an estimate for the best constant in the lp-wirtinger...

JOURNAL OF c© 2008, Scientific Horizon

FUNCTION SPACES AND APPLICATIONS http://www.jfsa.net

Volume 6, Number 1 (2008), 1-16

An estimate for the best constant in the Lp-Wirtinger

inequality with weights

Raffaella Giova

(Communicated by Carlo Sbordone)

2000 Mathematics Subject Classification. 26D15.

Keywords and phrases. Wirtinger weighted inequality, best constant.

Abstract. We prove an estimate for the best constant C in the followingWirtinger type inequality ∫ 2π

0

a|w|p ≤ C

∫ 2π

0

b|w′|p.

1. Introduction

In this paper we consider some Wirtinger-type inequalities. Moreprecisely, our aim is to give an estimate for the best constant C(a, b) inthe following type inequality (for p > 1):

(1.1)∫ 2π

0

a|w|p ≤ C(a, b)∫ 2π

0

b|w′|p

2 Best constant in the Lp -Wirtinger inequality

where w ∈ W 1,p(0, 2π) is a 2π−periodic function satisfying the constraint

(1.2)∫ 2π

0

a|w|p−2w = 0,

and a and b are two measurable and non negative functions belonging tosuitable spaces.

We note that when a = b = 1 and p = 2, inequality (1.1)-(1.2) reducesto the well-known Wirtinger inequality. In this case

(1.3) C(a, b) = C(1, 1) = 1.

The case b = a and p = 2 has been analyzed by Piccinini and Spagnolo in[3], who obtained the estimate

(1.4) C(a, a) ≤(

4π

arctan

√min a

max a

)−2

.

The case b = a−1 and p = 2 has been studied in [4] (when a is a measurablefunction bounded from above and away from 0) and in [2] (when a issimply a function belonging to L1(0, 2π)), where the following estimatewas obtained:

(1.5) C

(a,

1a

)=(

12π

∫ 2π

0

a

)2

.

The case a �= b and p = 2 has been studied in [4] (when a, b are positivemeasurable functions on (0, 2π) bounded from above and away from 0)and in [2] (when a and b are non negative measurable functions on (0, 2π)such that a and 1

b belong to L1(0, 2π)), where the following estimate wasobtained:

(1.6) C(a, b) ≤

⎛⎜⎝ 12π

∫ 2π

0

√ab−1

4π arctan

(inf absup ab

)1/4

⎞⎟⎠2

In this paper we will extend the estimates (1.3)-(1.6) to the more generalcase p > 1.

We will denote by p′ the conjugate exponent to p (i.e. 1p + 1

p′ = 1) andby B( 1

p , 1p′ ) the Beta function

B

(1p,

1p′

)= Γ(

1p

)Γ(

1p′

)=∫ 1

0

t1p−1(1 − t)

1p′ −1

dt.

R. Giova 3

Furthermore, if a : (0, 2π) → R is an integrable positive function wewill denote by Lp

(1

ap−1

)the set of all measurable functions u on [0, 2π]

such that∫ 2π

01

ap−1 |u|p < ∞ and by W 1,pper

(1

ap−1

)the set of all functions

w ∈ L1(0, 2π) such that the distributional derivative w′ belongs to Lp( 1ap−1 )

and w(0) = w(2π). If b : (0, 2π) → R is any nonnegative measurablefunction, W 1,p

per(b) is defined similarly.

We will prove the following two theorems:

Theorem 1.1. Let a ∈ L1(0, 2π) be a nonnegative function. Then thefollowing Wirtinger inequality holds:

(1.7)∫ 2π

0

a|w|p ≤ C

(a,

1ap−1

)∫ 2π

0

1ap−1

|w′|p

where

(1.8) C

(a,

1ap−1

)=

(4∫ 2π

0 a

(1p′

) 1p(

1p

) 1p′

B

(1p,

1p′

))−p

for every w ∈ W 1,pper(

1ap−1 ) such that

∫ 2π

0 a|w|p−2w = 0 .

