Approximation theorems in weighted Lorentz spaces

CARPATHIAN J. MATH. 26 (2010), No. 1, 108 - 119

Online version available at http://carpathian.ubm.ro Print Edition: ISSN 1584 - 2851 Online Edition: ISSN 1843 - 4401

Approximation theorems in weighted Lorentz spaces Y UNUS E. Y ILDIRIR and D ANIYAL M. I SRAFILOV

A BSTRACT. In this paper we deal with the converse and simultaneous approximation problems of functions possessing derivatives of positive orders by trigonometric polynomials in the weighted Lorentz spaces with weights satisfying the so called Muckenhoupt’s Ap condition. 1. I NTRODUCTION AND MAIN RESULTS Let T = [−π, π] and w : T → [0, ∞] be an almost everywhere positive, integrable function. Let fw∗ (t) be a decreasing rearrangement of f : T → [0, ∞] with respect to the Borel measure Z w(e) = w(x)dx, e

i.e., fw∗ (t) = inf {τ ≥ 0 : w (x ∈ T : |f (x)| > τ ) ≤ t} . Let 1 < p, s < ∞ and let Lps w (T) be a weighted Lorentz space, i.e., the set of all measurable functions for which 1/s  Z s s dt < ∞, kf kLps =  (f ∗∗ (t)) t p  w t T

where 1 f (t) = t ∗∗

Zt

fw∗ (u)du.

0

Lps w (T)

If p = s, is the weighted Lebesgue space Lpw (T) [5, p.20]. The weights w used in the paper are those which belong to the Muckenhoupt class Ap (T), i.e., they satisfy the condition  p−1 Z Z 0 1 1 p sup w(x)dx  w1−p (x)dx < ∞, p0 = |I| |I| p−1 I

I

where the supremum is taken with respect to all the intervals I with length ≤ 2π and |I| denotes the length of I. Received: 16.01.2009; In revised form: 25.07.2009; Accepted: 08.02.2010 2000 Mathematics Subject Classification. 41A10, 42A10. Key words and phrases. Best approximation, weighted Lorentz space, mean modulus of smoothness, fractional derivative and phrases. 108

Approximation theorems in weighted Lorentz spaces

109

Let ∞ X

f (x) ∼

(1.1)

∞

ck eikx =

k=−∞

a0 X + (ak cos kx + bk sin kx) 2 k=1

and f˜(x) ∼

∞ X

(ak sin kx − bk cos kx)

k=1

be the Fourier and the conjugate Fourier series of f ∈ L1 (T), respectively. In addition, we put Sn (x, f ) :=

n X

n

ck eikx =

a0 X + (ak cos kx + bk sin kx), 2

n = 1, 2, ....

k=1

k=−n

By L10 (T) we denote the class of L1 (T) functions f for which the constant term c0 in (1.1) equals zero. If α > 0, then α − th integral of f ∈ L10 (T) is defined as X Iα (x, f ) := ck (ik)−α eikx , k∈Z∗ −α (−1/2)πiα sign k

where (ik)−α := |k| For α ∈ (0, 1) let

e

and Z∗ := {±1, ±2, ±3, ...} .

d I1−α (x, f ), dx (r) dr+1 = r+1 I1−α (x, f ) f (α+r) (x) := f (α) (x) dx if the right hand sides exist, where r ∈ Z+ := {1, 2, 3, ...} [15, p. 347]. By c, c(α, ...) we denote the constants, which can be different in different place, such that they are absolute or depend only on the parameters given in their brackets. Let x, t ∈ R := (−∞, ∞), r ∈ R+ := (0, ∞) and let f (α) (x) :=

(1.2)

4rt f (x) :=

∞ X

(−1)k [Ckr ]f (x + (r − k)t), f ∈ L1 (T),

k=0

where [Ckr ] :=

r(r − 1)...(r − k + 1) for k > 1, [Ckr ] := r for k = 1 and [Ckr ] := 1 k!

for k = 0. Since [15, p. 14]

r(r − 1)...(r − k + 1) ≤ c(r) , k ∈ Z+ |[Ckr ]| = k r+1 k! we have

∞ X

|[Ckr ]| < ∞,

k=0

and therefore 4rt f (x) is defined a.e. on R. Furthermore, the series in (1.2) converges absolutely a.e. and 4rt f (x) is measurable [17]. If r ∈ Z+ , then the fractional difference 4rt f (x) coincides with usual forward difference.

