 1 (4) 1 ϕ ϕ ) ≤ ω2,4 (f1 , ) + ω2,4 (f2 , ) nσ nσ nσ 2 (Dσ−1 ln ln ln nσ + c) ≤ n2σ 3Dσ−1 ≤ ln ln ln nσ , n2σ

for sufficiently large σ. On the other hand, (5.22) and (5.23) imply that, if Pnσ ∈ P∗nσ , then kf − Pnσ k ≥ kf1 − (Pnσ − Qnσ )k − kf2 − Qnσ k C c∗ Dσ−1 c∗ Dσ−1 ln ln n − ≥ ln ln nσ ≥ σ 2 n6σ n6σ 4 n6σ for sufficiently large σ, where we used the fact that Pnσ − Qnσ ∈ P∗nσ . (2) Thus, taking into account that En (f ) ≥ inf pn ∈P∗n kf − pn k we have (2)

n4σ Enσ (f) c∗ ln ln nσ ≥ · → ∞ as σ → ∞ . ϕ (4) ω2,4 (f , 1/nσ ) 12 ln ln ln nσ This concludes the proof of Theorem 2.3.

¤ 21

6

Appendix

Lemma 6.1 Let n ∈ N and 1 ≤ j ≤ n. Denote tj (x) :=

Tn (x) |Ij |, x 6= x˜j , x − x˜j

and tj (˜ xj ) := Tn0 (˜ xj )|Ij |,

where Tn (x) := cos n arccos x is the Chebyshev polynomial of degree n, and x˜j := cos(j − 1/2) πn is the zero of Tn , lying in Ij . Then, for every x ∈ Ij , we have 4 < |tj (x)| < 4. 3

(6.1)

Moreover, the constants in (6.1) are exact and cannot be improved. Proof. First, we observe that |tn−j+1 (−x)| = |tj (x)| and {−x | x ∈ In−j+1 } = Ij . Hence, without loss of generality, one can assume that j ≤ b(n + 1)/2c. Note that tj is a polynomial of degree n − 1 having exactly n − 1 real zeros x˜i , 1 ≤ i ≤ n and i 6= j. Therefore, by Rolle’s theorem, t0j has exactly n − 2 distinct zeros. In particular, t0j has a unique zero in [˜ xj+1 , x˜j−1 ] ⊃ Ij if j ≥ 2, and so |tj (x)| ≥ min{|tj (xj )|, |tj (xj−1 )|} for x ∈ Ij and j ≥ 1. Hence, for x ∈ Ij , |tj (x)| ≥ |tj (xj )| =

|Ij | |I1 | 4 cos2 (π/4n) 4 ≥ = > , 2 x˜j − xj x˜1 − x1 4 cos (π/4n) − 1 3

which is the lower estimate in (6.1). To prove the upper estimate, we use (i) sin(αθ) ≤ α sin θ, for all α ≥ 1 and 0 ≤ θ ≤ π/α, and (ii) | sin θ| ≤ |θ|, for all θ ∈ R. Denoting τ := arccos x (and hence x ∈ Ij implies that (j − 1)π/n ≤ τ ≤ jπ/n) and τj := (j − 1/2) πn we have, for x ∈ Ij , ¯ ¯ ¯ sin n(τ − τj )/2 sin n(τ + τj )/2 ¯ n|Ij | n|Ij | ¯ ¯≤ |tj (x)| = |Ij | ¯ ≤ ¯ sin(τ − τj )/2 sin(τ + τj )/2 sin(τ + τj )/2 sin((j − 1) πn + τj )/2 π sin(j − 1/2) πn 2n sin 2n j − 1/2 . ≤π = π sin(j − 3/4) n j − 3/4 Therefore, if j 6= 1, then ktj kIj ≤ 6π/5 < 4. Finally, if j = 1, since t1 is positive and strictly increasing on [˜ x2 , 1], we have 0 < t1 (x) ≤ t1 (1) =

³π´ |I1 | = 4 cos2 < 4, 1 − x˜1 4n

x ∈ I1 .

The (6.1) are now verified. in (6.1) are exact since t1 (1) = ¡ inequalities ¡ ¢ The2constants ¡π¢ ¢ 2 π 2 π 4 cos 4n → 4 and t1 (x1 ) = 4 cos 4n /(4 cos 4n − 1) → 4/3 as n → ∞. ¤

22

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