Lectures On
Some Fixed Point Theorems
Of Functional Analysis
By
F.F.Bonsall
Tata Institute Of Fundamental Research,Bombay
1962
Lectures On
Some Fixed Point Theorems
Of Functional Analysis
By
F.F.Bonsall
Notes by
K.B.Vedak
No part of this book may be reproduced in any
formby print,microlmor any other means with
out written permission from the Tata Institute of
Fundamental Research,Colaba,Bombay 5
Tata Institute of Fundamental Research
Bombay
1962
Introduction
These lectures do not constitute a systematic account of xe d point the
orems.I have said nothing about these theorems where the interest is
essentially topological,and in particular have nowhere introduced the
important concept of mapping degree.The lectures have been con
cerned with the application of a variety of methods to both nonlinear
(xed point) problems and linear (eigenvalue) problems in i nnite di
mensional spaces.A wide choice of techniques is available for linear
problems,and I have usually chosen to use those that give something
more than existence theorems,or at least a promise of something more.
That is,I have been interested not merely in existence theorems,but also
in the construction of eigenvectors and eigenvalues.For this reason,I
have chosen elementary rather than elegant methods.
I would like to draw special attention to the Appendix in which I
give the solution due to B.V.Singbal of a problem that I raised in the
course of the lectures.
I am grateful to Miss K.B.Vedak for preparing these notes and
seeing to their publication.
Frank F.Bonsall
iii
Contents
1 The contraction mapping theorem 1
2 Fixed point theorems in normed linear spaces 13
3 The Schauder  Tychonoﬀ theorem 31
4 Nonlinear mappings in cones 43
5 Linear mapping in cones 51
6 Selfadjoint linear operator in a Hilbert space 79
7 Simultaneous xed points 95
8 A class of abstract semialgebras 103
v
Chapter 1
The contraction mapping
theorem
Given a mapping T of a set E into itself,an element u of E is called a 1
xed point of the mapping T if Tu = u.Our problem is to nd condi
tions on T and E suﬃcient to ensure the existence of a xed point of T
in E.We shall also be interested in uniqueness and in procedures for the
calculation of xed points.
Denition 1.1.Let E be a nonempty set.A real valued function d de
ned on E × E is called a distance function or metric in E if it satises
the following conditions
i) d(x,y) ≥ 0,x,yεE
ii) d(x,y) = 0 ⇐⇒ x = y
iii) d(x,y) = d(y,x)
iv) d(x,z) ≤ d(x,y) + d(y,z)
A nonempty set with a specied distance function is called a m etric
space.
1
2
Thecontractionmappingtheorem
Example.Let X be a set and E denote a set of bounded real valued
functions dened on X.Let d be dened on E × E by
d( f,g) = sup
 f (t) − g(t):tεX
,f,gεE.
Then d is a metric on E called the uniform metric or uniform dis
tance function.
Denition 1.2.A sequence {x
n
} in a metric space (E,d) is said to con2
verge to an element x of E if
lim
n→∞
d(x
n
,x) = 0
A sequence x
n
of elements of a metric space (E,d) is called a Cauchy
sequence if given ǫ > 0,there exists N such that for p,q ≥ N,d(x
p
,x
q
) <
ǫ.
A metric space (E,d) is said to be complete if every Cauchy se
quence of its elements converges to an element of E.It is easily veried
that each sequence in a metric space converges to at most one on point,
and that every convergent sequence is a Cauchy sequence.
Example.The space C
R
[0,1] of all continuous real valued functions on
the closed interval [0,1] with the uniform distance is a complete metric
space.It is not complete in the metric d
′
dened by
d
′
( f,g) =
Z
1
0
 f (x) − g(x)dx f,gεC
R
[0,1].
Denition 1.3.A mapping T of a metric space E into itself is said to
satisfy a Lipschitz condition with Lipschitz constant K if
d(Tx,Ty) ≤ Kd(x,y) (x,yεE)
If this conditions is satised with a Lipschitz constant K such that 0 ≤
K < 1 then T is called a contraction mapping.
Theorem 1.1 (The contraction mapping theorem).Let T be a contrac
tion mapping of a complete metric space E into itself.Then
Thecontractionmappingtheorem
3
i) T has a unique xed point u in E3
ii) If x
o
is an arbitrary point of E,and (x
n
) is dened inductively by
x
n+1
= Tx
n
(n = 0,1,2,...),
then lim
n→∞
x
n
= u and
d(x
n
,u) ≤
K
n
1 − K
d(x
1
,x
o
)
where K is a Lipschitz constant for T.
Proof.Let K be a Lipschitz constant for T with 0 ≤ K < 1.Let x
o
εE
and let x
n
be the sequence dened by
x
n+1
= Tx
n
(n = 0,1,2,...)
We have
d(x
r+1
,x
s+1
) = d(Tx
r
,Tx
s
) ≤ Kd(x
r
,x
s
) (1)
and so
d(x
r+1
,x
r
) ≤ K
r
(x
1
,x
o
) (2)
Given p q,we have by (1) and (2),
d(x
p
,x
q
) ≤ K
q
d(x
p−q
,x
o
)
≤ K
q
d(x
p−q
,x
p−q−1
) + d(x
p−q−1
,x
p−q−2
) + + d(x
1
,x
o
)
≤ K
q
K
p−q−1
+ K
p−q−2
+ + K + 1
d(x
1
,x
o
)
≤
K
q
1 − K
d(x
1
,x
o
) (3)
since the right hand side tends to zero as q →∞,it follows that (x
n
) is a 4
Cauchy sequence,and since E is complete,(x
n
) converges to an element
u of E.Since d(x
n+1
,Tu) ≤ Kd(x
n
,u) →0 as n →∞,
Tu = lim
n→∞
x
n+1
= u.
4
Thecontractionmappingtheorem
d(u,x
n
) ≤ d(u,x
p
) + d(x
p
,x
n
) ≤ d(u,x
p
) +
K
n
1 − K
d(x
1
,x
o
) for n < p,by
(3).Letting p →∞,we obtain
d(u,x
n
) ≤
K
n
1 − K
d(x
1
,x
o
)
Example.As an example of the applications of the contraction map
ping theorem,we prove Picard's theoremon the existence of s olution of
ordinary diﬀerential equation.
Let D denote an open set in R
2
,(x
o
,y
o
)εD.Let f (x,y) be a real val
ued function dened and continuous in D,and let it satisfy the Lipschitz
condition:
 f (x,y
1
) − f (x,y
2
) ≤ My
1
− y
2
 ((x,y
1
),(x,y
2
)εD)
Then there exists a t > 0,and a function φ(x) continuous and diﬀeren
tiable in [x
o
− t,x
o
+ t] such that i)φ(x
o
) = y
o
,(ii)y = φ(x) satises the
diﬀerential equation
dy
dx
= f (x,y) for xε[x
o
− t,x
o
+ t]
We show rst that there exists an ǫ > 0 and a function φ(x) contin
uous in [x
o
− t,x
o
+ t] such that iii)φ(x) = y
o
+
R
x
x
o
f (t,φ(t))dt (x
o
− t ≤5
x ≤ x
o
+t),and ǫ(x,φ(x))εD(x
o
− ≤ x ≤ x
o
+t).Then it follows fromthe
continuity of f (t,φ(t)) that φ(x) is in fact diﬀerentiable in [x
o
−t,x
o
+t]
and satises (i) and (ii).
Let Udenote a closed disc of centre (x
o
,y
o
) with positive radius and
contained in the open set D,and let mdenote the least upper bound of the
continuous function  f  on the compact set U.We now choose t,δ such
that 0 < t < M
−1
,the rectangle x−x
o
 ≤ δ is contained in U,and mt < δ.
Let E denote the set of all continuous functions mapping [x
o
− t,x
o
+t]
into [y
o
− δ,y
o
+ δ].With respect to the uniform distance function E is
a closed subset of the complete metric space C
R
[x
o
− t,x
o
+ t] and is
therefore complete.We dene a mapping Tφ = ψ for φ ∈ E by
ψ(x) = y
o
+
Z
x
x
o
f (t,φ(t))dt
Thecontractionmappingtheorem
5
Clearly ψ(x) is continuous in [x
o
− t,x
o
+ t].Also ψ(x) − y
o
 ≤
mx − x
o
 ≤ mt < δ whenever x − x
o
 ≤ t.Thus T maps E into it self.
Finally,T is a contraction mapping,for if φ
i
∈ E,ψ
i
= Tφ)i(i = 1,2),
then
ψ
1
(x) − ψ
2
(x) = 
Z
x
x
o
f (t
′
,φ
1
(t
′
)) − f (t
′
,φ
2
(t
′
))
dt
′

