The finite Hilbert transform and weighted Sobolev spaces

Math. Nachr. 266, 34 – 47 (2004) / DOI 10.1002/mana.200310142

The finite Hilbert transform and weighted Sobolev spaces

David Elliott∗1 and Susumu Okada∗∗2

1 School of Mathematics and Physics, University of Tasmania, Private Bag 37, Hobart, TAS 7001, Australia2 Math-Geogr. Fakultat, Katholische Universtitat Eichstatt, 85071 Eichstatt, Germany

Received 3 April 2002, revised 13 December 2002, accepted 16 February 2003Published online 5 February 2004

Key words Finite Hilbert transform, weighted Sobolev spacesMSC (2000) Primary 44A15, 46E35; Secondary 47G10

The boundedness of the finite Hilbert transform operator on certain weighted Lp spaces is well known. Weextend this result to give the boundedness of that operator on certain weighted Sobolev spaces.

c© 2004 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

1 Introduction

Throughout, we shall let Ω denote the open interval (0, 1). The finite Hilbert transform of a function f definedon Ω will be denoted and defined by

(Tf)(t) :=1π

∫ 1

0

− f(τ) dτ

τ − t:= lim

ε→0+

(∫ t−ε

0

+∫ 1

t+ε

)f(τ) dτ

π(τ − t), (1.1)

for t ∈ Ω, whenever the limit exists. If Lp(Ω), 1 < p < ∞, denotes the space of pth integrable functions definedon Ω then it is well known (see, for example, Butzer and Nessel [2, Proposition 8.1.9]), that T is a bounded linearoperator on Lp(Ω) into itself. That is, there exists a positive number c, independent of f ∈ Lp(Ω), such that

‖Tf‖p ≤ c ‖f‖p (1.2)

for 1 < p < ∞, where the norm ‖f‖p is defined in the usual way by

‖f‖p :=(∫ 1

0

|f(τ)|p dτ

)1/p

. (1.3)

Throughout this paper we shall let c denote a generic non-negative constant. Note, however, that the value of cmay change from line to line.

As is customary, with 1 < p < ∞ we shall define the conjugate number q, also in (1,∞), such that

1p

+1q

= 1 . (1.4)

In this paper we shall make considerable use of Holder’s inequalities for integrals and sums. These are given by∣∣∣∣∫ 1

0

f(t)g(t) dt

∣∣∣∣ ≤(∫ 1

0

|f(t)|p dt

)1/p(∫ 1

0

|g(t)|q dt

)1/q

(1.5)

and ∣∣∣∣∣N∑

k=1

akbk

∣∣∣∣∣ ≤(

N∑k=1

|ak|p)1/p( N∑

k=1

|bk|q)1/q

, (1.6)

∗ Corresponding author: e-mail: [email protected], Phone: +61 (0)3 6226 2442, Fax: +61 (0)3 6226 2867∗∗ e-mail: [email protected], Phone: +49 (0)8421 931176, Fax: +49 (0)8421 931789


Math. Nachr. 266 (2004) / www.interscience.wiley.com 35

respectively.In addition to (1.2) we also have the boundedness of the operator T on certain weighted Lp(Ω) spaces as

follows.

Theorem 1.1 Suppose 1 < p < ∞ and let α be a real number such that 0 < α + 1/q < 1. Then there existsa positive constant c independent of f such that

∫ 1

0

((t(1 − t))−α · |(Tf)(t)|)p dt ≤ c ·

∫ 1

0

((t(1 − t))−α · |f(t)|)p dt (1.7)

for every measurable function f on Ω for which the right-hand side of (1.7) is finite.

P r o o f. This is a particular case of a somewhat more general theorem whose proof is to be found in the textby Gohberg and Krupnik [3, §1.4, Theorem 4.1].

In this paper we shall generalise this theorem although, of course, differently from Gohberg and Krupnik. Weshall assume that, in addition to f , some of its derivatives also exist on Ω and are such that they lie in a weightedSobolev space (see Definition 2.1, below). Then we shall also show that T is a bounded linear operator on thisspace into itself (see Theorem 3.9).

In §2 we shall define the particular weighted Sobolev space we shall be using and derive some of its relevantproperties.

In §3 we shall state and prove the principal result of this paper.

2 A weighted Sobolev space

In addition to letting N denote the set of all natural numbers 1, 2, 3, . . . , we shall also use N0 to denote theset 0, 1, 2, . . . .

Let L1,loc(Ω) denote the space of all locally integrable functions on Ω; that is, f ∈ L1,loc(Ω) if, and only if,

we have that∫ b

a|f(t)| dt < ∞ whenever 0 < a < b < 1. Each function f ∈ L1,loc(Ω) is a distribution on Ω (see

Treves [6, Chapter 21]), and for each k ∈ N the kth distributional derivative will be denoted by f (k) or Dkf . Weshall also write f ′ = f (1).

We shall define a C∞-function ρ on Ω by

ρ(t) := t(1 − t) , t ∈ Ω . (2.1)

Definition 2.1 For 1 < p < ∞, N ∈ N0 and real α we denote by W(N)p,α (Ω) the space of all functions

f ∈ L1,loc(Ω) such that ρk−αf (k) ∈ Lp(Ω), that is,

∫ 1

0

((t(1 − t))k−α

∣∣f (k)(t)∣∣)p dt < ∞ , (2.2)

for every k ∈ 0, 1, . . . , N. A norm on this space will be denoted and defined by

‖f‖p,α,N := max0≤k≤N

(∫ 1

0

((t(1 − t))k−α |f (k)(t)|)p dt

)1/p

= max0≤k≤N

∥∥ρk−αf (k)∥∥

p. (2.3)

The following result is a special case of that given by Kufner [4, Theorem 3.6]. However, no explicit proof isgiven there. So, making use of the assumption that ρ is a C∞-function, we shall give a short proof by adaptingthat given by Treves [6, Proposition 31.1].

