Fixed Point Theorems and Applications VITTORINO PATA Dipartimento di Matematica “F. Brioschi”

Fixed Point Theorems and Applications VITTORINO PATA Dipartimento di Matematica “F. Brioschi”
Fixed Point Theorems
and Applications
VITTORINO PATA
Dipartimento di Matematica “F. Brioschi”
Politecnico di Milano
[email protected]
Contents
Preface
2
Notation
3
1. FIXED POINT THEOREMS
The Banach contraction principle .
Sequences of maps and fixed points
Fixed points of non-expansive maps
The Riesz mean ergodic theorem . .
The Brouwer fixed point theorem .
The Schauder-Tychonoff fixed point
The Markov-Kakutani theorem . . .
The Kakutani-Ky Fan theorem . . .
Notes on Chapter 1 . . . . . . . . . .
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2. SOME APPLICATIONS OF FIXED POINT THEOREMS
The implicit function theorem . . . . . . . . . . . . . . . . . .
Ordinary differential equations in Banach spaces . . . . . . .
Semilinear equations of evolution . . . . . . . . . . . . . . . .
An abstract elliptic problem . . . . . . . . . . . . . . . . . . . .
The invariant subspace problem . . . . . . . . . . . . . . . . .
Measure preserving maps on compact Hausdorff spaces . .
Invariant means on semigroups . . . . . . . . . . . . . . . . . .
Haar measures . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Game theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Notes on Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . .
Bibliography
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1
Preface
Fixed point theory is a fascinating subject, with an enormous number of applications
in various fields of mathematics. Maybe due to this transversal character, I have
always experienced some difficulties to find a book (unless expressly devoted to
fixed points) treating the argument in a unitary fashion. In most cases, I noticed
that fixed points pop up when they are needed. On the contrary, I believe that
they should deserve a relevant place in any general textbook, and particularly, in
a functional analysis textbook. This is mainly the reason that made me decide to
write down these notes. I tried to collect most of the significant results of the field,
and then to present various related applications.
This work consists of two chapters which, although rather self-contained, are
conceived to be part of a future functional analysis book that I hope to complete in
the years to come. Thus some general background is needed to approach the next
pages. The reader is in fact supposed to be familiar with measure theory, Banach
and Hilbert spaces, locally convex topological vector spaces and, in general, with
linear functional analysis.
Even if one cannot possibly find here all the information and the details contained
in a dedicated treatise, I hope that these notes could provide a quite satisfactory
overview of the topic, at least from the point of view of analysis.
Vittorino Pata
2
Notation
Given a normed space X, x ∈ X and r > 0 we denote
BX (x, r) = y ∈ X : ky − xk < r
B X (x, r) = y ∈ X : ky − xk ≤ r
∂BX (x, r) = y ∈ X : ky − xk = r .
Whenever misunderstandings might occur, we write kxkX to stress that the norm
is taken in X. For a subset Y ⊂ X, we denote by Y the closure of Y , by Y C the
complement of Y , by span(Y ) the linear space generated by Y , and by co(Y ) the
convex hull of Y , that is, the set of all finite convex combinations of elements of Y .
We will often use the notion of uniformly convex Banach space. Recall that a
Banach space X is uniformly convex if given any two sequences xn , yn ∈ X with
kxn k ≤ 1,
kyn k ≤ 1,
lim kxn + yn k = 2
n→∞
it follows that
lim kxn − yn k = 0.
n→∞
In particular, we will exploit the property (coming directly from the definition of uniform convexity) that minimizing sequences in closed convex subsets are convergent.
Namely, if C ⊂ X is (nonvoid) closed and convex and xn ∈ C is such that
lim kxn k = inf kyk
n→∞
y∈C
then there exists a unique x ∈ C such that
kxk = inf kyk
y∈C
and
lim xn = x.
n→∞
Clearly, being Hilbert spaces uniformly convex, all the results involving uniformly
convex Banach spaces can be read in terms of Hilbert spaces.
A weaker notion is strict convexity : a Banach space X is strictly convex if for
all x, y ∈ X with x 6= y the relation
kxk = kyk ≤ 1
implies
kx + yk < 2.
3
4
It is immediate to check from the definitions that uniform convexity implies strict
convexity.
Other spaces widely used here are locally convex spaces. A locally convex space
X is a vector space endowed with a family P of separating seminorms. Hence for
every element x ∈ X there is a seminorm p ∈ P such that p(x) = 0. Therefore
P gives X the structure of (Hausdorff) topological vector space in which there is a
local base whose members are covex. A local base B for the topology is given by
finite intersections of sets of the form
x ∈ X : p(x) < ε
for some p ∈ P and some ε > 0. Note that, given U ∈ B, there holds
U + U := x + y : x ∈ U, y ∈ U = 2U := 2x : x ∈ U .
Good references for these arguments are, e.g., the books [1, 4, 14]. Concerning
measure theory, we address the reader to [11, 13].
1. FIXED POINT THEOREMS
Fixed point theorems concern maps f of a set X into itself that, under certain
conditions, admit a fixed point, that is, a point x ∈ X such that f (x) = x. The
knowledge of the existence of fixed points has relevant applications in many branches
of analysis and topology. Let us show for instance the following simple but indicative
example.
1.1 Example Suppose we are given a system of n equations in n unknowns of
the form
gj (x1 , . . . , xn ) = 0,
j = 1, . . . , n
where the gj are continuous real-valued functions of the real variables xj . Let
hj (x1 , . . . , xn ) = gj (x1 , . . . , xn ) + xj , and for any point x = (x1 , . . . , xn ) define
h(x) = (h1 (x), . . . , hn (x)). Assume now that h has a fixed point x̄ ∈ Rn . Then
it is easily seen that x̄ is a solution to the system of equations.
Various application of fixed point theorems will be given in the next chapter.
The Banach contraction principle
1.2 Definition Let X be a metric space equipped with a distance d. A map
f : X → X is said to be Lipschitz continuous if there is λ ≥ 0 such that
d(f (x1 ), f (x2 )) ≤ λd(x1 , x2 ),
∀ x1 , x2 ∈ X.
The smallest λ for which the above inequality holds is the Lipschitz constant of f .
If λ ≤ 1 f is said to be non-expansive, if λ < 1 f is said to be a contraction.
1.3 Theorem [Banach] Let f be a contraction on a complete metric space X.
Then f has a unique fixed point x̄ ∈ X.
proof Notice first that if x1 , x2 ∈ X are fixed points of f , then
d(x1 , x2 ) = d(f (x1 ), f (x2 )) ≤ λd(x1 , x2 )
which imply x1 = x2 . Choose now any x0 ∈ X, and define the iterate sequence
xn+1 = f (xn ). By induction on n,
d(xn+1 , xn ) ≤ λn d(f (x0 ), x0 ).
5
6
1. fixed point theorems
If n ∈ N and m ≥ 1,
d(xn+m , xn ) ≤ d(xn+m , xn+m−1 ) + · · · + d(xn+1 , xn )
≤ (λn+m + · · · + λn )d(f (x0 ), x0 )
λn
d(f (x0 ), x0 ).
≤
1−λ
(1)
Hence xn is a Cauchy sequence, and admits a limit x̄ ∈ X, for X is complete.
Since f is continuous, we have f (x̄) = limn f (xn ) = limn xn+1 = x̄.
Remark Notice that letting m → ∞ in (1) we find the relation
d(xn , x̄) ≤
λn
d(f (x0 ), x0 )
1−λ
which provides a control on the convergence rate of xn to the fixed point x̄.
The completeness of X plays here a crucial role. Indeed, contractions on incomplete metric spaces may fail to have fixed points.
Example Let X = (0, 1] with the usual distance. Define f : X → X as
f (x) = x/2.
1.4 Corollary Let X be a complete metric space and Y be a topological space.
Let f : X × Y → X be a continuous function. Assume that f is a contraction on X
uniformly in Y , that is,
d(f (x1 , y), f (x2 , y)) ≤ λd(x1 , x2 ),
∀ x1 , x2 ∈ X, ∀ y ∈ Y
for some λ < 1. Then, for every fixed y ∈ Y , the map x 7→ f (x, y) has a unique
fixed point ϕ(y). Moreover, the function y 7→ ϕ(y) is continuous from Y to X.
Notice that if f : X × Y → X is continuous on Y and is a contraction on X
uniformly in Y , then f is in fact continuous on X × Y .
proof In light of Theorem 1.3, we only have to prove the continuity of ϕ. For
y, y0 ∈ Y , we have
d(ϕ(y), ϕ(y0 )) = d(f (ϕ(y), y), f (ϕ(y0 ), y0 ))
≤ d(f (ϕ(y), y), f (ϕ(y0 ), y)) + d(f (ϕ(y0 ), y), f (ϕ(y0 ), y0 ))
≤ λd(ϕ(y), ϕ(y0 )) + d(f (ϕ(y0 ), y), f (ϕ(y0 ), y0 ))
which implies
d(ϕ(y), ϕ(y0 )) ≤
1
d(f (ϕ(y0 ), y), f (ϕ(y0 ), y0 )).
1−λ
Since the above right-hand side goes to zero as y → y0 , we have the desired
continuity.
7
the banach contraction principle
Remark If in addition Y is a metric space and f is Lipschitz continuous in Y ,
uniformly with respect to X, with Lipschitz constant L ≥ 0, then the function
y 7→ ϕ(y) is Lipschitz continuous with Lipschitz constant less than or equal to
L/(1 − λ).
Theorem 1.3 gives a sufficient condition for f in order to have a unique fixed
point.
Example Consider the map
1/2 + 2x
g(x) =
1/2
x ∈ [0, 1/4]
x ∈ (1/4, 1]
mapping [0, 1] onto itself. Then g is not even continuous, but it has a unique
fixed point (x = 1/2).
The next corollary takes into account the above situation, and provides existence
and uniqueness of a fixed point under more general conditions.
Definition For f : X → X and n ∈ N, we denote by f n the nth -iterate of f ,
namely, f ◦ · · · ◦ f n-times (f 0 is the identity map).
1.5 Corollary Let X be a complete metric space and let f : X → X. If f n is a
contraction, for some n ≥ 1, then f has a unique fixed point x̄ ∈ X.
proof Let x̄ be the unique fixed point of f n , given by Theorem 1.3. Then
f n (f (x̄)) = f (f n (x̄)) = f (x̄), which implies f (x̄) = x̄. Since a fixed point of f is
clearly a fixed point of f n , we have uniqueness as well.
Notice that in the example g 2 (x) ≡ 1/2.
1.6 Further extensions of the contraction principle There is in the literature
a great number of generalizations of Theorem 1.3 (see, e.g., [6]). Here we point out
some results.
Theorem [Boyd-Wong] Let X be a complete metric space, and let f : X → X.
Assume there exists a right-continuous function ϕ : [0, ∞) → [0, ∞) such that
ϕ(r) < r if r > 0, and
d(f (x1 ), f (x2 )) ≤ ϕ(d(x1 , x2 )),
∀ x1 , x2 ∈ X.
Then f has a unique fixed point x̄ ∈ X. Moreover, for any x0 ∈ X the sequence
f n (x0 ) converges to x̄.
Clearly, Theorem 1.3 is a particular case of this result, for ϕ(r) = λr.
proof If x1 , x2 ∈ X are fixed points of f , then
d(x1 , x2 ) = d(f (x1 ), f (x2 )) ≤ ϕ(d(x1 , x2 ))
8
1. fixed point theorems
so x1 = x2 . To prove the existence, fix any x0 ∈ X, and define the iterate
sequence xn+1 = f (xn ). We show that xn is a Cauchy sequence, and the desired
conclusion follows arguing like in the proof of Theorem 1.3. For n ≥ 1, define
the positive sequence
an = d(xn , xn−1 ).
It is clear that an+1 ≤ ϕ(an ) ≤ an ; therefore an converges monotonically to some
a ≥ 0. From the right-continuity of ϕ, we get a ≤ ϕ(a), which entails a = 0. If
xn is not a Cauchy sequence, there is ε > 0 and integers mk > nk ≥ k for every
k ≥ 1 such that
dk := d(xmk , xnk ) ≥ ε,
∀ k ≥ 1.
In addition, upon choosing the smallest possible mk , we may assume that
d(xmk−1 , xnk ) < ε
for k big enough (here we use the fact that an → 0). Therefore, for k big enough,
ε ≤ dk ≤ d(xmk , xmk−1 ) + d(xmk−1 , xnk ) < amk + ε
implying that dk → ε from above as k → ∞. Moreover,
dk ≤ dk+1 + amk+1 + ank+1 ≤ ϕ(dk ) + amk+1 + ank+1
and taking the limit as k → ∞ we obtain the relation ε ≤ ϕ(ε), which has to be
false since ε > 0.
Theorem [Caristi] Let X be a complete metric space, and let f : X → X.
Assume there exists a lower semicontinuous function ψ : X → [0, ∞) such that
d(x, f (x)) ≤ ψ(x) − ψ(f (x)),
∀ x ∈ X.
Then f has (at least) a fixed point in X.
Again, Theorem 1.3 is a particular case, obtained for ψ(x) = d(x, f (x))/(1 − λ).
Notice that f need not be continuous.
proof We introduce a partial ordering on X, setting
xy
if and only if
d(x, y) ≤ ψ(x) − ψ(y).
Let ∅ 6= X0 ⊂ X be totally ordered, and consider a sequence xn ∈ X0 such that
ψ(xn ) is decreasing to α := inf{ψ(x) : x ∈ X0 }. If n ∈ N and m ≥ 1,
d(xn+m , xn ) ≤
≤
m−1
X
i=0
m−1
X
d(xn+i+1 , xn+i )
ψ(xn+i ) − ψ(xn+i+1 )
i=0
= ψ(xn ) − ψ(xn+m ).
9
the banach contraction principle
Hence xn is a Cauchy sequence, and admits a limit x∗ ∈ X, for X is complete.
Since ψ can only jump downwards (being lower semicontinuous), we also have
ψ(x∗ ) = α. If x ∈ X0 and d(x, x∗ ) > 0, then it must be x xn for large n.
Indeed, limn ψ(xn ) = ψ(x∗ ) ≤ ψ(x). We conclude that x∗ is an upper bound
for X0 , and by the Zorn lemma there exists a maximal element x̄. On the other
hand, x̄ f (x̄), thus the maximality of x̄ forces the equality x̄ = f (x̄).
If we assume the continuity of f , we obtain a slightly stronger result, even relaxing the continuity hypothesis on ψ.
Theorem Let X be a complete metric space, and let f : X → X be a continuous
map. Assume there exists a function ψ : X → [0, ∞) such that
d(x, f (x)) ≤ ψ(x) − ψ(f (x)),
∀ x ∈ X.
Then f has a fixed point in X. Moreover, for any x0 ∈ X the sequence f n (x0 )
converges to a fixed point of f .
proof Choose x0 ∈ X. Due the above condition, the sequence ψ(f n (x0 ))
is decreasing, and thus convergent. Reasoning as in the proof of the Caristi
theorem, we get that f n (x0 ) admits a limit x̄ ∈ X, for X is complete. The
continuity of f then entails f (x̄) = limn f (f n (x0 )) = x̄.
We conclude with the following extension of Theorem 1.3, that we state without
proof.
Theorem [Čirič] Let X be a complete metric space, and let f : X → X be such
that
d(f (x1 ), f (x2 ))
n
o
≤ λ max d(x1 , x2 ), d(x1 , f (x1 )), d(x2 , f (x2 )), d(x1 , f (x2 )), d(x2 , f (x1 ))
for some λ < 1 and every x1 , x2 ∈ X. Then f has a unique fixed point x̄ ∈ X.
Moreover, d(f n (x0 ), x̄) = O(λn ) for any x0 ∈ X.
Also in this case f need not be continuous. However, it is easy to check that it
is continuous at the fixed point. The function g of the former example fulfills the
hypotheses of the theorem with λ = 2/3.
1.7 Weak contractions We now dwell on the case of maps on a metric space
which are contractive without being contractions.
Definition Let X be a metric space with a distance d. A map f : X → X is a
weak contraction if
d(f (x1 ), f (x2 )) < d(x1 , x2 ),
∀ x1 6= x2 ∈ X.
10
1. fixed point theorems
Being a weak contraction is not in general a sufficient condition for f in order to
have a fixed point, as it is shown in the following simple example.
Example Consider the complete metric space X = [1, +∞), and let f : X → X
be defined as f (x) = x + 1/x. It is easy to see that f is a weak contraction with
no fixed points.
Nonetheless, the condition turns out to be sufficient when X is compact.
Theorem Let f be a weak contraction on a compact metric space X. Then f
has a unique fixed point x̄ ∈ X. Moreover, for any x0 ∈ X the sequence f n (x0 )
converges to x̄.
proof The uniqueness argument goes exactly as in the proof of Theorem 1.3.