≤ x − x
o
Msup φ
1
(t
′
) − φ
2
(t
′
) (x
o
− t ≤ t ≤ x
o
+ t)
≤ tMd(φ
1
,φ
2
)
Hence 6
d(ψ
1
,ψ
2
) ≤ tMd(φ
1
,φ
2
)
As tM < 1,this proves that T is a contraction mapping.By the contrac
tion mapping theorem,there exists φεE with Tφ = φ i.e.,with
φ(x) = y
o
+
Z
x
x
o
f (t,φ(t))dt
This complete the proof of Picard's theorem.
Asimilar method may be applied to prove the existence of solutions
of systems of ordinary diﬀerential equations of the form
dy
i
dx
= f
i
(x,y
1
,...,y
n
)(i = 1,2,...,n)
with given initial conditions.Instead of considering real valued func
tions dened on [ x
o
− t,x
o
+ t],one considers vector valued functions
mapping [x
o
− t,x
o
+ t] into R
n
.
In the following theorem we are concerned with the continuity of
the xed point.
Theorem 1.2.Let E be a complete metric space,and let T and T
n
(n =
1,2,...) be contraction mappings of E into itself with the same Lipschitz
constant K < 1,and with xed points u and u
n
respectively.Suppose
that lim
n→∞
T
n
x = Tx for every xεE.Then lim
n→∞
u
n
= u.By the inequality
in Theorem 1.1,we have for each r = 1,2,...,
d(u
r
,T
n
r
x
o
) ≤
K
n
1 − K
d(T
r
x
o
,x
o
),x
o
εE
6
Thecontractionmappingtheorem
setting n = 0 and x
o
= u,we have 7
d(u
r
,u) ≤
1
1 − K
d(T
r
u,u) =
1
−1
d(T
r
u,Tu)
But d(T
r
u,Tu) →0 as r →∞.Hence
lim
r→∞
d(u
r
,u) = 0
Example.In the notation of the last example,suppose that y
n
is a real
sequence converging to y
o
and let T
n
be the mapping dened on E by
(T
n
φ)(x) = y
n
+
Z
x
x
o
f (t,φ(t))dt
Then (T
n
φ)(x) − y
o
 ≤ y
n
− y
o
 + mε < δ for n suﬃciently large
i.e.T
n
map E into itself for n suﬃciently large.Also the mapping
T
n
,T have the same Lipschitz constant εM < 1.Obviously for each
φεE,lim
n→∞
T
n
φ = Tφ.Hence if φ
n
is the unique xed point of T
n
(n =
1,2,...) then lim
n→∞
φ
n
= φ.In other words,if φ
n
is the solution of the
diﬀerential equation
dy
dx
= f (x,y)
in [x
o
−t,x
o
+t] with the initial condition φ
n
(x
o
) = y
n
,then φ
n
converges
uniformly to the solution φ with φ(x
o
) = y
o
.
Remark.The contraction mapping theorem is the simplest of the xed
point theorems that we shall consider.It is concerned with mappings of a
complete metric space into itself and in this respect is very general.The8
theoremis also satisfactory in that the xed point is always unique and is
obtained by an explicit calculation.Its disadvantage is that the condition
that the mapping be a contraction is a somewhat severe restriction.In the
rest of this chapter we shall obtain certain extension of the contraction
mapping theorem in which the conclusion is obtained under modied
conditions.
Denition 1.4.A mapping T of a metric space E into a metric space E
′
is said to be continuous if for every convergent sequence (x
n
) of E,
lim
n→∞
T
x
n
= T( lim
n→∞
x
n
).
Thecontractionmappingtheorem
7
Theorem 1.3.Let T be a continuous mapping of a complete metric
space E into itself such that T
k
is a contraction mapping of E for some
positive integer k.Then T has a unique xed point in E.
Proof.T
k
has a unique xed point u in E and u = lim
n→∞
(T
k
)
n
x
o
,x
o
εE
arbitrary.
Also lim
n→∞
(T
k
)
n
Tx
o
= u.Hence
u = lim
n→∞
(T
k
)
n
Tx
o
= lim
n→∞
T(T
k
)
n
x
o
= T lim
n→∞
(T
k
)
n
x
o
(by the continuity of ) T
= Tu.
The uniqueness of the xed point of T is obvious,since each xed
point of T is also a xed point of T
k
.
Example.We consider the nonlinear integral equation 9
f (x) = λ
Z
x
a
K(x,y,f (y))dy + g(x) (1)
where g is continuous in [a,b] and K(x,y,z) is continuous in the region
[a,b] × [a,b] × R and satises the Lipschitz condition.
K(x,y,z
1
) − K(x,y,z
2
) ≤ Mz
1
− z
2
.
(The classical Volterra equation is obtained by taking K(x,y,z) =
H(x,y).z,with H continuous in [a,b] × [a,b]).Let E = C
K
[a,b] and T
be the mapping of E into itself given by
(T f )(x) = λ
Z
x
a
K(x,y,f (u))dy + g(x) ( f εE,a ≤ x ≤ b).
Given f
1
,f
2
in E it is easy to prove by induction on n that
(T
n
f
1
)(x) − (T
n
f
2
)(x) ≤
1
n!
λ
n
M
n
d( f
1
,f
2
)(x − a)
n
,(a ≤ x ≤ b)
8
Thecontractionmappingtheorem
Then
d(T
n
f
1
,T
n
f
2
) ≤
1
n!
λ
n
M
n
(b − a)
n
d( f
1
,f
2
)
This proves that all T
n
and in particular T,are continuous and,for
n suﬃciently large
1
n!
λ
n
M
n
(b − a)
n
< 1,so that T
n
is a contraction
mapping for n large.Applying the theorem,we have a unique f εE with
T f = f which is the required unique solution of the equation (1).
Denition 1.5.Let (E,d) be a metric space and ε > 0.Anite sequence10
x
o
,x
1
,...,x
n
of points of E is called an ε  chain joining x
0
and x
n
if
d(x
i−1
,x
i
) < ε (i = 1,2,...,n)
The metric space (E,d) is said to be ε chainable if for each pair
(x,y) of its points there exists an ε chain joining x and y.
Theorem1.4 (Edelstein).Let T be a mapping of a complete ε chainable
metric space (E,d) into itself,and suppose that there is a real number K
with 0 ≤ K < 1 such that
d(x,y) < ε ⇒d(Tx,Ty) ≤ Kd(x,y)
Then T has a unique xed point u in E,and u = lim
n→∞
T
n
x
o
where x
o
is an arbitrary element of E.
Proof.(E,d) being ε chainable we dene for x,yεE,
d
ε
(x,y) = inf
n
X
i=1
d(x
i−1
,x
i
)
where the inmum is taken over all ε  chains x
0
,...,x
n
joining x
0
= x
and x
n
= y.Then d is a distance function on E satisfying
i) d(x,y) ≤ d
ε
(x,y)
ii) d(x,y) = d
ε
(x,y) for d(x,y) < ε
Thecontractionmappingtheorem
9
From (ii) it follows that a sequence (x
n
),x
n
εE is a Cauchy sequence
with respect to d
ε
if and only if it is a Cauchy sequence with respect11
to d and is convergent with respect to d
ε
if and only if it converges
with respect with respect to d.Hence (E,d) being complete,(E,d
ε
) is
also a complete metric space.Moreover T is a contraction mapping with
respect to d
ε
.Given x,yεE,and any εchain x
o
,...,x
n
with x
o
= x,x
n
=
y,we have
d(x
i−1
,x
i
) < ε (i = 1,2,...,n),
so that
d(Tx
i−1
,Tx
i
) ≤ Kd(x
i−1
,x
i
) < ε (i = 1,2,...,n)
Hence Tx
o
,...,Tx
n
is an ε chain joining T
x
and T
y
and
d
ε
(Tx,Ty) ≤
n
X
i=1
d(Tc
i−1
,Tx
i
) ≤ K
n
X
i=1
d(x
i−1
,x
i
)
x
o
,...,x
n
being an arbitrary ε chain,we have
d
ε
(Tx,Ty) ≤ Kd
ε
(x,y)
and T has a unique xed point uεE given by
lim
n→∞
d
ε
(T
n
x
0
,u) = 0 for x
o
εE arbitrary (1)
But in view of the observations made in the beginning of this proof,(1)
implies that
lim
n→∞
d(T
n
x
o
,u) = 0
Example.Let E be a connected compact subset of a domain D in the
complex plane.Let f be a complex holomorphic function in D which 12
maps E into itself and satises  f
′
(z) < 1 (z,εE).Then there is a unique
point z in E with f (z) = z.Since f
′
is continuous in the compact set E,
there is a construct K with 0 < K < 1 such that  f
′
(z) < K εE).For
each point ωεE there exists ρ
ω
> 0 such that f (x) is holomorphic in the
disc S(ω,2ρ
ω
) of center ωand radius 2ρ
ω
and satises  f
′
(z) < K there.
10
Thecontractionmappingtheorem
E being compact,we can choose ω
1
,...,ω
n
εE such that E is covered
by
S(ω
1
,2ρ
ω
1
),...,S(ω
n
,2ρ
ω
n
)
Let ε = min{ρω
i
,i = 1,2,...,n}.If z,z
′
εE and z − z
′
 < ε then
z,z
′
εS(ω
i
,2ρ
ω
1
) for some i and so
 f (z) − f (z
′
) = 
Z
z
z
′
f
′
(ω)dω ≤ Kz − z
′
.
This proves that Theorem1.4 is applicable to the mapping z → f (z)
and we have a unique xed point.
Denition 1.6.A mapping T of a metric space E into itself is said to be
contractive if
d(Tx,Ty) < d(x,y) (x y,x,yεE)
and is said to be εcontractive if
0 < d(x,y) < ε ⇒d(Tx,Ty) < d(x,y)
Remark.A contractive mapping of a complete metric space into itself
need not have a xed point.e.g.let E = {x/x ≥ 1} with the usual13
distance d(x,y) = x − y,let T:E →E be given by Tx = x +
1
x
.
Theorem1.5 (Edelsten).Let T be an εcontractive mapping of a metric
space E into itself,and let x
o
be a point of E such that the sequence
(T
n
x
o
) has a subsequence convergent to a point u of E.Then u is a
periodic point of T,i.e.there is a positive integer k such that
T
k
u = u
Proof.Let (n
i
) be a strictly increasing sequence of positive integers such
that lim
i→∞
T
n
i
x
o
= u and let x
i
= T
n
i
x
o
.There exists N such that d(x
i
,u) <
ε/4 for i ≥ N.Choose any i ≥ N and let k = n
i+1
− n
i
.Then
d(x
i+1
,T
k
u) = d(T
k
x
i
,T
k
u) ≤ d(x
i
,u) < ε/4
and
d(T
k
u,u) ≤ d(T
k
u,x
i+1
) + d(x
i+1
,u) < ε/2
Thecontractionmappingtheorem
11
Suppose that v = T
k
u u.Then T being ε  contractive,
d(Tu,Tv) < d(u,v) or
d(Tu,Tv)
d(u,v)
< 1.
The function (x,y) →
Tx,Ty
d(u,v)
is continuous at (u,v).So there exist
δ,K > 0 with 0 < K < 1 such that d(x,u) < δ,d(y,v) < δ implies that
d(Tx,Ty) < Kd(x,y).As lim
r→∞
T
K
x
r
= T
K
u = v,there exists N
′
≥ N
such that d(x
r
,u) < δ,d(Tx
r
,v) < δ for r ≥ N
′
and so 14
d(Tx
r
,TT
K
x
r
) < Kd(x
r
,T
K
x
r
) (1)
d(x
r
,T
K
x
r
) ≤ d(x
r
,u) + d(u,T
K
u) + d(T
K
u,T
K
x
r
)
<
ε
4
+
ε
2
+
ε
4
= ε for r ≥ N
′
> N (2)
From(1) and (2)
d(Tx
r
,TT
K
x
r
) < Kd(x
r
,T
K
x
r
) < ε for r ≥ N
′
and so T being εcontractive,
d(T
p
x
r
,T
p
T
k
x
r
) < Kd(x
r
,T
K
x
r
) for n ≥ N
′
,p > 0 (3)
Setting p = n
r+1
− n
r
in (3)
d(x
r+1
,T
k
x
r+1
) < Kd(x
r
,T
k
x
r
) for any r ≥ N
′
Hence d(x
s
,T
k
x
s
) < K
s−r
d(x
r
,T
k
x
r
) < K
s−r
ε and d(u,v) < d(u,x
s
) +
d(x
s
,T
k
x
s
) + d(T
k
x
s
,v) →0 as s →∞This contradicts the assumption
that d(u,v) > 0.Thus u = v = T
k
u.
Theorem 1.6 (Edelstein ).Let T be a contractive mapping of a metric
space E into itself,and let x
o
be a point of E such that the sequence T
n
x
o
has a subsequence convergent to a point u of E.Then u is a xed point
of T and is unique.
Proof.By Theorem 1.5,there exists an integer k > 0 such that T
k
u = 15
u.Suppose that v = Tu ,u.Then T
k
u = u,T
k
v = v and d(u,v) =
d(T
k
u,T
k
v) < d(u,v),since T is contractive.As this is impossible,
u = v is a xed point.The uniqueness is also immediate.
12
Thecontractionmappingtheorem
Corollary.If T is a contractive mapping of a metric space E into a
compact subset of E,then T has a unique xed point u in E and u =
lim
n→∞
T
n
x
o
where x
o
is an arbitrary point of E.
Chapter 2
Fixed point theorems in
normed linear spaces
In Chapter 1,we proved xed point theorems in metric spaces w ithout 16
any algebraic structure.We now consider spaces with a linear structure
but nonlinear mappings in them.In this chapter we restrict our attention
to normed spaces,but our main result will be extended to general locally
convex spaces in Chapter 3
Denition 2.1.Let E be a vector space over.A mapping of E into R is
called a normon E if it satises the following axioms.
i) p(x) ≥ 0 (x ∈ E)
ii) p(x) = 0 if and only if x = 0
iii) p(x + y) ≤ p(x) + p(y) (x,y ∈ E).
Avector space E with a specied normon it called a normed space.
The normof an element x ∈ E will usually be denoted by x.Anormed
space is a metric space with the metric d(x,y) = x − y (x,yεE) and the
corresponding metric topology is called the normed topology.Anormed
linear space complete in the metric dened by the norm is call ed a Ba
nach space.We now recall some denitions and well known prop erties
13
14
Fixedpointtheoremsinnormedlinearspaces
of linear spaces.Two norms p
1
and p
2
on a vector space E are said to
be equivalent if there exist positive constants k,k
′
such that
p
1
(x) ≤ kp
2
(x),p
2
(x) ≤ k
′
p
1
(x) (x ∈ E)
Two norms are equivalent if and only if they dene the same top ology.17
Denition 2.2.Amapping f of a vector space E into R is called a linear
functional on E if it satises
i) f (x + y) = f (x) + f (y) (x,y ∈ E)
ii) f (αx) = αf (x) (x ∈ E,α ∈ R).
A mapping p:E →R is called a sublinear functional if
i)
′
p(x + y) ≤ p(x) + p(y) (x,y ∈ E)
ii)
′
p(αx) = αp(x) (x ∈ E,α ≥ 0).
Hahn Banach Theorem.Let E
0
be a subspace of a vector space E over
R;let p be a sublinear functional on E and let f
o
be a linear functional
on E
o
that satises
f
o
(x) ≤ p(x) (x ∈ E
0
).
Then there exists a linear functional f on E that satises
i) f (x) ≤ p(x) (x ∈ E),
ii) f (x) = f
o
(x) (x ∈ E
0
).
[For the proof refer to Dunford and Schwartz ([14],p.62) or Day [13,
p.9]].
Corollary.Given a sublinear functional on E and x
o
∈ E,there exists a
linear functional f such that
f (x
0
) = p(x
0
),f (x) ≤ p(x) (x ∈ E).
Fixedpointtheoremsinnormedlinearspaces
15
In particular,a norm being a sublinear functional,given a point x
0
of a normed space E,there exists a linear functional f on E such that18
 f (x) ≤ x (x ∈ E) and f (x
o
) = x
o