110

Y. E. Yildirir and D. M. Israfilov

Now let σδr f (x)

1 := δ

Zδ

|4rt f (x)| dt,

0

for f ∈ Lps w (T), 1 < p, s < ∞, w ∈ Ap (T). Since the series in (1.2) converges absolutely a.e., we have σδr f (x) < ∞ a.e. and using boundedness of the HardyLittlewood Maximal function [3, Th. 3] in Lps w (T), w ∈ Ap (T), we get kσδr f (x)kLps ≤ c kf kLps < ∞. w w

(1.3)

Hence, if r ∈ R+ and w ∈ Ap (T), 1 < p, s < ∞, we can define the r-th mean modulus of smoothness of a function f ∈ Lps w (T) as := sup kσδr f (x)kLps . Ωr (f, h)Lps w w

(1.4)

|δ|≤h

Remark 1.1. Let f, f1 , f2 ∈ Lps w (T), w ∈ Ap (T), 1 < p, s < ∞. The r-th mean modulus of smoothness Ωr (f, h)Lps , r ∈ R+ , has the following properties: w (i) Ωr (f, h)Lps is non-negative and non-decreasing function of h ≥ 0. w ps + Ω (f , ·) ps . (ii) Ωr (f1 + f2 , ·)Lps ≤ Ω (f , ·) r 1 r 2 Lw Lw w (iii) lim Ωr (f, h)Lps = 0. w h→0

The best approximation of f ∈ Lps w (T) in the class Πn of trigonometric polynomials of degree not exceeding n is defined by En (f )Lps = inf kf − Tn kLps : Tn ∈ Πn . w w A polynomial Tn (x, f ) := Tn (x) of degree n is said to be a near best approximant of f if kf − Tn kLps ≤ cEn (f )Lps , n = 0, 1, 2, .... w w α (α) Let Wps,w (T), α > 0, be the class of functions f ∈ Lps ∈ Lps w (T) such that f w (T). α Wps,w (T), 1 < p, s < ∞, α > 0, becomes a Banach space with the norm

kf kW α (T) := kf kLps + f (α) Lps . w ps,w

w

In this paper we deal with the converse and simultaneous approximation problems of functions possessing derivatives of positive orders by trigonometric polynomials in the weighted Lorentz spaces Lps w (T) with weights satisfying so called Muckenhoupt’s Ap condition. Our new results are the following Theorem 1.1. Let 1 < p < ∞ and 1 < s ≤ 2 or p > 2 and s ≥ 2. Then for a given + f ∈ Lps w (T), w ∈ Ap (T), and r ∈ R we have !1/γ n X c γ rγ−1 Ωr (f, π/(n + 1))Lps ≤ (k + 1) Ek (f )Lps , n = 0, 1, 2, ... w w (n + 1)r k=0

with a positive constant c independent of n, where γ = min(s, 2). In case of r ∈ Z+ this result was proved in [10]. In the space Lp (T), 1 ≤ p ≤ ∞, using the usual modulus of smoothness, it was obtained in [17] without γ. In case of r ∈ Z+ in the spaces Lpw (T), w ∈ Ap (T), 1 < p < ∞, this theorem was proved


111

+

in [12] without γ. In case of r ∈ Z ,Theorem 1 without γ in term of ButzerWehrens’s type modulus of smoothness in the spaces Lpw (T), w ∈ Ap (T), 1 < p < ∞, and in the weighted Orlicz spaces was obtained in [6] and [8], respectively. Note that the above defined modulus of smoothness is more general than ButzerWehrens’s type modulus of smoothness and in special case, when r ∈ Z+ is even, it coincides with Butzer-Wehrens’s type modulus of smoothness. Theorem 1.2. Let 1 < p < ∞ and 1 < s ≤ 2 or p > 2 and s ≥ 2. Let w ∈ Ap (T) and f ∈ Lps w (T). Assume that ∞ X 2 and s ≥ 2. Let w ∈ Ap (T) and f ∈ Lps w (T). If ∞ X k αγ−1 Ekγ (f )Lps 0 independent of n and f.