From the compactness of X, the continuous function x 7→ d(x, f (x)) attains its
minimum at some x̄ ∈ X. If x̄ 6= f (x̄), we get
d(x̄, f (x̄)) = min d(x, f (x)) ≤ d(f (x̄), f (f (x̄))) < d(x̄, f (x̄))
x∈X
which is impossible. Thus x̄ is the unique fixed point of f (and so of f n for all
n ≥ 2). Let now x0 6= x̄ be given, and define dn = d(f n (x0 ), x̄). Observe that
dn+1 = d(f n+1 (x0 ), f (x̄)) < d(f n (x0 ), x̄) = dn .
Hence dn is strictly decreasing, and admits a limit r ≥ 0. Let now f nk (x0 ) be a
subsequence of f n (x0 ) converging to some z ∈ X. Then
r = d(z, x̄) = lim dnk = lim dnk +1 = lim d(f (f nk (x0 )), x̄) = d(f (z), x̄).
k→∞
k→∞
k→∞
But if z 6= x̄, then
d(f (z), x̄) = d(f (z), f (x̄)) < d(z, x̄).
Therefore any convergent subsequence of f n (x0 ) has limit x̄, which, along with
the compactness of X, implies that f n (x0 ) converges to x̄.
Obviously, we can relax the compactness of X by requiring that f (X) be compact
(just applying the theorem on the restriction of f on f (X)).
Arguing like in Corollary 1.5, it is also immediate to prove the following
Corollary Let X be a compact metric space and let f : X → X. If f n is a weak
contraction, for some n ≥ 1, then f has a unique fixed point x̄ ∈ X.
1.8 A converse to the contraction principle Assume we are given a set X and
a map f : X → X. We are interested to find a metric d on X such that (X, d) is a
complete metric space and f is a contraction on X. Clearly, in light of Theorem 1.3,
a necessary condition is that each iterate f n has a unique fixed point. Surprisingly
enough, the condition turns out to be sufficient as well.
11
sequences of maps and fixed points
Theorem [Bessaga] Let X be an arbitrary set, and let f : X → X be a map
such that f n has a unique fixed point x̄ ∈ X for every n ≥ 1. Then for every
ε ∈ (0, 1), there is a metric d = dε on X that makes X a complete metric space,
and f is a contraction on X with Lipschitz constant equal to ε.
proof Choose ε ∈ (0, 1). Let Z be the subset of X consisting of all elements z
such that f n (z) = x̄ for some n ∈ N. We define the following equivalence relation
on X \ Z: we say that x ∼ y if and only if f n (x) = f m (y) for some n, m ∈ N.
0
0
0
0
Notice that if f n (x) = f m (y) and f n (x) = f m (y) then f n+m (x) = f m+n (x).
But since x 6∈ Z, this yields n + m0 = m + n0 , that is, n − m = n0 − m0 . At
this point, by means of the axiom of choice, we select an element from each
equivalence class. We now proceed defining the distance of x̄ from a generic
x ∈ X by setting d(x̄, x̄) = 0, d(x, x̄) = ε−n if x ∈ Z with x 6= x̄, where
n = min{m ∈ N : f m (x) = x̄}, and d(x, x̄) = εn−m if x 6∈ Z, where n, m ∈ N are
such that f n (x̂) = f m (x), x̂ being the selected representative of the equivalence
class [x]. The definition is unambiguous, due to the above discussion. Finally,
for any x, y ∈ X, we set
d(x, x̄) + d(y, x̄)
if x 6= y
d(x, y) =
0
if x = y.
It is straightforward to verify that d is a metric. To see that d is complete,
observe that the only Cauchy sequences which do not converge to x̄ are ultimately
constant. We are left to show that f is a contraction with Lipschitz constant
equal to ε. Let x ∈ X, x 6= x̄. If x ∈ Z we have
d(f (x), f (x̄)) = d(f (x), x̄) ≤ ε−n = εε−(n+1) = εd(x, x̄).
If x 6∈ Z we have
d(f (x), f (x̄)) = d(f (x), x̄) = εn−m = εεn−(m+1) = εd(x, x̄)
since x ∼ f (x). The thesis follows directly from the definition of the distance. Sequences of maps and fixed points
Let (X, d) be a complete metric space. We consider the problem of convergence of
fixed points for a sequence of maps fn : X → X. Corollary 1.5 will be implicitly
used in the statements of the next two theorems.
1.9 Theorem Assume that each fn has at least a fixed point xn = fn (xn ). Let
f : X → X be a uniformly continuous map such that f m is a contraction for some
m ≥ 1. If fn converges uniformly to f , then xn converges to x̄ = f (x̄).
proof We first assume that f is a contraction (i.e., m = 1). Let λ < 1 be the
Lipschitz constant of f . Given ε > 0, choose n0 = n0 (ε) such that
d(fn (x), f (x)) ≤ ε(1 − λ),
∀ n ≥ n0 , ∀ x ∈ X.
12
1. fixed point theorems
Then, for n ≥ n0 ,
d(xn , x̄) = d(fn (xn ), f (x̄))
≤ d(fn (xn ), f (xn )) + d(f (xn ), f (x̄))
≤ ε(1 − λ) + λd(xn , x̄).
Therefore d(xn , x̄) ≤ ε, which proves the convergence.
To prove the general case it is enough to observe that if
d(f m (x), f m (y)) ≤ λm d(x, y)
for some λ < 1, we can define a new metric d0 on X equivalent to d by setting
d0 (x, y) =
m−1
X
k=0
1
d(f k (x), f k (y)).
k
λ
Moreover, since f is uniformly continuous, fn converges uniformly to f also with
respect to d0 . Finally, f is a contraction with respect to d0 . Indeed,
d0 (f (x), f (y)) =
m−1
X
1
d(f k+1 (x), f k+1 (y))
k
λ
k=0
m−1
X
= λ
≤ λ
k=1
m−1
X
k=0
1
1
d(f k (x), f k (y)) + m+1 d(f m (x), f m (y))
k
λ
λ
1
d(f k (x), f k (y)) = λd0 (x, y).
λk
So the problem is reduced to the previous case m = 1.
The next result refers to a special class of complete metric spaces.
1.10 Theorem Let X be locally compact. Assume that for each n ∈ N there
is mn ≥ 1 such that fnmn is a contraction. Let f : X → X be a map such that
f m is a contraction for some m ≥ 1. If fn converges pointwise to f , and fn is an
equicontinuous family, then xn = fn (xn ) converges to x̄ = f (x̄).
proof Let ε > 0 be sufficiently small such that
K(x̄, ε) := x ∈ X : d(x, x̄) ≤ ε ⊂ X
is compact. As a byproduct of the Ascoli theorem, fn converges to f uniformly on
K(x̄, ε), since it is equicontinuous and pointwise convergent. Choose n0 = n0 (ε)
such that
d(fnm (x), f m (x)) ≤ ε(1 − λ),
∀ n ≥ n0 , ∀ x ∈ K(x̄, ε)
13
fixed points of non-expansive maps
where λ < 1 is the Lipschitz constant of f m . Then, for n ≥ n0 and x ∈ K(x̄, ε)
we have
d(fnm (x), x̄) = d(fnm (x), f m (x̄))
≤ d(fnm (x), f m (x)) + d(f m (x), f m (x̄))
≤ ε(1 − λ) + λd(x, x̄) ≤ ε.
Hence fnm (K(x̄, ε)) ⊂ K(x̄, ε) for all n ≥ n0 . Since the maps fnmn are contractions, it follows that, for n ≥ n0 , the fixed points xn of fn belong to K(x̄, ε),
that is, d(xn , x̄) ≤ ε.
Fixed points of non-expansive maps
Let X be a Banach space, C ⊂ X nonvoid, closed, bounded and convex, and let
f : C → C be a non-expansive map. The problem is whether f admits a fixed point
in C. The answer, in general, is false.
Example Let X = c0 with the supremum norm. Setting C = B X (0, 1), the
map f : C → C defined by
f (x) = (1, x0 , x1 , . . .),
for x = (x0 , x1 , x2 , . . .) ∈ C
is non-expansive but clearly admits no fixed points in C.
Things are quite different in uniformly convex Banach spaces.
1.11 Theorem [Browder-Kirk] Let X be a uniformly convex Banach space and
C ⊂ X be nonvoid, closed, bounded and convex. If f : C → C is a non-expansive
map, then f has a fixed point in C.
We provide the proof in the particular case when X is a Hilbert space (the general
case may be found, e.g., in [6], Ch.6.4).
proof Let x∗ ∈ C be fixed, and consider a sequence rn ∈ (0, 1) converging to
1. For each n ∈ N, define the map fn : C → C as
fn (x) = rn f (x) + (1 − rn )x∗ .
Notice that fn is a contractions on C, hence there is a unique xn ∈ C such that
fn (xn ) = xn . Since C is weakly compact, xn has a subsequence (still denoted by
xn ) weakly convergent to some x̄ ∈ C. We shall prove that x̄ is a fixed point of
f . Notice first that
lim kf (x̄) − xn k2 − kx̄ − xn k2 = kf (x̄) − x̄k2 .
n→∞
14
1. fixed point theorems
Since f is non-expansive we have
kf (x̄) − xn k ≤ kf (x̄) − f (xn )k + kf (xn ) − xn k
≤ kx̄ − xn k + kf (xn ) − xn k
= kx̄ − xn k + (1 − rn )kf (xn ) − x∗ k.
But rn → 1 as n → ∞ and C is bounded, so we conclude that
lim sup kf (x̄) − xn k2 − kx̄ − xn k2 ≤ 0
n→∞
which yields the equality f (x̄) = x̄.
Proposition In the hypotheses of Theorem 1.11, the set F of fixed points of f
is closed and convex.
proof The first assertion is trivial. Assume then x0 , x1 ∈ F , with x0 6= x1 , and
denote xt = (1 − t)x0 + tx1 , with t ∈ (0, 1). We have
kf (xt ) − x0 k = kf (xt ) − f (x0 )k ≤ kxt − x0 k = tkx1 − x0 k
kf (xt ) − x1 k = kf (xt ) − f (x1 )k ≤ kxt − x1 k = (1 − t)kx1 − x0 k
that imply the equalities
kf (xt ) − x0 k = tkx1 − x0 k
kf (xt ) − x1 k = (1 − t)kx1 − x0 k.
The proof is completed if we show that f (xt ) = (1 − t)x0 + tx1 . This follows from
a general fact about uniform convexity, which is recalled in the next lemma. Lemma Let X be a uniformly convex Banach space, and let α, x, y ∈ X be such
that
kα − xk = tkx − yk,
kα − yk = (1 − t)kx − yk,
for some t ∈ [0, 1]. Then α = (1 − t)x + ty.
proof Without loss of generality, we can assume t ≥ 1/2. We have
k(1 − t)(α − x) − t(α − y)k = k(1 − 2t)(α − x) − t(x − y)k
≥ tkx − yk − (1 − 2t)kα − xk
= 2t(1 − t)kx − yk.
Since the reverse inequality holds as well, and
(1 − t)kα − xk = tkα − yk = t(1 − t)kx − yk
from the uniform convexity of X (but strict convexity would suffice) we get
kα − (1 − t)x − tyk = k(1 − t)(α − x) + t(α − y)k = 0
as claimed.
15
the riesz mean ergodic theorem
The Riesz mean ergodic theorem
If T is a non-expansive linear map of a uniformly convex Banach space, then all the
fixed points of T are recovered by means of a limit procedure.
1.12 Projections Let X be a linear space. A linear operator P : X → X is
called a projection in X if P 2 x = P P x = P x for every x ∈ X. It is easy to check
that P is the identity operator on ran(P ), and the relations ker(P ) = ran(I − P ),
ran(P ) = ker(I − P ) and ker(P ) ∩ ran(P ) = {0} hold. Moreover every element
x ∈ X admits a unique decomposition x = y + z with y ∈ ker(P ) and z ∈ ran(P ).
Proposition If X is a Banach space, then a projection P is continuous if and
only if X = ker(P ) ⊕ ran(P ).
The notation X = A ⊕ B is used to mean that A and B are closed subspaces of
X such that A ∩ B = {0} and A + B = X.
proof If P is continuous, so is I − P . Hence ker(P ) and ran(P ) = ker(I − P )
are closed. Conversely, let xn → x, and P xn → y. Since ran(P ) is closed,
y ∈ ran(P ), and therefore P y = y. But P xn − xn ∈ ker(P ), and ker(P ) is closed.
So we have x − y ∈ ker(P ), which implies P y = P x. From the closed graph
theorem, P is continuous.
1.13 Theorem [F. Riesz] Let X be a uniformly convex Banach space. Let
T : X → X be a linear operator such that
kT xk ≤ kxk,
∀ x ∈ X.
Then for every x ∈ X the limit
x + T x + · · · + T nx
n→∞
n+1
px = lim
exists. Moreover, the operator P : X → X defined by P x = px is a continuous
projection onto the linear space M = {y ∈ X : T y = y}.
proof Fix x ∈ X, and set
C = co({x, T x, T 2 x, T 3 x, . . .}).
C is a closed nonvoid convex set, and from the uniform convexity of X there is
a unique px ∈ C such that
µ = kpx k = inf kzk : z ∈ C .
Select ε > 0. Then,Pfor px ∈ C, there are m ∈ N and nonnegative constants
α0 , α1 , . . . , αm with m
j=0 αj = 1 such that, setting
z=
m
X
j=0
αj T j x
16
1. fixed point theorems
there holds
kpx − zk < ε.
In particular, for every n ∈ N,
z + T z + · · · + T nz ≤ kzk ≤ µ + ε.
n+1
Notice that
z + T z + · · · + T n z = (α0 x + · · · + αm T m x) + (α0 T x + · · · + αm T m+1 x)
+ · · · + (α0 T n x + · · · + αm T m+n x).
Thus, assuming n m, we get
z + T z + · · · + T nz = x + T x + · · · + T nx + r
where
r = (α0 − 1)x + · · · + (α0 + α1 + · · · + αm−1 − 1)T m−1 x
+(1 − α0 )T 1+n x + · · · + (1 − α0 − α1 − · · · − αm−1 )T m+n x.
Therefore
z + T z + · · · + T nz
r
x + T x + · · · + T nx
=
−
.
n+1
n+1
n+1
Since
r 2mkxk
n + 1 ≤ n + 1
upon choosing n enough large such that 2mkxk < ε(n + 1) we have
x + T x + · · · + T nx z + T z + · · · + T nz r ≤
+
n + 1 ≤ µ + 2ε.
n+1
n+1
On the other hand, it must be
x + T x + · · · + T nx ≥ µ.
n+1
Then we conclude that
x + T x + · · · + T nx = µ.
lim
n→∞ n+1
This says that the above is a minimizing sequence in C, and due to the uniform
convexity of X, we gain the convergence
x + T x + · · · + T nx
= px .
n→∞
n+1
lim
17
the brouwer fixed point theorem
We are left to show that the operator P x = px is a continuous projection onto
M. Indeed, it is apparent that if x ∈ M then px = x. In general,
T x + T 2 x + · · · + T n+1 x
T n+1 x − x
= px + lim
= px .
n→∞
n→∞
n+1
n+1
T px = lim
Finally, P 2 x = P P x = P px = px = P x. The continuity is ensured by the
relation kpx k ≤ kxk.
When X is a Hilbert space, P is actually an orthogonal projection. This follows
from the next proposition.
Proposition Let H be a Hilbert space, P = P 2 : H → H a bounded linear
operator with kP k ≤ 1. Then P is an orthogonal projection.
proof Since P is continuous, ran(P ) is closed. Let then E be the orthogonal
projection having range ran(P ). Then
P = E + P (I − E).
Let now x ∈ ran(P )⊥ . For any ε > 0 we have
kP (P x + εx)k ≤ kP x + εxk
which implies that
ε
kxk2 .
2+ε
Hence P x = 0, and the equality P = E holds.
kP xk2 ≤
The role played by uniform convexity in Theorem 1.13 is essential, as the following example shows.
Example Let X = `∞ , and let T ∈ L(X) defined by
T x = (0, x0 , x1 , . . .),
for x = (x0 , x1 , x2 , . . .) ∈ X.
Then T has a unique fixed point, namely, the zero element of X. Nonetheless, if
y = (1, 1, 1, . . .), for every n ∈ N we have
y + T y + · · · + T n y k(1, 2, . . . , n, n + 1, n + 1, . . .)k
=
= 1.
n+1
n+1
The Brouwer fixed point theorem
Let Dn = {x ∈ Rn : kxk ≤ 1}. A subset E ⊂ Dn is called a retract of Dn if there
exists a continuous map r : Dn → E (called retraction) such that r(x) = x for every
x ∈ E.
