(1)http://jipam.vu.edu.au/ Volume 5, Issue 1, Article 18, 2004 ASYMPTOTIC BEHAVIOR OF THE APPROXIMATION NUMBERS OF THE HARDY-TYPE OPERATOR FROM Lp INTOLq (cases 1&lt

http://jipam.vu.edu.au/

Volume 5, Issue 1, Article 18, 2004

ASYMPTOTIC BEHAVIOR OF THE APPROXIMATION NUMBERS OF THE HARDY-TYPE OPERATOR FROM L^p INTOL^q

(cases 1< p ≤q≤2,2≤p≤q <∞ and 1< p ≤2≤q <∞)

J. LANG, O. MENDEZ, AND A. NEKVINDA

DEPARTMENT OFMATHEMATICS, 100 MATHTOWER, 231 WEST18THAVE. COLUMBUS, OH 43210-1174, USA.

lang@math.ohio-state.edu

W. UNIVERSITYAVE., 124 BELLHALL, UNIVERSITY OFTEXAS ATELPASO

EL, PASO, TX 79968, USA.

mendez@math.utep.edu FACULTY OFCIVILENGINEERING, CZECHTECHNICALUNIVERSITY, THAKUROVA7,

166 29 PRAGUE6, CZECHREPUBLIC. nekvinda@fsv.cvut.cz

Received 17 December, 2003; accepted 04 February, 2004 Communicated by D. Hinton

ABSTRACT. We consider the Hardy-type operator (T f) (x) :=v(x)

Z x

a

u(t)f(t)dt, x > a,

and establish properties ofTas a map fromL^p(a, b)intoL^q(a, b)for1< p≤q≤2,2≤p≤ q <∞and1< p ≤2 ≤q <∞. The main result is that, with appropriate assumptions onu andv, the approximation numbersa_n(T)ofTsatisfy the inequality

c₁ Z b

a

|uv|^rdt≤lim inf

n→∞ na^r_n(T)≤lim sup

n→∞

na^r_n(T)≤c₂ Z b

a

|uv|^rdt when1< p≤q≤2or2≤p≤q <∞, and in the case1< p≤2≤q <∞we have

lim sup

n→∞

na^r_n(T)≤c3

Z d

0

|u(t)v(t)|^rdt and

c₄ Z d

0

|u(t)v(t)|^rdt≤lim inf

n→∞ n(1/2−1/q)r+1a^r_n(T),

wherer= _p^p0+q^0q and constantsc₁, c₂, c₃, c₄. Upper and lower estimates for thel^sandl^s,knorms of{a_n(T)}are also given.

Key words and phrases: Approximation numbers, Hardy operator, Voltera operator.

2000 Mathematics Subject Classification. Primary 46E30; Secondary 47B38.

ISSN (electronic): 1443-5756 c

2004 Victoria University. All rights reserved.

J. Lang wishes to thank the Leverhulme Foundation for its support and also to the Ohio State University for support via SRA..

A. Nekvinda was supported by MSM 210000010.

172-03

1. INTRODUCTION

The operatorT :L^p(a, b)→L^q(a, b)(where0≤a≤b≤d <∞) defined by

(1.1) T f(x) =v(x)

Z x 0

u(t)f(t)dt

was studied in [1] and [5], in the case 1 ≤ p ≤ q ≤ ∞, for real–valued functions u ∈ L^p0(0, c), v ∈ L^p(c, d), for any c ∈ (0, d) and p⁰ = p/(p−1). In the aformentioned works, the following estimates for the approximation numbersa_n(T)ofT were obtained:

(1.2) aN(ε)+3 ≤σpε,

(1.3) a_N(ε)−1 ≥ν_q(N(ε)−1)^1q−1pε, forp < q <∞ and

(1.4) a_N(ε)/2−1 ≥ε/2, forp=q,

whereσ_p, ν_q, are constants depending onq, andN(ε)is anε-depending natural number . In the case p = q, these results are sharp and are used in [2] and [5] to obtain asymptotic results for the approximation numbers.

Specifically, it was proved in [2] that forp=q= 2

(1.5) lim

n→∞na_n(T) = 1 π

Z d 0

|u(t)v(t)|dt and that for1< p=q <∞,

(1.6) 1

4αp

Z d 0

|u(t)v(t)|dt≤lim inf

n→∞ nan(T)≤lim sup

n→∞

nan(T)≤αp

Z d 0

|u(t)v(t)|dt.

The endpoint cases were studied in [5]: it was shown there that forp= q = ∞(and similarly forp=q= 1)

(1.7) 1

4 Z d

0

|u(t)v_s(t)|dt ≤lim inf

n→∞ na_n(T)≤lim sup

n→∞

na_n(T)≤ Z d

0

|u(t)v_s(t)|dt, where

v_s(t) = lim

ε→0₊kv χ(t−ε,t+ε)k_L^∞. Ifp < q, the estimates (1.2) and (1.3) are not sharp.

The estimates (1.2) and (1.3) were used in [7] to obtain the following asymptotic results for the approximation numbers in the case1< p < q <∞:

(1.8) lim sup

n→∞

na^r_n(T)≤c_p,q Z d

0

|u(t)v(t)|^rdt and

(1.9) ≤d_p,q

Z d 0

|u(t)v(t)|^rdt≤lim inf

n→∞ n^(1p−1q)r+1a^r_n(T) wherer =pq⁰/(q+p⁰).