Theorem 1.2. Let a, b : (0, 2π) → [0,∞[ be measurable functionssuch that a, 1

b and p√

ab−1 belong to L1(0, 2π) . Assume that 0 < I =inf p

√ap−1b ≤ S = sup p

√ap−1b < ∞ and set, for all x, y ≥ 0 , A(x, y) =∫ y

x1

1+tp

p−1dt . Then for every w ∈ W 1,p

per(b) such that∫ 2π

0a|w|p−2w = 0

(1.9)∫ 2π

0

a|w|p ≤

⎛⎜⎝ 12π

∫ 2π

0p

√a(t)b(t)−1

2π

1

(p−1)p−1

p

[cp′ − Ψ(S, I)]

⎞⎟⎠p ∫ 2π

0

b|w′|p,

where

Ψ(S, I) = A

⎛⎝ I

S

((S − I)S

1p−1

I(S1

p−1 − I1

p−1 )

) p−1p

,

((S − I)S

1p−1

I(S1

p−1 − I1

p−1 )

) p−1p

⎞⎠and, for 1 < q < ∞

cq =∫ +∞

0

11 + xq

dx =1qB

(1q,

1q′

).


Note that when a is equal to 1, Theorem 1.1 yields

C(1, 1) =

(2π

(1p′

) 1p(

1p

) 1p′

B

(1p,

1p′

))−p

which is the optimal constant in the Wirtinger inequality, as proved byCroce and Dacorogna in [1].

In order to prove Theorem 1.2, we need Lemmas 3.1 and 3.2. The firstone gives an estimate for C(a, a) (when a = b) that reduces to the estimate(1.5) obtained by Piccinini and Spagnolo when p = 2; the second one givesan estimate for C(a, b) for arbitrary weight functions a, b belonging toL∞ .

2. Proof of Theorem 1.1

First we proceed with a lemma and then with the proof of the theorem.

Lemma 2.1. Let a be a measurable function which is bounded from aboveand away from zero. If C(a, 1

ap−1 ) is defined by (1.8), then

(2.1)∫ 2π

0

a|w|p ≤ C

(a,

1ap−1

)∫ 2π

0

1ap−1

|w′|p

for every w ∈ W 1,pper(0, 2π) such that

(2.2)∫ 2π

0

a|w|p−2w = 0.

Proof. Let us still denote by a the 2π−periodic extension of the functiona originally defined only in [0, 2π] and set

y(x) =∫ x

0

a(t)dt and W (y) = w(x(y)).

Then W ∈ W 1,p(R) is β−periodic where β =∫ 2π

0 adt.

It easily seen that (2.1) and (2.2) may be rewritten as

(2.3)∫ β

0

|W |pdy ≤ C

∫ β

0

|W ′|pdy

and

(2.4)∫ β

0

|W |p−2Wdy = 0.

R. Giova 5

Therefore, applying Theorem 1.1 of [1] we conclude the proof. �

Proof of the Theorem 1.1. The proof is similar in spirit to the one of [2]but it differs in technical aspects. It will be divided into 3 steps.

Step 1. Let us first assume that w ∈ W 1,pper(0, 2π) ∩ W 1,∞(0, 2π) and

(2.5) a(t) ≥ 1.

For h > 1 we set

(2.6) ah(t) ={

a(t) if 1 ≤ a(t) ≤ hh if a(t) > h

and define

(2.7) wh(x) = w(x) + εh

where εh is a constant, hence wh(0) = wh(2π).A simple continuity argument yields that for all h there exists εh such

that

(2.8)∫ 2π

0

ah|w + εh|p−2(w + εh) =∫ 2π

0

ah|wh|p−2wh = 0

hence

−1 − max w ≤ εh ≤ 1 − min w,

i.e., the sequence {εh} is bounded, we claim that εh → 0. In fact if thiswere not true there would exist a subsequence, still denoted by εh , suchthat εh → ε with ε �= 0.

