www.elsevier.com/locate/anihpc
An inverse function theorem in Fréchet spaces
Ivar Ekeland
1Canada Research Chair in Mathematical Economics, University of British Columbia, Canada Received 23 September 2010; received in revised form 23 October 2010; accepted 23 October 2010
Available online 6 November 2010
Abstract
I present an inverse function theorem for differentiable maps between Fréchet spaces which contains the classical theorem of Nash and Moser as a particular case. In contrast to the latter, the proof does not rely on the Newton iteration procedure, but on Lebesgue’s dominated convergence theorem and Ekeland’s variational principle. As a consequence, the assumptions are substantially weakened: the mapFto be inverted is not required to beC2, or evenC1, or even Fréchet-differentiable.
©2010 Elsevier Masson SAS. All rights reserved.
Résumé
Je présente un théorème d’inversion pour des applications différentiables entre espaces de Fréchet, qui contient le théorème classique de Nash et Moser. Contrairement à ce dernier, la démonstration donnée ici ne repose pas sur l’algorithme itératif de Newton, mais sur le théorème de convergence dominée de Lebesgue et le principe variationnel d’Ekeland. Comme conséquence, les hypothèses sont substantiellement affaiblies : on ne demande pas que l’applicationFà inverser soit de classeC2, ni mêmeC1, ni même différentiable au sens de Fréchet.
©2010 Elsevier Masson SAS. All rights reserved.
Keywords:Inverse function theorem; Implicit function theorem; Fréchet space; Nash–Moser theorem
1. Introduction
Recall that a Fréchet spaceXisgradedif its topology is defined by an increasing sequence of norms k,k0:
∀x∈X, xkxk+1.
Denote byXkthe completion ofXfor the norm k. It is a Banach space, and we have the following scheme:
→Xk+1→ikXk→ik−1Xk−1→ · · · →X0
E-mail address:ekeland@math.ubc.ca.
URL:http://www.pims.math.ca/~ekeland.
1 The author thanks Eric Séré, Louis Nirenberg, Massimiliano Berti and Philippe Bolle, who were the first to see this proof. Particular thanks go to Philippe Bolle for spotting a significant mistake in an earlier version, and to Eric Séré, whose careful reading led to several improvements and clarifications. Their contribution and friendship is gratefully acknowledged.
0294-1449/$ – see front matter ©2010 Elsevier Masson SAS. All rights reserved.
doi:10.1016/j.anihpc.2010.11.001
where each identity mapik is injective and continuous, in factik1. By definition,Xis a dense subspace ofXk, we haveX=∞
k=0Xk, andxj→ ¯xinXif and only ifxj− ¯xk→0 for everyk0.
Our main example will beX=C∞(Ω,¯ Rd), whereΩ¯ ⊂Rn is compact, is the closure of its interiorΩ, and has smooth boundary. It is well known that the topology ofC∞(Ω,¯ Rd)can be defined in two equivalent ways. On the one hand, we can writeC∞(Ω,¯ Rd)=
Ck(Ω,¯ Rd), whereCk(Ω,¯ Rd)is the Banach space of all functions continuously differentiable up to orderk, endowed with the sup norm:
xk:= max
p1+···+pnpmax
ω∈ ¯Ω
∂p1+···+pnx
∂p1ω1· · ·∂pnωn(ω)
. (1.1)
On the other, we can also writeC∞(Ω,¯ Rd)=
Hk(Ω,Rd), whereHk(Ω,Rd)is the Sobolev space consisting of all functions with square-integrable derivatives, up to orderk, endowed with the Hilbert space structure:
x2k:=
p1+···+pnp
Ω
∂p1+···+pnx
∂p1ω1· · ·∂pnωn
2dω. (1.2)
We will prove an inverse function theorem between graded Fréchet spaces. Let us give a simple version inC∞; in the following statement, either definition of thek-norms, (1.1) or (1.2), may be used:
Theorem 1.LetF be a map from a graded Fréchet spaceX=
k0Xk intoC∞. Assume that there are integersd1 andd2, and sequencesmk>0,mk>0,k∈N, such that, for allx is some neighborhood of0inX, we have:
(1) F (0)=0.