Denition.A norm p on a vector space E said to be strictly convex if
p(x + y) = p(x) + p(y) only when x and y are linearly dependent.
Theorem 2.1 (Clarkson ).If a normed space E has a countable every
where dense subset,then there exists a strictly convex normon E equiv
alent to the given norm.
Proof.Let S denote the surface of the unit ball in E,
S = {x:x = 1}
Then there exists a countable set (x
n
) of points of S that is dense in S.
For each n,there exists a linear functional f
n
on E such that
f
n
(x
n
) = x
n
 = 1 and  f
n
(x) ≤ x (x ∈ E).
If x 0,then f
n
(x) 0 for some n.For,by homogeneity,it is
enough to consider x with x = 1,and for such x there exists n with
x − x
n
 <
1
2
.But then
f
n
(x) = f
n
(x
n
) + f
n
(x − x
n
) ≥ 1 − f
n
(x − x
n
)
≥ 1 − x − x
n
 >
1
2
We now take p(x) = x +
(
∞
P
n=1
2
−n
( f
n
(x))
2
)
1
2
.It is easily veried 19
that p is a norm on E and that
x ≤ p(x) ≤ 2x.
Finally p is strictly convex.To see this,suppose that
p(x + y) = p(x) + p(y),
16
Fixedpointtheoremsinnormedlinearspaces
and write ξ
n
= f
n
(x),y
n
= f
n
(y).Then
∞
X
n=1
2
−n
(ξ
n
+ η
n
)
2
1
2
=
∞
X
n=1
s
−n
ξ
2
n
1
2
+
∞
X
n=1
2
−n
η
2
n
1
2
and we have the case of equality in Minkowsiki's inequality.It follows
that the sequence (ξ
n
) and (η
n
) are linearly dependent.Thus there exist
λ,ν,not both zero,such that
λξ
n
+ η
n
= 0 (n = 1,2,...)
But this implies that
f
n
(λx + y) = 0 (n = 1,2,...),
and so λx + y = 0.This completes the proof.
Lemma 2.1.Let K be a compact convex subset of a normed space E
with a strictly convex norm.Then to each point x of E corresponds a
unique point Px of K at K at minimum distance from x,i.e.,with
x − Px = inf{x − y:yεK}
and the mapping x →Px is continuous in E.20
Proof.Let x ∈ E,and let the function f be dened on K by f (y) =
x − y.Then f is a continuous mapping of the compact set K into R
and therefore attains its minimumat a point z say of K
x − z = inf{x − y:y ∈ K}.
Evidently for x ∈ K,z = x is uniquely determined.If x K,suppose
that z
′
is such that
0 x − z = x − z
′
 (1)
since K is convex,y =
1
2
(z + z
′
) ∈ K and therefore
x − y ≥ x − z = x − z
′
 =
1
2
x − z +
1
2
x − z
′

Fixedpointtheoremsinnormedlinearspaces
17
But
(x − y) =
1
2
(x − z) +
1
2
(x − z
′
),
so that x − y ≤
1
2
x − z
′
 +
1
2
x − z
′

Hence x − y = 
1
2
(x − z) +
1
2
(x − z
′
) = 
1
2
(x − z) + 
1
2
(x − z
′
)
As the norm is strictly convex,
λ(x − z) + (x − z
′
) = 0
for λ, not both zero.By (1),λ =  and so x − z = ±(x − z
′
).If
x − z = −(x − z
′
),then x =
z + z
′
2
∈ K,which is not true.Hence
x − z = x − z
′
or z = z
′
.This proves that the mapping x → Px = z is
uniquely on E.Given x,x
′
∈ E,
x − Px ≤ x − Px
′
 ≤ x − x
′
 + x
′
− Px
′
,
and similarly x
′
− Px
′
 ≤ x − x
′
 + x − Px.So 21
 x − Px − x
′
− Px
′
  ≤ x − x
′
 (2)
Let x
n
∈ E(n = 1,2,...) converge to x ∈ E.Then the sequence Px
n
in the compact metric space K has a subsequence Px
n
k
converging to
y ∈ K.Then
lim
k→∞
x
n
k
− Px
n
k
 = x − y (3)
By (2) x
n
− Px
n
 − x − Px  ≤ x
n
− x → 0 as n → ∞,and
so x − y = x − Px.Hence Px = yi.e.lim
k→∞
Px
n
k
= Px.Thus if (x
n
)
converges to x,(Px
n
) has a subsequence converging to Px ans so every
subsequence of (Px
n
) has a subsequence converging to Px.Therefore
(Px
n
) converges to Px and P is continuous.
Denition 2.3.The mapping P of Lemma 2.1 is called the metric pro
jection onto K.
18
Fixedpointtheoremsinnormedlinearspaces
Denition 2.4.A subset A of a normed space is said to be bounded if
there exists a constant M such that x ≤ M (x ∈ A).
We now state without proof three properties of nite dimensi onal
normed spaces.
Lemma 2.2.Every nite dimensional normed space is complete.
Lemma 2.3.Every bounded closed subset of a nite dimensional norm
ed space is compact.
Lemma 2.4 (Brouwer xed point theorem).Let K be a nonempty com
pact convex subset of a nite dimensional normed space,and l et T be a22
continuous mapping of K into itself.Then T has a xed point in K.
The proofs of the rst two of these Lemmas are elementary.(Re fer
to Dunford and Schwartz [14,p.244245].) The Brouwer xed p oint
theorem on the other hand is far from trivial.For a proof using some
elements of algebraic topology refer to P.Alexandroﬀ and H.Hopf ([1],
p.376378).A proof of a more analytical kind is given by Dunford ans
Schwartz ([14],p.467).
Theorem 2.2 (Schauder).Let K be a nonempty closed convex subset
of a normed space.Let T be a continuous mapping of K into a cumpact
subset of K.Then T has xed point in K.
Proof.Let E denote the normed space and let TK ⊂ A,a compact subset
of K.A is contained in a closed convex bounded subset of E.
T(B ∩ K) ⊂ T(K) ⊂ A ⊂ B
so T(B ∩ K) is contained in a compact subset of B,K and there is no
loss of generality in supposing that K is bounded.If A
o
is a countable
dense subset of the compact metric space A,then the set of all rational
linear combinations of elements of A
o
is a countable dense subset of the
closed linear subspace E
o
spanned by A
o
and A ⊂ E
0
.Then T(K ∩
E
0
) ⊂ T(K) ⊂ A,a compact subset of E
0
,and K ∩ E
0
is closed and
convex.Hence without loss of generality we may assume that K is a
bounded closed convex subset of a separable normed space E with a
strictly convex norm (Theorem 2.1).
Fixedpointtheoremsinnormedlinearspaces
19
Given a positive integer n,there exists a
1
n
net Tx
1
,...,Tx
m
say in 23
TK,so that
min
1≤k≤n
Tx − Tx
k
 <
1
n
(x ∈ K) (1)
Let E
n
denote the linear hull of Tx
1
,...,Tx
m
.K
n
= K ∩ E
n
is a
closed bounded subset of E
n
and therefore compact (Lemma 2.3).Since
the norm is strictly convex,the metric projection Pn of E onto the con
vex compact subset Kn exists.T
n
= P
n
T is a continuous mapping of the
nonempty convex compact subset Kn into itself,and therefore by the
Brouwer xed point theorem,it has a xed point u
n
εK
n
,
T
n
u
n
= u
n
(2)
By (1),since Tx
k
∈ K
n
(k = 1,2,...,m),we have
Tx − T
n
x <
1
n
(3)
The sequence {Tu
n
} of TK has a subsequence Tu
n
k
converging to a
point v ∈ K.By (2) and (3),u
n
k
−v = T
n
k
u
n
k
−v ≤ T
n
k
u
n
k
−Tu
n
k
 +
Tu
n
k
−v <
1
n
+Tu
n
k
−v.Therefore,lim
k→∞
u
n
k
= u,and by continuity
of T,lim
k→∞
Tu
n
k
= Tv or Tv = v.
Example.Suppose that a function f (x,y) of two real variables is con
tinuous on a neighbourhood of (x
o
,y
o
).Then we can choose ε > 0 such
that f is continuous in the rectangle
x − x
o
 ≤ ε,y − y
o
 ≤ mε
and satises there the inequality 24
 f (x,y) ≤ m.
Let E denote the Banach space C
R
[x
o
−ε,x
o
+ε],which is a Banach
space with the uniform norm
ϕ = sup
Φ(t):t − x
o
 ≤ ε
20
Fixedpointtheoremsinnormedlinearspaces
Let K be the subset of E consisting of all continuous mappings of
[x
o
− ε,x
o
+ ε] into [y
o
− mε,y
o
+ mε].Then K is a bounded closed
convex subset of E.Let T be the mapping dened on K by
(Tφ)(x) = y
o
+
Z
x
xo
f (t,φ(t))dt (x − x
o
 ≤ ε)
Then TK ⊂ K.Also since
(Tφ)(x) − (Tφ)(x
′
)
≤
Z
x
x
′
f (t,(t))dt
≤ mx − x
′
 (φ ∈ K),
TK is an equicontinuous set.Since also TK is bounded,TK is contained
in a compact set by the Ascoli  Arzela theorem.Therefore,by Theorem
2.2,T has a xed point φ in K i.e.,
φ(x) = y
o
+
Z
x
x
o
f (t,φ(t)) dt (x − x
o
 ≤ ε).
Then φ is diﬀerentiable in [x
o
− ε,x
o
+ ε] and provides a solution
y = φ(x) there of the diﬀerential equation
dy
dx
= f (x,y)
with φ(x
o
) = y
o
.This is Peano's theorem.As a particular case of25
Schauder's theorem,we have
Theorem2.3.Let K be a nonempty compact convex subset of a normed
space,and let T be a continuous mapping of K into itself.Then T has a
xed point in K.
Remark.Theorem 2.2 and 2.3 are almost equivalent,in the sense that
Theorem2.2,with the additional hypothesis that K be complete,follows
from Theorem 2.3.For,if K is a complete convex set and TK is con
tained in a compact subset A of K,then the closed convex hull of A is a
compact convex subset K
o
of K,and TK
0
⊂ K
0
.
Denition 2.5.A mapping T which is continuous and maps each boun
ded set into a compact set is said to be completely continuous.
Fixedpointtheoremsinnormedlinearspaces
21
Theorem 2.4.Let T be a completely continuous mapping of a normed
space E into itself and let TE be bounded.Then T has a xed point.
Proof.Let K be the closed convex hull of TE.Then K is bounded and
so TK is contained in a compact subset of K.By Theorem 2.2,T has a
xed point in K.
The Theorem 2.4 implies Theorem 2.3 is seen as follows.Let K
be a compact convex set and let T be continuous mapping of K into
itself.There is no loss of generality is supposing that the norm in E
is strictly convex.Let P be the metric projection of E onto K,and let 26
T = TP.Then
T satises the conditions of theorem 24,and so there
exists u in E with Tu = u.Since T maps E into K,we have u ∈ K and
so Pu = uTu = TRu = u.
Lemma 2.5.Let K be a nonempty complete convex subset of a normed
space E,let A be a continuous mapping of K into a compact subset of
E,and let F be a mapping of K × K into K such that
(i) F(x,y) − F(x,y) ≤ ky − y
′
 (x,y,y
′
∈ K),where k is a constant
with 0 < k < 1,
(ii) F(x,y) − F(x
′
,y) ≤ Ax − Ax
′
 (x,x
′
,y ∈ K).Then there exists
a point u in K with
F(u,u) = u.
Proof.For each xed x,the mapping y →F(x,y) is a contraction map
ping of the complete metric space K into itself,and it therefore has a
unique xed point in K which we denote by Tx,
Tx = F(x,Tx) (x ∈ K).
We have Tx − Tx
′
 = F(x,Tx) − F(x
′
,Tx
′
)
≤ F(x,Tx) − F(x
′
,Tx)
+ F(x
′
,Tx) − F(x
′
,Tx
′
)
≤ Ax − Ax
′
 + kTx − Tx
′

