Norm inequalities for certain classes of functions and their Fourier transforms

J. Math. Anal. Appl. 335 (2007) 1416–1433 www.elsevier.com/locate/jmaa

Norm inequalities for certain classes of functions and their Fourier transforms Barry Booton, Yoram Sagher ∗ Department of Mathematical Sciences, Florida Atlantic University, 777 Glades Road, Boca Raton, FL 33431, USA Received 11 August 2006 Available online 15 February 2007 Submitted by Richard M. Aron

Abstract We apply tools of interpolation theory and a commutative property of the Hilbert transform to prove necessary and sufficient conditions related to trigonometric series. These results extend and improve related theorems proven by several authors, summarized by Boas. In addition, we explore inequalities and operators, both connected to Hardy’s inequalities, on certain classes of functions, including quasimonotone functions. © 2007 Elsevier Inc. All rights reserved. Keywords: Hardy’s inequalities; Quasi-monotonicity; Fourier transforms; Interpolation of operators

1. Introduction For a sequence {λn }, we denote by λ∗n the nonincreasing rearrangement of |λn |. Hardy and Littlewood [5] showed that for q  2, a necessary and sufficient condition that  λn be, for every rearrangement, the Fourier coefficients of a function f ∈ Lq (0, π) is that (λ∗n )q nq−2 < ∞; and then 1   q q f Lq (0,π)  C λ∗n nq−2 (1) for every such f . (Throughout we denote by C a generic constant that does not depend on the functions in the inequalities.) They also showed that if f (x) = λn cos nx ∈ Lq , 1 < q  2, * Corresponding author. Fax: +1 (561) 297 2436.

E-mail addresses: [email protected] (B. Booton), [email protected] (Y. Sagher). 0022-247X/$ – see front matter © 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2007.02.022

B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433

then

  {λn }

l(q  ,q)

1417

 Cf Lq .

 Conversely, if {λn }l(q  ,q) < ∞ and 1 < q < ∞, then f , defined as f (x) = λ∗n cos nx, satisfies f q  C{λn }l(q  ,q) . These inequalities were precursors to the Lorentz spaces L(p, q) (l(p, q) for sequences), defined in full generality by G.G. Lorentz [10], A.P. Calderón [3], and R. Hunt [8]. Note that  ∗ q q−2 )1/q can be taken as the l(q  , q) norm of {λ }, and so (1) can be written as ( ∞ n n=1 (λn ) n f Lq (0,π)  C{λn }l(q  ,q) for q  2. Paley [11] generalized some of Hardy and Littlewood’s results on trigonometric series to ones on more general bounded orthonormal systems  on (0, π). Specifically, forq  2, if {φn } is a bounded orthonormal system on (0, π), and (λ∗n )q nq−2 < ∞, then f = λn φn ∈ Lq (0, π), and f Lq (0,π)  C{λn }l(q  ,q) . Also, for 1 < q  2, if {φn } is a bounded orthonormal sys ∗ q q−2  (λn ) n < ∞, and {λn }l(q  ,q)  tem on (0, π), and f = λn φn ∈ Lq (0, π), then Cf Lq (0,π) . Applying Paley’s theorem to appropriate permutations of the trigonometric system gives the Hardy and Littlewood theorem. Zygmund [14] introduced interpolation methods into this subject by proving Paley’s theorem using the Marcinkiewicz interpolation theorem. Work along these lines was continued in [12,13]. Note that classes of functions with nonincreasing Fourier transforms do not form linear spaces or even groups, and so the application of interpolation theory required recasting the theory to interpolated semi-groups. Many additional results may be found in Boas [2]. Boas’s presentation in [2] does not use interpolation techniques. We will show how results in [2] can be strengthened, and their proofs simplified, using interpolation theory. We will also extend results that Boas proved for monotone functions to quasi-monotone ones. 2. Inequalities for series and integrals We recall the definition of L(p, q) spaces. Definition 1. Let (M, Σ, μ) be a σ -finite measure space. Assume f is a measurable function and that lima→∞ μ{|f | > a} = 0. Let f ∗ denote the nonincreasing rearrangement of |f |. For 0 < p < ∞, 0 < q  ∞, define μ(M) 1

 1 ∗ q dx q x p f (x) . (2) f L(p,q) = x 0

In the special case where (M, Σ, μ) is the space of positive integers with μ({n}) = 1, we use an equivalent norm: ∞ 1  1  1 q q   {an } n p an∗ = . l(p,q) n n=1

We define L(p, q) = {f : f L(p,q) < ∞} and l(p, q) = {{an }: {an }l(p,q) < ∞}. Definition 2. Let (M, Σ, μ) be a measure space. Let ω be a weight function; i.e., an a.e. positive p measurable function. For 0 < p  ∞ and f measurable, define f Lpω = ωf Lp and Lω = {f : f Lpω < ∞}.