Lemma 2.2 If 1 < p < ∞, N ∈ N0 and α ∈ R then the weighted Sobolev space W(N)p,α (Ω) equipped with

norm ‖ · ‖p,α,N is a Banach space.


36 Elliott and Okada: Finite Hilbert transform

P r o o f. Letfj

∞j=1

be a Cauchy sequence in W(N)p,α (Ω).

Given k∈0, 1, . . . , N the sequenceρk−αf

(k)j

∞j=1

in the Banach space Lp(Ω) has a limit gk. In particular,

limj→∞ ρ−αfj = g0 in Lp(Ω), and hence in the space D′(Ω) of all distributions on Ω because Lp(Ω) is contin-uously embedded into D′(Ω) which is equipped with the usual topology (see Treves [6, Example I, Chapter 23]).Let k ∈ 1, 2, . . . , N. Since the differential operator Dk and multiplications by the C∞-functions ρα and ρk−α

are all continuous on D′(Ω), we have

limj→∞

ρk−αf(k)j = lim

j→∞ρk−αDk

(ρα(ρ−αfj

))= ρk−αDk(ραg0) in D′(Ω) .

Again, since Lp(Ω) is continuously embedded into D′(Ω), it follows that ρk−αDk(ραg0) = gk ∈ Lp(Ω). So we

have that ραg0 ∈ W(N)p,α (Ω) and consequently that limj→∞ fj = ραg0 in the norm ‖ · ‖p,α,N . Thus, W

(N)p,α (Ω) is

complete and therefore a Banach space.

We shall now consider some properties of the space W(N)p,α (Ω) which will be required in the next section. The

following lemma determines which power of the function ρ belongs to this space.

Lemma 2.3 The function ρ has the following properties:(i) Suppose γ ∈ R and j ∈ N0. Then for any t ∈ Ω we have∣∣(ργ)(j)(t)

∣∣ ≤ c · ργ−j(t) , (2.4)

for some constant c independent of t.(ii) Let 1 < p < ∞ and α, γ ∈ R . If γ satisfies γ > α + 1/q − 1 then

ργ ∈∞⋂

N=0

W (N)p,α (Ω) . (2.5)

(iii) Let g be a C∞-function on Ω such that g(t) = 1/ ln t for every t ∈ (0, 1/2) with g vanishing on theinterval (3/4, 1). Then, for 1 < p < ∞ and α ∈ R we have

ρα+1/q−1 · g ∈∞⋂

N=0

W (N)p,α (Ω) . (2.6)

P r o o f. (i) This is trivially true when γ = 0, and we assume that γ = 0. Since ργ(t) = tγ(1 − t)γ we haveby Leibniz’ theorem that

(ργ)(j)(t) =j∑

k=0

(j

k

)(tγ)(k) · ((1 − t)γ

)(j−k). (2.7)

On using

(tγ)(k) =

Γ(γ + 1)Γ(γ − k + 1)

tγ−k , (2.8)

we find, on multiplying each side of (2.7) by ρj−γ(t), that

ρj−γ(t) · (ργ)(j)(t) =

j∑k=0

(j

k

)(−1)j−k

(Γ(γ + 1)

)2Γ(γ − j + k + 1)Γ(γ − k + 1)

tj−k(1 − t)k . (2.9)

Consider the sum. Since both k ≥ 0 and j − k ≥ 0 and since 0 < t < 1 it follows at once that

ρj−γ(t) · ∣∣(ργ)(j)(t)∣∣ ≤ c , (2.10)



where c is a positive constant, depending upon γ and j, but independent of t. From (2.10), the inequality (2.4)follows at once, which proves (i).

(ii) Fix N ∈ N0 and let j ∈ N0 be such that j ≤ N . Consider

Ij :=∫ 1

0

ρ(γ−α)p(t) · (ρj−γ(t) · ∣∣(ργ)(j)(t)∣∣)p dt .

On recalling (2.10) we have

Ij ≤ c ·∫ 1

0

ρ(γ−α)p(t) dt < ∞ (2.11)

since (γ −α)p > −1, i.e. γ > α + 1/q− 1. Thus for every N ∈ N0 we have that ργ ∈ W(N)p,α (Ω), which is (2.5)

and proves (ii).(iii) Fix N ∈ N0. In order to show that ρα+1/q−1 · g ∈ W

(N)p,α (Ω) we need to prove that

Kj :=∫ 1

0

∣∣∣ρj−α(t) · (ρα+1/q−1 · g)(j)(t)∣∣∣p dt < ∞ (2.12)

for j ∈ 0, 1, . . . , N. First consider K0 since this illustrates how to proceed in general. We have

K0 =∫ 1/2

0

ρ−1(t) · |g(t)|p dt +∫ 1

1/2

ρ−1(t) · |g(t)|p dt . (2.13)

Now the second integral is finite since the integrand is bounded and continuous on [1/2, 1). The first integral isalso finite since∫ 1/2

0

ρ−1(t) · |g(t)|p dt ≤ 2∫ 1/2

0

dt

t | ln t|p =2

(p − 1) · (ln 2)p−1, (2.14)

and we have chosen 1 < p < ∞. Thus K0 is finite.Suppose now that j ∈ 1, 2, . . . , N. Leibniz’ theorem gives

Dj(ρα+1/q−1 · g) =

j∑k=0

(j

k

)Dj−k

(ρα+1/q−1

) · Dkg . (2.15)

An appeal to mathematical induction shows that given k ∈ 0, 1, . . . , N there exists a polynomial hk, of degreek, satisfying (

Dkg)(t) =

(tk ln t

)−1 · hk(1/ ln t) for t ∈ (0, 1/2) . (2.16)