(2) F is continuous and Gâteaux-differentiable, with derivative DF(x).
(3) For everyu∈X, we have
∀k0, DF(x)u
kmkuk+d1. (1.3)
(4) DF(x)has a right-inverseL(x):
∀v∈C∞, DF(x)L(x)v=v.
(5) For everyv∈C∞, we have:
∀k0, L(x)v
kmkvk+d2. (1.4)
Then for everyy∈C∞such that yk0+d2< R
mk
0
and everym > mk
0 there is somex∈Xsuch that:
xk0< R,
xk0myk0+d2, and:
F (x)=y.
The full statement of our inverse function theorem and of its corollaries, such as the implicit function theorem, will be given in the text (Theorems 4 and 5). Note the main feature of our result: there is aloss of derivativesboth for DF(x), by condition (1.3), and forL(x), by condition (1.4).
Since the pioneering work of Andrei Kolmogorov and John Nash [10,2–4,13], this loss of derivatives has been overcome by using the Newton procedure: the equation F (x)=y is solved by the iteration schemexn+1=xn− L(xn)F (xn)([15,11,12]; see [9,1,5] for more recent expositions). This method has two drawbacks. The first one is
that it requires the functionF to beC2, which is quite difficult to satisfy in infinite-dimensional situations. The second is that it gives a set of admissible right-hand sidesywhich is unrealistically small: in practical situations, the equation F (x)=ywill continue to have a solution long after the Newton iteration starting fromyceases to converge.
Our proof is entirely different. It gives the solution ofF (x)=y directly by using Ekeland’s variational principle ([7,8]; see [6] for later developments). Since the latter is constructive, so is our proof, even if it does not rely on an iteration scheme to solve the equation. To convey the idea of the method in a simple case, let us now state and prove an inverse function theorem in Banach spaces:
Theorem 2.LetXandYbe Banach spaces. LetF :X→Ybe continuous and Gâteaux-differentiable, withF (0)=0.
Assume that the derivative DF(x)has a right-inverseL(x), uniformly bounded in a neighborhood of0:
∀v∈Y, DF(x)L(x)v=v, xR ⇒ L(x)m.
Then, for everyy¯such that:
¯y<R m
and everyμ > mthere is somex¯such that:
¯x< R, ¯xμ ¯y, and:
F (x)¯ = ¯y.
Proof. Consider the functionf:X→Rdefined by f (x)=F (x)− ¯y.
It is continuous and bounded from below, so that we can apply Ekeland’s variational principle. For everyr >0, we can find a pointx¯such that:
f (x)¯ f (0), ¯xr,
∀x, f (x)f (x)¯ −f (0)
r x− ¯x. Note thatf (0)= ¯y. Taker=μ ¯y, so that:
f (x)¯ f (0), ¯xμ ¯y,
∀x, f (x)f (x)¯ − 1
μx− ¯x. (1.5)
The pointx¯satisfies the inequality ¯xμ ¯y. Sincem < R ¯y−1, we can assume without loss of generality that μ < R ¯y−1, so ¯x< R. It remains to prove thatF (x)¯ = ¯y.
We argue by contradiction. Assume not, soF (x)¯ = ¯y. Then, write the inequality (1.5) withx= ¯x+t u. We get:
∀t >0, ∀u∈X, F (x¯+t u)− ¯y − F (x)¯ − ¯y
t −1
μu. (1.6)
As we shall see later on (Lemma 1), the functiont→ F (x¯+t u)− ¯y is right-differentiable att =0, and its derivative is given by
tlim→0 t >0
F (x¯+t u)− ¯y − F (x)¯ − ¯y
t =
y∗,DF(x)u¯
for somey∗∈Y∗withy∗∗=1 andy∗, F (x)¯ − ¯y = F (x)¯ − ¯y. In the particular case whenY is a Hilbert space, we have
y∗= F (x)¯ − ¯y F (x)¯ − ¯y. Lettingt→0 in (1.6), we get:
∀u,
y∗,DF(x)u¯
−1
μu.
