Báo cáo hóa học: " Research Article Generalizations of the Lax-Milgram Theorem" doc

Rabier We prove a linear and a nonlinear generalization of the Lax-Milgram theorem.. In partic-ular, we give sufficient conditions for a real-valued function defined on the product of a re

Volume 2007, Article ID 87104, 9 pages

doi:10.1155/2007/87104

Research Article

Generalizations of the Lax-Milgram Theorem

Dimosthenis Drivaliaris and Nikos Yannakakis

Received 12 December 2006; Revised 8 March 2007; Accepted 19 April 2007

Recommended by Patrick J Rabier

We prove a linear and a nonlinear generalization of the Lax-Milgram theorem In partic-ular, we give sufficient conditions for a real-valued function defined on the product of a reflexive Banach space and a normed space to represent all bounded linear functionals of the latter We also give two applications to singular differential equations

Copyright © 2007 D Drivaliaris and N Yannakakis This is an open access article dis-tributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is prop-erly cited

1 Introduction

The following generalization of the Lax-Milgram theorem was proved recently by An et al

in [1]

Theorem 1.1 Let X be a reflexive Banach space overR, let { X n } n ∈N be an increasing se-quence of closed subspaces of X and V =n ∈N X n Suppose that

is a real-valued function on X × V for which the following hold:

(a)A n = A | X n × X n is a bounded bilinear form, for all n ∈ N ;

(b)A( ·,v) is a bounded linear functional on X, for all v ∈ V;

(c)A is coercive on V, that is, there exists c > 0 such that

for all v ∈ V.

Then, for each bounded linear functional v ∗ on V, there exists x ∈ X such that

A(x,v) =v ∗,v

for all v ∈ V.

In this paper our aim is to prove a linear extension and a nonlinear extension of

Theorem 1.1 In the linear case, we use a variant of a theorem due to Hayden [2,3], and thus manage to substitute the coercivity condition in (c) of the previous theorem with a more general inf-sup condition In the nonlinear case, we appropriately modify the notion of typeM operator and use a surjectivity result for monotone,

hemicontinu-ous, coercive operators We also present two examples to illustrate the applicability of our results

All Banach spaces considered are overR Given a Banach spaceX, X ∗will denote its dual and·,·will denote their duality product Moreover, ifM is a subset of X, then

M ⊥will denote its annihilator inX ∗and ifN is a subset of X ∗, then⊥ N will denote its

preannihilator inX.

2 The linear case

To prove our main result for the linear case, we need the following lemma which is a variant of [2, Theorem 12] and [3, Theorem 1]

Lemma 2.1 Let X be a reflexive Banach space, let Y be a Banach space and let

be a bounded, bilinear form satisfying the following two conditions:

(a)A is nondegenerate with respect to the second variable, that is, for each y ∈ Y \ {0} , there exists x ∈ X with A(x, y) 0;

(b) there exists c > 0 such that

sup

 y =1

for all x ∈ X.

Then, for every y ∗ ∈ Y ∗ , there exists a unique x ∈ X with

A(x, y) =y ∗,y

for all y ∈ Y.

Proof Let T : X → Y ∗with Tx, y  = A(x, y), for all x ∈ X and all y ∈ Y Obviously ,T

is a bounded linear map Since, by (b), Tx  ≥ c  x , for allx ∈ X, T is one to one To

complete the proof, we need to show thatT is onto.

SinceA is nondegenerate with respect to the second variable, we have that

⊥ T(X) =y ∈ Y | A(x, y) =0,∀ x ∈ X

= {0} (2.4)

⊥

T(X) ⊥

and so by [4, Proposition 2.6.6],

Thus to show thatT maps X onto Y ∗, we need to prove thatT(X) is w ∗-closed inY ∗ To see that, let{ Tx λ } λ ∈Λbe a net inT(X) and let y ∗be an element ofY ∗such that

Without loss of generality, we may assume, using the special case of the Krein-ˇSmulian theorem onw ∗-closed linear subspaces (see [4, Corollary 2.7.12]), the proof of which

is originally due to Banach [5, Theorem 5, page 124] for the separable case and due to Dieudonn´e [6, Theorem 23] for the general case, that{ Tx λ } λ ∈Λis bounded Thus, since