22
Fixedpointtheoremsinnormedlinearspaces
Therefore Tx−Tx
′
 ≤
1
1 − k
Ax−Ax
′
,(1) which shows that Tk is
continuous and that TK ⊂ K is precompact since AK is compact,since
K is complete,
TK ⊂ K is compact.By the Schander theorem,T has
xed point u in K,
Tu = u.
But then27
F(u,u) = F(u,Tu) = Tu = u
Theorem 2.5 (Kranoselsku ).Let K be a nonempty complete convex
subset of a normed space E,let A be a continuous mapping of K into a
compact subset of E,let B map K and satisfy a Lipschitz condition
Bx − Bx
′
 ≤ kx − x
′
 (x,x
′
∈ k)
with 0 < k < 1 and let Ax + By ∈ K for all x,y in K.Then there is a
point u ∈ K with
Au + Bu = u
Proof.Take F(x,y) = Ax + By and apply Lemma 2.5.
Corollary.Let K be a nonempty complete convex subset of a normed
space,let A be a continuous of K into a compact subset of K,let B map
K into itself ans satisfy the Lipschitz condition
Bx − Bx
′
 ≤ x − x
′
 (x,x
′
εK),
and let 0 < α < 1.Then there exists a point u ∈ K with
αAu + (1 − α)Bu = u
In general,under the condition of Schauder's theorem,we have no
method for the calculation of a xed point of a mapping.Howev er there
is a special case in which this can be done using a method due to Kras
noselsku.
Fixedpointtheoremsinnormedlinearspaces
23
Denition 2.6.A norm p is uniformly convex if it satises
p(x
n
) = p(y
n
) = 1 (n = 1,2,...),lim
n→∞
p(x
n
+ y
n
)
= 2 =⇒ lim
n→∞
p(x
n
− y
n
) = 0.
28
Lemma.Let p be a uniformly convex norm,and let εM be positive
constants.Then there exists a constant δ with 0 < δ < 1 such that
p(x) ≤ M,p(y) ≤ M,p(x − y) ≥ ε ⇒ p(x + y) ≤ 2δ max(p(x),p(y)).
Proof.For all x,y,we have
p
1
2
(x + y)
!
≤
1
2
p(x) +
1
2
p(y) ≤ max(p(x),p(y)).(1)
If there is no constant δ with the stated properties,there exist se
quences (x
n
),(y
n
) with p(x
n
) ≤ M,p(y
n
) ≤ M,
p(x
n
− y
n
) ≥ ε,(2)
and
p
1
2
(x
n
+ y
n
)
!
>
1 −
1
n
!
max(p(x
n
),p(y
n
)).(3)
Let α
n
= p(x
n
),β
n
= p(y
n
),γ
n
= max(α
n
,β
n
).By (1) and (2),
γ
n
≥
1
2
,(4)
and so,by (1) and (3)
lim
n→∞
1
γ
n
p
1
2
(x
n
+ y
n
)
!
= 1.(5)
It follows from (1) and (5),that
lim
n→∞
α
n
+ β
n
2γ
n
= 1.(6)
24
Fixedpointtheoremsinnormedlinearspaces
Since (γ
n
) is bounded,there exists a convergent sequence (γ
n
k
),and
by (4)
lim
k→∞
γ
n
k
= γ ≥ ε
γ
2
(7)
lim
k→∞
(γ
n
k
− α
n
k
) + (γ
n
k
− β
n
k
) = 0,
and,since each bracket is nonnegative,each tends to zero.Therefore29
lim
k→∞
α
n
k
= lim
k→∞
β
n
α
= γ (8)
By discarding some terms of the subsequence if necessary,we may
suppose that α
n
k
≥ 0 and β
n
k
≥ 0 for all k.Since
p
1
α
n
k
x
n
k
+
1
β
n
k
y
n
k
!
− p
1
γ
n
k
x
n
k
+
1
γ
n
k
y
n
k
!
≤ p
1
α
n
k
−
1
γ
n
k
!
x
n
k
+
1
β
n
k
−
1
γ
n
k
!
y
n
k
≤ M
(
1
α
n
k
−
1
γ
n
k
+
1
β
n
k
−
1
γ
n
k
)
,
it follows from(5),(7),(8),that
lim
k→∞
p
1
α
n
k
x
n
k
+
1
β
n
k
y
n
k
!
= 2.
Therefore,
lim
k→∞
p
1
α
n
k
x
n
k
−
1
β
n
k
y
n
k
!
= 0.
and so
lim
k→∞
1
γ
n
k
p(x
n
k
− y
n
k
) = 0,
which contradicts (2).
Theorem 2.6 (Krashoselsku).Let K be a bounded closed convex set in30
a Banach space E with a uniformly convex norm.Let T be a mapping
of K into a compact subset of K that satises a Lipschitz condition with
Fixedpointtheoremsinnormedlinearspaces
25
Lipschitz constant 1,and let x
o
be an arbitrary point of K.Then the
sequence dened by
x
n+1
=
1
2
(x
n
+ Tx
n
) (n = 0,1,2,...)
converges to a xed point of T in K.
Proof.By Schauder's theorem,there is a nonempty set F of xed points
of T in K.We prove rst that
x
n+1
− y ≤ x
n
− y (y ∈ F,n = 0,1,2,...)
In fact if y = T
y
,then
x
n+1
− y = 
1
2
(x
n
+ Tx
n
) −
1
2
(y + Ty)
= 
1
2
(x
n
− y) +
1
2
(Tx
n
− Ty)
≤
1
2
x
n
− y +
1
2
Tx
n
− Ty
≤ x
n
− y
which is (1).
Suppose that there exist an ε > 0 and N,such that
x
n
− Tx
n
 ≥ ε for all n ≥ N (2)
Then x
n
− y − (Tx
n
− T
y
) ≥ ε for all n ≥ N,y ∈ F.
Also Tx
n
− T
y
 ≤ x
n
− y ≤ x
o
− y,by (1).
Since the norm is uniformly convex,this implies that there exists a
constant δ,0 < δ < 1,such that
x
n+1
− y = 
1
2
(x
n
+ Tx
n
) −
1
2
(y + Ty)
= 
1
2
(x
n
− y) +
1
2
(Tx
n
− Ty)
≤ max {x
n
− y,Tx
n
− Ty}
26
Fixedpointtheoremsinnormedlinearspaces
≤ x
n
− y for n ≥ N.
Therefore lim
n→∞
x
n
= y where T
y
= y.31
If there does not exist an ε > 0 for which (2) holds,there exists a
sequence n
k
of integers such that lim
k→∞
(x
n
k
− Tx
n
k
) = 0,and such that
(Tx
n
k
) converges.But this implies that lim
k→∞
x
n
k
= u = lim
k→∞
Tx
n
k
and so
Tu = u.
Hence x
n+1
− u ≤ x
n
− u,by (1).Since lim
k
x
n
k
− u = 0,we
have lim
n→∞
x
n
− u = 0 and the theorem is proved.
The following theorem was proved by Altam by means of the con
cept of`degree of a mapping',but we can easily deduce it from schan
der's theorem.
Theorem 2.7 (Altman).Let E be a normed space,let Q be the closed
ball of radius r > 0,
Q = {x:x ≤ r}
and let T be a continuous mapping of Qinto a compact subset of E such
that
Tx − x
2
≥ Tx
2
− x
2
(x = r)
Then T has xed point in Q.
Proof.Suppose T has no xed point in Q then
Tx − x + x > Tx (x = r) (1)
For
(Tx − x + x)
2
− Tx
2
= Tx − x
2
+ x
2
− Tx
2
+2x Tx − x ≥ 2rTx − x > 0
Let P be the mapping dened by32
Px =
x (x ∈ Q)
r
x
x (x Q)
Plainly P is a continuous projection of E onto Q.
Fixedpointtheoremsinnormedlinearspaces
27
Let
T = PT
Then T maps Qcontinuously into a compact subset of Q.Hence,by
the Schauder theorem,
T has a xed point u in Q,
PTu = u
If Tu ∈ Q,then PTu = Tu,and
Tu = u
If Tu Q,then Tu > r and
u = PTu =
r
Tu
Tu
If follows that u = r,and we have
Tu − u + u = 
Tu
r
u − u + u
=
Tu
r
− 1 + 1
!
u = Tu
which contradicts (1)
supplementary results and exercises
(1) For further results connected with Theorem2.6,see [19]
(2) Let A be a continuous mapping of a normed space E into itself
which maps bounded sets into compact sets and satises
lim
x→∞
Ax
x
= 0
Then given arbitrary real λ > 0 and y in E,the equation
x = λAx + y
has a solution x in E 33
Consider the mapping Tx = λAx + y
28
Fixedpointtheoremsinnormedlinearspaces
Clearly T has all the properties of A
Let S
n
= {x:x ∈ E,x ≤ n} (n = 1,2,...)
Then
TS
n
⊂ S
n
for some n (1)
Otherwise Tx
n
 > n,for some x
n
∈ S
n
n = 1,2,...(2)
If {x
n
} were bounded,then {Tx
n
} will be contained in a compact set
and therefore Tx
n
 will be bounded which contradicts (2).Hence
x
n
 →∞as n →∞But
Tx
n