Now, we show that the value ε is equal to zero, thus proving the claim.The function

φ(ε) =∫ 2π

0

a|w + ε|p−2(w + ε)

is strictly increasing and from (1.2)

(2.9) φ(0) = 0.

On the other hand

(2.10)∫ 2π

0

ah|w + εh|p−2(w + εh) →∫ 2π

0

a|w + ε|p−2(w + ε) = 0.

Combining (2.9) and (2.10) we have

(2.11) εh → 0 as h → ∞.


By contradiction, we have proved the claim.From Lemma 2.1 we have the following inequality for wh and ah

(2.12)∫ 2π

0

ah|wh|p ≤ C

(ah,

1ap−1

h

)∫ 2π

0

1ap−1

h

|w′h|p.

where

C

(ah,

1ap−1

h

)=

(4∫ 2π

0ah

(1p′

) 1p(

1p

) 1p′

B

(1p,

1p′

))−p

.

Since εh → 0, wh → w uniformly in (0, 2π); moreover ah → a in L1(0, 2π)and we have

(2.13)∫ 2π

0

ah|wh|p →∫ 2π

0

a|w|p.

On the other hand (2.6) and (2.7) imply∫ 2π

0

1ap−1

h

|w′h|p =

∫a≤h

1ap−1

|w′|p +1

hp−1

∫a>h

|w′|p.

Notice that since w′ is bounded, the last integral is infinitesimal ash → ∞ , hence

(2.14)∫ 2π

0

1ap−1

h

|w′h|p →

∫ 2π

0

1ap−1

|w′|p

and (1.7) is established.Step 2. Now, still assuming that w ∈ W 1,p

per(0, 2π)∩W 1,∞(0, 2π), we makeno special assumptions on a .Notice, however, that we may always assume, without loss of generality,that

∫ 2π

0a(x)dx > 0, otherwise (1.7) would be trivial.

For δ > 0 we set

(2.15) aδ ={

a if a ≥ δδ if 0 ≤ a < δ

and define

(2.16) wδ(x) = w(x) + εδ

where εδ is a constant.

R. Giova 7

As in the previous step it is possible to prove that for any δ there existsεδ such that

(2.17)∫ 2π

0

aδ|w + εδ|p−2(w + εδ) =∫ 2π

0

aδ|wδ|p−2wδ = 0

and

(2.18) εδ → 0 as δ → 0.

From the previous step we have the following inequality for wδ and aδ

(2.19)∫ 2π

0

aδ|wδ|p ≤ C

(aδ,

1ap−1

δ

)∫ 2π

0

1ap−1

δ

|w′δ|p ≤ C

(aδ,

1ap−1

δ

)∫ 2π

0

1ap−1

|w′|p

where

C

(aδ,

1ap−1

δ

)=

(4∫ 2π

0 aδ

(1p′

) 1p(

1p

) 1p′

B

(1p,

1p′

))−p

.

As in (2.13), from (2.18) we have,

(2.20)∫ 2π

0

aδ|wδ|p →∫ 2π

0

a|w|p.

Combining (2.19) and (2.20) we obtain (1.7).Step 3. Finally we assume that w ∈ W 1,p

per(1

ap−1 ).For h > 0 we set

(2.21) Th(w′) =

⎧⎨⎩h if w′ > hw′ if − h < w′ < h−h if w′ < −h .

We define

(2.22) wh(x) = w(0) +∫ x

0

σh(t)Th(w′)dt + εh

where εh is a constant and

(2.23) σh(x) ≡ 1 if∫{Th(w′)>0}

Th(w′)dx = −∫{Th(w′)<0}

Th(w′)dx


otherwise, if ∫{Th(w′)>0}

Th(w′)dx > −∫{Th(w′)<0}

Th(w′)dx,

then

(2.24) σh(x) =

⎧⎨⎩ 1 if x ∈ {Th(w′) < 0}−∫{Th(w′)<0} Th(w′)dx∫

{Th(w′)>0} Th(w′)dxif x ∈ {Th(w′) > 0}.