Since the estimates upon which they are based are not sharp, these results are not sharp either, in contrast to (1.5), (1.6). Our research is directed toward finding alternative, refined versions of (1.2) and (1.3) in the casep < q, aiming to get better asymptotic results than (1.8) and (1.9).

In this paper, we succeed in showing that for1≤p≤q≤ ∞,

(1.10) a_N(ε)+1 ≤2ε,

and for1≤p≤q ≤2or2≤p≤q≤ ∞

(1.11) a_N(ε)/4−1 ≥cε,

and for1< p≤2≤q <∞

(1.12) a_N(ε)/4−1 ≥cεN(ε)^12−1q,

where cis a constant independent of ε andN(ε). And under some condition on u andv we show that for1≤p≤q≤2or2≤p≤q≤ ∞

c₁ Z b

a

|uv|^r ≤lim inf

n→∞ na^r_n(T)≤lim inf

n→∞ na^r_n(T)≤c₂ Z b

a

|uv|^r, and for1< p≤2≤q <∞

lim sup

n→∞

na^r_n(T)≤c_p,q Z d

0

|u(t)v(t)|^rdt and

d_p,q Z d

0

|u(t)v(t)|^rdt≤lim inf

n→∞ n^(12−1q)r+1a^r_n(T), wherer = _p^p0⁰+q^q . We also describel^randl^r,snorms of{a_n}^∞_n=1.

Under much stronger conditions on u and v in neighborhood of boundary points of I this problem was also studied in [6] by using different techniques.

2. PRELIMINARIES

Throughout this paper we will suppose that 1 < p ≤ q ≤ 2. In what follows we shall be concerned with the operator T defined in (1.1) as a map fromL^p(0, d)intoL^q(0, d) where 0< d≤ ∞. The functionsu, vare subject to the following restrictions: for allx∈(0, d)

(2.1) u∈L^p0(0, x),

and

(2.2) v ∈L^q(x, d).

It is well-known that these assumptions guarantee thatT is well defined (see (1.9)). Moreover, the norm of this operator is equivalent to:

J := sup

x∈(0,d)

Z x 0

|u(t)|^p0dt

_p¹0 Z d x

|v(t)|^qdt ¹_q

,

(see [4], [8] and [5]). We define the operatorT_I by (2.3) T_If(x) :=v(x)χ_I(x)

Z x 0

u(t)f(t)χ_I(t)dt, x >0, whereI = (a, b)⊂(0, d), and the quantity

(2.4) J(I)≡J(a, b) := sup

x∈I

Z x a

|u(t)|^p0dt

_p¹0 Z d x

|v(t)|^qdt ¹q

.

It is obvious that J(I) ≈ kT_Ikp→q, where the symbol≈ indicates that the quotient of the two sides is bounded above and below by positive constants.

Proposition 2.1. There are two positive constantsK₁,K₂ such that for anyI = (a, b)⊂(0, d) the inequality

K1J(a, b)≤ kTIk ≤K2J(a, b) holds.

We start by proving an important continuity property ofJ:

Lemma 2.2. Suppose that (2.1) and (2.2) are satisfied. Then the functionJ(·, b)is continuous and non–increasing on(0, b), for anyb≤ ∞.

Proof. It is easy to verify thatJ(·, b)is non-increasing on (0, b). To prove the continuity ofJ, fixx∈(0, b)anε >0. By (2.1) and (2.2) there exists0< h₀ <min{x, b−x}such that

Z x x−h0

|u(t)|^p0dt _p¹0

kvkq,(x−h₀,x)< ε.

It follows that forh,0< h < h0, J(x, b)≤J(x−h, b)

= sup

x−h<z<b

Z z x−h

|u(t)|^p0dt _p¹0

kvk_q,(z,b)

= max

sup

x−h<z<x

Z z x−h

|u(t)|^p0dt _p¹0

kvk_q,(z,b),

sup

x<z<b

Z x x−h

+ Z z

x

|u(t)|^p0dt _p¹0

kvk_q,(z,b)

≤max{ε, ε+J(x, d)}=ε+J(x, d), (2.5)

which yields0< J(x−h, b)−J(x, b)< ε. The inequality0< J(x, b)−J(x+h, b)< εcan

be proved analogously.

For the sake of completeness, we include the following known result (see [4] and [9]):

Proposition 2.3. The operator T defined by (1.1), with1 < p < ∞ andu, v satisfying (2.1), (2.2) andJ <∞is a compact map fromL^p(0, d)intoL^q(0, d)if and only if lim_c→0+J(0, c) = lim_c→d−J(c, d) = 0.

In what followsA(I) is a function defined on all sub-intervalsI = (a, b) ⊂ (0, d), defined by

(2.6) A(I) =A(a, b) := sup

kfk_p,I=1

α∈<inf kT f −αvk_p,I.

A similar function can be found in [5]. Next, we prove some basic properties ofA(I). Choosing α= 0in (2.6) we immediately obtain for anyI = (a, b),0≤a < b ≤d,

(2.7) A(I)≤ kTIk.