By part (i) there is a constant c > 0 such that∣∣Dj−k(ρα+1/q−1

)(t)∣∣ ≤ c · ρα+1/q−1−j+k(t) (2.17)

for t ∈ Ω and k ∈ 0, 1, . . . , j. It follows from (2.15), (2.16) and (2.17) that

∣∣ρj−α(t) · Dj(ρα+1/q−1 · g)(t)∣∣ ≤ c · ρ−1/p(t)

| ln t|j∑

k=0

(j

k

)(ρ(t)t

)k

· ∣∣hk(1/ ln t)∣∣ (2.18)

for t ∈ (0, 1/2]. Given k ∈ 0, 1, . . . , j, the function t → hk(1/ ln t) is bounded on (0, 1/2] because−1/ ln 2 < 1/ ln t < 0 for t ∈ (0, 1/2] and hk is a polynomial. This together with (2.18) and the fact that1/2 ≤ ρ(t)/t ≤ 1 for t ∈ (0, 1/2] give

∣∣ρj−α(t) · Dj(ρα+1/q−1 · g)(t)∣∣p ≤ c

t | ln t|p , (2.19)



for some positive constant c independent of t ∈ (0, 1/2]. Returning to Kj if, as in the case when j = 0, wesplit the integration to be over (0, 1/2) and [1/2, 1), then we find that the integrand over [1/2, 1) is bounded andcontinuous so that the integral is finite. From (2.19) we have that∫ 1/2

0

∣∣ρj−α(t) · Dj(ρα+1/q−1 · g)(t)∣∣p dt ≤ c ·

∫ 1/2

0

dt

t | ln t|p < ∞ , (2.20)

since 1 < p < ∞. Thus Kj < ∞ for j ∈ 0, 1, . . . , N so that ρα+1/q−1 · g ∈ W(N)p,α (Ω) for every N ∈ N0,

which proves (iii).

In the next theorem, C(r)(Ω) for r ∈ N0 denotes the space of all r times continuously differentiable functionson the open interval Ω. We shall write C(0)(Ω) as C(Ω). Moreover, when 1 < p < ∞, we shall let Lp,loc(Ω)denote the space of all locally pth integrable functions on Ω.

Theorem 2.4 Suppose 1 < p < ∞ and α and β are real numbers. Then the following statements hold:(i) If M, N ∈ N0 are such that M ≤ N then

W (N)p,α (Ω) ⊆ W (M)

p,α (Ω) and ‖f‖p,α,M ≤ ‖f‖p,α,N for every f ∈ W (N)p,α (Ω) .

(ii) If β ≤ α and N ∈ N0 then W(N)p,α (Ω) ⊆ W

(N)p,β (Ω) with

‖f‖p,β,N ≤ (1/4)α−β ‖f‖p,α,N for every f ∈ W (N)p,α (Ω) . (2.21)

(iii) Let N ∈ N and f ∈ W(N)p,α (Ω). Then, for j ∈ 0, 1, . . . , N,

f (j) ∈

C(N−1−j)(Ω), j ∈ 0, 1, . . . , N−1 ,

Lp,loc(Ω), j = N .

(iv) The inequality α + 1/q > 0 is equivalent to the inclusion⋂∞N=0

W (N)p,α (Ω) ⊆ L1(Ω) . (2.22)

P r o o f. (i) This is an immediate consequence of Definition 2.1.(ii) For any f ∈ W

(N)p,α (Ω) and k ∈ N0 with k ≤ N we have∫ 1

0

((t(1 − t))k−β · ∣∣f (k)(t)

∣∣)p dt =∫ 1

0

((t(1 − t))k−α · ∣∣f (k)(t)

∣∣)p(t(1 − t))(α−β)p dt

≤ (1/4)(α−β)p ·∫ 1

0

((t(1 − t))k−α · ∣∣f (k)(t)

∣∣)p dt .

Since this is true for k ∈ 0, 1, . . . , N we have that f ∈ W(N)p,β (Ω) and furthermore that

‖f‖p,β,N ≤ (1/4)α−β ‖f‖p,α,N

as required.

(iii) If 0 < a < b < 1 then we have∫ b

a

∣∣f (N)(t)∣∣p dt =

∫ b

a

ρ(N−α)p(t) · ∣∣f (N)(t)∣∣p · ρ−(N−α)p(t) dt

≤ ‖f‖pp,α,N · max

a≤t≤bρ−(N−α)p(t)

< ∞ ,

which shows that f (N) ∈ Lp,loc(Ω). On fixing j ∈ 0, 1, . . . , N − 1 it follows immediately from Definition 2.1

that f (j) ∈ W(N−j)p,α−j (Ω). Let t0 ∈ Ω and let V := (t0/2, (t0+1)/2). For each k ∈ 0, 1, . . . , N−j the restriction



f (j+k)∣∣V

of f (j+k) to V is pth integrable on V since ρj+k−αf (j+k) ∈ Lp(Ω) and supt∈V ρ−(j+k−α)(t) < ∞.That is, f (j)

∣∣V

belongs to the usual Sobolev space over V of order (N − j) with respect to p. Thus Sobolev’sembedding theorem (see, for example, Adams [1, Theorem 5.4]) implies that f (j) is (N − j − 1) times con-tinuously differentiable on the open interval V containing t0. Since t0 is an arbitrary point of Ω we have thatf (j) ∈ C(N−j−1)(Ω).

(iv) Assume first that α + 1/q > 0. Given f ∈ W(0)p,α(Ω), we have, by Holder’s inequality (1.5), that

∫ 1

0

|f(t)| dt ≤ ‖f‖p,α,0

(∫ 1

0

(t(1 − t))αq dt

)1/q

.

But, since αq > −1, the integral on the right-hand side is finite, which implies that f ∈ L1(Ω). In other words,

the space L1(Ω) contains W(0)p,α(Ω), and consequently (2.22) holds.

Next assume that α + 1/q = 0. Take a function g satisfying the assumption given in part (iii) of Lemma 2.3.

