M7180 Funkcionální analýza II
M7180 Functional analysis II
Michal Veselý
Masarykova univerzita, Pˇrírodovˇedecká fakulta
Ústav matematiky a statistiky
Se sazbou v systému LATEX pomáhal Jakub Juránek
Tato publikace vznikla v rámci projektu Fond rozvoje
Masarykovy univerzity (projekt MUNI/FR/1133/2018)
realizovaného v období 1/201912/2019.
Contents
0 Linear operators 1
0.1 Preliminaries and examples . . . . . . . . . . . . . . . . . . . . . . . 1
0.2 Continuity and boundedness . . . . . . . . . . . . . . . . . . . . . . 2
0.3 Inverse operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
0.4 Adjoint operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
0.5 Spectrum of operator . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1 Completely continuous operators 18
1.1 Preliminaries and examples . . . . . . . . . . . . . . . . . . . . . . . 18
1.2 Basic properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
1.3 Self-adjoint operator in Hilbert space . . . . . . . . . . . . . . . . . . 28
2 Derivative in Banach spaces 32
2.1 Weak and strong derivative . . . . . . . . . . . . . . . . . . . . . . . 32
2.2 Convex function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
2.3 Tangent functional . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3 Strictly and uniformly convex spaces 40
3.1 Strictly convex space . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.2 Uniformly convex space . . . . . . . . . . . . . . . . . . . . . . . . 42
3.3 Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.4 Smooth space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
4 Fixed point theorems 49
4.1 Topological degree . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.2 Brouwer and 1. and 2. Shauder theorem . . . . . . . . . . . . . . . . 50
5 Integration in Banach spaces 53
5.1 Preliminaries and basic definitions . . . . . . . . . . . . . . . . . . . 53
5.2 Bochner integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
5.3 Gelfand and Pettis integral . . . . . . . . . . . . . . . . . . . . . . . 56
0. Linear operators
0.1 Preliminaries and examples
Let X,Y be linear spaces.
Definition 0.1. Any map F : X → Y satisfying
F(λx+ µy) = λF(x)+ µF(y)
for all x,y ∈ X and scalars λ,µ is called a linear operator.
Remark 0.2. For operator F, we can write Fx instead of F(x).
Example 0.3. The operator I : X → X given by x → x is called the identical operator.
Obviously, I is linear. If X is normed, then I is continuous (by the Heine definition of
continuity: xn → x =⇒ Fxn → Fx).
Example 0.4. Let H be a Hilbert space and H1 its (closed) subspace. We know that
H = H1 ⊕ H⊥
1 , i.e., all x ∈ H can be uniquely expressed as x = x1 + x2, where x1 ∈
H1, x2 ∈ H⊥
1 . We put Px = x1 for x ∈ H. The operator P is called the (orthogonal)
projection. Evidently, it is linear and continuous.
Example 0.5. Let us consider the linear space C[a,b] and define the operator
T : C[a,b] → C[a,b]
by the formula
T f(t) =
b
a
k(t,s)f(s)ds, f ∈ C[a,b], t ∈ [a,b],
with the so-called core k ∈ C([a,b]×[a,b]). We can see that T is linear. If we consider
the norm
f = max
t∈[a,b]
|f(t)|, f ∈ C[a,b], (0.1)
i.e., the norm of the uniform convergence, then T is continuous.
Example 0.6. Let us consider the linear space C1
[a,b] of functions with continuous
derivatives on [a,b] and define the operator
D: C1
[a,b] → C[a,b], Df(t) = f (t), f ∈ C1
[a,b], t ∈ [a,b].
This operator is called the differential operator and it is linear. In spacesC1
[a,b],C[a,b],
let us consider the norm from (0.1). We prove that D is not continuous: the sequence
of
fn(t) =
sin(nt)
n
, n ∈ N,
satisfies fn → 0 as n → ∞, but the sequence of
Dfn(t) = fn(t) = cos(nt)
does not converge to 0.
1
0.2 Continuity and boundedness
Throughout this section, let
X = (X, · ) = (X, · X ),Y = (Y, · ) = (Y, · Y )
be normed linear spaces.
Definition 0.7. An operator L: X → Y is called bounded if it maps any bounded set
into a bounded set.
Theorem 0.8. Any continuous linear operator L: X → Y is bounded.
Proof. We assume the opposite. Let the operator L not be bounded. Then, there exists
a sequence {xn}∞
n=1 ⊆ X such that xn X ≤ c for all n ∈ N and some c ∈ R and
Lxn Y > n, n ∈ N. (0.2)
Since L is continuous, there exists δ ∈ (0,1) such that, for all x ∈ X satisfying x X < δ,
we have Lx Y ≤ 1. Let us choose n0 ∈ N such that c < δn0. The inequality
xn X
n
< δ
is valid for all n ≥ n0. Hence, we get Lxn Y ≤ n for all n ≥ n0, which is a contradiction
with (0.2).
Theorem 0.9. Any bounded linear operator L: X → Y is continuous at 0 ∈ X.
Proof. Let us assume the opposite. Let the operator L not be continuous at 0 ∈ X.
Then, there exist ε > 0 and a sequence {xn}∞
n=1 ⊆ X such that
xn X <
1
n
, Lxn Y ≥ ε, n ∈ N.
Let us denote yn = nxn for n ∈ N. The sequence {yn}∞
n=1 ⊆ X is bounded, but the
sequence {Lyn}∞
n=1 ⊆ Y is not bounded, because Lyn Y ≥ nε, n ∈ N, which is a con-
tradiction.
Theorem 0.10. If a linear operator L: X → Y is continuous at x0 ∈ X, then L is continuous
at any point (vector) of X.
Proof. For all ε > 0, there exists δ > 0 such that, for x ∈ X satisfying
x−x0 < δ,
we have
Lx−Lx0 < ε.
Let x1 ∈ X be arbitrarily given and y ∈ X satisfy y−x1 < δ. Then,
(y−x1 +x0)−x0 < δ,
and thus
L(y−x1 +x0)−Lx0 < ε,
i.e., Ly−Lx1 < ε, which proves that L is continuous at x1.
2
Remark 0.11. The definition of bounded linear operators can be reformulated as follows.
A linear operator L: X → Y is called bounded if there exists c > 0 such that
Lx Y ≤ c x X , x ∈ X.
Definition 0.12. Let L: X → Y be a bounded linear operator. The number
inf{c ∈ R; Lx Y ≤ c x X for x ∈ X}
is denoted by L and called the norm of L.
Theorem 0.13. Let L: X → Y be a bounded linear operator. Then,
L = sup
x X ≤1
Lx Y = sup
x=0
Lx Y
x X
.
Proof. We denote
λ = sup{ Lx Y ; x ∈ X, x X ≤ 1}.
Firstly, we can see that
sup
x X ≤1
Lx Y = sup
x=0
Lx Y
x X
,
where it suffices to consider the linearity of L and x X = 1. Therefore,
Lx Y
x X
≤ λ, x = 0,
i.e., Lx Y ≤ λ x X for all x ∈ X. Considering Definition 0.12, we have L ≤ λ.
Let ε > 0 be arbitrary. Then, there exists xε ∈ X such that xε = 0 and
λ −ε ≤
Lxε Y
xε X
.
However, from Definition 0.12, we obtain Lx Y ≤ L · x X for all x ∈ X. Thus,
λ −ε ≤ L . The arbitrariness of ε gives λ = L .
Remark 0.14. The set of all continuous linear operators L: X → Y is denoted by
L (X,Y). For L1,L2 ∈ L (X,Y) and a scalar k, we put
(L1 +L2)(x) = L1x+L2x, x ∈ X,
(kL1)(x) = kL1(x), x ∈ X.
Evidently, L (X,Y) forms a linear space.
This space is normed (with respect to the norm of operators introduced above).
Indeed, for the triangular inequality, it is sufficient to consider
(L1 +L2)x Y ≤ L1x Y + L2x Y ≤ L1 + L2
for all L1,L2 ∈ L (X,Y) and x X ≤ 1, x ∈ X. Moreover, from Definition 0.12, we have
the inequality
Lx Y ≤ L · x X , x ∈ X, L ∈ L (X,Y).
In the case when X = Y, we can write only L (X) (instead of L (X,X)).
3
Remark 0.15. Recall that norms − 1 and − 2 on X are called equivalent if there
exist α,β > 0 such that
α x 1 ≤ x 2 ≤ β x 1, x ∈ X.
Two norm are equivalent if and only if the topologies generated by them are same. For
any space with a finite dimension, all norms are equivalent.
Theorem 0.16. Let X have a finite dimension. Any linear operator L: X → Y is con-
tinuous.
Proof. According to Remark 0.15, we can choose a norm of X. Let e1,...,en be a base
of X. For x ∈ X,x = λ1e1 +···+λnen, we put
x X = |λ1|+···+|λn|.
It is seen that
Lx Y = λ1Le1 +···+λnLen Y ≤ max{ Le1 Y ,..., Len Y }· x X .
Remark 0.17. A basic example of continuous linear operators is given by functionals f
on X with the norm
f = sup
x ≤1
|f(x)|.
For details, we refer to the course Functional Analysis I. We add:
1. If X has a finite dimension, then f is realized. In addition, the norm of any
continuous linear functional is realized on the closed unit ball with the center 0 if
and only if the Banach space X is reflexive (the so-called James characteristic).
2. If any linear functional on X is bounded, then the dimension of X is finite.
3. The so-called Bishop–Phelps theorem says that the set of all continuous linear
functionals on a Banach space X, whose norms are realized on the closed unit
ball with the center 0, is a dense subset of the dual space X .
Example 0.18. We mention a series of examples.
a) Let us consider the space C[−1,1] with the norm
f = max
t∈[−1,1]
|f(t)|
and the functional
L f = 9 f(−1)−2f(0)+ f
1
4
.
If f ≤ 1, then
|L f| ≤ 9 f +2 f + f ≤ 12.
Especially, L ≤ 12. On the contrary, let us consider a function g ∈C[−1,1] for
which
max
t∈[−1,1]
|g(t)| = 1
and
g(−1) = 1, g(0) = −1, g
1
4
= 1.
We get Lg = 12. Therefore, L = 12.
4
b) For the space from a), we consider the functional
L f =
1
−1
sgn(t)f(t)dt.
Since
|L f| ≤
1
−1
|f(t)|dt ≤ f
1
−1
dt = 2 f ,
we have L ≤ 2. Let choose ε ∈ (0,1) and put gε (t) = t/ε for t ∈ (−ε,ε) and
gε (t) = sgn t for others t. Then, gε = 1 and |Lgε | = 2−ε. Now, we see that
L = 2.
c) For the space l2
, we consider L{xn}∞
n=1 = x1 +x2. For x = {xn}∞
n=1 ∈ l2
satisfying
x = |x1|2 +|x2|2 +··· ≤ 1,
we get
|Lx|2
≤ (|x1|+|x2|)2
≤ 2 |x1|2
+|x2|2
≤ 2 |x1|2
+|x2|2
+··· ≤ 2.
Therefore, L ≤
√
2. For
x =
1
√
2
,
1
√
2
,0,0, ... ,
we have x = 1 and |Lx| =
√
2. Thus, L =
√
2.
We also know (from the Riesz theorem) that there exists h ∈ l2
such that
Lx = x,h , x ∈ l2
.
We see that h = {1,1,0,0, ...}. In addition, we know that L = h , which
gives L =
√
1+1 =
√
2. Our knowledge of the dual space gives a powerful
tool to compute the norm of L.
d) We compute the norm of the functional
L: {xn}∞
n=1 →
∞
∑
n=1
xn
n
in l1
and l2
.
For x = {xn}∞
n=1 ∈ l1
, we have
|Lx| =
∞
∑
n=1
xn
n
≤
∞
∑
n=1
|xn| = x .
Therefore, L ≤ 1. For x = {1,0,0, ...}, we have x = 1, |Lx| = 1, consequently,
L = 1.
5
We can also use our knowledge of the dual space. The dual space of l1
is l∞
.
There exists just one {an}∞
n=1 ∈ l∞
satisfying
Lx =
∞
∑
n=1
anxn, x = {xn}∞
n=1 ∈ l1
,
where L = {an}∞
n=1 l∞ . Obviously, an = 1/n, n ∈ N. We get
L =
1
n
∞
n=1 l∞
= 1.
Now, we consider L in l2
. Since {1/n}∞
n=1 ∈ l2
, for x = {xn}∞
n=1 ∈ l2
, we have
|Lx| =
∞
∑
n=1
xn
n
≤
∞
∑
n=1
1
n2
·
∞
∑
n=1
|xn|2 =
π
√
6
· x .
Thus, L ≤ π/
√
6. For y =
√
6/(πn)
∞
n=1
∈ l2
, we have y = 1 and
|Ly| =