We now takeu= −L(x)(F (¯ x)¯ − ¯y), so thatDF(x)u¯ = −(F (x)¯ − ¯y). The preceding inequality yields:
F (x)¯ − ¯y 1
μum
μF (x)¯ − ¯y which is a contradiction sinceμ > m. 2
The reader will have noted that this is much stronger than the usual inverse function theorem in Banach spaces: we do not require thatF be Fréchet-differentiable, nor that the derivativeDF(x)or its inverseL(x)depend continuously onx. All that is required is an upper bound onL(x). Note that it is very doubtful that, with such weak assumptions, the usual Euler or Newton iteration schemes would converge.
Our inverse function theorem will extend this idea to Fréchet spaces. Ekeland’s variational principle holds for any complete metric space, hence on Fréchet spaces. Our first result, Theorem 3, depends on the choice of a non-negative sequence βk which converges rapidly to zero. An appropriate choice ofβk will lead to the facts that F is a local surjection (Corollary 1), that the (multivalued) local inverseF−1satisfies a Lipschitz condition (Corollary 2), and that the result holds also with finite regularity (ify¯does not belong toYk for everyk, but only toYk0+d2, then we can still solveF (x)¯ = ¯ywithx¯∈Xk0). These results are gathered together in Theorem 4, which is our final inverse function theorem. As usual, an implicit function theorem can be derived; it is given in Theorem 5.
The structure of the paper is as follows. Section 2 introduces the basic definitions. There will be no requirement onX, whileY will be asked to belong to a special class of graded Fréchet spaces. This class is much larger than the one used in the Nash–Moser literature: it is not required thatXorY be tame in the sense of Hamilton [9] or admit smoothing operators. Section 3 states Theorem 3 and derives the other results. The proof of Theorem 3 is given in Section 4.
Before we proceed, it will be convenient to recall some well-known facts about differentiability in Banach spaces.
LetXbe a Banach space. LetU be some open subset ofX, and letF be a map fromU into some Banach spaceY. Definition 1.F isGâteaux-differentiableatx∈U if there exists some linear continuous map fromXtoY, denoted byDF(x), such that:
∀u∈X, lim
t→0
F (x+t u)−F (x)
t −DF(x)u
=0.
Definition 2.F isFréchet-differentiableatx∈U if it is Gâteaux-differentiable and:
ulim→0
F (x+u)−F (x)−DF(x)u u =0.
Fréchet-differentiability implies Gâteaux-differentiability and continuity, but Gâteaux-differentiability does not even imply continuity.
IfY is a Banach space, then the normy→ yis convex and continuous, so that it has a non-empty subdifferential N (y)⊂Y∗at every pointy:
y∗∈N (y) ⇐⇒ ∀z∈Y, y+zy + y∗, z
.
Wheny=0 we have the alternative characterization:
y∗∈N (y) ⇐⇒ y∗∗=1 and y∗, y
= y.
N (y) is convex and compact in the σ (X∗, X)-topology. It is a singleton if and only if the norm is Gâteaux- differentiable aty. Its unique element then is the Gâteaux-derivative ofF aty:
N (y)= y∗
⇐⇒ DF(x)=y∗.
IfN (y)contains several elements, the norm is not Gâteaux-differentiable aty, and the preceding relation is then replaced by the following classical result (see for instance Theorem 11, p. 34 in [14]):
Proposition 1.TakeyandzinY. Then there is somey∗∈N (y)(depending possibly onz)such that:
tlim→0 t >0
y+t z − y
t =
y∗, z .
The following result will be used repeatedly:
Lemma 1.AssumeF :X→Y is Gâteaux-differentiable. Takexandξ inX, and define a functionf: [0,1] →Rby f (t ):=F (x+t ξ ), 0t1.