When∫{Th(w′)>0} Th(w′)dx < − ∫{Th(w′)<0} Th(w′)dx , σh(x) is defined

similarly to (2.24).Also in this case with similar considerations to those of the previous steps

we obtain:for any h there exists εh such that

(2.25)∫ 2π

0

a|wh|p−2wh = 0;

and

(2.26) εh → 0 as h → ∞.

Moreover, for any h , wh(0) = wh(2π).From the previous case, we have the Wirtinger inequality for a and wh

(2.27)∫ 2π

0

a|wh|p ≤ C

(a,

1ap−1

)∫ 2π

0

1ap−1

|w′h|p.

Since wh → w uniformly, we get

(2.28)∫ 2π

0

a|wh|p →∫ 2π

0

a|w|p.

On the other hand, for any h

(2.29)∫ 2π

0

1ap−1

|w′h|p ≤

∫ 2π

0

1ap−1

|w′|p

and (1.7) is proved. �

R. Giova 9

3. Proof of the Theorem 1.2

In order to prove the Theorem 1.2 we need the following two lemmas. Thefirst lemma extends an analogous one proved in Piccinini-Spagnolo (Lemma1. of [3]). Here we will follow closely their proof by adapting to the casep �= 2.

Lemma 3.1. Let a(t) be a periodic measurable function, with period2π , such that 1 ≤ a(t) ≤ L ; let w(t) be a periodic function, belonging toW 1,p

loc (R) , with period 2π , such that∫ 2π

0 a|w|p−2w = 0 . Then the followinginequality holds:

(3.1)∫ 2π

0

a(t)|w(t)|pdt ≤ (C(L))−p

∫ 2π

0

a|w′(t)|pdt

where

(3.2) C(L) =2π

1

(p − 1)p−1

p

[cp′ −

∫ β(L)

α(L)

1

1 + tp

p−1dt

]with

(3.3) cp′ =∫ +∞

0

11 + xp

dx =1p′

B

(1p,

1p′

);

(3.4) β(L) =[

L1

p−1 (L − 1)

L1

p−1 − 1

] p−1p

and α(L) =β(L)

L;

Proof. Consider for any a(t) such that 1 ≤ a(t) ≤ L , the eigenvalueproblem

(3.5)

{(a|w′|p−2w′)′ + λa|w|p−2w = 0

w periodic of period 2π.

When p = 2, (3.5) becomes the problem (13) in [3].It is easy to prove that the values of λ for which this problem has not

constant solutions form a sequence λn with 0 < λ1 < λ2 < ..., and that forany function w(t), periodic of period 2π , such that

∫ 2π

0a|w|p−2w = 0, the

following estimate holds∫ 2π

0

a(t)|w(t)|pdt ≤ 1λ1

∫ 2π

0

a(t)|w′(t)|pdt.


Therefore, in order to prove (3.1) it is sufficient to show that, if λ �= 0 andw �= 0 satisfy (3.5), then necessarily

λ ≥[

2π

1

(p − 1)p−1

p

[cp′ −

∫ β(L)

α(L)

1

1 + tp

p−1dt

]]p.

It is easily seen that a solution of (3.5) in each period has at least two zeros,and that between any pair of zeros of the function there is one and only onezero of its derivative. Let t0, t2, t4 be three consecutive zeros of w and lett1 and t3 be two zeros of w′ in such a way that t0 < t1 < t2 < t3 < t4 .Without loss of generality we may suppose that w(t1) > 0 and w(t3) < 0.It is obvious that

(3.6) t4 − t0 ≤ 2π.

We define, for t0 < t ≤ t1 , the function

(3.7) f(t) =a|w′|p−2w′

|w|p−2w.

According to (3.5) this function satisfies the following first order differentialequation:

(3.8) f ′(t) = −λa(t) − |f | pp−1

a1

p−1(p − 1).