18
1. fixed point theorems
1.14 Lemma
The set Sn−1 = {x ∈ Rn : kxk = 1} is not a retract of Dn .
The lemma can be easily proved by means of algebraic topology tools. Indeed,
a retraction r induces a homomorphism r∗ : Hn−1 (Dn ) → Hn−1 (Sn−1 ), where Hn−1
denotes the (n−1)-dimensional homology group (see, e.g., [8]). The natural injection
j : Sn−1 → Dn induces in turn a homomorphism j∗ : Hn−1 (Sn−1 ) → Hn−1 (Dn ), and
the composition rj is the identity map on Sn−1 . Hence (rj)∗ = r∗ j∗ is the identity
map on Hn−1 (Sn−1 ). But since Hn−1 (Dn ) = 0, j∗ is the null map. On the other
hand, Hn−1 (Sn−1 ) = Z if n 6= 1, and H0 (S0 ) = Z ⊕ Z, leading to a contradiction.
The analytic proof reported below is less evident, and makes use of exterior
forms. Moreover, it provides a weaker result, namely, it shows that there exist no
retraction of class C 2 from Dn to Sn−1 . This will be however enough for our scopes.
proof Associate to a C 2 function h : Dn → Dn the exterior form
ωh = h1 dh2 ∧ · · · ∧ dhn .
The Stokes theorem (cf. [12], Ch.10) entails
Z
Z
Z
Z
Dh :=
ωh =
dωh =
dh1 ∧ · · · ∧ dhn =
Sn−1
Dn
Dn
det[Jh (x)] dx
Dn
where Jh (x) denotes the (n × n)-Jacobian matrix of h at x. Assume now that
there is a retraction r of class C 2 from Dn to Sn−1 . From the above formula, we
see that Dr is determined only by the values of r on Sn−1 . But r|Sn−1 = i|Sn−1 ,
where i : Dn → Dn is the identity map. Thus Dr = Di = vol (Dn ). On the other
hand krk ≡ 1, and this implies that the vector Jr (x)r(x) is null for every x ∈ Dn .
So 0 is an eigenvalue of Jr (x) for every x ∈ Dn , and therefore det[Jr ] ≡ 0 which
implies Dr = 0.
Remark Another way to show that det[Jr ] ≡ 0 is to observe that the determinant is a null lagrangian (for more details see, e.g., [5], Ch.8)
1.15 Theorem [Brouwer]
has a fixed point x̄ ∈ Dn .
Let f : Dn → Dn be a continuous function. Then f
proof Since we want to rely on the analytic proof, let f : Dn → Dn be of class
C 2 . If f had no fixed point, then
r(x) = t(x)f (x) + (1 − t(x))x
where
t(x) =
kxk2 − hx, f (x)i −
p
(kxk2 − hx, f (x)i)2 + (1 − kxk2 )kx − f (x)k2
kx − f (x)k2
is a retraction of class C 2 from Dn to Sn−1 , against the conclusion of Lemma 1.14.
Graphically, r(x) is the intersection with Sn−1 of the line obtained extending the
the schauder-tihonov fixed point theorem
19
segment connecting f (x) to x. Hence such an f has a fixed point. Finally, let
f : Dn → Dn be continuous. Appealing to the Stone-Weierstrass theorem, we
find a sequence fj : Dn → Dn of functions of class C 2 converging uniformly to f
on Dn . Denote x̄j the fixed point of fj . Then there is x̄ ∈ Dn such that, up to a
subsequence, x̄j → x̄. Therefore,
kf (x̄) − x̄k ≤ kf (x̄) − f (x̄j )k + kf (x̄j ) − fj (x̄j )k + kx̄j − x̄k −→ 0
as j → ∞, which yields f (x̄) = x̄.
Remark An alternative approach to prove Theorem 1.15 makes use of the
concept of topological degree (see, e.g., [2], Ch.1).
The Schauder-Tychonoff fixed point theorem
We first extend Theorem 1.15 to a more general situation.
1.16 Lemma Let K be a nonvoid compact convex subset of a finite dimensional
real Banach space X. Then every continuous function f : K → K has a fixed point
x̄ ∈ K.
proof Since X is homeomorphic to Rn for some n ∈ N, we assume without
loss of generality X = Rn . Also, we can assume K ⊂ Dn . For every x ∈ Dn ,
let p(x) ∈ K be the unique point of minimum norm of the set x − K. Notice
that p(x) = x for every x ∈ K. Moreover, p is continuous on Dn . Indeed, given
xn , x ∈ Dn , with xn → x,
kx − p(x)k ≤ kx − p(xn )k ≤ kx − xn k + inf kxn − kk −→ kx − p(x)k
k∈K
as n → ∞. Thus x − p(xn ) is a minimizing sequence as xn → x in x − K, and
this implies the convergence p(xn ) → p(x). Define now g(x) = f (p(x)). Then
g maps continuously Dn onto K. From Theorem 1.15 there is x̄ ∈ K such that
g(x̄) = x̄ = f (x̄).
As an immediate application, consider Example 1.1. If there is a compact and
convex set K ⊂ Rn such that h(K) ⊂ K, then h has a fixed point x̄ ∈ K.
Other quite direct applications are the Frobenius theorem and the fundamental
theorem of algebra.
Theorem [Frobenius] Let A be a n × n matrix with strictly positive entries.
Then A has a strictly positive eigenvalue.
proof The matrix A can be viewed as a linear transformation from Rn to Rn .
Introduce the compact convex set
n
n
o
X
n
K= x∈R :
xj = 1, xj ≥ 0 for j = 1, . . . , n
j=1
20
1. fixed point theorems
and define f (x) = Ax/kAxk1 (where k · k1 is the euclidean 1-norm). Notice
that if x ∈ K, then all the entries of x are nonnegative and at least one is
strictly positive, hence all the entries of Ax are strictly positive. Then f is a
continuous function mapping K into K, and therefore there exists x̄ ∈ K such
that Ax̄ = kAx̄k1 x̄.
Theorem [Fundamental of algebra] Let p(z) = a0 + a1 z + · · · + an z n be a
complex polynomial of degree n ≥ 1. Then there exits z0 ∈ C such that p(z0 ) = 0.
proof For our purposes, let us identify C with R2 . Suppose without loss of
generality an = 1. Let r = 2 + |a0 | + · · · + |an−1 |. Define now the continuous
function g : C → C as

p(z) i(1−n)ϑ

 z−
e
|z| ≤ 1
r
g(z) =

 z − p(z) z (1−n)
|z| > 1
r
where z = ρeiϑ with ϑ ∈ [0, 2π). Consider now the compact and convex set
C = {z : |z| ≤ r}. In order to apply the Brouwer fixed point theorem we need
to show that g(C) ⊂ C. Indeed, if |z| ≤ 1,
|g(z)| ≤ |z| +
1 + |a0 | + · · · + |an−1 |
|p(z)|
≤1+
≤ 2 ≤ r.
r
r
Conversely, if 1 < |z| ≤ r we have
p(z) z a0 + a1 z + · · · an−1 z n−1 |g(z)| ≤ z − n−1 = z − −
rz
r
rz n−1
|a0 | + · · · + |an−1 |
r−2
≤ r−1+
≤r−1+
≤ r.
r
r
Hence C is invariant for g, and so g has a fixed point z0 ∈ C, which is clearly a
root of p.
1.17 Partition of the unity Suppose V1 , . . . , Vn are open subsets of a locally
compact Hausdorff space X, K ⊂ X is compact, and
K ⊂ V1 ∪ · · · ∪ Vn .
Then for every j = 1, . . . , n there exists ϕj ∈ C(X), 0 ≤ ϕj ≤ 1, supported on Vj
such that
ϕ1 (x) + · · · + ϕn (x) = 1,
∀ x ∈ K.
The collection ϕ1 , . . . , ϕn is said to be a partition of the unity for K subordinate to
the open cover {V1 , . . . , Vn }.
The existence of a partition of the unity is a straightforward consequence of the
Urysohn lemma (see for instance [7]). We are often interested to find partitions of
21
the schauder-tihonov fixed point theorem
the unity for a compact set K ⊂ X whose members are continuous functions defined
on K. Clearly, in this case X need not be locally compact.
1.18 Theorem [Schauder-Tychonoff ] Let X be a locally convex space, K ⊂ X
nonvoid and convex, K0 ⊂ K, K0 compact. Given a continuous map f : K → K0 ,
there exists x̄ ∈ K0 such that f (x̄) = x̄.
proof Denote by B the local base for the topology of X generated by the
separating family of seminorms P on X. Given U ∈ B, from the compactness of
K0 , there exist x1 , . . . , xn ∈ K0 such that
K0 ⊂
n
[
(xj + U ).
j=1
Let ϕ1 , . . . , ϕn ∈ C(K0 ) be a partition of the unity for K0 subordinate to the
open cover {xj + U }, and define
fU (x) =
n
X
ϕj (f (x))xj ,
∀ x ∈ K.
j=1
then
fU (K) ⊂ KU := co({x1 , . . . , xn }) ⊂ K
and Lemma 1.16 yields the existence of xU ∈ KU such that fU (xU ) = xU . Then
xU − f (xU ) = fU (xU ) − f (xU ) =
n
X
ϕj (f (xU ))(xj − f (xU )) ∈ U
(2)
j=1
for ϕj (f (xU )) = 0 whenever xj −f (xU ) 6∈ U . Appealing again to the compactness
of K0 , there exists
\ f (xU ) : U ∈ B, U ⊂ W ⊂ K0 .
(3)
x̄ ∈
W ∈B
Select now p ∈ P and ε > 0, and let
V = x ∈ X : p(x) < ε ∈ B.
Since f is continuous on K, there is W ∈ B, W ⊂ V , such that
f (x) − f (x̄) ∈ V
whenever x − x̄ ∈ 2W , x ∈ K. Moreover, by (3), there exists U ∈ B, U ⊂ W ,
such that
x̄ − f (xU ) ∈ W ⊂ V.
(4)
Collecting (2) and (4) we get
xU − x̄ = xU − f (xU ) + f (xU ) − x̄ ∈ U + W ⊂ W + W = 2W
22
1. fixed point theorems
which yields
f (xU ) − f (x̄) ∈ V.
(5)
Hence (4)-(5) entail
p(x̄ − f (x̄)) ≤ p(x̄ − f (xU )) + p(f (xU ) − f (x̄)) < 2ε.
Being p and ε arbitrary, we conclude that p(x̄ − f (x̄)) = 0 for every p ∈ P, which
implies the equality f (x̄) = x̄.
The following two propositions deal with the existence of zeros of maps on Banach
spaces. For a Banach space X and r > 0, let Br = B X (0, r), and consider a
continuous map g : Br → X, such that g(Br ) is relatively compact.
Proposition Let g(x) 6∈ {λx : λ > 0} for every x ∈ ∂Br . Then there is x0 ∈ Br
such that g(x0 ) = 0.
proof If not so, the function f (x) = rg(x)/kg(x)k is continuous from Br to
Br and f (Br ) is relatively compact. From Theorem 1.18, f admits a fixed point
x̄ ∈ Br . Then g(x̄) = kg(x̄)kx̄/r with kx̄k = r, against the hypotheses.
Proposition Assume that for every x ∈ ∂Br there exists Λx ∈ X ∗ , Λx x = 1,
such that Λx g(x) ≥ 0. Then there is x0 ∈ Br such that g(x0 ) = 0.
proof Define f (x) = −rg(x)/kg(x)k. If g has no zeros, reasoning as above, f
has a fixed point x̄ ∈ Br , and the relation −g(x̄) = kg(x̄)kx̄/r, with kx̄k = r,
holds. Taking Λ ∈ X ∗ such that Λx̄ = 1, we obtain Λg(x̄) = −kg(x̄)k/r < 0.
Contradiction.
Let us see an interesting corollary about surjectivity of maps from Rn to Rn , that
extends a well-known result for matrices to the more general situation of continuous
maps.
Corollary Let f : Rn → Rn be a continuous function satisfying
hf (x), xi
= ∞.
kxk→∞
kxk
lim
Then f (Rn ) = Rn .
proof Fix y0 ∈ Rn , and set g(x) = f (x) − y0 . Then, for r > 0 big enough,
hg(x), x/kxki ≥ 0,
∀ x ∈ ∂Br .
Hence for every x ∈ ∂Br , the functional Λx := h·, x/kxki fulfills the hypotheses
of the last proposition. Therefore there is x0 ∈ Br such that g(x0 ) = 0, that is,
f (x0 ) = y0 .
the schauder-tihonov fixed point theorem
23
Remark These kind of conditions concerning the behavior of g on ∂Br are
known as Leray-Schauder boundary conditions. The above results can be generalized to continuous functions defined on the closure of open subsets of X
with values in X satisfying certain compactness properties (see for instance [2],
pp.204–205, and [6], Ch.5.3).
In the applications it is somehow easier to work with functions defined on the
whole space X, and rather ask more restrictive conditions, such as compactness, on
the maps.
1.19 Definition Let X, Y be Banach spaces, and C ⊂ X. A map f : C → Y is
said to be compact provided it transforms bounded sets into relatively compact sets.
If f ∈ L(X, Y ) it is the same as saying that the image of the unit closed ball under
f is relatively compact. In terms of sequences, f is compact if for every bounded
sequence xn , the sequence f (xn ) has a convergent subsequence.
1.20 Theorem [Schaefer] Let X be a Banach space, f : X → X continuous
and compact. Assume further that the set
F = x ∈ X : x = λf (x) for some λ ∈ [0, 1]
is bounded. Then f has a fixed point.
Remark Theorem 1.20 holds if we can prove a priori estimates on the set of
all the possible fixed point of λf . This technique is typical in partial differential
equations, where one proves estimates on a possible solution to a certain equation, and then, in force of those estimates, concludes that such solution actually
exists.
proof Let r > supx∈F kxk, and define the map

if kf (x)k ≤ 2r
 f (x)
g(x) =
2rf (x)

if kf (x)k > 2r.
kf (x)k
Then g : B X (0, 2r) → B X (0, 2r) is continuous and compact. Theorem 1.18 yields
the existence of x0 ∈ B X (0, 2r) for which g(x0 ) = x0 . Notice that kf (x0 )k ≤ 2r,
for otherwise
2r
x0 = λ0 f (x0 )
with
λ0 =
<1
kf (x0 )k
which forces kx0 k = 2r, against the fact that x0 ∈ F . Hence we get that
g(x0 ) = f (x0 ) = x0 .
1.21 Theorem [Krasnoselskii] Let X be a Banach space, C ⊂ X nonvoid,
closed and convex. Let f, g : C → X be such that
(a) f (x1 ) + g(x2 ) ∈ C,
∀ x1 , x2 ∈ C;
24
1. fixed point theorems
(b) f is continuous and compact;
(c) g is a contraction (from C into X).
Then there is x̄ ∈ C such that f (x̄) + g(x̄) = x̄.
proof Notice first that I−g maps homeomorphically C onto (I−g)(C). Indeed,
I − g is continuous and
k(I − g)(x1 ) − (I − g)(x2 )k ≥ kx1 − x2 k − kg(x1 ) − g(x2 )k ≥ (1 − λ)kx1 − x2 k
(λ < 1 is the Lipschitz constant of g) so (I − g)−1 is continuous. For any y ∈ C,
the map x 7→ f (y) + g(x) is a contraction on C, hence by Theorem 1.3 there is a
unique z = z(y) ∈ C such that z = f (y) + g(z). Thus z = (I − g)−1 (f (y)) ∈ C.
On the other hand, the map (I−g)−1 ◦f is continuous and compact from C to C,
being the composition of a continuous map with a continuous and compact map.
Then Theorem 1.18 entails the existence of x̄ ∈ C such that (I − g)−1 (f (x̄)) = x̄,
that is, f (x̄) + g(x̄) = x̄.
In general, it is not possible to extend Theorem 1.18 to noncompact settings.
This fact was already envisaged in our previous discussion about non-expansive
maps. Let us recall another famous example in Hilbert spaces.
Example [Kakutani] Consider the Hilbert space `2 . For a fixed ε ∈ (0, 1], let
fε : B `2 (0, 1) → B `2 (0, 1) be given by
fε (x) = (ε(1 − kxk), x0 , x1 , . . .),
for x = (x0 , x1 , x2 , . . .) ∈ `2 .
Then fε has no fixed points in B `2 (0, 1), but it is Lipschitz continuous with
Lipschitz constant slightly greater than 1. Indeed,
√
kfε (x) − fε (y)k ≤ 1 + ε2 kx − yk
for all x, y ∈ B `2 (0, 1).