112

Y. E. Yildirir and D. M. Israfilov +

In case of α ∈ Z , this theorem in the Lebesgue spaces Lp (T), 1 ≤ p ≤ ∞, was proved in [4]. We prove also the following inequality of Jackson type in the weighted Lorentz space Lps w (T). r Theorem 1.4. If f ∈ Wps,w (T), r ∈ R+ , 1 < p, s < ∞, and w ∈ Ap (T), then

r (r) ≤ ch Ωr (f, h)Lps

ps , 0 < h ≤ π

f w Lw

with a constant c independent of h and f. This Theorem in case of r ∈ R+ in the Lebesgue spaces Lp (T), 1 ≤ p ≤ ∞, was obtained in [2] (See also [17]), and in case of r ∈ Z+ , in the weighted Lebesgue spaces Lpw (T) with w ∈ Ap (T) and 1 < p < ∞, was proved in [12]. 2. A UXILIARY RESULTS Lemma 2.1. Let w ∈ Ap (T) and r ∈ R+ , 1 < p, s < ∞. If Tn ∈ Πn ,n ≥ 1, then there exists a constant c > 0 depending only on r, p and s such that

r (r) Ωr (Tn , h)Lps ≤ ch

T

ps , 0 < h ≤ π/n. n w Lw

Proof. Since r X [r] t 4rt Tn x − t = 2i sin ν cν eiνx , 2 2 ∗ ν∈Zn

[r] X [r] t [r] r−[r] 4t Tn(r−[r]) x − t = (iν) cν eiνx 2i sin ν 2 2 ∗ ν∈Zn

with Z∗n := {±1, ±2, ±3, ...} , [r] ≡ integer part of r, putting [r] r−[r] t 2 t r−[r] ϕ(z) := 2i sin z (iz) , g(z) := sin z , −n ≤ z ≤ n, g(0) := tr−[r] , 2 z 2 we get [r] 4t Tn(r−[r])

X X [r] [r] iνx r x− t = ϕ(ν)cν e , 4t Tn x − t = ϕ(ν)g(ν)cν eiνx . 2 2 ∗ ∗ ν∈Zn

ν∈Zn

Taking into account the fact that [17] g(z) =

∞ X

dk eikπz/n

k=−∞ k+1

uniformly in [−n, n] , with d0 > 0, (−1) dk ≥ 0, d−k = dk (k = 1, 2, ...) , we have ∞ X kπ r − [r] [r] 4rt Tn (·) = dk 4t Tn(r−[r]) · + + t . n 2 k=−∞


Consequently we get

δ

Z

1 r

δ |4t Tn (·)| dt

ps

0

δ

Z

1 =

δ

∞ X kπ r − [r] [r] (r−[r]) dk 4t Tn ·+ + t dt n 2

k=−∞

0

Lw

113

Lps w

δ

Z ∞ X

1 ≤ |dk |

δ

k=−∞

0

[r] (r−[r]) kπ r − [r] 4t Tn ·+ + t dt

n 2

Lps w

and since [20, p. 103] [r] 4t Tn(r−[r]) (·)

Zt =

Zt ... Tn(r) (· + t1 + ...t[r] )dt1 ...dt[r]

0

0

we find

δ

Z

1 [r] (r−[r]) kπ r − [r] Ωr (Tn , h)Lps ≤ sup |dk | ·+ + t dt w

δ 4t Tn

n 2 |δ|≤hk=−∞

ps 0 Lw

δ t t

Z Z Z ∞ X

1 r − [r] kπ (r)

|dk | = sup ... Tn + t + t1 + ...t[r] dt1 ...dt[r] dt ·+

n 2 |δ|≤hk=−∞

δ ∞ X

0

≤

h[r] sup

∞ X

0

Lps w

0

|dk |

|δ|≤hk=−∞

δ

Z Zδ Zδ

1 1

× ... [r]

δ δ 0

≤ h[r] sup

0

∞ X

0

(r) r − [r] kπ Tn + t + t1 + ...t[r] dt1 ...dt[r] dt ·+

n 2

Lps w

|dk |

|δ|≤hk=−∞



Zδ Zδ  Zδ

1 1 (r)