Thenf has a right-derivative everywhere, and there is somey∗∈N (F (x+t ξ ))such that:
hlim→0 h>0
f (t+h)−f (t )
h =
y∗,DF(x+t ξ )ξ .
Proof. We have:
h−1
f (t+h)−f (t )
=h−1F
x+(t+h)ξ−F (x+t ξ )
=h−1F (x+t ξ )+hz(h)−F (x+t ξ ) (1.7)
with:
z(h)=F (x+(t+h)ξ )−F (x+t ξ )
h .
SinceF is Gâteaux-differentiable, we have:
hlim→0z(h)=DF(x+t ξ )ξ :=z(0).
By the triangle inequality, we have:
F (x+t ξ )+hz(h)−F (x+t ξ )+hz(0)hz(h)−z(0). Writing this into (1.7) and using Proposition 1, we find:
hlim→0 h>0
[f (t+h)−f (t )]
h = lim
h→0 h>0
F (x+t ξ )+hz(0) − F (x+t ξ ) h
=
y∗,DF(x+t ξ )ξ
for somey∗∈N (F (x+t ξ )). This is the desired result. 2 One last word. Throughout the paper, we shall use the following:
Definition 3.A sequenceαkhasunbounded supportif sup{k|αk=0} = ∞.
2. The inverse function theorem LetX=
k0Xkbe a graded Fréchet space. The following result is classical:
Proposition 2. Let αk0 be a sequence with unbounded support such that
αk<∞. Let r >0 be a positive number. Then the topology ofXis induced by the distance:
d(x, y):=
k
αkmin r,x−yk
(2.1)
andxnis a Cauchy sequence ford if and only if it is a Cauchy sequence for all thek-norms. It follows that(X, d)is a complete metric space.
The main analytical difficulty with Fréchet spaces is that, given x∈X, there is no information on the sequence k→ xk, except that it is positive and increasing. For instance, it can grow arbitrarily fast. So we single out some elementsx such thatxk has at most exponential growth ink.
Definition 4.A pointx∈Xiscontrolledif there is a constantc0(x)such that:
xkc0(x)k. (2.2)
Definition 5.A graded Fréchet space isstandardif, for everyx∈X, there is a constantc3(x)and a sequencexnsuch that:
∀k, lim
n→∞xn−xk=0, (2.3)
∀n, xnkc3(x)xk (2.4)
and eachxnis controlled:
xnkc0(xn)k. (2.5)
Proposition 3.The graded Fréchet spacesC∞(Ω,¯ Rd)=
Ck(Ω,¯ Rd)andC∞(Ω,¯ Rd)=
Hk(Ω,Rd)are both standard.
Proof. In fact, much more is true. It can be proved (see [9], [1, Proposition II.A.1.6]) that both admit a family of smoothing operatorsSn:X→Xsatisfying:
(1) Snx−xk→0 whenn→ ∞,
(2) Snxk+dc1ndxk,∀d0,∀k0,∀n0, (3) (I−Sn)xkc2n−dxk+d,∀d0,∀k0,∀n0,
wherec1andc2are positive constants. For everyx∈X, condition (2) withk=0 implies that:
Snxdc1ndx0.
So (2.2) is satisfied and the pointxn=Snx is controlled. Condition (2.3) follows from (1) and condition (2.4) follows from (2) withd=0. 2
Now consider two graded Fréchet spaces X=
k0Xk andY =
k0Yk. We are given a map F :X→Y, a numberR∈(0,∞]and an integerk00. We consider the ball:
BX(k0, R):= x∈Xxk0< R .
Note thatR= +∞is allowed; in that case,BX(k0, R)=X.
Theorem 3.AssumeY is standard. Let there be given two integersd1,d2and two non-decreasing sequencesmk>0, mk>0. Assume that, onBX(k0, R), the mapF satisfies the following conditions:
(1) F (0)=0.
(2) F is continuous, and Gâteaux-differentiable.
(3) For everyu∈Xthere is a numberc1(u) >0for which:
∀k∈N, DF(x)u
kc1(u)
mkuk+d1+F (x)
k
.