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B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433 1

It will be convenient to write ω(p, q) for the weight ω(x) = x p

− q1

on R+ .

In the sequel we shall apply interpolation with change of measure. Theorem 1. (See [4].) Assume that 0 < θ < 1, 0 < p, q  ∞. Then f (Lp ,Lpω )θ,q ∞  1 ∞ 1  −θ q dt q  −θ q dt q t f · tωI{tω<1} Lp t f I{tω1} Lp ∼ max , . t t 0

(3)

0

The statement of Theorem 1 in [4] is for 1  p, q  ∞. The extension for 0 < p, q  ∞ is routine and so we shall skip its proof. Lemma 1. (See [7].) Assume that (M, Σ, μ) is a measure space, and that 0  f, g are measurable. For any E ∈ Σ , μ(E)



f ∗g∗.

f g dμ  E

0

 f L(p,q) for q  p and f L(p,q)  f Lq As a consequence, we have that f Lq ω(p,q) ω(p,q) for q  p. We will make frequent use of Hardy’s inequalities. These inequalities have several equivalent formulations and for the convenience of the reader, we include the forms that we shall use. Throughout, when the domain of a function is not specified, we take it to be (0, a), where 0 < a  ∞. Theorem 2. Assume that f  0, α > 0, and 1  q  ∞. Then 1 a q  1 a

x  −α q q dx q dt 1 −α x f (x) f (t)  x t x α 0

0

(4)

0

and a

a x

0

α x

dt f (t) t

q

dx x

1 q

1  α

a

 α q x f (x)

1 q

.

(5)

0

Hardy, Littlewood, and Paley considered norm inequalities for the Fourier transforms when the functions or the Fourier transforms are monotone. It turns out that their results also hold for quasi-monotone functions. This generalization is particularly interesting when one considers Besov norm inequalities, since the class of functions with quasi-monotone Fourier transforms is closed under differentiation; see [12]. Definition 3. Define for β  0: f ∈ QD(β) if x −β f (x) 0; and f ∈ QD if f ∈ QD(β) for some β  0. Similarly, {an } ∈ QD(β) if n−β an 0, and {an } ∈ QD if {an } ∈ QD(β) for some β  0.

B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433

1419

t 1/q and Theorem 3. Assume that f ∈ QD(β). For 0 < q  ∞, denote Bq (t) = ( 0 (f (s))q ds s ) a 1/q . Then for 0 < q < r  ∞, Dq (t) = ( t (f (s))q ds s ) and

f (t)  C(β, r)Br (t)  C(β, r, q)Bq (t)

(6)

    t t f (t)  C(β, r)Dr  C(β, r, q)Dq . 2 4

(7)

Proof.

t

 β −β q ds s s f (s) s

Bq (t) =

t

1 q

t

−β

f (t)

s

0

βq ds

1 q

− q1

= (βq)

s

f (t).

0

Hence, t Br (t) =

 r−q  q ds f (s) f (s) s

0

 (βq)

1 1 q−r

t



 (βq)

r

q ds r−q  Bq (s) f (s) s

0 1 1 q−r

1

 1− q r Bq (t)

t

 q ds f (s) s

1 r

1 r

1

= (βq) q

− 1r

Bq (t)

0

proving (6). Similarly,  1 t 1   a  q ds q  β −β q ds q t Dq = f (s) s s f (s)  2 s s t 2

t

 t −β f (t)

s βq

ds s

1

t 2

q

= C(β, q)f (t).

t 2

Therefore, a

1  r−q  q ds r f (s) f (s) Dr (t) = s t a    1 r−q  q ds r s Dq f (s)  C(β, q, r) 2 s t 1   1− q a r  q ds r t f (s)  C(β, q, r) Dq 2 s t     1− q r  q t t r . Dq (t)  C(β, q, r)Dq = C(β, q, r) Dq 2 2

2

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B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433

Assume that f and g are two nonnegative functions. We write f ∼ g if there exists a constant C so that for all x, 1 f (x)  g(x)  Cf (x). C Theorem 4. Assume that f ∈ QD, 0 < q  ∞, α > 0. Then a t

−α

0

t

ds f (s) s

q

dt t

a

1 q

∼

0

 −α q dt t f (t) t

1 q

(8)

0

and a

a tα

ds f (s) s

q

dt t

a

1 q

∼

t

0

α q dt t f (t) t

1 q

(9)

.

