generalized characterization theorem for the k functional gm zapryanova
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General Mathematics Vol. 20, No. 5 (2012), Special Issue, 157–178
Generalized characterization theorem for theK−functional associated with the algebraic version of
trigonometric Jackson Integrals 1
Teodora D. Zapryanova
Abstract
The purpose of this paper is to present a characterization of a certain PeetreK−functional in Lp[−1, 1] norm, for 1 ≤ p ≤ 1/λ, λ ∈ (0, 1] by means ofmodulus of smoothness. We generalize the earlier characterization of Ivanov [3]and extend the result to a more general setting.
2010 Mathematics Subject Classification: 41A25, 41A36.
Key words and phrases: K−functional, Modulus of smoothness, Jacksonintegral.
1 Introduction
1.1 Notations
Let X be a normed space. For a given ”differential” operator D we set X ∩D−1(X) = {g ∈ X : Dg ∈ X} . Let X be one of the spaces Lp[−1, 1], 1 ≤ p < ∞ orC[−1, 1]. In this case we denote the norm in X by ∥.∥p , 1 ≤ p ≤ ∞, where ∥.∥∞means the uniform norm. Two examples of the operator D are
D1g(x) : =1− x2
φ(x)(φ(x)g′(x))′ , where φ(x) =
(1− x2
)λ, λ ∈ (0, 1],
D2g(x) : =(1− x2
)g′′(x).
We define for every f ∈ X and t > 0 the K-functionals
1Received 06 June, 2012Accepted for publication (in revised form) 05 September, 2012
157
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158 T. D. Zapryanova
(1) K(f, t;X,Y,D1) := inf{∥f − g∥p + t ∥D1g∥p : g ∈ Y
},
(2) K(f, t;X,Y,D2) := inf{∥f − g∥p + t ∥D2g∥p : g ∈ Y
},
where Y is a given subspace of X ∩D−11 (X) or X ∩D−1
2 (X), respectively.The K-functional (1) with X = C[−1, 1], Y = C2[−1, 1] and λ = 1/2 is
equivalent to the approximation error of Jackson type operator Gs,n in uniformnorm (such equivalence was established in [5]), while the K-functional (2) withX = Lp[−1, 1], Y = C2[−1, 1] is well-known and is equivalent to the approximationerror of Bernstain polynomials in the interval [0, 1] (p = ∞ ) and characterizes thebest polynomial approximations (1 ≤ p ≤ ∞).
We recall that the operator Gs,n : C[−1, 1] → Πsn−s is defined by (see [4])
Gs,n(f, x) = π−1
π∫−π
f(cos(arccosx+ v))Ks,n(v)dv,
where
Ks,n(v) = cn,s
(sin(nv/2)
sin(v/2)
)2s
, π−1
π∫−π
Ks,n(v)dv = 1.
Πr denotes the set of all algebraic polynomials of degree not exceeding r (r isnatural number).
Notation Φ(f, t) ∼ Ψ(f, t) means that there is a positive constant γ, independentof f and t, such that γ−1Ψ(f, t) ≤ Φ(f, t) ≤ γΨ(f, t).
By c we denote positive constants, independent of f and t, that may differ ateach occurrence.
For r - natural number we denoteCr[a, b] =
{f : f, f
′, ..., f (r) ∈ C[a, b] (continuous function in [a, b])
}.
1.2 Known results
The idea for the equivalence of the approximation errors of a given sequence ofoperators and the values of proper K-functionals was studied systematically in [1].
Such equivalence was established for the algebraic version of trigonometric Jacksonintegrals Gs,n and K-functionals (1) with X = C[−1, 1], Y = C2[−1, 1], p = ∞ andλ = 1/2 in [5] (see Theorem A).
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Generalized characterization theorem for the K−functional... 159
Theorem A. For s > 2, λ = 1/2 and every f ∈ C[−1, 1] we have
∥f −Gs,nf∥∞ ∼ K(f,1
n2;C[−1, 1], C2, D1).
Using a linear transform of functions in [3] Ivanov compares the K-functional
(3) K(f, t;X,Y,D3) := inf{∥f − g∥p + t
∥∥(ψg′)′∥∥p: g ∈ Y
},
with the already characterized K-functional
(4) K(f, t;X,Y,D4) := inf{∥f − g∥p + t
∥∥ψg′′∥∥p: g ∈ Y
},
where ψ(x) = x(1 − x); X is one of the spaces Lp[0, 1], 1 ≤ p < ∞ or C[0, 1];Yis a given subspace of X ∩ D−1
3 (X) or X ∩ D−14 (X), respectively; D3g := (ψg′)′,
D4g := ψg′′.Ivanov proved in [3, Theorem 1] the following
Theorem B. For every t ∈ (0, 1] and f ∈ L1[0, 1] we have
K(f, t;L1[0, 1], C2, D3) ∼ K(Bf, t;L1[0, 1], C
2, D4) + tE0(f)1,
where (Bf)(x) = f(x) +x∫1/2
(xy2
− 1−x(1−y)2
)f(y)dy
and E0(f)1 denotes the best approximation of f in L1[0, 1] by constant.
1.3 New results
The aim of this paper is to define a modulus that is equivalent to the K-functional (1) for 1 ≤ p ≤ 1/λ. We apply the method presented in [3].
First, let us note that the K-functionals K(f, t;Lp[−1, 1], C2, D1) for λ = 1/2and K(f, t;Lp[−1, 1], C2, D2) are not equivalent. The inequalityK(f, t;Lp[−1, 1], C2, D2) ≤ cK(f, t;Lp[−1, 1], C2, D1) is not true for a fixed c, everyf , every t ∈ (0, 1] and 1 ≤ p ≤ 1/λ because of functions like (with small positive ε)
fε(x) =
arcsinx, x ∈ [−1 + ε, 1− ε];
ax3 + bx+ d,ax3 + bx− d,
x ∈ [1− ε, 1];x ∈ [−1,−1 + ε];
where a, b, d are chosen such that fε ∈ C2.But these K−functionals can become equivalent if in the one of them instead f staysAf for appropriate operator A.
Let f ∈ L1 [−1, 1] . For every −1 < x < 1 we define the value of the operator Aby
(5) (Af)(x) := f(x) +
x∫0
f(y)
[(y − x)
φ′(y)
φ(y)
]′dy,
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160 T. D. Zapryanova
where the derivative is on y, φ(y) = (1− y2)λ, λ ∈ (0, 1].
Using operator (5) we prove
Theorem 1 a) For every t ∈ (0, 1] and f ∈ Lp[−1, 1], λ ∈ (0, 1), 1 ≤ p ≤ 1/λ, we
have
K(f, t;Lp[−1, 1], C2, D1) ∼ K(Af, t;Lp[−1, 1], C2, D2).
b) For every t ∈ (0, 1], f ∈ L1[−1, 1] and λ = 1 we have
K(f, t;L1[−1, 1], C2, D1) ∼ K(Af, t;L1[−1, 1], C2, D2) + tE0(f)1.