(4) There exists a linear mapL(x):Y→Xsuch that DF(x)L(x)=IY:
∀v∈Y, DF(x)L(x)v=v.
(5) For everyv∈Y:
∀k∈N, L(x)v
kmkvk+d2.
Letβk0be a sequence with unbounded support satisfying:
∀n∈N, ∞ k=0
βkmkmk+d
1nk<∞. (2.6)
Then, for everyy¯∈Y such that:
∞ k=0
βk ¯yk<βk0+d2 mk
0
R (2.7)
there exists a pointx¯such that:
F (x)¯ = ¯y, (2.8)
¯xk0 < R. (2.9)
The proof is given in the next section. We will now derive the consequences. Before we do that, note that condition (3) is equivalent to the following, seemingly more general, one:
∀k, DF(x)u
kc1(u)
mkuk+d1+F (x)
k+1
. (2.10)
Indeed, condition (3) clearly implies (2.10) withc1(u)=c1(u). Conversely, if (2.10) holds, foru=0 we set:
c1(u)=c1(u)
1+ 1
m0u0
and then:
c1(u)
mkuk+d1+F (x)
k
c1(u)
mkuk+d1+F (x)
k+1 , so condition (3) holds as well.
Corollary 1(Local surjection). LetX=
k0XkandY =
k0Yk be graded Fréchet spaces, withY standard, and letF :X→Y satisfy conditions(1)to(5). Suppose:
yk0+d2< R mk
0
. (2.11)
Then, for everyμ > mk
0, there is somex∈BX(k0, R)such that:
xk0 μyk0+d2, (2.12)
and:
F (x)=y.
Proof. Givenμ > mk
0, takeR< Rsuch thatmk
0yk0+d2< R< μyk0+d2and chooseh >0 so small that:
yk0+d2+h ∞ k=0
k−k< R mk
0
. (2.13)
Define a sequenceβk by βk=
⎧⎨
⎩
0 fork < k0+d2,
1 fork=k0+d2,
hk−k[max{yk, mkmk+d
1}]−1 fork > k0+d2.
(2.14) Then (2.6) is satisfied. On the other hand:
∞ k=0
βkykyk0+d2+h k−k
and using (2.13) we find that:
∞ k=0
βkyk< 1 mk
0
R<∞. (2.15)
We now apply Theorem 3 withRinstead ofR, and we find somex∈XwithF (x)=yandxk0R. The result follows. 2
Consider the two balls:
BX(k0, R)= xxk0< R , BY
k0+d2, R mk
0
=
yyk0+d2 < R mk
0
.
The preceding corollary tells us thatF maps the first ball onto the second. The inverse map F−1:BY
k0+d2, R mk
0
→BX(k0, R)
is multivalued:
F−1(y)= x∈BX(k0, R)F (x)=y
(2.16) and has non-empty values:F−1(y)= ∅for everyy. The following result shows that it satisfies a Lipschitz condition.
Corollary 2 (Lipschitz inverse). For every y0 and y1 in BY(k0+d2, R(mk
0)−1), every x0∈F−1(y0), and every μ > mk
0, we have:
inf x0−x1k0 x1∈F−1(y1)
=inf x0−x1k0F (x1)=y1
μy0−y1k0+d2.
Proof. Take someR< R with max{y0k0+d2,y1k0+d2}< R(mk
0)−1and consider the line segmentyt =y0+ t (y1−y0), 0t1, joiningy0toy1. We haveyt< R(mk
0)−1for everyt, so that, by Corollary 1, there exists somext∈F−1(yt)withxt< R. The functionFt(x)=F (x+xt)−yt then satisfies conditions (1) to (5) withR replaced byρ=R−R.
Pick somex0∈F−1(y0). By Corollary 1 applied toF0we find that, for everyysuch thaty−y0k0+d2ρ(mk
0)−1 we have somex∈F−1(y)with:
x0−xk0μy0−yk0.