x
n

> 1 so lim
x
n
→∞
Tx
n

x
n

≥ 1
As this is not true,T maps some S
n
into its compact subset;Schau
der's theorem then a gives a xed point x which is the required
solution.
(3) Let
∞
P
k=1
a
k
be a convergent series of nonnegative real numbers and
let ( f
k
) be a sequence of continuous mappings of the real line R into
itself such that
 f
k
(t) ≤ a
k
(t ∈ R,k = 1,2,...)
Given α ∈ R,there exists a convergent real sequence (ξ
k
) such that
(i) ξ = α
(ii) ξ
k+1
− ξ
k
= f
k
(ξ
k
) (k = 1,2,...)
consider the mapping T of (c) into itself given by
(Tx)
1
= α
(Tx)
n+1
= α +
n
X
k=1
f
k
(ξ
k
) (n = 1,2,...)
where x = (ξ
k
).34
Fixedpointtheoremsinnormedlinearspaces
29
(4) Let E be a Banach space with a uniformly convex norm,and let K
be a bounded closed convex subset of E.Then the metric projection
E
onto
−−−−→ K exists and is uniformly continuous on each bounded
subset of E.
(5) Brodsku and Milman [10],give conditions under which a convex
set in a Banach space has a point invariant under all isometric self
mappings.In this connection see also Dunford and Schwartz ([14],
p.459).
(6) Browder (11) gives some generalization of the Schauder theorem
which appear to lie rather deep.Perhaps the most striking of these
results is the following generalization of theorem 2.4.Let T be
a continuous mapping of a Banach space E into itself that maps
bounded sets into compact sets.If,for some positive integer m,T
m
E
is bounded,then T has a xed point.For a generalization of the
Schauder theorem of a diﬀerent kind see Stepaneek (32).
(7) Aronszajn [2] gives general regularity condition on T suﬃcient to
establish that the set of its xed points is an R
δ
i.e.is a homeomor
phic image of the intersection of decreasing sequence of absolute
retracts.
Chapter 3
The Schauder  Tychonoﬀ
theorem
It this chapter we are concerned with nonlinear operators in general 35
locally convex spaces.
Denition 3.1.A vector space E over R which is also a topological
space is called a linear topological space (l.t.s) if the mappings
(x,y) → x + y
(α,x) →αx
from E × E and R × E respectively into E are continuous.If also every
open set in E is a union of convex open sets,then F is said to be locally
convex.
We establish the elementary properties of a l.t.s E.Since the map
ping (α,x) → αx is continuous,the mapping x → αx,with xed α,
is continuous.Therefore,if α is a nonzero constant then the mapping
x →αx is a homeomorphism,and so
a) G open,α 0 =⇒αG open.
In particular
b) G open implies that  G is open.
31
32
TheSchauderTychonoﬀtheorem
c) Similarly,G open ⇐⇒y +G open,and so
d) V is neighbourhood of 0 if and only if y + V is neighbourhood of y.
Let V be a neighbourhood of 0,and let x ∈ E.Since the mapping36
α → αx is continuous,and 0x = 0,we have
1
λ
x ∈ V for all suﬃ
ciently large λ,i.e.,
e) x ∈ λV for all suﬃciently large λ.
We prove next that
f) The closure of a convex set is convex.
For 0 ≤ α ≤ 1,the mapping f:E × E →E given by
(x,y) →αx + (1 − α)y
is continuous and f (K ×K) ⊂ K.Therefore f (
K × K) ⊂
¯
K,where
¯
K
denotes the closure of K.But
K × K =
¯
K ×
¯
K and so f (
¯
K ×
¯
K) ⊂
¯
K
i.e.,αa + (1 − α)b ∈
¯
K for a,b ∈
¯
K.
g) The interior of a convex set is convex.
Let K
0
be the interior of a convex set K,let a,b ∈ K
0
and 0 < α < 1.
By (a),αK
0
,(1 − α)K
0
are open sets.By (c) αK
0
+ (1 − α)K
0
is a
union of open sets and is therefore open.Since
αa + (1 − α)b ∈ K
0
+ (1 − α)K
0
⊂ K,
it follows that αa + (1 − α)b ∈ K
0
.A subset A of a vector space E
over R is said to be symmetric if −A = A.
h) Let U be a neighbourhood of 0 in a locally convex l.t.s.Then there
exists a closed convex symmetric neighbourhood V of 0 with V ⊂ U.
Since 0 is an interior point of U and the space is locally convex,
there exists a convex open set G with 0 ∈ G ⊂ U.Let H =
1
2
(G ∩37
−G),and V =
¯
H.By (b) and ( f ) V is a closed convex symmetric
neighbourhood of 0.Finally V ⊂ U;for if v ∈ V,then v + H is an
TheSchauderTychonoﬀtheorem
33
open set containing and therefore has nonempty intersection with H,
ie there exists h,h
′
in H with v + h = h
′
.Since H is convex and
symmetric,
v = h
′
− h ∈ 2H ⊂ G.Thus V ⊂ G ⊂ U.
Denition.Given an l.t.s.E over K,a subset A of E is said absorb
points if for every x in E,
x ∈ λA
for all suﬃciently large λ.
Denition.Given a convex set K that absorbs points,the
Minkowski functional p
K
is dened by
p
K
(x) = inf{λ;λ > 0,and x ∈ λK}
Denition.A mapping p of E into R is called a seminorm on E if it
satises the axioms.
i) p(x) ≥ 0 (x ∈ E)
ii) p(αx) = α p(x) (x ∈ E,α ∈ R)
iii) p(x + y) ≤ p(x) + p(y)
Given a seminorm p,the seminorm topology determined by p is
the class of unions of open balls
S(x,∈) = {y:p(y − x) < ε} (ε > 0)
With this topology E is a locally convex l.t.s.which is not in general a
Hausdorﬀ space.
The Minkowski functional of a convex set K that absorbs points is 38
sublinear,and if K is also symmetric,then it is a seminorm.Also if
x ∈ λK and > λ,then x ∈ K,for 0 ∈ K since K absorbs points and
1
x =
λ
1
λ
x
!
+
1 −
λ
!
0 ∈ K
34
TheSchauderTychonoﬀtheorem
i) If K is a closed convex symmetric neighbourhood of 0 in a l.t.s,the
p Minkowski functional p
K
is a continuous seminorm in E,and
K = {x;p
K
(x) ≤ 1}
Conversely,if p is a continuous seminorm in E,then {x:p(x) ≤ 1}
is a closed convex symmetric neighbourhood K of 0,and p
K
= p.
Proof.Let K be a closed convex symmetric neighbourhood of 0.
Then p
K
is a seminorm on E,and so
 p
K
(x
′
) − p
K
(x) ≤ p
K
(x
′
− x) (x
′
,x ∈ E)
Given ε > 0,
x
′
ε,x + εK ⇒ x
′
− xεK
⇒ p
K
(x
′
− x) ≤ ε
⇒ p
K
(x
′
) − p
K
(x) ≤ ε
since x + εK is a neighbourhood of x,this shows that p
K
is continuous.
If xεK,then p
K
(x) ≤ 1,by the denition of p
k
.On the other hand,if
p
K
(x) ≤ 1,then x ε λK(λ > 1),
1
λ
x ∈ K(λ > 1),
and,since K is closed,x ∈ K.
Thus39
K = {x:p
k
(x) ≤ 1}.
Conversely,let p be a continuous seminorm on E,and let K = {x:
p(x) ≤ 1}.That K is a closed convex symmetric neighbourhood of the
origin is evident.We have
p(x) ≤ 1 ⇔ x ∈ K ⇔ p
k
(x) ≤ 1,
and,since p and p
k
are both positivehomogeneous,it follows that p =
p
k
.
TheSchauderTychonoﬀtheorem
35
(j) Let x be a nonzero point of a Hausdorﬀ locally convex l.t.s.E.Then
there exists a continuous seminorm p on E with p(x) > 0.
Proof.Since x 0 and E is a Hausdorﬀ space,there exists a neigh
bourhood U of 0 such that x U.By (h) there exists a closed con
vex symmetric neighbourhood U of 0 with V ⊂ U.By (i),there
exists a continuous seminorm p on E such that
V = {y:p(y) ≤ 1}
Hence p(x) > 1.
(k) Let E be a vector space over K.Let p be a seminorm on E,and let
N = {x:p(x) = 0}.Then N is a subspace of E,and the functional
q dened on the quotient space
E
N
by
q( x) = p(x) (x ∈ x,x
E
N
)
is a normon E/N
Proof.If x,y ∈ N,then 40
0 ≤ p(x + y) ≤ p(x) + p(y) = 0,
and so x + y ∈ N.Also p(x) = 0 implies p(λx) = 0,and so N is
a linear subspace of E.The denition of q( x) is in fact free from
ambiguity,for if x,x
′
∈ x,then x − x
′
∈ N,and so
 px − p(x
′
) ≤ p(x − x
′
) = 0,
p(x) = p(x
′
).
Finally that q satises the axioms of a norm is entirely straight
forward.
36
TheSchauderTychonoﬀtheorem
Lastly,among these preliminary results,we need a proposition
which is a special case of a general theorem on uniform spaces.How
ever,it is more convenient for our purposes to prove the special case
than to invoke the general theory.
(b) Let E,F be linear topological spaces,let K be a compact subset
of E and let T be a continuous mapping of K into F.Given a neigh
bourhood U of 0 in F,there exists a neighbourhood V of 0 in E such
that
x,x
′
∈ K,x − x
′
∈ V ⇒Tx − Tx
′
∈ U.
Proof.Let H be an open set containing 0 such that
H − H ⊂ U