Then the function ρ−1 ·g = ρα+1/q−1 ·g belongs to W(N)p,α (Ω) for every N ∈ N0 by (2.6), while ρ−1 ·g /∈ L1(Ω)

because∫ 1/2

0 |t(ln t)|−1 dt = ∞. This violates (2.22).Finally assume that α + 1/q < 0, and choose a number γ satisfying α + 1/q − 1 < γ < −1. It follows from

Lemma 2.3 (ii) that ργ is an element of W(N)p,α (Ω) for every N ∈ N0 while the assumption γ < −1 gives that

ργ /∈ L1(Ω). So (2.22) is violated in this case also and hence (iv) holds.

We shall now give bounds for∣∣f (j)

∣∣.Theorem 2.5 Suppose f ∈ W

(N)p,α (Ω) with 1 < p < ∞, α ∈ R is such that 0 < α + 1/q < 1 and N ∈ N.

Then for all t ∈ Ω and j ∈ 0, 1, . . . , N−1∣∣f (j)(t)∣∣ ≤ c ‖f‖p,α,N · (t(1 − t))α−1/p−j , (2.23)

where c is a positive constant independent of both t and f .

P r o o f. Fix j ∈ 0, 1, . . . , N−1. From Theorem 2.4 (iii) we know that f (j) is certainly continuous on Ωso that f (j)(a) exists for any a ∈ Ω. Fix a ∈ Ω and first suppose that 0 < a ≤ t < 1. Then

f (j)(t) − f (j)(a) =∫ t

a

f (j+1)(s) ds =∫ t

a

ρj+1−α(s) f (j+1)(s) ρα−j−1(s) ds . (2.24)

By Holder’s inequality (1.5) we have that∣∣f (j)(t)∣∣ ≤ ∣∣f (j)(a)

∣∣+ ‖f‖p,α,N

(Hj(t)

)1/q, (2.25)

say, where the function Hj is defined by

Hj(t) :=∫ t

a

ds

(s(1 − s))(j+1−α)q=∫ t

a

(1s

+1

1 − s

)(j+1−α)q

ds . (2.26)

Note that since α + 1/q < 1 then (j + 1 − α)q > 1 for all j ∈ 0, 1, . . . , N−1.By Holder’s inequality for sums (1.6) and (2.26) it follows that

Hj(t) ≤ c

∫ t

a

(1

s(j+1−α)q+

1(1 − s)(j+1−α)q

)ds , (2.27)

c being independent of a and t. Evaluating this integral gives

Hj(t) ≤ c

(1

(1 − t)(j+1−α)q−1− 1

(1 − a)(j+1−α)q−1

)+

(1

a(j+1−α)q−1− 1

t(j+1−α)q−1

), (2.28)

where c is independent of a and t. Since we have assumed that 0 < a ≤ t < 1 it follows that

Hj(t) ≤ c

(a(1 − t))(j+1−α)q−1, (2.29)



with c being independent of a and t. From (2.25) and (2.29) we find that

∣∣f (j)(t)∣∣ ≤ ∣∣f (j)(a)

∣∣+ c ‖f‖p,α,N

(a(1 − t))j−α+1/p, (2.30)

on using (1.4). We may proceed similarly when 0 < t ≤ a < 1 to give in this case

∣∣f (j)(t)∣∣ ≤ ∣∣f (j)(a)

∣∣+ c ‖f‖p,α,N

(t(1 − a))j−α+1/p. (2.31)

We may combine (2.30) and (2.31) to give for any a and t in Ω

∣∣f (j)(t)∣∣ ≤ ∣∣f (j)(a)

∣∣+ c ‖f‖p,α,N

(t(1 − t))j−α+1/p · (a(1 − a))j−α+1/p, (2.32)

with c now being independent of a, t and f .Let us fix t in (2.32) and consider a as a variable over Ω. Applying Holder’s inequality for sums (see (1.6)) to

(2.32) we have

∣∣f (j)(t)∣∣p ≤ c

∣∣f (j)(a)∣∣p +

‖f‖pp,α,N

(t(1 − t))p(j−α+1/p) · (a(1 − a))p(j−α+1/p)

, (2.33)

where c is independent of a, t and f . Choose β < α, multiply each side of (2.33) by (a(1 − a))p(j−β) andintegrate with respect to a over (0, 1) to give

∣∣f (j)(t)∣∣p ·∫ 1

0

(a(1 − a))p(j−β) da

≤ c ·‖f‖p

p,β,N +‖f‖p

p,α,N

(t(1 − t))p(j−α+1/p)·∫ 1

0

(a(1 − a))p(α−β)−1 da

.

(2.34)

The integral on the left of (2.34) exists since, for j ∈ 0, 1, . . . , N −1, we shall have p(j − β) > −1; it isobviously non-zero. Again, the integral on the right-hand side of (2.34) exists since p(α− β)− 1 > −1. Finally,since ‖f‖p,β,N ≤ (1/4)α−β · ‖f‖p,α,N by Theorem 2.4 (ii), we find from (2.34) that∣∣f (j)(t)

∣∣ ≤ c ‖f‖p,α,N · (t(1 − t))α−1/p−j (2.35)

for t ∈ Ω, where c is independent of both f and t. This proves the theorem.

One further result is required before starting the analysis of §3.

Theorem 2.6 Suppose f ∈ W(N)p,α (Ω) with 1 < p < ∞, α ∈ R is such that 0 < α + 1/q < 1 and N ∈ N.

Then the following statements hold:(i) For r ∈ 0, 1, . . . , N−1 the equalities

(ρr+1f (r)

)(0+) =

(ρr+1f (r)

)(1−) = 0 are valid.

(ii) For all r ∈ 0, 1, . . . , N we have that ρrf (r) ∈ W(N−r)p,α (Ω) with∥∥ρrf (r)

∥∥p,α,N−r

≤ c ‖f‖p,α,N , (2.36)

for some constant c which is independent of f .

