Trigonometric approximation in generalized Lebesgue ... - Ele-Math

J ournal of Mathematical I nequalities Volume 4, Number 2 (2010), 285–299

TRIGONOMETRIC APPROXIMATION IN GENERALIZED LEBESGUE SPACES L p(x)

A LI G UVEN AND DANIYAL M. I SRAFILOV (Communicated by S. Samko) Abstract. The approximation properties of Nörlund (Nn ) and Riesz (Rn ) means of trigonometric Fourier series are investigated in generalized Lebesgue spaces L p(x) . The deviations f − Nn ( f ) p(x) and f − Rn ( f ) p(x) are estimated by n−α for f ∈ Lip (α , p (x)) (0 < α 1) .

1. Introduction and main results Let p : R → [1, ∞) be a measurable 2π − periodic function, that is p (x + 2π ) = p(x). Denote by L p(x) = L p(x) ([0, 2π ]) the set of all measurable 2π − periodic functions f such that m p (λ f ) < ∞ for some λ = λ ( f ) > 0 , where m p ( f ) :=

2π

| f (x)| p(x) dx.

0

L p(x) becomes a Banach space with respect to the norm f f p(x) = inf λ > 0 : m p 1 . λ If p(x) ≡ p is constant (1 p < ∞) , then the space L p(x) is isometrically isomorphic to the Lebesgue space L p . If the function p satisfies 1 < p− := ess inf p(x), x∈[0,2π ]

then the function p (x) :=

p+ := ess sup p(x) < ∞ x∈[0,2π ]

(1)

p(x) p(x) − 1

is well defined and satisfies (1) itself. Mathematics subject classification (2010): 41A25, 42A10, 46E30. Keywords and phrases: Generalized Lebesgue space, Lipschitz class, modulus of continuity, Nörlund mean, Riesz mean. c , Zagreb Paper JMI-04-25

285

286

A LI G UVEN AND D ANIYAL M. I SRAFILOV

The space L p(x) consists of all measurable 2π − periodic functions f such that 2π

| f (x)g (x)| dx < ∞

0

for all measurable g with m p (g) 1 and ⎧ ⎫ ⎨2π ⎬ f ∗p(x) = sup | f (x)g(x)| dx : m p (g) 1 ⎩ ⎭ 0

is also a norm on L p(x) . It is known that the inequalities f p(x) f ∗p(x) r p f p(x) satisfied for all functions f ∈ L p(x) , where r p := 1 +

1 1 − , p− p+

and hence the norms f p(x) and f ∗p(x) are equivalent. We refer to [13], [9], [10] and [7] for properties above and for more general information about L p(x) spaces. Denote by M the Hardy-Littlewood maximal operator, defined for f ∈ L1 by M( f )(x) = sup I

1 |I|

| f (t)| dt,

x ∈ [0, 2π ],

I

where the supremum is taken over all intervals I with x ∈ I. The boundedness problem of the operator M on the space L p(x) was studied by many authors ([5], [8], [17], [18], etc.). In [8] it was proved that if the function p satisfies (1) and the condition |p(x) − p (y)|

C , − ln |x − y|

1 0 < |x − y| , 2

(2)

then the maximal operator M is bounded on L p(x) , that is, M( f ) p(x) c f p(x)

(3)

for all f ∈ L p(x) , where c is a constant depends only on p. The set of all measurable 2π − periodic functions p : R → [1, ∞) satisfies the conditions (1) and (2) will be denoted by M . Let p ∈ M and f ∈ L p(x) . The modulus of continuity of the function f is defined by Ω p(x) ( f , δ ) = sup Th ( f ) p(x) , |h|δ

δ > 0,

(4)

T RIGONOMETRIC APPROXIMATION IN GENERALIZED L EBESGUE SPACES L p(x)

where 1 Th ( f )(x) := h

h

| f (x + t) − f (x)| dt.