0

Proof. Let us show that there exists a constant C = C(α, β, q) so that for all f ∈ QD(β), a t

−α

0

t

ds f (s) s

q

dt t

a

1 q

C

0

 −α q dt t f (t) t

1 q

.

0

For 1  q  ∞ this is (4). For 0 < q < 1, using (7),

a t

−α

t

0

ds f (s) s

q

dt = t

0

a

 −α q dt C t B1 (t) t

0

a

 −α q dt t Bq (t) t

0

a =C

 q f (s)

a t

−αq dt



t

ds C s

s

0

a

 −α q ds . s f (s) s

0

Conversely, by (6), a



t

−α

q dt f (t) t

a

1 q

β

0

 −α q dt t B1 (t) t

a

1 q

=β

0

t 0

proving (8). Note that the preceding inequality holds for all α ∈ R. Let us see that for all f ∈ QD(β), a

a t

0

α t

ds f (s) s

q

dt t

a

1 q

C

α q dt t f (t) t

0

For 1  q  ∞ this is (5). For 0 < q < 1, using (7),

1 q

.

−α

t 0

ds f (s) s

q

dt t

1 q

B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433

a

a tα

0

ds f (s) s

q

dt t

1421

1 q

t

a

=

α q dt t D1 (t) t

0

a =C

a t

αq

a =C

q

C

 q  q1 t dt t Dq 2 t α



0

 q ds dt f (s) s t

t 2

0

a 

1

a

1 q

C

 q f (s)

0

 α q ds s f (s) s

2s t

αq dt

t



ds s

1 q

0

1 q

.

0

Conversely, by (6), a

α q dt t f (t) t

a 

1 q

C

0

2 a q  1  q  q1 t dt ds dt q α α t D1 =C f (s) t 2 t s t a

0

a C

a tα

0

f (s)

ds s

q

dt t

0

1

t

q

t

proving (9). Observe that this inequality holds for all α ∈ R.

2

Our next goal is to show that a QD function can be written as the difference of two nonq increasing functions whose Lω(p,q) and L(p, q) norms are equivalent to those of the original function. From the corresponding inequality for the sum we get: Lemma 2. Assume that f, g are measurable. For a, b  0, (fg)∗ (a + b)  f ∗ (a)g ∗ (b). Corollary 1. Assume that f ∈ QD(β) and is right continuous. For s  t,  ∗ s −β f (s) = u−β f (u) (s)  t −β f ∗ (s − t). In particular, f (s)  2β f ∗

  s . 2

(10)

The next two theorems were proved in [13] for 1 < p < ∞ and 1  q  ∞. a Theorem 5. Assume that f ∈ QD, and that x f (t) dtt < ∞ for all 0 < x  a. Then there exist nonincreasing functions f1 , f2 so that f = f2 − f1 .

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B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433 q

If furthermore f ∈ Lω(p,q) , 0 < p < ∞, 0 < q  ∞, then for j = 1, 2, fj Lq

∼ f Lq

ω(p,q)

ω(p,q)

Proof. We define f1 (x) = β 0 < x0 < x1 . Then

(11)

.

a x

f (t) dtt and f2 = f + f1 . Let us see that f2 is nonincreasing. Let

x1

f2 (x0 ) − f2 (x1 ) = f (x0 ) − f (x1 ) + β

f (t)

dt t

x0

x1  −β  dt = f (x0 ) − f (x1 ) + β t f (t) t β t x0

x1

−β  f (x0 ) − f (x1 ) + βx1 f (x1 )

 = f (x0 ) −

x0 x1

tβ x0

β f (x1 )  0.

q

Assume that f ∈ Lω(p,q) . From (9) it follows that ∼ f Lq , and thus furthermore f1 Lq ω(p,q)

f Lq

ω(p,q)

so that f Lq

a x

f (t) dtt < ∞ for all 0 < x  a, and

ω(p,q)

 f2 Lq

ω(p,q)

ω(p,q)

dt t

∼ f2 Lq

  C f Lq

ω(p,q)

ω(p,q)

.



+ f1 Lq

ω(p,q)

 Cf Lq

ω(p,q)

2

Theorem 6. Assume that f ∈ QD, 0 < p < ∞, and 0 < q  ∞. Then f L(p,q) ∼ f Lq

ω(p,q)

.