We mention that in Theorem 1 a) there is no additional term tE0(f)p in theequivalence relation, while in Theorem B there is. Moreover, the equivalence inTheorem B is valid only for p = 1, while Theorem 1 a) holds for 1 ≤ p ≤ 1/λ, λ ∈(0, 1). Although the operators D1 for λ = 1/2 and D3 are similar we cannot reduceone to another. We can write the operator D1g(x) for λ = 1/2 of the form :
(D1g)(x) = (1− x2)g′′(x)− xg′(x).
On the other hand, the analogue of D3 for the interval [−1, 1] is
D3G(y) = (1− y2)G′′(y)− 2yG′(y),
i. e. D3G differs from D1G by constant multiplier 2 in the term containing G′.
From Theorem 1 and characterizations of some weighted Peetre K-functionalsin terms of weighted moduli established in [2, Ch. 2, Theorem 2.1.1] we get
Corollary 1 a) For f ∈ Lp[−1, 1], t ∈ (0, 1] and λ ∈ (0, 1), 1 ≤ p ≤ 1/λ, withϕ =
√1− x2 we have
K(f, t;Lp[−1, 1], C2, D1) ∼ ω2ϕ(Af,
√t)p,
where ω2ϕ is Ditzian-Totik modulus of smoothness, introduced in [2].
b) For f ∈ L1[−1, 1], t ∈ (0, 1] and λ = 1 with ϕ =√1− x2 we have
K(f, t;L1[−1, 1], C2, D1) ∼ ω2ϕ(Af,
√t)1 + tω1(f, 1)1.
The equivalence in Theorem 1 is no longer true for 1/λ < p <∞ as the followingexample shows. Let F (x) = arcsinx, λ = 1/2. We have E0(F )p ∼ 1 and thusct ≤ K(F, t;Lp[−1, 1], C2, D1) for 2 < p < ∞ (see Lemma 4). On the other handK(AF, t;Lp[−1, 1], C2, D2) = 0 for every p because AF (x) = x, i. e. AF ∈ C2[−1, 1]and D2(AF ) =
(1− x2
)(AF )′′ = 0.
The connection between the K-functionals of f and Af with D1 and D2 as dif-ferential operators respectively, is not so satisfactory when 1/λ < p <∞. We have
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Generalized characterization theorem for the K−functional... 161
Theorem 2 For every t ∈ (0, 1] and f ∈ Lp[−1, 1], λ ∈ (0, 1], 1/λ < p < ∞, wehave
K(f, t;Lp[−1, 1], C2, D1) ≤ c[K(Af, t
1p+(1−λ)
;Lp[−1, 1], C2, D2) + t1p+(1−λ)
E0(f)p
],
K(Af, t;Lp[−1, 1], C2, D2) + tE0(f)p ≤ cK(f, t;Lp[−1, 1], C2, D1).
The proof of Theorem 1 follows the scheme. First we establish in Lemma 2 theequivalence
K(f, t;Lp[−1, 1], Z1, D1) ∼ K(Af, t;Lp[−1, 1], Z2, D2) for 1 ≤ p <∞,
where Z1 and Z2 are suitable subspaces of C2 (see Definition 2). On the otherhand these variations of Y produce K- functionals equal to the K - functionals wecompare in Theorem 1. In Lemma 1 and Lemma 3 b) respectively we prove that
K(f, t;Lp[−1, 1], Z1, D1) = K(f, t;Lp[−1, 1], C2, D1) for 1 ≤ p <∞ and
K(F, t;Lp[−1, 1], Z2, D2) = K(F, t;Lp[−1, 1], C2, D2) for 1 ≤ p ≤ 1/λ, λ ∈ (0, 1).
The last two relations we obtain using Lemma 2 from [3, p.116]. We state thislemma, as we use it several times.
Definition 1 For given Y ⊂ X∩D−1(X) and a positive number γ we define Sγ(Y )as the set of all g ∈ X ∩D−1(X) such that for every ε > 0 there is h ∈ Y such that∥g − h∥ < ε and ∥Dh∥ < γ ∥Dg∥+ ε.
Lemma B Let Y1, Y2 ⊂ X ∩D−1(X) and ρ > 0. Then for a given positive γ thefollowing statements are equivalent:
i) K(f, t;X,Y1, D) ≤ K(f, γt;X,Y2, D) for every f ∈ X, 0 < t.ii) K(f, t;X,Y1, D) ≤ K(f, γt;X,Y2, D) for every f ∈ X, 0 < t ≤ ρ.iii) Y2 ⊂ Sγ(Y1).
In particular, i) with γ = 1 holds when Y2 ⊂ Y1.
Theorems 1 and 2 are proved in Section 3.
2 Properties of the operators
In the next statement we collect some properties of operator A.
Theorem 3 a) A is a linear operator, satisfying ∥Af∥p ≤ c ∥f∥p for every1 ≤ p ≤ ∞.
b) Af = f for every f ∈ Π0.
c) If f , f ′ ∈ ACloc(−1, 1) , then (Af)(0) = f(0), (Af)′(0) = f ′(0) and
(1− x2
)(Af)′′(x) =
1− x2
φ(x)(φ(x)f ′(x))′, −1 < x < 1,
i.e. D2Af = D1f for − 1 < x < 1.
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162 T. D. Zapryanova
Proof. We write (5) as
(Af)(x) = f(x) +
1∫−1
R(x, y)f(y)dy,
where the kernel R : (−1, 1)× (−1, 1) → R is given as follows: R(0, y) = 0; for x ∈(−1, 0) we have R(x, y) = −
[(y − x)φ
′(y)φ(y)
]′y= −φ′(y)
φ(y) − (y−x)(φ′(y)φ(y)
)′if y ∈ (x, 0)
and R(x, y) = 0 if y /∈ (x, 0); for x ∈ (0, 1) we have R(x, y) =[(y − x)φ
′(y)φ(y)
]′y=
φ′(y)φ(y) + (y − x)
(φ′(y)φ(y)
)′if y ∈ (0, x) and R(x, y) = 0 if y /∈ (0, x).
We set φ(x) = (1− x)λ φ1(x), where φ1(x) = (1 + x)λ . Then φ′(x)φ(x) = −λ
1−x+g1(x),
g1(x) :=φ′1(x)
φ1(x). As φ1(x) ≥ c > 0 for x ∈ [0, 1] we have g1(x) ∈ C1[0, 1], |g1(x)| ≤ c
and |g′1(x)| ≤ c.Thus for x > 0 and y ∈ (0, x) we have
R(x, y) =−λ1− y
+ g1(y) + (y − x)
[−λ
(1− y)2+ g′1(y)
]=
−λ(1− y)2
[(1− y) + (y − x)] + g1(y) + (y − x)g′1(y)
=−λ(1− x)
(1− y)2+ g1(y) + (y − x)g′1(y).