287

(5)

0

The existence of Ω p(x) ( f , δ ) follows from (3), and also the inequality Ω p(x) ( f , δ ) c f p(x) satisfied for all δ > 0. The modulus Ω p(x) ( f , ·) is nonnegative, continuous function such that lim Ω p(x) ( f , δ ) = 0,

Ω p(x) ( f1 + f2 , ·) Ω p(x) ( f1 , ·) + Ω p(x) ( f2 , ·) .

δ →0

In the Lebesgue spaces L p (1 < p < ∞) , the classical modulus of continuity ω p ( f , ·) is defined by

ω p ( f , δ ) = sup Th ( f ) p , δ > 0, (6) |h|δ

where

Th ( f )(x) := f (x + h) − f (x) .

It is known that in the Lebesgue spaces L p the moduli of continuity (4) and (6) are equivalent (see [14]). We define in the spaces L p(x) the modulus of continuity by using the shift (5), because the space L p(x) is not translation invariant, in general (see, for example [13, Example 2.9]). Let p ∈ M and 0 < α 1. We define the Lipschitz class Lip(α , p(x)) as Lip(α , p(x)) = f ∈ L p(x) : Ω p(x) ( f , δ ) = O (δ α ) , δ > 0 . Let f ∈ L1 has the Fourier series f (x) ∼

∞ a0 + ∑ (ak cos kx + bk sin kx) . 2 k=1

(7)

Denote by Sn ( f )(x), n = 0, 1, ... the n th partial sums of the series (7) at the point x, that is, Sn ( f )(x) =

n

∑ Ak ( f )(x),

k=0

where A0 ( f )(x) =

a0 , 2

Ak ( f ) (x) = ak cos kx + bk sin kx,

k = 1, 2, ....

Let {pn }∞ 0 be a sequence of positive real numbers. We consider two means of the series (7) defined by 1 n Nn ( f )(x) = ∑ pn−m Sm ( f )(x) Pn m=0

288


and Rn ( f )(x) =

1 n ∑ pm Sm ( f )(x), Pn m=0

n

where Pn := ∑ pm , p−1 = P−1 := 0. The means Nn ( f ) and Rn ( f ) are called the m=0

Nörlund and Riesz means of the series (7), respectively. In the case pn = 1, n 0, both of Nn ( f ) and Rn ( f ) are equal to the Cesàro mean

σn ( f )(x) = β −1

If we take pn = An

β

1 n ∑ Sm ( f )(x). n + 1 m=0

(β > 0) , where

A0 := 1,

β

Ak :=

β (β + 1)... (β + k) , k!

k 1,

β

the mean Nn ( f ) be the generalized Cesàro mean σn ( f )(x), that is Nn ( f )(x) =

1

n

β −1

∑ An−mSm ( f )(x).

β An m=0

The approximation properties of the Cesàro means σn in Lipschitz classes Lip(α , p), 1 p < ∞, 0 < α 1 were investigated by Quade in [19]. The generalizations of Quade’s results were studied by Mohapatra and Russell [16], Chandra ([1], [2], [3], [4]) and Leindler [15]. In [1], Chandra obtained estimates for f − Nn ( f ) p , where 1 < p < ∞. Chandra also gave estimates for the difference f − Rn ( f ) p , where f ∈ Lip (α , p) , 1 < p < ∞, 0 < α 1 [2]. In the paper [4], Chandra gave some conditions on the sequence {pn }∞ 0 and obtained very satisfactory results about approximation by the means Nn ( f ) and Rn ( f ) in Lip (α , p) , 1 p < ∞, 0 < α 1. Later, Leindler in [15] weakened the conditions given by Chandra on the sequence {pn }∞ 0 and generalized his results. In [11], the analogues of Chandra’s results was obtained for weighted Lebesgue spaces. In the present paper we give L p(x) analogues of the results obtained by Leindler in [15] and Chandra in [4]. We shall use the notations Δgn := gn − gn+1,

Δm g (n, m) := g (n, m) − g (n, m + 1).

A sequence of positive real numbers {pn }∞ 0 is called almost monotone decreasing (increasing) if there exists a constant c, depending only on the sequence {pn }∞ 0 such that for all n m the inequality pn c pm

(c pn pm )

∞ holds. Such sequences will be denoted by {pn }∞ 0 ∈ AMDS ({pn }0 ∈ AMIS). Our main results are the following.