We can connecty0andy1by a finite chainy0 =y0, y2, . . . yN =y1of aligned points, such that the distance between ynandyn+1is always less thanρ(mk
0)−1, and for eachynchoose somexn∈F−1(yn)such that:
xn−xn+1k0μyn −yn+1
k0+d2. Summing up:
x0−x1k0μ N n=0
yn −yn+1
k0+d2 =μy1−y0k0+d2. 2
Note that we are not claiming that the multivalued mapF−1has a Lipschitz section overBY(k0+d2, R(mk
0)−1), or even a continuous one.
As a consequence of Corollary 2, we can solve the equationF (x)=ywhen the right-hand side no longer is inY, but in some of theYk, withkk0+d2.
Corollary 3(Finite regularity). Suppose F extends to a continuous mapF¯ :Xk0 →Yk0−d1. Take somey∈Yk0+d2 withyk0+d2< R(mk
0)−1. Then there is somex∈Xk0 such thatxk0< RandF (x)¯ =y.
Proof. Let yn∈Y be such that yn−yk0+d2−n. By Corollary 2, we can find a sequence xn ∈X such that F (xn)=ynand
xn−xn+1k0μyn−yn+1k0+dμ2−n.
Soxn−xpk0μ2−n+1forp > n, and the sequencexn is Cauchy inXk0. It follows thatxnconverges to some x∈Xk0, withxk0< R, and we getF (x)=y by continuity. 2
Let us sum up our results in a single statement:
Theorem 4 (Inverse function theorem). Let X=
k0Xk and Y =
k0Yk be graded Fréchet spaces, with Y standard, and letF be a map fromXtoY. Assume there exist some integerk0, someR >0 (possibly equal to+∞), integersd1,d2, and non-decreasing sequencesmk>0,mk>0such that, forxk0< R, we have:
(1) F (0)=0.
(2) F is continuous, and Gâteaux-differentiable with derivative DF(x).
(3) For everyu∈Xthere is a numberc1(u)such that:
∀k, DF(x)u
kc1(u)
mkuk+d1+F (x)
k
.
(4) There exists a linear mapL(x):Y→Xsuch that DF(x)L(x)=IY
∀u∈X, DF(x)L(x)v=v.
(5) For everyv∈Y, we have:
∀k∈N, L(x)v
kmkvk+d2.
ThenF maps the ball{xk0< R}inXonto the ball{yk0+d2 < R(mk
0)−1}inY, and for everyμ > mk
0 the inverseF−1satisfies a Lipschitz condition:
∀x1∈F−1(y1), inf x1−x2k0x2∈F−1(y2)
μy1−y2k0+d2.
If F extends to a continuous map F¯ :Xk0 →Yk0−d1, then F¯ maps the ball {xk0 < R}in Xk0 onto the ball {yk0+d2< R(mk
0)−1}inYk0+d2, and the inverseF¯−1satisfies the same Lipschitz condition.
We conclude by rephrasing Theorem 4 as an implicit function theorem.
Theorem 5. Let X=
k0Xk and Y =
k0Yk be graded Fréchet spaces, withY standard, and let F (ε, x)= F0(x)+εF1(x)be a map fromX×RtoY. Assume there exist integersk0,d1,d2, sequencesmk>0,mk>0, and numbersR >0,ε0>0such that, for every(x, ε)such thatxk0Rand|ε|< ε0, we have:
(1) F0(0)=0andF1(0)=0.
(2) F0andF1are continuous, and Gâteaux-differentiable.
(3) For everyu∈Xthere is a numberc1(u)such that:
∀k0, DF(ε, x)u
kc1(u)
mkuk+d1+F (ε, x)
k
.
(4) There exists a linear mapL(ε, x):Y→Xsuch that:
DF(ε, x)L(ε, x)=IY. (5) For everyv∈Y, we have:
∀k0, L(ε, x)v
kmkvk+d2. Then, for everyεsuch that:
|ε|<min R
mk
0
F1(0)−1
k0+d2, ε0
and everyμ > mk
0, there is anxεsuch that:
F (ε, xε)=0 and:
xεk0μ|ε|F1(0)
k0+d2.