√
6
π
∞
∑
n=1
1
n2
=
π
√
6
,
which yields that L = π/
√
6.
e) Let us consider X = C1
[a,b], Y = C[a,b], and the operator (see Example 0.6)
D: X → Y, Df(t) = f (t), f ∈ C1
[a,b], t ∈ [a,b].
We consider the standard norm (see (0.1)) in Y and the norm
f C1 = max
t∈[a,b]
|f(t)|+ max
t∈[a,b]
f (t)
in X. Since
Df = max
t∈[a,b]
|f (t)| ≤ max
t∈[a,b]
|f(t)|+ max
t∈[a,b]
|f (t)| = f C1 , f ∈ C1
[a,b],
the operator D is continuous and D ≤ 1. To prove D = 1, it is sufficient to
consider
fn(t) =
sin(nt)
n
, n ∈ N,
for which
fn C1 = 1+
1
n
, Dfn = 1,
where n is sufficiently large.
f) Let us consider the operators L1,L2 ∈ L l2
given by
L1x = 0,x1,
x2
2
,
x3
3
, ... , x = {xn}∞
n=1 ∈ l2
,
L2x = {0,x1,x2,x3, ...}, x = {xn}∞
n=1 ∈ l2
.
Since the operator L2 is isometric, we see that L2 = 1.
For L1, we consider arbitrary x = {xn}∞
n=1 ∈ l2
. We have
L1x 2
=
∞
∑
n=1
xn
n
2
≤
∞
∑
n=1
|xn|2
= x 2
,
which gives L1 ≤ 1. For y = {1,0,0, ...}, we have y = 1 and L1y = 1.
Altogether, L1 = 1.
6
0.3 Inverse operator
In this section, let X,Y be linear spaces.
Definition 0.19. Let L : X → Y be an arbitrary operator. We put
R(L) = {y ∈ Y; there exists x ∈ X : Lx = y}.
We say that the operator L has an inverse if, for all y ∈ R(L), there exists just one x ∈ X
such that Lx = y. In this case, the map Y → X given by y → x is called the inverse
operator of L and it is denoted by L−1
.
Theorem 0.20. Let L : X → Y be linear and have an inverse. Then, L−1
: Y → X is
linear as well.
Proof. Note that the range R(L) of the operator L, i.e., the domain D L−1
of the
inverse operator L−1
, is a linear space. Let y1,y2 ∈ R(L). It suffices to prove the
identity
L−1
(α1y1 +α2y2) = α1L−1
y1 +α2L−1
y2 (0.3)
for all scalars α1,α2. Put Lx1 = y1, Lx2 = y2. We know that
L(α1x1 +α2x2) = α1y1 +α2y2. (0.4)
According to Definition 0.19, we see that L−1
y1 = x1 and L−1
y2 = x2. Thus, we have
α1L−1
y1 +α2L−1
y2 = α1x1 +α2x2.
At the same time, from Definition 0.19 and (0.4), we get
α1x1 +α2x2 = L−1
(α1y1 +α2y2),
which gives (0.3).
Lemma 0.21. Let M be a dense subset of a Banach space Y. Any y = 0, y ∈ Y, can be
expressed in the form
y =
∞
∑
n=1
yn, i.e., y = lim
n→∞
(y1 +y2 +···+yn),
where yn ∈ M and
yn ≤
3 y
2n
, n ∈ N.
Proof. At first, we choose y1 ∈ M in such a way that the inequality
y−y1 ≤
y
2
is valid. Next, we choose y2 ∈ M so that
y−y1 −y2 ≤
y
4
.
In general, we choose yn ∈ M so that
y−y1 −···−yn ≤
y
2n
, n ∈ N.
7
Such a choice is possible, because M is dense in Y.
Now,
y−
m
∑
n=1
yn → 0 as m → ∞,
i.e.,
y =
∞
∑
n=1
yn,
and we have
y1 = y1 −y+y ≤ y1 −y + y ≤
3 y
2
,
y2 = y2 +y1 −y+y−y1 ≤ y−y1 −y2 + y−y1 ≤
3 y
4
,
...
yn = yn +yn−1 +···+y1 −y+y−y1 −···−yn−1
≤ y−y1 −···−yn + y−y1 −···−yn−1 ≤
y
2n
+
y
2n−1
=
3 y
2n
.
Theorem 0.22 (Banach). Let X,Y be Banach spaces and let L : X → Y be a bounded,
bijective, and linear operator. Then, the inverse operator L−1
is bounded as well.
Proof. In Y, we consider the sets Mk ⊆ Y, k ∈ N, of all elements y ∈ Y for which
L−1
y ≤ k y .
All element y ∈ Y belongs to some Mk. Thus,
Y =
∞
k=1
Mk.
According to the Baire theorem, at least one of Mk, say Mn, is dense in some ball B.
Inside B, we consider the set P of all elements z such that β < z − y0 < α, where
0 < β < α, y0 ∈ Mn. We move the set P so that the center is 0, i.e., let us consider the
set
P0 = {z ∈ Y; β < z < α}.
We prove that some MN is dense in P0. If z ∈ P∩Mn, then z−y0 ∈ P0 and
L−1
(z−y0) ≤ L−1
z + L−1
y0
≤ n( z + y0 ) ≤ n( z−y0 +2 y0 )
= n z−y0 1+
2 y0
z−y0
≤ n z−y0 1+
2 y0
β
.
(0.5)
The term
n 1+
2 y0
β
8
does not depend on z. So, we put
N = 1+ n 1+
2 y0
β
.
From (0.5), we have z − y0 ∈ MN. Since Mn is dense in P, we have that MN is dense
in P0.
Let y = 0, y ∈ Y, be arbitrary. We can choose such a number λ that
β < λy < α,
i.e., λy ∈ P0. Since MN is dense in P0, we can construct a sequence of yk ∈ MN which
converges to λy. Then, yk/λ converges to y. It is obvious that {xk/µ}∞
k=1 ⊆ MN for
arbitrary µ = 0 if {xk}∞
k=1 ⊆ MN. Therefore, the set MN is dense in Y {0} and,
consequently, it is dense in Y.
We consider arbitrary y = 0, y ∈ Y. According to Lemma 0.21, we can expand y
into the series y = y1 +y2 +··· , where yk ∈ MN and
yk ≤
3 y
2k
, k ∈ N.
In X, we consider the series L−1
y1 +L−1
y2 +···, where we put xk = L−1
yk. This series
converges to some x ∈ X, because
xk = L−1
yk ≤ N yk ≤ N
3 y
2k
, k ∈ N,
x ≤
∞
∑
k=1
xk ≤ 3N y
∞
∑
k=1
1
2k
= 3N y .
Since the series of xk is convergent and the operator L is continuous, we have
Lx = Lx1 +Lx2 +··· = y1 +y2 +··· = y.
Therefore, x = L−1
y. We also know that
L−1
y = x ≤ 3N y ,
where N does not depend on y. This estimation is valid for arbitrary y = 0. Thus, the
operator L−1
is bounded.
Remark 0.23. Let X,Y be Banach spaces. The symbol ˜L (X,Y) denotes the set of all
bijective, continuous, and linear operators X → Y.
Theorem 0.24. Let X,Y be Banach spaces. Let L0 ∈ ˜L (X,Y) and L ∈ L (X,Y), where
L ≤
1
L−1
0
.
Then, the bounded operator (L0 +L)−1
exists on Y, i.e., L1 = L0 +L ∈ ˜L (X,Y).
Proof. We choose y ∈Y and consider the map B: X → X, Bx = L−1
0 y−L−1
0 (Lx). From
L < L−1
0
−1
, it follows that B is a contraction. Indeed,
Bx1 −Bx2 ≤ L−1
0 · L · x1 −x2 , x1,x2 ∈ X.
9
Since X is complete, according to the Banach theorem, there exists just one x ∈ X such
that
x = Bx = L−1
0 y−L−1
0 (Lx),
i.e.,
L1x = L0x+Lx = y.
If L1 ¯x = y for some ¯x ∈ X, then ¯x is also a fixed point of B, and therefore ¯x = x. Thus, for
all y ∈ Y, there exists only one solution of the equation L1x = y in X, i.e., for L1, there
exists the inverse operator L−1
1 on Y. Considering Theorem 0.22, L−1
1 is bounded.
Remark 0.25. According to the previous theorem, ˜L (X,Y) forms an open subset in
L (X,Y), where X,Y are Banach spaces.
Theorem 0.26 (Neumann). Let X be a Banach space, let I be the identical operator
on X, and let L: X → X be a bounded linear operator, where L ≤ 1. Then, the
operator (I −L)−1
exists on X, it is bounded, and it can be expressed in the form
(I −L)−1
=
∞
∑
k=0
Lk
,
where
Lk
= L◦L◦···◦L
k
.
Proof. The existence on X and the boundedness of (I −L)−1
come from Theorem 0.24
(also from treatments below). Because of L < 1, we have
∞
∑
k=0
Lk
≤
∞
∑
k=0
L k
< ∞. (0.6)
The space X is complete. Thus, considering (0.6), the treated infinite sum of Lk
is a
bounded linear operator. For arbitrary n ∈ N∪{0}, we have
(I −L)
n
∑
k=0
Lk
=
n
∑
k=0
Lk
(I −L) = I −Ln+1
.
Taking into account to Ln+1
≤ L n+1
→ 0 as n → ∞, we get
(I −L)
∞
∑
k=0
Lk
=
∞
∑
k=0
Lk
(I −L) = I,
which yields
(I −L)−1
=
∞
∑
k=0
Lk
.
10
0.4 Adjoint operator
Throughout this section, let X,Y be normed linear spaces.
Definition 0.27. Let L ∈ L (X,Y) and let g ∈ Y , i.e., let g be a continuous linear
functional on Y. Let us consider the continuous linear functional f = g◦L ∈ X and the
map g ∈ Y → f ∈ X . This map L : Y → X is called the adjoint operator of L.
Remark 0.28. We also denote f(x) = f,x . Thus, we can write
f,x = g,Lx , i.e., L g,x = g,Lx .
Remark 0.29. Let L,L1,L2 ∈ L (X,Y) and let k be a scalar. Immediately, from Definition
0.27, we see:
1. L is linear;
2. (L1 +L2) = L1 +L2;
3. (kL) = kL ;
4. L is continuous.
Example 0.30. Let us consider a linear (continuous) operator L: Rn
→ Rm
given by a
matrix (lij). The map y = Lx can be expressed as the system
yi =
n
∑
j=1
lijxj, i ∈ {1,...,m},
and any functional f : Rn
→ R as
f(x) =
n
∑
j=1
fjxj,
where fj = f(ej) for the standard base e1,...,en of Rn
. From
f(x) = g(Lx) =
m
∑
i=1
giyi =
m
∑
i=1
n
∑
j=1
gilijxj =
n
∑
j=1
xj
m
∑
i=1
gilij,
we obtain
fj = f(ej) =
m
∑
i=1
gilij, j ∈ {1,...,n}.
Since f = L g, the operator L is given by the transpose matrix.
Theorem 0.31. If L ∈ L (X,Y), then
L = L .
Proof. Obviously, it holds
L g,x = | g,Lx | ≤ g · Lx ≤ g · L · x
for all x ∈ X and g ∈ Y . Thus,
L g ≤ g · L , i.e., L ≤ L .
11
Now, we prove the opposite inequality. Let x0 ∈ X, Lx0 = 0. We put
y0 =
Lx0
Lx0
∈ Y.
It is seen that y0 = 1. Due to a well-known corollary of the Hahn–Banach theorem,
there exists a functional g such that g = 1 and g,y0 = 1, i.e.,
g,Lx0 = Lx0 .
From
Lx0 = g,Lx0 = | L g,x0 | ≤ L g · x0
≤ L · g · x0 = L · x0 ,
we get that L ≤ L .
Let H be a Hilbert space and let L : H → H be a bounded linear operator. We know
that there exists a map τ which assigns to any element y ∈ H the continuous linear
functional (τy)(x) = x,y ∈ H . Moreover, this map is an isometry. For the operator
L , we consider the map ˜L = τ−1
L τ, which is a bounded linear operator on H. One
can easily show that
Lx,y = x, ˜L y , x,y ∈ H.
Since L = L and the maps τ and τ−1
are isometries, we have the identity
˜L = L .
Definition 0.32. In a Hilbert space H, the above mentioned operator ˜L : H → H is
called the adjoint operator of L: H → H.
Remark 0.33. It should be emphasized that Definition 0.32 differs from Definition 0.27.
For a general Banach space X and a bounded linear operator L: X → X, the adjoint
operator of L is defined on X .
The operator ˜L is sometimes called the Hermitian adjoint. We write only L (instead
of ˜L ) and speak about the adjoint operator of L. It should be remembered that,
in Hilbert spaces, the concept of adjoint operators differs from the one in general Banach
spaces. For H, it is seen that the adjoint operator of a bounded linear operator
L: H → H can be defined as the operator L : H → H which satisfies
Lx,y = x,L y , x,y ∈ H.
Definition 0.34. Let H be a Hilbert space. A bounded linear operator L: H → H is
called self-adjoint if
Lx,y = x,Ly , x,y ∈ H.
Definition 0.35. Let H be a Hilbert space and let L: H → H be a linear operator.
A (closed) subspace H1 of H is called invariant with respect to L: H → H if x ∈ H1
implies Lx ∈ H1.
Remark 0.36. Let H be a Hilbert space and let L: H → H be a bounded linear operator.
If H1 is a (closed) subspace of H, which is invariant with respect to L, then its
orthogonal complement H⊥
1 is invariant with respect to L . Indeed, if y ∈ H⊥
1 , then
x,L y = Lx,y = 0, x ∈ H1,
because Lx ∈ H1. Especially, if L is self-adjoint, then the orthogonal complement of
any invariant subspace is invariant with respect to L as well.
12
0.5 Spectrum of operator
Let L: Cn
→ Cn
be a linear operator. A number λ ∈ C is called an eigenvalue of L if