289

T HEOREM 1. Let p ∈ M , 0 < α < 1, f ∈ Lip(α , p(x)) and {pn }∞ 0 be a sequence of positive numbers. If {pn }∞ 0 ∈ AMDS, or

{pn }∞ 0 ∈ AMIS

then

and

(n + 1) pn = O (Pn ) ,

f − Nn ( f ) p(x) = O n−α .

T HEOREM 2. Let p ∈ M , f ∈ Lip (1, p(x)) and {pn }∞ 0 be a sequence of positive numbers. If n−1

∑ k |Δpk | = O (Pn ) ,

k=1

or

then the estimate

n−1

Pn ∑ |Δpk | = O n k=0

,

f − Nn ( f ) p(x) = O n−1

holds for n = 1, 2, .... T HEOREM 3. Let p ∈ M , 0 < α 1, f ∈ Lip(α , p(x)) and {pn }∞ 0 be a sequence of positive numbers. If Pm Pn Δ =O , m+1 n+1

n−1

∑

m=0

(8)

then for n = 1, 2, ... the estimate f − Rn ( f ) p(x) = O n−α holds. In the classical Lebesgue spaces L p , the analogues of Theorem 1 and Theorem 2 were proved by Leindler in [15], and Theorem 3 in L p spaces was obtained by Chandra [4]. 2. Some auxiliary results In this section c will denote the constants (in general, different in different relations) depend only on quantities that are not important for the questions of interest. Let p ∈ M . Denote by En ( f ) p(x) (n = 0, 1, ...) the best approximation of f ∈ p(x) L in Πn (the set of trigonometric polynomials of degree at most n ), that is En ( f ) p(x) = inf f − tn p(x) : tn ∈ Πn .

290


It follows that, for example from Theorem 1.1 in [6], there exists a trigonometric polynomial tn∗ ∈ Πn such that En ( f ) p(x) = f − tn∗ p(x) for n = 0, 1, .... By W p(x) = W p(x) ([0, 2π ]) we denote the set of absolutely continuous functions f such that f ∈ L p(x) . L EMMA 1. Let p ∈ M and f ∈ W p(x) . Then the estimate

1

f En ( f ) p(x) = O p(x) n

(9)

holds for n = 1, 2, .... Proof. It follows from Theorem 6.1 of [21] that f − Sn( f ) p(x) → 0,

n → ∞.

It is easy to see that Ak f (x) = kAk ( f ) (x),

k = 1, 2, ...,

where f is the conjugate function of f . By considering the uniform boundedness of ∞ {Sn }0 and the boundedness of the conjugation operator in the space L p(x) (see [21]), we get

∞

∞ 1

= ∑ Ak f f − Sn ( f ) p(x) = ∑ Ak ( f )

k=n+1

k=n+1 k p(x) p(x)

∞ 1 1 1

− Sk f − f + Sn f − f = ∑

k=n+1 k k + 1

n+1 p(x)

∞ 1 1 1

f − f f − f − ∑ +

Sk

Sn k k + 1 p(x) n + 1 p(x) k=n+1 ∞

1 1 1 c ∑ k − k + 1 f p(x) + n + 1 f p(x) k=n+1

c f p(x) , n and hence (9) follows.

L EMMA 2. If p ∈ M , the Jackson type inequality 1 En ( f ) p(x) = O Ω p(x) f , , n = 1, 2, ... n

291


holds for f ∈ L p(x) . Proof. Let f ∈ L p(x) . Consider the transform ⎛ ⎞ δ h 2 ⎝1 f (x + t)dt ⎠ dh, Uδ ( f )(x) := δ h δ /2

δ > 0.