Proof. Fixεwith|ε|< ε0. Consider the function:
Gε(x):=F0(x)+ε
F1(x)−F1(0) .
It satisfies conditions (1) to (5) of Theorem 4. The equationF (ε, xε)=0 can be rewrittenGε(x)= −εF1(0):=y. By Theorem 4, we will be able to solve it provided:
yk0+d2= |ε|F1(0)
k0+d2< R 1 mk
0
and the solutionxεthen satisfies
xεk0μyk0+d2=μ|ε|F1(0)
k0+d2. 2 3. Proof of Theorem 3
The proof proceeds in three steps. Using Ekeland’s variational principle, we first associate withy¯∈Y a particular point x¯∈X. We then argue by contradiction, assuming that F (x)¯ = ¯y. We then identify for every na particular direction un∈X, and we investigate the derivative of x →
βkF (x)− ¯yk in the direction un. We finally let n→ ∞and derive a contradiction.
3.1. Step 1
We define a new sequenceαkby αk=βk+d2
mk
and we endowXwith the distanceddefined by d(x1, x2):=
k
αkmin R,x1−x2k
. (3.1)
By Proposition 2,(X, d)is a complete metric space. Now consider the functionf :X→R∪ {+∞}(the value +∞is allowed) defined by
f (x)= ∞ k=0
βkF (x)− ¯y
k. (3.2)
It is obviously bounded from below, and 0inff f (0)=∞
k=0
βk ¯yk<∞. (3.3)
It is also lower semi-continuous. Indeed, letxn→x in(X, d). Thenxn→x in everyXk. By Fatou’s lemma, we have:
lim inf
n f (xn)=lim inf
n
k
βkF (xn)− ¯y
k
k
βklimF (xn)− ¯y
k
and sinceF :X→Y is continuous, we get:
k
βklimF (xn)− ¯y
k=
k
βkF (x)− ¯y
k=f (x) so that lim infnf (xn)f (x), as desired.
By assumption (2.7), we can take someR< Rwith:
∞ k=0
βk ¯yk<βk0+d2 mk
0
R. (3.4)
We now apply Ekeland’s variational principle tof (see [7,8]). We find a pointx¯∈Xsuch that:
f (x)¯ f (0), (3.5)
d(x,¯ 0)Rαk0, (3.6)
∀x∈X, f (x)f (x)¯ − f (0)
Rαk0d(x,x).¯ (3.7)
Replacef (x)by its definition (3.2) in inequality (3.5). We get:
βkF (x)¯ − ¯y
k
∞ k=0
βk ¯yk<∞ (3.8)
and from the triangle inequality it follows that:
∞ k=0
βkF (x)¯
k2 ∞ k=0
βk ¯yk<∞. (3.9)
If ¯xk0 > R, thend(x,¯ 0) > Rαk0, contradicting formula (3.6). So we must have ¯xk0 R< R, and (2.9) is proved.
We now work on (3.7). To simplify notations, we set:
A= ∞
k=0βk ¯yk
Rαk0 = f (0)
Rαk0. (3.10)
It follows from (3.7) that, for everyu∈Xandt >0, we have:
−
f (x¯+t u)−f (x)¯
Ad(x¯+t u,x)¯
and hence, dividing byt:
−1 t
∞
k=0
βky¯−F (x¯+t u)
k−∞
k=0
βky¯−F (x)¯
k
A
t
k0
αkmin R, tuk
. (3.11)
3.2. Step 2
IfF (x)¯ = ¯y, the proof is over. If not, we set:
v=F (x)¯ − ¯y, u= −L(x)v¯ so that:
DF(x)u¯ = −
F (x)¯ − ¯y .
