Introduction to [624989]

Part I
Introduction to
variational inequalities

1
Preliminaries on functional analysis
This chapter presents preliminary material from functional analysis which
will be used in subsequent chapters. Some of the results are stated without
proofs, since they are standard and can be found in many references. Nev-
ertheless, we pay particular attention to the results which are repeatedlyused in the following chapters of the book, for which we present the details
in proofs. They include the Banach fixed point theorem, the projection
lemma, the Riesz representation theorem and the Weiersterass theorem,among others. All the linear spaces considered in this book including ab-
stract normed spaces, Banach spaces, Hilbert spaces and various function
spaces are assumed to be real linear spaces.
1.1 Normed spaces
We start this section with basic definitions, notation and results concerning
the normed spaces. Then we recall two main fixed point results: the Banach
fixed point theorem and the Schauder fixed point theorem.
1.1.1 Basic definitions
Given a linear space X, we recall that a norm/bardbl·/bardblXis a function from X
toRwith the following properties.
(1)/bardblu/bardblX≥0∀u∈X,a n d/bardblu/bardblX=0i ffu=0X.

4 Preliminaries on functional analysis
(2)/bardblαu/bardblX=|α|/bardblu/bardblX∀u∈X,∀α∈R.
(3)/bardblu+v/bardblX≤/bardblu/bardblX+/bardblv/bardblX∀u,v∈X.
The pair ( X,/bardbl·/bardblX) is called a normed space . Here and everywhere in this
book 0 Xwill denote the zero element of X. Also, we will simply say Xis
a normed space when the definition of the norm is understood from the
context.
On a linear space various norms can be defined. Sometimes it is desirable
to know if two norms are related. Let /bardbl·/bardbl(1)and/bardbl·/bardbl(2)be two norms over
a linear space X. The two norms are said to be equivalent if there exist two
constants c1,c2>0 such that
c1/bardblu/bardbl(1)≤/bardblu/bardbl(2)≤c2/bardblu/bardbl(1)∀u∈X. (1.1)
We recall that a sequence {un}⊂Xis said to converge (strongly) to
u∈Xif
/bardblun−u/bardblX→0a sn→∞. (1.2)
In this case uis called the (strong) limit of the sequence {un}and we write
u= lim
n→∞unorun→uinX.
It is straightforward to verify that the limit of a sequence, if it exists,
is unique. The adjective “strong” is introduced in the previous definition
to distinguish this convergence from the weak convergence which will beintroduced on page 6.
A sequence {u
n}⊂Xis said to be bounded if there exists M> 0 such
that
/bardblun/bardblX≤M∀n∈N (1.3)
or, equivalently, if
sup
n/bardblun/bardblX<∞.
To test the convergence of a sequence without knowing its limit, it is
usually convenient to refer to the notion of a Cauchy sequence. Let Xbe
a normed space. A sequence {un}⊂Xis called a Cauchy sequence if
/bardblum−un/bardblX→0a sm,n→∞.
Obviously, a convergent sequence is a Cauchy sequence, but in a general
infinite dimensional space, a Cauchy sequence may fail to converge. Thisjustifies the following definition.
Definition 1.1 A normed space is said to be complete if every Cauchy
sequence from the space converges to an element in the space. A completenormed space is called a Banach space .

1.1 Normed spaces 5
From this definition it follows that ( X,/bardbl·/bardblX)i saB a n a c hs p a c ei fff o r
every Cauchy sequence {un}⊂Xthere exists an element u∈Xsuch that
un→uinX. Moreover, using (1.1) and (1.2) it is easy to see that, for
two equivalent norms, convergence in one norm implies convergence in theother norm. As a consequence, if /bardbl·/bardbl
(1)and/bardbl·/bardbl(2)are equivalent norms
on the linear space Xthen (X,/bardbl·/bardbl(1))i saB a n a c hs p a c ei ff( X,/bardbl·/bardbl(2))i s
a Banach space.
We introduce in what follows a particular type of normed space, in which
the norm is defined in a special way. Given a linear space Xwe recall that
aninner product (·,·)Xis a function from X×XtoRwith the following
properties.
(1) (u,u)X≥0∀u∈X, and (u,u)X=0i ffu=0X.
(2) (u,v)X=(v,u)X∀u,v∈X.
(3) (αu+βv,w)X=α(u,w)X+β(v,w)X∀u,v,w∈X,∀α,β∈R.
The pair ( X,(·,·)X) is called an inner product space . When the definition
of the inner product ( ·,·)Xis clear from the context, we simply say Xis
an inner product space.
Next, it is well known that an inner product ( ·,·)Xinduces a norm
through the formula
/bardblu/bardblX=/radicalbig
(u,u)X∀u∈X (1.4)
and we note that everywhere in this book the norm in an inner product
space is the one induced by the inner product through the above formula.
For an inner product space we have the Cauchy–Schwarz inequality :
|(u,v)X|≤/bardblu/bardblX/bardblv/bardblX∀u,v∈X, (1.5)
with the equality holding iff uandvare linearly dependent. Moreover, the
following identity holds:
/bardblu+v/bardbl2
X+/bardblu−v/bardbl2
X=2/parenleftbig
/bardblu/bardbl2
X+/bardblv/bardbl2
X/parenrightbig
∀u,v∈X. (1.6)
Identity (1.6) is called the parallelogram identity or the parallelogram law .
Among the inner product spaces, of particular importance are the Hilbert
spaces.
Definition 1.2 A complete inner product space is called a Hilbert space .
From the definition, we see that an inner product space Xis a Hilbert
space ifXis a Banach space under the norm induced by the inner product.

6 Preliminaries on functional analysis
1.1.2 Linear continuous operators
Let (X,/bardbl·/bardblX) and (Y,/bardbl·/bardblY) be two normed spaces and let L:X→Ybe
an operator. We recall that L:X→Yislinear if
L(α1v1+α2v2)=α1L(v1)+α2L(v2)∀v1,v2∈X,∀α1,α2∈R.
The operator Lis said to be continuous if
un→uinX=⇒L(un)→L(u)i nY.
It can be proved that, if Lis linear, then Lis continuous if and only if it
isbounded , i.e., there exists M>0 such that
/bardblL(v)/bardblY≤M/bardblv/bardblX∀v∈X.
We will use the notation L(X,Y) for the set of all linear continuous
operators from XtoY.F o rL∈L(X,Y), the quantity
/bardblL/bardblL(X,Y )=s u p
0/negationslash=v∈X/bardblLv/bardblY
/bardblv/bardblX(1.7)
is called the operator norm ofL,a n dL/mapsto→/bardblL/bardblL(X,Y )defines a norm on
the space L(X,Y). Moreover, if Yis a Banach space then L(X,Y) is also
a Banach space. For a linear operator L, we usually write L(v)a sLv, but
sometimes we also write Lveven when Lis not linear.
For a normed space X, the space L(X,R) is called the dualspace of X
and is denoted by X/prime. The elements of X/primeare linear continuous functionals
onX. Recall that a linear functional /lscript:X→Rbelongs to X/primeiff it is
continuous , i.e.
un→uinX=⇒/lscript(un)→/lscript(u)i n R
or, equivalently, if it is bounded , i.e. there exists M>0 such that
|/lscript(v)|≤M/bardblv/bardblX∀v∈X.
The duality pairing between X/primeandXis usually denoted by /lscript(v)o r
/angbracketleftv/prime,v/angbracketrightor/angbracketleftv/prime,v/angbracketrightX/prime×Xfor/lscript,v/prime∈X/primeandv∈X. It follows from (1.7) that a
norm on X/primeis
/bardbl/lscript/bardblX/prime=s u p
0/negationslash=v∈X|/lscript(v)|
/bardblv/bardblX. (1.8)
Moreover, ( X/prime,/bardbl·/bardblX/prime) is always a Banach space.
We can now introduce another kind of convergence in a normed space. A
sequence {un}⊂Xis said to converge weakly tou∈Xif for each /lscript∈X/prime,
/lscript(un)→/lscript(u)a sn→∞.