Proof. For a < ∞, we extend f by defining f (x) = 0 for x  a. Assume that f ∈ QD(β), and let g(x) = limt→x + (t −β f (t)). Then g is right-continuous and nonincreasing. Therefore, h(x) := x β g(x) ∈ QD and is right-continuous, and x β g(x) = f (x) nearly everywhere, so that f L(p,q) = hL(p,q)

and f Lq

ω(p,q)

= hLq

ω(p,q)

.

We can assume therefore that f is right-continuous. By (10), f Lq

ω(p,q)

∞ ∞ 1  q  q1 1 q dx q  1 x dx β+ 1 β ∗ xpf = 2 = 2 p f L(p,q) . x p f (x) x 2 x 0

0

For the other inequality, let f1 , f2 be as in Theorem 5. Then f L(p,q)  f2 L(p,q) = f2 Lq

ω(p,q)

 Cf Lq

ω(p,q)

.

2

B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433

1423

Corollary 2. Assume that f ∈ L(p, q) ∩ QD, where 0 < p < ∞ and 0 < q  ∞. Then there exist nonincreasing functions f1 , f2 so that f = f2 − f1 and for j = 1, 2, fj L(p,q) ∼ f L(p,q) .

(12)

Clearly Theorem 5 and Corollary 2 hold for sequence spaces as well. The series version of the following lemma, with η = δ, is proved differently in [2, Lemma 6.18]. q For 0 < q  ∞, we define q  = q−1 , so that q1 + q1 = 1. Note that for 0 < q < 1, q  < 0. Lemma 3. Assume that G is nondecreasing and right-continuous on (a, b), where 0  a < b  ∞. For s > 0, δ > 0, 0 < η  δ, and 1  q  ∞, b  q  q1

dx x −δ y s dG(y) x a

(a,x]

b   C(s, η, q)

x s−η

a

(x,b)

q  q1

dx K(a)a s η−δ y dG(y) + y −δ dG(y) 1 x q (δq) (a,b)

(13)

and b 

x

δ [x,b)

a

q  q1 dx dG(y) x b 

 C(s, η, q)

x

η−s

a

y

s+δ−η

(a,x)

q  q1

K( b1 ) dx dG(y) + y s+δ dG(y), (14) 1 x s q (δq) b (a,b)

where K(0) = 0 and K(a) = 1 for a > 0. If a = 0, then (13) holds for 0 < q  ∞, and if b = ∞, then (14) holds for 0 < q  ∞. In both (13) and (14), if q  1, then C(s, η, q)  ηs . Proof. It suffices to consider q < ∞. b  q  q1

dx −δ s x y dG(y) x a

(a,x]

b

 q  1

y dt dx q s t s + K(a)a s dG(y) t x

x −δ

= a

a

(a,x]

b s

x −δ

a

y

dt ts t

q dG(y)

dx x

1

a

(a,x]

b 



a

(a,x]

x −δ

+ K(a)a s



q  q1 dx dG(y) . x

q

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B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433

We consider each of the last two terms. b  q  1

y dx q −δ s dt s t x dG(y) t x a

a

(a,x]

b =s

x

−δ

t

a

a

b

x

=s

x

−δ

a

s (t,x]



t

s

a

b s

x

−η

a

x

s

x

−η

0

y

η−δ

y



t

a

y

δ−η η−δ

 q  q1 dx dt dG(y) t x

(t,x]

x

b

 q  q1 dt dx dG(y) t x



x

s

 q  q1 dx dt dG(y) t x

(t,x]



 q  q1 dx dt η−δ t I(a,b) (t) y dG(y) . t x s

0

(t,b)

We apply Hardy’s inequality to the last term. The case q  1 requires no comment. If a = 0,

y η−δ dG(y) ∈ QD(s) ts (t,b)

and so by (8), the application of Hardy’s inequality is valid for 0 < q  ∞. Thus, b x

s

−η

0

x



 q  q1 dx dt η−δ t I(a,b) (t) y dG(y) t x s

0

(t,b)

b   C(s, η, q)

x

s−η



q  q1 dx η−δ I(a,b) (x) y dG(y) x

0

(x,b)

b  = C(s, η, q)

x s−η

a

q  q1 dx y η−δ dG(y) . x

(x,b)