SinceY is standard, there is a sequencevnsuch that:
∀k, vn−vk→0, (3.12)
∀n, vnkc3(v)vk, (3.13)
vnkc0(vn)k. (3.14)
Setun= −L(x)v¯ n. Clearlyun−uk→0 for everyk. We have, using condition (5):
unkmkvnk+d2
mkc0(vn)k+d2. (3.15)
We now substituteuninto formula (3.11), always under the assumption thatF (x)¯ − ¯y=0, and we lett→0. If convergence holds, we get:
−lim
t→0 t >0
1 t
∞
k=0
βky¯−F (x¯+t un)
k− ∞ k=0
βky¯−F (x)¯
k
Alim
t→0 t >0
k0
αk
t min R, tunk
. (3.16)
We shall treat the right- and the left-hand side separately, leavingnfixed throughout.
We begin with the right-hand side. We have:
αk
t min R, tunk
=αkmin R
t ,unk
=:γk(t ).
We haveγk(0)0,γk(t)γk(t )fortt, andγk(t )→αkunk when t→0. By the monotone convergence theorem:
tlim→0 t >0
∞ k=0
1
tαkmin R, tunk
=∞
k=0
αkunk. (3.17)
Now for the left-hand side of (3.16). Rewrite it as
− ∞ k=0
βk
gk(t )−gk(0)
t (3.18)
where:
gk(t ):=y¯−F (x¯+t un)
k.
We have ¯x+t unk ¯xk+tunk. We have seen that ¯xk0 R< R, so there is somet >¯ 0 so small that, for 0< t <t¯, we have ¯x+t unk0 < R. Without loss of generality, we can assume t¯1. Since F is Gâteaux- differentiable, by Lemma 1gkhas a right derivative everywhere, and:
(gk)+(t )= lim
h→0 h>0
gk(t+h)−gk(t ) h
DF(x¯+t un)un
k. (3.19)
Introduce the functionfk(t )= F (x¯+t un)k. It has a right derivative everywhere, and (fk)+(t )DF(x¯+ t un)unk, still by Lemma 1. We shall henceforth write the right derivativesgkandfkinstead of(gk)+and(fk)+. By condition (3), we have:
fk(t )c1(un)
mkunk+d1+fk(t ) , fk(t )−c1(un)fk(t )c1(un)mkunk+d1. Integrating, we get:
e−t c1(un)fk(t )−fk(0)
1−e−t c1(un)
mkunk+d1, mkunk+d1+fk(t )et c1(un)
mkunk+d1+F (x)¯
k
.
Substituting this into condition (3) and using (3.15), we get:
DF(x¯+t un)un
kc1(un)
mkunk+d1+fk(t ) c1(un)et c1(un)
mkunk+d1+F (x)¯
k
C1(un)mkmk+d
1c0(vn)k+d1+d2+C1(un)F (x)¯
k=:k
where the termC1(un):=c1(un)ec1(un)depends onun, but not onk. We have used the fact that 0< t <1.
It follows from (3.19) that the functiongk isk-Lipschitzian. So we get, for everykand 0< t <t¯: βk
gk(t )−gk(0) t
C1(un)βkmkmk+d
1c0(vn)k+d1+d2+C1(un)βkF (x)¯
k.
By assumption (2.6), the first term on the right-hand side belongs to a convergent series. By inequality (3.9), the second term is also summable. So we can apply Lebesgue’s dominated convergence theorem to the series (3.18), yielding:
∞ k=0
−βkgk(0)=lim
t→0 t >0
∞ k=0
−βk
gk(t )−gk(0)
t . (3.20)
Writing (3.17) and (3.20) into formula (3.11) yields:
∞ k=0
−βkgk(0)A
k0
αkunk. (3.21)
We now apply Lemma 1. Denote byNkthe subdifferential of the norm inYk:
∀y=0, y∗∈Nk(y) ⇐⇒ y∗∗
k=1 and
y∗, y
k= yk. There is someyk∗(n)∈Nk(F (x)¯ − ¯y)such that:
gk(0)=
yk∗(n),DF(x)u¯ n
k. (3.22)
Substituting into (3.21) we get:
∞ k=0
−βk
yk∗(n),DF(x)u¯ n
kA
k0
αkunk. (3.23)