1.1 Normed spaces 7
In this case uis called the weak limit of{un}and we write
un/arrowrighttophalfu inX.
It follows from the Hahn–Banach theorem that the weak limit of a sequence,
if it exists, is unique. Moreover, it is easy to see that strong convergence
implies weak convergence, i.e., if un→uinX, thenun/arrowrighttophalfuinX.T h e
converse of this property is not true in general.
Assume now that ( X,(·,·)X) is an inner product space and note that in
this case u/mapsto→(u,v)Xis a linear continuous functional on X, for allv∈X.
Therefore, it follows from the definition of the weak convergence that
un/arrowrighttophalfu inX=⇒(un,v)X→(u,v)Xasn→∞,∀v∈X.(1.9)
We shall see in Section 1.2 that, in the case when ( X,(·,·)X) is a Hilbert
space, the converse of (1.9) is true.
The convergence of sequences is used to define closed subsets in a normed
space.
Definition 1.3 LetXbe a normed space. A subset K⊂Xis called:
(i)(strongly) closed if the limit of each convergent sequence of elements
ofKbelongs to K, that is
{un}⊂K, u n→uinX=⇒u∈K;
(ii)weakly closed if the limit of each weakly convergent sequence of ele-
ments of Kbelongs to K,t h a ti s
{un}⊂K, u n/arrowrighttophalfu inX=⇒u∈K.
Evidently, every weakly closed subset of Xis (strongly) closed, but the
converse is not true, in general. An exception is provided by the class of
convex subsets of a Banach space, as shown in the following result.
Theorem 1.4 (Mazur’s theorem) A convex subset of a Banach space is
(strongly) closed if and only if it is weakly closed.
Here, for the convenience of the reader, we recall that a subset Kof a
linear space is said to be convex if it has the property
u,v∈K=⇒(1−t)u+tv∈K∀t∈[0,1].
LetXandYbe linear spaces. A mapping a:X×Y→Ris called a
bilinear form if it is linear in each argument, that is for any u1,u2,u∈X,
v1,v2,v∈Y,a n dα1,α 2∈IR,
a(α1u1+α2u2,v)=α1a(u1,v)+α2a(u2,v),
a(u,α 1v1+α2v2)=α1a(u,v 1)+α2a(u,v 2).

8 Preliminaries on functional analysis
For the case X=Ywe say that a bilinear form is symmetric if
a(u,v)=a(v,u)∀u,v∈X.
Let now ( X,/bardbl·/bardblX) and (Y,/bardbl·/bardblY) be two normed spaces. A bilinear form
a:X×Y→Ris said to be continuous if there exists a constant M> 0
such that
a(u,v)≤M/bardblu/bardblX/bardblv/bardblY∀u∈X,∀v∈Y.
For the case X=Ywe say that a bilinear form is positive if
a(u,u)≥0∀u∈X
andX-elliptic if there exists a constant m>0 such that
a(u,u)≥m/bardblu/bardbl2
X∀u∈X.
It is easy to see that if the bilinear form aisX-elliptic then it is positive.
The converse of this property is not true, in general.
1.1.3 Fixed point theorems
LetXbe a Banach space with the norm /bardbl·/bardblX,a n dKa subset of X. Also,
let Λ :K→Xbe an operator defined on K. We are interested in the
existence of a solution u∈Kof the operator equation
Λu=u. (1.10)
An element u∈Kwhich satisfies (1.10) is called a fixed point of the oper-
ator Λ.
We introduce in what follows two main theorems which state the ex-
istence of the fixed points of nonlinear operators: the Banach fixed pointtheorem and the Schauder fixed point theorem. Both of them represent
major results of functional analysis.
Theorem 1.5 (The Banach fixed point theorem) LetKbe a nonempty
closed subset of a Banach space (X,/bardbl·/bardbl
X). Assume that Λ:K→Kis a
contraction , i.e. there exists a constant α∈[0,1)such that
/bardblΛu−Λv/bardblX≤α/bardblu−v/bardblX∀u,v∈K. (1.11)
Then there exists a unique u∈Ksuch that Λu=u.
Proof Letu0∈Kbe an arbitrary element of Kand let{un}be the
sequence defined by
un+1=Λun∀n=0,1,2,…

1.1 Normed spaces 9
Since Λ : K→K, the sequence {un}is well-defined. Let us first prove that
{un}is a Cauchy sequence. Using the contractivity of the operator Λ, we
have
/bardblun+1−un/bardblX≤α/bardblun−un−1/bardblX≤···≤αn/bardblu1−u0/bardblX.
Then for any m>n≥1,
/bardblum−un/bardblX≤m−n−1/summationdisplay
j=0/bardblun+j+1−un+j/bardblX
≤m−n−1/summationdisplay
j=0αn+j/bardblu1−u0/bardblX
≤αn
1−α/bardblu1−u0/bardblX.
Sinceα∈[0,1), it follows from the previous inequalities that
/bardblum−un/bardblX→0a sm,n→∞.
Thus,{un}is a Cauchy sequence and has a limit u∈K, sinceKis a closed
subset of the Banach space X. Moreover, since un→uinX, it follows from
(1.11) that Λ un→ΛuinX. Therefore, taking the limit in un+1=Λunit
follows that u=Λu,which concludes the existence part of the theorem.
Suppose now that u1,u2∈Kare fixed points of Λ. Then from u1=Λu1
andu2=Λu2we obtain
u1−u2=Λu1−Λu2.
Hence, (1.11) yields
/bardblu1−u2/bardblX=/bardblΛu1−Λu2/bardblX≤α/bardblu1−u2/bardblX,
which implies /bardblu1−u2/bardblX= 0, since α∈[0,1). So, if Λ has a fixed point it
follows that this fixed point is unique, which concludes the proof.
We also need a version of the Banach fixed point theorem which we
recall in what follows. To this end, for an operator Λ, we define its powers
inductively by the formula Λm=Λ ( Λm−1)f o rm≥2.
Theorem 1.6 Assume that Kis a nonempty closed subset of a Banach
spaceXand let Λ:K→K. Assume also that Λm:K→Kis a contraction
for some positive integer m.T h e n Λhas a unique fixed point.
Proof By Theorem 1.5, the mapping Λmhas a unique fixed point u∈K.
From
Λm(u)=u,

10 Preliminaries on functional analysis
we obtain
Λm(Λ(u)) = Λ(Λm(u)) = Λu.
Thus Λu∈Kis also a fixed point of Λm. Since Λmhas a unique fixed
point, we must have
Λu=u,
i.e.uis a fixed point of Λ. The uniqueness of a fixed point of Λ follows
easily from that of Λm.
We turn now to the Schauder fixed point theorem. To introduce it we
need the following preliminaries.
Definition 1.7 LetXbe a normed space. A subset K⊂Xis called:
(i)bounded if there exists M>0 such that
/bardblu/bardblX≤M∀u∈K;
(ii)relatively sequentially compact if each sequence in Khas a convergent
subsequence in X.
It follows from Definition 1.7 (i) that if K⊂Xis bounded, then every
sequence {un}⊂Ksatisfies (1.3) and, therefore, is bounded. Moreover,
Definition 1.7 (ii) shows that a subset K⊂Xis relatively sequentially
compact if for each {un}⊂Kthere exists a subsequence {unk}and an
element u∈Xsuch that unk→uinX. Note also that, for simplicity,
everywhere below we shall use the terminology relatively compact instead
ofrelatively sequentially compact .
Definition 1.8 LetXandYbe normed spaces. The operator Λ : K⊂
X→Yis called:
(i)continuous at the point u∈Kif for each sequence {un}⊂Kwhich
converges in Xtou, the sequence {Λun}⊂Yconverges to Λ uinY,
that is
{un}⊂K, u n→uinX=⇒Λun→ΛuinY;
(ii)continuous if it is continuous at each point u∈K;
(iii)compact if it is continuous and maps bounded sets into relatively
compact sets.
It follows from Definitions 1.8 (iii) and 1.7 (ii) that a continuous operator
Λ:K⊂X→Yis compact if and only if for each bounded sequence
{un}⊂Kthere exists a subsequence {unk}such that the sequence {Λunk}
is convergent in Y.
We proceed with the following result.