Similarly, if we set φ(x) = (1 + x)λ φ2(x), where φ2(x) = (1− x)λ we haveφ′(x)φ(x) = λ
1+x + g2(x), g2(x) :=φ′2(x)
φ2(x)and g2(x) ∈ C1[−1, 0].Thus for x < 0 and
y ∈ (x, 0) we have
R(x, y) = −[λ(1 + x)
(1 + y)2+ g2(y) + (y − x)g′2(y)
].
Then for x ∈ (0, 1) we get
1∫−1
|R(x, y)| dy =
x∫0
|R(x, y)| dy ≤x∫0
λ(1− x)
(1− y)2dy +
x∫0
∣∣g1(y) + (y − x)g′1(y)∣∣ dy.
The last integral is bounded as |g1(y) + (y − x)g′1(y)| is bounded. The first integralon the right side
λ(1− x)
x∫0
dy
(1− y)2= λ− λ(1− x) ≤ λ.
Therefore1∫−1
|R(x, y)| dy ≤ c, for x ∈ (0, 1).
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Generalized characterization theorem for the K−functional... 163
By analogy1∫−1
|R(x, y)| dy ≤ c, for x ∈ (−1, 0).
Hence ∥Af∥∞ ≤ c ∥f∥∞ .The function R(x, y) is bounded in every fixed, closed interval [a, b] ⊂ (−1, 1).
We investigate R(x, y) when y is close to 1 and −1. Let first y be close to 1. Asy ∈ (0, x) we have 0 < y < x < 1 and 1 − y > 1 − x > 0. We notice that as
φ′(x)φ(x) = −λ
1−x + g1(x) and g1(x) ∈ C1[0, 1] we have∣∣∣φ′(x)φ(x)
∣∣∣ ≤ c1−x ,
∣∣∣∣(φ′(x)φ(x)
)′∣∣∣∣ ≤ c(1−x)2
for x ∈ [0, 1). Then
∣∣∣∣(y − x)(φ′(y)φ(y)
)′∣∣∣∣ ≤ c∣∣∣(y − x) 1
(1−y)2
∣∣∣ = c∣∣∣ (y−1)+(1−x)
(1−y)2
∣∣∣ ≤ c1−y +
c(1−x)(1−y)2
< c1−y + c(1−y)
(1−y)2≤ c
1−y and
|R(x, y)| ≤ c
1− yfor y > 0.
By analogy
|R(x, y)| ≤ c
1 + yfor y < 0.
Therefore|R(x, y)| ≤ c
1− y2for y ∈ (−1, 1), x ∈ (−1, 1)
and
1∫−1
|R(x, y)| dx =
−|y|∫−1
|R(x, y)| dx+
1∫|y|
|R(x, y)| dx ≤ c
1− y2(1− |y|) ≤ c.
Hence ∥Af∥1 ≤ c ∥f∥1 . Now the Riesz-Thorin theorem proves a).
Part b) follows from1∫−1
R(x, y)dy = 0. Indeed, for x > 0
1∫−1
R(x, y)dy =
x∫0
[(y − x)
φ′(y)
φ(y)
]′y
dy = xφ′(0)
φ(0)= 0.
By analogy for x < 01∫−1
R(x, y)dy = 0.
Part c) follows from (5) by direct computation.The operator A is invertible and we give an explicit formula for its inverse oper-
ator A−1 . Let every f ∈ L1[−1, 1] and −1 < x < 1 we set
(A−1f)(x) = f(x) +
x∫0
φ′′(y)
x∫y
dt
φ(t)− φ′(y)
φ(y)
f(y)dy.
In the next statement we collect some properties of A−1.
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164 T. D. Zapryanova
Theorem 4a) A−1 is a linear operator,
∥∥A−1f∥∥p≤ c ∥f∥p for every 1 ≤ p <∞.
b) A−1f = f for every f ∈ Π0.
c) A−1Af = AA−1f = f for every f ∈ L1[−1, 1].
d) If f , f ′ ∈ ACloc(−1, 1), then (A−1f)(0) = f(0), (A−1f)′(0) = f ′(0) and
1− x2
φ(x)(φ(x)(A−1f)′(x))′ =
(1− x2
)f ′′(x), −1 < x < 1,
i.e. D1A−1f = D2f for − 1 < x < 1.
Proof. a) Let x > 0. First we estimate φ′′(y)x∫y
dtφ(t) . As 0 < y < x we have∣∣∣∣∣x∫y dt
φ(t)
∣∣∣∣∣ ≤∣∣∣∣∣ 1∫y dt
φ(t)
∣∣∣∣∣ ≤ c1∫y
dt(1−t)λ
≤ c(1− y)1−λ for λ ∈ (0, 1).Let λ ∈ (0, 1). We have
1∫0
∣∣∣∣∣∣x∫0
φ′′(y)
x∫y
dt
φ(t)
f(y)dy
∣∣∣∣∣∣p
dx
≤ c
1∫0
x∫0
(1− y)λ−2(1− y)1−λ |f(y)| dy
p
dx = c
1∫0
x∫0
1
1− y|f(y)| dy
p
dx
=
1
c
∫0
1∫1−x
1
y|f(1− y)| dy
p
dx (y → 1− y) (Hardy inequality)
≤ c
1∫0
(y|f(1− y)|
y
)p
dy = c
1∫0
|f(1− y)|p dy = c ∥f∥pp[0,1] .
Similarly, using Hardy inequality we get
1∫0
∣∣∣∣∣∣x∫0
φ′(y)
φ(y)f(y)dy
∣∣∣∣∣∣p
dx ≤ c
1∫0
x∫0
|f(y)|1− y
dy
p
dx ≤ c ∥f∥pp[0,1] .
Similar estimates hold for x < 0
0∫−1
∣∣∣∣∣∣x∫0
φ′′(y)
x∫y
dt
φ(t)
f(y)dy
∣∣∣∣∣∣p
dx ≤ c ∥f∥pp[−1,0] and
0∫−1
∣∣∣∣∣∣x∫0
φ′(y)
φ(y)f(y)dy
∣∣∣∣∣∣p
dx ≤ c ∥f∥pp[−1,0] .
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Generalized characterization theorem for the K−functional... 165
From these inequalities we get1∫−1
∣∣∣∣∣∣x∫0
φ′′(y)
x∫y
dt
φ(t)− φ′(y)
φ(y)
f(y)dy∣∣∣∣∣∣p
dx
1p
≤ c ∥f∥p[−1,1] for λ ∈ (0, 1).
This proves a) in the case λ ∈ (0, 1).