Also, by Minkowski’s inequality, b  K(a)a s a

x −δ



1 q  q1

b q dx dx dG(y)  K(a)a s x −δq dG(y) x x

(a,x] − q1

 (δq)

(a,b)

y

K(a)a

s



(a,b)

proving (13).

y −δ dG(y)

B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433

1425

We use −v −s dG( v1 ) as the measure in (13) to prove (14), b 

x

δ [x,b)

a

q  q1 dx dG(y) x

a1 



u−δ

= 1 b

 q  q1 du 1 −v s v −s dG v u

( b1 ,u]

a1   C(s, η, q)

us−η 1 b

+

K( b1 ) 1

(δq) q bs

 

 1 −δ −s −v v dG v

( b1 , a1 )

= C(s, η, q)

(δq)

1 q



x η−s a

+

v η−δ



q  q1 dx y δ+s−η dG(y) x

(a,x)

y δ+s dG(y).

bs

 q  q1 du 1 −v −s dG v u

(u, a1 )

b 

K( b1 )





2

(a,b)

3. Norm inequalities for Fourier transforms  In this section, I denotes the interval (0, π), E(x) = ∞ n=1 an cos nx, and S(x) = ∞ a sin nx. n n=1 The following is a generalization to quasi-monotone sequences of a theorem of Hardy and Littlewood for monotone sequences. Theorem 7. (See [12,13].) Assume that f is either the Fourier sine or Fourier cosine series associated with {an }. If either {an } ∈ QD or f ∈ QD then for 1 < p < ∞ and 0 < q  ∞,   {an }  ∼ f L(p,q)(I) (15) l(p ,q) and for 1 < p < ∞ and 1  q  ∞,   {an } q ∼ f Lq (I) . l ω(p  ,q)

ω(p,q)

(16)

Theorem 8. (See [9].) Let H denote the Hilbert transform. Assume that for some α ∈ R, we have f (x)−α ∈ L1 (I). Then x  H

f (x) − α x

 =

Hf (x) − Hf (0) . x

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B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433

Definition 4. (See [2].) For f so that xf (x) ∈ L1 (I), define the sequence of generalized sine coefficients {bn } of f by 2 bn = π

π f (x) sin nx dx. 0

For f so that x 2 f (x) ∈ L1 (I), define the sequence of generalized cosine coefficients {an } of f by 2 an = π

π f (x)(1 − cos nx) dx. 0

The case s = 1 in the next theorem is Theorem 6.11 in [2]. Theorem 9. Assume that g  0 on I, xg(x) ∈ L1 (I), and 1 < p < ∞. Assume that {λn } is the sequence of generalized sine coefficients or cosine coefficients of g. For p1 < s and 0 < q  ∞,  x  1     s t s g(t) dt  x  0

and for 1  q  ∞,  x  1    s  s t g(t) dt  x  0

q

Lω(p,q) (I)

q Lω(p,q) (I)

   λn    C  n   l(p ,q)

   λn    C  n q l

(17)

(18)

.

ω(p  ,q)

For s > 0 and 1  q  ∞,   x   1   λn    s    C t g(t) dt    n   s   x l(p ,q) 0



π +

t

s+ p1

t

s+ p1

(19)

g(t) dt

q

Lω(p,q) (I)

0

and    λn     n q l

ω(p  ,q)

  x 1    s  C  s t g(t) dt  x 

+

q Lω(p,q) (I)

0



π

g(t) dt .

(20)

0

π Proof. Let G(x) = x g(t) dt. Note that G ∈ L1 (I). If λn are the generalized sine coefficients of g, then λnn are the cosine coefficients of G, and if λn are the generalized cosine coefficients of g then λnn are the sine coefficients of G. By (13), for 0 < q  ∞,  x  1     s t s g(t) dt  x  0

 CGLq q Lω(p,q) (I)

ω(p,q) (I)

= CGL(p,q)(I) .

B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433

1427

By Theorem 7, since G , for 1 < p < ∞, 0 < q  ∞,    λn   GL(p,q)(I) ∼   n   l(p ,q) and for 1 < p  ∞, 1  q  ∞,    λn    ∼ GLq (I)  n q ω(p,q) l

ω(p  ,q)

proving (17) and (18). By (14), for 1  q  ∞,  x  1    s  ps t g(t) dt GLq   ω(p,q) (I)  xs 

q

Lω(p,q) (I)

0

 1

π p q 1 s+ 1 + t p g(t) dt s q π 0

2

proving (19) and (20).