1.2 Hilbert spaces 11
Theorem 1.9 (The Schauder fixed point theorem) LetKbe a nonempty
closed convex bounded subset of a Banach space Xand let Λ:K→Xbe a
compact operator such that Λ(K)⊂K.T h e n Λhas at least one fixed point.
The proof of Theorem 1.9 requires a number of preliminary results and,
therefore, we skip it. Nevertheless, we indicate that such a proof can be
found in [155, p. 56]. Also, note that, unlike the Banach fixed point the-
orem, the Schauder fixed point theorem does not provide the uniquenessof the fixed point of the operator Λ. We shall use Theorem 1.9 in Section 2.3
in order to prove the solvability of a class of elliptic quasivariational
inequalities.
1.2 Hilbert spaces
We introduce in what follows some useful results which are valid in Hilbert
spaces. This concerns the projection operators, some properties related
to orthogonality and the Riesz representation theorem, together with itsconsequences.
1.2.1 Projection operators
The projection operators represent an important class of nonlinear oper-ators defined in a Hilbert space; to introduce them we need the following
existence and uniqueness result.
Theorem 1.10 (The projection lemma) LetKbe a nonempty closed con-
vex subset of a Hilbert space X. Then, for each f∈Xthere exists a unique
elementu∈Ksuch that
/bardblu−f/bardbl
X= min
v∈K/bardblv−f/bardblX. (1.12)
Proof Letf∈X, denote
d=i n f
v∈K/bardblv−f/bardblX, (1.13)
and let{un}be a sequence of elements of Ksuch that
/bardblun−f/bardblX→d asn→∞. (1.14)
SinceKis a convex subset of Xwe deduce thatum+un
2∈Kfor all
m, n∈Nand, therefore, (1.13) implies that
/vextenddouble/vextenddouble/vextenddoubleu
m+un
2−f/vextenddouble/vextenddouble/vextenddouble
X≥d∀m, n∈N. (1.15)

12 Preliminaries on functional analysis
We use identity (1.6) with u=um−fandv=un−fto see that
4/vextenddouble/vextenddouble/vextenddoubleum+un
2−f/vextenddouble/vextenddouble/vextenddouble2
X+/bardblum−un/bardbl2
X=2/parenleftbig
/bardblum−f/bardbl2
X+/bardblun−f/bardbl2
X/parenrightbig
for allm, n∈Nand, therefore, (1.15) yields
/bardblum−un/bardbl2
X≤2/parenleftbig
/bardblum−f/bardbl2
X+/bardblun−f/bardbl2
X/parenrightbig
−4d2(1.16)
for allm, n∈N. We combine now (1.14) and (1.16) to obtain
/bardblum−un/bardblX→0a s m, n→∞,
which shows that {un}is a Cauchy sequence in X. Therefore, since Xis a
Hilbert space and Kis a closed subset of X,t h e r ee x i s t sa ne l e m e n t u∈K
such that
/bardblun−u/bardblX→0a sn→∞. (1.17)
We use (1.13) again to see that
d≤/bardblu−f/bardblX≤/bardblu−un/bardblX+/bardblun−f/bardblX∀n∈N
and, by (1.14) and (1.17), we obtain that
d≤/bardblu−f/bardblX≤d.
We conclude that /bardblu−f/bardblX=dwhich proves the existence part of the
theorem.
Assume now that uandvare two elements of Ksuch that
/bardblu−f/bardblX=/bardblv−f/bardblX=d=i n f
w∈K/bardblw−f/bardblX. (1.18)
By the convexity of Kwe haveu+v
2∈Kand, therefore, (1.18) yields
d≤/vextenddouble/vextenddouble/vextenddoubleu+v
2−f/vextenddouble/vextenddouble/vextenddouble
X=/vextenddouble/vextenddouble/vextenddouble1
2(u−f)+1
2(v−f)/vextenddouble/vextenddouble/vextenddouble
X
≤1
2/bardblu−f/bardblX+1
2/bardblv−f/bardblX=d.
It follows from this inequality that
/vextenddouble/vextenddouble/vextenddoubleu+v
2−f/vextenddouble/vextenddouble/vextenddouble
X=d. (1.19)
Using now the identity (1.6) with u−fandv−finstead of uandv,
respectively, we have
/bardblu−v/bardbl2
X=2/parenleftbig
/bardblu−f/bardbl2
X+/bardblv−f/bardbl2
X/parenrightbig
−4/vextenddouble/vextenddouble/vextenddoubleu+v
2−f/vextenddouble/vextenddouble/vextenddouble2
X. (1.20)
We now combine (1.18)–(1.20) to see that /bardblu−v/bardbl2
X= 0, which implies that
u=vand concludes the proof of the theorem.
Theorem 1.10 allows us to introduce the following definition.

1.2 Hilbert spaces 13
Definition 1.11 LetKbe a nonempty closed convex subset of a Hilbert
spaceX. Then, for each f∈Xthe element uwhich satisfies (1 .12) is
called the projection offonKand is usually denoted PKf. Moreover, the
operator PK:X→Kis called the projection operator onK.
It follows from Definition 1.11 that
f=PKf⇐⇒f∈K. (1.21)
We conclude from (1.21) that the element f∈Xis a fixed point of the
projection operator PKifff∈K.
Next, we present the following characterization of the projection.
Proposition 1.12 LetKbe a nonempty closed convex subset of a Hilbert
spaceXand letf∈X.T h e nu=PKfif and only if
u∈K,(u,v−u)X≥(f,v−u)X∀v∈K. (1.22)
Proof Assume that u=PKfand letv∈K. Then, by the definition of
the projection it follows that u∈Kand, for all t∈(0,1), we have
/bardblu−f/bardbl2
X≤/bardbl(1−t)u+tv−f/bardbl2
X=/bardbl(u−f)+t(v−u)/bardbl2
X
=/bardblu−f/bardbl2
X+2t(u−f,v−u)X+t2/bardblv−u/bardbl2
X.
This inequality implies that
2(u−f,v−u)X+t/bardblv−u/bardbl2
X≥0∀t∈(0,1).
We pass to the limit in the previous inequality as t→0 to obtain (1.22).
Conversely, assume that usatisfies (1.22) and, again, let v∈K.U s i n g
(1.4) we have
/bardblv−f/bardbl2
X−/bardblu−f/bardbl2
X=/bardblv/bardbl2
X−/bardblu/bardbl2
X−2(f,v−u)X
and, using (1.22), it follows that
/bardblv−f/bardbl2
X−/bardblu−f/bardbl2
X≥/bardblv/bardbl2
X−/bardblu/bardbl2
X−2(u,v−u)X=/bardblv−u/bardbl2
X≥0.
The last inequality implies that
/bardblu−f/bardblX≤/bardblv−f/bardblX
and, since u∈Kandvis an arbitrary element in K, we deduce that
usatisfies (1.12). Therefore, by Definition 1.11 it follows that u=PKf,
which concludes the proof.
Note that, besides the characterization of the projection in terms of in-
equalities, Proposition 1.12 provides, implicitly, the existence of a unique
solution to the inequality (1.22). Moreover, using this proposition it is easyto prove the following results.