0

It is clear that Uδ ( f ) ∈ W p(x) for each δ > 0 and 2 (Uδ ( f )) (x) = δ

δ

δ /2

1 ( f (x + h) − f (x)) dh h

for almost all x. Since ⎛ ⎞ δ 4 1 (Uδ ( f )) (x) ⎝ | f (x + h) − f (x)| dh⎠ , δ δ 0

it follows from definition of Ω p(x) ( f , δ ) that

(Uδ ( f ))

p(x)

δ

4 1

| f (· + h) − f | dh

δ δ

0

p(x)

4 Ω p(x) ( f , δ ) . δ On the other hand, since 2 Uδ ( f )(x) − f (x) = δ

δ δ /2

⎛ ⎝1 h

h

⎞ ( f (x + t) − f (x)) dt ⎠ dh,

0

we get

Uδ ( f ) − f p(x)

| f (· + t) − f | dt

dh

0 δ /2 p(x)

δ h

2 1

sup

h | f (· + t) − f | dt δ δ /2hδ

2 δ

2 δ

h δ

1

h

δ δ /2

⎛

δ /2

0

dh p(x)

h

1 ⎜

| f (· + t) − f | dt ⎝ sup

δ /2hδ h

0

⎞ ⎟ dh⎠

p(x)

292


h

1

= sup | f (· + t) − f | dt

δ /2hδ h

0

Ω p(x) ( f , δ ) . p(x)

Hence, by subadditivity of the best approximation and (9) we obtain En ( f ) p(x) En f − U1/n ( f ) p(x) + En U1/n ( f ) p(x)

c

f − U1/n( f ) p(x) + U1/n ( f ) n p(x) 1 1 c Ω p(x) f , + 4nΩ p(x) f , , n n n which finishes the proof.

L EMMA 3. Let p ∈ M and 0 < α 1. Then for every f ∈ Lip (α , p(x)) the estimate f − Sn( f ) p(x) = O n−α , n = 1, 2, ... holds. Proof. Let tn∗ (n = 0, 1, ...) be the trigonometric polynomial of best approximation to f ∈ Lip (α , p(x)) . By Lemma 2 f − tn∗ p(x) = O Ω p(x) ( f , 1/n) , and hence

f − tn∗ p(x) = O n−α .

By the uniform boundedness of the partial sums Sn ( f ) in the space L p(x) (see [21]), we get f − Sn( f ) p(x) f − tn∗ p(x) + tn∗ − Sn ( f ) p(x) = f − tn∗ p(x) + Sn (tn∗ − f ) p(x) = O f − tn∗ p(x) = O n −α . L EMMA 4. Let p ∈ M . If f ∈ Lip (1, p(x)) , then f is absolutely continuous and f ∈ L p(x) , that is f ∈ W p(x) . Proof. Let f ∈ Lip (1, p(x)) and δ > 0. Since p− p(x) almost everywhere, by Theorem 2.8 of [13] the space L p(x) is continuously embedded in L p− . Hence we have Th ( f ) p− c Th ( f ) p(x) for every h with |h| δ . This inequality and equivalence of ω p− ( f , ·) and Ω p− ( f , ·) yield ω p− ( f , δ ) c Ω p(x) ( f , δ ) .


293

Hence, f ∈ Lip (1, p(x)) implies ω p− ( f , δ ) = O (δ ) , and this implies that f is absolutely continuous and f ∈ L p− ([6, pp. 51–54]). Since the relation f (x + t) − f (x) → f (x) , t

t →0

holds almost everywhere, for almost all x we get 2 δ

δ δ /2

| f (x + t) − f (x)| dt → f (x) , t

δ → 0+ .

By Fatou Lemma, for every measurable function g with m p (g) 1, 2π

f (x) |g(x)| dx =

0

2π 0

⎛ 2 ⎜ ⎝ lim+ δ →0 δ

lim inf δ →0+

2π 0

4 lim inf + δ →0 δ = lim inf δ →0+

4 δ

⎛

δ /2

⎜2 ⎝ δ 2π 0 2π

⎞

δ

⎛

| f (x + t) − f (x)| ⎟ dt ⎠ |g(x)| dx t ⎞

δ δ /2

⎝1 δ

| f (x + t) − f (x)| ⎟ dt ⎠ |g(x)| dx t

δ

⎞ | f (x + t) − f (x)| dt ⎠ |g(x)| dx

0

Tδ ( f ) (x) |g(x)| dx

0

4 4 lim inf Tδ ( f ) p(x) lim inf Ω p(x) ( f , δ ) δ →0+ δ δ →0+ δ 4 = lim inf O (δ ) = O (1) , δ →0+ δ and this means that f ∈ L p(x) .