 Tx  ≥ c  x for allx ∈ X, the net { x λ } λ ∈Λis also bounded Hence, sinceX is reflexive,

there exist a subnet{ x λ μ } μ ∈ Mand an elementx of X such that { x λ μ } μ ∈ Mconverges weakly

tox Since T is w − w ∗continuous, Tx λ μ w → ∗ Tx Hence Tx = y ∗, and soT(X) is w ∗

Remark 2.2 An alternative proof of the previous lemma can be obtained using the closed

range theorem

We are now in a position to prove our main result for the linear case

Theorem 2.3 Let X be a reflexive Banach space, let Y be a Banach space, let Λ be a directed set, let { X λ } λ ∈Λ be a family of closed subspaces of X, let { Y λ } λ ∈Λ be an upwards directed family of closed subspaces of Y, and let V =λ ∈Λ Y λ Suppose that

is a function for which the following hold:

(a)A λ = A | X λ × Y λ is a bounded bilinear form, for all λ ∈ Λ;

(b)A( ·,v) is a bounded linear functional on X, for all v ∈ V;

(c)A λ is nondegenerate with respect to the second variable, for all λ ∈ Λ;

(d) there exists c > 0 such that for all λ ∈ Λ,

sup

y ∈ Y λ, y =1

for all x ∈ X λ

Then, for each bounded linear functional v ∗ on V, there exists x ∈ X such that

A(x,v) =v ∗,v

for all v ∈ V.

Proof Let v ∗ ∈ V ∗, and for eachλ ∈ Λ, let v ∗

λ = v ∗ | Y λ For allλ ∈ Λ, v ∗

λ is a bounded linear functional onY λ By hypothesis, for allλ ∈ Λ, A λ is a bounded bilinear form on

X λ × Y λsatisfying the two conditions ofLemma 2.1 Since for allλ ∈ Λ, X λis a reflexive Banach space, we get that for eachλ ∈ Λ, there exists a unique x λsuch thatA λ(x λ,y) =

 v λ ∗,y , for ally ∈ Y λ SinceA satisfies condition (d), we get that for all λ ∈Λ,

c  x λ  ≤ sup

y ∈ Y λ, y =1

A λ(x λ,y)  = sup

y ∈ Y λ, y =1

v ∗

λ,y  ≤  v ∗  (2.11)

So{ x λ } λ ∈Λ is a bounded net inX Since X is reflexive, there exist a subnet { x λ μ } μ ∈ M of

{ x λ } λ ∈Λandx in X such that { x λ μ } μ ∈ Mconverges weakly tox.

We are going to prove that A(x,v) =  v ∗,v , for allv ∈ V Take v ∈ V Then there

exists someλ0∈ Λ with v ∈ Y λ0 Since{ x λ μ } μ ∈ Mis a subnet of{ x λ } λ ∈Λ, there exists some

μ0∈ M with λ μ0≥ λ0 Hence, since the family{ Y λ } λ ∈Λis upwards directed,

for allμ ≥ μ0 Thus, for allμ ≥ μ0,

A λ μ

x λ μ,v

=v ∗ λ μ,v

Therefore

lim

μ ∈ M A

x λ μ,v

=v ∗,v

SinceA( ·,v) is a bounded linear functional on X,

lim

μ ∈ M A

x λ μ,v

The following example illustrates the possible applicability ofTheorem 2.3

Example 2.4 Let a ∈ C1(0, 1) be a decreasing function with limt →0 a(t) = ∞anda(t) ≥0, for allt ∈(0, 1) We will establish the existence of a solution for the following Cauchy problem:

u +a(t)u = f a.e on (0, 1),

where f ∈ L2(0, 1)

LetX = { u ∈ H1(0, 1)| u(0) =0}be equipped with the norm u  =(01| u  |2dt)1/2, which is equivalent to the original Sobolev norm, andY = L2(0, 1) Note thatX is a

re-flexive Banach space, being a closed subspace ofH1(0, 1) Let{ α n } n ∈N be a decreasing sequence in (0, 1) with limn →∞ α n =0 Define

X n =u ∈ H1 

α n, 1

| u

α n

=0

, Y n = L2 

α n, 1

(2.17)

(we can considerX nand Y n as closed subspaces ofX and Y, resp., by extending their

elements by zero outside (α n, 1)) Also letV =∞ n =1 Y n

LetA : X × V → Rbe the bilinear map defined by

A(u,v) =

1

0u  v dt +

1

A is well defined and A( ·,v) is a bounded linear functional on X for any v ∈ V.