14 Preliminaries on functional analysis
Proposition 1.13 LetKbe a nonempty closed convex subset of a Hilbert
spaceX. Then the projection operator PKsatisfies the following inequali-
ties:
(PKu−PKv,u−v)X≥0∀u,v∈X, (1.23)
/bardblPKu−PKv/bardblX≤/bardblu−v/bardblX∀u,v∈X. (1.24)
Proof Letu, v∈X. We use (1.22) to obtain
(PKu,PKv−PKu)X≥(u,PKv−PKu)X,
(PKv,PKu−PKv)X≥(v,PKu−PKv)X.
We add these inequalities to see that
(PKu−PKv,PKv−PKu)X≥(u−v,PKv−PKu)X
and, therefore,
(PKu−PKv,u−v)X≥/bardbl PKu−PKv/bardbl2
X. (1.25)
Inequality (1.23) follows from (1.25) and inequality (1.24) follows from
(1.25) and the Cauchy–Schwarz inequality.
Proposition 1.14 LetKbe a nonempty closed convex subset of a Hilbert
spaceXand letGK:X→Xbe the operator defined by
GKu=u−PKu∀u∈X. (1.26)
Then, the following properties hold:
(GKu−GKv,u−v)X≥0∀u,v∈X, (1.27)
/bardblGKu−GKv/bardblX≤2/bardblu−v/bardblX∀u, v∈X, (1.28)
(GKu,v−u)X≤0∀u∈X, v∈K. (1.29)
GKu=0Xiffu∈K. (1.30)
Proof Letu, v∈X. We use (1.26), the Cauchy–Schwarz inequality (1.5)
and (1.24) to see that
(GKu−GKv,u−v)X=(u−PKu−v+PKv,u−v)X
=/bardblu−v/bardbl2
X−(PKu−PKv,u−v)X
≥/bardblu−v/bardbl2
X−/bardblPKu−PKv/bardblX/bardblu−v/bardblX≥0,
which shows that (1.27) holds. Also, since
/bardblGKu−GKv/bardblX=/bardbl(u−v)+(PKv−PKu)/bardblX
≤/bardblu−v/bardblX+/bardblPKv−PKu/bardblX,
it follows from (1.24) that (1.28) holds, too.

1.2 Hilbert spaces 15
Next, let u∈Xandv∈K.Then, from (1.26) and Proposition 1.12 we
see that
(GKu,v−u)X=(u−PKu,v−u)X
=(u−PKu,v−PKu)X+(u−PKu,PKu−u)X
≤(u−PKu,PKu−u)X=−/bardblu−PKu/bardbl2
X≤0,
and, therefore, we deduce (1.29). And, finally, (1.30) is a direct consequence
of (1.21), which concludes the proof.
1.2.2 Orthogonality
The inner product allows us to introduce the concept of orthogonality of
elements in an inner product space.
Definition 1.15 LetXbe an inner product space. Two elements u, v∈X
are said to be orthogonal if (u,v)X=0 .T h e orthogonal complement of a
subsetA⊂Xis defined by
A⊥={u∈X:(u,v)X= 0 for all v∈A}. (1.31)
Ifu, v∈Xare two orthogonal elements we usually write u⊥v.M o r e –
over, it is known that if Ais an arbitrary subset of an inner product space X,
then its orthogonal complement A⊥is a closed linear subspace of X. Indeed,
by using the previous definition and the properties of the inner product itis easy to see that if u
1,u2∈A⊥andλ1,λ2∈Rthenλ1u1+λ2u2∈A⊥
and, moreover, if {un}⊂A⊥is such that un→uinXthenu∈A⊥.
An important role in the theory of Hilbert spaces is played by the fol-
lowing result.
Theorem 1.16 LetMbe a closed linear subspace of a Hilbert space X
and letM⊥denote its orthogonal complement. Then any element u∈X
can uniquely be decomposed as u=m+m/prime,w h e r em∈Mandm/prime∈M⊥.
In this case we write X=M⊕M⊥and we say that Xis the direct sum
of the spaces MandM⊥.
Proof Letu∈Xand letm,m/primebe the elements of Xgiven by m=PMu,
m/prime=u−PMuwherePM:X→Mis the projection operator on the closed
subspace M. It follows from this that m∈Mand, clearly, u=m+m/prime.
Next, we use Proposition 1.12 to see that
(m,v−m)X≥(u,v−m)X∀v∈M
and, therefore,
(m/prime,v−m)X≤0∀v∈M.

16 Preliminaries on functional analysis
We take in the previous inequality v=m±w, wherewis an arbitrary
element of M, to obtain
(m/prime,w)X=0∀w∈M.
It follows from this that m/prime∈M⊥, which proves the existence of the
decomposition.
To prove the uniqueness, we assume in what follows that u=m+m/prime=
p+p/prime, wherem, p∈Mandm/prime,p/prime∈M⊥. These equalities yield
m−p=p/prime−m/prime. (1.32)
On the other hand, since MandM⊥are linear subspaces of Xit follows
thatm−p∈Mandp/prime−m/prime∈M⊥. We now use (1.31) to see that
(m−p,p/prime−m/prime)X= 0 and, therefore, (1.32) yields ( m−p,m−p)X=0
which implies that m=p. It follows now from (1.32) that m/prime=p/prime,w h i c h
proves the uniqueness of the decomposition.
Theorem 1.16 allows us to prove the following result which will be useful
later in this book.
Theorem 1.17 LetMbe a closed linear subspace of a Hilbert space Xand
letM⊥⊥be the orthogonal complement of the space M⊥, that is M⊥⊥=
(M⊥)⊥.T h e nM⊥⊥=M.
Proof Letu∈M⊥⊥and letu=m+m/primebe the decomposition of u
provided by Theorem 1.16. We have
(u,m/prime)X=(m+m/prime,m/prime)X=(m,m/prime)X+(m/prime,m/prime)X=(m/prime,m/prime)X(1.33)
sincem∈Mandm/prime∈M⊥and, therefore, ( m,m/prime)X= 0. On the other
hand, since u∈M⊥⊥andm/prime∈M⊥we deduce that ( u,m/prime)X= 0 and,
therefore, (1.33) implies that ( m/prime,m/prime)X= 0, i.e. m/prime=0X. It follows from
this that u=mwhich shows that u∈M. We conclude from above that
M⊥⊥⊂M. (1.34)
Next, assume in what follows that u∈Mand consider an arbitrary
elementv∈M⊥. Definition 1.15 implies that ( v,u)X=(u,v)X= 0 and,
therefore, u∈(M⊥)⊥. We conclude that
M⊂M⊥⊥. (1.35)
Theorem 1.17 is now a consequence of the inclusions (1.34) and (1.35).

1.2 Hilbert spaces 17
1.2.3 Duality and weak convergence
On Hilbert spaces, linear continuous functionals are limited in the forms
they can take. The following theorem makes this more precise.
Theorem 1.18 (The Riesz representation theorem) Let(X,(·,·)X)be a
Hilbert space and let /lscript∈X/prime. Then there exists a unique u∈Xsuch that
/lscript(v)=(u,v)X∀v∈X. (1.36)
Moreover,
/bardbl/lscript/bardblX/prime=/bardblu/bardblX. (1.37)
Proof If/lscript=0X/primeit is easy to see that the element u=0Xis the unique
element of Xwhich satisfies (1.36) and (1.37).
Assume now that /lscript/negationslash=0X/primeand denote by Mthekernel of the functional
/lscript,t h a ti s
M={v∈X:/lscript(v)=0}.
Since/lscript∈X/primeit follows that Mis a closed subspace of Xand, since /lscript/negationslash=0X/prime
it follows that M/negationslash=X, i.e. there exists w∈Xsuch that w/∈M.L e t
w=m+m/primebe the orthogonal decomposition of wprovided by Theorem
1.16 and note that, since w/∈M,w eh a v e m/prime/negationslash=0X.
Letvbe an arbitrary element of Xand consider the elements uandzof
Xdefined by the equalities
u=/lscript(m/prime)
/bardblm/prime/bardbl2
Xm/prime, (1.38)
z=/lscript(v)m/prime−/lscript(m/prime)v. (1.39)
It is easy to see that /lscript(z) = 0 and, therefore, z∈M. Next, since m/prime∈M⊥
we have
(m/prime,z)X=0. (1.40)
We now use (1.39) and (1.40) to deduce that
/lscript(v)/bardblm/prime/bardbl2
X=/lscript(m/prime)(m/prime,v)X.
This last equality implies that
/lscript(v)=/lscript(m/prime)
/bardblm/prime/bardbl2
X(m/prime,v)X
and, using (1.38), we obtain (1.36).
Assume now that u1andu2are two elements of Xsuch that (1.36) holds.
We take v=u1−u2in (1.36) to obtain
/lscript(u1−u2)=(u1,u1−u2)X=(u2,u1−u2)X,