Lx = λx for some non-zero x ∈ Cn
. Any such solution x is called an eigenvector of L.
The set of all eigenvalues is called the spectrum of the operator L and all other values λ
are called regular; i.e., λ is a regular value if the operator L − λI has an inverse. In
this case, the operator (L−λI)−1
is defined on the entire space Cn
and (as well as any
linear operator on a space with a finite dimension) it is bounded (continuous).
In the space with a finite dimension, there are two possibilities:
1. the equation Lx = λx has a non-zero solution, i.e., λ is an eigenvalue of L—the
operator (L−λI)−1
does not exist;
2. the bounded operator (L−λI)−1
is defined in the whole space, i.e., λ is a regular
value.
If the operator L is defined on a space whose dimension is infinite, then we have the
third possibility:
3. the operator (L−λI)−1
exists, i.e., the equation Lx = λx has only the zero solution,
but this operator is not defined in the whole space (and it is not necessarily
bounded).
Let X be a complex Banach space.
Definition 0.37. A number λ ∈ C is called a regular value of a bounded linear operator
L: X → X if the operator Rλ = (L−λI)−1
, called the resolvent of L, is defined in the
whole space X. The set of all non-regular values is called the spectrum of L and it is
denoted by σ(L). The spectrum includes all eigenvalues of the operator L. Indeed, if
(L−λI)x = 0 for some x = 0, then (L−λI)−1
does not exist. The set of all eigenvalues
of L is called the point (or discrete) spectrum and the corresponding x = 0 are called
eigenvectors. The remaining part of the spectrum, i.e., the set of all λ, for which the
inverse operator (L−λI)−1
exists, but is not defined in the whole space X, is called the
continuous spectrum.
Theorem 0.38. The set of all regular values of a bounded linear operator L: X → X
is open, i.e., the spectrum is a closed set.
Proof. Let λ be a regular value of L. Then,
L−λI ∈ ˜L (X) = ˜L (X,X).
Let δ ∈ C satisfy
|δ| <
1
||(L−λI)−1
||
.
From Theorem 0.24, we have L − (λ +δ)I ∈ ˜L (X). Hence, λ + δ is a regular value
of L and the set of all regular values is open.
Theorem 0.39. Let L: X → X be a bounded linear operator and let |λ| > L . Then,
λ is a regular value of L.
13
Proof. Because of L < |λ|, applying Theorem 0.26, we know that the operator
Rλ = (L−λI)−1
= −
1
λ
I −
L
λ
−1
exists in X and it is bounded. Therefore, λ is a regular value.
Remark 0.40. Theorem 0.39 can be specified as follows. Let
r = lim
n→∞
n
Ln ,
where it is possible to show that this limit exists for any bounded linear operator
L: X → X. The spectrum of L is in the closed circle with the radius r and the centre
in 0. The value r is called the spectral radius of L and r ≤ L .
For example, the operator L: C[0,1] → C[0,1] (we consider the norm from (0.1))
given by
L: f(t) → t
1
0
f(x)dx, t ∈ [0,1], f ∈ C[0,1],
has the norm L = 1 and
r = lim
n→∞
n
||Ln|| =
1
2
.
We add that, for any bounded linear operator L: X → X, it holds
r = sup{|λ|; λ ∈ σ(L)}.
Example 0.41. We define the operator T : C[0,1] → C[0,1] by the formula
T : f(t) → f t2
, t ∈ [0,1], f ∈ C[0,1],
where we consider the norm from (0.1). At first, we determine T , which is easy. We
have
T = sup{ T f ; f ≤ 1} ≤ 1.
For f0 ≡ 1, we obtain f0 = 1 and also T f0 = 1. Thus, T = 1.
We find eigenvalues of T. We find such λ that the equation T f = λ f has non-zero
solutions. We know that
σ(T) ⊆ {z ∈ C; |z| ≤ T = 1}.
For λ = 0 and T f = λ f = 0, we have f ≡ 0 and, therefore, 0 is not an eigenvalue. For
λ = 1 and T f = λ f = f, all constant functions are solutions of the equation
f t2
= f(t), t ∈ [0,1].
It remains to investigate such λ that |λ| ∈ (0,1],λ = 1. From
T f(t) = f t2
= λ f(t), t ∈ [0,1],
we obtain
f(t) =
1
λ
f t2
=
1
λ2
f t4
= ··· =
1
λn
f t2n
14
for all t ∈ [0,1] and n ∈ N. We see that
f(t) = λn
f t2−n
.
If |λ| < 1, then f ≡ 0, because λn
→ 0 as n → ∞ and f is bounded on [0,1].
Let us consider the last case when |λ| = 1 and λ = 1. In this case, we get
f(0) = λ f(0), f(1) = λ f(1)
and, consequently, f(0) = f(1) = 0. Let t ∈ (0,1). From
f(t) =
1
λn
f t2n
and t2n
→ 0 as n → ∞, it follows that f ≡ 0. Therefore, the set of all eigenvalues is {1}.
It remains to find out other values belonging to the spectrum of T. Thus, we analyse
solutions of the equation T f −λ f = g in C[0,1]. We know, when the operator T −λI
is not injective. So, we are interested in the case, when it is not surjective. For t ∈ [0,1],
we obtain
f t2
= g(t)+λ f(t)
and
f(t) = g t1/2
+λ f t1/2
= g t1/2
+λg t1/4
+λ2
f t1/4
.
By induction, one can obtain
f(t) =
n−1
∑
j=0
λ j
g t2−1−j
+λn
f t2−n
, t ∈ [0,1].
Because of the boundedness of g for |λ| < 1, the series above converges and
λn
f t2−n
→ 0 as n → ∞
for all f. Therefore, the continuous function
f(t) =
∞
∑
j=0
λ j
g t2−1−j
, t ∈ [0,1],
is a solution of the equation T f −λ f = g. If |λ| = 1,λ = 1, then the equation
T f −λ f = g
has no solution for all g ∈ C[0,1]. For example, for λ = −1, there exists a continuous
function g ∈ C[0,1] such that
g t2−1−j
=
(−1)j
j
for given t ∈ (0,1), j ∈ N.
Altogether, we have
σ(T) = {λ ∈ C; |λ| = 1}.
15
Example 0.42. Now, on the space C[0,1] with the norm from (0.1), we consider the
operator
˜T : f(t) →
t
0
f(x)dx, t ∈ [0,1], f ∈ C[0,1].
Its norm is ˜T = 1.
Let us identify the point spectrum of the operator ˜T. We know that λ is an eigenvalue
if there exists a non-zero solution of the equation ˜T f = λ f. So, we are looking
for a non-trivial continuous function f ∈ C[0,1] with the property that
t
0
f(x)dx = λ f(t), t ∈ [0,1].
We see that λ f(0) = 0 and that the function f has to have a continuous derivative if
λ = 0. If λ = 0, then
t
0
f(x)dx = 0, t ∈ [0,1], i.e., f ≡ 0.
Therefore, 0 is not an eigenvalue of ˜T. If λ = 0, from the mentioned identity, we obtain
f(t) = λ f (t), t ∈ [0,1].
Solutions of this equation are functions f(t) = Ket/λ
. Since f(0) = 0, we get K = 0.
Thus, the operator ˜T has no eigenvalues.
Now, we determine the whole spectrum of ˜T. We need to find the values λ for
which the considered operator is surjective, i.e., we need to determine when the equation
˜T f − λ f = g has solutions for all g ∈ C[0,1]. Let λ = 0. Since ˜T f(0) = 0, for
a function g such that g(0) = 0, any solution does not exist. Hence, 0 ∈ σ( ˜T). In the
case when λ = 0, we are looking for solutions of the equation
t
0
f(x)dx−λ f(t) = g(t), t ∈ [0,1],
where g ∈ C[0,1] is a given function. Let h be the appropriate primitive function of f.
The aim is to solve the differential equation h−λh = g. Of course, this equation has
solutions. Thus, we get
σ ˜T = {0}.
Example 0.43. On the space l∞
, we consider the operator
R: {x1,x2,x3, ...} → {x2,x3, ...}.
It is easy to verify that R ∈ L (l∞
) and that R = 1. We know that
σ(R) ⊆ {λ ∈ C; |λ| ≤ 1}.
If |λ| ≤ 1, the equation Rx = λx has the non-trivial solution
xλ = 1,λ,λ2
, ... .
16
Since xλ ∈ l∞
for |λ| ≤ 1, we have σ(R) = {λ ∈ C; |λ| ≤ 1}.
Let us consider the same operator R, but on l1
. Of course, R ∈ L l1
and R = 1.
The equation Rx = λx has (again) the non-trivial solution xλ . But, for |λ| = 1, this
element is not in l1
. However, xλ ∈ l1
for all λ satisfying |λ| < 1. Therefore, we get
that all such λ are in the point spectrum. Since the spectrum σ(R) is a closed set which
contains the point spectrum, we have again
σ(R) = {λ ∈ C; |λ| ≤ 1}.
17
1. Completely continuous
operators
1.1 Preliminaries and examples
Throughout this section, let X,Y be Banach spaces.
Definition 1.1. An operator L: X → Y is called completely continuous if it maps any
bounded set into a precompact set.
Remark 1.2. If X has a finite dimension, then any linear operator L: X → Y is completely
continuous (see also Theorem 0.16). For spaces whose dimension is infinite,
the complete continuity differs from the continuity.
Theorem 1.3. Let x1,x2, ... be linearly independent vectors in X a let Xn be the subspace
of X generated by x1,...,xn. Then, there exists a sequence {yn}∞
n=1 such that
yn = 1, yn ∈ Xn, n ∈ N,
and
inf
x∈Xn−1
yn −x >
1
2
, n ≥ 2, n ∈ N.
Proof. Since x1,x2, ... are linearly independent, xn /∈ Xn−1 and the distance between
xn and Xn−1 is positive. Let us denote it by α and let x∗
be an element of Xn−1 for
which xn −x∗
< 2α. Then,
yn =
xn −x∗
xn −x∗
, n ≥ 2, n ∈ N,
because
α = inf
x∈Xn−1
xn −x = inf
x∈Xn−1
xn −x∗
−x xn −x∗
.
We add that we can easily put
y1 =
x1
x1
.
Example 1.4. Let the dimension of X be infinite and let us consider the identical operator
I on X. Using Theorem 1.3, in B[0,1], one can construct a sequence {yn}∞
n=1 such
that
yi −yn >
1
2
, i ∈ {1,2,...,n−1}, n ≥ 2, n ∈ N.
Obviously, such a sequence cannot have a convergent subsequence. Therefore, B[0,1]
is not (pre)compact and I is not completely continuous.
Example 1.5. Let L be a continuous linear operator which maps X into a subspace of X
with a finite dimension. The operator L is evidently completely continuous. Especially,
in a Hilbert space, the (orthogonal) projection is completely continuous if and only if
the considered subspace has a finite dimension. Note that an operator, which maps X
into a subspace of X having a finite dimension, is called degenerate.
18
Example 1.6. In the space l2
, we consider the operator L: l2
→ l2
given by
x = {x1,x2,...,xn, ...} → Lx = x1,
x2
2
,...,
xn
2n−1
, ... .
This operator is completely continuous. It suffices to consider that the image of the
unit ball is precompact and to use the linearity.
Example 1.7. We consider C[a,b] with the norm
f = max
t∈[a,b]
|f(t)|, f ∈ C[a,b],
and the operator L: C[a,b] → C[a,b] defined by
Lx = y(s) =
b
a
k(s,t)x(t)dt, x ∈ C[a,b], s ∈ [a,b]. (1.1)
It is possible to prove the following implication. If k is bounded for s ∈ [a,b], t ∈
[a,b] and all points of the discontinuity of k are on finitely many curves t = ϕk(s),
k = {1,...,n}, where ϕk are continuous functions on [a,b], then the operator L given
by (1.1) is completely continuous. We remark that this operator L is called the Fredholm
operator.
The requirement, that the points of the discontinuity of k are only on finitely many
curves which intersect the lines s = const. in only one point, is essential. For example,
for the function
k(s,t) =