It is therefore a natural question to ask when a continuous map f : C → C,
where C is a closed, bounded and convex subset of a Banach space X, admits fixed
points. We know from Theorem 1.18 that if X is finite-dimensional the answer is
positive, since finite-dimensional Banach spaces have the Heine-Borel property. The
analogous result in infinite-dimensional Banach spaces turns out to be false, without
a compactness assumption.
We first report a characterization of noncompact closed bounded sets.
Lemma Let X be a Banach space, C ⊂ X a closed, bounded noncompact set.
Then there are ε > 0 and a sequence xn of elements of C such that
dist xn+1 , span({x0 , . . . , xn }) ≥ ε.
(6)
the schauder-tihonov fixed point theorem
25
proof Clearly, X has to be infinite-dimensional, otherwise no such C exists.
We first show that there is ε > 0 such that, for any finite set F ⊂ X,
C \ [span(F ) + BX (0, ε)] 6= ∅.
If not, for every ε > 0 we find a finite set F ⊂ X such that C ⊂ span(F ) +
BX (0, ε). Since C is bounded, C ⊂ BX (0, r) for some r > 0. Therefore,
C ⊂ [span(F ) + BX (0, ε)] ∩ BX (0, r) ⊂ [span(F ) ∩ BX (0, r + ε)] + BX (0, ε).
But span(F ) ∩ BX (0, r + ε) is totally bounded, and so it admits a finite cover
of balls of radius ε, which in turn implies that C admits a finite cover of balls
of radius 2ε, that is, C is totally bounded (hence compact), contradicting the
hypotheses.
To construct the required sequence xn , we proceed inductively. First we select
an arbitrary x0 ∈ C. If we are given x0 , . . . , xn+1 satisfying (6), we choose
xn+2 ∈ C \ [span({x0 , . . . , xn+1 }) + BX (0, ε)].
1.22 Theorem [Klee] Let X be an infinite-dimensional Banach space, C ⊂ X
a closed, bounded, convex noncompact set. Then there exists a continuous map
f : C → C which is fixed point free.
proof Let xn be a sequence in C satisfying (6). Without loss of generality, we
may assume that 0 ∈ C and kx0 k ≥ ε. We construct a piecewise linear curve
joining the points x0 , x1 , x2 , . . ., by setting
Γ=
∞
[
[xn , xn+1 ],
where [xn , xn+1 ] = co({xn , xn+1 }).
n=0
Then Γ ⊂ C is closed and can be parameterized by a function γ : [0, ∞) → Γ,
given by
γ(t) = (1 − s)xn + sxn+1
where n = [t] and s = t − n. By means of (6), γ is one-to-one, onto and open,
since for every open set O ⊂ [0, ∞), γ(O) contains the intersection of Γ with an
open ball of X of radius ν, for some ν < ε. Hence its inverse γ −1 : Γ → [0, ∞)
is a continuous function. Applying a slightly modified version of the Tietze
extension theorem (see, e.g., [7]), we can extend γ −1 to a continuous function
g : C → [0, ∞). Now a fixed point free continuous map f : C → C can be
defined as
f (x) = γ(g(x) + 1).
Indeed, if f (x̄) = x̄, then x̄ ∈ Γ. Hence γ(γ −1 (x̄) + 1) = γ(γ −1 (x̄)), against the
injectivity of γ.
Remark Notice however that Klee’s map f : C → C cannot be uniformly continuous, since the restriction of g on Γ is easily seen to be uniformly continuous,
whereas g ◦ f maps the bounded convex set C onto [1, ∞).
26
1. fixed point theorems
In particular, Theorem 1.22 says that in an infinite-dimensional Banach space X
there is a continuous map f : B X (0, 1) → B X (0, 1) without fixed points. This allow
us to provide another interesting characterization of infinite-dimensional Banach
spaces.
Corollary Let X be an infinite-dimensional Banach space. Then the closed unit
sphere is a retract of the closed unit ball.
proof Let f : B X (0, 1) → B X (0, 1) be a
Extend the map to the doubled ball B X (0, 2)
f (x)
f1 (x) =
(2 − kxk)f (x/kxk)
continuous fixed point free map.
by defining
if kxk ≤ 1
if 1 < kxk ≤ 2
and consider the new map f2 : B X (0, 1) → B X (0, 1) as
1
f2 (x) = f1 (2x).
2
Observe that f2 is fixed point free, and for all x ∈ ∂B X (0, 1) we have f2 (x) = 0.
Then the function r : B X (0, 1) → ∂B X (0, 1) given by
r(x) =
x − f2 (x)
kx − f2 (x)k
is the desired retraction.
The Markov-Kakutani theorem
The following theorem is concerned with common fixed points of a family of linear
maps.
1.23 Theorem [Markov-Kakutani] Let X be a locally convex space, and let
K ⊂ X be a nonvoid, compact and convex set. Assume G is a family of bounded
linear operators from X into X such that
(a) G is abelian, that is, T S = ST for every T, S ∈ G;
(b) T K ⊂ K for every T ∈ G.
Then there exists x̄ ∈ K such that T x̄ = x̄ for every T ∈ G.
proof For any T ∈ G and any n ∈ N, define the operator
Tn =
I + T + ··· + Tn
.
n+1
Notice that (b) and the convexity of K imply that
Tn K ⊂ K.
27
the markov-kakutani theorem
Given T (1) , . . . , T (k) ∈ G and n1 , . . . , nk ∈ N, it follows that
k
\
Tn(j)
K 6= ∅.
j
j=1
Indeed, from (a), for any T, S ∈ G and any n, m ∈ N,
Tn K ∩ Sm K ⊃ Tn Sm K = Sm Tn K 6= ∅.
Invoking the compactness of K, the finite intersection property yields
\
F =
Tn K 6= ∅.
T ∈G, n∈N
We claim that every x ∈ F is a fixed point for all T ∈ G. Let then x̄ ∈ F ,
and select any T ∈ G. Then for every n ∈ N there is y = y(n) ∈ K such that
x̄ = Tn y. Hence
T x̄ − x̄ =
=
y + T y + · · · + T ny
T y + T 2 y + · · · + T n+1 y
−
n+1
n+1
T n+1 y − y
1
∈
K −K .
n+1
n+1
Thus
T x̄ − x̄ ∈
1
K −K .
n+1
n∈N
\
The set K −K is clearly compact, being the image of K ×K under the continuous
map Φ(x1 , x2 ) = x1 − x2 . Let then p be a seminorm on X. Then, for any ε > 0,
there is n ∈ N such that
1
K − K ⊂ x ∈ X : p(x) < ε .
n+1
We conclude that p(T x̄ − x̄) = 0 for every seminorm on X, which entails the
equality T x̄ = x̄.
Remark With no changes in the proof, the results holds more generally if G is
an abelian family of continuous affine maps from K to K. A map f : C → C, C
convex, is said to be affine if for all x1 , x2 ∈ C and t ∈ [0, 1]
f (tx1 + (1 − t)x2 ) = tf (x1 ) + (1 − t)f (x2 ).
28
1. fixed point theorems
The Kakutani-Ky Fan theorem
Along the section, let X be locally convex space.
Definition Let C ⊂ X be a convex set. A function f : C → (−∞, ∞] is said
to be convex if
f (λx1 + (1 − λ)x2 ) ≤ λf (x1 ) + (1 − λ)f (x2 )
for all ∀ x1 , x2 ∈ X and every λ ∈ [0, 1]. A function g : C → [−∞, ∞) is said to
be concave if −g is convex (it is understood that −(∞) = −∞).
We now recall the well-known definition of lower and upper semicontinuous real
functions.
Definition Let Y be a topological space. A function f : Y → (−∞, ∞] is said
to be lower semicontinuous if f −1 ((α, ∞]) is open for every α ∈ R. Similarly,
a function g : Y → [−∞, ∞) is said to be upper semicontinuous if −g is lower
semicontinuous.
It is an immediate consequence of the definition that the supremum of any collection of lower semicontinuous functions is lower semicontinuous. Moreover, if f is
lower semicontinuous and Y is compact, then f attains its minimum on Y . Indeed, if
it is not so, denoting m = inf y∈Y f (y) ∈ [−∞, ∞), the sets f −1 ((α, ∞]) with α > m
form an open cover of Y that admits no finite subcovers.
The next result is the famous Ky Fan inequality.
1.24 Theorem [Ky Fan] Let K ⊂ X be a nonvoid, compact and convex. Let
Φ : K × K → R be map such that
(a) Φ(·, y) is lower semicontinuous ∀ y ∈ K;
(b) Φ(x, ·) is concave
∀ x ∈ K.
Then there exists x0 ∈ K such that
sup Φ(x0 , y) ≤ sup Φ(y, y).
y∈K
y∈K
proof Fix ε > 0. In correspondence with every x ∈ K there are yx ∈ K and
an open neighborhood Ux of x such that
Φ(z, yx ) > sup Φ(x, y) − ε,
∀ z ∈ Ux ∩ K.
y∈K
Being K compact, for some x1 , . . . , xn ∈ K there holds
K ⊂ Ux1 ∪ · · · ∪ Uxn .
29
the kakutani-ky fan theorem
Let ϕ1 , . . . , ϕn ∈ C(K) be a partition of the unity for K subordinate to the open
cover {Uxj }, and define
f (x) =
n
X
∀ x ∈ K.
ϕj (x)yxj ,
j=1
The map f is clearly continuous, and
f (co({yx1 , . . . , yxn })) ⊂ co({yx1 , . . . , yxn }).
Hence by Lemma 1.16 f admits a fixed point x̄ ∈ K. Therefore,
sup Φ(y, y) ≥ Φ(x̄, x̄) ≥
y∈K
≥
n
X
n
X
ϕj (x̄)Φ(x̄, yxj )
j=1
ϕj (x̄) sup Φ(xj , y) − ε
j=1
y∈K
≥ inf sup Φ(x, y) − ε
x∈K y∈K
= sup Φ(x0 , y) − ε
y∈K
for some x0 ∈ K. Letting ε → 0 we get the desired conclusion.
The aim of this section is to consider a fixed point theorem for maps carrying
points into sets. Let K ⊂ X, and consider a map f : K → 2K := {Y : Y ⊂ K}. A
fixed point for f is a point x ∈ K such that x ∈ f (x).
Definition The map f is upper semicontinuous if for every x ∈ K and every
open set U ⊃ f (x), there exists a neighborhood V of x such that if y ∈ V then
f (y) ⊂ U .
Upper semicontinuous point-closed maps from a compact set K with values in
2 can be characterized in terms of graphs.
K
Proposition Let K ⊂ X be compact, and let f : K → 2K be such that f (x) is
closed for every x ∈ K. Then f is upper semicontinuous if and only if the set
G(f ) = (x, y) ∈ K × K : y ∈ f (x)
is closed in K × K.
proof Let f be upper semicontinuous, and (x0 , y0 ) ∈ (K × K) \ G(f ). Then
y0 6∈ f (x0 ). Since K is compact, we find two disjoint open sets U1 , U2 such that
y0 ∈ U1 and f (x0 ) ⊂ U2 . By the upper semicontinuity of f , there is an open set
V 3 x0 such that f (x) ⊂ U2 for all x ∈ V . Thus the neighborhood V × U1 of
(x0 , y0 ) does not intersect G(f ), which is henceforth closed.
Conversely, let G(f ) be closed. Let x be an arbitrary point of K, and U be an
30
1. fixed point theorems
arbitrary open set containing f (x). If f is not upper semicontinuous at x, for
every neighborhood V 3 x there is y ∈ V such that f (y) 6⊂ U . So we have
G(f ) ∩ (V × U C ) 6= ∅.
But G(f ) and V × U C are both compact, being closed in K × K, and since the
finite intersection property holds, we conclude that there is (x0 , y0 ) ∈ K × K
such that
\
(x0 , y0 ) ∈
G(f ) ∩ (V × U C ).
V 3x
This implies that x0 = x, and since y0 ∈ f (x0 ) = f (x), it follows that y0 ∈ U .
Contradiction.
1.25 Theorem [Kakutani-Ky Fan] Let K be a nonvoid, compact and convex
subset of a locally convex space X. Let f : K → 2K be upper semicontinuous, such
that f (x) is nonvoid, convex and closed for every x ∈ K. Then f has a fixed point
x̄ ∈ K.
proof If we assume the theorem false, from the Hahn-Banach theorem for every
x ∈ K there are αx ∈ R and Λx ∈ X ∗ such that
sup Re Λx z < αx < Re Λx x.
z∈f (x)
The set
Ux = z ∈ X : Re Λx z < αx
is open, and contains f (x). Hence there is an open neighborhood Vx of x such
that f (y) ⊂ Ux whenever y ∈ Vx ∩ K. We conclude that there exists an open
neighborhood Wx ⊂ Vx of x such that
Re Λx y > αx ,
∀ y ∈ Wx ∩ K
and
Re Λx z < αx , ,
∀ y ∈ Wx ∩ K, ∀ z ∈ f (y).
From the compactness of K, there exist x1 , . . . , xn ∈ K such that
K ⊂ W x1 ∪ · · · ∪ W xn .
Let ϕ1 , . . . , ϕn ∈ C(K) be a partition of the unity for K subordinate to the open
cover {Wxj }, and define Φ : K × K → R as
Φ(x, z) =
n
X
ϕj (x)Re Λxj x −
j=1
n
X
ϕj (x)Re Λxj z.
j=1
Clearly, Φ fulfills the hypotheses Theorem 1.24. Hence there is x0 ∈ K such that
sup Φ(x0 , z) ≤ sup Φ(z, z) = 0.
z∈K
z∈K
31
notes on chapter 1
Therefore, for z ∈ f (x0 ),
n
X
ϕj (x0 )αxj <
j=1
n
X
ϕj (x0 )Re Λxj x0 ≤
j=1
n
X
ϕj (x0 )Re Λxj z <
j=1
n
X
ϕj (x0 )αxj
j=1
which is a contradiction. Notice that in the above sums we have nonzero contributions only for those j for which x0 ∈ Wxj .
Notes on Chapter 1
The construction of the approximating sequence xn of Theorem 1.3 is known as Picard’s method [J. de
Math. 6, 145–210 (1890)]. The abstract proof is due to Banach [Fund. Math. 3, 133–181 (1922)] and
Caccioppoli [Rend. Accad. Naz. Lincei 11, 794–799 (1930)]. Concerning 1.6, Boyd-Wong’s theorem is in
Proc. Amer. Math. Soc. 20, 458–464 (1969). Caristi’s theorem has been proved more or less explicitly
by many authors. We refer to Caristi [Trans. Amer. Math. Soc. 215, 241–251 (1976)], where some
applications of the theorem are also provided. Caristi’s original proof involves an intricate transfinite
induction argument, whereas the direct proof reported here is taken from [2], p.198. Čirič’s theorem is in
Proc. Amer. Math. Soc. 45, 267–273 (1974). Bessaga’s theorem in 1.8 appears in Colloq. Math 7, 41–43
(1969). An alternative proof can be found in [2], pp.191–192. The argument we presented, indeed much
simpler, is due to Peirone [personal communication] . There is an interesting related result for compact
metric spaces due to Janos [Proc. Amer. Math. Soc. 18, 287–289 (1967)], where an equivalent metric that
makes f a contraction is constructed.
Theorem 1.9 and Theorem 1.10 are due to Nadler [Pacif. J. Math. 27, 579–585 (1968)] and Fraser
and Nadler [Pacif. J. Math. 31, 659–667 (1969)]. Uniform convergence in Theorem 1.9 cannot in general
be replaced by pointwise convergence. For instance, in every infinite-dimensional separable or reflexive
Banach space there is a pointwise convergent sequence of contractions whose sequence of fixed points has
no convergent subsequences (see, e.g. [6], Ch.7).
Theorem 1.11 has been proved by Kirk [Amer. Math. Monthly 72, 1004–1006 (1965)], for the more
general case when X is a reflexive Banach spaces and C has normal structure, and by Browder [Proc. Nat.
Acad. Sci. USA 54, 1041–1044 (1965)]. The proof for Hilbert spaces reported here is contained in a paper
of Browder and Petryshyn [J. Math. Anal. Appl. 20, 197–228 (1967)].
Theorem 1.13 is in J. London Math. Soc. 13, 274–278 (1938). See also [10].
Brouwer’s Theorem 1.15 is in Math. Ann. 71, 97–115 (1910) (see also [3]). A different proof that
requires only elementary notions of continuity and compactness has been given by Kuga [SIAM J. Math.
Anal. 5, 893–897 (1974)].
Schauder’s original proof of Theorem 1.18 for Banach spaces can be found in Studia Math. 2, 171–180
(1930), whereas Tychonoff’s generalization to locally convex spaces is in Math. Ann. 111, 767–776 (1935).