Let now λ = 1. We have
∣∣∣∣∣x∫y dtφ(t)
∣∣∣∣∣ ≤ c
∣∣∣∣∣x∫y dt1−t
∣∣∣∣∣ = c∣∣∣ln 1−y
1−x
∣∣∣ , |φ′′(y)| ≤ c for y ∈
[−1, 1]. We estimate
1∫0
∣∣∣∣∣∣x∫0
φ′′(y)
x∫y
dt
φ(t)
f(y)dy
∣∣∣∣∣∣p
dx ≤ c
1∫0
∣∣∣∣∣∣x∫0
∣∣∣∣ln 1− y
1− x
∣∣∣∣ |f(y)| dy∣∣∣∣∣∣p
dx
≤ c
1∫0
∣∣∣∣∣∣x∫0
(|ln(1− y)|+ |ln(1− x)|) |f(y)| dy
∣∣∣∣∣∣p
dx
≤ c
1∫0
∣∣∣∣∣∣x∫0
|ln(1− y)| |f(y)| dy
∣∣∣∣∣∣p
dx+
1∫0
∣∣∣∣∣∣x∫0
|ln(1− x)| |f(y)| dy
∣∣∣∣∣∣p
dx
= I1 + I2. Using Hardy inequality as above we get
I1 =
1∫0
∣∣∣∣∣∣x∫0
|ln(1− y)| |f(y)| dy
∣∣∣∣∣∣p
dx ≤ c
1∫0
∣∣∣∣∣∣x∫0
1
1− y|f(y)| dy
∣∣∣∣∣∣p
dx ≤ c ∥f∥pp[0,1] .
I2 =
1∫0
∣∣∣∣∣∣x∫0
|ln(1− x)| |f(y)| dy
∣∣∣∣∣∣p
dx =
1∫0
|ln(1− x)|p∣∣∣∣∣∣x∫0
|f(y)| dy
∣∣∣∣∣∣p
dx
≤1∫0
|ln(1−x)|p dx
∣∣∣∣∣∣1∫0
|f(y)| dy
∣∣∣∣∣∣p
≤ c ∥f∥p1,[0,1]≤c ∥f∥pp[0,1] , as ∥f∥1≤c ∥f∥p , p≥1.
The case x ∈ (−1, 0), y ∈ (x, 0) is similar. This proves a).Part b) follows from
x∫0
φ′′(y)
x∫y
dt
φ(t)− φ′(y)
φ(y)
dy = 0.
Finally, c) and d) can be obtained by direct computation.The action of the operators A and A−1 for λ = 1
2 on the function f(x) = x isgiven bellow:
(A(.))(x) = 12(1 + x) ln(1 + x)− 1
2(1− x) ln(1− x),(A−1(.))(x) = arcsinx.
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166 T. D. Zapryanova
Definition 2 Set
Z1 ={f ∈ C2[−1, 1] : f ′(−1) = 0, f ′(1) = 0
},
Z2 =
f ∈ C2[−1, 1] :
0∫−1
φ′(x)f ′(x)dx = 0,
1∫0
φ′(x)f ′(x)dx = 0
.
Theorem 5 Let λ ∈ (0, 1]. Thena) (Af)′′(x) is continuous at x = −1 and at x = 1 for every f ∈ Z1.
b)0∫−1
φ′(x)(Af)′(x)dx =1∫0
φ′(x)(Af)′(x)dx = 0 for every f ∈ Z1.
c) (A−1f)′′(x) is continuous at x = −1 and at x = 1 for every f ∈ Z2.
d) (A−1f)′(−1) = (A−1f)′(1) = 0 for every function f ∈ Z2.
e) A(Z1) = Z2 and A−1(Z2) = Z1.
Proof. For every function f ∈ Z1 we have that f ′(x) = (x− 1)f ′′(1) + o(1− x)and f ′(x) = (x+ 1)f ′′(−1) + o(1 + x). From Theorem 3 c) we have
(Af)′′(x) = f ′′(x) +φ′(x)
φ(x)f ′(x).
As in the proof of Theorem 3 we use presentations φ′(x)φ(x) = −λ
1−x + g1(x) and φ′(x)φ(x) =
λ1+x + g2(x) . Then
limx→1−0
φ′(x)
φ(x)f ′(x) = lim
x→1−0
(−λ1− x
+ g1(x)
)[(x− 1)f ′′(1) + o(1− x)
]= λf ′′(1),
and
limx→−1+0
φ′(x)
φ(x)f ′(x) = lim
x→−1+0
(λ
1 + x+ g2(x)
)[(x+ 1)f ′′(−1) + o(1 + x)
]= λf ′′(−1)
which together with the above representations gives for x close to 1
(Af)′′(x) = (1 + λ)f ′′(1) + o(1)
and(Af)′′(x) = (1 + λ)f ′′(−1) + o(1)
for x close to −1. This proves a).
We write the derivative of Af as
(Af)′(x) = f
′(x) +
x∫0
φ′(y)
φ(y)f
′(y)dy.
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Generalized characterization theorem for the K−functional... 167
Then we have
1∫0
φ′(x)(Af)′(x)dx
= limε→0+
1−ε∫0
φ′(x)(Af)′(x)dx
= limε→0+
1−ε∫0
φ′(x)f′(x)dx+
1−ε∫0
φ′(x)
x∫0
φ′(y)
φ(y)f
′(y)dydx
.
We consider the last term.
1−ε∫0
φ′(x)
x∫0
φ′(y)
φ(y)f
′(y)dydx =
1−ε∫0
φ′(y)
φ(y)f
′(y)
1−ε∫y
φ′(x)dx
dy
=
1−ε∫0
φ′(y)
φ(y)(φ(1− ε)− φ(y)) f
′(y)dy
= φ(1− ε)
1−ε∫0
φ′(y)
φ(y)f
′(y)dy −
1−ε∫0
φ′(y)f′(y)dy.
Then1−ε∫0
φ′(x)(Af)′(x)dx = φ(1− ε)
1−ε∫0
φ′(y)
φ(y)f
′(y)dy.
For every f ∈ Z1,
∣∣∣∣ 1∫0
φ′(y)φ(y) f
′(y)dy
∣∣∣∣ ≤ c1∫0
∣∣∣f ′(y)
∣∣∣1−y dy ≤ c as f ′(1) = 0 and f ′(y) =
(y − 1)f ′′(1) + o(1− y). We use that limε→0
φ(1− ε) = φ(1) = 0 and hence eventually
limε→0+
1−ε∫0
φ′(x)(Af)′(x)dx =
1∫0
φ′(x)(Af)′(x)dx = 0.
Similar arguments prove the other claim of b).
We have
(6) (A−1f)′(x) = f
′(x)− 1
φ(x)
x∫0
φ′(y)f′(y)dy.
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168 T. D. Zapryanova
(A−1f)′′(x) = f
′′(x) +
φ′(x)
φ2(x)
x∫0
φ′(y)f′(y)dy − φ′(x)
φ(x)f ′(x)
= f′′(x)− φ′(x)
φ(x)(A−1f)
′(x).(7)
For every f ∈ Z2 , x ∈ [0, 1]x∫0
φ′(y)f ′(y)dy +1∫xφ′(y)f ′(y)dy = 0 and we rewrite (6)
as
(A−1f)′(x) = f
′(x) +
1
φ(x)
1∫x
φ′(y)f ′(y)dy.
Using Taylor,s expansion of f′around 1 we get from the above
(A−1f)′(x) = f
′(x) +
1
φ(x)
1∫x
φ′(y)(f′(1) + (y − 1)f
′′(1) + o(1− y))dy.