Given xǫK,there exists a neighbourhood G(x) of 0 such that x
′
∈ K ∩41
(X +G(x)) ⇒Tx
′
∈ Tx + H.
Let V(x) be an open neighbourhood of 0 in E such that
V(x) + V(x) ⊂ G(x),
since K is compact and is covered by open sets x + V(x),it has a nite
covering
x
1
+ V(x
1
),...,x
n
+ V(x
n
).
Let V =
n
T
i=1
V(x
i
).
Then V is a neighbourhood of 0 in E.Suppose x,x
′
∈ K and x−x
′
∈
V.Then there exists j with
x
′
∈ x
j
+ V(x
j
) ⊂ x
j
+G(x
j
)
x − x
j
= x − x
′
+ x
′
− x
j
∈ V + V(X
j
) ⊂ V(x
j
) + V(x
j
) ⊂ G(x
j
)
since x,x
′
∈ x
j
+G(x
j
),
we have Tx ∈ Tx
j
+ H,Tx
′
∈ Tx
j
+ H,
and so Tx − Tx
′
∈ H − H ⊂ U.
We are nowready to prove the main theoremby which we are able to de
duce properties of operators in a locally convex linear topological space
TheSchauderTychonoﬀtheorem
37
from the corresponding properties of operators in normed spaces.The
main idea of this theorem was derived from the proof of the Schauder42
Tychnoﬀ theorem in Dunford and Schwartz [14] p.454.
Theorem3.1.Let K be a compact subset of a locally convex l.t.s E,T a
continuous mapping of K into itself,p
◦
a continuous seminorm on E.
Then there exists a seminorm q on the linear L(K) of K such that
i) q(x) ≥ p
◦
(x) (x ∈ L(K));
ii) q is continuous on K − K;
iii) K is compact with respect to the seminorm topology given by q;
iv) T is uniformly continuous in K with respect to q i.e.,given ǫ > 0,
there exists δ > 0,such that
x,x
′
∈ K,q(x − x
′
) < δ ⇒q(Tx − Tx
′
) < ǫ
Remark.It would be better if one could prove the existence of a con
tinuous seminorm q on E satisfying (i) and (iv).
Proof.Since p
◦
is bounded on K there is no real loss of generality in
supposing that
p
◦
(x) ≤ 1 (x ∈ K).
It is convenient to introduce the following denition.We sa y that a set Γ
of continuous seminorm dominates a set Γ of continuous seminorms
if the following two conditions are satised.
a) p
′
(x) ≤ 1 (x ∈ K,p
′
∈ Γ
′
) 43
b) given p ∈ Γ and ǫ > 0,there exists p
′
∈ Γ
′
and δ > 0 such that
x,x
′
∈ K,p
′
(x − x
′
) < δ ⇒ p(Tx − Tx
′
) < ǫ.
We construct a countable selfdominating set containing p
◦
.Given
a continuous seminorm p,and a positive integer n,the set
(
x:p(x) <
1
n
)
38
TheSchauderTychonoﬀtheorem
is a neighbourhood of 0.Therefore,by proposition (2),there exists a
neighbourhood V of 0 in E such that
x,x
′
∈ K,x − x
′
∈ V ⇒ p(Tx − Tx
′
) <
1
n
.
By (h),we may suppose that V is a closed convex symmetric neigh
bourhood of 0,and then by (i),p
V
is a continuous seminorm and
V = {x:p
V
(x) ≤ 1}.
Multiplying p by an appropriate positive constant δ
n
,we obtain a
continuous seminorm q
n
such that
q
n
(x) ≤ 1 (x ∈ K),
and such that x,x
′
∈ K,q
n
(x − x
′
) < δ
n
⇒ p(Tx − Tx
′
) <
1
n
.
Plainly the set of seminorms q
n
is a countable set dominating the set44
(p).
It follows that given a countable set Γ of continuous seminorms,
there exists a countable set Γ
′
that dominates Γ.Now the set (p
◦
) is
dominated by a countable set Γ
1
,Γ
1
is dominated by a countable set Γ
2
,
and so on.Finally,we take
Γ = (p
◦
) ∪
∞
[
n=1
Γ
n
.
Then Γ is a countable selfdominating set.Let (p
n
)
∞
◦
be an enumer
ation of Γ and take
q(x) =
∞
X
n=0
2
−n
p
n
(x) (1)
since
p
n
(x) ≤ 2 (x ∈ K − K),
the series (1) converges uniformly on K − K,and so q is continuous on
K − K.Also the series converges on L(K) (linear hull of K) and q is a
seminorm there satisfying (i).Given xǫK,let
S(x,ρ) = {x
′
;x
′
∈ K and q(x − x
′
) < ρ}
TheSchauderTychonoﬀtheorem
39
since q is continuous on K − K,S(x,ρ) is an open subset of K in the
topology τ on L(K) induced fromthe initial topology on E.Hence each
open subset of K in the topology induced by τ
q
(topology on L(K) de 45
ned by q) is also open in the topology induced by τ.(iii) is now an
immediate consequence of the τcompactness of K.
Given ǫ > 0,we choose N with 2
−N
<
ǫ
4
since we have
∞
X
n=N+1
1
2
n
p
n
(x − x
′
) ≤
∞
X
n=N+1
1
2
n+1
<
ǫ
2
(x,x
′
∈ K),
and so
q(x − x
′
) ⊂
N
X
n=0
1
2
n
p
n
(x − x
′
) +
ǫ
2
(x,x
′
∈ K) (2)
since T maps K into itself,(2) gives
q(Tx − Tx
′
) <
N
X
n=0
1
2
n
p
n
(Tx − Tx
′
) +
ǫ
2
(x,x
′
∈ K) (3)
since Γ is selfdominated,for each n,there exists k
n
and δ
n
> 0 such that
p
kn
(x − x
′
) < δ
n
⇒ p
n
(Tx − Tx
′
) <
ǫ
4
(x,x
′
∈ K) (4)
Let N
′
= max(k
◦
,...,k
N
),and
= 2
−N
′
min(δ
◦
,...,δ
N
).
Then since p
n
≤ 2
N
′
q for n ≤ N
′
,we have
q(x − x
′
) < δ ⇒
p
k
n
(x − x
′
) < δ
n
(n ≤ N)
and so,by (4)
x,x
′
ǫk,q(x − x
′
),< δ ⇒ p
n
(Tx − Tx
′
) <
ǫ
4
(n = 0,1,...,N).
Therefore,by (3),46
x,x
′
∈ K,q(x − x
′
) < δ ⇒q(Tx − Tx
′
) < ε.
40
TheSchauderTychonoﬀtheorem
Theorem 3.2 (SchauderTychonoﬀ).Let K be a non empty compact
convex subset of a locally convex Hausdorﬀ l.t.sE,and let T be a con
tinuous mapping of K into itself.Then T has a xed point in K.
Proof.There is no loss of generality in supposing that L(K) = E.Sup
pose that T has no xed point in K.Then Tx − Tx 0(x ∈ K).
It follows by proposition ( j) that for each point x of K there exists a
continuous seminorm p
x
such that
p
x
(Tx − x) > 0
By continuity of T and p
x
,there exists a neighbourhood U
x
of x such
that
p
x
(T
y
− y) > 0 (y ∈ U
x
)
Since K is compact,there is a nite covering of K by such neigh
bourhood say
U
x
1
,...,U
x
m
.
Let p = p
x
1
+ p
x
2
+ + p
x
m
.
Then p is a continuous seminorm and47
p(Tx − x) > 0 (x ∈ K) (1)
Let q be the seminorm constructed as in Theorem 3.1 with p
0
= p.
Then q is dened on L(K) = E,q ≥ p,K is compact in the seminorm
topology τ
q
,and given ε > 0,there exists δ > 0 such that
x,x
′
ǫK,q(x − x
′
) < δ ⇒q(Tx − Tx
′
) < ε (2)
Let N = {x:q(x) = 0}.By lemma 3.4,E/N is a normed space with
the norm given by
 x = q(x)
where x is the coset of x.Let
K = {x:x ∈ K}.Then since mapping
x → x is a continuous homomorphism from E with the topology τ
q
to
TheSchauderTychonoﬀtheorem
41
E/N with the norm topology,K is a compact convex set in E/N.Also
by (2),
x,x
′
∈ K,q(x − x
′
) = 0 ⇒q(Tx − Tx
′
) = 0 (3)
For each x in
K there exists a point x in x ∩ K,and we dene
T x by
taking
T x =
f
Tx
By (3),this denition is unambiguous,and
T maps
K into itself.
Also,by (2),given ε > 0,there exists δ > 0 such that x,
x
′
∈ K, x−x
′
 <
δ ⇒ 
T x −
T x
′
 < ǫ.For given x,x
′
ǫK,there exist x ∈ K ∩ x and 48
x
′
ǫK ∩ x
′
and q(x − x
′
) =  x − x
′
.Hence
T is a continuous mapping of
the compact convex subset
K of the normed space E/N.Applying the
Schauder xed point theorem,
T has a xed point u say
T u = u.
Since uǫ
K,there exists u ∈ K ∩ u,and we have
T u =
f
Tu.Thus
Tu − u ∈ N,
i.e.,q(Tu − u) = 0
It follows that p(Tu − u) = 0,which contradicts (1) since u ∈ K.
Problem.It will be noticed that Theorem 3.2 generalizes theorem 2.3
rather than the full force of the Schauder theorem (2.2).It is not known
whether the following proposition is true.
Q.Let K be a closed convex subset of a locally convex Hausdorﬀ
l.t.s.E,and let T be a continuous mapping of K into a compact subset
of K.Then T has a xed point in K.
It is obvious that if T maps K into a compact convex subset H of K,
then T has a xed point.For
TH ⊂ TK ⊂ H
and we can apply theorem 3.2 to H instead of K.In particular,Q will 49
hold if every compact subset of K is contained in compact convex sub
set of K.By an elementary theorem of Bourbaki (Espaces Vectoriels
42
TheSchauderTychonoﬀtheorem
Topologiques,Ch.II,p.80) the convex hull of a precompact subset of a
locally convex Hausdorﬀ l.t.s is precompact.Thus we can obtain a true
theoremfrom Qby supposing that K be complete instead of closed or E
quasicomplete.However,this is certainly unnecessarily restrictive.By
the KreinSmulian Theorem [21],if E is a branch space with the weak
topology as the specied topology,then the closed convex hu ll of each