× [r] ... T ·+ δ n

δ 0 0 0

δ

Z ∞ X

1 [r] ≤ ch sup |dk |

δ |δ|≤h





kπ r − [r] + t + t1 + ...t[r] dt dt1 ...dt[r]

 n 2

Lps w

(r) r − [r] kπ Tn + t dt ·+

n 2

ps k=−∞ 0 Lw

r−[r]

+ δ ·+ kπ n Z 2 ∞

1 X (r)

≤ ch[r] sup |dk | r−[r] . Tn (u) du

|δ|≤hk=−∞

2 δ

·+ kπ ps n

On the other hand [17] ∞ X k=−∞

|dk | < 2g(0) = 2tr−[r] , 0 < t ≤ π/n

Lw

114


and for 0 < t < δ < h ≤ π/n we have ∞ X |dk | < 2g(0) = 2hr−[r] . k=−∞

Therefore the boundedness of Hardy-Littlewood maximal function in Lps w (T) implies that

r (r) Ωr (Tn , h)Lps ≤ ch

ps .

T n w Lw

By similar way for 0 < −h ≤ π/n, the same inequality also holds and the proof of Lemma 1 is completed. Lemma 2.2. Let w ∈ Ap (T), 1 < p, s < ∞. If Tn ∈ Πn and α > 0, then there exists a constant c > 0 depending only on α, p and ssuch that

(α) .

Tn ps ≤ cnα kTn kLps w Lw

Proof. Since w ∈ Ap (T), 1 < p, s < ∞, we have [21, Chap. VI] kSn (f )kLps

w

˜

f ps Lw

≤ c kf kLps , w ≤ c kf kLps . w

Now, following the method given in [13] we obtain the request result.

Definition 2.1. For f ∈ Lps w (T), 1 < p, s < ∞, δ > 0 and r = 1, 2, 3, ..., the Peetre K-functional is defined as

(r) ps r (2.6) K δ, f ; Lw (T), Wps,w (T) := inf kf − gkLps + δ g ps . w r g∈Wps,w (T)

Lw

Lemma 2.3. Let w ∈ Ap (T), 1 < p, s < ∞. If f ∈ Lps w (T) and r = 1, 2, 3, ..., then (i) the K-functional (2.6) and the modulus (1.4) are equivalent and (ii) there exists a constant c > 0 depending only on r, p and s such that En (f )Lps ≤ cΩr (f, 1/n)Lps . w w Proof. (i) can be proved by the similar way to that of Theorem 1 in [12] and later (ii) is proved by standard way (see for example, [12], [8]). 3. P ROOFS OF THE MAIN RESULTS Proof of Theorem 1. Let Sn be the n − th partial sum of the Fourier series of f ∈ Lpw (T), w ∈ Ap (T) and let m ∈ Z+ . By Remark 1 (ii), (1.3) and [10, prop.3.4] Ωr (f, π/(n + 1))Lps w

≤

+ Ωr (S2m , π/(n + 1))Lps Ωr (f − S2m , π/(n + 1))Lps w w

≤

+ Ωr (S2m , π/(n + 1))Lps cE2m (f )Lps w w

and by Lemma 1, Ωr (S2m , π/(n + 1))Lps ≤c w Since (r) S2m (x)

=

(r) S1 (x)

+

π n+1

m−1 Xn

r

(r)

S2m

Lps w

, n + 1 ≥ 2m .

o (r) (r) S2ν+1 (x) − S2ν (x) ,

ν=0


115

we have

≤

(3.7)

Ωr (S2m , π/(n + 1))Lps w 

m−1

r 

X h

i π

(r) (r) (r) + S2ν+1 − S2ν c

S

n + 1  1 Lps w

Lps w

ν=0

 

.



Following the method used in [10, Proof of Theorem 1] , we obtain

m−1

X h i

(r) (r) S2ν+1 (x) − S2ν (x)

ps ν=0

Lw

m−1 X

γ

(r) (r)

S2ν+1 (x) − S2ν (x) ps

≤ c

, γ = min(s, 2).