Lemma 2.4. LetI = (a, b)andkuk_p⁰_,I <∞,kvk_q,I <∞. Set A(I) =e sup

kfkp,I=1

|α|≤2kukinf _p0,I

kT f −αvkp,I. ThenA(I) = A(I).e

Proof. Hölder’s inequality yields kT_Ik= sup

kfk_p,I=1

Z b a

Z x a

f(t)u(t)dt

q

dx ¹_q

≤ sup

kfk_p,I=1

Z b a

|v(x)|^q Z x

a

|f(t)|^pdt

^q_pZ x a

|u(t)|^p0dt _p^q0

dx

!¹_q

≤ Z b

a

|v(x)|^q Z b

a

|u(t)|^p0dt

q p0

dx

!¹_q

=kuk_p⁰_,Ikvk_q,I.

Ifkvk_q,I = 0 thenA(I) = A(I) = 0. Assumee kvk_q,I > 0. Letkfk_p,I = 1 and suppose that

|α| > 2kuk_p⁰_,I. Then|α| ≥ 2_kvk^kTIk

q,I and using the trivial inequality|a−b|^q ≥ 2^1−q|a|^q− |b|^q valid for any real numbersa,bwe obtain for eachα∈ <

Z b a

α−

Z x a

f(t)u(t)dt

v(x)

q

dx≥ Z b

a

|αv(x)| −

Z x a

f(t)u(t)dt

q

dx

≥2^1−q|α|^q Z b

a

|v(x)|^qdx− Z b

a

v(x) Z x

a

f(t)u(t)dt

q

dx

>2^1−q

2 kT_Ik kvk_q,I

qZ b a

|v(x)|^qdx− kT_Ik^q =kT_Ik^q. In conjunction with (2.7), the above yields

kTIk ≥A(I)

= sup

kfk_p,I=1

min (

inf

|α|≤2kuk_p0,I

Z b a

α−

Z x a

f(t)u(t)dt

v(x)

q

1 q

,

inf

|α|>2kuk_p0,I

Z b a

α−

Z x a

f(t)u(t)dt

v(x)

q

1 q)

= inf

|α|≤2kuk_p0,I

Z b a

α−

Z x a

f(t)u(t)dt

v(x)

q

1 q

=A(I),e

which finishes the proof.

Lemma 2.5. Let u andv satisfy (2.1) and (2.2) respectively. Then A(I₁) ≤ A(I₂), provided I₁ ⊂I₂. Moreover, given0< b < dthe functionA(·, b)is continuous on(0, b).

Proof. Let0≤a₁ ≤a₂ < b₂ ≤b₁ ≤d,I₁ = (a₁, b₁),I₂ = (a₂, b₂). Then A(I₁) = sup

kfk_p,I1=1 α∈<inf

Z b1

a1

v(x) Z x

a1

(f(t)u(t)dt−α)

q

dx ¹_q

≥ sup

kf χ_I

2k_p,I

1=1 α∈<inf

Z b1

a1

v(x) Z x

a1

(f(t)u(t)dt−α)

q

dx ¹q

≥ sup

kfk_p,I2=1 α∈<inf

Z b2

a2

v(x) Z x

a2

(f(t)u(t)dt−α)

q

dx ¹_q

=A(I₂) which proves the first part of Lemma 2.5.

For the remaining statement, fixb ∈ (0, d) and0 < y < b. Letε > 0. By (2.1) and (2.2) there exists0< h₀such that0< y−h₀and

Z y y−h₀

|u|^p0 < εand Z y

y−h₀

|v|^q < ε.

Set Dh = 2kuk_p⁰,(y−h,b) for any 0 ≤ h < y. Recall that by (2.1), one has Dh < ∞ for 0 ≤ h < d. Using the trivial inequality (a+b)^1q ≤ a^1q +b^1q, the triangle inequality and the Hölder inequality, it follows that

A(y, b)≤A(y−h, b)

= sup

kfk_p,(y−h,b)=1 α∈<inf

Z b y−h

α−

Z x y−h

f(t)u(t)dt

v(x)

q

dx ¹_q

= sup

kfk_p,(y−h,b)=1

inf

|α|≤Dh

Z y y−h

α−

Z x y−h

f(t)u(t)dt

v(x)

q

dx

+ Z b

y

Z y y−h

f(t)u(t)dt+ Z x

y

f(t)u(t)dt−α

v(x)

q

dx ¹_q

≤ sup

kfk_p,(y−h,b)=1

inf

|α|≤D_h







"

Z y y−h

|v(x)|^q Z x

y−h

|u(t)|^p0dt

_p^q0 Z x y−h

|f(t)|^pdt ^q_p

dx

#¹_q

+

|α|^q Z y

y−h

|v(x)|^qdx ¹_q

+

"

Z b y

|v(x)|^qdx Z y

y−h

|u(t)|^p0dt

_p^q0 Z y y−h

|f(t)|^pdt _p^q#¹_q

+ Z b

y

v(x) Z x

y

f(t)u(t)dt−α

q

dx ¹q)

≤ (

ε^1+p10 +D_hε^1q +kvk_q,(y,b)ε^p10

+ sup

kfk_p,(y−h,b)=1

inf

|α|≤Dh

Z b y

Z x y

f(t)u(t)dt−α

v(x)

q

dx

1 q )

.