P r o o f. (i) From (1.4) and (2.23) we have at once

ρr+1(t) · ∣∣f (r)(t)∣∣ ≤ c ‖f‖p,α,N · (t(1 − t))α+1/q , t ∈ Ω , (2.37)

with c being independent of t. Since we are assuming that α + 1/q > 0 it follows immediately that

limt→0+

ρr+1(t) · ∣∣f (r)(t)∣∣ = 0 ;



that is,(ρr+1f (r)

)(0+) = 0. A similar result holds at the end-point 1 so that (i) is proved.

(ii) In order to prove that ρrf (r) ∈ W(N−r)p,α (Ω) and (2.36) we need to show that

Ik :=(∫ 1

0

(ρk−α(t) ·

∣∣∣(ρrf (r))(k)(t)

∣∣∣)p

dt

)1/p

≤ c ‖f‖p,α,N ,

for k ∈ 0, 1, . . . , N−r. Fix t ∈ Ω. On using Leibniz’ theorem we have

(ρrf (r)

)(k)(t) =k∑

j=0

(k

j

)(ρr)(j)(t) · f (r+k−j)(t) . (2.38)

From Lemma 2.3 (i) we have that

ρk−α(t) ·∣∣∣(ρrf (r)

)(k)(t)∣∣∣ ≤ c

k∑j=0

(k

j

)ρr+k−j−α(t) · ∣∣f (r+k−j)(t)

∣∣ , (2.39)

c being independent of f . On using Holder’s inequality for sums (1.6) we find that

(ρk−α(t) ·

∣∣∣(ρrf (r))(k)(t)

∣∣∣)p

≤ c

k∑j=0

(ρr+k−j−α(t) · ∣∣f (r+k−j)(t)

∣∣)p

, (2.40)

the positive constant c again being independent of f . Since f ∈ W(N)p,α (Ω) it follows from (2.40) firstly that Ik is

finite for k ∈ 0, 1, . . . , N−r) so that ρrf (r) ∈ W(N−r)p,α (Ω) and secondly that

∥∥ρrf (r)∥∥

p,α,N−r≤ c ‖f‖p,α,N ,

the constant c being independent of f , as required.

In the light of this discussion it is useful to restate Theorem 1.1.

Theorem 2.7 Let 1 < p < ∞ and α ∈ R be such that 0 < α + 1/q < 1. Then T is a bounded lin-

ear operator from W(0)p,α(Ω) into itself; that is, there exists a positive constant c, independent of f , such that

‖Tf‖p,α,0 ≤ c ‖f‖p,α,0 for all f ∈ W(0)p,α(Ω).

In the next section we shall prove a generalisation of this result, this being that if 1 < p < ∞, 0 < α+1/q < 1and N ∈ N0 then T is a bounded linear operator from W

(N)p,α (Ω) into itself.

3 T as an operator on W (N)p,α (Ω)

Before stating a well known lemma (Lemma 3.2) we need to introduce the (unweighted) Sobolev space whichwe shall denote by W

(1)r (Ω).

Definition 3.1 When 1 < r < ∞, a function g ∈ Lr(Ω) is said to be in the Sobolev space W(1)r (Ω) if its

distributional derivative g′ also belongs to Lr(Ω). In this case

∫ 1

0

|g(t)|r dt < ∞ and∫ 1

0

|g′(t)|r dt < ∞ . (3.1)

We now come to the lemma.

Lemma 3.2 Suppose g ∈ W(1)r (Ω) for 1 < r < ∞ with g(0+) = g(1−) = 0. Then

(Tg)′ = Tg′ on Ω . (3.2)

P r o o f. See, for example, Mikhlin and Prossdorf [5, Chapter 2, Lemma 6.1].

We shall apply this lemma to the function ρf but in order to do so we first need the following result.



Theorem 3.3 Suppose f ∈ W(N)p,α (Ω) where 1 < p < ∞, 0 < α + 1/q < 1 and N ∈ N. Then there exists

an r ∈ (1,∞) with

0 < max1 − (α + 1/q), 1/p

< 1/r < 1 , (3.3)

such that ρf ∈ W(1)r (Ω).

P r o o f. We need to show (see(3.1)) that both

I0 :=∫ 1

0

∣∣(ρf)(t)∣∣r dt < ∞ (3.4)

and

I1 :=∫ 1

0

∣∣(D(ρf))(t)∣∣r dt < ∞ , (3.5)

for some r ∈ (1,∞) satisfying (3.3). Now

I0 =∫ 1

0

∣∣ρ−α(t) f(t)∣∣r · ρ(1+α)r(t) dt

≤(∫ 1

0

∣∣ρ−α(t) f(t)∣∣p dt

)r/p(∫ 1

0

ρr(1+α)p/(p−r)(t) dt

)(p−r)/p

by Holder’s inequality (1.5), since r is chosen (see (3.3)) so that 1 < p/r < ∞. Now the second integral on theright-hand side is finite provided that

r(1 + α)pp − r

> −1 or r(α + 1/q) > −1 . (3.6)

Since r > 1 and α + 1/q > 0 the last inequality is trivially true so that

I0 ≤ c ‖f‖rp,α,N < ∞ , (3.7)

where c is independent of f . Consider now I1 (see(3.5)). We have

I1 =∫ 1

0

∣∣ρ′(t)f(t) + ρ(t)f ′(t)∣∣r dt .

Since supt∈Ω |ρ′(t)| = 1 we have from (1.6) that

I1 ≤ 2r−1

(∫ 1

0

∣∣ρ−α(t) f(t)∣∣r · ραr(t) dt +

∫ 1

0

∣∣ρ1−α(t) f ′(t)∣∣r · ραr(t) dt

).