18 Preliminaries on functional analysis
which implies that /bardblu1−u2/bardbl2
X= 0. It follows that u1=u2,which concludes
the first part of the theorem.
Next, to prove (1.37) we note that (1.36) and the Cauchy–Schwarz in-
equality (1.5) yield
|/lscript(v)|≤/bardblu/bardblX/bardblv/bardblX∀v∈X
and, therefore, (1.8) shows that
/bardbl/lscript/bardblX/prime≤/bardblu/bardblX. (1.41)
Moreover, (1.8) implies that
|/lscript(u)|≤/bardbl/lscript/bardblX/prime/bardblu/bardblX
and, since by (1.36) we have /lscript(u)=/bardblu/bardbl2
X, we deduce that
/bardblu/bardblX≤/bardbl/lscript/bardblX/prime. (1.42)
Equality (1.37) is now a consequence of inequalities (1.41) and (1.42).
We proceed with some important consequences of the Riesz representa-
tion theorem concerning the weak convergence in a Hilbert space. First, we
recall that if ( X,(·,·)X) is an inner product space then, as shown on page
7, (1.9) holds.
Assume now that ( X,(·,·)X) is a Hilbert space. Then, it follows from
Riesz’s representation theorem that the converse of (1.9) is true, i.e.
(un,v)X→(u,v)Xasn→∞,∀v∈X=⇒un/arrowrighttophalfu inX.
We conclude from this that a sequence {un}⊂Xconverges weakly to
u∈Xiff
(un,v)X→(u,v)Xasn→∞,∀v∈X.
The Riesz representation theorem also allows to identify a Hilbert space
with its dual and, therefore, with its bidual which, roughly speaking, shows
that each Hilbert space is reflexive. Based on this result we have the follow-ing important property which represents a particular case of the well-known
Eberlein–Smulyan theorem.
Theorem 1.19 IfXis a Hilbert space, then any bounded sequence in X
has a weakly convergent subsequence.
It follows that if Xis a Hilbert space and the sequence {u
n}⊂X
is bounded, that is, supn/bardblun/bardblX<∞, then there exists a subsequence
{unk}⊂{un}and an element u∈Xsuch that unk/arrowrighttophalfuinX. Further-
more, if the limit uis independent of the subsequence, then the whole
sequence {un}converges weakly to u, as stated in the following result.

1.3 Elements of nonlinear analysis 19
Theorem 1.20 LetXbe a Hilbert space and let {un}be a bounded se-
quence of elements in Xsuch that each weakly convergent subsequence of
{un}converges weakly to the same limit u∈X.T h e nun/arrowrighttophalfuinX.
Proof Arguing by contradiction, we assume in what follows that the se-
quence{un}does not converge weakly to uinX. Then, there exists an
element v∈Xsuch that ( un,v)X/negationslash→(u,v)XinR,a sn→∞.T h i s ,i n
turn, implies that there exists ε0>0 such that for all k∈Nthere exists
unk∈Xwhich satisfies
|(unk,v)X−(u,v)X|≥ε0∀k∈N. (1.43)
Since{unk}is a subsequence of the bounded sequence {un}it follows that
{unk}is bounded in Xand, therefore, by Theorem 1.19 there exists a
subsequence {unkp}⊂{unk}which is weakly convergent in X. Using the
assumption of Theorem 1.20 it follows that unkp/arrowrighttophalfuinX, which yields
(unkp,v)X→(u,v)X asp→∞. (1.44)
On the other hand, {unkp}⊂{unk}and, therefore, by (1.43) we have
|(unkp,v)X−(u,v)X|≥ε0∀p∈N. (1.45)
Inequality (1.45) is in contradiction to the convergence (1.44), which con-
cludes the proof.
1.3 Elements of nonlinear analysis
In the study of variational inequalities presented in Chapters 2 and 3 we
need several results on nonlinear operators and convex functions that weintroduce in this section.
1.3.1 Monotone operators
The projection operator on a convex subset Kof a Hilbert space is, in
general, a nonlinear operator on X. Its properties (1.23) and (1.24) can be
extended as follows.
Definition 1.21 LetXbe a space with inner product ( ·,·)Xand norm
/bardbl·/bardblXand letA:X→Xbe an operator. The operator Ais said to be
monotone if
(Au−Av,u−v)X≥0∀u,v∈X.
The operator Aisstrictly monotone if
(Au−Av,u−v)X>0∀u,v∈X, u/negationslash=v,

20 Preliminaries on functional analysis
andstrongly monotone if there exists a constant m>0 such that
(Au−Av,u−v)X≥m/bardblu−v/bardbl2
X∀u, v∈X. (1.46)
The operator Aisnonexpansive if
/bardblAu−Av/bardblX≤/bardblu−v/bardblX∀u, v∈X
andLipschitz continuous if there exists M>0 such that
/bardblAu−Av/bardblX≤M/bardblu−v/bardblX∀u, v∈X. (1.47)
Finally, the operator Aishemicontinuous if the real-valued function
θ/mapsto→(A(u+θv),w)Xis continuous on R,∀u, v, w∈X
and, recalling Definition 1 .8,Aiscontinuous if
un→uinX=⇒Aun→AuinX.
It follows from the definition above that each strongly monotone operator
is strictly monotone and a nonexpansive operator is Lipschitz continuous,
with Lipschitz constant M= 1. Also, it is easy to check that a Lipschitz
continuous operator is continuous and a continuous operator is hemicon-tinuous. Moreover, it follows from Proposition 1.13 that the projection
operators are monotone and nonexpansive.
In many applications it is not necessary to define nonlinear operators
on the entire space X. Indeed, in the study of the variational inequalities
presented in Chapters 2 and 3 we shall consider strongly monotone Lip-schitz continuous operators defined on a subset K⊂X. For this reason we
complete Definition 1.21 with the following one.
Definition 1.22 LetXbe a space with inner product ( ·,·)
Xand norm
/bardbl·/bardblXand letK⊂X. An operator A:K→Xis said to be strongly
monotone if there exists a constant m>0 such that
(Au−Av,u−v)X≥m/bardblu−v/bardbl2
X∀u, v∈K. (1.48)
The operator AisLipschitz continuous if there exists M>0 such that
/bardblAu−Av/bardblX≤M/bardblu−v/bardblX∀u, v∈K. (1.49)
The following result involving monotone operators will be used in Chap-
ter 2, in the analysis of elliptic variational inequalities.
Proposition 1.23 Let(X,(·,·)X)be an inner product space and let A:
X→Xbe a monotone hemicontinuous operator. Assume that {un}is a