Concerning Leray-Schauder boundary conditions, we refer the reader to the work of Leray and Schauder
[Ann. Sci. École Norm. Sup. 51, 45–78 (1934)]. Schaefer’s Theorem 1.20 is taken from Math. Ann. 129,
415–416 (1955). Theorem 1.21 is in [Amer. Math. Soc. Transl. 10, 345–409 (1958)]. Theorem 1.22 has
been proved by Klee [Trans. Amer. Math. Soc. 78, 30–45 (1955)]. The result actually holds for metrizable
locally convex spaces. In a subsequent paper [Proc. Amer. Math. Soc. 93, 633–639 (1985)], Lin and
Sternfeld have shown that the map f of Theorem 1.22 can be taken to be Lipschitz continuous, with
Lipschitz constant greater than (but arbitrarily close to) 1.
32
1. fixed point theorems
Theorem 1.23 has been proved with different techniques by Markov [Dokl. Akad. Nauk. SSSR 10, 311–
314 (1936)], and Kakutani [Proc. Imp. Acad. Tokyo 14, 242–245 (1938)]. The same kind of result does not
hold in general for commuting nonlinear continuous maps. Indeed Boyce [Trans. Amer. Math. Soc. 137,
77–92 (1969)] and Huneke [Trans. Amer. Math. Soc. 139, 371–381 (1969)] have provided examples of two
commuting nonlinear continuous maps of [0, 1] onto itself without a common fixed point.
Theorem 1.25 for finite dimensional spaces is due to Kakutani [Duke Math. J. 8, 457–459 (1941)]. The
generalization to infinite dimensional spaces has been given by Ky Fan, by means of Theorem 1.24 [Proc.
Nat. Acad. Sci. USA 88, 121–126 (1952)].
2. SOME APPLICATIONS OF FIXED POINT
THEOREMS
The implicit function theorem
2.1 Fréchet differentiability Let X, Y be (real or complex) Banach spaces,
U ⊂ X, U open, x0 ∈ U , and f : U → Y .
Definition f is Fréchet differentiable at x0 is there exists T ∈ L(X, Y ) and
σ : X → Y , with
kσ(x)kY
−→ 0
kxkX
uniformly as
kxkX → 0
such that
f (x) − f (x0 ) = T (x − x0 ) + σ(x − x0 ),
∀ x ∈ U.
The operator T is called the Fréchet derivative of f at x0 , and is denoted by
f 0 (x0 ). The function f is said to be Fréchet differentiable in U if it is Fréchet
differentiable at every x0 ∈ U .
It is straightforward to verify the Fréchet derivative at one point, if it exists, is
unique.
2.2 Lemma Let X, Y be Banach spaces, let f : BX (0, r) → Y be Fréchet differentiable and kf 0 (x)kL(X,Y ) ≤ λ for every x ∈ BX (0, r) and some λ ≥ 0. Then f is
Lipschitz continuous with Lipschitz constant less than or equal to λ.
proof Let x1 , x2 ∈ BX (0, r). By the Hahn-Banach theorem, there is Λ ∈ Y ∗
of unit norm such that
kf (x1 ) − f (x2 )kY = |Λ(f (x1 ) − f (x2 ))|.
For t ∈ [0, 1] set
Φ(t) = Λf (tx1 + (1 − t)x2 ).
Applying the Lagrange mean value theorem to Φ, there is τ ∈ (0, 1) such that
|Λf (x1 ) − Λf (x2 )| = |Φ(1) − Φ(0)| ≤ |Φ0 (τ )| = |Λf 0 (τ x1 + (1 − τ )x2 )(x1 − x2 )|
33
34
2. some applications of fixed point theorems
(the chain holds as in the classical case; if X is real equality holds). Hence
kf (x1 ) − f (x2 )kY ≤ kf 0 (τ x1 + (1 − τ )x2 )(x1 − x2 )kY ≤ λkx1 − x2 kX
as claimed.
2.3 Given two Banach spaces X and Y , the vector space X × Y is a Banach space
with any of the (equivalent) euclidean norms
k(x, y)kp =
kxkpX
+
kykpY
1/p
,
k(x, y)k∞ = max kxkX , kykY
(p ≥ 1).
In the sequel, we will always use the ∞-norm, so that
BX×Y ((x0 , y0 ), r) = BX (x0 , r) × BY (y0 , r).
For X, Y, Z Banach spaces, given T ∈ L(X, Z) and S ∈ L(Y, Z), the operator
R : X × Y → Z defined by
R(x, y) = T x + Sy
belongs to L(X × Y, Z). Conversely, any R ∈ L(X × Y, Z), has the above representation with T x = R(x, 0) and Sy = R(0, y). It is then immediate to see that
L(X, Z) × L(Y, Z) and L(X × Y, Z) are isomorphic Banach spaces. Given then
f : U ⊂ X × Y → Z, U open, f Fréchet differentiable at u0 = (x0 , y0 ) ∈ U ,
one easily checks that the partial derivatives Dx f (u0 ) and Dy f (u0 ) exist (that is,
the Fréchet derivatives of f (·, y0 ) : X → Z in x0 and of f (x0 , ·) : Y → Z in y0 ,
respectively), and
f 0 (u0 )(x, y) = Dx f (u0 )(x) + Dy f (u0 )(y).
2.4 Theorem [Dini] Let X, Y, Z be Banach spaces, U ⊂ X × Y be an open set,
u0 = (x0 , y0 ) ∈ U , and F : U → Z. Assume that
(a) F is continuous and F (u0 ) = 0;
(b) Dy F (u) exists for every u = (x, y) ∈ U ;
(c) Dy F is continuous at u0 and Dy F (u0 ) is invertible.
Then there exists α, β > 0 for which B X (x0 , α) × B Y (y0 , β) ⊂ U and a unique
continuous function f : B X (x0 , α) → B Y (y0 , β) such that the relation
F (x, y) = 0
⇐⇒
holds for all (x, y) ∈ B X (x0 , α) × B Y (y0 , β).
y = f (x)
35
the implicit function theorem
proof Without loss of generality, we assume x0 = 0 and y0 = 0. Define
Φ(x, y) = y − [Dy F (0, 0)]−1 F (x, y),
(x, y) ∈ U.
By (a) Φ is continuous from U into Y . Since
[Dy Φ(0, 0)]−1 Dy Φ(0, 0) − Dy Φ(x, y) ,
by (c) there is γ > 0 small enough such that
1
kDy Φ(x, y)kL(Y ) ≤ ,
2
∀ (x, y) ∈ BX (0, γ) × BY (0, γ) ⊂ U.
Thus Lemma 2.2 and the continuity of Φ entail the inequality
1
kΦ(x, y1 ) − Φ(x, y2 )kY ≤ ky1 − y2 kY ,
2
kxkX , ky1 kY , ky2 kY ≤ β < γ.
Using now (a), we find 0 < α < β such that
kΦ(x, 0)kY ≤
β
,
2
kxkX ≤ α.
Then, for kxkX ≤ α and kykY ≤ β,
kΦ(x, y)kY ≤ kΦ(x, 0)kY + kΦ(x, y) − Φ(x, 0)kY ≤
1
β + kykY ≤ β.
2
Therefore the continuous map Φ : B X (0, α) × B Y (0, β) → B Y (0, β) is a contraction on B Y (0, β) uniformly in B X (0, α). From Corollary 1.4, there exists a unique
continuous function f : B X (0, α) → B Y (0, β) such that Φ(x, f (x)) = f (x), that
is, F (x, f (x)) = 0.
Obviously, the thesis still holds replacing in the hypotheses closed balls with
open balls.
Corollary Let the hypotheses of Theorem 2.4 hold. If in addition F is Fréchet
differentiable at u0 = (x0 , y0 ), then f is Fréchet differentiable at x0 , and
f 0 (x0 ) = −[Dy F (u0 )]−1 Dx F (u0 ).
proof Applying the definition of Fréchet differentiability to F (x, f (x)) at the
point (x0 , f (x0 )), we get
0 = Dx F (u0 )(x − x0 ) + Dy F (u0 )(f (x) − f (x0 )) + σ(x − x0 , f (x) − f (x0 )).
Notice that the above relation implies that f is locally Lipschitz at x0 . Hence
kσ(x − x0 , f (x) − f (x0 ))kZ
−→ 0
kx − x0 kX
which yields the thesis.
uniformly as
kx − x0 kX → 0
36
2. some applications of fixed point theorems
A consequence of Theorem 2.4 is the inverse function theorem.
Theorem Let X, Y be Banach spaces, V ⊂ Y open, y0 ∈ V . Let g : V → X be
Fréchet differentiable in a neighborhood of y0 , g(y0 ) = x0 , g 0 continuous at y0 ,
and g 0 (y0 ) invertible. Then there are α, β > 0 and a unique continuous function
f : BX (x0 , α) → BY (y0 , β) such that x = g(f (x)) for every x ∈ BX (x0 , α).
Moreover, f is Fréchet differentiable at x0 and f 0 (x0 ) = g(y0 )−1 .
proof Apply Theorem 2.4 and the subsequent corollary to F (x, y) = g(y) − x,
keeping in mind the considerations made in 2.3.
Theorem 2.4 can also be exploited to provide an alternative proof to the wellknown fact that the set of invertible bounded linear operators between Banach spaces
is open.
Theorem Let X, Y be Banach spaces, and let Lreg (X, Y ) ⊂ L(X, Y ) be the set
of invertible bounded linear operators from X onto Y . Then Lreg (X, Y ) is open
in L(X, Y ). Moreover, the map T 7→ T −1 is continuous.
proof Let F : L(X, Y ) × L(Y, X) → L(X) defined by F (T, S) = IY − T S.
Let T0 ∈ Lreg (X, Y ), and set S0 = T0−1 . Notice that DS (T, S)(R) = −T R. In
particular, DS (T0 , S0 )(R) = −T0 R. Then the hypotheses of Theorem 2.4 are
satisfied; therefore there is a continuous function f : BL(X,Y ) (T0 , α) → L(Y, X)
such that IY − T f (T ) = 0, that is T f (T ) = IY . Analogously, we can find a
continuous function f1 : BL(X,Y ) (T0 , α) → L(Y, X) (perhaps for a smaller α)
such that f1 (T )T = IX . It is straightforward to verify that f ≡ f1 , that is,
f (T ) = T −1 for all T ∈ BL(X,Y ) (T0 , α).
2.5 Location of zeros Let X, Y be Banach spaces, and f : BX (x0 , r) → Y be
a Fréchet differentiable map. In order to find a zero for f , the idea is to apply an
iterative method constructing a sequence xn (starting from x0 ) so that xn+1 is the
zero of the tangent of f at xn . Assuming that f 0 (x)−1 ∈ L(Y, X) on BX (x0 , r), one
has
xn+1 = xn − f 0 (xn )−1 f (xn )
(1)
provided xn ∈ BX (x0 , r) for every n. This procedure is known as the Newton method.
However, for practical purposes, it might be complicated to invert f 0 at each step.
So one can try the modification
xn+1 = xn − f 0 (x0 )−1 f (xn ).
(2)
Clearly, using (2) in place of (1), a lower convergence rate is to be expected.
The following result is based on (2).
Theorem Let X, Y be Banach spaces, and f : BX (x0 , r) → Y be a Fréchet
differentiable map. Assume that, for some λ > 0,
(a) f 0 (x0 ) is invertible;
37
the implicit function theorem
(b) kf 0 (x) − f 0 (x0 )kL(X,Y ) ≤ λkx − x0 kX ,
(c)
(d)
∀ x ∈ BX (x0 , r);
0
µ := 4λkf (x0 )−1 k2L(Y,X) kf (x0 )kY ≤ 1;
s := 2kf 0 (x0 )−1 kL(Y,X) kf (x0 )kY < r.
Then there exists a unique x̄ ∈ B X (x0 , s) such that f (x̄) = 0.
proof Define Φ : B X (x0 , s) → X as Φ(x) = x − f 0 (x0 )−1 f (x). Then
µ
.
2
Hence Φ is Lipschitz, with Lipschitz constant less than or equal to µ/2 ≤ 1/2.
Moreover,
s
kΦ(x0 ) − x0 kX ≤ kf 0 (x0 )−1 kL(Y,X) kf (x0 )kY =
2
which in turn gives
µ
s
kΦ(x) − x0 kX ≤ kΦ(x) − Φ(x0 )kX + kΦ(x0 ) − x0 kX ≤ kx − x0 kX + ≤ s.
2
2
kΦ0 (x)kL(X) ≤ kf 0 (x0 )−1 kL(Y,X) kf 0 (x0 ) − f 0 (x)kL(X,Y ) ≤ λskf 0 (x0 )−1 kL(Y,X) =
Hence Φ is a contraction on B X (x0 , s). From Theorem 1.3 there exists a unique
x̄ ∈ B X (x0 , s) such that Φ(x̄) = x̄, which implies f (x̄) = 0.
Concerning the convergence speed of xn to x̄, by virtue of the remark after
Theorem 1.3, we get
sµn
.
kxn − x̄kX ≤
(2 − µ)2n
Also, since
xn+1 − x̄ = f 0 (x0 )−1 (f 0 (x0 ) − f 0 (xn ))(xn − x̄) + o(kxn − z̄kX )
it follows that
kxn+1 − x̄kX =
µ
kxn − x̄kX + o(kxn − z̄kX ).
2
Hence
kxn+1 − x̄kX ≤ ckxn − x̄kX
for some c ∈ (0, 1). for all large n. This is usually referred to as linear convergence
of the method.
Remark If we take µ < 2, and we assume that f 0 is Lipschitz continuous on
BX (x0 , r) with Lipschitz constant λ, we can still obtain the thesis with an entirely
different proof (see, e.g., [2], pp.157–159), exploiting the iterative method (1).
In this case we get the much better estimates
s µ 2n −1
kxn − x̄kX ≤ n
2 2
and
kxn+1 − x̄kX ≤ ckxn − x̄k2X
for some c > 0 (i.e., we have quadratic convergence).
38
2. some applications of fixed point theorems
Ordinary differential equations in Banach spaces
2.6 The Riemann integral Let X be a Banach space, I = [α, β] ⊂ R. The
notion of Riemann integral and the related properties can be extended with no
differences from the case of real-valued functions to X-valued functions on I. In
particular, if f ∈ C(I, X) then f is Riemann integrable on I,
Z β
Z β
kf (t)kX dt
f (t) dt
≤
α
and
d
dt
Z
X
α
t
f (y) dy = f (t),
∀ t ∈ I.
α
Recall that a function h : I → X is differentiable at t0 ∈ I if the limit
lim
t→t0
h(t) − h(t0 )
t − t0
exists in X. This limit is the derivative of h at t0 , and is denoted by h0 (t0 ) or dtd h(t0 ).
If t0 ∈ (α, β) we recover the definition of Fréchet differentiability.
It is easy to see that if h0 (t) = 0 for all t ∈ [α, β], then h(t) is constant on
[α, β]. Indeed, for every Λ ∈ X ∗ , we have (Λ ◦ h)0 (t) = Λ0 h(t) = 0, that implies
Λ(h(t) − h(α)) = 0, and from the Hahn-Banach theorem there is Λ ∈ X ∗ such that
Λ(h(t) − h(α)) = kh(t) − h(α)k.
2.7 The Cauchy problem Let X be a Banach space, U ⊂ R × X, U open,
u0 = (t0 , x0 ) ∈ U , f : U → X continuous. The problem is to find a closed interval
I, with t0 belonging to the interior of I and a differentiable function x : I → X such
that
0
x (t) = f (t, x(t)),
t∈I
(3)
x(t0 ) = x0 .
It is apparent that such x is automatically of class C 1 on I. Also, it is readily seen
that (3) is equivalent to the integral equation
Z t
f (y, x(y)) dy,
t ∈ I.
(4)
x(t) = x0 +
t0
Namely, x is a solution to (3) if and only if it is a solution to (4).
2.8 Theorem [Local solution]
Assume the following hypotheses:
(a) f is continuous;
(b) The inequality
kf (t, x1 ) − f (t, x2 )kX ≤ k(t)kx1 − x2 kX ,
holds for some k(t) ∈ [0, ∞];
∀ (t, x1 ), (t, x2 ) ∈ U
ordinary differential equations in banach spaces
39
(c) k ∈ L1 ((t0 − a, t0 + a)) for some a > 0;
(d) There exist m ≥ 0 and B R ×X (u0 , s) ⊂ U such that
kf (t, x)kX ≤ m,
∀ (t, x) ∈ B R ×X (u0 , s).
Then there is τ0 > 0 such that, for any τ < τ0 , there exists a unique solution
x ∈ C 1 (Iτ , X) to the Cauchy problem 2.7, with Iτ = [t0 − τ, t0 + τ ].