LetA n = A | X n × Y n.A nbe a bounded bilinear form since

A n(u,v)  ≤ 1 +M n

 u  X n  v  Y n, (2.19) whereM nis the bound ofa on [α n, 1] It should be noted thatA is not bounded on the

whole ofX × V.

To show thatA nis nondegenerate, letv ∈ Y nand assume thatA n(u,v) =0 for allu ∈

X n, that is,

1

α n



u +a(t)u

v dt =0, ∀ u ∈ X n (2.20)

It is easy to see that the above implies that

1

α n

for any continuous functionw, and therefore v =0

We next show that

sup

 v =1, v ∈ Y n

A n(u,v)  ≥  u  X n (2.22)

DefineT n:X n → Y n ∗by T n u,v  = A n(u,v) T nis a well-defined bounded linear operator andT n u = u +a(t)u Hence

T n u 2

=

1

α n

u +a(t)u 2

dt

=

1

α n

| u  |2dt +

1

α n

a2(t) | u |2dt +

1

α n

a(t)(u2) dt

=

1

α n

| u  |2

dt +

1

α n



a2(t) − a (t)

| u |2

dt + a(1)u2(1)≥  u 2

X n,

(2.23)

sinceu(α n)=0,a is decreasing and a(t) ≥0 for allt ∈(0, 1)

All the hypotheses ofTheorem 2.3are hence satisfied and so ifF ∈ V ∗is defined by

F(v) = 1

0 f v dt, then there exists u ∈ X such that

Thusu satisfies (2.16)

3 The nonlinear case

We start by recalling some well-known definitions

Definition 3.1 Let T : X → X ∗be an operator ThenT is said to be

(i) monotone if Tx − T y,x − y  ≥0, for allx, y ∈ X;

(ii) hemicontinuous if for allx, y ∈ X, T(x + ty) → w Tx as t →0+;

(iii) coercive if

lim

 x →∞

 Tx,x 

We also need the following generalization of the notion of typeM operator (for the

classical definition, see [7] or [8])

Definition 3.2 Let X be a Banach space, let V be a linear subspace of X, and let

be a function ThenA is said to be of type M with respect to V if for any net { v λ } λ ∈Λin

V,x ∈ X and v ∗ ∈ V ∗;

(a)v λ → w x;

(b)A(v λ,v) →  v ∗,v , for allv ∈ V;

(c)A(v λ,v λ) v ∗,x , wherev ∗is the extension ofv ∗on the closure ofV,

imply thatA(x,v) =  v ∗,v , for allv ∈ V.

Our result is the following

Theorem 3.3 Let X be a reflexive Banach space, let Λ be a directed set, let { X λ } λ ∈Λ be an upwards directed family of closed subspaces of X, and let V =λ ∈Λ X λ Suppose that

is a function for which the following hold:

(a)A is of type M with respect to V;

(b) lim x →∞ A(x,x)/  x  = ∞ ;

(c)A λ(x, ·)∈ X λ ∗ , for all λ ∈ Λ and all x ∈ X λ , where A λ is the restriction of A on

X λ × X λ ;

(d) the operator T λ:X λ → X λ ∗ , defined by  T λ x, y  = A λ(x, y) for all x, y ∈ X λ , is mono-tone and hemicontinuous for all λ ∈ Λ.

Then for each v ∗ ∈ V ∗ , there exists x ∈ X such that

A(x,v) =v ∗,v

for all v ∈ V.

Proof As in the proof ofTheorem 2.3, for eachλ ∈ Λ, let v ∗

λ = v ∗ | X λ By the Browder-Minty theorem (see [8, Theorem 26.A]), a monotone, coercive, and hemicontinuous op-erator, from a real reflexive Banach space into its dual, is onto Thus, by (b) and (d), for

eachλ ∈ Λ, the operator T λis onto and so there existsx λ ∈ X λsuch that

A λ

x λ,y

=v ∗ λ,y

for all y ∈ X λ In particularA λ(x λ,x λ)=  v ∗ λ,x λ , and hence by (b), we get that the net

{ x λ } λ ∈Λ is bounded Continuing as in the proof ofTheorem 2.3and applying the fact thatA is of type M with respect to V, we get the required result. 