Applying Holder’s inequality (1.5) to each integral we find that

I1 ≤ c ‖f‖rp,α,N

(∫ 1

0

ραrp/(p−r)(t) dt

)(p−r)/p

, (3.8)

where c is independent of f . Again, the integral on the right-hand side is finite provided that

αrp

p − r> −1 or

1r

> 1 − (α + 1/q) > 0 .

This is true by assumption, so we also have that

I1 ≤ c ‖f‖rp,α,N < ∞ ,

for some c independent of f . Thus (3.4) and (3.5) are satisfied and the theorem is proved.



This theorem, together with Theorem 2.6 (i) and Lemma 3.2, gives the following important result.

Theorem 3.4 Suppose f ∈ W(N)p,α (Ω) where 1 < p < ∞, 0 < α + 1/q < 1 and N ∈ N. Then

DT (ρf) = T((ρf)′

)on Ω . (3.9)

P r o o f. From Theorem 2.6 (i) we have that (ρf)(0+) = (ρf)(1−) = 0, and from Theorem 3.3 we have thatρf ∈ W

(1)r (Ω) for some 1 < r < ∞. Since ρf satisfies the conditions of Lemma 3.2, Equation (3.9) follows at

once.

At this stage, one further definition is required.

Definition 3.5 For r ∈ 1, 2, . . . , N, we define on Ω

∆rf := T(ρrf (r)

)− ρrDr(Tf) , (3.10)

provided it exists.

We now need to identify ∆rf for each f ∈ W(N)p,α (Ω) and this we shall do in stages. First let us consider ∆1f

which will be shown to be a constant function.

Theorem 3.6 Suppose f ∈ W(N)p,α (Ω) for 1 < p < ∞, 0 < α + 1/q < 1 and N ∈ N. Then

∆1f =1π

∫ 1

0

f(τ) dτ (3.11)

and

|∆1f | ≤ c ‖f‖p,α,N , (3.12)

for some positive constant c independent of f .

P r o o f. From (3.10) we have

∆1f = T (ρf ′) − ρD(Tf) . (3.13)

From Theorem 3.4,

DT (ρf) = T ((ρf)′) = T (ρ′f + ρf ′) . (3.14)

Consider, for t ∈ Ω,

ρ′(t)(Tf)(t) − (T (ρ′f))(t) =1π

∫ 1

0

ρ′(t) − ρ′(τ)τ − t

f(τ) dτ =2π

∫ 1

0

f(τ) dτ , (3.15)

since ρ(t) = t − t2 (see (2.1)). Hence

T (ρ′f) = ρ′T (f)− 2π

∫ 1

0

f(τ) dτ

and consequently from (3.14) we have

T (ρf ′) = DT (ρf)− ρ′T (f) +2π

∫ 1

0

f(τ) dτ .

Substituting this equation into (3.13) gives

∆1f =2π

∫ 1

0

f(τ) dτ + D[T (ρf) − ρT (f)

]. (3.16)



Again, since ρ(t) = t − t2 we have

T (ρf)(t) − ρ(t)(Tf)(t) =(1 − t)

π

∫ 1

0

f(τ) dτ − 1π

∫ 1

0

τf(τ) dτ (3.17)

for every t ∈ Ω, so that

D[T (ρf)− ρT (f)

]= − 1

π

∫ 1

0

f(τ) dτ . (3.18)

From (3.16) and (3.18) we obtain (3.11). Now

|∆1f | ≤ 1π

∫ 1

0

(τ(1 − τ))−α · |f(τ)| · (τ(1 − τ))α dτ

≤ 1π

(∫ 1

0

((τ(1 − τ))−α · |f(τ)|)p dτ

)1/p(∫ 1

0

(τ(1 − τ))αq dτ

)1/q

by Holder’s inequality (1.5). Again, since 0 < α + 1/q < 1 we have αq > −1 so that the second integral on

the right-hand side is finite. Choosing c to be (1/π)( ∫ 1

0(τ(1 − τ))αq dτ

)1/qgives |∆1f | ≤ c ‖f‖p,α,N which is

(3.12). This completes the proof of the theorem.

Before proceeding we shall make one further observation. From Theorem 2.6 (ii) we have that if f ∈ W(N)p,α (Ω)

for some N ∈ N then ρrf (r) ∈ W(N−r)p,α (Ω) for r ∈ 0, 1, . . . , N so that from Theorem 3.6 we would have that∣∣∆1

(ρrf (r)

)∣∣ ≤ c∥∥ρrf (r)

∥∥p,α,N−r

≤ c ‖f‖p,α,N . (3.19)

We shall need this result when considering ∆2f , ∆3f, . . . .

Theorem 3.7 Suppose f ∈ W(N)p,α (Ω) for 1 < p < ∞, 0 < α + 1/q < 1 and N ∈ N with N ≥ 2. Then

∆2f = 3∆1(ρf ′) − ρ′∆1f (3.20)

on Ω and

supt∈Ω

|(∆2f)(t)| ≤ c ‖f‖p,α,N , (3.21)


P r o o f. On replacing f by ρf ′ in (3.10) with r = 1, we find

∆1(ρf ′) = T(ρ(ρf ′)′

)− ρDT (ρf ′) = T(ρ2f ′′ + ρρ′f ′)− ρD

[∆1f + ρD(Tf)

]on using (3.13) for T (ρf ′). Consequently

∆1(ρf ′) = T (ρ2f ′′) − ρ2D2(Tf) + T (ρρ′f ′) − ρρ′D(Tf) (3.22)

since ∆1f is a constant function. Recalling the definition of ∆2f from (3.10) we see from (3.22) that

∆2f = ∆1(ρf ′) + ρρ′D(Tf)− T (ρρ′f ′) . (3.23)

But again from (3.10) with r = 1 using the expression for ρD(Tf) we find

∆2f = ∆1(ρf ′) + ρ′(T (ρf ′) − ∆1f

)− T (ρρ′f ′) ,

or

∆2f = ∆1(ρf ′) − ρ′∆1f +[ρ′T (ρf ′) − T (ρρ′f ′)

]. (3.24)

From (3.11) and (3.15) with ρf ′ in place of f , it follows that

ρ′T (ρf ′) − T (ρρ′f ′) = 2∆1(ρf ′) . (3.25)



From (3.24) and (3.25) we recover (3.20).We observe from (3.20) that ∆2f is a polynomial of degree 1. With (3.12) and (3.19) we see from (3.20)

that supt∈Ω |(∆2f)(t)| ≤ c ‖f‖p,α,N for some constant c independent of f , which completes the proof of thetheorem.