Remark Notice first that from (b), since U is open, k is defined in a neighborhood of zero. If k is constant then (c)-(d) are automatically satisfied. Indeed,
for (x, t) ∈ B R ×X (u0 , s), we have
kf (t, x)kX ≤ ks + max kf (t, x0 )kX .
|t−t0 |≤s
Also, (d) is always true if X is finite-dimensional, for closed balls are compact.
In both cases, setting
s0 = sup σ > 0 : B R ×X (u0 , σ) ⊂ U
we can choose any s < s0 .
proof Let r = min{a, s}, and set
n ro
.
τ0 = min r,
m
Select then τ < τ0 , and consider the complete metric space Z = B C(Iτ ,X) (x0 , r)
with the metric induced by the norm of C(Iτ , X) (here x0 is the constant function
equal to x0 ). Since τ < r, if z ∈ Z then (t, z(t)) ∈ B R ×X (u0 , r) ⊂ U for all t ∈ Iτ .
Hence, for z ∈ Z, define
Z
t
F (z)(t) = x0 +
f (y, z(y)) dy,
t ∈ Iτ .
t0
Observe that
Z t
sup kF (z)(t) − x0 kX ≤ sup kf (y, z(y))kX dy ≤ mτ ≤ r.
t∈I
t∈I
τ
τ
t0
We conclude that F maps Z into Z. The last step is to show that F n is a
contraction on Z for some n ∈ N . By induction on n we show that, for every
t ∈ Iτ ,
Z
n
1 t
n
n
kF (z1 )(t) − F (z2 )(t)kX ≤ k(y) dy kz1 − z2 kC(Iτ ,X) .
(5)
n! t0
40
2. some applications of fixed point theorems
For n = 1 it holds easily. So assume it is true for n − 1, n ≥ 2. Then, taking
t > t0 (the argument for t < t0 is analogous),
kF n (z1 )(t) − F n (z2 )(t)kX
= kF (F n−1 (z1 ))(t) − F (F n−1 (z2 ))(t)kX
Z t
kf (y, F n−1 (z1 )(y)) − f (y, F n−1 (z2 )(y))kX dy
≤
t0
Z
t
≤
k(y)kF n−1 (z1 )(y) − F n−1 (z2 )(y)kX dy
t0
Z t
Z y
n−1 1
≤
k(y)
k(w) dw
dy kz1 − z2 kC(Iτ ,X)
(n − 1)! t0
t0
Z t
n
1
=
k(y) dy kz1 − z2 kC(Iτ ,X) .
n!
t0
Therefore from (5) we get
kF n (z1 ) − F n (z2 )kC(Iτ ,X) ≤
1
kkknL1 (Iτ ) kz1 − z2 kC(Iτ ,X)
n!
which shows that for n big enough F n is a contraction. By means of Corollary 1.5,
we conclude that F admits a unique fixed point, which is clearly the (unique)
solution to the integral equation (4) and hence to (3).
2.9 Proposition [Continuous dependence] The solution to the Cauchy problem 2.7 depends with continuity on the initial data.
proof Assume that xj ∈ C(Iτ , X) are two solutions to (3) with initial data x0j
(j = 1, 2). Setting x = x1 − x2 and x0 = x01 − x02 , from (b) we get
Z t
kx(t)kX ≤ kx0 kX + k(y)kx(y)kX dy ,
∀ t ∈ Iτ .
(6)
t0
The positive function
ϕ(t) = kx0 kX
Z t
exp k(y) dy t0
satisfies the equation
Z t
ϕ(t) = kx0 kX + k(y)ϕ(y) dy ,
∀ t ∈ Iτ .
(7)
t0
By comparing (6) with (7), we conclude that
Z t
kx(t)kX ≤ kx0 kX exp k(y) dy ,
t0
∀ t ∈ Iτ .
(8)
41
ordinary differential equations in banach spaces
Indeed, defining ψ = kxkX − ϕ, addition of (6) and (7) entails
Z
t
ψ(t) ≤ sgn(t − t0 )
∀ t ∈ Iτ .
k(y)ψ(y) dy,
t0
Let us show that the above inequality implies ψ ≤ 0 on [t0 , t0 + ρ] for any ρ < τ ,
an thus on [t0 , t0 + τ ) (the argument for (t0 − τ ] is the same). Choose n big
enough such that
Z
t0 +
(j+1)ρ
n
k(y) dy < 1,
∀ j = 0, 1, . . . , n − 1.
t0 + jρ
n
To finish the proof, we use an inductive argument. Suppose we proved that
ψ ≤ 0 on [t0 , t0 + jρ/n] for some j ≤ n − 1, and let t∗ be such that
ψ(t∗ ) = max ψ(t) : t ∈ [t0 + jρ/n, t0 + (j + 1)ρ/n] .
Then
∗
Z
t∗
ψ(t ) ≤
∗
k(y)ψ(y) dy ≤ ψ(t )
t0
Z
t0 +
(j+1)ρ
n
k(y) dy.
t0 + jρ
n
So if ψ(t∗ ) is strictly positive, we may cancel ψ(t∗ ) in the above inequality, getting
that the integral exceeds 1. Therefore ψ ≤ 0 on [t0 , t0 + (j + 1)ρ/n].
Remark The implication (6)⇒(8) is known as the Gronwall lemma, which can
also be proved via differential techniques.
2.10 Theorem [Global solution]
Theorem 2.8, and replace (c) with
Let U = (α, β) × X. Assume (a) and (b) of
(c 0 ) k ∈ L1 ((α, β)).
Then there exists a unique solution x ∈ C 1 (I, X) to the Cauchy problem 2.7, for
every I ⊂ (α, β).
proof Proceed like in the proof of Theorem 2.8, taking now Z = C(I, X). The
details are left to the reader.
When f is merely continuous and fulfills a compactness property, it is possible to
provide an existence result, exploiting the Schauder-Tychonoff fixed point theorem.
2.11 Theorem [Peano] Assume f be a continuous function, and f (V ) be relatively compact in X for some open neighborhood V ⊂ U of u0 . Then there exists
τ > 0 such that there is a (possibly nonunique) solution x ∈ C 1 (Iτ , X) to the Cauchy
problem 2.7, with Iτ = [t0 − τ, t0 + τ ].
42
2. some applications of fixed point theorems
proof Choose r > 0 such that B R ×X (u0 , r) ⊂ V , and let
m = sup kf (t, x)kX : (x, t) ∈ B R ×X (u0 , r) .
Finally, set τ = r/m. With the same notation of the proof of Theorem 2.8, F
maps the closed and convex set Z = B C(Iτ ,X) (u0 , r) into itself. Moreover, F is
continuous, since for zn , z ∈ Z,
Z t0 +τ
kf (y, zn (y)) − f (y, z(y))kX dy
sup kF (zn )(t) − F (z)(t)kX ≤
t∈Iτ
t0 −τ
which vanishes as zn → z, due to the dominated convergence theorem. The
Cauchy problem 2.7 has a solution if F has fixed point, which is guaranteed by
Theorem 1.18 once we show that F (Z) is relatively compact. To establish this
fact, we appeal to the Ascoli theorem for X-valued continuous functions (see,
e.g., Theorem 0.4.11 in [4]). First we observe that F (Z) is an equicontinuous
family, since for any z ∈ Z and t1 , t2 ∈ Iτ ,
Z t2
kF (z)(t1 ) − F (z)(t2 )kX = f (y, z(y)) dy ≤ m|t1 − t2 |.
t1
X
S
Next we prove that, for every fixed t ∈ Iτ , z∈Z F (z)(t) is relatively compact.
From the hypotheses, there is a compact set K ⊂ X such that f (y, z(y)) ∈ K
for all y ∈ Iτ and all z ∈ Z. Let H = co(K). It is well-known that in a
Banach space the convex hull of a compact set is relatively compact (see, e.g.,
[14], Theorem 3.25). Then, for a fixed t ∈ Iτ and every z ∈ Z we have
Z t
F (z)(t) = x0 +
f (y, z(y)) dy ∈ x0 + (t − t0 )H
t0
Rt
which isPcompact. Indeed, t0 f (y, z(y)) dy is the limit of the Riemann sums
(t − t0 ) i f (yi , z(yi ))(yi − yi−1 )/(t − t0 ). Hence Theorem 1.18 applies, and the
thesis follows.
Definition The Peano theorem is said to hold in a Banach space X if for an
arbitrary continuous function f and an arbitrary x0 ∈ X the Cauchy problem 2.7
admits a solution x(t) in some neighborhood of t0 .
It is clear from Theorem 2.11 that if X is finite-dimensional the Peano theorem
holds, since finite-dimensional Banach spaces have the Heine-Borel property. What
about infinite-dimensional Banach spaces? Let us examine a famous counterexample
to the existence problem.
2.12 Example [Dieudonné] Take the Banach space c0 with the supremum norm
and define the function f : c0 → c0 by
(f (x))n =
p
|xn | +
1
,
1+n
for x = (x0 , x1 , x2 , . . .) ∈ c0 .
43
semilinear equations of evolution
Consider the Cauchy problem
0
x (t) = f (x(t)),
x(0) = 0.
t ∈ (−ε, ε)
p
Due to uniform continuity of the real map s 7→ |s|, f is uniformly continuous on
c0 . Assume now that the Cauchy problem admits a solution y ∈ C 1 ((−ε, ε), c0 ) for
some ε > 0. In particular, each component yn is differentiable in (−ε, ε) and fulfills
the Cauchy problem
(
p
1
,
t ∈ (−ε, ε)
yn0 (t) = |yn (t)| +
1+n
yn (0) = 0.
Then yn (t) > 0 for t ∈ (0, ε), and
p
1 p
2
1 2
2 yn (t) −
log
yn (t) +
+
log
= t,
1+n
1+n
1+n
1+n
∀ t ∈ (0, ε).
On the other hand, y(t) ∈ c0 , that is, limn→∞ yn (t) = 0. Therefore the limit as
n → ∞ of the left-hand side of the above equality must be 0 for all t ∈ (0, ε). This
is a contradiction, and we conclude that no such solution y exists.
In fact, it has been shown quite recently that the Peano theorem provides a
characterization of finite-dimensional Banach spaces, namely,
2.13 Theorem [Godunov]
then X is finite-dimensional.
If the Peano theorem holds in a Banach space X,
Semilinear equations of evolution
2.14 Strongly continuous semigroups Let X be a Banach space. A oneparameter family S(t) (t ≥ 0) of bounded linear operators on X is said to be a
strongly continuous semigroup (C0 -semigroup, for short) if
(a) S(0) = I (identity operator on X);
(b) S(t + s) = S(t)S(s) for every t, s ≥ 0;
(c) lim S(t)x = x for every x ∈ X (strong continuity).
t→0
As a quite direct application of the uniform boundedness theorem, there exist ω ≥ 0
and M ≥ 1 such that
kS(t)kL(X) ≤ M eωt ,
∀ t ≥ 0.
This in turn entails the continuity of the map t 7→ S(t)x from [0, ∞) to X, for every
fixed x ∈ X (cf. [9]).
44
2. some applications of fixed point theorems
The linear operator A of domain
S(t)x − x
exists
D(A) = x ∈ X : lim
t→0
t
defined by
S(t)x − x
,
∀ x ∈ D(A)
t
is the infinitesimal generator of the semigroup S(t).
We now recall some basic facts on A (see, e.g., [9]).
Ax = lim
t→0
Proposition A is a closed linear operator with dense domain. For every fixed
x ∈ D(A), the map t 7→ S(t)x belongs to C 1 ([0, ∞), D(A)) and
d
S(t)x = AS(t)x = S(t)Ax.
dt
2.15
We consider the following semilinear Cauchy problem in X:
0
x (t) = Ax(t) + f (t, x(t)),
0<t≤T
x(0) = x0 ∈ X.
(9)
where A is the infinitesimal generator of a strongly continuous semigroup S(t), and
f : [0, T ] × X → X is continuous and uniformly Lipschitz continuous on X with
Lipschitz constant λ ≥ 0.
Definition A function x : [0, T ] → X is said to be a classical solution to (9) if
it is differentiable on [0, T ], x(t) ∈ D(A) for every t ∈ [0, T ], and (9) is satisfied
on [0, T ].
If x is a classical solution, it is necessarily unique, and it is given by
Z t
x(t) = S(t)x0 +
S(t − s)f (s, x(s)) ds.
(10)
0
This can be easily proved integrating in ds on [0, t] the derivative with respect to s of
the (differentiable) function S(t−s)x(s) and using (9). Notice that above (Riemann)
integral is well-defined, since if x ∈ C([0, T ], X) the map t 7→ f (t, x(t)) belong to
C([0, T ], X) as well (see 2.6). Of course, there is no reason why there should exist a
classical solution for a certain initial value x0 . However, formula (10) makes sense
for every x0 ∈ X. This motivates the following definition.
Definition A function x : [0, T ] → X is said to be a mild solution to (9) if it
continuous on [0, T ] and fulfills the integral equation (10).
Theorem For every x0 ∈ X the Cauchy problem (9) has a unique mild solution.
Moreover the map x0 7→ x(t) is Lipschitz continuous from X into C([0, T ], X).
45
an abstract elliptic problem
proof For a given x0 ∈ X define the map F : C([0, T ], X) → C([0, T ], X) by
Z
t
S(t − s)f (s, x(s)) ds.
F (x)(t) = S(t)x0 +
0
Then we have
kF (x)(t) − F (y)(t)kX ≤ λM tkx − ykC([0,T ],X)
where M = supt∈[0,T ] kS(t)kL(X) . By an inductive argument, analogous to the
one used in the proof of Theorem 2.8, we get
kF n (x)(t) − F n (y)(t)kX ≤
(λM T )n
kx − ykC([0,T ],X) .
n!
Hence for n ∈ N big enough F n is a contraction, so by Corollary 1.5 F has a
unique fixed point in C([0, T ], X) which is clearly the desired mild solution to
the Cauchy problem (9).
To complete the proof, let y be the unique mild solution corresponding to the
initial value y0 . Then
Z t
kx(t) − y(t)kX ≤ M kx0 − y0 kX + λM
kx(s) − y(s)kX ds.
0
By the Gronwall Lemma (cf. Proposition 2.9), we get at once
kx(t) − y(t)kX ≤ M eλM T kx0 − y0 kX ,
∀ t ∈ [0, T ]
which entails the Lipschitz continuity of the map x0 7→ x(t)
An abstract elliptic problem
Let X, V be Banach spaces with compact and dense embeddings V ,→ X ,→ V ∗ .
Assume we are given a bounded linear operator A : V → V ∗ and a (nonlinear)
continuous map B : X → V ∗ which carries bounded sets into bounded sets, such
that
hAu, ui ≥ εkuk2V ,
∀u∈V
(11)
and
hB(u), ui ≥ −c(1 + kukαV ),
∀u∈X
(12)
for some ε > 0, c ≥ 0 and α ∈ [0, 2), where h·, ·i denotes the duality pairing between
V ∗ and V .
Problem Given g ∈ V ∗ , find a solution u ∈ V to the abstract equation
Au + B(u) = g.
(13)
46
2. some applications of fixed point theorems
For v ∈ X, let w be the solution to the equation
Aw = g − B(v).
From (11) A is injective onto V ∗ , and by the open mapping theorem, A−1 is a
bounded linear operator from V ∗ onto V . Therefore
w = A−1 (f − B(v)) ∈ V.
Define then the map f : X → X as f (v) = A−1 (g − B(v)). Notice that f is
continuous and compact. Suppose then that, for some λ, there is uλ such that
uλ = λf (uλ ). This means that uλ solves the equation
Auλ + λB(uλ ) = λg.
Taking the duality pairing of the above equation and uλ , and exploiting (11)-(12),
we get
εkuλ k2V ≤ λc(1 + kuλ kαV ) + λkgkV ∗ kuλ kV .
Recalling now that λ ∈ [0, 1], and using the Young inequality
ab ≤ νap + K(ν, p)bq
(a, b ≥ 0, ν > 0)
where K(ν, p) = (νp)−q/p q −1 (1 < p, q < ∞, 1/p + 1/q = 1), we find the a priori
estimate
1
2
c + K(ε/4, 2/α) c2/(2−α) + kgk2V ∗ .
kuλ k2V ≤
ε
ε
A direct application of Theorem 1.20 entails the existence of a fixed point u for f ,
which is clearly a solution to (13). Notice that the solution might not be unique.
Example Let Ω ⊂ R2 be a bounded domain with smooth boundary ∂Ω. Find
a weak solution to the nonlinear elliptic problem
−∆u + u5 = g
u|∂Ω = 0.