L EMMA 5. Let p ∈ M and f ∈ Lip(1, p(x)). Then for n = 1, 2, ... the estimate Sn ( f ) − σn ( f ) p(x) = O n−1 holds. Proof. By Lemma 4, f ∈ W p(x) . If f has the Fourier series f (x) ∼

∞

∑ Ak ( f ) (x),

k=0

294


then the Fourier series of the conjugate function f be f (x) ∼

∞

∑ kAk ( f )(x).

k=1

On the other hand, Sn ( f )(x) − σn ( f ) (x) =

n

k

∑ n + 1 Ak ( f ) (x)

k=1

=

1 Sn f (x). n+1

Hence, by considering the uniform boundedness of the partial sums and the conjugation operator in the space L p(x) (see [21]), we obtain Sn ( f ) − σn ( f ) p(x) = O n−1 for n = 1, 2, ....

The following Lemma was proved in [15]. ∞ L EMMA 6. Let {pn }∞ 0 be a sequence of positive numbers. If {pn }0 ∈ AMDS, or ∈ AMIS and (n + 1)pn = O (Pn ) , then

{pn }∞ 0

n

∑ m−α pn−m = O

n−α Pn

m=1

for 0 < α < 1. 3. Proofs of the main results Proof of Theorem 1. Since f (x) = we have f (x) − Nn ( f )(x) =

1 n ∑ pn−m f (x), Pn m=0

1 n ∑ pn−m { f (x) − Sm ( f )(x)} . Pn m=0

By Lemma 3 and Lemma 6 we obtain f − Nn ( f ) p(x)

1 n ∑ pn−m f − Sm ( f ) p(x) Pn m=0

pn 1 n pn−m O m−α + f − S0( f ) p(x) ∑ Pn m=1 Pn 1 1 −α = O n Pn + O Pn n+1 −α =O n . =


295

Proof of Theorem 2. It is clear that Nn ( f )(x) =

1 n ∑ Pn−mAm ( f )(x). Pn m=0

By Abel transform, 1 n ∑ (Pn − Pn−m) Am ( f )(x) Pn m=1 m 1 n 1 n Pn −Pn−m Δm kAk ( f )(x) + = ∑ ∑ ∑ kAk ( f )(x), Pn m=1 m n+1 k=1 k=1

Sn ( f )(x) − Nn ( f )(x) =

and hence Sn ( f ) − Nn ( f ) p(x)

m

Pn − Pn−m 1 n

kA ( f ) Δ

∑ m ∑ k

Pn m=1 m k=1 p(x)

n 1

+

∑ kAk ( f ) .

n + 1 k=1 p(x)

Since Sn ( f )(x) − σn ( f ) (x) =

1 n ∑ kAk ( f ) (x), n + 1 k=1

by Lemma 5 we get

n 1

∑ kAk ( f )

n + 1 k=1

= Sn ( f ) − σn ( f ) p(x) = O n−1 . p(x)

Hence, Sn ( f ) − Nn ( f ) p(x) = O

1 Pn

Pn − Pn−m Δ + O n−1 . m ∑ m m=1 n

Suppose the condition n−1

∑ k |Δpk | = O (Pn )

k=1

holds. This implies that (see [15]) n Pn − Pn−m Pn ∑ Δm =O n , m m=1 and hence by (10) we have Sn ( f ) − Nn ( f ) p(x) = O n−1 .

(10)

296


This relation and Lemma 3 yield

f − Nn ( f ) p(x) = O n−1 .

Now let

n−1

Pn ∑ |Δpk | = O n k=0

.