Now we compute the last integral, which is equal to
f′(1)
1∫x
φ′(y)dy + f′′(1)
1∫x
(y − 1)φ′(y)dy +
1∫x
φ′(y)o(1− y))dy.
As1∫xφ′(y)dy = φ(1)− φ(x) = −φ(x) we have
(A−1f)′(x) = f
′(x)− f
′(1) +
f′′(1)
φ(x)
1∫x
(y − 1)φ′(y)dy +1
φ(x)
1∫x
φ′(y)o(1− y))dy.
Applying the L’Hopital’s rule we obtain
limx→1−0
1∫x(y − 1)φ′(y)dy
(1− x)φ(x)
= limx→1−0
(1− x)φ′(x)
(1− x)φ′(x)− φ(x)
= limx→1−0
(1− x)λ(1− x2
)λ−1(−2x)
(1− x)λ (1− x2)λ−1 (−2x)− (1− x2)λ=
λ
1 + λ
and we can write1∫x(y − 1)φ′(y)dy = λ
1+λ(1− x)φ(x) + o ((1− x)φ(x)) for x close to
1. Similarly1∫xφ′(y)o(1− y)dy = o
(1− x2
)λ+1for x close to 1. Above computations
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Generalized characterization theorem for the K−functional... 169
and Taylor,s expantion of f′around 1 imply
(8) (A−1f)′(x) = − 1
1 + λ(1− x)f
′′(1) + o(1− x).
We use φ′(x)φ(x) = −λ
1−x + g1(x). Equations (7), (8) and φ′(x)φ(x) (1 − x) = −λ + o(1) give
(A−1f)′′(x) = 1
λ+1f′′(1) + o(1) for x close to 1. In a similar way we get
(9) (A−1f)′(x) =
1
1 + λ(1 + x)f
′′(−1) + o(1 + x).
Hence (A−1f)′′(x) = 1
1+λf′′(−1) + o(1) for x close to −1, which proves c).
Part d) follows from (8) and (9).For every f ∈ Z1 from a) we get (Af)
′, Af ∈ AC[−1, 1] and hence Af ∈ C2.
Now using b) we get Af ∈ Z2, i.e., A(Z1) ⊂ Z2. Similarly, from c) and d) weget A−1(Z2) ⊂ Z1. Using Theorem 4 c) we get Z1 = A−1(A(Z1)) ⊂ A−1(Z2) andZ2 = A(A−1(Z2)) ⊂ A(Z1). Hence A
−1(Z2) = Z1 and A(Z1) = Z2.
3 Proofs of the Theorems
Lemma 1 a) For every t > 0 and f ∈ Lp[−1, 1], 1 ≤ p <∞, we have
K(f, t;Lp[−1, 1], C2, D1) = K(f, t;Lp[−1, 1], Z1, D1).
b) For every t > 0 and f ∈ C[−1, 1] we have
K(f, t;C[−1, 1], C2, D1) ∼ K(f, t;C[−1, 1], Z1, D1).
Proof. Let µ ∈ C∞(R) be such that µ(x) = 1 for x ≤ 0, µ(x) = 0 for x ≥ 1and 0 < µ(x) < 1 for 0 < x < 1. For given δ ∈ (0, 12) we set µ−1(x) = µ(1+x
δ )and µ1(x) = µ(1−x
δ ) for every x ∈ [−1, 1]. Thus, supp µ−1(x) = [−1,−1 + δ], supp
µ1(x) = [1− δ, 1] and∥∥∥µ(k)j
∥∥∥∞
= O(δ−k) for j = −1, 1 and k = 1, 2.
Let g ∈ C2[−1, 1]. For x ∈ [−1, 1] set
(10) G(x) = [1− µ−1(x)− µ1(x)]g(x) + µ−1(x)g(−1) + µ1(x)g(1).
Then G ∈ Z1. From G(x)− g(x) = µ−1(x)[g(−1)− g(x)] +µ1(x)[g(1)− g(x)] we get∥G− g∥p ≤ 21/p ∥G− g∥∞ ≤ 21/pω1(g, δ)∞ = O(δ).From (10) and the form of the
operator (D1g) (x) =(1− x2
)g′′(x)− 2λxg
′(x) we obtain
(D1G)(x) = [1− µ−1(x)− µ1(x)](D1g)(x)− 2(1− x2)[µ′−1(x) + µ′1(x)]g′(x)(11)
+(D1µ1)(x)[g(1)− g(x)] + (D1µ−1)(x)[g(−1)− g(x)].
From (11) for 1 ≤ p <∞ we get ∥D1G∥p ≤ ∥D1g∥p +O(δ1/p), which proves part a)in view of Lemma 2 in [3, p.116].
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170 T. D. Zapryanova
For p = ∞ (11) implies
∥D1G∥∞ ≤ ∥D1g∥∞ + c[∣∣g′(−1)
∣∣+ ∣∣g′(1)∣∣] +O(δ) ≤ c ∥D1g∥∞ +O(δ),
because of |(D1g)(−1)| = c |g′(−1)| and |(D1g)(1)| = c |g′(1)| . Applying Lemma 2again in [3, p.116] we prove part b).
Lemma 2 For every t > 0 and f ∈ Lp[−1, 1], 1 ≤ p <∞, we have
K(f, t;Lp[−1, 1], Z1, D1) ∼ K(Af, t;Lp[−1, 1], Z2, D2).
Proof. For a given g ∈ Z1 we set G = Ag ∈ Z2 (see Theorem 5 e)). Then Theorem4 c), a) and d) implies ∥f − g∥p =
∥∥A−1(Af −Ag)∥∥p≤ c ∥Af −G∥p and ∥D1g∥p =∥∥D1A
−1G∥∥p= ∥D2G∥p . Hence,
∥f − g∥p + t ∥D1g∥p ≤ c(∥Af −G∥p + t ∥D2G∥p),
which gives K(f, t;Lp, Z1, D1) ≤ cK(Af, t;Lp, Z2, D2).For a given G ∈ Z2 we set g = A−1G ∈ Z1 (see Theorem 5 e)). Using Theorem
3 a), c) and Theorem 4 c), we get
∥Af −G∥p =∥∥A(f −A−1G)
∥∥p≤ c ∥f − g∥p , ∥D2G∥p = ∥D2Ag∥p = ∥D1g∥p .
Hence,∥Af −G∥p + t ∥D2G∥p ≤ c(∥f − g∥p + t ∥D1g∥p),
which gives K(Af, t;Lp[−1, 1], Z2, D2) ≤ cK(f, t;Lp[−1, 1], Z1, D1).
From Lemmas 1, 2 we obtain
Corollary 2 For every t > 0 and f ∈ Lp[−1, 1], 1 ≤ p <∞, we have
K(f, t;Lp[−1, 1], C2, D1) ∼ K(Af, t;Lp[−1, 1], Z2, D2).