In this case V = H01 (Ω), X = L6 (Ω), and g ∈ H01 (Ω)∗ = H −1 (Ω) (see, e.g.,
[1, 5] for the definitions and the properties of the Sobolev spaces H01 and H −1 ;
in particular, we recall that the embedding H01 (Ω) ,→ L6 (Ω) is compact when
Ω ⊂ R2 ). Then A = −∆ and B(u) = u5 are easily seen to fulfill the required
hypotheses.
The invariant subspace problem
The invariant subspace problem is probably the problem of operator theory. The
question, that attracted the attention of a great deal of mathematicians, is quite
simple to state. Given a Banach space X and an operator T ∈ L(X), find a closed
nontrivial subspace M of X (i.e., M 6= X and M 6= {0}) for which T M ⊂ M . Such
47
the invariant subspace problem
M is said to be an invariant subspace for T . It is known that not all continuous
linear operators on Banach spaces have invariant subspaces. The question is still
open for Hilbert spaces.
The most general, and at the same time most spectacular, result on the subject,
is the Lomonosov theorem, that provides the existence of hyperinvariant subspaces
for a vast class of operators. The proof is relatively simple, and the role played by
the Schauder-Tychonoff theorem is essential. In order to state the result, we need
first a definition.
Definition Let X be a Banach space. An invariant subspace M for T ∈ L(X)
is said to be hyperinvariant if it is invariant for all operators commuting with T
(that is, for all T 0 ∈ L(X) such that T T 0 = T 0 T ).
Remark If T ∈ L(H) is nonscalar, i.e., is not a multiple of the identity, and it
has an eigenvalue λ, then the eigenspace M corresponding to λ is hyperinvariant
for T . Indeed, if x ∈ M and T 0 commutes with T , we have that
λT 0 x = T 0 T x = T T 0 x.
Therefore T 0 x ∈ M .
2.16 Theorem [Lomonosov] Let X be a Banach space. Let T ∈ L(X) be a
nonscalar operator commuting with a nonzero compact operator S ∈ L(X). Then T
has a hyperinvariant subspace.
We anticipate an observation that will be used in the proof.
Remark Let S ∈ L(X) be a compact operator (cf. Definition 1.19). If λ 6= 0 is
an eigenvalue of S, then the eigenspace
F := {x ∈ X : Sx = λx}
relative to λ has finite dimension. Indeed, the restriction of S on F is a (nonzero)
multiple of the identity on F , and the identity is compact if and only if the space
is finite-dimensional.
proof We proceed by contradiction. Let A be the algebra of operators commuting with T . It is immediate to see that if T has no hyperinvariant subspaces,
then Ax = X for every x ∈ X, x 6= 0.
Without loss of generality, let kSkL(X) ≤ 1. Choose then x0 ∈ X such that
kSx0 k > 1 (which implies kx0 k > 1) and set B = B X (x0 , 1). For x ∈ SB (notice
that x cannot be the zero vector), there is T 0 ∈ A such that kT 0 x − x0 k < 1.
Hence every x ∈ SB has an open neighborhood Vx such that T 0 Vx ⊂ B for some
T 0 ∈ A. Exploiting the compactness of SB, we find a finite cover V1 , . . . , Vn and
T10 , . . . , Tn0 ∈ A such that
Tj0 Vj ⊂ B,
∀ j = 1, . . . , n.
48
2. some applications of fixed point theorems
Let ϕ1 , . . . , ϕn ∈ C(SB) be a partition of the unity for SB subordinate to the
open cover {Vj }, and define, for x ∈ B,
f (x) =
n
X
ϕj (Sx)Tj0 Sx.
j=1
Then f is a continuous function from B into B. Since Tj0 S is a compact map for
every j, it is easily seen that f (B) is relatively compact. Hence Theorem 1.18,
yielding the existence of x̄ ∈ B such that f (x̄) = x̄. Defining the operator Te ∈ A
as
n
X
Te =
ϕj (S x̄)T 0
j
j=1
we get the relation
TeS x̄ = x̄.
But TeS is a compact operator, hence the eigenspace F of TeS relative to the
eigenvalue 1 is finite-dimensional. Since TeS commutes with T , we conclude
that F is invariant for T , which means that T has an eigenvalue, and thus a
hyperinvariant subspace, contrary to our assumption.
Measure preserving maps on compact Hausdorff spaces
Let X be a compact Hausdorff space, and let P (X) be the set of all Borel probability
measures on X. By means of the Riesz representation theorem, the dual space of
C(X) can be identified with the space M (X) of complex regular Borel measures on
X. Recall that the norm kµk of an element µ ∈ M (X) is given by the total variation
of µ. It is straightforward to check that P (X) is convex and closed in the weak∗
topology. Moreover, P (X) is weak∗ compact. Indeed, it is a weak∗ closed subset of
the unit ball of M (X), which is weak∗ compact by the Banach-Alaoglu theorem.
Definition Let µ ∈ P (X). A µ-measurable map f : X → X is said to be
measure preserving with respect to µ if µ(B) = µ(f −1 (B)) for every Borel set
B ⊂ X. Such µ is said to be an invariant measure for f .
Notice that, if f is a µ-measurable map, the measure f˜µ defined by
f˜µ(B) = µ(f −1 (B)),
∀ Borel set B
belongs to P (X). In particular, if f is continuous (and therefore measurable with
respect to every µ ∈ P (X)), we have a map f˜ : P (X) → P (X) defined by µ 7→ f˜µ.
In addition, applying the monotone convergence theorem to an increasing sequence
of simple functions, it is easy to see that
Z
Z
˜
g d(f µ) =
g ◦ f dµ,
∀ g ∈ C(X).
X
X
49
invariant means on semigroups
Lemma Assume f : X → X be continuous. Then the map f˜ : P (X) → P (X)
is continuous in the weak ∗ topology.
proof Let {µι }ι∈I ⊂ P (X) be a net converging to some µ ∈ P (X). Then, for
every g ∈ C(X),
Z
Z
Z
Z
g d(f˜µι ) = lim
g ◦ f dµι =
g ◦ f dµ =
g d(f˜µ)
lim
ι∈I
ι∈I
X
X
X
X
which entails the claimed continuity.
We are now interested to find elements of P (X) that are invariant measures for
f . This is the same as finding a fixed point for the map f˜.
Theorem Let f : X → X be a continuous map. Then there exists µ ∈ P (X)
for which f is measure preserving.
proof On account of the above discussion, f˜ is a continuous map of a compact
convex subset of M (X) into itself, and the existence of a fixed point is then
guaranteed by Theorem 1.18.
Clearly, the invariant measure for f might not be unique. In particular, notice
that if x is a fixed point for f , then the Dirac measure δx is invariant for f .
Invariant means on semigroups
Let S be a semigroup, that is, a set endowed with an associative binary operation,
and consider the (real) Banach space of all real-valued bounded functions on S,
namely,
n
o
∞
` (S) = f : S → R : kf k := sup |f (s)| < ∞ .
s∈S
An element f ∈ `∞ (S) is positive if f (s) ≥ 0 for every s ∈ S. A linear functional
Λ : `∞ (S) → R is positive if Λf ≥ 0 for every positive element f ∈ `∞ (S). We agree
to denote a constant function on S by the value of the constant.
We now recall a result that actually holds for more general situations.
Lemma Let Λ ∈ `∞ (S)∗ , with kΛk = Λ1 = 1. Then Λ is positive.
proof Assume not. Then there is f ∈ `∞ (S), f ≥ 0, such that Λf = β < 0.
For ε > 0 small, we have
k1 − εf k = sup |1 − εf (s)| ≤ 1.
s∈S
Hence
1 < 1 − εβ ≤ |1 − εβ| = |Λ(1 − εf )| ≤ k1 − εf k ≤ 1
leading to a contradiction.
50
2. some applications of fixed point theorems
If t ∈ S, we can define the left t-translation operator Lt : `∞ (S) → `∞ (S) to be
(Lt f )(s) = f (ts),
∀ s ∈ S.
In an analogous manner, we can define the right t-translation operator Rt .
Definition A (left) invariant mean on S is a positive linear functional Λ on
`∞ (S) satisfying the following conditions:
(a) Λ1 = 1;
(b) Λ(Ls f ) = Λf for every s ∈ S and every f ∈ `∞ (S).
When such a functional exists, S is said to be (left) amenable.
Clearly, we can give the above definition replacing left with right or two-sided.
The distinction is relevant if S is not abelian.
2.17 Example [Banach] Let S = N. Then `∞ (N) = `∞ . An invariant mean in
this case is called a Banach generalized limit. The reason is that if Λ is an invariant
mean on N and x = {xn }n∈N ∈ `∞ is such that limn→∞ xn = α ∈ R, then Λx = α.
Indeed, for any ε > 0, we can choose n0 such that α + ε ≤ xn ≤ α + ε for every
n ≥ n0 . Hence, if we define y = {yn }n∈N ∈ `∞ by yn = xn+n0 , we have Λx = Λy,
and
α − ε = Λ(α − ε) ≤ Λy ≤ Λ(α + ε) ≤ α + ε
which yields the equality Λx = α. To prove the existence of an invariant mean, one
has to consider the subspace M of `∞ given by
x0 + · · · + xn
∞
= αx ∈ R
M = x = {xn }n∈N ∈ ` : lim
n→∞
n+1
and define the linear functional Λ0 on M as Λ0 x = αx . Setting for every x ∈ `∞
x0 + · · · + xn
p(x) = lim sup
n+1
n→∞
it is then possible, by means of the Hahn-Banach theorem, to extend Λ0 to a functional Λ defined on the whole space, in such a way that −p(−x) ≤ Λx ≤ p(x) for
every x ∈ `∞ . In particular, Λ is continuous. A remarkable consequence of this fact
is that not every continuous linear functional on `∞ can be given the representation
{xn }n∈N 7→
∞
X
cn xn
n=0
for some numerical sequence cn . Indeed, for every k ∈ N, taking ek = {δnk }n∈N , we
have Λek = 0. Hence if Λ has the above representation, all the cn must be zero, i.e.,
Λ is the null functional, contrary to the fact that Λ1 = 1.
The next result, based on the Markov-Kakutani theorem, provides an elegant
generalization of Example 2.17.
2.18 Theorem [Day]
Let S be an abelian semigroup. Then S is amenable.
51
haar measures
proof Denote
K = Λ ∈ `∞ (S)∗ : kΛk = Λ1 = 1 .
In particular, if Λ ∈ K then Λ is positive. K is convex, and from the BanachAlaoglu theorem is compact in the weak∗ topology of `∞ (S)∗ . We define the
family of linear operators Ts : `∞ (S)∗ −→ `∞ (S)∗ , s ∈ S, as
(Ts Λ)(f ) = Λ(Ls f ),
∀ f ∈ `∞ (S).
First we show that Ts is continuous in the weak∗ topology for every s ∈ S. Of
course, it is enough to show the continuity at zero. Thus let V be a neighborhood
of zero of the local base for the weak∗ topology, that is,
V = Λ ∈ `∞ (S)∗ : |Λfj | < εj , j = 1, . . . , n
for some ε1 , . . . , εn > 0 and f1 , . . . , fn ∈ `∞ (S). Then
Ts−1 (V ) = Λ ∈ `∞ (S)∗ : |(Ts Λ)(fj )| < εj , j = 1, . . . , n
= Λ ∈ `∞ (S)∗ : |Λ(Ls fj )| < εj , j = 1, . . . , n
is an open neighborhood of zero as well.
The second step is to prove that Ts K ⊂ K. Indeed,
(Ts Λ)(1) = Λ(Ls 1) = Λ1 = 1
and
kTs Λk = sup |(Ts Λ)(f )| = sup |Λ(Ls f )| ≤ sup |Λf | = kΛk = 1
kf k≤1
kf k≤1
kf k≤1
since kLs f k ≤ kf k.
Finally, for every s, t ∈ S,
Ts Tt Λ = Ts (Λ ◦ Lt ) = Λ ◦ Lt ◦ Ls = Λ ◦ Lst = Λ ◦ Lts = Tt Ts Λ.
Hence by Theorem 1.23 there is Λ ∈ K such that Ts Λ = Λ for every s ∈ S, which
means that Λ(Ls f ) = Λf for every s ∈ S and every f ∈ `∞ (S).
Haar measures
2.19 Topological groups A topological group is a group G endowed with a
Hausdorff topology that makes the group operations continuous; namely, the map
(x, t) 7→ xy −1 is continuous for every x, y ∈ G.
For any y ∈ G the maps x 7→ xy, x 7→ yx, and x 7→ x−1 are homeomorphisms
of G onto G. Hence the topology of G is uniquely determined by any local base
at the identity element e. Indeed, if U is a neighborhood of some x ∈ G, the sets
xU = {xy : y ∈ U } and U x = {yx : y ∈ U } are neighborhoods of e.
The next topological lemma will be used in the proof of the main result of the
section.
52
2. some applications of fixed point theorems
Lemma Let G be a compact topological group, and let K1 , K2 ⊂ G be disjoint
compact sets. Then there exists an open neighborhood U of e such that no translate of U meets both K1 and K2 .
proof Let O ⊂ G be an open set such that O ⊃ K1 and O ∩ K2 = ∅. For every
x ∈ K1 , x−1 O ∈ U. Using the continuity of the group operations, select Ux ∈ U
such that Ux = S
Ux−1 and Ux Ux Ux ⊂ x−1 O, and let Vx = xUx ∩ Ux x. Since Vx ∈ U,
it follows that x∈K1 Vx is an open
Sn cover of K1 , and by compactness there are
x1 , . . . , xn ∈ K1 such that K1 ⊂ i=1 Vxi . Finally define
U=
n
\
Vxi ∈ U.
i=1
Assume now that some translate of U meets both K1 and K2 . Then there are
k1 ∈ K1 and k2 ∈ K2 such that either k2 = k1 u−1 v or u−1 vk2 = k1 for some
u, v ∈ U . Let us examine the first case (being the second analogous). We have
that k1 ∈ Vxi ⊂ xi Uxi for some i ∈ {1, . . . , n}. Hence
k2 ∈ xi Uxi U −1 U ⊂ xi Uxi (xi Uxi )−1 xi Uxi = xi Uxi Uxi Uxi ⊂ xi x−1
i O ⊂ O
which is a contradiction.
Notice that the proof holds in fact for locally compact topological groups.
2.20 Definition Let G be a compact topological group. A Haar measure on G
is a regular Borel probability measure µ which is simultaneously left invariant, i.e.,
Z
Z
f (x) dµ(x) =
G
f (yx) dµ(x),
∀ y ∈ G, f ∈ C(G)
(14)
f (xy) dµ(x),
∀ y ∈ G, f ∈ C(G)
(15)
∀ f ∈ C(G).
(16)
G
and right invariant, i.e.,
Z
Z
f (x) dµ(x) =
G
G
and satisfies the relation
Z
Z
f (x) dµ(x) =
f (x−1 ) dµ(x),
G
G
It is readily seen that a Haar measure on G, if it exists, is unique. Indeed, if µ
and ν are two Borel probability measures on G, with µ left invariant and ν right
53
haar measures
invariant, for every f ∈ C(G) the Fubini theorem yields
Z
Z
f (x) dµ(x) =
G
f (yx) dµ(x)
Z Z
f (yx) dµ(x) dν(y)
G
G
Z Z
f (yx) dν(y) dµ(x)
ZG G
f (yx) dν(y)
G
Z
f (y) dν(y)
G
=
=
=
=
G
forcing the equality µ = ν.
2.21 Theorem Let G be a compact topological group. Then there exists a unique
Haar measure on G.
proof By means of the above uniqueness argument, it is enough to prove the
existence of two regular Borel probability measure µ and ν satisfying (14) and
(15), respectively. Then (16) follows at once for if µ is a Haar measure on G, so
is ν defined by dν(x) = dµ(x−1 ).
We then proceed by proving the existence of µ (the proof for ν being the same).
Let K be the family of compact subsets of G, and U be the family of open
subsets of G containing e. If K ∈ K and U ∈ U, we can always cover K by a
finite number of translates of U . We define the covering number [K : U ] of K by
U the smallest number of translates of U required to cover K. For every K ∈ K,
we introduce the normalized covering ratio
ξK (U ) =
[K : U ]
,
[G : U ]
∀ U ∈ U.
It is immediate to verify that 0 ≤ ξK (U ) ≤ 1, ξK1 (U ) ≤ ξK2 (U ) for K1 ⊂ K2 ,
and ξxK (U ) = ξK (U ) for all x ∈ G. Moreover, if K1 , K2 ∈ K are disjoint sets,
exploiting the lemma we find U ∈ U such that no translate of U meets both K1
and K2 . Hence any covering of K1 ∪ K2 by translates of U is the disjoint union
of coverings of K1 and K2 . The same holds replacing U with any V ∈ U, V ⊂ U .