1.3 Elements of nonlinear analysis 21
sequence of elements in Xwhich converges weakly to the element u∈X,
i.e.
un/arrowrighttophalfu inXasn→∞. (1.50)
Moreover, assume that
limsup
n→∞(Aun,un−u)X≤0. (1.51)
Then, for all v∈X, the following inequality holds:
liminf
n→∞(Aun,un−v)X≥(Au,u−v)X. (1.52)
Proof By the monotonicity of the operator Ait follows that
(Aun,un−u)X≥(Au,un−u)X∀n∈N
and, passing to the lower limit as n→∞, by using (1.50) and (1.9) we find
that
liminf
n→∞(Aun,un−u)X≥0. (1.53)
We combine the inequalities (1.51) and (1.53) to see that the sequence
(Aun,un−u)Xconverges as n→∞ and, moreover,
lim
n→∞(Aun,un−u)X=0. (1.54)
Letv∈X,θ>0 and denote w=( 1−θ)u+θv. We use the monotonicity
of the operator Ato obtain
(Aun−Aw,un−w)X≥0∀n∈N
which implies that
(Aun−Aw,un−u+θ(u−v))X≥0∀n∈N
or, equivalently,
(Aun,un−u)X+θ(Aun,u−v)X
≥(Aw,un−u)X+θ(Aw,u−v)X∀n∈N.(1.55)
We pass to the lower limit as n→∞ in (1.55), use (1.9) and (1.54) and
then divide the resulting inequality by θ. We obtain
liminf
n→∞(Aun,u−v)X≥(Aw,u−v)X
and, therefore,
liminf
n→∞(Aun,u−v)X≥(A(u+θ(v−u)),u−v)X.

22 Preliminaries on functional analysis
Next, we pass to the limit in this last inequality as θ→0 and use the
hemicontinuity of the operator Ato see that
liminf
n→∞(Aun,u−v)X≥(Au,u−v)X. (1.56)
We write now
(Aun,un−v)X=(Aun,un−u)X+(Aun,u−v)X∀n∈N,
then we pass to the lower limit as n→∞ in this equality and use (1.54)
and (1.56). As a result we obtain (1.52), which concludes the proof.
An operator A:X→Xfor which (1.50) and (1.51) imply (1.52) for all
v∈Xis called a pseudomonotone operator. We conclude from Proposition
1.23 that every monotone hemicontinuous operator on a Hilbert space is a
pseudomonotone operator.
We proceed with the following existence and uniqueness result in the
study of nonlinear equations involving monotone operators.
Theorem 1.24 LetXbe a Hilbert space and let A:X→Xbe a strongly
monotone Lipschitz continuous operator. Then, for each f∈Xthere exists
a unique element u∈Xsuch that Au=f.
Proof SinceAis strongly monotone and Lipschitz continuous it follows
from Definition 1.21 that there exist two constants m>0a n dM>0 such
that (1.46) and (1.47) hold. Moreover, we have
M≥m. (1.57)
Letf∈Xand letρ>0 be given. We consider the operator Sρ:X→X
defined by
Sρu=u−ρ(Au−f)∀u∈X.
It follows from this definition that
/bardblSρu−Sρv/bardblX=/bardbl(u−v)−ρ(Au−Av)/bardblX∀u, v∈X
and, using (1.46), (1.47) yields
/bardblSρu−Sρv/bardbl2
X=/bardbl(u−v)−ρ(Au−Av)/bardbl2
X
=/bardblu−v/bardbl2
X−2ρ(Au−Av,u−v)X+ρ2/bardblAu−Av/bardbl2
X
≤(1−2ρm+ρ2M2)/bardblu−v/bardbl2
X∀u, v∈X.
Next, using (1.57) it is easy to see that if 0 <ρ<2m
M2then
0≤1−2ρm+ρ2M2<1.

1.3 Elements of nonlinear analysis 23
Therefore, with this choice of ρ, it follows that
/bardblSρu−Sρv/bardblX≤k(ρ)/bardblu−v/bardblX∀u, v∈X, (1.58)
wherek(ρ)=( 1−2ρm+ρ2M2)1
2∈[0,1). Inequality (1.58) shows that Sρ
is a contraction on the space Xand, using Theorem 1.5, we obtain that
there exists u∈Xsuch that
Sρu=u−ρ(Au−f)=u. (1.59)
Equality (1.59) yields Au=f, which proves the existence part of the
theorem.
Next, consider two elements u∈Xandv∈Xsuch that Au=fand
Av=f. It follows that
(Au,u−v)X=(f,u−v)X,(Av,u−v)X=(f,u−v)X.
We subtract these equalities to obtain
(Au−Av,u−v)X=0,
then we use assumption (1.46) to find that u=v, which proves the unique-
ness part.
Theorem 1.24 shows that if A:X→Xis a strongly monotone Lipschitz
continuous operator defined on a Hilbert space X, thenAis invertible. The
properties of its inverse, denoted A−1, are given by the following result.
Proposition 1.25 LetXbe a Hilbert space and let A:X→Xbe a
strongly monotone Lipschitz continuous operator. Then, A−1:X→Xis
a strongly monotone Lipschitz continuous operator.
Proof Letu1,u2∈Xand denote
Au 1=v1,A u 2=v2. (1.60)
It follows that
A−1v1=u1,A−1v2=u2. (1.61)
We use now (1.46) and (1.60) to see that
(Au 1−Au 2,u1−u2)X=(v1−v2,u1−u2)X≥m/bardblu1−u2/bardbl2
X,
which implies that
/bardblu1−u2/bardblX≤1
m/bardblv1−v2/bardblX.
This inequality and (1.61) lead to
/bardblA−1v1−A−1v2/bardblX≤1
m/bardblv1−v2/bardblX,
which shows that A−1is a Lipschitz continuous operator.

24 Preliminaries on functional analysis
On the other hand, using (1.46), (1.47), (1.60) and (1.61) we have
(v1−v2,A−1v1−A−1v2)X≥m/bardblA−1v1−A−1v2/bardbl2
X,
1
M/bardblv1−v2/bardblX≤/bardblA−1v1−A−1v2/bardblX.
These inequalities imply that
(A−1v1−A−1v2,v1−v2)X≥m
M2/bardblv1−v2/bardbl2
X,
which shows that the operator A−1is strongly monotone.
1.3.2 Convex lower semicontinuous functions
Convex lower semicontinuous functions represent a crucial ingredient in
the study of variational inequalities. To introduce them, we start with thefollowing definitions.
Definition 1.26 LetXbe a linear space and let Kbe a nonempty convex
subset of X. A function ϕ:K→Ris said to be convex if
ϕ((1−t)u+tv)≤(1−t)ϕ(u)+tϕ(v) (1.62)
for allu, v∈Kandt∈[0,1]. The function ϕisstrictly convex if the
inequality in (1 .62) is strict for u/negationslash=vandt∈(0,1).
We note that if ϕ,ψ:K→Rare convex and λ≥0, then the functions
ϕ+ψandλϕare also convex.
Definition 1.27 Let (X,/bardbl·/bardbl
X) be a normed space and let Kbe a nonempty
closed convex subset of X. A function ϕ:K→Ris said to be lower semi-
continuous (l.s.c.) atu∈Kif
liminf
n→∞ϕ(un)≥ϕ(u) (1.63)
for each sequence {un}⊂Kconverging to uinX. The function ϕisl.s.c.
if it is l.s.c. at every point u∈K. When inequality (1 .63) holds for each
sequence {un}⊂Kthat converges weakly to u, the function ϕis said to
beweakly lower semicontinuous atu. The function ϕis weakly l.s.c. if it is
weakly l.s.c. at every point u∈K.
We note that if ϕ,ψ:K→Rare l.s.c. functions and λ≥0, then
the functions ϕ+ψandλϕare also lower semicontinuous. Moreover, if
ϕ:K→Ris a continuous function then it is also lower semicontinuous.
The converse is not true and a lower semicontinuous function can be dis-
continuous. Since strong convergence in Ximplies weak convergence, it
follows that a weakly lower semicontinuous function is lower semicontinu-ous. Moreover, the following results hold.