We remark that, since f ′ < 0, f is strictly decreasing. Furthermorelimt→t+0

f(t) = +∞ , f(t1) = 0. Hence, there is one and only one point, say

τ , in the interval (t0, t1) such that f(τ)=[

λL1

p−1 (L−1)

(p−1)(L1

p−1 −1)

] p−1p

. Since f is

decreasing, the following inequalities hold:(3.9)⎧⎪⎨⎪⎩

−λa(t) − |f |p

p−1

a1

p−1(p − 1) ≥ −λ − |f | p

p−1 (p − 1) for t0 < t ≤ τ

−λa(t) − |f |p

p−1

a1

p−1(p − 1) ≥ −λL − |f |

pp−1

L1

p−1(p − 1) for τ ≤ t ≤ t1.

Thus, calling f0(t) the function such that

(3.10)

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩f ′0(t) =

⎧⎨⎩ −λ − |f0|p

p−1 (p − 1) for t0 < t ≤ τ

−λL − |f0|p

p−1

L1

p−1(p − 1) for τ ≤ t < t1.

f0(τ) =[

λL1

p−1 (L−1)

(p−1)(L1

p−1 −1)

]p−1p

R. Giova 11

we get

(3.11)

{f0(t) ≥ f(t) for t ≤ τ

f0(t) ≤ f(t) for t ≥ τ.

Let us set

(3.12) F (z) =∫ z

0

1

1 + xp

p−1dx

and

(3.13) F (∞) =∫ +∞

0

1

1 + xp

p−1dx = cp′ .

The solution of (3.10) is given by

(3.14) f0(t) =

⎧⎨⎩ P (λ)F−1[

λP (λ) (τ − t) + F

(f0(τ)P (λ)

)]for t ≤ τ

P (λ, L)F−1[

λLP (λ,L)(τ − t) + F

(f0(τ)

P (λ,L)

)]for t ≥ τ,

where P (λ) =(

λp−1

) p−1p

and P (λ, L) =(

λLp

p−1

p−1

) p−1p

.

Therefore f0(t) tends to infinity for t converging toτ−P (λ)

λ

(cp′ − F

(f0(τ)P (λ)

))and vanishes for t equal to τ+ P (λ,L)

λL F(

f0(τ)P (λ,L)

).

It follows that

t1 − t0 ≥ P (λ, L)λL

F

(f0(τ)

P (λ, L)

)+

P (λ)λ

(cp′ − F

(f0(τ)P (λ)

)).

In a similar way we can prove that

ti+1− ti ≥ P (λ, L)λL

F

(f0(τ)

P (λ, L)

)+

P (λ)λ

(cp′ − F

(f0(τ)P (λ)

))for i = 1, 2, 3;

hence by adding the relations above we get thatt4 − t0 ≥ 4

[P (λ,L)

λL F(

f0(τ)P (λ,L)

)+ P (λ)

λ

(cp′ − F

(f0(τ)P (λ)

))]. So recalling

(3.6), we can state

2π ≥ 4[P (λ, L)

λLF

(f0(τ)

P (λ, L)

)+

P (λ)λ

(cp′ − F

(f0(τ)P (λ)

))],


that is

λ ≥[

2π

1

(p − 1)p−1

p

[F

(f0(τ)

P (λ, L)

)− F

(f0(τ)P (λ)

)+ cp′

]]p.

Since f0(τ)P (λ,L) = α(L) and f0(τ)

P (λ) = β(L) the proof is complete. �

Lemma 3.2. Suppose a and b ∈ L∞(R) and inf a > 0 , inf b > 0 .Assume that S = sup p

√ap−1b , I = inf p

√ap−1b > 0 and set A(x, y) =∫ y

x1

1+tp

p−1dt . Then

(3.15) C(a, b) ≤

⎛⎜⎝ 12π

∫ 2π

0p

√a(t)b(t)−1

2π

1

(p−1)p−1

p

[cp′ − Ψ(S, I)]

⎞⎟⎠p

,

where

(3.16) Ψ(S, I) = A

⎛⎝ I

S

((S − I)S

1p−1

I(S1

p−1 − I1

p−1 )

) p−1p

,

((S − I)S

1p−1

I(S1

p−1 − I1

p−1 )

) p−1p

⎞⎠and

(3.17) cp′ =∫ +∞

0

11 + xp′ dx =

1p′

B

(1p,

1p′

).