Lw

ν=0

By Lemma 2, we get

(r)

(r)

S2ν+1 (x) − S2ν (x)

≤ c2νr kS2ν+1 (x) − S2ν (x)kLps ≤ c(p, r)2νr+1 E2ν (f )Lps w w

Lps w

and

! γ1

(r)

S1

Lps w

(r) (r) = S1 − S0

Lps w

≤ cE0 (f )Lps . w

Then from (3.7) we have Ωr (S2m , π/(n + 1))Lps ≤c w

 r  π E0 (f )Lps + w n+1 

m−1 X

2(ν+1)rγ E2γν (f )Lps w

! γ1  

.



ν=0

It is easily seen that ν

(3.8)

2

(ν+1)rγ

E2γν (f )Lps w

≤c

2 X

µγr−1 Eµγ (f )Lps , ν = 1, 2, 3, .... w

µ=2ν−1 +1

Therefore, Ωr (S2m , π/(n + 1))Lps w    γ1  ν  r  m−1 2   X X π r γr−1 γ ps + c  ps  E0 (f )Lps ≤ c + 2 E (f ) µ E (f ) 1 Lw Lw µ w  n+1    ν=0 µ=2ν−1 +1   ! γ1 r  2m  X π γr−1 γ ps ≤ c E0 (f )Lps + µ E (f ) Lw µ w  n+1  µ=1

≤ c

π n+1

r

! γ1

m 2X −1

(ν + 1)γr−1 Eνγ (f )Lps w

.

ν=0

If we choose 2m ≤ n + 1 ≤ 2m+1 , then Ωr (S2m , π/(n + 1))Lps w

c ≤ r (n + 1)

n X

! γ1 (ν + 1)

ν=0

γr−1

Eνγ (f )Lps w

.

116


Taking also the relation E2m (f )Lps ≤ E2m−1 (f )Lps w w

n X

c ≤ r (n + 1)

! γ1 (ν + 1)γr−1 Eνγ (f )Lps w

ν=0

into account we obtain the required inequality of Theorem 1.

Proof of Theorem 2. If Tn is the best approximating polynomial of f, then by Lemma 2

(α) (α)

T2m+1 − T2m ps ≤ c2(m+1)α E2m (f )Lps w Lw

and hence by this inequality, (3.8) and hypothesis of Theorem 2 we have ∞ X

kT2m+1 − T2m kWps,w α (T)

∞ X

=

m=1

+

∞

X

(α) (α)

T2m+1 − T2m

≤c

∞ X

≤ c

Lps w

m=1

j α−1 Ej (f )Lps ≤c w

m=1 j=2m−1 +1

Therefore

∞ X

2(m+1)α E2m (f )Lps w

m=1

m

2 X

kT2m+1 − T2m kLps w

m=1 ∞ X

∞ X j α−1 Ej (f )Lps < ∞. w j=2

kT2m+1 − T2m kW α

ps,w (T)

< ∞,

m=1 α which implies that {T2m } is a Cauchy sequence in Wps,w (T). Since T2m → f in α ps the Banach space Lw (T), we have f ∈ Wps,w (T). It is clear that

(α) (α) En (f (α) )Lps ≤ − S f

f

ps n w Lw

≤ S2m+2 f (α) − Sn f (α)

Lps w

∞

X h i

(α) (α) + S2k+1 f − S2k f

k=m+2

.

Lps w

By Lemma 2

S2m+2 f (α) − Sn f (α)

Lps w

≤ c2(m+2)α En (f )Lps w ≤ c(n + 1)α En (f )Lps w

for 2m < n < 2m+1 . On the other hand, following the method used in [10, Proof of Theorem 1], we get

∞

! γ1 ∞

X h

γ i X

(α)

(α) (α) (α) , S2k+1 f − S2k f ≤c

S2k+1 (x) − S2k (x) ps

ps Lw k=m+2

Lw

k=m+2

where γ = min(s, 2). Since by Lemma 2

(α) (α) ≤ c2kα+1 E2k (f )Lps ,

S2k+1 (x) − S2k (x) ps ≤ c2kα kS2k+1 (x) − S2k (x)kLps w w Lw


we get

∞

X h i

(α) (α) − S2k f S2k+1 f

k=m+2

k=m+2

γkα+1

2

E2γk (f )Lps w

.

k=m+2

Lps w

Therefore, we have

∞

X h i

(α) (α) S2k+1 f − S2k f

! γ1

∞ X

≤c

117

≤c

Lps w

∞ X

! γ1 k

γα−1

Ekγ (f )Lps w

k=n+1

for 2m < n < 2m+1 . This completes the proof.