This now paves the way for a more general result.

Theorem 3.8 Suppose f ∈ W(N)p,α (Ω) where 1 < p < ∞, 0 < α + 1/q < 1 and N ∈ N. If r ∈ N and

r ≤ N then ∆rf, as defined in (3.10), is a polynomial of degree (r − 1) such that

supt∈Ω

|(∆rf)(t)| ≤ c ‖f‖p,α,N , (3.26)


P r o o f. We shall first show that for r ∈ 1, 2, . . . , N−1,∆r+1f = ∆r(ρf ′) − rρ′ · ∆rf + r(r − 1)ρ · ∆r−1f

+(

3r + 12

)∆1

(ρrf (r)

)+

r(r − 1)2

ρ′ · ∆1

(ρr−1f (r−1)

),

(3.27)

with the understanding that ∆0f = 0. Here, we might observe that putting r = 1 in (3.27) recovers (3.20) for∆2f . We also observe, since ρ is a polynomial of degree 2, that if ∆r−1f and ∆rf are polynomials of degree(r − 2) and (r − 1) respectively then ∆r+1f becomes a polynomial of degree r. On replacing f by ρf ′ in (3.10)we obtain that

∆r(ρf ′) = T(ρr(ρf ′)(r)

)− ρrDr[T (ρf ′)

]= T

(ρr+1f (r+1) + rρ′ρrf (r) − r(r − 1)ρrf (r−1)

)− ρrDr[∆1f + ρD(Tf)

],

by using Leibniz’ theorem for the derivative of a product (recall that ρ′′ = −2) and (3.13). On using Leibniz’theorem again on the term Dr(ρD(Tf)) and recalling from (3.10) the definition of ∆r+1f we find that

∆r+1f = ∆r(ρf ′) + r[ρrρ′Dr(Tf) − T

(ρ′ρrf (r)

)]+ r(r − 1)

[T(ρrf (r−1)

)− ρrDr−1(Tf)].

(3.28)

It follows from (3.10) that

∆r+1(f) = ∆r(ρf ′) + r[ρ′T(ρrf (r)

)− T(ρ′ρrf (r)

)− ρ′∆rf]

+ r(r − 1)[T(ρrf (r−1)

)− ρT(ρr−1f (r−1)

)+ ρ∆r−1f

].

(3.29)

Now, on using (3.11) and (3.15) with f replaced by ρrf (r) we have, for each t ∈ Ω,

ρ′(t) · T (ρrf (r))(t) − T

(ρ′ρrf (r)

)(t) = 2∆1

(ρrf (r)

)(t) . (3.30)

Similarly applying (3.11) and (3.17) with f replaced by ρr−1f (r−1) gives, for each t ∈ Ω,

T(ρrf (r−1)

)(t) − ρT

(ρr−1f (r−1)

)(t)

= (1 − t)∆1

(ρr−1f (r−1)

)(t) − ∆1

(τρr−1f (r−1)

)(t) .

(3.31)

On substituting (3.30) and (3.31) for the corresponding terms in (3.29) we obtain that(∆r+1f

)(t) = ∆r

(ρf ′)(t) − rρ′(t) · (∆rf)(t)

+ r(r − 1)ρ(t) · (∆r−1f)(t) + 2r∆1

(ρrf (r)

)(t)

+ r(r − 1)(1 − t) · ∆1

(ρr−1f (r−1)

)(t) − r(r − 1)∆1

(τρr−1f (r−1)

)(t) ,

(3.32)



for every t ∈ Ω. But consider ∆1

(ρrf (r)

)= (1/π)

∫ 1

0 ρr(τ)f (r)(τ) dτ (see (3.11)). Integration by parts gives

∆1

(ρrf (r)

)= 2r ∆1

(τρr−1f (r−1)

)− r∆1

(ρr−1f (r−1)

), (3.33)

on using Theorem 2.6 (i). On substituting for ∆1

(τρr−1f (r−1)

)from (3.33) into (3.32) we recover (3.27).

To prove (3.26) we shall proceed by induction. Let us assume that if either s = N−1 or s = N then

supt∈Ω

|(∆kf)(t)| ≤ c ‖f‖p,α,s , f ∈ W (s)p,α(Ω) , (3.34)

for k ∈ 1, 2, . . . , r. In Theorems 3.6 and 3.7 we have shown that (3.34) is true with r = 2. Since by Theorem

2.6 (ii) we have ρf ′ ∈ W(N−1)p,α (Ω), it follows from the above assumption and (2.36) that

supt∈Ω

|(∆r(ρf ′))(t)| ≤ c ‖ρf ′‖p,α,N−1 ≤ c ‖f‖p,α,N , (3.35)

(cf. (3.19)). Since supt∈Ω |ρ(t)| = 1/4 and supt∈Ω |ρ′(t)| = 1 we find from (3.19), (3.27), (3.34) and (3.35)that supt∈Ω

∣∣(∆r+1f)(t)∣∣ ≤ c ‖f‖p,α,N for some c independent of f , so that by the principle of mathematical

induction (3.26) is satisfied for r ∈ 1, . . . , N.

We now come to the principal result of this paper.