SinceD₀ ≤D_h ≤D_h0 we have by Lemma 2.4 inf

|α|≤D_h

Z b y

Z x y

f(t)u(t)dt−α

v(x)

q

dx

1 q

≤ inf

|α|≤D0

Z b y

Z x y

f(t)u(t)dt−α

v(x)

q

dx ¹q

=A(y, b) and thus

A(y, b)≤A(y−h, b)≤2^q−1

ε^1+p10 +D_h0ε^1q +kvk_q,(y,b)ε^1/p0 +A(y, b)

which proves that

h→0lim+

A(y−h, b) = A(y, b).

Analogously,

h→0lim+

A(y+h, b) = A(y, b),

which finishes the proof of our lemma.

Lemma 2.6. Suppose u, v > 0 satisfy (2.1) and (2.2) and that T : L^p(a, b) → L^q(a, b) is compact. Let I₁ = (c, d) and I₂ = (c⁰, d⁰) be subintervals of (a, b), withI₂ ⊂ I₁, |I₂| > 0,

|I₁−I₂|>0,Rb

a v^q(x)dx <∞. Then0< A(I₂)< A(I₁).

Proof. Let0≤f ∈L^p(I₂),0<kfk_p,I2 ≤ kfk_p,I1 ≤1withsuppf ⊂I₂. Lety ∈I₂ then kT_(c⁰_,y)k_p,I2 >0 and kT_(y,d⁰₎k_p,I2 >0

and then by simple modification of [5, Lemma 3.5] for the casep < q we have min{kT_(c⁰_,y)kq,I2,kT_(y,d⁰₎kq,I2} ≤min

x∈J kTx,Jkq,I2

which meansA(I₂)>0.

Next, suppose that c = c⁰ < d⁰ < d.A slight modification of [5, Theorem 3.8] for p < q, yields x₀ ∈ I₂ and x₁ ∈ I₁ such that A(I₂) = kT_x0,I2k_q,I2 and A(I₁) = kT_x1,I1k_q,I1. Since u, v >0onI₁, it is then quite easy to see thatx₀ ∈I₂^o andx₁ ∈I₁^o.

Ifx₀ =x₁, then, sinceu, v >0onI₁, we get

A(I₁) = kT_x1,I1k_q,I1 >kT_x1,I1k_q,I2 =kT_x1,I2k_q,I2 =A(I₂).

On the other hand, ifx0 6=x1, then

A(I₁) = kT_x1,I1k_q,I1 ≥ kT_x1,I1k_q,I2 ≥ kT_x1,I2k_q,I2 >kT_x0,I2k_q,I2 =A(I₂).

The casec < c⁰ < d⁰ =dcould be proved similarly and the casec < c⁰ < d⁰ < dfollows from

previous cases and the monotonicity ofA(I₁).

LetI = (a, b) ⊂ (0, d)and I_i = (a_i, b_i) ⊂ I, i = 1,2, . . . , k. Say that{I_i}^k_i=1 ∈ P(I)if Sk

i=1I_i ⊃I and assume the intervals{I_i}^k_i=1to be non-overlapping.

Now, for any intervalI ⊆(0, d)andε >0, we define the numbersM andN, as follows:

(2.8) M(I, ε) := inf{n:J(I_i)≤ε,{I_i}ⁿ_i=1 ∈ P(I)}

and

(2.9) N(I, ε) := inf{n;A(I_i)≤ε,{I_i}ⁿ_i=1 ∈ P(I)}.

Since by Proposition 2.1,A(I)≤ kT_Ik ≤K₂J(I), we have

(2.10) N(I, ε)≤M(I, K2ε).

PutN(ε) =N((0, d), ε)andM(ε) =M((0, d), ε). From Proposition 2.3 and the definition of J(I)one gets the following:

Remark 2.7. Suppose that (2.1) and (2.2) are satisfied. Then T : L^p(0, d) → L^q(0, d) is compact if and only ifM(ε)<∞for eachε >0.

Lemma 2.8. LetT be a compact operator. Then

x→0lim+

A(0, x) = 0and lim

x→d−

A(x, d) = 0.

Lemma 2.9. Suppose that T is a compact operator, ε > 0 and I = (a, b) ⊂ (0, d). Let m = N(I, ε). Then there exists a sequence of non-overlapping intervals {I_i}^m_i=1 covering I, such thatA(I_i) = εfori∈ {2, . . . , m−1},A(I₁)≤ε, andA(I_m)≤ε.

Proof. From Remark 2.8 and (2.10), one hasm < ∞. Define a systemS ={I_j}j∈J,I_j ⊂ I, of intervals as follows: Setb₁ = inf{x∈ I;A(x, b)≤ε}. By Lemma 2.8 we havea≤ b₁ < b.