Proof of Theorem 3. We set Wn (f ) := Wn (x, f ) :=

2n 1 X Sν (x, f ), n = 0, 1, 2, ... . n + 1 ν=n

Since Wn (·, f (α) ) = Wn(α) (·, f ), we have

(α)

f (·) − Tn(α) (·, f )

Lps w

+ Tn(α) (·, Wn (f )) − Tn(α) (·, f )

≤ f (α) (·) − Wn (·, f (α) )

+

Lps w

(α)

(α) + (·, f ) − T (·, W (f ))

W

n n n ps

Lps w

Lw

=: I1 +I2 +I3 .

Let Tn (x, f ) be the best approximating polynomial of degree at most n to f in ps Lps w (T). From the boundedness of Wn in Lw (T) we have

I1 ≤ f (α) (·) − Tn (·, f (α) ) ps + Tn (·, f (α) ) − Wn (·, f (α) ) ps Lw

≤ cEn (f (α) )Lps w

Lw

+ Wn (·, Tn (f (α) ) − f (α) )

Lps w


and by Lemma 2 I2 ≤ cnα kTn (·, Wn (f )) − Tn (·, f )kLps w and I3

≤ c(2n)α kWn (·, f ) − Tn (·, Wn (f ))kLps w ≤ c(2n)α En (Wn (f ))Lps . w

Taking into account that kTn (·, Wn (f )) − Tn (·, f )kLps w ≤

+ kWn (·, f ) − f (·)kLps + kf (·) − Tn (·, f )kLps kTn (·, Wn (f )) − Wn (·, f )kLps w w w

≤

cEn (Wn (f ))Lps + cEn (f )Lps + cEn (f )Lps w w w

and En (Wn (f ))Lps ≤ cEn (f )Lps , w w

118


we get

(α)

f (·) − Tn(α) (·, f )

Lps w

≤ cEn (f (α) )Lps + cnα En (Wn (f ))Lps + cnα En (f )Lps + c(2n)α En (Wn (f ))Lps w w w w . + cnα En (f )Lps ≤ cEn (f (α) )Lps w w Since [1] ≤ En (f )Lps w

(3.9)

c (α) ps )Lw , α En (f (n + 1)

we conclude that

(α)

f (·) − Tn(α) (·)

Lps w


and the proof is completed.

Proof of Theorem 4. Let Tn ∈ Πn be the trigonometric polynomial of the best approximation of f in Lps w (T) metric. By Remark 1 (ii), Lemma 1 and (1.3) we get Ωr (f, h)Lps ≤ Ωr (Tn , h)Lps + Ωr (f − Tn , h)Lps w w w

≤ chr Tn(r)

Lps w

+ cEn (f )Lps , 0 < h < π/n. w

Using (3.9), Lemma 3 (ii) and the inequality

l Ωl (f, h)Lps ≤ chl f (l) ps , f ∈ Wps,w (T), l = 1, 2, 3, ..., w Lw

which can be showed using the judgements given in [12, Theorem 1], we have 2π c c (r−[r]) ps (r−[r]) ps En (f )Lw ≤ E (f )Lw ≤ Ω f , r−[r] n r−[r] [r] n + 1 Lps (n + 1) (n + 1) w [r]

2π c

(r) ≤

f ps . r−[r] n+1 Lw (n + 1) On the other hand, by Theorem 3 we find

(r)

Tn ps ≤ Tn(r) − f (r) Lw

≤ cEn (f (r) )Lps w

(r) +

f

ps Lps Lw w

(r)

+ f ps ≤ c f (r) Lw

Lps w

.

Choosing h with π/(n + 1) < h ≤ π/n, (n = 1, 2, 3, ...) , we obtain

r (r) Ωr (f, h)Lps ≤ ch

f

ps w Lw

and we are done.


119

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