1, s <
1
2
;
0, s ≥
1
2
,
the operator L maps x ≡ 1 into a discontinuous function.
We prove the complete continuity of the operator L only in the case, when the
function k is continuous on [a,b] × [a,b]. It is easy to see that Lx is defined correctly,
Lx ∈ C[a,b], and that L is a linear and bounded operator (see also Example 0.5). We
consider B[0,1] ⊆ C[a,b]. It suffices to show that the set L(B[0,1]) is precompact. We
apply the Arzel`a–Ascoli theorem. Of course, L(B[0,1]) is a bounded set, because L
is a bounded operator. It remains to show that L(B[0,1]) is a set of equicontinuous
functions. For an arbitrarily given ε > 0, there exists δ > 0 such that
|k(s1,t)−k(s2,t)| < ε
if
t ∈ [a,b], |s1 −s2| < δ, s1,s2 ∈ [a,b].
Then,
|Lx(s1)−Lx(s2)| ≤ (b−a) max
t∈[a,b]
|k(s1,t)−k(s2,t)|· x ≤ ε(b−a)
for all x ∈ B[0,1] and s1,s2 ∈ [a,b] satisfying |s1 −s2| < δ.
If we put k(s,t) = 0 for t > s, then L takes the form
Lx = y(s) =
s
a
k(s,t)x(t)dt, x ∈ C[a,b], s ∈ [a,b]. (1.2)
19
If the function k is continuous, then the operator defined in (1.2) is completely continuous.
This operator is called the Volterra operator.
Remark 1.8. For a completely continuous operator, the image of the closed unit ball
B[0,1] does not need to be compact, although it is precompact. As at the end of Example
1.7, on C[−1,1] with the uniform norm, we consider the completely continuous
operator
Jx(s) =
s
−1
x(t)dt, s ∈ [−1,1], x ∈ C[−1,1].
For n ∈ N, we put
xn(t) =