Put I₁ = [b₁, b]. Then A(I₁) ≤ ε. If a = b₁ write S = {I₁}, otherwise set b₂ = inf{x ∈ I;A(x, b₁) ≤ ε} andI₂ = [b₂, b₁]. Observe that by Lemma 2.5 we have A(I₂) = ε. We can now proceed by mathematical induction to construct a (finite or infinite) system of intervals S = {Ij}^α_j=1. Note that we have only A(Iα) ≤ ε (not A(Iα) = ε) provided α < ∞ and A(I_β) = εforβ < α. Writingb₀ =bwe can setI_j = [b_j, bj−1],1≤j ≤α.

Our next step is to show that α = m. By the definition ofm one hasα ≥ m and a finite sequence of numbersa =a_m< am−1 < . . . a₀ =band intervalsJ_i = [a_i, ai−1],i= 1,2, . . . , m such that A(J_i) ≤ ε. Notice that b₁ ≤ a₁, for if not, we can take λ : 0 < λ < b₁, which, from Lemma 2.5 and the definition of I, would yield ε < A(λ, b₀) ≤ A(J₁) ≤ ε, which is a contradiction. Assume now that for some α > 1, b_k > a_k. If bk−1 ≤ ak−1, then talking a_k < λ < b_k, Lemma 2.5 and the definition ofI_kyieldε < A(λ, bk−1) ≤A(J_k) ≤ε, which is a contradiction, so thatak−1 ≤bk−1. Repeating this reasoning, one arrives atb₁ > a₁, which is again a contradiction. Thus,b_k ≤a_k for allk = 1,2, . . . m. Choosingk =mwe haveb_m =a and consequently,α=mandS coversI which finishes the proof.

For future reference (see the proof of (1.11) in the next section) we include the following lemmas and remarks.

Let X be a Banach space and M ⊂ X. Recall the definition of the distance function dist(·, M),

dist(x, M) = inf{kx−yk;y ∈M}, x ∈X.

Lemma 2.10. LetT be a compact operator, u, v > 0, ε > 0, I = (a, b) ⊂ (0, d)and m = N(I, ε).

(i) Then there exists 0 < ε₁ < ε and a sequence of non-overlapping intervals {I_i}^m_i=1 coveringI, such thatA(I_i) =ε₁ fori∈ {1, . . . , m}.

(ii) There existsε₂ : 0< ε₂ < εsuch thatm+ 1 =N(I, ε₂).

Proof. The proof follows from the strict monotonicity and the continuity ofA(I).

Lemma 2.11. LetHbe an infinite dimensional separable Hilbert space. LetY ={u₁, . . . , u_2n} be any orthonormal set with2n vectors and letX be anym-dimensional subspace ofH with m≤n. Then there exists an integerj,1≤j ≤2n, such that

dist(u_j, X)≥ 1

√2.

Proof. Denote the inner product inH by(u, v). ExtendY to an orthonormal topological basis {u_i}^∞_i=1 ofH. Choose an orthonormal basis ofX, sayv₁, . . . , v_m. Denote byP the orthogonal projection ofHintoX. Then

P u=

m

X

j=1

(u, v_j)v_j for anyu∈H.

SinceP is a self-adjoint projection we obtain

2n

X

k=1

ku_k−P u_kk² =

2n

X

k=1

(1−2(u_k, P u_k) + (P u_k, P u_k))

= 2n−

2n

X

k=1

(u_k, P u_k)

= 2n−

2n

X

k=1 m

X

j=1

(u_k, v_j)²

= 2n−

m

X

j=1 2n

X

k=1

(uk, vj)². The Parseval identity yields

∞

X

k=1

(u_k, v_j)² =kv_jk² = 1, which implies

2n

X

k=1

(u_k, v_j)² ≤1.

Consequently,

2n

X

k=1

ku_k−P u_kk² ≥2n−m≥n,

which guarantees the existence of an integerj,1≤j ≤2n, withku_j −P u_jk² ≥ ¹₂. Then dist(u_j, X) = ku_j−P u_jk ≥ 1

√2,

which finishes the proof.

Lemma 2.12. Let 1 ≤ p ≤ 2 and X be any n-dimensional subspace of l_p. Set e_j ∈ l_p, e_j ={δ_ij}^∞_i=1whereδ_ij is Kronecker’s symbol. Then there exists an integerj,1≤j ≤2n, such that

distp (ej, X)≥ 1

√2.

Proof. Denote byk · k_p the norm and bydist_pthe distance function inl_p. Sincek · k_l2 ≤ k · k_lp we can considerX as ann-dimensional subspace of l₂. Thus, using the previous lemma there isj,1≤j ≤2nwithdist₂(e_j, X)≥ ^√1

2 from which immediately follows that distp (e_j, X) = inf{ke_j −xk_p;x∈X}

≥inf{ke_j −xk₂;x∈X}

= dist

2 (e_j, X)≥ 1

√2.

Lemma 2.13. Let 2 < p ≤ ∞, n ∈ N and X be any n-dimensional subspace of l^p. Set e_j ={δ_ij}^∞_i=1 ∈l_p whereδ_ij is the Kronecker’s symbol. Then there isj,1≤j ≤2nsuch that

(2.11) dist

p (e_j, X)≥2^1p−1n^1p−12.