The following two theorems are related to Theorem 7 in [1]. We state that theorem using the notation of this paper. If λn  0 are the Fourier sine or cosine coefficients of f , then for 0  a < π , 1 < p < ∞ and 1 < q < ∞,  ∞           f (x)  f (x) − f (a) 1        λk   C .  q q  C x −a x Lq L k=n

1−1 |x−a| p q

lω(p ,q)

ω(p,q)

We generalize this theorem in several ways. We relax the condition λn  0, extend the range of the parameter q, and prove related  L(p, q) inequalities. ∞ Given a sequence {λk } so that ∞ k=1 λk converges, we denote Λn = k=n λk . A summation by parts argument proves the following: Lemma 4. Assume that x  π, ∞ 

Λn cos nx =

n=1

∞

n=1 λn ,

∞

n=1 λn sin nx,

and

∞

n=1 λn cos nx

converge. Then for 0 <

∞ ∞ cos x2  1 λ sin nx − λn (1 − cos nx) n 2 sin x2 2 n=1

n=1

and ∞  n=1

Λn sin nx =

∞ ∞ cos x2  1 λ (1 − cos nx) + λn sin nx. n 2 sin x2 2 n=1

n=1

∞ ∞ Lemma 5. Assumethat n=1 λn converges, {Λn } ∈ QD, and n=1  ∞ ∞ n=1 λn sin nx and n=1 λn cos nx converge.

Λn n

< ∞. Then

Proof. By the decomposition of QD sequences as in Theorem 5,there exist {Λ1,n }, {Λ2,n } ∞ 0 so that Λn = Λ1,n − Λ2,n .  Since Λj,n 0, sin nx converge. with Λj,n  n=1 Λj,n ∞ ∞ ∞ Λ sin nx converges, as does Λ sin nx. Therefore, Therefore, n n+1 n=1 n=1 n=1 λn sin nx = ∞ (Λ − Λ ) sin nx converges. The proof for the cosine series is the same. 2 n n+1 n=1

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B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433

Theorem 10. (a) Assume thatf (x) x ∈ L(p, q)(I), 1 < p < ∞. Let {λk } be the sequence of sine coefficients of f . Then ∞ k=1 λk converges. Assume that {Λn } ∈ QD. Then for 0 < q  ∞,       {Λn }   C  f (x)  (21)  x  l(p ,q) L(p,q)(I) and for 1 < q < ∞,

   f (x)     C  x  q l(p  ,q) L

  {Λn }

(22)

.

ω(p,q) (I)

(b) Assume that for some A ∈ R, f (x)−A x ∈ L(p, q)(I), 1 < p < ∞. Let {λk } be the sequence of cosine coefficients of f . Then ∞ k=1 λk converges. Assume that {Λn } ∈ QD. Then for 0 < q  ∞,       {Λn }   C  f (x) − A  (23)   l(p ,q) x L(p,q)(I) and for 1 < q < ∞,

   f (x) − A     C  q  l(p  ,q) x L

  {Λn }

.

(24)

ω(p,q) (I)

 Proof. Let us see that ∞ k=1 λk converges in either case. Assume that {λn } are the cosine coefficients of f . Let Dn be the Dirichlet kernel,  n   π     2       λk  =  f (x) Dn (x) − Dm (x) dx     π  k=m+1

0

 π  2      = f (x) − A Dn (x) − Dm (x) dx  π  0

 π  2   sin(n + 12 )x − sin(m + 12 )x   = f (x) − A dx  π  2 sin x2 0

and the last expression converges to 0 by the Riemann–Lebesgue Lemma. The proof for the sine coefficients is similar. Observe that for this part of the proof we need to assume only f (x)−A ∈ L1 x f (x) or x ∈ L1 . We proceed to prove (21). Of course f (x) x ∈ L(p, q) implies f ∈ L(p, q). By Theorem 7, for 1 < p < ∞, 0 < q  ∞,  ∞        {Λn }   C  Λn cos nx  . l(p ,q)   n=1

By Lemma 4,

L(p,q)(I)

B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433

1429

 ∞      Λn cos nx     n=1 L(p,q)(I)  ∞     ∞  cos 1 x     1     2 C  λn sin nx  +  λn (1 − cos nx)  2 sin 1 x   2 2 n=1 n=1 L(p,q)(I) L(p,q)(I)  ∞        f (x)     C  + λn (1 − cos nx) .  x    L(p,q)(I) n=1

Since f ∈ L(p, q), Hf (x) = −

L(p,q)(I)