compact subset of E is compact,and so the proposition Q holds,even
though K need not be completes (in the weak topology).
Example.Let E be a reexive Banach space,K a closed convex subset
of E,T a weakly continuous mapping of K into a bounded subset of K.
Then T has a xed point in K.
For since K is norm closed and convex it is also weakly closed.
Also since E is reexive,each bounded weakly closed subset of E is
weakly compact.Hence the weakly closed convex hull of TK is weakly
compact.
Chapter 4
Nonlinear mappings in cones
The theorems in this chapter are mainly due to Krein and Rutman [20] 50
and to Schaefer [28].They may be regarded as a further step in the
transition from nonlinear to linear problems.We will be content with
considering normed spaces only,through theorems of the kind studied
here have been proved for general locally convex spaces by H.Schaeﬀer.
Denition 4.1.Asubset C of a vector space E over R is called a positive
cone if it satises
(i) x,y ∈ C ⇒ x + y ∈ C
(ii) x ∈ C,α ≥ 0 ⇒α ∈ C
(iii) x,−x ∈ C ⇒ x = 0
(iv) C contains nonzero vectors.
A vector space E over R with a specied positive cone is called a
partially ordered vector space,and we write x ≤ y( or y ≥ x) to denote
that y − x ∈ C.It is easily veried that this relation ≤ is a relation of
partial order in the usual sense i.e.,
(v) x ≤ x (x ∈ E),
(vi) x ≤ y,y ≤ z ⇒ x ≤ z,
43
44
Nonlinearmappingsincones
(vii) x ≤ y,y ≤ x ⇒ x = y.
Also the partial ordering and the linear structure are related by the prop
erties:51
(viii) x
i
≤ y
i
(i = 1,2) ⇒ x
1
+ x
2
≤ y
1
+ y
2
,
(ix) x ≤ y,0 ≤ α ≤ β ⇒αx ≤ βy.
Conversely,given a nontrivial relation ≤ in E satisfying (v),...,
(xi),the set {x:0 ≤ x} is a positive cone in E to which the given relation
corresponds in the above manner.
Denition 4.2.Let C be a positive cone in a normed space E.A map
ping T of a subset D of C into C is said to be strictly positive in D
if
x
n
∈ D,lim
n→∞
Tx
n
= 0 ⇒ lim
n→∞
x
n
= 0
A mapping T dened on C is said to be completely continuous in C if it
is continuous in C and maps each bounded subset of C into a compact
set.
Theorem 4.1 (Morgenstern [23]).Let C be a closed positive cone in a
normed vector space E such that the norm is additive on Ci.e.
x + y = x + y (x,y ∈ C)
Let c > 0,let K = {x:x ∈ C,x = c},and let T be a continuous
and positive mapping on K and strictly positive on K and map K into
a compact subset of C.Then there exists u in K and λ > 0 such that
Tu = λ
u
.
Proof.Since the norm is additive on C,K is a convex set.Since T is
strictly positive on K,
inf{Tx,x ∈ K} > 0
and therefore the mapping A dened on K by
Ax = cTx
−1
Tx
is continuous and maps K into a compact subset A itself.By the52
Schauder theorem,there exists u ∈ K with Au = u.
Nonlinearmappingsincones
45
Corollary.Let C be a closed positive cone in a normed vector space
E such that the norm is additive on C.Let T be a strictly positive and
completely continuous mapping of C into itself.Then for each c > 0,
there exists u
c
in C and λ
c
> 0 such that Tu
c
= λ
c
u
c
and u
c
 = c.
Denition 4.3.A positive cone C in a normed vector space is said be
normal if there exists a positive constant γ such that
x + y ≥ γx (x,y ∈ C).
Theorem 4.2 (Schaefer).Let C be a closed normal positive cone in a
normed space.Let c > 0 and let K =
n
x
x∈C
x≤c
o
.Let T be a continuous
and strictly positive on K and map K into a compact set.Then there
exists u ∈ C,and λ > 0,such that Tu = λ
u
and u = c.
Proof.Since TK is contained in a compact set we can choose > 0,
such that Tk ⊂ K.Let A = T,let y be a point of K with y = c,and let
B be the mapping dened on K by Bx = c
−1
xAx + c
−1
(c − x)y (x ∈ k),
since K is convex,we have BK ⊂ K.Also B is continuous in K,and
maps K into a compact set.Since T is strictly positive on K,there exists
c > 0 such that
x ∈ K,x ≥
1
2
c ⇒Ax ≥ ε
since C is a normal cone,it follows that 53
x ∈ K,x ≥
1
2
c ⇒Bx ≥ γc
−1
1
2
cε =
1
2
γε
On the other hand,
x ∈ K,x ≤
1
2
c ⇒c
−1
(c − x)y ≥
1
2
c ⇒Bx
1
2
γc
Therefore Bx ≥
1
2
γ min(ε,c) > 0 (x ∈ K).
It follows that the mapping
x →cBx
−1
Bx
46
Nonlinearmappingsincones
is a continuous mapping of K into a compact subset of itself.Therefore,
there exists u ∈ K with
u = cBu
−1
Bu
Plainly u = c,and so Bu = Au,and we have
Au = λ,with λ = c
−1
Au > 0.
The following theorem due to Krein and Rutman [20,Theorem 9.1]
marks a further transition towards a linear problem.
Denition 4.4.Let E be a partially ordered vector space with positive
cone C,let T be a mapping of C into itself and let c be be positive real
number.T is said to be
(i) positivehomogeneous of54
T(αx) = αTx (α ≥ 0,x ∈ C)
(ii) monotonic increasing if
x,y ∈ C,x ≤ y ⇒Tx ≤ Ty
(iii) cdominant if there exists a nonzero vector u in C with Tu ≥ cu.
Lemma 4.1.Let C be a closed positive cone in a normed space E,
and let u be a point that does not belong to −C.Then there exists a
continuous linear functional f on E such that
(i) f (u) = d(u,−C) > 0,
(ii) f (x) ≥ 0 (x ∈ C),
(iii)  f  ≤ 1
Proof.Let
p(x) = d(x,−C) = inf{x + y:y ∈ C}
Then p is a sublinear functional on E,with the properties
Nonlinearmappingsincones
47
(a) p(u) = d(u,−C) > 0,
(b) p(x) = 0(x ∈ −c),
(c) p(x) ≤ x (x ∈ E).
By the HahnBanach theoremthere exists a linear functional f on E
with f (u) = p(u) and with
f (x) ≤ p(x) (x ∈ E).
Plainly f has the required properties.55
Theorem4.3 (Krein and Rutman).Let E be a partially ordered normed
vector space with a closed positive cone C.Let T be a completely con
tinuous mapping of Gin to itself which is positivehomogeneous,mono
tonic increasing,and cdominant for some c > 0.Then there exists a
nonzero vector v in C and a real number λ ≥ c such that Tv = λv.
Proof.Since T is positivehomogeneous and cdominant,there exists a
vector u in C with u = 1 and
Tu ≥ cu (1)
since u −C,Lemma 4.1 establishes the existence of a continuous
linear functional f on E with
f (u) > 0,f (x) ≥ 0 (x ∈ C) (2)
and  f  = 1 (3)
We now prove that
x ∈ C,α > 0,β > 0,Tx = αx − βu ⇒α > c.(4)
Let Γ denote the set of positive real numbers t with x ≥ tu.Since
x =
β
α
u +
1
α
Tx ≥
β
α
u,we have
β
α
∈ Γ.Also Γ is bounded above,
for otherwise
1
n
x ≥ u (n = 1,2,...),
48
Nonlinearmappingsincones
and since C is closed,this gives 0 ≥ u,u = 0 with is not true.
Let mdenote the least upper bound of Γ.Using again the fact that C 56
is closed,we have
x ≥ mu
and therefore
Tx ≥ T(mu) = mTu ≥ mcu
Since Tx = αx − βu,this gives
x ≥
β + mc
α
u,
and therefore
β + mc
α
≤ m,
m(α − c) ≥ β > 0,
α > c
In the rest of the proof ε will denote a real number with
0 < ε <
1
2
(5)
Let K
ε
= {x:x ∈ E, x ≤ 1,xgeqε  x  u,f (x) ≥ εf (u) Clearly K
is a closed,convex,bounded subset of E.Next we note that,for some
δ > 0,
 Tx ≥ δ ∈ f (u)0(x ∈ K
∈
) (6)
For x ∈ K
ε
,x ≥ ε  x  u gives
Tx ≥ ε  x  Tu
since T is positive homogeneous and monotonic increasing.57
By (3), Tx ≥ f (Tx)
and by (2) f (Tx) ≥ f (εc  x  u) = εc  x  f (u) (xǫK
∈
)
i.e., Tx ≥ εc  x  f (u) ≥ εe f (x) f (u)
Nonlinearmappingsincones
49
≥ δεf (u) (x ∈ K
ε
)
with δ = εc f (u) > 0.
Let V
ǫ
be the mapping dened by taking
V
ǫ
(0) = 0,
V
ε
(x) = x . x + 2ε  x  u 
−1
(x + 2ε  x  u),x 0
V is well dened since
 x + 2ǫ  x  u ≥ x  −2ε  x   u 
= x  (1 − 2ε) > 0 if  x  0.
Plainly V
ǫ
is continuous in E and
 V
ǫ
x = x  (7)
Also
x ∈ C, x = 1 V
ε
x ∈ K
ε
(8)
For f (V
ε
x) = xx + 2εxu
−1
{ f (x) + 2εx f (u)}
≤
2ǫ
1 + 2ǫ
f (u) ≥ εf (u)
Let A
ε
be the mapping dened on K
ǫ
by 58
A
ε
= V
ε
LT
when Lx =
x
x
x 0
Then by (6) and (8)
A
ε
K
ε
⊂ K
ε
By (6),V
ε
L is continuous in
TK
ε
,and A
ε
continuously into a compact
subset of K
ε
.Applying the Schauder theorem,we see that there exists a
point x
ε
in K
ε
such that
A
ε
x
ε
= x
ε
,
i.e.,V
ε
(
Tx
Tx
ε