Proof. LetR :l^p →l^pbe the restriction operator given by R(a) = (a₁, a₂, . . . , a_2n,0,0, . . .),

wherea = (a₁, a₂, . . .) ∈ l^p. Choose u_i ∈ X such thatdist_p(e_i, X) = ke_i −u_ik. Using the well-known inequality

kR(a)k₂ ≤(2n)^12−1pkR(a)k_p for all a∈l^p it follows that for each1≤i≤2n,

distp (e_i, X) = ke_i−u_ik_p

≥ kR(e_i)−R(u_i)k_p

≥(2n)^12−1pkR(e_i)−R(u_i)k₂

≥(2n)^12−1pdist

2 (e_i, R(X)).

SinceR(X)is a linear subspace ofl²,by Lemma 2.11 there existsj with dist2 (e_j, X)≥ 1

√2,

which finishes the proof of the lemma.

It is shown in the appendix that the power ofnin (2.11) is the best possible if2< p≤ ∞.

With the aid of the last lemmas we can get a modified version Lemma 2.11 withH replaced byL^p(0, d).

We start by recalling some lemmas referring to the properties of the map takingx∈X to its nearest elementMA(x)∈A⊂X.

Lemma 2.14. Assume thatXis a strictly convex Banach space,V ⊂X is a finite dimensional subspace of X andx₀ ∈ X. SetA = {x₀ +v;v ∈ V}. Then for anyx ∈ X there exists a unique elementvsuch that

kx−vk= inf{kx−yk;y∈A}.

Denote byM_Athe mapping which assigns tox∈Xthe nearest element ofA.

Lemma 2.15. For anyα ∈R,x∈X andv ∈V, one has

(2.12) M_V(αx) = αM_V(x),

(2.13) M_V(x+v) =M_V(x) +v

and

(2.14) kx−vk ≥ 1

2kMV(x)−vk.

The proof of these last two lemmas can be found in [10].

Recall thatP :X →X is called a projection ifP is linear,P² =P andkPk<∞.

Lemma 2.16. LetXbe a strictly convex Banach space andV ⊂X be a subspace,dim(V) =

√nis finite. Then there exists a projectionP :X →V which is onto such thatkPk ≤√ n.

For a proof of the above lemma, see [10, III. B, Theorem 10].

The following lemma, whose proof is included for the sake of completeness, plays a critical role in the sequel, since it provides an approximation to the mapM_Aabove by a linear operator of at most one dimensional range. The proof can also be found in [5].

Lemma 2.17. LetI ⊂(0, d),1≤q ≤ ∞and letR

I|g(t)v(t)|^qdt <∞. Set

ω_I(g) =











0 if R

I|v(t)|^qdt = 0;

R

Ig(t)|v(t)|^qdt R

I|v(t)|^qdt if 0<R

I|v(t)|^qdt <∞;

0 if R

I|v(t)|^qdt =∞.

Then

(2.15) inf

α∈<k(g−α)vk_q,I ≤ k(g−ω_I(g))vk_q,I ≤2 inf

α∈<k(g−α)vk_q,I. Proof. It suffices to prove the second inequality. Fixgsuch thatR

Ig(t)|v(t)|^qdt <∞.

Assume first thatR

I|v(t)|^qdt= 0. Thenv(t) = 0almost everywhere inI and all members in (2.15) are equal zero.

Let R

I|v(t)|^qdt = ∞. We claim that kαvk_q,I ≤ k(α −g)vk_q,I. If α = 0 the inequality is clear. Let α 6= 0, otherwise kαvkq,I = ∞ and by the triangle inequality, it follows that k(α−g)vk_q,I ≥ kαvk_q,I − kgvk_q,I =∞and hence the claim. Thus, for eachα ∈R

k(g−ωI(g))vkq,I =k(g−α+α)vkq,I ≤2k(g−α)vkq,I

which gives

k(g−ωI(g))vkq,I ≤2 inf

α∈<k(g−α)vkq,I. Assume now0<R

I|v(t)|^qdt <∞. By the Hölder’s inequality, we obtain, for anyα ∈ <

k(α−w_I(g))vk^q_q,I = Z

I

α−

R

Ig(t)|v(t)|^qdt R

I|v(t)|^qdt

v(x)

q

dx

= Z

I

|v(x)|^q

R

I(α−g(t))|v(t)|^qdt R

I|v(t)|^qdt

q

dx

= Z

I

|v(t)|^q (R

I|v(t)|^qdt)^q Z

I

(α−g(t))|v(t)|^qdt

q

dx

= Z

I

|v(t)|^qdx 1−q

Z

I

(α−g(t))|v(t)||v(t)|^q−1dt

q

≤ Z

I

|v(x)|^qdx ^1−qZ

I

|(α−g(t))v(t)|^qdt Z

I

|v(t)|^q0(q−1)dt _q^q0

= Z

I

|(α−g(t))v(t)|^qdt=k(α−g)vk^q_q,I which provesk(α−w_I(g))vk_q,I ≤ k(α−g)vk_q,I.

Now, using this inequality, for any realαone has:

k(g−w_I(g))vk_q,I ≤ k(g−α)vk_q,I +k(α−w_I(g))vk_q,I ≤2k(α−w_I(g))vk.

The lemma follows by taking the infimum overαon the right hand side.