Lemma 3 a) For every t ∈ (0, 1] and F ∈ Lp[−1, 1], 1λ < p <∞, λ ∈ (0, 1) and for
λ = 1, 1 ≤ p <∞ we have
K(F, t;Lp[−1, 1], Z2, D2) ≤ c[K(F, t
1p+(1−λ)
;Lp[−1, 1], C2, D2) + t1p+(1−λ)
E0(F )p
].
b) For every t ∈ (0, 1] and F ∈ Lp[−1, 1], 1 ≤ p ≤ 1λ , λ ∈ (0, 1) we have
K(F, t;Lp[−1, 1], Z2, D2) = K(F, t;Lp[−1, 1], C2, D2).
Proof. Let λ ∈ (0, 1). For δ ∈ (0, 1/2) we set µ(x) = (1− xδ−1)3+ , where (y)+ = yif y ≥ 0 and (y)+ = 0 if y ≤ 0. For g ∈ C2[−1, 1] we set
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Generalized characterization theorem for the K−functional... 171
(12) G(x) = g(x) + αµ(x+ 1) + βµ(1− x)
where
α =δ
3
0∫−1
φ′(y)g′ (y)dy
0∫−1
φ′(y)(1− y+1δ )2+dy
, β = −δ3
1∫0
φ′(y)g′ (y)dy
1∫0
φ′(y)(1− 1−yδ )2+dy
.
Parameters α, β and δ are chosen in such way that G ∈ Z2. From (12) we get∥G− g∥p ≤ cδ1/p(|α|+ |β|) and
G′′(x) = g′′(x) + 6δ−2
[α
(1− x+ 1
δ
)+
+ β
(1− 1− x
δ
)+
].
Hence∥∥(1− x2
)G′′∥∥
p≤∥∥(1− x2
)g′′∥∥p+ cδ−1+1/p(|α|+ |β|), and
K(F, t;Lp, Z2, D2) ≤ ∥F −G∥p + t ∥D2G∥p≤ ∥F − g∥p + t ∥D2g∥p + cδ1/p(1 + tδ−1)(|α|+ |β|).
In order to estimate |α|+ |β| we calculate∣∣∣∣∣∣1∫0
φ′(y)(1− 1− y
δ)2+dy
∣∣∣∣∣∣ =1
δ2
1∫1−δ
φ′(y)(δ − 1 + y)2dy
≥ c
δ2
1∫1−δ
1
(1− y)1−λ(δ − (1− y))2dy
=c
δ2
δ∫0
(δ − t)2
t1−λdt = cδλ,
becauseδ∫0
(δ−t)2
t1−λ dt = δ2+λ1∫0
(1−t)2
t1−λ dt, (t→ δt) .
In a similar way we get ∣∣∣∣∣∣0∫−1
φ′(y)(1− y + 1
δ)2+dy
∣∣∣∣∣∣ ≥ cδλ.
Then
|α| ≤ cδ1−λ
∣∣∣∣∣∣0∫−1
φ′(y)g′(y)dy
∣∣∣∣∣∣ , |β| ≤ cδ1−λ
∣∣∣∣∣∣1∫0
φ′(y)g′(y)dy
∣∣∣∣∣∣ .
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172 T. D. Zapryanova
Using Holder inequality we estimate
|α| ≤ cδ1−λ
∣∣∣∣∣∣0∫−1
φ′(y)g′(y)dy
∣∣∣∣∣∣≤ cδ1−λ
0∫−1
∣∣φ′(y)∣∣q dy
1/q
0∫−1
∣∣g′(y)∣∣p dy
1/p
≤ cδ1−λ
0∫−1
∣∣g′(y)∣∣p dy
1/p
= cδ1−λ∥∥g′∥∥
p[−1,0]for p >
1
λ,1
p+
1
q= 1.
We used that0∫−1
|φ′(y)|q dy ≤ c0∫−1
∣∣∣ 1(1+y)1−λ
∣∣∣q dy ≤ c, as q = pp−1 = 1
1− 1p
< 11−λ for
p > 1λ , i.e. q(1− λ) < 1. Similarly
|β| ≤ cδ1−λ∥∥g′∥∥
p[0,1]for p >
1
λ.
Hence
|α|+ |β| ≤ cδ1−λ∥∥g′∥∥
pfor p >
1
λ.
In order to estimate the norm of g′ we apply the inequality∥∥g′∥∥p≤ c
(∥D2g∥p + E0(g)p
),
which, for instance, follows from [2, p.135, assertion (a)]. Then we get
|α|+ |β| ≤ cδ1−λ(∥D2g∥p + E0(g)p
)≤ cδ1−λ
(∥D2g∥p + ∥F − g∥p + E0(F )p
).
Now we take δ = t/2 . Thus
K(F, t;Lp, Z2, D2)
≤ ∥F − g∥p + t ∥D2g∥p + ct1/p+1−λ(∥D2g∥p + ∥F − g∥p + E0(F )p
)≤ c
(∥F − g∥p + t1/p+1−λ ∥D2g∥p + t1/p+1−λE0(F )p
),
for every g ∈ C2[−1, 1], as 0 < 1p + 1 − λ = 1+ (1p − λ) < 1 and then for t ∈ (0, 1]
we have t1/p+1−λ ≥ t which proves part a) in the case λ ∈ (0, 1), 1λ < p <∞.Let λ = 1. We have as above∣∣∣∣∣∣
1∫0
φ′(y)(1− 1− y
δ)2+dy
∣∣∣∣∣∣ ≥ cδ,
∣∣∣∣∣∣0∫−1
φ′(y)(1− 1 + y
δ)2+dy
∣∣∣∣∣∣ ≥ cδ.
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Generalized characterization theorem for the K−functional... 173
Then as |φ′(y)| ≤ c for y∈ [−1, 1], λ = 1 we have
|α|+ |β| ≤ c
∣∣∣∣∣∣0∫−1
φ′(y)g′(y)dy
∣∣∣∣∣∣+∣∣∣∣∣∣1∫0
φ′(y)g′(y)dy
∣∣∣∣∣∣
≤ c(∥∥g′∥∥
1,[−1,0]+∥∥g′∥∥
1,[0,1]
)≤ c
∥∥g′∥∥1≤ c
∥∥g′∥∥pfor p ≥ 1.
The proof completes as in the case λ ∈ (0, 1).
In order to prove part b) it is sufficient to show (see Lemma 2 in [3, p.116]) thatfor every g ∈ C2[−1, 1] and every ε > 0 there exists G ∈ Z2 such that ∥G− g∥p < ε
and∥∥∥(1− x2)G
′′∥∥∥p<∥∥(1− x2)g′′
∥∥p+ ε. For 1 ≤ p < 1
λ we can define G by (12).We
have
∥G− g∥p ≤ cδ1/p(|α|+ |β|)
≤ cδ1/p+1−λ
∣∣∣∣∣∣0∫−1
φ′(y)g′(y)dy
∣∣∣∣∣∣+∣∣∣∣∣∣1∫0
φ′(y)g′(y)dy
∣∣∣∣∣∣ →
δ→00.