∞

n=1 λn cos nx

∈ L(p, q). By Theorem 8,

 ∞  f (y) (x) = Hf (x) − Hf (0) = xH λn (1 − cos nx). y 

n=1

Therefore,  ∞      λn (1 − cos nx)    n=1

and so

L(p,q)(I)

      f (y)  = xH (x)  y

 ∞       C Λ cos nx   n  l(p ,q)  

  {Λn }

n=1

L(p,q)(I)

   f (y)     C(p) y L(p,q)(I) L(p,q)(I)

   f (x)     C . x L(p,q)(I)

Statement (22) is proved similarly by using the continuity of the Hilbert transform with respect q to the Lω(p,q) norm; see [6].  ∞ Assume that f (x) = ∞ n=1 λn cos nx. Since f ∈ L(p, q), Hf (x) = n=1 λn sin nx ∈ L(p, q) and Hf (0) = 0. By Theorem 8,   ∞ Hf (x) − Hf (0) 1  f (y) − A (x) = = H λn sin nx. y x x n=1

Therefore, by the first part of this theorem,  ∞     1      f (y) − A     {Λn }   C   H λ sin nx = C  n   l(p ,q) x  y L(p,q)(I) n=1 L(p,q)(I)    f (y) − A    C   y L(p,q)(I) proving (23). Statement (24) is proved similarly.

2

Clearly, Lemma 6. Assume that f , defined on (0, a), 0 < a  ∞, has a nonincreasing rearrangement. Then for all 0 < b < a,   ∗ f |s − b| (x)  2f ∗ (x)I(0,max{b,a−b}) (x).

1430

B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433

Proof.  ∗   ∗ f |s − b| (x) = f (b − s)I(0,b) (s) + f (s − b)I(b,a) (s) (x)  ∗ = f (t)I(0,b) (b − t) + f (t)I(b,a) (b + t) (x)  ∗ = f (t)I(0,b) (t) + f (t)I(0,a−b) (t) (x)  ∗  2f (t)I(0,max{b,a−b}) (t) (x)  2f ∗ (x)I(0,max{b,a−b}) (x).

2

Theorem 11. Assume that {λk } ∈ l 1 , 0  a < π , 1 < p < ∞, 0 < q  ∞, and g(x) = ∞  ∞ k=1 λk sin kx or g(x) = k=0 λk cos kx. Then  ∞

       g(x) − g(a)       C |λk |  . (25)  x −a  q    L (I) k=n

1−1 |x−a| p q

Also,

l(p ,q)

 ∞

       g(x) − g(a)       C |λ |   k  x −a    L(p,q)(I) k=n

.

(26)

l(p  ,q)

 y Proof.In the sequel we agree that if y < 1, then k=1 αk = 0. Let g(x) = ∞ k=1 λk sin kx and |λ |. Bn = ∞ k k=n Inequality (25) is a part of Theorem 7 in [1]; we give an alternate proof, using interpolation techniques, that is both simpler and at the same time yields (26):   ∞  g(x) − g(a)   | sin kx − sin ka|   |λk |  x −a  |x − a| k=1

so that   ∞   g(x) − g(a)       k|λk | = {λk }l 1  x −a  ∞ k L

(27)

k=1

as well as    g(x) − g(a)     x −a  ∞ L

2

|x−a|

∞ 

  |λk | = 2{λk }l 1 .

(28)

k=1

By Lemma 6,      g(x) − g(a)  g(x) − g(a) ∗   = sup t (t)  x −a  x−a 0
p ,q

Using Theorem 1, and later in the argument the continuity of g,

(29)

B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433

1431

   g(x) − g(a)    C x − a (L∞ ,L∞

|x−a| ) 1 ,q p

∞ 



t

− p1   g(x) − g(a)



x −a

0

∞ =

t

q p −1

0

  I{|x−a|> 1 } (x)  t

q L∞

dt t

1 q

1   q  g(x) − g(a)  q   dt I (x) + I (x) (a+t,π) (0,a−t)  x −a  ∞ L

π−a 1 q

q  q   g(x) − g(a) −1  dt , I  max tp  (x) (a+t,π)  x−a  ∞ L 0

a t

q p −1

0

1  q q  g(x) − g(a)    dt I (x) (0,a−t)  x−a  ∞ L

π−a  1 a 1   

q  q q q  g(a + t) − g(a) q  g(a − t) − g(a) q −1 −1  dt ,  dt  p  max t p  t    t −t 0