1.3 Elements of nonlinear analysis 25
Proposition 1.28 Let(X,/bardbl·/bardblX)be a Banach space, Ka nonempty closed
convex subset of Xandϕ:K→Ra convex function. Then ϕis lower
semicontinuous if and only if it is weakly lower semicontinuous.
Proposition 1.29 Let(X,/bardbl·/bardblX)be a normed space, Ka nonempty closed
convex subset of Xandϕ:K→Ra convex lower semicontinuous function.
Thenϕis bounded from below by an affine function, i.e. there exist /lscript∈X/prime
andα∈Rsuch that ϕ(v)≥/lscript(v)+αfor allv∈K.
The proof of Proposition 1 .28 is a consequence of Mazur’s theorem. It
follows from this proposition that a convex continuous function ϕ:X→R
defined on the Banach space Xis weakly lower semicontinuous. In particu-
lar, the norm function v/mapsto→/bardblv/bardblXis weakly lower semicontinuous. A second
example of a lower semicontinuous function is provided by the followingresult.
Proposition 1.30 Let(X,/bardbl·/bardbl
X)be a normed space and let a:X×X→R
be a bilinear symmetric continuous and positive form. Then the functionv/mapsto→a(v,v)is strictly convex and lower semicontinuous.
Proof The strict convexity is straightforward to show. To prove the lower
semicontinuity consider a sequence {u
n}⊂Xsuch that un/arrowrighttophalfu∈X. Since
ais positive it follows that
a(un−u,un−u)≥0∀n∈N
and, therefore,
a(un,un)≥a(u,un)+a(un,u)−a(u,u)∀n∈N. (1.64)
For a fixed v∈X, the mappings u/mapsto→a(v,u)a n du/mapsto→a(u,v) define linear
continuous functionals on Xand, since {un}converges weakly to u,w e
have
lim
n→∞a(u,un) = lim
n→∞a(un,u)=a(u,u). (1.65)
We pass to the lower limit in (1.64) and use (1.65) to see that
liminf
n→∞a(un,un)≥a(u,u),
which concludes the proof.
In particular, it follows from Proposition 1.30 that, if ( X,(·,·)X)i sa n
inner product space then the function v/mapsto→/bardblv/bardbl2
X=(v,v)Xis strictly convex
and lower semicontinuous.
We now recall the definition of Gˆ ateaux differentiable functions.

26 Preliminaries on functional analysis
Definition 1.31 Let (X,(·,·)X) be an inner product space, ϕ:X→R
andu∈X. ThenϕisGˆateaux differentiable atuif there exists an element
∇ϕ(u)∈Xsuch that
lim
t→0ϕ(u+tv)−ϕ(u)
t=(∇ϕ(u),v)X∀v∈X. (1.66)
The element ∇ϕ(u) which satisfies (1 .66) is unique and is called the gradient
ofϕatu. The function ϕ:X→Ris said to be Gˆateaux differentiable if
it is Gˆ ateaux differentiable at every point of X. In this case the operator
∇ϕ:X→Xwhich maps every element u∈Xinto the element ∇ϕ(u)i s
called the gradient operator ofϕ.
The convexity of Gˆ ateaux differentiable functions can be characterized
as follows.
Proposition 1.32 Let(X,(·,·)X)be an inner product space and let ϕ:
X→Rbe a Gˆ ateaux differentiable function. Then the following statements
are equivalent:
(i)ϕis a convex function;
(ii)ϕsatisfies the inequality
ϕ(v)−ϕ(u)≥(∇ϕ(u),v−u)X∀u, v∈X; (1.67)
(iii) the gradient of ϕis a monotone operator, that is
(∇ϕ(u)−∇ϕ(v),u−v)X≥0∀u, v∈X. (1.68)
Proof Proposition 1.32 is a direct consequence of the four implications we
state and prove below.
(i)=⇒(ii)Ifϕis a convex function, then for all u, v∈Xandt∈(0,1),
from (1.62) we deduce that
t(ϕ(v)−ϕ(u))≥ϕ(u+t(v−u))−ϕ(u).
We divide both sides of the inequality by t>0, pass to the limit as t→0
and use (1.66) to obtain (1.67).
(ii)=⇒(i)Conversely, assume that (1.67) holds and let u, v∈X,t∈
[0,1]. Letw=( 1−t)u+tv; it follows from (1.67) that
ϕ(v)−ϕ(w)≥(∇ϕ(w),v−w)X,
ϕ(u)−ϕ(w)≥(∇ϕ(w),u−w)X.
We multiply the inequalities above by tand (1−t), respectively, then add
the results to obtain (1.62), which shows that ϕis a convex function.

1.3 Elements of nonlinear analysis 27
(ii)=⇒(iii)Assume that (1.67) holds and let u, v∈X.W eh a v e
ϕ(u)−ϕ(v)≥(∇ϕ(v),u−v)X
and, adding this inequality to (1.67), we obtain (1.68).
(iii)=⇒(ii)Assume that (1.68) holds and let u, v∈X. For allt∈[0,1]
we consider the element w(t)∈Xdefined by w(t)=( 1−t)u+tvand let
gbe the real-valued function defined by g(t)=ϕ(w(t)). First, we see that
w(t)−u=t(v−u) and, therefore, we use (1.68) to write
(∇ϕ(w(t)),v−u)X≥(∇ϕ(u),v−u)X∀t∈[0,1].
On the other hand w(t+h)=w(t)+h(v−u) and, using the definition
(1.66) of the gradient operator, it is easy to see that
lim
h→0ϕ(w(t+h))−ϕ(w(t))
h=(∇ϕ(w(t)),v−u)X∀t∈[0,1],
where the limit is taken with respect to hwitht+h∈[0,1]. We conclude
from here that gis derivable on [0,1] and, moreover,
g/prime(t)≥(∇ϕ(u),v−u)X∀t∈[0,1].
We integrate this inequality on [0 ,1] to obtain
g(1)−g(0)≥(∇ϕ(u),v−u)X. (1.69)
Next, since w(0) =uandw(1) =vit follows that g(0) =ϕ(u)a n dg(1) =
ϕ(v). We use these equalities in (1.69) to obtain (1.67).
From the previous proposition we easily deduce the following result.
Corollary 1.33 Let(X,(·,·)X)be an inner product space and let ϕ:X→
Rbe a convex Gˆ ateaux differentiable function. Then ϕis lower semicon-
tinuous.
Proof Let{un}be a sequence of elements of Xconverging to uinX.I t
follows from Proposition 1.32 that
ϕ(un)−ϕ(u)≥(∇ϕ(u),un−u)X∀n∈N.
We pass to the lower limit as n→∞ in the previous inequality to obtain
(1.63), which concludes the proof.
A large number of boundary value problems in Contact Mechanics lead
to variational formulations in which the frictional term is associated with a
continuous seminorm. For this reason we collect at the end of this subsection
some results on continuous seminorms defined on a normed space, which
we shall need in the rest of the book.
Given a linear space X, we recall that a seminorm jis a function from
XtoRsatisfying the following properties:

28 Preliminaries on functional analysis
(1)j(u)≥0∀u∈X;
(2)j(αu)=|α|j(u)∀u∈X,∀α∈R;
(3)j(u+v)≤j(u)+j(v)∀u,v∈X.
It follows from the above that a seminorm satisfies the properties of a
norm except that j(u) = 0 does not necessarily imply u=0X. In particular,
note that j(0X) = 0. The continuity of a seminorm defined on a normed
space is characterized by the following result.
Proposition 1.34 Letjbe a seminorm on the normed space (X,/bardbl·/bardblX).
Thenjis continuous if and only if there exists m>0such that
j(v)≤m/bardblv/bardblX∀v∈X. (1.70)
Proof Assume that jis continuous and, arguing by contradiction, assume
that (1.70) does not hold. Then, for each n∈Nthere exists vn∈Xsuch
thatj(vn)>n/bardblvn/bardblX. It follows that vn/negationslash=0Xand, moreover,
j/parenleftBigvn
n/bardblvn/bardblX/parenrightBig
>1. (1.71)
For each n∈Ndenote
un=vn
n/bardblvn/bardblX. (1.72)
We use (1.72) and (1.71) to see that un→0XinXasn→∞ butj(un)>1
for alln∈N, which contradicts the continuity of jin 0X.
Conversely, assume that (1.70) holds. Then, using the properties of the
seminorm jwe have
|j(u)−j(v)|≤j(u−v)≤m/bardblu−v/bardblX∀u, v∈X,
which proves that jis continuous.
It is easy to see that a seminorm defined on a linear space is a convex
function. Therefore, a direct consequence of Proposition 1.34 is given by
the following result.
Corollary 1.35 Let(X,/bardbl·/bardblX)be a normed space and let jbe a seminorm
onXwhich satisfies (1.70) with some m>0.T h e njis a convex lower
semicontinuous function.
Note that in convex analysis it is usual to consider functions ϕdefined
on a normed space Xwith values on ( −∞,∞]. For such functions, concepts
such as convexity and lower semicontinuity are introduced in Definitions
1.26 and 1.27, respectively, by using the convention that ∞+∞=∞
while an expression of the form ∞−∞ is undefined. Nevertheless, in many
applications in mechanics the domain of definition of various functions

1.3 Elements of nonlinear analysis 29
(like the energy function) is subjected to constraints. For this reason, above,
we restricted ourselves to considering only functions ϕ:K→Rdefined
on a subset Kof a normed or inner product space X.This choice is not
restrictive. Indeed, assume that Kis a nonempty closed convex subset of
X,ϕ:K→Ris a given function and let /tildewideϕ:X→(−∞,∞] be the function
given by
/tildewideϕ(v)=/braceleftBigg
ϕ(v)i fv∈K,
∞ ifv/∈K.
Then, it is easy to see that /tildewideϕis convex and l.s.c. if and only if ϕis convex
and l.s.c.
1.3.3 Minimization problems
An important property of convex lower semicontinuous functions is given
by the following well-known theorem.
Theorem 1.36 (The Weierstrass theorem) LetXbe a Hilbert space and
Ka nonempty closed convex subset of X.L e tJ:K→Rbe a convex
lower semicontinuous function. Then Jis bounded from below and attains
its infimum on Kwhenever one of the following two conditions hold:
(i)Kis bounded;
(ii)Jiscoercive , i.e.J(u)→+∞as/bardblu/bardblX→+∞.
Moreover, if Jis a strictly convex function, then Jattains its infimum on
Kat only one point.
Proof Letθdenote the infimum of JonK,t h a ti s
θ=i n f
v∈KJ(v)<∞. (1.73)
Consider a sequence {un}of elements in Ksuch that
J(un)→θasn→∞. (1.74)
We claim that the sequence {un}is bounded. Indeed, this claim is obviously
satisfied if condition (i)holds. Assume now that (ii)holds, i.e. Jis coercive.
Arguing by contradiction, if the sequence {un}is not bounded then, passing
to a subsequence again denoted {un}, we can assume that
/bardblun/bardblX→∞ asn→∞. (1.75)
Using now the coercivity of Jit follows that
J(un)→∞ asn→∞. (1.76)

30 Preliminaries on functional analysis
The convergences (1.76) and (1.74) are in contradiction to the inequality
θ<∞in (1.73). It follows from the above that the sequence {un}is
bounded and, therefore, our claim holds in this case, too.
We now use Theorem 1.19 to see that there exists an element u∈K
such that, passing to a subsequence again denoted {un},w eh a v e
un/arrowrighttophalfu asn→∞. (1.77)
Moreover, since Kis a closed convex subset it follows from Theorem 1.4
thatKis weakly closed and, therefore,
u∈K. (1.78)
We use Proposition 1.28 to see that J:K→Ris a weakly lower semi-
continuous function. Hence, (1.77) and (1.74) imply that θ≥J(u). On the
other hand, the converse inequality θ≤J(u) is a consequence of (1.73) and
(1.78) and, therefore, we obtain
θ=J(u). (1.79)
We now combine (1.78), (1.73) and (1.79) to see that
u∈K, J (v)≥J(u)∀v∈K, (1.80)
which concludes the existence part of the theorem.
Assume now that Jis strictly convex. Arguing by contradiction, assume
thatu1andu2satisfy (1.80) and, moreover, u1/negationslash=u2. Then, J(u1)=
J(u2)=θand by Definition 1.26 it follows that
J/parenleftBigu1+u2
2/parenrightBig
<1
2J(u1)+1
2J(u2)=θ.
This inequality is in contradiction to the definition (1.73) and proves the
uniqueness of the minimizer of the function J.
We use Theorem 1.36 to prove the following result.
Theorem 1.37 LetXbe a Hilbert space, Ka nonempty closed convex
subset of X,a:X×X→Ra bilinear continuous symmetric and X-
elliptic form, j:K→Ra convex lower semicontinuous function and
f∈X. Denote by J:K→Rthe function given by
J(v)=1
2a(v,v)+j(v)−(f,v)X∀v∈K (1.81)
and consider the minimization problem
u∈K, J (v)≥J(u)∀v∈K. (1.82)

1.3 Elements of nonlinear analysis 31
Then:
(1) an element uis a solution to problem (1.82) iff
u∈K, a (u,v−u)+j(v)−j(u)≥(f,v−u)X∀v∈K; (1.83)
(2) there exists a unique solution to problem (1.82).
Proof (1) Assume that usatisfies (1.82) and let v∈Kandt∈(0,1). It
follows that
J(u+t(v−u))≥J(u)
and, using (1.81), the properties of aand the convexity of j, we deduce
that
ta(u,v−u)+t2
2a(v−u,v−u)+t(j(v)−j(u))≥t(f,v−u)X.
Dividing both sides of the inequality by t>0 and passing to the limit as
t→0, we conclude that usatisfies inequality (1.83).
Conversely, suppose uis a solution of the inequality (1.83). Using this
inequality and the properties of the form a,f o re a c h v∈Kwe have
J(v)−J(u)=a(u,v−u)+j(v)−j(u)−(f,v−u)X
+1
2a(v−u,v−u)≥1
2a(v−u,v−u)≥0.
Thus,uis a solution of the minimization problem (1.82).
(2) We use Proposition 1.30 to see that the functional Jis strictly convex
and l.s.c. Therefore, in the case when Kis bounded, Theorem 1.37 follows
from Theorem 1.36.
Assume now that Kis not bounded. Since the bilinear form aisX-
elliptic, it follows that there exists m>0 such that
J(v)≥m
2/bardblv/bardbl2
X+j(v)−(f,v)X∀v∈K. (1.84)
Next, Proposition 1.29 and Riesz’s representation theorem guarantee that
there exist g∈Xandα∈Rsuch that
j(v)≥(g,v)X+α∀v∈K. (1.85)
We combine inequalities (1.84) and (1.85) and then use the Cauchy–Schwarz
inequality to see that the functional Jis coercive. And, again, Theorem 1.37
is a consequence of Theorem 1.36.
We conclude from the first part of Theorem 1.37 that the minimization
problem (1.82) and inequality (1.83) are equivalent. Since the existence of
a unique minimizer for Jis guaranteed by the second part of this theorem,
we deduce the following existence and uniqueness result.

32 Preliminaries on functional analysis
Corollary 1.38 LetXbe a Hilbert space, Ka nonempty closed convex
subset of X,a:X×X→Ra bilinear continuous symmetric and X-elliptic
form and j:K→Ra convex lower semicontinuous function. Then, for
eachf∈Xthere exists a unique element usuch that (1.83) holds.