0, −1 ≤ t ≤ 0;
nt, 0 < t ≤
1
n
;
1,
1
n
< t ≤ 1.
For n ∈ N, it is seen that xn ∈ C[−1,1], xn = 1, and that
yn(s) = Jxn(s) =



0, −1 ≤ s ≤ 0;
ns2
2
, 0 < s ≤
1
n
;
s−
1
2n
,
1
n
< s ≤ 1.
We immediately see that, in C[−1,1], {yn}∞
n=1 converges to
y(s) =
0, −1 ≤ s ≤ 0;
s, 0 < s ≤ 1.
But, for the operator J, the function y is not the image of any function from C[−1,1],
because y is not continuous. However, it is possible to prove that, for any completely
continuous linear operator, the image of B[0,1] is compact if the considered space is
reflexive.
1.2 Basic properties
Throughout this section, let X be a Banach space.
Theorem 1.9. If {Ln}∞
n=1 is a sequence of completely continuous operators on X,
which converges to an operator L: X → X, i.e., Ln − L → 0 as n → ∞, then L is
completely continuous as well.
Proof. It is sufficient to prove that, for an arbitrarily given bounded sequence {xk}∞
k=1 ⊂
X, one can extract a convergent subsequence from {Lxk}∞
k=1.
Since the operator L1 is completely continuous, {L1xk}∞
k=1 has a convergent subsequence.
Let x1
k
∞
k=1
be a subsequence of {xk}∞
k=1 such that L1x1
k
∞
k=1
converges.
Now, let us consider L2x1
k
∞
k=1
. From this sequence, we can extract a convergent
subsequence as well. Let x2
k
∞
k=1
be a subsequence of x1
k
∞
k=1
such that L2x2
k
∞
k=1
converges. We proceed in the same way. An the end, we consider the diagonal sequence
xk
k
∞
k=1
. Any of the operators L1,L2,L3, ... transforms this sequence into a
20
convergent one. We show that L also transforms it into a convergent sequence. Since
X is complete, it suffices to show that Lxk
k
∞
k=1
satisfies the Cauchy criterion. For
n,k,l ∈ N, it holds
Lxk
k −Lxl
l ≤ Lxk
k −Lnxk
k + Lnxk
k −Lnxl
l + Lnxl
l −Lxl
l . (1.3)
Let ε > 0 be given and c > 0 be such that xk ≤ c, k ∈ N. Let n ∈ N be such that
L−Ln <
ε
3c
.
Then, we consider so large N ∈ N that
Lnxk
k −Lnxl
l <
ε
3
for all k,l > N, k,l ∈ N. Now, from (1.3), it follows that
Lxk
k −Lxl
l < ε
for all sufficiently large k,l.
Remark 1.10. Since any linear combination of completely continuous operators is a
completely continuous operator, from Theorem 1.9, we know that completely continuous
linear operators form a closed subspace of L (X).
Theorem 1.11. Let L1,L2 ∈ L (X) and let L1 be completely continuous. Then, the
operators L1 ◦L2 and L2 ◦L1 are completely continuous as well.
Proof. If a set M ⊆ X is bounded, then
L2(M) = {y ∈ X; y = L2x,x ∈ M}
is bounded as well. Thus, the set L1(L2(M)) is precompact and L1 ◦ L2 is completely
continuous. If M ⊆ X is bounded, then L1(M) is precompact. Since L2 is continuous,
L2(L1(M)) is precompact and L2 ◦L1 is completely continuous.
Corollary 1.12. In the space X whose dimension is infinite, any linear completely
continuous operator L: X → X does not have a bounded inverse L−1
.
Proof. It is enough to consider Theorem 1.11 and the identical operator I = L ◦ L−1
(see Example 1.4).
Theorem 1.13 (Schauder). The adjoint operator of a completely continuous operator
L ∈ L (X) is completely continuous as well.
Proof. We want to prove that L : X → X maps any ball into a precompact set. Due to
the linearity of L , it suffices to show that the image L B of the closed unit ball with
the centre in 0 ∈ X is precompact. We point out that B is the unit ball in X . Elements
of X can be considered as functions defined on L(B[0,1]). We show that the set Φ of
all functions assigned to the functionals belonging to the ball B is a set of uniformly
bounded and equicontinuous functions. If a functional ϕ ∈ X satisfies ϕ ≤ 1, then
sup
x∈L(B[0,1])
|ϕ(x)| = sup
x∈L(B[0,1])
|ϕ(x)| ≤ ϕ sup
x∈B[0,1]
Lx ≤ L
21
and
ϕ(x )−ϕ(x ) ≤ ϕ · x −x ≤ x −x .
Thus (consider the Arzel`a–Ascoli theorem), the set Φ is precompact in
U = C L(B[0,1]) .
But, the set Φ with the metric of the uniform convergence is isometric with the set
L B with the metric given by the norm of X , because, for g1,g2 ∈ B , we have
L g1 −L g2 = sup
x∈B[0,1]
L g1 −L g2,x = sup
x∈B[0,1]
| g1 −g2,Lx |
= sup
z∈L(B[0,1])
| g1 −g2,z | = sup
z∈L(B[0,1])
| g1 −g2,z | = g1 −g2 .
Since Φ is precompact, it is totally bounded. Hence, the set L B , which is isometric
with it, is totally bounded as well. Therefore, the set L B is precompact in X .
Remark 1.14. One can prove that the set Φ (from the proof of Theorem 1.13) is closed
in U. Hence, Φ is compact and, consequently, the set L B is compact. By Remark
1.8, for a completely continuous linear operator, the image of the closed unit ball
does not need to be bounded. But, for any completely continuous linear operator on X ,
the image of the set B is compact.
Theorem 1.15. Let L ∈ L (X) be a completely continuous operator. For arbitrary
δ > 0, there exists only a finite number of linearly independent eigenvectors associated
with eigenvalues of L whose absolute values are greater than δ.
Proof. By contradiction, let us consider a sequence λ1,λ2,...λn, ... of eigenvalues
of L such that |λn| > δ, n ∈ N, and a sequence x1,x2,...xn, ... of associated linearly
independent eigenvectors. According to Theorem 1.3, we can construct a sequence
y1,y2,...yn, ... such that
yn = 1, yn ∈ Xn, n ∈ N,
and that
inf
x∈Xn−1
yn −x >
1
2
, n ≥ 2, n ∈ N,
where Xn is the subspace generated by x1,...,xn. The sequence {yn/λn}∞
n=1 is bounded,
because |λn| > δ, n ∈ N. Now, we prove that, from the sequence of the images
{L(yn/λn)}∞
n=1, one cannot choose a convergent subsequence. If
yn =
n
∑
k=1
αkxk,
then
L
yn
λn
=
n−1
∑
k=1
αkλk
λn
xk +αnxn = yn +zn,
where
zn =
n−1
∑
k=1
αk
λk
λn
−1 xk ∈ Xn−1.
22
Therefore, for any p,q ∈ N, q < p, it holds
L
yp
λp
−L
yq
λq
= yp +zp −(yq +zq) = yp −(yq +zq −zp) >
1
2
,
because yq +zq −zp ∈ Xp−1. We have a contradiction.
Remark 1.16. Especially, from Theorem 1.15, it follows that the number of linearly
independent eigenvectors associated with an eigenvalue λ = 0 of a completely continuous
linear operator is finite.
Remark 1.17. Let the dimension of X be infinite. For any completely continuous
operator L ∈ L (X), we see that 0 ∈ σ(L). Indeed, if 0 /∈ σ(L), then L−1
is defined
on X, which is a contradiction with Theorem 1.11 (Corollary 1.12). In the case of a
linear completely continuous operator, the spectrum is non-empty. Using the so-called
Fredholm alternative, one can prove that σ(L) of a completely continuous operator
L ∈ L (X) can contain only eigenvalues and 0. Thus, the spectrum of a completely
continuous linear operator has a very simple structure. In particular, we recall that it is
a closed set.
Example 1.18. We consider the following series of examples.
a) Let us consider the operator T : C[0,1] → C[0,1] given by
T : f(t) → f t2
, t ∈ [0,1], f ∈ C[0,1].
See Example 0.41. Especially, we consider the norm from (0.1). The operator T
is not completely continuous. The spectrum
σ(T) = {λ ∈ C; |λ| = 1}
is an uncountable set. In the case of a space whose dimension is infinite, we
also know that 0 ∈ σ(L) for any completely continuous linear operator L. We
can also use directly Definition 1.1 and consider the sequence {tn
}∞
n=1 ⊆ C[0,1].
From the sequence
{Ttn
}∞
n=1 = t2n ∞
n=1
,
one cannot choose a convergent subsequence.
b) Let us consider the operator ˜T : C[0,1] → C[0,1] given by
˜T : f(t) →
t
0
f(x)dx, t ∈ [0,1], f ∈ C[0,1].
See Example 0.42. Especially, we consider the norm from (0.1). We show that ˜T
is completely continuous. We consider an arbitrary bounded sequence { fn}∞
n=1 ⊆
C[0,1] and, using the Arzel`a–Ascoli theorem, we prove that ˜T fn
∞
n=1
is precompact.
Let K > 0 be such that
fn ≤ K, n ∈ N.
We have
˜T fn = max
t∈[0,1]
t
0
fn(x)dx = max
t∈[0,1]
t
0
|fn(x)| dx ≤
1
0
K dx = K, n ∈ N.
23
At the same time, for t,s ∈ [0,1], n ∈ N, we have
˜T fn(t)− ˜T fn(s) ≤
s
t
|fn(x)| dx ≤ K |t −s|.
Thus, the sequence ˜T fn
∞
n=1
is uniformly bounded and equicontinuous.
We remark that if we use the complete continuity of ˜T, then it is easy to determine
the spectrum σ( ˜T) = {0} (cf. Example 0.42; see Remark 1.17).
c) We consider L1,L2 ∈ L l2
from Example 0.18, f). For these operators, we
determine σ(L1) and σ(L2) and we decide whether L1,L2 are completely continuous.
We recall that
L1x = 0,x1,
x2
2
,
x3
3
, ... , x = {xn}∞
n=1 ∈ l2
,
L2x = {0,x1,x2,x3, ...}, x = {xn}∞
n=1 ∈ l2
.
At first, we analyse L1. For a constant λ, we want to find x = {xn}∞
n=1 ∈ l2
such
that L1x = λx. We get
0,x1,
x2
2
,
x3
3
, ... = {λx1,λx2,λx3, ...}.
For λ = 0, we have x1 = 0, x2 = 0, x3 = 0, ...; and, for λ = 0, we also have
x1 = x2 = x3 = ··· = 0.
The equation L1x = λx does not have any non-zero solution for any λ. Therefore,
any complex number is not an eigenvalue of the operator L1.
We show that L1 is completely continuous. For all k ∈ N, we define Lk
1 : l2
→ l2
by
Lk
1 : x = {xn}∞
n=1 → 0,x1,
x2
2
,...,
xk
k
,0,0, ... .
Any of the operators Lk
1 is linear, bounded, with a finite dimension of its range.
Therefore, they are completely continuous. Since, for all k ∈ N and x = {xn}∞
n=1 ∈
l2
, x ≤ 1, it holds
Lk
1x−L1x
2
=
∞
∑
n=k+1
xn
n
2
≤
1
k2
x 2
≤
1
k2
,
we have
Lk
1 −L1
2
≤
1
k2
.
Hence, the completely continuous operators Lk
1 converge to L1 (as k → ∞), which
proves that L1 is completely continuous (see Theorem 1.9). From Remark 1.17,
we get that σ(L1) = {0}.
Now, we prove that any λ satisfying |λ| ≤ 1 lies in σ(L2). In this case, the
operator L2 − λI does not map l2
into l2
, because there is no z = {zn}∞
n=1 ∈ l2
such that
(L2 −λI)z = {−λz1,z1 −λz2,z2 −λz3, ...} = {−1,0,0, ...}.
24
If we exclude the trivial case λ = 0, we have
z =
1
λ
,
1
λ2
,
1
λ3
, ... .
Such z (for |λ| ≤ 1) is not in l2
. Therefore, the operator L2 is not completely
continuous. It follows from the fact that its spectrum is the uncountable set
{λ ∈ C; |λ| ≤ 1}.
d) On the space C[0,1] with the uniform norm, we consider the operator given by
F : f(t) → t2
f(0), t ∈ [0,1], f ∈ C[0,1].
This operator is completely continuous. Obviously, it is bounded, linear, and its
range is a one-dimensional subspace of C[0,1]. The estimation
|F f(t)| = t2
f(0) ≤ |f(0)| ≤ f , t ∈ [0,1], f ∈ C[0,1],
gives F ≤ 1. Then, the choice f ≡ 1 shows that F = 1.
The complete continuity of the operator F can be proved also directly. Let
{fn}∞
n=1 ⊂ B[0,1] ⊂ C[0,1]. From the estimations
F fn ≤ F · fn ≤ fn ≤ 1, n ∈ N,
|F fn(t)−F fn(s)| ≤ t2
−s2
fn(0)
≤ |(t −s)(t +s)| ≤ 2|t −s|, n ∈ N, t,s ∈ [0,1],
and from the Arzel`a–Ascoli theorem, it follows the complete continuity of F.
Now, we identify eigenvalues of the operator F. We look for non-trivial solutions
of the equation
t2
f(0) = λ f(t).
For λ = 0, e.g., f(t) = t is a solution. Therefore, 0 is an eigenvalue. For λ = 0,
we obtain
f(t) =
1
λ
f(0)t2
.
Hence, f(0) = 0 and, consequently, f ≡ 0. Since F is completely continuous,
σ(F) = {0}.
e) On the complex space l2
, we define the operator R: l2
→ l2
by
Rx = {in
xn}∞
n=1, x = {xn}∞
n=1 ∈ l2
,
where i is the imaginary unit. For x = {xn}∞
n=1 ∈ l2
, we have
Rx 2
=
∞
∑
n=1
|xn|2
= x 2
,
which gives R = 1. If λ is an eigenvalue, then there exists a non-zero element
x = {xn}∞
n=1 ∈ l2
such that Rx = λx, i.e., in
xn = λxn, n ∈ N. It is seen that the
eigenvalues are i,i2
,i3
,i4
, i.e., i,−1,−i,1. For example, for i, the corresponding
25
eigenvector is {1,0,0, ...}. For λ /∈ {i,−1,−i,1}, the operator R − λI has an
inverse. The inverse is
Sx =
1
in −λ
xn
∞
n=1
, x = {xn}∞
n=1 ∈ l2
,
where
Sx ≤ k x
for
k = max
1
|i−λ|
,
1
|i2 −λ|
,
1
|i3 −λ|
,
1
|i4 −λ|
.
Note that the operator R is not completely continuous, because 0 /∈ σ(R) (the
dimension of l2
is infinite).
f) Let the operator T1 ∈ L L2
[0,1] be defined by
T1 f(x) = x·
1
0
f(t)dt, x ∈ [0,1], f ∈ L2
[0,1].
For f ∈ L2
[0,1], we have
T1 f = x·
1
0
f(t)dt =