We used that
∣∣∣∣∣ 0∫−1
φ′(y)g′(y)dy
∣∣∣∣∣ ≤ c0∫−1
1(1+y)1−λ |g′(y)| dy ≤ c and∣∣∣∣ 1∫
0
φ′(y)g′(y)dy
∣∣∣∣ ≤ c1∫0
1(1−y)1−λ |g′(y)| dy ≤ c for λ > 0.
∥∥∥(1− x2)G′′∥∥∥p
≤∥∥(1− x2)g′′
∥∥p+ cδ−1+1/p(|α|+ |β|)
≤∥∥(1− x2)g′′
∥∥p+ cδ1/p−λ
∣∣∣∣∣∣0∫−1
φ′(y)g′(y)dy
∣∣∣∣∣∣+∣∣∣∣∣∣1∫0
φ′(y)g′(y)dy
∣∣∣∣∣∣ .
When 1p − λ > 0 the last term tends to zero as δ → 0, which proves part b) in case
1 ≤ p < 1λ .
The case p = 1λ needs special consideration and different definition of G.
Let δ ∈ (0, 12). We set
ψ′′δ (x) =
0 for x ∈ [−1, 0];
x(1−x2)1+λ for x ∈ (0, 1− δ];
1−δ(2δ−δ2)1+λ for x ∈ (1− δ, 1].
By integration we have
ψ′δ(x) =
0 for x ∈ [−1, 0];
12λ(1−x2)λ
− 12λ for x ∈ (0, 1− δ];
12λ(2δ−δ2)λ
− 12λ + 1−δ
(2δ−δ2)1+λ [x− (1− δ)] for x ∈ (1− δ, 1],
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174 T. D. Zapryanova
ψδ(x) =
0 for x ∈ [−1, 0];
12λ
x∫0
dt
(1−t2)λ− 1
2λx for x ∈ (0, 1− δ];
12λ
1−δ∫0
dt
(1−t2)λ− 1
2λ(1− δ)
+(
12λ(2δ−δ2)λ
− 12λ
)[x− (1− δ)]
+ 1−δ2(2δ−δ2)1+λ [x− (1− δ)]2
for x ∈ (1− δ, 1];
ψ′′δ (x), ψ
′δ(x) and ψδ(x) are continuous and increasing functions.
We set now µ(x) = ψδ(x). For g ∈ C2[−1, 1] we set
(13) G(x) = g(x) + αµ(x) + βµ(−x).
Parameters α and β are chosen in such way that G ∈ Z2 :
0 =
1∫0
φ′(x)G′(x)dx =
1∫0
φ′(x)g′(x)dx+ α
1∫0
φ′(x)ψ′δ(x)dx. Hence
α = −
1∫0
φ′(x)g′(x)dx
1∫0
φ′(x)ψ′δ(x)dx
. Similarly β =
0∫−1
φ′(x)g′(x)dx
0∫−1
φ′(x)ψ′δ(−x)dx
=
0∫−1
φ′(x)g′(x)dx
1∫0
φ′(−x)ψ′δ(x)dx
.
From (13) we get
∥G− g∥ 1λ
≤ (|α|+ |β|) ∥ψδ∥ 1λ
and∥∥∥(1− x2)G′′∥∥∥
1λ
≤∥∥(1− x2)g′′
∥∥1λ+ (|α|+ |β|)
∥∥(1− x2)ψ′′δ
∥∥1λ.
In order to estimate the last expressions we use some properties of ψδ given in thefollowing
Assertion 1
Let δ ∈ (0, 12). Then we have
a)
∣∣∣∣ 1∫0
φ′(x)ψ′δ(x)dx
∣∣∣∣ ∼ ln 1δ .
b)∥∥∥(1− x2)ψ
′′δ
∥∥∥1λ
∼(ln 1
δ
)λ.
c) ∥ψδ∥ 1λ∼ 1.
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Generalized characterization theorem for the K−functional... 175
Using Assertion 1 we obtain
∥G− g∥ 1λ
≤ (|α|+ |β|) ∥ψδ∥ 1λ≤ c(|α|+ |β|)
≤ c
∣∣∣∣ 1∫0
φ′(x)g′(x)dx
∣∣∣∣∣∣∣∣ 1∫0
φ′(x)ψ′δ(x)dx
∣∣∣∣ +∣∣∣∣∣ 0∫−1
φ′(x)g′(x)dx
∣∣∣∣∣∣∣∣∣ 1∫0
φ′(−x)ψ′δ(x)dx
∣∣∣∣
≤ c
ln 1δ
∣∣∣∣∣∣0∫−1
φ′(x)g′(x)dx
∣∣∣∣∣∣+∣∣∣∣∣∣1∫0
φ′(x)g′(x)dx
∣∣∣∣∣∣ .
∥∥(1− x2)G′′∥∥1λ
≤∥∥(1− x2)g′′
∥∥1λ+ (|α|+ |β|)
∥∥(1− x2)ψ′′δ
∥∥1λ
≤∥∥(1− x2)g′′
∥∥1λ+
∣∣∣∣ 1∫0
φ′(x)g′(x)dx
∣∣∣∣∣∣∣∣ 1∫0
φ′(x)ψ′δ(x)dx
∣∣∣∣ +∣∣∣∣∣ 0∫−1
φ′(x)g′(x)dx
∣∣∣∣∣∣∣∣∣ 1∫0
φ′(−x)ψ′δ(x)dx
∣∣∣∣
(ln
1
δ
)λ
≤∥∥(1− x2)g′′
∥∥1λ+
(ln 1
δ
)λln 1
δ
∣∣∣∣∣∣1∫0
φ′(x)g′(x)dx
∣∣∣∣∣∣+∣∣∣∣∣∣0∫−1
φ′(x)g′(x)dx
∣∣∣∣∣∣ .
Let g ∈ C2[−1, 1] , ε > 0 is a small number. For a given function g and ε > 0we may choose δ > 0 such that
c(ln 1
δ
)1−λ
∣∣∣∣∣∣0∫−1
φ′(x)g′(x)dx
∣∣∣∣∣∣+∣∣∣∣∣∣1∫0
φ′(x)g′(x)dx
∣∣∣∣∣∣ ≤ ε,
which proves b) in case p = 1λ in view of [3, Lemma 2, p.116]).
Lemma 4 For every t ∈ (0, 1] and f ∈ Lp[−1, 1], 1λ < p < ∞ for λ ∈ (0, 1) and
1 ≤ p <∞ for λ = 1 we have
tE0(f)p ≤ cK(f, t;Lp[−1, 1], C2, D1).
Proof. Let first λ ∈ (0, 1). For every g ∈ C2[−1, 1] we have
|g(x)− g(0)| ≤
∣∣∣∣∣∣x∫0
dt
φ(t)
∣∣∣∣∣∣ ∥∥φg′∥∥∞ , x ∈ (−1, 1).