Lemma 2.18. Let X = L^p(0, d), p > 1. Letv₁, v₂, . . . , v_n be functions in X with pairwise disjoint supports andkv_ik_p = 1fori = 1,2, . . . , n. SetV = span{v₁, v₂, . . . , v_n}. Then there is a projectionP_V withrankP_v ≤n, such that

kf−M_V(f)k_p,(0,d)≤ kf −P_V(f)k_p,(0,d)≤2kf−M_V(f)k_p,(0,d), whereM_V is as defined in Lemma 2.14.

Proof. Denote S_i = suppv_i, V_i = span{v_i}. Given any f ∈ X, with suppf ⊂ S_i, let M_i(f) = M_vi(f). PutP_if =ω_i(f χ_Si)χ_Si, andP f =Pn

i=1P_i(f χ_Si)χ_Si.

From the definition ofM_v andP_v we havekf −M_V(f)k_p,(0,d) ≤ kf −P_V(f)k_p,(0,d), which is the first inequality. Also

kf−M_V(f)k^p_p =

n

X

i=1

kf χ_Si +M_v(f)χ_Sik^p_p,S

i

≤

n

X

i=1

kf χ_Si −M_i(f χ_Si)χ_Sik^p_p,S

i

≤2^−1p

n

X

i=1

kf χSi −Pi(f χSi)χSik^p_p,S

i

= 2^−p1

f−

n

X

i=1

P_i(f χ_Si)χ_Si

p

p

≤2^−1pkf −P(f)χ_Sik^p_p,

which gives the second inequality and completes the proof.

Lemma 2.19. Let 1 < p ≤ 2 and let u₁, ..., u_2n be a system of functions from L^p(0, d)with disjoint supports. LetX ⊂L^p(0, d)be a subspace,dimX ≤ n. Then there exists an integerj, 1≤j ≤2n, such that

distp (u_j, X)≥ 1 3√

2ku_jk_p.

Proof. Ifku_ik_p = 0for somei, it suffices to choosej = i. Letku_ik_p >0for all1 ≤ i ≤ 2n.

Setv_i = _ku^ui

ikp. LetV = span{v₁, v₂, . . . , v_2n} and letP_V be the projection from the previous lemma. LetY =P_V(X). ThenY ⊂V,dimY ≤n. Denote byZ the subspace ofl^p consisting of all sequences {a_i}^∞_i=1 such that a_k = 0 for all k > 2n. Lete_i be the canonical basis of Z.

Define a linear mappingI :Y →Z by I

2n

X

i=1

α_iv_i

!

=

2n

X

i=1

α_ie_i.

Since kv_ik = 1 and the functions v_i have pairwise disjoint supports, it follows that I is an isometry betweenY andZ. According to Lemma 2.12 there exists1≤j ≤2nsuch that

(2.16) dist

p (e_j, I(Y))≥ 1

√2, and from Lemma 2.14 there is a uniquex∈X with

(2.17) dist

p (v_j, X) =kv_j −xk_p. By the definition ofPV andMV, we have

1

2kx−M(x)k_p ≤ 1

2kx−P_V(x)k_p ≤ kx−M_V(x)k_p ≤ kv_j −xk_p which yields, with the triangle inequality,

kP_V(x)−v_jk_p ≤ kP_V(x)−xk_p+kx−v_jk_p

≤2kx−vjkp

≤2kx−v_jk_p+kx−v_jk_p ≤3kx−v_jk_p.

This together with (2.16) and (2.17), gives distp (v_j, X) =kv_j−xk_p

≥ 1

3kvj−PV(x)kp

≥ 1 3dist

p (v_j, Y) = 1 3dist

p (e_j, I(Y))≥ 1 3√

2.

Denoting byM₁ the mapping which assigns to anyf ∈L^p(0, d)the element ofX nearest tof and using (2.12) we can rewrite the previous inequality as

distp (u_j, X) =ku_j −M₁(u_j)k_p

=ku_jk_pkv_j−M₁(v_j)k_p

=ku_jk_p dist

p (v_j, X)≥ 1 3√

2ku_jk_p

which yields the claim.

Lemma 2.20. Let 2 < p ≤ ∞and let u₁, ..., u_2n be a system of functions fromL^p(0, d)with disjoint supports. LetX ⊂L^p(0, d)be a subspace,dimX ≤ n. Then there exists an integerj, 1≤j ≤2n, such that

distp (uj, X)≥ 1 2√

2kujkpn^1p−12.

Proof. LetV, M_V, P_V, Y, Z and I have the same meanings as in Lemma 2.19. Proceeding as before, Lemma 2.13 yieldsj : 1≤j ≤2nsuch that

distp (ej, I(Y))≥ 1 2n^1p−12. Letx∈X be the element given by Lemma 2.14 so that

dist(v_j, X) =kv_j −xk_p. In exactly the same way as in Lemma 2.19, one gets

distp (v_j, X)≥ 1 3n^1p−12, which can be written as

distp (u_j, X)≥ 1

3ku_jk_pn^p1−12,

and the proof is complete.