1
0
x·
1
0
f(t)dt
2
dx



1
2
=
1
0
f(t)dt ·


1
0
x2
dx


1
2
=
1
√
3
1
0
f(t)dt
≤
1
√
3


1
0
|f(t)|2
dt


1
2


1
0
1dt


1
2
=
1
√
3
f .
Therefore, T1 ≤ 1/
√
3. For f ≡ 1, we see that f = 1, T1 f = 1/
√
3. Thus,
T1 =
1
√
3
.
We determine the discrete spectrum of T1. We want to find a non-trivial function
f such that T1 f(x) = λ f(x), x ∈ [0,1]. For λ = 0, it suffices to consider an
arbitrary identically non-zero function f ∈ C[0,1] for which
1
0
f(t)dt = 0.
For example, f(x) = sin(2πx). For λ = 0 and
T1 f(x) = x
1
0
f(t)dt = λ f(x),
26
we see that f has to be linear, i.e., f(x) = kx. Then,
x
1
0
kt dt = λkx, x ∈ [0,1],
i.e., λ = 1/2. Thus, the point spectrum is {0,1/2}. Obviously, the operator T1
is linear and continuous and its range has a finite dimension. Therefore, it is
completely continuous and σ (T1) = {0,1/2}.
Let the operator T2 ∈ L L2
[−1,1] be defined by
T2 : f(x) →
1
−1
x2
t f(t)dt, x ∈ [−1,1], f ∈ L2
[−1,1].
For any f ∈ L2
[−1,1], we have
T2 f 2
=
1
−1
|T2 f(x)|2
dx =
1
−1
x4


1
−1
t f(t)dt


2
dx
=
2
5


1
−1
t f(t)dt


2
≤
2
5
t 2
· f 2
=
4
15
f 2
.
For f(x) = x, we have
f =
2
3
, T2 f 2
=
2
5
2
3
2
.
Thus,
T2 =
2
√
15
.
Let us find eigenvalues of T2. We look for non-trivial solutions of the equation
(T2 −λI)f = 0. From the identity
x2
1
−1
t f(t)dt = λ f(x),
we see that the function f has to be a multiple of the function x2
, i.e., f(x) = kx2
.
From
x2
1
−1
tkt2
dt = λkx2
,
it follows
λ =
1
−1
t3
dt = 0.
The point spectrum is {0}. Since T2 is completely continuous, σ(T2) = {0}.
27
Let the operator T3 ∈ L L2
[0,1] be defined by
T3 f(x) = x· f(x), x ∈ [0,1], f ∈ L2
[0,1].
Since
T3 f =