Theorem 3.9 Suppose 1 < p < ∞, α ∈ R is such that 0 < α + 1/q < 1 and N ∈ N0. Then T is a bounded

linear operator from W(N)p,α (Ω) into itself.

P r o o f. The case when N = 0 is, of course, the well known Theorem 2.7 (or Theorem 1.1). We needto show that for every N ∈ N there exists a positive constant c, independent of f ∈ W

(N)p,α (Ω), such that

‖Tf‖p,α,N ≤ c ‖f‖p,α,N .Choose any r ∈ 1, 2, . . . , N. Then, by (3.10),∫ 1

0

∣∣ρr−α(t)Dr(Tf)(t)∣∣p dt =

∫ 1

0

ρ−αp(t)∣∣T (ρrf (r)

)(t) − (∆rf)(t)

∣∣p dt

≤ c ·∫ 1

0

ρ−αp(t)[ ∣∣T (ρrf (r)

)(t)∣∣p +

∣∣(∆rf)(t)∣∣p] dt ,

(3.36)

on using (1.6). From Theorem 2.6 (ii) we see that if f ∈ W(N)p,α (Ω) then ρrf (r) ∈ W

(N−r)p,α (Ω) ⊆ W

(0)p,α(Ω) so

that by Theorem 2.7 we have T(ρrf (r)

) ∈ W(0)p,α(Ω) and

∫ 1

0

∣∣ρ−α(t) · T (ρrf (r))(t)∣∣p dt ≤ c

∥∥ρrf (r)∥∥p

p,α,0≤ c ‖f‖p

p,α,N . (3.37)

On the other hand since, from Theorem 3.8, the inequality supt∈Ω |(∆rf)(t)| ≤ c ‖f‖p,α,N holds, we have that

∫ 1

0

ρ−αp(t) |(∆rf)(t)|p dt ≤ c ‖f‖pp,α,N

∫ 1

0

ρ−αp(t) dt ≤ c ‖f‖pp,α,N . (3.38)

Here recall that α + 1/q < 1 implies that −αp > −1 so that∫ 1

0 ρ−αp(t) dt < ∞. From (3.36)–(3.38) we findthat for all r ∈ 1, 2, . . . , N

(∫ 1

0

∣∣ρr−α(t) · Dr(Tf)(t)∣∣p dt

)1/p

≤ c ‖f‖p,α,N , (3.39)

which proves the theorem.

There is an immediate consequence of this theorem which is worth noting but first we must define theHadamard finite-part integral of integer order.



Definition 3.10 For any m ∈ N, the Hadamard finite-part integral will be denoted and defined by

(Hmf)(t) :=1π

∫ 1

0

=f(τ) dτ

(τ − t)m:=

1(m − 1)!

(Dm−1Tf

)(t) (3.40)

for t ∈ Ω.In particular we note that H1f = Tf .

Corollary 3.11 Suppose 1 < p < ∞, α ∈ R, 0 < α + 1/q < 1 and N ∈ N. If m ∈ N and m ≤ N then

Hm is a bounded linear operator from W(N)p,α (Ω) into W

(N−m+1)p,α−m+1 (Ω) with

‖Hmf‖p,α−m+1,N−m+1 ≤ c ‖f‖p,α,N (3.41)

for every f ∈ W(N)p,α (Ω) and for some constant c independent of f .

P r o o f. Let us recall that f ∈ W(N)p,α (Ω) implies that∫ 1

0

((t(1 − t))k−α · ∣∣f (k)(t)

∣∣)p dt ≤ ‖f‖pp,α,N (3.42)

for k ∈ 0, 1, . . . , N. Fix j ∈ 0, 1, . . . , N. Writing k = l + j, we may express (3.42) as∫ 1

0

((t(1 − t))l−(α−j) ·

∣∣∣(f (j))(l)(t)∣∣∣)p

dt ≤ ‖f‖pp,α,N (3.43)

for l ∈ 0, 1, . . . , N − j. We observe that (3.43) immediately implies that f (j) ∈ W(N−j)p,α−j (Ω) and furthermore

that ∥∥f (j)∥∥

p,α−j,N−j≤ ‖f‖p,α,N . (3.44)

On the other hand, Theorem 3.9 gives that if f ∈ W(N)p,α (Ω) then

‖Tf‖p,α,N ≤ c ‖f‖p,α,N , (3.45)

for some c independent of f . Since Dm−1(Tf) = (m − 1)! Hmf we see from (3.44) and (3.45) that

‖Hmf‖p,α−m+1,N−m+1 =1

(m − 1)!

∥∥Dm−1(Tf)∥∥

p,α−m+1,N−m+1

≤ 1(m − 1)!

‖Tf‖p,α,N

≤ c ‖f‖p,α,N ,

for some c independent of f . This establishes (3.41).

Acknowledgements The authors wish to thank an unknown referee for his/her excellent comments on a previous draft ofthis paper. The second author’s research, leading to the publication of this paper, was carried out when he was a staff memberof the University of New South Wales, supported by a grant from the Australian Research Council (ARC).

References

[1] R. A. Adams, Sobolev Spaces (Academic Press, New York, 1975).[2] P. L. Butzer and R. J. Nessel, Fourier Analysis and Approximation, Vol. 1 (Birkhauser-Verlag, Basel, 1971).[3] I. Gohberg and N. Krupnik, One-dimensional Linear Singular Integral Equations, Vol. 1 (Birkhauser-Verlag, Basel, 1992).[4] A. Kufner, Weighted Sobolev Spaces (Wiley-Interscience, Chichester, 1985).[5] S. G. Mikhlin and S. Prossdorf, Singular Integral Operators (Springer-Verlag, Berlin, 1986).[6] F. Treves, Topological Vector Spaces, Distributions and Kernels (Academic Press, New York, 1967).


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The finite Hilbert transform and weighted Sobolev spaces