3. BOUNDS FOR THEAPPROXIMATIONNUMBERS

We recall that, given anym ∈N,them^thapproximation numbera_M(S)of a bounded oper- atorS fromL^p intoL^q, is defined by

a_m(S) := inf

F kS−Fkp→q,

where the infimum is taken over all bounded linear mapsF : L^p(0, d) → L^q(0, d) with rank less thanm. Futher discussions on approximation numbers may be found in [3]. An operatorS is compact if and only ifa_m(S)→0asm→ ∞. The first two lemmas of this section provide estimates for a_m(T)withT as in (1.1), which are analogous to those obtained in [1] and [5].

Hereafter, we shall always assume (2.1) and (2.2).

Lemma 3.1. Let1 ≤ p ≤ q ≤ ∞ and suppose thatT : L^p(0, d) → L^q(0, d)is bounded. Let ε > 0and suppose that there exist N ∈ Nand numbers c_k, k = 0,1, . . . , N, with 0 = c₀ <

c₁ <· · ·< c_N =d, such thatA(I_k) ≤εfork = 0,1, . . . , N −1, where I_k = (c_k, c_k+1). Then a_N+1(T)≤2ε.

Proof. Consider for f ∈ L^p(a, b)and0≤ k ≤ N −1one-dimensional linear operators given by

PI_kf(x) :=χI_k(x)v(x) Z x

ck

uf dt+ωI_k

Z x ck

uf dt

,

whereωI_k is the functional from Lemma 2.17. We claim thatPk is bounded fromL^p(0, d)into L^q(0, d)for eachk.

Assume first that either 0 = kvk_q,Ik or kvk_q,Ik = ∞. Then P_k = 0 and consequently, it is bounded.

Assume now0<kvk_q,Ik < ∞and fixf,kfk_p,(0,d) = 1. Then using Hölder’s inequality, we obtain

ω_Ik Z x

ck

u(t)f(t)dt

= R

Ik

Rx

cku(t)f(t)dt|v(x)|^qdx R

Ik|v(x)|^qdx

≤ R

Ik|v(x)Rx

cku(t)f(t)dt| |v(x)|^q−1dx R

Ik|v(x)|^qdx

≤ R

Ik|v(x)Rx

cku(t)f(t)dt|^qdx¹_q R

Ik|v(x)|^(q−1)q0dx_q¹0

R

Ik|v(x)|^qdx

≤ kT_Ikfk_q

kvk_q,Ik ≤ kTk kvk_q,Ik and consequently,

Z d 0

|(P_kf)(x)|^qdx= Z

I_k

v(x) Z x

c_k

uf dt+ω_Ik Z x

c_k

uf dt

q

dx

≤2^q−1 Z

Ik

v(x) Z x

ck

uf dt

q

+ω^q_I

k

Z x ck

uf dtdx

≤2^q−1

kTkfkq+ kTk kvk_q,Ik

≤ kTk

1 + 1 kvk_q,Ik

.

SetP =PN−1

k=0 P_k. ThenP is a linear bounded operator fromL^p(0, d)intoL^q(0, d). Moreover, we have by Lemma 2.17 and the well-known inequality

∞

X

k=1

|a_k|^q

!¹_q

≤

∞

X

k=1

|a_k|^p

!¹_p ,

kT f −P fk^q_q=

N−1

X

k=0

kT f −P_Ikfk^q_q,I

k

=

N−1

X

k=0

v(x) Z x

ck

uf dt−ω_Ik Z x

ck

uf dt

q

q,Ik,µ

≤2^q−1

N−1

X

k=0

α∈<inf kT_Ikf−αfk^q_q,I

k

≤2^q

N−1

X

k=0

A^q(I_k)kfk^q_p,I

k

≤(2ε)^q

N−1

X

k=0

kfk^q_p,I

k

≤(2ε)^q

N−1

X

k=0

kfk^p_p,I

k

!^q_p

≤(2ε)^q

by Lemma 2.5. SincerankP ≤N, the proof of the lemma is complete.

Lemma 3.2. Let1< p≤q <∞,T be bounded fromL^p(0, d)toL^q(0, d),0≤a < b < c < d and denoteI = [a, b], andJ = [b, c]. Further, letf, g∈L^p(0, d)withsuppf ⊂I,suppg ⊂J, kfk_p =kgk_p = 1.

Letr,sbe real numbers and set

h(x) = v(x) Z d

0

u(t)(rf(t) +sg(t))dt.

AssumeRc

au(t)h(x) = 0. Then

khk_q ≥

|r|^q inf

α∈<kT_If −αvk^q+|s|^q inf

α∈<kT_Jg −αvk^{q 1}_q

.

Proof. Sincesuppf ⊂Iandsuppg ⊂J we have (3.1)

Z a 0

v(x) Z x

0

u(t)(rf(t) +sg(t))dt

q

dx= 0.

Ifx∈(c, d)we have (recall thatRc

a u(t)h(x) = 0) that

(3.2)

Z d c

v(x) Z x

0

u(t)(rf(t) +sg(t))dt

q

dx= Z d

c

v(x) Z c

a

u(t)h(t)dt

q

dx= 0.

Assume firsts6= 0. Then it follows from (3.1) and (3.2) that khk^q_q=

Z d 0

v(x) Z x

0

u(t)(rf(t) +sg(t))dt

q

dx

= Z a

0

+ Z b

a

+ Z c

b

+ Z d

c