1
0
|x f(x)|2
dx


1
2
≤


1
0
|f(x)|2
dx


1
2
= f , f ∈ L2
[0,1],
we have the inequality T3 ≤ 1. Let us determine the discrete spectrum. We
look for non-trivial solutions of the equation (T3 −λI) f = 0. As a solution of
the equation (x − λ)f(x) = 0, x ∈ [0,1], we get only f = 0. Thus, the discrete
spectrum is empty. Now, we consider the continuous spectrum. We consider a
function g ∈ L2
[0,1] and find λ for which the equation (T3 −λI)f = g has solutions
in L2
[0,1]. If λ = 0, then the equation x f(x) = g(x) has only the solution
f(x) =
g(x)
x
.
For example, for g(x) =
√
x ∈ L2
[0,1], we have f(x) = 1/
√
x /∈ L2
[0,1]. Thus,
0 ∈ σ(T3). In the case when λ = 0, we get the solution
f(x) =
g(x)
x−λ
, x ∈ [0,1].
For λ ∈ (0,1] and g ≡ 1, we obtain that f /∈ L2
[0,1]. For others λ ∈ C, i.e.,
λ /∈ [0,1], the function 1/(x−λ) is continuous on [0,1]. Therefore, the equation
T3 f −λ f = g has a solution in L2
[0,1] for any function g ∈ L2
[0,1]. The spectrum
is σ (T3) = [0,1]. The operator T3 cannot be completely continuous, because its
spectrum is an uncountable set. Moreover, T3 = 1 (see Theorem 0.39).
1.3 Self-adjoint operator in Hilbert space
For self-adjoint linear operators in Hilbert spaces with finite dimensions, we have the
well-known theorem about the existence of an operator matrix in the diagonal form.
Now, we extend this theorem to completely continuous self-adjoint operators in Hilbert
spaces. Let H be a Hilbert space.
Theorem 1.19. All eigenvalues λ of a self-adjoint operator L: H → H are real.
Proof. Let Lx = λx for some non-zero x ∈ H. Then,
λ x,x = λx,x = Lx,x = x,Lx = x,λx = ¯λ x,x .
We see that λ = ¯λ.
Theorem 1.20. Eigenvectors of a self-adjoint operator L: H → H, which correspond
to different eigenvalues, are orthogonal.
Proof. If Lx = λx and Ly = µy for λ = µ, then
λ x,y = λx,y = Lx,y = x,Ly = x,µy = µ x,y ,
which gives x,y = 0.
28
Lemma 1.21. If a sequence {ξn}∞
n=1 ⊆ H and ξ ∈ H satisfy
sup
n∈N
ξn < +∞
and
L(ξn −ξ) → 0 as n → ∞
for a self-adjoint operator L: H → H, then
Q(ξn) = Lξn,ξn → Lξ,ξ = Q(ξ) as n → ∞.
Proof. For all n ∈ N, we have
| Lξn,ξn − Lξ,ξ | ≤ | Lξn,ξn − Lξ,ξn |+| ξ,Lξn − ξ,Lξ |
together with
| Lξn,ξn − Lξ,ξn | = | L(ξn −ξ),ξn | ≤ ξ · L(ξn −ξ)
and
| ξ,Lξn − ξ,Lξ | = | ξ,L(ξn −ξ) | ≤ ξ · L(ξn −ξ) .
Since the set of all numbers ξn for n ∈ N is bounded and L(ξn −ξ) → 0 as n → ∞,
we get
| Lξn,ξn − Lξ,ξ | → 0 as n → ∞.
Lemma 1.22. If the functional
ξ → |Q(ξ)| = | Lξ,ξ |, ξ ∈ H,
where L: H → H is a self-adjoint operator, assumes a maximum on B[0,1] ⊆ H at an
element ξ0, then
ξ0,η = 0
implies that
Lξ0,η = ξ0,Lη = 0.
Proof. Obviously, ξ0 = 1. Let η = 0 and ξ0,η = 0. We set
ξ =
ξ0 +aη
1+|a|2 η 2
,
where |a| > 0 is a sufficiently small number. From ξ0 = 1, we get ξ = 1. It holds
Q(ξ) =
1
1+|a|2 η 2
Q(ξ0)+ ¯a Lξ0,η +a Lξ0,η +|a|2
Q(η)
=
1
1+|a|2 η 2
Q(ξ0)+2 ¯a Lξ0,η +|a|2
Q(η)
for such a number a that ¯a Lξ0,η is real. From the expression above, we see the
following implication. If Lξ0,η = 0, then (consider |a| ≈ 0+
)
|Q(ξ)| > |Q(ξ0)|,
which is a contradiction.
29
Remark 1.23. From Lemma 1.22, it follows that ξ0 is an eigenvector of the operator L
if the functional |Q(ξ)| assumes a maximum at ξ = ξ0.
Theorem 1.24 (Hilbert–Schmidt). For every completely continuous self-adjoint operator
L: H → H, there exists an orthonormal system of eigenvectors ϕ1,ϕ2, ... corresponding
to non-zero eigenvalues λ1,λ2, ... such that any element ξ ∈ H can be
uniquely written as
ξ =
N
∑
k=1
ckϕk + ¯ξ,
where N ∈ N∪{∞}, ¯ξ satisfies L ¯ξ = 0, and
Lξ =
N
∑
k=1
λkckϕk.
If the system of ϕk is infinite, then
lim
k→∞
λk = 0.
Proof. By induction, we construct eigenvectors ϕn so that the absolute values of the
corresponding eigenvalues satisfy
|λ1| ≥ |λ2| ≥ ··· ≥ |λn| ≥ ···
In the construction of ϕ1, we investigate the functional |Q(ξ)| = | Lξ,ξ | and we
prove that it assumes a maximum on B[0,1]. We denote
S = sup
ξ ≤1
| Lξ,ξ |.
Let {ξk}∞
k=1 be a sequence such that ξk ≤ 1 for k ∈ N and
| Lξk,ξk | → S as k → ∞.
By Remark 1.14 (or Remark 1.8), from the sequence {ξk}∞
k=1, one can extract a subsequence
{ξ1
k }∞
k=1 such that
Lξ1
k −Lη → 0 as k → ∞
for some η ∈ H, where η ≤ 1. By Lemma 1.21, it holds | Lη,η | = S. We put
ϕ1 = η. We add that η = 1. Indeed, for η < 1, the element η1 = η/ η satisfies
η1 = 1, | Lη1,η1 | > S.
We know that (see Remark 1.23)
Lϕ1 = λ1ϕ1,
where
|λ1| =
| Lϕ1,ϕ1 |
ϕ1,ϕ1
= | Lϕ1,ϕ1 | = S.
Let eigenvectors ϕ1,ϕ2,...,ϕn correspond to eigenvalues λ1,λ2,...,λn from our construction.
Let Hn be the subspace of H generated by ϕ1,ϕ2,...,ϕn. We consider the
functional | Lξ,ξ | on the set H⊥
n ∩ B[0,1] ⊆ H. Since the subspace Hn is invariant
and L is self-adjoint, the set H⊥
n is invariant with respect to L (see Remark 0.36). According
to the considerations above for H⊥
n , we get that, in H⊥
n ∩ B[0,1], one can find
the required element (denoted as ϕn+1) which is an eigenvector of the operator L. The
following two cases are possible:
30
1. after a finite number of the steps, we get a subspace H⊥
n0
, where Lξ,ξ ≡ 0;
2. Lξ,ξ ≡ 0 on H⊥
n for all n ∈ N.
In the first case, from Lemma 1.22, it follows that L maps the subspace H⊥
n0
into {0},
i.e., H⊥
n0
is composed of eigenvectors corresponding to the eigenvalue λ = 0. In this
case, the set {ϕk} is finite. In the second case, we obtain a sequence {ϕk}∞
k=1 of eigenvectors
for which λk = 0, k ∈ N. We know that λk → 0 as k → ∞ (see Theorem 1.15).
Let
¯H =
∞
n=1
H⊥
n = {0}.
If ξ ∈ ¯H, then
| Lξ,ξ | ≤ |λn|· ξ 2
, n ∈ N,
i.e., Lξ,ξ = 0. Therefore, by Lemma 1.22 (for the subspace ¯H), the operator L maps
¯H into {0}.
From the construction of {ϕk}N
k=1, it follows that any element ξ ∈ H can be uniquely
expressed as
ξ =
N
∑
k=1
ckϕk + ¯ξ,
where N ∈ N∪{∞}, L ¯ξ = 0, and
Lξ =
N
∑
k=1
λkckϕk.
Remark 1.25. Theorem 1.24 says that, for any completely continuous self-adjoint operator
on H, there exists an orthogonal base of the space H which is composed of eigenvectors
of this operator. To obtain such a base, it is enough to consider {ϕk}N
k=1 with
an arbitrary orthogonal base of H⊥
n0
or ¯H (see the proof of Theorem 1.24). In other
words, we get a result entirely analogous to the theorem about the existence of an
operator matrix in the diagonal form for self-adjoint operators in a space with a finite
dimension.
31
2. Derivative in Banach spaces
Let X,Y be real Banach spaces.
2.1 Weak and strong derivative
Let f be a map defined on an open set G ⊆ X with values in Y.
Definition 2.1. For x ∈ G, we define the directional derivative of f in the direction of
h ∈ X as the limit
lim
λ→0
f(x+λh)− f(x)
λ
if it exists. We denote it by Dh f(x).
Definition 2.2. If f has the directional derivative in the direction of any h ∈ X at x ∈ G
and df(x): h → Dh f(x) is a continuous linear map from X to Y, we say that f has
the weak derivative d f(x) at x. The weak derivative is sometimes called the Gâteaux
derivative.
Definition 2.3. If there exists a continuous linear map L: X → Y such that
lim
h→0
f(x+h)− f(x)−L(h)
h
= 0, (2.1)
we say that f has the strong derivative at x ∈ G. If f has the strong derivative at x ∈ G,
the map L from (2.1), which is uniquely determined, is called the Fréchet derivative
of f at x and is denoted by f (x).
Remark 2.4. If f (x) exists, f has also the weak derivative at x and f (x) = d f(x). In
this case, there exists the directional derivative of f in the direction of any h and
Dh f(x) = f (x)(h).
Remark 2.5. Let f be a real function (i.e., a functional) on X (i.e., G = X). The function