)
= x
ε
(9)
50
Nonlinearmappingsincones
i.e.Tx
ε

−1
Tx
ε
+2εU =  Tx
ε

−1
Tx
ε
+2εux
ε
This can be written in
the form
Tx
ε
= α
ε
x
ε
− β
ε
u,(10)
where c < α
ε
< (1 + 2ε)Tx
ε
.
We now choose a sequence (ε
n
) such that lim
n→∞
ε
n
= 0,and such
that the sequences (Tx
ε
n
) and (αε
n
) converges.Let v = lim
n→∞
Tx
ε
n
and
λ = lim
n→∞
α
ǫ
n
.Then λ ≥ c,and since limn →∞β
ε
n
= 0,(10) gives59
lim
n→∞
x
ε
n
=
1
λ
v
By continuity and positive homogeneity of T,
1
λ
Tv = T
1
λ
v
!
= lim
n→∞
Tx
ǫ
n
= v
Finally by (7) and (9),x
ǫ
 = 1 and so v 0.
Remark.The theorems in this chapter are unsatisfactory in that each of
them involves an adhoc condition (strict positivity and c dominance).
It turns out that for linear mappings such an ad hoc condition can be
avoided,and I think that there is still scope for proving a better theorem
on nonlinear mappings also.
Chapter 5
Linear mapping in cones
If A is a linear operator in R
(n)
with a matrix (a
i j
) with nonnegative 60
elements,a
i j
≥ 0,then,by a famous theorem of Perron and Frobenius
(see for example Gantmacher,The theory of matrices),there exists an
eigen vector of A with nonnegative coordinates and with eigenvalue ρ,
such that all other eigenvalues satisfy λ ≤ ρ.
If we take E = R
(n)
and C to be the set of all vectors in E with non
negative coordinates,then C is a positive cone in E,and n × n matrices
with nonnegative elements correspond to linear operators in E that map
C into itself.Then theorems of the present chapter may be regarded as
generalizations of the PerronFrobenius Theorem.A great many such
generalizations with various methods of proof have been published dur
ing recent years,and our list of references is far from complete.
The idea of the method of proof adopted here is the use of the simple
concept of'topological divisor of zero'.
Let U be a Banach algebra with a unit element e,and let a be a
frontier point of the set of invertible elements.Then a is a topological
divisor of zero,i.e.,there exists a sequence (x
n
) with x
n
 = 1(n =
1,2,...) such that
lim
n→∞
ax
n
= 0
61
51
52
Linearmappingincones
Proof.There exists a sequence (a
n
) of invertible elements such that
lim
n→∞
a
n
= a
Then the sequence (a
−1
n
) is unbounded.For otherwise
lim
n→∞
(a
n
− a)a
−1
n
= 0
i.e.,lim
n→∞
(e − aa
−1
n
) = 0
But this implies that aa
−1
n
is invertible for some n,and therefore a has a
right inverse.Similarly a has a left inverse,and a is invertible,which is
absurd since the set of invertible elements is open.
We may therefore suppose that a
−1
n
 → ∞,and take x
n
= a
−1
n

−1
a
−1
n
.Then x
n
 = 1,and
lim
n→∞
ax
n
= lim
n→∞
(a − a
n
)x
n
+ a
n
x
n
= 0.
In particular,if λ is a frontier point of the spectrum of an element b,
then λe − b is a frontier point of the set of invertible elements of Uand
so there exists (x
n
),with x
n
 = 1 and
lim
n→∞
(λe − b)x
n
= 0
If further we know that λ 0,and bx
n
k
→u for some subsequence62
(x
n
k
).Then
λx
n
k
→u,
λbx
n
k
→bu,
so that bu = λu and u 0,since u = λ.
Actually,our method is not quite so simple as this,for our Banach
algebra is a Banach algebra of operation on a Banach space X,and we
have to replace the sequence (x
n
) of operators by a sequence of elements
of X.
Until we reach the statement of our main theorem(Theorem5.1) we
shall use the following notation.
Linearmappingincones
53
E will denote a normed and partially ordered vector space with norm
x and positive cone C.We suppose that C is complete with respect to
x,and that
E = C −C
We do not suppose that E is complete with respect to x.We denote by
B the intersection of C and the closed unit of E,i.e.,
B = {x:x ∈ C and x ≤ 1}
We denote by B
0
the convex symmetric hull of B,i..e.
B
0
= {αx − βy:x,y ∈ B,α ≥ 0,β ≥ 0,α + β = 1},
and by x
c
the Mindowski functional of B
0
,i.e.,63
x
c
= inf{λ:λ > 0,x ∈ λB
0
}
Lemma 5.1.(α) x
c
is a norm on E and satises
x
c
= x (x ∈ C),x
c
≥ x (x ∈ E)
(β) E is complete and C is a closed subset of E with respect to x
c
.
Proof.(α) since E = C − C,B
0
is an absorbing set for E,and so the
Mindowski functional x
c
is dened on E.Since B
0
is convex and
symmetric,x
c
is a seminorm on E.If z ∈ E and z
c
< 1,then z ∈ B
0
i.e.z = αx − βy with x,y ∈ B and α ≥ 0,β ≥ 0,α + β = 1.Therefore
z = αx − βy ≤ αx + βy ≤ α + β = 1.
This proves that
x ≤ x
c
(x ∈ E)
and completes the proof that x
c
is a norm.Since B ⊂ B
0
,we have
x
c
≤ 1 (x ∈ B),
and therefore
x
c
≤ x (x ∈ C)
54
Linearmappingincones
This completes the proof of (α).64
(β) Let (z
n
) be a Cauchy sequence in E with respect to x
c
.Then
there exists a strictly increasing sequence (n
k
) of positive integers such
that
p,q ≥ n
k
⇒z
p
− z
q

c
< 2
−k
Let w
k
= z
n
k
(k = 1,2,...).Then,in particular,
w
k+1
− w
k

c
< 2
−k
(k = 1,2,...)
Therefore
w
k+1
− w
k
∈ 2
−k
B
0
,
and so w
k+1
− w
k
= α
k
x
k
− β
k
y
k
,with
α
k
≥ 0,β ≥ 0,α
k
+ β
k
= 1,x
k
,y
k
∈ 2
−k
B
Let
s
n
=
n
X
k=1
α
k
x
k
,t
n
=
n
X
k=1
β
k
y
k
.
Then p > q gives
s
p
− s
q
 ≤
p
X
q+1
α
k
x
k
 ≤
p
X
k=q+1
2
−k
< 2
−q
,
and similarly
t
p
− t
q
 < 2
−q
.
Since C is complete,there exist s,t in C such that65
lim
n→∞
s − s
n
 = 0,lim
n→∞
t − t
n
 = 0
Also,lim
p→∞
s
p
− s
q
= s − s
q
,and s
p
− s
q
∈ C whenever p > q.Hence
s − s
n
∈ C (n = 1,2,...).Therefore
s − s
n

c
= s − s
n
 (n = 1,2,...),
Linearmappingincones
55
and so lim
n→∞
s − s
n

c
= 0.
Similarly,lim
n→∞
t − t
n

c
= 0,and so (w
n
) converges with respect to
x
c
to w
1
+s−t.It is noweasily seen that (z
n
) converges with respect to
x
c
,and so E is complete with respect to x
c
.That C is a closed subset
of E with respect to x
c
is a simple consequence of the inequality
x ≤ x
c
(x ∈ E)
and the closeness of C with respect to x,(in fact a larger normgives a
stronger topology).
Denition 5.1.A linear operator in E is said to be positive if it maps C
into C and to be partially bounded if it maps B into a bounded set.The
partial bound p(T) of a partially bounded linear operator T is dened
by
p(T) = sup{T:x ∈ B}
Given partially bounded positive linear operators S and T,we have 66
p(ST) ≤ p(S)p(T),
and therefore the limit
lim
n→∞
p(T)
n
1
n
exists.It is called the partial spectral radius of T.
Lemma 5.2.A positive linear operator T is partially bounded if and
only if it is a bounded linear operator in the Banach space (E,x 
c
).
For such an operator T,
p(T) = T
c
= sup {Tx
c
:x ∈ E and x
c
≤ 1},
and the partial spectral radius of T is equal to its spectral radius as
an operator in (E,x
c
),i.e.
= lim
n→∞
{T
n

c
}
1
n
56
Linearmappingincones
Finally,if λ > ,there exists a partially bounded positive linear opera
tor R
λ
,such that
(λI − T)R
λ
= R
λ
(λI − T) = I,
where I is the identity operator in E.
Proof.Let T be a partially bounded positive linear operator in E and let
x ∈ E with x
c
< 1.Then x ∈ B
0
,and so
x = αy − βz
with α ≥ 0,β ≥ 0,α + β = 1,y,z ∈ B.Then67
Tx = αTy − βTz,
and so
Tx
c
≤ αTy
c
+ βTz
c
= αTy + βTz
≤ (α + β)p(T) = p(T)
Thus T is a bounded linear operator in (E,x
c
) and
T
c
≤ p(T)
For the converse and the reversed inequality it is enough to note that
T
c
≥sup {Tx
c
:x ∈ C and x
c
≤ 1}
sup {Tx:x ∈ B} = p(T).
That = lim
n→∞
{T
n

c
}
1
n
is an obvious consequence of the fact that
T
c
= p(T) for each partially bounded linear operator.
If λ > ,the series
1
λ
I +
1
λ2
T +
1
λ
3
T
2
+
Linearmappingincones
57
converges with respect to the operator normfor bounded linear operators68
in the Banach space (E,x
c
) to a bounded linear operator R
λ
,and
(λI − T)R
λ
= R
λ
(λI − T) = I.
Since C is closed with respect to x
c
,and the partial sums of the
series are obviously positive operators,it follows that R
λ
is a positive
operator.
Denition 5.2.A positive linear operator T is said to be a normalising
operator if it satises the following condition:
α
n
≥ 0,y
n
∈ B,α
n
x ≥ Ty
n
,lim
n→∞
α
n
= 0 lim
n→∞
Ty
n
 = 0
If C is a normal cone,then every positive linear operator is obvi
ously a normalising operator.We shall see later that a certain compact
ness condition on T suﬃces to make T a normalising operator (without
restriction on C).
Lemma 5.3.Let T be a normalizing partially bounded positive linear
operator with partial spectral radius .Then
lim
λ→+0
p(R
λ
) = ∞
Corollary.For each such operator T,the partial spectral radius is in
the spectrumof T regarded as an operator in the Banach space (E,x
c
).
Suppose that the conditions of the lemma are satised but tha t p(R
λ
)
does not tend to innity as λ decreases to .Then there exists a positive 69
constant M such that p(R
ν
) ≤ M for some ν gather than and arbitrarily
close to .
The case = 0 is easily settled.For if = 0,then,fromthe formula
(λI − T)R
λ
= I,it follows that
λR − λx ≥ x (λ > 0,x ∈ C) (1)
If we let λ tend to zero through values for which p(R
λ
) ≤ M,the left
hand side of (1) tends to zero,and,since C is closed,we obtain
−x ∈ C (x ∈ C).
58
Linearmappingincones
But this implies that C = (0) which was excluded by our axioms on C.
Suppose now that > 0.Then we may choose λ,ν with
0 < λ < < ν < λ + M
−1
,
and with p(R
ν
) ≤ M.Since R
c
= p(R
ν
),it follows that the series
R
ν
+ (ν − λ)R
2
ν
(ν − λ)
2
R
3
ν
+
converges with respect to the operator norm  
c
to a partially bounded
positive linear operator S which is easily seen to satisfy
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