Thus
ξK1 ∪K2 (V ) = ξK1 (V ) + ξK2 (V ),
∀ V ∈ U, V ⊂ U.
(17)
Notice that (U, ∩) is an abelian semigroup, so Theorem 2.18 applies, yielding
the existence of an invariant mean Λ on U. Since ξK : U → R belongs to `∞ (U),
we can define
ψ(K) = ΛξK ,
∀ K ∈ K.
54
2. some applications of fixed point theorems
The following hold:
ψ(∅) = 0, ψ(G) = 1;
ψ(K1 ) ≤ ψ(K2 ),
if K1 ⊂ K2 ;
ψ(xK) = ψ(K),
∀ x ∈ G.
ψ(K1 ∪ K2 ) = ψ(K1 ) + ψ(K2 ),
if K1 ∩ K2 = ∅.
(18)
(19)
(20)
(21)
Properties (18)-(20) are direct consequences of the definition of ξK (U ), whereas
(21) follows from (17) and the fact that if ξ ∈ `∞ (U) and there is U ⊂ U such
that ξ(V ) = 0 for every V ⊂ U, V ⊂ U , then Λξ = 0. Indeed, Λ(RU ξ) = Λξ,
and (RU ξ)(W ) = ξ(U ∩ W ) = 0 for every W ⊂ U.
For each open set O ⊂ G define
µ∗ (O) = sup{ψ(K) : K ⊂ O, K compact}
and for every E ⊂ G let
µ∗ (E) = inf{µ∗ (O) : O ⊃ E, O open}.
By (18)-(20), µ∗ (∅) = 0, µ∗ (G) = 1, µ∗ (E1 ) ≤ µ∗ (E2 ) if E1 ⊂ E2 ⊂ G, and µ∗ is
left invariant. Moreover µ∗ is countably subadditive, i.e.,
µ
∗
∞
[
Ej ≤
j=1
∞
X
µ∗ (Ej ).
(22)
j=1
To see that, let ε > 0, andSchoose open sets Oj ⊂ Ej with µ∗ (Oj )S≤ µ∗ (Ej )+ε/2j ,
n
and a compact set K ⊂ ∞
j=1 Oj for some
j=1 Oj . Due to compactness, K ⊂
n. Consider a partition of the unity ϕ1 , . . . , ϕn for K subordinate to the open
cover
Sn {O1 , . . . , On }, and set Kj = {x ∈ G : ϕj (x) > 1/n}. The Kj are compact,
j=1 Kj ⊃ K, and Kj ∩ K ⊂ Oj . Thus
ψ(K) ≤
n
X
j=1
ψ(Kj ∩ K) ≤
n
X
µ∗ (Oj ) ≤
j=1
∞
X
µ∗ (Ej ) + ε.
j=1
Taking the supremum over all such K, and letting ε → 0, we get (22). We
conclude that µ∗ is an outer measure. Applying then the Carathéodory extension
process, we build a measure µ, which coincides with µ∗ on the measurable sets,
namely, those sets E ⊂ G such that
µ∗ (T ) = µ∗ (T ∩ E) + µ∗ (T ∩ E C ),
∀ T ⊂ G.
The proof is finished if we show that the open sets are measurable, since µ is
then a Borel outer regular measure (by construction) and hence regular (being
finite). So let O be an open set, and T be any set. Let ε > 0, and take an open
set A ⊃ T such that µ∗ (A) ≤ µ∗ (T ) + ε. Choose a compact set K ⊂ A ∩ O with
55
haar measures
µ∗ (A ∩ O) ≤ ψ(K) + ε, and a compact set K0 ⊂ A ∩ K C with µ∗ (A ∩ K C ) ≤
ψ(K0 ) + ε. Then
µ∗ (T ∩ O) + µ∗ (T ∩ OC ) ≤ µ∗ (A ∩ O) + µ∗ (A ∩ K C ) ≤ ψ(K) + ψ(K0 ) + 2ε
= ψ(K ∪ K0 ) + 2ε ≤ µ∗ (A) + 2ε ≤ µ∗ (T ) + 3ε.
Since ε is arbitrary, the desired conclusion is immediate.
A different proof of Theorem 2.21, that directly applies the Markov-Kakutani
theorem, can be found in [14]. The advantage of the approach presented here is that
it can be easily modified to extend the result to the locally compact case.
2.22 When G is a locally compact topological group it is still possible to talk of
Haar measures. In this case we shall distinguish between left and right invariant
ones.
Definition Let G be a locally compact topological group. A left Haar measure
on G is a nonnull Radon measure µ on G satisfying (14). Analogously, a right
Haar measure on G is a nonnull Radon measure ν on G satisfying (15).
Recall that a Radon measure is a positive linear functional on Cc (G) (the space
of compactly supported continuous functions on G). From the Riesz representation
theorem every such functional can be uniquely represented by a positive outer regular
Borel measure, finite on compact sets, for which open sets are inner regular (cf. [12]).
Theorem Let G be a locally compact topological group. Then there exists left
Haar measures and right Haar measures on G. Moreover, any two left (right)
Haar measure differ by a multiplicative positive constant.
The proof of the above theorem can be found in [11]. We list some important
properties of Haar measures, that generalize (14)-(16) (cf. [4, 11]).
Theorem Let G be a locally compact topological group.
(a) There exists a continuous strictly positive function ∆ on G, with ∆(e) = 1
and ∆(xy) = ∆(x)∆(y) for all x, y ∈ G, such that if µ is any left Haar
measure on G, then
Z
Z
f (xy) dµ(x) = ∆(y) f (x) dµ(x),
∀ y ∈ G, f ∈ Cc (G).
G
G
(b) For any left Haar measure µ on G there holds
Z
Z
−1
f (x ) dµ(x) =
f (x) ∆(x)dµ(x),
G
∀ f ∈ Cc (G).
G
(c) For any left Haar measure µ and any right Haar measure ν on G there is
c > 0 such that
Z
Z
f (x) dν(x) = c f (x)∆(x) dµ(x),
∀ f ∈ Cc (G).
G
G
56
1. fixed point theorems
(d) Given any any left or right Haar measure on G, every nonvoid open set has
nonnull measure, and G has finite measure if and only if it is compact.
Notice from (a)-(c) it follows that if ν is any right Haar measure on G then
Z
Z
1
f (yx) dν(x) =
f (x) dν(x),
∀ y ∈ G, f ∈ Cc (G).
∆(y) G
G
Remark The function ∆ is termed the modular function. A locally compact
topological group is called unimodular if there exists a bi-invariant Radon measure. This happens if and only if ∆ ≡ 0. Besides compact groups, abelian groups
are clearly unimodular.
Game theory
We consider a game with n ≥ 2 players, under the assumption that the players do
not cooperate among themselves. Each players pursue a strategy, in dependence of
the strategies of the other players. Denote the set of all possible strategies of the
k th player by Kk , and set K = K1 × · · · × Kn . An element x ∈ K is called a strategy
profile. For each k, let fk : K → R be the loss function of the k th player. If
n
X
fk (x) = 0,
∀x∈K
(23)
k=1
the game is said to be of zero-sum. The aim of each player is to minimize his loss,
or, equivalently, to maximize his gain.
Definition A Nash equilibrium is a strategy profile with the property that no
player can benefit by changing his strategy, while all other players keep their
strategies unchanged. In formulas, it is an element x̄ = (x̄1 , . . . , x̄n ) ∈ K such
that
fk (x̄) ≤ fk (x̄1 , . . . , x̄k−1 , xk , x̄k+1 , . . . , x̄n ),
∀ x k ∈ Kk
(24)
for every k = 1, . . . , n.
Strictly speaking, a Nash equilibrium suggests a convenient “cautious” strategy
to be adopted by each player in the game. We said a strategy rather than the
strategy, since a Nash equilibrium (if it exists) might not be unique.
We need of course further hypotheses on the sets Kk and on the maps fk . It
is reasonable to assume that, with all the other strategies fixed, the loss function
fk has a small variation in correspondence of a small variation of xk . Also, loosely
speaking, it is assumed that the average of losses corresponding to two different
strategies of the k th player is grater than the loss corresponding to the “average”
strategy. Convexity can suitably translate this issue.
The fundamental result of game theory is the following.
57
game theory
2.23 Theorem [Nash] For every k = 1, . . . , n, let Kk be a nonvoid, compact
and convex subset of a locally convex space Xk . Assume that, for every k, the loss
function fk is continuous on K. In addition, for every fixed xj ∈ Kj with j 6= k, the
map
fk (x1 , . . . , xk−1 , ·, xk+1 , . . . , xn ) : Kk → R
is convex. Then there exists x̄ ∈ K satisfying (24), i.e., there is a Nash equilibrium.
proof Define Φ : K × K → R as
Φ(x, y) =
n
X
fk (x) − fk (x1 , . . . , xk−1 , yk , xk+1 , . . . , xn ) .
k=1
Then Φ is continuous, and Φ(x, ·) is concave for every fixed x ∈ K. From
Theorem 1.24 there exists x̄ ∈ K such that
sup Φ(x̄, y) ≤ sup Φ(y, y) = 0.
y∈K
y∈K
In particular, if we set ȳ = (x̄1 , . . . , x̄k−1 , xk , x̄k+1 , . . . , x̄n ), for xk ∈ Kk , we get
Φ(x̄, ȳ) ≤ 0,
∀ x k ∈ Kk
which is nothing but (24).
The hypotheses can be weakened if we consider a two-player zero-sum game
(sometimes called a duel). In this case, on account of (23), we have
Ψ(x1 , x2 ) := f1 (x1 , x2 ) = −f2 (x2 , x2 ).
Therefore we can repeat the above proof taking Ψ to be convex and lower semicontinuous in the first variable, and concave and upper semicontinuous in the second.
Now Ψ is the loss function of the first player or, equivalently, the gain function of
the second one.
2.24 Theorem [von Neumann] Let K1 ⊂ X1 and K2 ⊂ X2 be as in Theorem 2.23. Let Ψ : K1 × K2 → R be such that
(a) Ψ(·, x2 ) is lower semicontinuous and convex ∀ x2 ∈ K2 ;
(b) Ψ(x1 , ·) is upper semicontinuous and concave ∀ x1 ∈ K1 .
Then there exists a Nash equilibrium (x̄1 , x̄2 ) ∈ K1 × K2 .
proof In this case Φ : (K1 × K2 ) × (K1 × K2 ) → R has the form
Φ((x1 , x2 ), (y1 , y2 )) = −Ψ(y1 , x2 ) + Ψ(x1 , y2 )
and the argument of the proof of Theorem 2.23 applies.
58
1. fixed point theorems
Theorem 2.24 is known in the literature as the minimax theorem. The reason is
clear from the next corollary.
Corollary In the hypotheses of Theorem 2.24, the equality
inf
sup Ψ(x1 , x2 ) = Ψ(x̄1 , x̄2 ) = sup
x1 ∈K1 x2 ∈K2
inf Ψ(x1 , x2 )
x2 ∈K2 x1 ∈K1
holds true.
proof Define g(x1 ) = supx2 ∈K2 Ψ(x1 , x2 ) and h(x2 ) = inf x1 ∈K1 Ψ(x1 , x2 ). Then
for all x1 ∈ K1 and x2 ∈ K2 we have
h(x2 ) ≤ Ψ(x1 , x2 ) ≤ g(x1 )
which entails
sup h(x2 ) ≤ inf g(x1 ).
x2 ∈K2
x1 ∈K1
On the other hand, by Theorem 2.24,
h(x̄2 ) = inf Ψ(x1 , x̄2 ) = Ψ(x̄1 , x̄2 ) = sup Ψ(x̄1 , x2 ) = g(x̄1 ).
x1 ∈K1
x2 ∈K2
Hence
sup h(x2 ) ≥ h(x̄2 ) = g(x̄1 ) ≥ inf g(x1 ) ≥ sup h(x2 )
x1 ∈K1
x2 ∈K2
x2 ∈K2
so that all the above inequalities are in fact equalities.
We conclude the section considering a duel game where the sets K1 and K2 of
all possible strategies of each player are finite. We also assume that, for k = 1, 2,
player k plays randomly the strategy xk ∈ Kk with probability pk (xk ). In this case
the players are said to adopt a mixed strategy. Denoting the loss function (of the
first player) by Ψ, the average loss function is given by
ΨP (p1 , p2 ) =
X X
p1 (x1 )p2 (x2 )Ψ(x1 , x2 )
x1 ∈F1 x2 ∈F2
defined on the set K1P × K2P , where
X
P
Kk = pk : Fk → [0, 1] :
pk (xk ) = 1 .
xk ∈Fk
Theorem Any duel with a finite numbers of strategy profiles admits a Nash
equilibrium made of mixed strategies.
proof Just observe that KkP and ΨP fulfill the hypotheses of Theorem 2.24. notes on chapter 2
59
Notes on Chapter 2
The implicit function Theorem 2.4 is due to Dini [Opere vol.II, Ed. Cremonese, Roma (1954)]. Concerning
the Newton method, we refer to the paper of Kantorovich [Acta Math. 71, 63–97 (1939)].
Theorem 2.8 and Theorem 2.8 are refined versions of the “method of the successive approximants”,
envisaged by Cauchy and Liouville, and developed in the most general form by Picard [J. de Math. 6,
145–210 (1890)]. Theorem 2.11 (for finite-dimensional Banach spaces) is due to Peano [Math. Ann. 37,
182–228 (1890)]. Dieudonné’s Example 2.12 is in Acta Sci. Math. Szeged 12, 38–40 (1950). Theorem 2.13
is due to Godunov [Functional Anal. Appl. 9, 53–55 (1975)]. The same result for nonreflexive Banach
spaces has been obtained by Cellina in a previous paper [Bull. Amer. Math. Soc. 78, 1069–1072 (1972)].
Lomonosov’s Theorem 2.16 has appeared in 1973 [the English translation is in Functional Anal. Appl.
7, 213-214 (1974)]. For some time it was not clear whether there could exist operators to which the
theorem does not apply. An example in that direction has been found by Hadvin, Nordgren, Radjavi and
Rosenthal [J. Funct. Anal. 38, 410–415 (1980)]. However, the problem of invariant subspaces in Banach
spaces has been solved (negatively) by Enflo some years later [Acta Math. 158, 213–313 (1987)]. A good
reference for the subject is the book of Beauzamy Introduction to operator theory and invariant subspaces,
North-Holland, Amsterdam (1988).
More details on measure preserving maps and their link with ergodic theory can be found in Walter’s
book An introduction to ergodic theory, Springer-Verlag, New York (1982).
Example 2.17 is due to Banach [Opérations linéaires, Monografje Matematyczne I, Warszawa (1932)].
Day’s Theorem 2.18 is in Illinois J. Math. 1, 509–544 (1957).
A complete reference for Haar measures is Nachbin’s book The Haar integral, Van Nostrand, Princeton
(1965).
The celebrated Theorem 2.23 on non-cooperative games is due to Nash [Ann. of Math. 54, 286–295
(1951)], who obtained for the result the Nobel Prize in Economics. The first striking result in game theory
is von Neumann’s Theorem 2.24 [Ergebnisse eines Math. Colloq. 8, 73–83 (1937)].
Bibliography
[1] H. Brezis, Analyse fonctionnelle - Théorie et applications, Masson, Paris
(1983)
[2] K. Deimling, Nonlinear functional analysis, Springer-Verlag, Berlin (1980)
[3] N. Dunford, J.T. Schwartz, Linear operators, Interscience Publishers, New
York (pt.I 1958, pt.II 1963)
[4] R.E. Edwards, Functional analysis, Holt, Rinehart and Winston, Inc., New
York (1965)
[5] L.C. Evans, Partial differential equations, Amer. Math. Soc., Providence
(1998)
[6] V.I. Istrăţescu, Fixed point theory, D. Reidel Publishing Co., Dordrecht
(1981)
[7] J.L. Kelley, General Topology, Van Nostrand Co., Princeton (1955)
[8] W.S. Massey, A basic course in algebraic topology, Springer-Verlag, New York
(1991)
[9] A. Pazy, Semigroups of linear operators and applications to partial differential
equations, Springer-Verlag, New York (1983)
[10] F. Riesz, B. Sz-Nagy, Functional analysis, Frederick Ungar Publishing Co.,
New York (1955)
[11] H.L. Royden, Real analysis, 3 rd ed., Macmillan Publishing Company, New
York (1988)
[12] W. Rudin, Principles of mathematical analysis, 3 rd ed., McGraw-Hill Book
Company, New York (1976)
[13] W. Rudin, Real and complex analysis, 3 rd ed., McGraw-Hill Book Company,
New York (1986)
[14] W. Rudin, Functional analysis, McGraw-Hill Book Company, New York (1973)
60
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