(11)

A simple calculation yields n Pn − Pn−m 1 = Δm ∑ pk − (m + 1) pn−m , m m (m + 1) k=n−m and by induction one can easily get n m ∑ pk − (m + 1) pn−m ∑ k |pn−k+1 − pn−k | . k=n−m k=1 Thus, n

∑

m=1

Δm Pn − Pn−m m

n

1 ∑ m (m + 1) m=1

m

∑ k |pn−k+1 − pn−k |

k=1

1 = ∑ k |pn−k+1 − pn−k | ∑ k=1 m=k m (m + 1) n ∞ 1 ∑ k |pn−k+1 − pn−k | ∑ k=1 m=k m (m + 1) n

n

n

n−1

k=1

k=0

=

∑ |pn−k+1 − pn−k | =

∑ |Δpk | .

Combining this, the assumption (11) and (10) we get Sn ( f ) − Nn ( f ) p(x) = O n−1 , and considering Lemma 3 again we obtain the desired result.

Proof of Theorem 3. Let 0 < α < 1. By definition of Rn ( f )(x), f (x) − Rn ( f )(x) =

1 n ∑ pm { f (x) − Sm ( f ) (x)} . Pn m=0

From Lemma 3, we get 1 n ∑ pm f − Sm( f ) p(x) Pn m=0 n 1 p0 =O pm m−α + f − S0( f ) p(x) ∑ Pn m=1 Pn n 1 =O ∑ p m m− α . Pn m=1

f − Rn ( f ) p(x)

(12)


By Abel transform, n

n−1

m=1

m=1 n−1

∑ p m m− α =

∑ Pm

! m−α − (m + 1)−α + n−α Pn Pm

∑ m−α m + 1 + n−α Pn,

m=1

and n−1

∑m

m=1

−α

Pm = m+1

n−1

Pm ∑ Δ m+1 m=1 = O n−α Pn

m

∑k

−α

k=1

+

Pn n−1 −α ∑m n + 1 m=1

by condition (8). This yields n

∑ p m m− α = O

−α n Pn

m=1

and from this and (12) we get f − Rn ( f ) p(x) = O n−α . Let’s consider the case α = 1. By Abel transform, Rn ( f )(x) = =

1 n−1 ∑ {Pm (Sm ( f ) (x) − Sm+1( f )(x)) + PnSn ( f ) (x)} Pn m=0 1 n−1 ∑ Pm (−Am+1( f )(x)) + Sn( f ) (x) , Pn m=0

and hence Rn ( f )(x) − Sn ( f )(x) = −

1 n−1 ∑ Pm Am+1 ( f ) (x). Pn m=0

Using Abel transform again yields n−1

n−1

m=0

m=0

∑ Pm Am+1 ( f ) (x) =

Pm

∑ m + 1 (m + 1)Am+1( f )(x)

n−1

Pm = ∑Δ m +1 m=0 +

m

∑ (k + 1)Ak+1 ( f )(x)

k=0

Pn n−1 ∑ (k + 1)Ak+1( f )(x). n + 1 k=0

297

298


Thus, by considering Lemma 5 and (8) we obtain

n−1

n−1 m P

m ∑ (k + 1)Ak+1 ( f ) ∑ Δ

∑ Pm Am+1 ( f )

m=0

m + 1 k=0 m=0 p(x) p(x)

n−1 Pn

+

∑ (k + 1)Ak+1 ( f )

n + 1 k=0 p(x) n−1 Pm (m + 2)Sm+1 ( f ) − σm+1 ( f ) = ∑ Δ p(x) m +1 m=0 +Pn Sn ( f ) − σn ( f ) p(x) n−1 Pm Pn = O (1) ∑ Δ +O n . m + 1 m=0 This gives Rn ( f ) − Sn ( f ) p(x)

n−1 1

=

∑ Pm Am+1 ( f )

Pn m=0 p(x) Pn 1 1 = O =O . Pn n n

Combining this estimate with Lemma 3 yields f − Rn( f ) p(x) = O n−1 .

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(Received February 16, 2009)

Ali Guven Department of Mathematics Faculty of Art and Science Balikesir University 10145, Balikesir Turkey e-mail: ag [email protected] Daniyal M. Israfilov Department of Mathematics Faculty of Art and Science Balikesir University 10145, Balikesir Turkey e-mail: [email protected]

Journal of Mathematical Inequalities

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