Proof. Under the change of variable

y(x) =1k

∫ x

0

p

√a(t)b(t)

dt,

where k is defined by

(3.18) k =12π

∫ 2π

0

p

√a(t)b(t)

dt

setting a(x) = α(y(x)), b(x) = β(y(x)), w(x) = W (y(x)), we obtain

dx = k p

√b(x(y))a(x(y))

dy, w′(x) = W ′(y(x))1k

p

√a(x)b(x)

,

R. Giova 13

and therefore: ∫ 2π

0

a|w|pdx = k

∫ 2π

0

p√

αp−1β |W |pdy

∫ 2π

0

b|w′|pdx =1

kp−1

∫ 2π

0

p√

αp−1β |W ′|pdy∫ 2π

0

a|w|p−2wdx = k

∫ 2π

0

p√

αp−1β |W |p−2Wdy = 0.

By the above equalities, (1.1)-(1.2) take the form:

(3.19)∫ 2π

0

p√

αp−1β |W |pdy ≤ C(a, b)kp

∫ 2π

0

p√

αp−1β |W ′|pdy

with the constraint

(3.20)∫ 2π

0

p√

αp−1β |W |p−2Wdy = 0.

If p√

αp−1β = p√

ap−1b ∈ L1 and 1 ≤ p√

αp−1β ≤ L an application ofLemma 3.1 enables us to write

C(a, b)kp

=C(a, b)(

12π

∫ 2π

0p

√a(t)b(t) dt

)p

= C( p√

αp−1β, p√

αp−1β)

≤(

2π

1

(p − 1)p−1

p

[cp′ − C(sup p

√ap−1b)

])−p

,

where C is defined as in (3.2).If I = inf p

√ap−1b �= 1 we set

a =a

Iand b =

b

I

and from (1.1) we consider∫ 2π

0

a

I|w|p ≤ C(a, b)

∫ 2π

0

b

I|w′|p.

Since I = inf p√

ap−1b = 1 from the previous case we obtain

C(a, b)(12π

∫ 2π

0p

√a(t)b(t) dt

)p = C( p√

ap−1b,p√

ap−1b)


≤(

2π

1

(p − 1)p−1

p

[cp′ − C(sup p

√ap−1b)

])−p

where C(sup p√

ap−1b) = C(

SI

)and (3.15) is proved. �

Proof of Theorem 1.2. The proof of this Theorem is analogous to the oneof Theorem 3.1 in [2].

We divide the proof into 3 steps.Step 1. Let us first assume w ∈ W 1,p

per(0, 2π) ∩ W 1,∞(0, 2π) and

(3.21) a(t) ≥ 1 b(t) ≤ γ

for some constant γ > 0.Let ah and wh be defined by (2.6) and (2.7) respectively, and for h > 1

(3.22) bh ={

b if b ≥ 1h

1h if b < 1

h

then

(3.23)12π

∫ 2π

0

p

√ah

bh≤ 1

2π

∫ 2π

0

p

√a

b.

From Lemma 3.2 we deduce the Wirtinger inequality for ah, bh and wh

(3.24)∫ 2π

0

ah|wh|p ≤

⎛⎜⎝ 12π

∫ 2π

0p

√ah(t)bh(t)−1

2π

1

(p−1)p−1

p

[cp′ − Ψ(Sh, Ih)]

⎞⎟⎠p ∫ 2π

0

bh|w′h|p,

where Sh = sup p

√ap−1

h bh and Ih = inf p

√ap−1

h bh . In view of (2.7) and(3.22) we have∫ 2π

0

bh|w′h|p =

∫ 2π

0

bh|w′|p =∫

b≥ 1h

b|w′|p +1h

∫b< 1

h

|w′|p.