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176 T. D. Zapryanova
We obtain the last inequality using the Generalized Mean Value Theorem
g(x)− g(0)x∫0
dtφ(t)
=g′(ξ)
1φ(ξ)
= φ(ξ)g′(ξ) for some ξ in (0, x).
∣∣∣∣x∫0
dtφ(t)
∣∣∣∣ ≤ max
{∣∣∣∣−1∫0
dtφ(t)
∣∣∣∣ , ∣∣∣∣ 1∫0
dtφ(t)
∣∣∣∣} ≤ c , for λ ∈ (0, 1).Hence ∥g − g(0)∥p ≤ c ∥φg′∥∞ .
Since φ(1)g′(1) = φ(−1)g′(−1)= 0 using Holder inequality we get for everyx ∈ [−1, 1]
∣∣φ(x)g′(x)∣∣ =
∣∣∣∣∣∣x∫−1
(φ(t)g′(t)
)′dt
∣∣∣∣∣∣=
∣∣∣∣∣∣x∫−1
φ(t)
1− t21− t2
φ(t)
(φ(t)g′(t)
)′dt
∣∣∣∣∣∣ =∣∣∣∣∣∣x∫−1
φ(t)
1− t2(D1g) (t)dt
∣∣∣∣∣∣≤
x∫−1
(1
(1− t2)1−λ
)q
dt
1/q
x∫−1
|(D1g) (t)|p dt
1/p
≤ c ∥D1g∥p for p >1
λand
1
p+
1
q= 1, as q(1− λ) < 1. Thus,
tE0(f)p ≤ t ∥f − g(0)∥p ≤ t ∥f − g∥p + t ∥g − g(0)∥p ≤ c[∥f − g∥p + t ∥D1g∥p
],
which proves Lemma 4 in the case p > 1λ , λ ∈ (0, 1).
Now we consider the case λ = 1. For every g ∈ C2[−1, 1] we have
|g(x)− g(0)| ≤
∣∣∣∣∣∣x∫0
dt
φ(t)
∣∣∣∣∣∣ ∥∥φg′∥∥∞ =
∣∣∣∣∣∣x∫0
dt
1− t2
∣∣∣∣∣∣ ∥∥φg′∥∥∞=
1
2
∣∣∣∣ln ∣∣∣∣1 + x
1− x
∣∣∣∣∣∣∣∣ ∥∥φg′∥∥∞ , x ∈ (−1, 1).
Then ∥g − g(0∥p ≤ 12 ∥φg
′∥∞∥∥∥ln ∣∣∣1+x
1−x
∣∣∣∥∥∥p≤ c ∥φg′∥∞ for 1 ≤ p < ∞. Using that
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Generalized characterization theorem for the K−functional... 177
φ(1)g′(1) = φ(−1)g′(−1)= 0 and Holder inequality we get for every x ∈ [−1, 1]
∣∣φ(x)g′(x)∣∣ =
∣∣∣∣∣∣x∫−1
(φ(t)g′(t)
)′dt
∣∣∣∣∣∣=
∣∣∣∣∣∣x∫−1
φ(t)
1− t21− t2
φ(t)
(φ(t)g′(t)
)′dt
∣∣∣∣∣∣ =∣∣∣∣∣∣x∫−1
φ(t)
1− t2(D1g) (t)dt
∣∣∣∣∣∣=
∣∣∣∣∣∣x∫−1
1. (D1g) (t)dt
∣∣∣∣∣∣ ≤
x∫−1
(1)q dt
1/q
x∫−1
|(D1g) (t)|p dt
1/p
≤ c ∥D1g∥p for p ≥ 1 and1
p+
1
q= 1. Thus,
tE0(f)p ≤ t ∥f − g(0)∥p ≤ t ∥f − g∥p + t ∥g − g(0)∥p ≤ c[∥f − g∥p + t ∥D1g∥p
],
which proves the lemma.
Proof of Theorems 1 and 2. From parts a) and b) of Theorems 3 and 4 we getE0(f)p ∼ E0(Af)p for λ ∈ (0, 1], 1 ≤ p <∞. Using Corollary 2 and Lemma 3 parta) with F = Af we get
K(f, t;Lp[−1, 1], C2, D1) ∼ K(Af, t;Lp[−1, 1], Z2, D2)
≤ c[K(Af, t
1p+1−λ
;Lp[−1, 1], C2, D2) + t1p+1−λ
E0(f)p
],
for1
λ< p <∞ , λ ∈ (0, 1) and for 1 ≤ p <∞ , λ = 1,
which proves the first inequality of Theorem 2.From Corollary 2 and Lemma 3 part b) we obtain
K(f, t;Lp[−1, 1], C2, D1) ∼ K(Af, t;Lp[−1, 1], C2, D2), for 1 ≤ p ≤ 1
λ, λ ∈ (0, 1),
which proves Theorem 1 a).From Corollary 2 and Lemma 4 we obtain for 1
λ < p < ∞, λ ∈ (0, 1) and for1 ≤ p <∞ , λ = 1
K(Af, t;Lp[−1, 1], C2, D2) + tE0(f)p ≤ K(Af, t;Lp[−1, 1], Z2, D2) + tE0(f)p
≤ cK(f, t;Lp[−1, 1], C2, D1),
which proves the second inequality of Theorem 2 and in the case λ = 1, p = 1Theorem 1 b).
Remark 1 Theorem 1 a) is not true for λ = 1 (see Theorem B). To make it trueon the right hand side of the relation we have to add the term tE0(f)1, what is
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178 T. D. Zapryanova
exactly the result in [3] for p = 1. That is not strange, because for λ = 1 after weintegrate by parts the integral conditions (describing the space Z2 ) we obtain theconditions considered by Ivanov in [3, p.120] and for λ = 1 the differential operator(D1g)(x) = (1 − x2)g
′′ − 2xg′(x) what is exactly the analogue of D3 in the inteval
[−1, 1].
References
[1] Z. Ditzian, K. Ivanov, Strong Converse Inequalities, J. Anal. Math. 61 (1993),61-111.
[2] Z. Ditzian, Totik, Moduli of Smoothness, Springer-Verlag, Berlin, 1987.
[3] K. Ivanov, A Characterization Theorem for the K-functional for Kantorovichand Durrmeyer Operators, Approximation Theory: A volume dedicated toBorislav Bojanov, Marin Drinov Academic Publishing House, Sofia, 2004, 112-125.
[4] Jia-Ding Cao, H. Gonska, Approximation by Boolean Sums of Pozitive LinearOperators. Gopengauz-Type Estimates, J. Approx. Theory 57 (1989), 77-89.
[5] T. Zapryanova, Approximation by Operators of Cao-Gonska type Gs,n and G∗s,n.
Direct and Converse Theorem, Math. and Education in Math. 33 (2004),189-194.
Teodora Dimova ZapryanovaUniversity of EconomicsDepartment of Math. SciencesBlvd. Kniaz Boris I ,77, 9002 Varna, Bulgariae-mail:[email protected]