π = max a

0

 1 a 1     q q q q  g(x) − g(a) q  g(x) − g(a) q −1 −1  dx ,  dx  p (x − a) p  (a − x)  x −a  x−a  0

π 1   q q  g(x) − g(a) q 1 −1  dx .  |x − a| p  2 x−a  0

Again using Theorem 1,   {λk }

(lk1 ,l 1 ) 1 ,q

  = {λk }(l 1 ,l 1 ) k

p

∞ ∼

t

− p1

 

1  C

t

=C



1 p

q  q1 dt |λk | t 1

 k t

q  q1 ∞ q  q1  dt dt − p1  t k|λk | + |λk | t t 1 1

1k< t

0

∞ t

− p1

1

=C

|λk |tk +

1k< 1t

0



1 ,q p

 1k
q dt k|λk | t

q

+

t 0

n+1   q  q1 ∞

 − pq −1 t k|λk | dt n=1 n

k t

0

∞

1

1k
1 p

 kt

q  q1  dt |λk | t

1432

B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433

+

∞ n 

t

q −1 p

C =C

∞ 

n



− pq −1

n=1 ∞ 

q

n p

∞ 

q |λk |

1  q

dt

k=n

n=1 n−1





−1

n=1

n 

q  1 k|λk |

k=1

1 n



q

n 

+ q  1 q

k|λk |

∞  n=1

n

q −1 p



∞ 

q  1  q

|λk |

k=n



  + {Bk }l(p ,q) .

k=1

Using Lemma 3, we shall see that

∞ 

n

q −1 p



n=1

q  1 n q   1 k|λk |  C {Bk }l(p ,q) . n k=1

Let G(y) =

y 

|λk |.

k=1

Using (13),

∞ 

n

q −1 p



n=1



=

q  1 n q 1 k|λk | n k=1

∞  

n

n=1

x

1 p

=

∞  n=1

q

∞

dG(y)

dx x



1 q

∼

q

n p

−1

∞ 

q  1 q

|λk |

0

q  q1 dG(y)

∞  1 1

 − n p q n=1

x

0



0

(0,n]

∞ C

q  1 q  q1 ∞

x dx q − p1 y dG(y) ∼ y dG(y) x x



− p1 − q1

[n,∞)

  = {Bk }l(p ,q) .

k=n

This gives us (25). The proof for the cosine series is similar. Interpolating between (27) and (29), we obtain    g(x) − g(a)       C {λk }(l 1 ,l 1 )  x −a  ∞ 1 k p ,q (L ,L(1,∞)) 1 p ,q

that is to say,      g(x) − g(a)     C {Bk }l(p ,q)  x −a  L(p,q) proving (26). Again, the proof for the cosine series is similar.

2

B. Booton, Y. Sagher / J. Math. Anal. Appl. 335 (2007) 1416–1433

1433

  Theorem 12. Assume  that ∞ k=1 λk converges ∞ and {Λn } ∈ QD ∩ l(p , q), where 1 < p < ∞ ∞ nx and n=1 λn cos nx converge. For all 0  a < π , and for and 0 < q  ∞. Then n=1 λn sin ∞ λ sin nx or g(x) = g(x) = ∞ n=1 n n=1 λn cos nx, we have    g(x) − g(a)       C(p){Λn }l(p ,q) . (30)  x −a  L(p,q)(I) Also,

   g(x) − g(a)     x −a  q L

1−1 |x−a| p q

(I)

   C(p){Λn }l(p ,q) .

(31)

Proof. Let us assume first that Λn 0. Then {λn } ∈ l 1 , and we obtain (30) by an application of Theorem 11. If {Λn } ∈ QD, then by Corollary 2, there exist sequences {Λ j,n } so that Λj,n 0, Λn = ∞  ,q) ∼ {Λj,n }l(p  ,q) . Let g(x) = − Λ , and {Λ } Λ 2,n 1,n n l(p n=1 λn sin nx and gj (x) = ∞ n=1 (Λj,n − Λj,n+1 ) sin nx. Then      g(x) − g(a)   g2 (x) − g2 (a) g1 (x) − g1 (a)      − =  x −a   x − a x − a L(p,q)(I) L(p,q)(I)       g2 (x) − g2 (a)   g1 (x) − g1 (a)      C  +   x −a x−a L(p,q)(I) L(p,q)(I)         C {Λ2,n }l(p ,q) + {Λ1,n }l(p ,q)  C {Λn }l(p ,q) . The same argument proves (30) for cosine series; (31) is proved in the same way.

2

References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14]

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