Therefore, as h → ∞ we have

(3.25)∫ 2π

0

bh|w′h|p →

∫ 2π

0

b|w′|p.

Furthermore, one can easily check that

(3.26) limh→∞

Sh = S

R. Giova 15

(3.27) limh→∞

Ih = I

Combining (2.13), (3.23), (3.24), (3.25), (3.26) and (3.27) we obtain (1.9).Step 2. Let us assume now, w ∈ W 1,p

per(0, 2π) ∩ W 1,∞(0, 2π), and

a ≥ 0 b < ∞.

For δ > 0, let aδ and wδ be defined by (2.15) and (2.16) respectively, and

(3.28) bδ ={

b if b ≤ 1δ

1δ if b > 1

δ .

From the step 1, we deduce the Wirtinger inequality for aδ, bδ and wδ

(3.29)∫ 2π

0

aδ|wδ|p ≤

⎛⎜⎝ 12π

∫ 2π

0p

√aδ(t)bδ(t)

−1

2π

1

(p−1)p−1

p

[cp′ − Ψ(Sδ, Iδ)]

⎞⎟⎠p ∫ 2π

0

bδ|w′δ|p,

where Sδ = sup p

√ap−1

δ bδ and Iδ = inf p

√ap−1

δ bδ .From the definitions (2.15),(2.16) and (3.28) of aδ, wδ and bδ we have

(3.30)∫ 2π

0

p

√aδ

bδ=∫{δ<a}∩{b< 1

δ }p

√a

b+∫{δ<a}∩{b> 1

δ }p√

aδ

=∫{a<δ}∩{b< 1

δ }p

√δ

b+∫{a<δ}∩{b> 1

δ }p√

δ2

and

(3.31)∫ 2π

0

bδ|w′δ|p ≤

∫ 2π

0

b|w′|p.

Notice that the last three integrals on the right side of (3.32) are allinfinitesimal, therefore

(3.32) limδ→0+

∫ 2π

0

p

√aδ

bδ= lim

δ→0+

∫{δ<a}∩{b< 1

δ }p

√a

b=∫ 2π

0

p

√a

b.

As in the previous case, we get

(3.33) limδ→0+

Sδ = S

(3.34) limδ→0+

Iδ = I.


Combining (2.20), (3.29), (3.31), (3.32), (3.33) and (3.34) we obtain (1.9).Step 3. Finally we assume that w ∈ W 1,p

per(0, 2π).We define wh as in (2.22). By using previous case, we deduce the

Wirtinger inequality for a, b and wh

(3.35)∫ 2π

0

a|wh|p ≤

⎛⎜⎝ 12π

∫ 2π

0p√

ab−1

2π

1

(p−1)p−1

p

[cp′ − Ψ(S, I)]

⎞⎟⎠p ∫ 2π

0

b|w′h|p.

Than the result follows since

(3.36)∫ 2π

0

b|w′h|p ≤

∫ 2π

0

b|w′|p.

References

[1] G. Croce and B. Dacorogna, On a generalized Wirtinger inequality,Discrete and Continuous Dynamical System, 9 (5) (2003), 1329–1341.

[2] R. Giova, A weighted Wirtinger inequality, Ricerche di Matematica, LIV,fasc. 1◦ (2005), 293–302.

[3] L. C. Piccinini and Spagnolo, On the Holder continuity of solutions ofsecond order elliptic equations in two variables, Ann. Scuola Norm. Sup.Pisa, 26 (2) (1972), 391–402.

[4] T. Ricciardi, A sharp weighted Wirtinger inequality, Boll. Unione Mat.Ital. Sez. B Artic. Ric. Mat(8), 8 (1) (2005), 259–267.

[5] T. Ricciardi, A sharp Holder estimate for elliptic equations in twovariables, Proceedings of the Royal Society of Edinburgh, 135 A (1)(2005), 165–173.

Dipartimento di Matematica e Applicazioni “R. Caccioppoli”Universita di Napoli “Federico II”Italy(E-mail : [email protected])

(Received : April 2007 )

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