Remark 3.4 It should be noted that since a crucial point in the above proof is the existence

and boundedness of the net{ x λ } λ ∈Λ, variants of the previous theorem could be obtained using in (b) and (d) alternative conditions corresponding to other surjectivity results

We now applyTheorem 3.3to a singular Dirichlet problem

Example 3.5 LetΩ be a bounded domain inRN We consider the Dirichlet problem

−N

i =1

∂

∂x i



a(x) ∂u

∂x i



+f (x,u) =0 a.e onΩ,

u =0 on∂Ω,

(3.6)

wherea ∈ L ∞loc(Ω) and there exists c1> 0 such that a(x) ≥ c1a.e onΩ, and f : Ω × R →

Ris a monotone increasing (with respect to its second variable for each fixed x ∈Ω) Carath´eodory function, for which there existh ∈ L2(Ω) and c2> 0 such that

f (x,u)  ≤ h(x) + c2| u |, ∀ x ∈ Ω, u ∈ R (3.7)

We will show that if the above hypotheses ona and f hold, then problem (3.6) has a weak solution, that is, that there exists a functionu ∈ H1(Ω) with

Ωa(x) ∇ u ∇ v dx +

Ωf (x,u)vdx =0, ∀ v ∈ C ∞0(Ω). (3.8)

To this end, letX = H1(Ω), let{Ωn } n ∈Nbe an increasing sequence of open subsets of

Ω such that Ωn ⊆Ωn+1and

∞



n =1

and X n = H1(Ωn), for eachn ∈ N Observe that we can consider each X n as a closed subspace ofX by extending its elements by zero outside Ω nand let

V =

∞



n =1

Finally, let

be the function defined by

A(u,v) =

Ωa(x) ∇ u ∇ v dx +

Bya(x) ≥ c1a.e onΩ, the monotonicity of f , and the growth condition (3.7), we have

A(u,u) =

Ωa(x) |∇ u |2dx +

Ωf (x,u)udx

=

Ωa(x) |∇ u |2

dx +

Ω



f (x,u) − f (x,0)

udx +

Ωf (x,0)udx

≥ c1∇ u 2

L2 (Ω)−  h  L2 (Ω) u  H1 (Ω).

(3.13)

Since by the Poincar´e inequality∇ u  L2 (Ω)is equivalent to the norm ofX, it follows that

A is coercive.

LetA n = A | X n × X n Then, sincea ∈ L ∞loc(Ω), it follows that a∈ L ∞(Ωn), for alln ∈ N Combining this with (3.7), we have that

A n(u,v)  ≤ c(u,n)  v  X n, (3.14) wherec(u,n) is a positive constant depending on n and u So the operator

T n:X n −→ X n ∗, (3.15) with T n u,v  X n = A n(u,v), is well defined for all n ∈ N Let

T1,n,T2,n:X n −→ X n ∗ (3.16)

be the operators defined by



T1,n u,v

X n =

Ωn

a(x) ∇ u ∇ v dx, 

T2,n u,v

X n =

Ωn

f (x,u)v dx. (3.17)

ThenT1,nis a monotone bounded linear operator Using the monotonicity of f , it is easy

to see thatT2,nis monotone Finally, recalling that the Nemytskii operator corresponding

to f is continuous (see, e.g., [8, Proposition 26.7]) and that the embedding ofX ninto

L2(Ωn) is compact, we have thatT2,nis hemicontinuous ThusT n = T1,n+T2,nis mono-tone and hemicontinuous for alln ∈ N

To finish the proof, letu n → w u in X Then since for all v ∈ V,

u −→

is a bounded linear functional and, by the continuity of the Nemytskii operator and the compactness of the embedding ofX into L2(Ω),

Ωf

x,u n

v dx −→

for allv ∈ V, we get that

A

u n,v

ThusA is of type M with respect to V Applying nowTheorem 3.3we get that there exists

u ∈ X such that A(u,v) =0 for allv ∈ V Observing that C0∞(Ω) is contained in V, we get thatu is the required weak solution of (3.6)

Acknowledgments

The authors would like to thank Professor A Katavolos for pointing out an error in an earlier version of this paper and the two referees for comments and suggestions which improved both the content and the presentation of this paper

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1990.

Dimosthenis Drivaliaris: Department of Financial and Management Engineering,

University of the Aegean, 31 Fostini Street, 82100 Chios, Greece

Email address:d.drivaliaris@fme.aegean.gr

Nikos Yannakakis: Department of Mathematics, School of Applied Mathematics and Natural Sciences, National Technical University of Athens, Iroon Polytexneiou 9, 15780 Zografou, Greece

Email address:nyian@math.ntua.gr

Thusu satisfies (2.16)

3...