f has the Gâteaux derivative at x ∈ X if there exists L ∈ X such that
lim
λ→0
f(x+λh)− f(x)
λ
= Lh (2.2)
for all h ∈ X.
Example 2.6. On the Banach space X = C[0,1] (with the norm of the uniform convergence),
we consider the functional
F : f →
1
0
f2
.
32
For any ϕ ∈ X, we compute the directional derivative of F in the direction of ϕ as
(see (2.2))
Dϕ F(f) = lim
λ→0
F(f +λϕ)−F(f)
λ
= lim
λ→0
1
λ
1
0
(f +λϕ)2
− f2
= lim
λ→0
1
0
2 fϕ +λϕ2
=
1
0
2 fϕ.
We introduce
L: ϕ →
1
0
2 fϕ.
It is immediately seen that L is a linear functional on X. From
|L(ϕ)| ≤ 2
1
0
|fϕ| ≤ 2 max
t∈[0,1]
|ϕ(t)|·
1
0
|f| = 2 ϕ
1
0
|f|,
it follows that L is bounded. Thus, F has the Gâteaux derivative at f (which is given
by L).
If the Fréchet derivative of F at f exists, then it is L (see Remark 2.4). Its existence
is guaranteed by the limit
lim
ϕ→0
F(f +ϕ)−F(f)−L(ϕ)
ϕ
= lim
ϕ→0
1
ϕ
1
0
(f +ϕ)2
− f2
−2 fϕ
= lim
ϕ→0
1
ϕ
1
0
ϕ2
≤ lim
ϕ→0
ϕ = 0.
We see that the functional F has also the strong derivative at f and its Fréchet derivative
is equal to L.
Before the next example, we recall that continuous linear functionals on l1
have the
form
∞
∑
n=1
anxn
for {an}∞
n=1 ∈ l∞
. More precisely, if {an}∞
n=1 ∈ l∞
and
Ψ(x) =
∞
∑
n=1
anxn
for x = {xn}∞
n=1 ∈ l1
, then Ψ ∈ l1
(= l∞
).
Example 2.7. In the Banach space l1
, we consider the function f : t → t on l1
and
x = {xn}∞
n=1 ∈ l1
. We show that f has the Gâteaux derivative at x if and only if xn = 0
for all n ∈ N. In this case,
df(x) = {sgnxn}∞
n=1 ∈ l∞
.
33
Then, we show that the Fréchet derivative of the norm in l1
does not exist at any point.
Let {xn}∞
n=1 be a sequence in l1
such that xk = 0 for some k ∈ N. We put
ek = {0,...,0,1,0, ...},
where 1 is in the k-th position. The non-existence of the directional derivative of f in
the direction of ek comes from
x+λek − x
λ
=
1
λ
∞
∑
n=1
|xn|+|λ|−
∞
∑
n=1
|xn| =
|λ|
λ
.
For λ → 0, we see that the limit of the considered term does not exist. Therefore, the
function f does not have the weak (and also the strong) derivative at any point with a
zero element.
Let {xn}∞
n=1 be a sequence in l1
such that xn = 0 for all n ∈ N. We consider h =
{hn}∞
n=1 ∈ l1
and ε > 0. Let k ∈ N be such that
∞
∑
n=k+1
|hn| < ε.
Obviously, there exists δ > 0 such that
sgn(xn +λhn) = sgnxn, |λ| < δ, n ∈ {1,...,k}.
For λ ∈ (−δ,δ), we obtain
x+λh − x
λ
−
∞
∑
n=1
hn sgnxn
≤
1
λ
k
∑
n=1
|xn +λhn|−|xn|−λhn sgnxn +
∞
∑
n=k+1
|xn +λhn|−|xn|−λhn sgnxn
≤
1
|λ|
∞
∑
n=k+1
(|xn|+|λ|·|hn|−|xn|+|λ|·|hn|) ≤
1
|λ|
∞
∑
n=k+1
2|λ|·|hn| < 2ε.
Due to the fact that ε > 0 is arbitrary, the function f has the weak derivative at x which
is equal to
{sgnxn}∞
n=1 ∈ l∞
,
because the map
h →
∞
∑
n=1
hn sgnxn
is a continuous linear functional on l1
.
Now, we show that f does not have the strong derivative at any point. We assume
that f has the derivative f (x) at x = {xn}∞
n=1. If f (x) exists, it is {sgnxn}∞
n=1 ∈ l∞
. We
consider
hj
= 0,0,...,0,−2xj,−2xj+1,−2xj+2, ... , j ∈ N,
where the value −2xj is in the j-th position. Obviously,
hj
= 2
∞
∑
n=j
|xn| → 0 as j → ∞
34
and (see (2.1))
x+hj
− x − f (x) hj
= x+hj
− x −
∞
∑
n=j
(−2xn)sgnxn
=
∞
∑
n=1
|xn|−
∞
∑
n=1
|xn|+
∞
∑
n=j
2|xn| = hj
.
Now, it is seen that f cannot have the strong derivative at x.
2.2 Convex function
Now, we study derivatives of convex functions.
Definition 2.8. A real function f : D ⊆ X → R is called convex on a convex set D ⊆ X
if
f(λx+(1−λ)y) ≤ λ f(x)+(1−λ)f(y), x,y ∈ D, λ ∈ [0,1].
An elementary example of convex functions on a Banach space is the norm.
Theorem 2.9. Let f be a real convex function on an open convex set D ⊆ X which is
continuous at x0 ∈ D. Then, there exist K > 0 and δ > 0 such that
|f(x)− f(y)| ≤ K x−y , x,y ∈ B(x0,δ),
where B(x0,δ) is the open ball with the center in x0 and the radius δ.
Proof. The continuity of f at x0 guarantees the existence of M > 0 and δ > 0 such that
|f(t)| ≤ M for t ∈ B(x0,2δ) ⊆ D. We consider x,y ∈ B(x0,δ), where x = y. We denote
α = x−y , z = y+
δ
α
(y−x).
We see that z ∈ B(x0,2δ). Since y is a convex combination of z and x, where
y =
α
α +δ
z+
δ
α +δ
x,
using the convexity of f, we have
f(y)− f(x) ≤
α
α +δ
f(z)+
δ
α +δ
f(x)− f(x)
=
α
α +δ
(f(z)− f(x)) ≤
α
α +δ
2M ≤
2M
δ
x−y .
Analogously,
f(x)− f(y) ≤
2M
δ
y−x .
Remark 2.10. On spaces whose dimension is infinite, there exist discontinuous linear
functionals and all linear functional is a convex function. Thus, there exist discontinuous
convex functions. Note that convex functions on open subsets of a space with a
finite dimension are continuous.
35
Remark 2.11. From the proof of Theorem 2.9, it follows that it suffices to assume only
the boundedness of f on some neighbourhood of x0. Thus, we know that a convex
function is continuous on an open convex set D ⊆ X if and only if it is locally bounded
on D.
Before the following theorem, we recall that a real function f defined on a metric
space M is called upper semi-continuous on M if the set
{x ∈ M; f(x) < α}
is open for all α ∈ R.
Theorem 2.12. Let f be a convex function on an open convex set D ⊆ X. Then, the
following conditions are equivalent:
i) f is continuous on D;
ii) f is upper semi-continuous on D;
iii) f is upper bounded on some neighbourhood of a point of D;
iv) f is continuous at a point of D.
Proof. The implication i) ⇒ ii) is obvious.
If a real function f is upper semi-continuous on D and x ∈ D, then the set
{t ∈ D; f(t) < f(x)+1}
is a neighbourhood of x, where f is upper bounded. We have proved the implication
ii) ⇒ iii).
For the implication iii) ⇒ iv), we consider that f is upper bounded on a neighbourhood
of x0 ∈ D. If f ≤ K on B(x0,δ), for t ∈ B(x0,δ), we have also 2x0 −t ∈ B(x0,δ).
Thus,
f(x0) = f
2x0 −t
2
+
t
2
≤
1
2
f(2x0 −t)+
1
2
f(t) ≤
1
2
(K + f(t)).
Then, we have
−f(t) ≤ K −2 f(x0) ≤ K +2|f(x0)|.
Since also
f(t) ≤ K ≤ K +2|f(x0)|,
we obtain
|f(t)| ≤ K +2|f(x0)|, t ∈ B(x0,δ).
Therefore, f is bounded on B(x0,δ) and it suffices to use Remark 2.11.
Now, we consider the implication iv) ⇒ i). Let f be continuous at x0 ∈ D and y ∈ D
be arbitrarily given. There exist z ∈ D and λ ∈ (0,1) such that y = λx+(1−λ)z. We
consider δ > 0 and K > 0 such that f ≤ K on B(x,δ) ⊆ D. We show that f is upper
bounded on the set B(y,λδ) which implies that the function f is continuous at y. We
consider t ∈ B(y,λδ). Since
t = λ x+
t −y
λ
+(1−λ)z ∈ D,
36
where
x+
t −y
λ
∈ B(x,δ),
the convexity of f gives the estimation
f(t) ≤ λ f x+
t −y
λ
+(1−λ)f(z) ≤ λK +(1−λ)f(z).
Remark 2.13. It is known that linear functionals are continuous if and only if they
are continuous at 0, which is if and only if they are bounded. So, continuous linear
functionals are bounded on the unit ball, but continuous convex functions do not need
to be bounded on the unit ball (although they are locally bounded). In addition, on any
separable Banach space whose dimension is infinite, there exists continuous convex
function, which is not bounded on the unit ball.
Before the following theorem, we recall that a real function p on X is called a
convex functional on X if:
a) p(λx) = λ p(x), λ ≥ 0, x ∈ X;
b) p(x+y) ≤ p(x)+ p(y), x,y ∈ X.
Theorem 2.14. Let D ⊆ X be an open convex set, let f be a convex function on D, and
let x ∈ D. Then,
d+
f(x)(h) = lim
t→0+
f(x+th)− f(x)
t
exists for all h ∈ X and the map d+
f(x): h → d+
f(x)(h) is a convex functional on X.
Proof. Let h ∈ X. The function
t →
f(x+th)− f(x)
t
is non-decreasing on a right neighbourhood of 0. If t,s, where 0 < t < s, are sufficiently
small (so that x+sh ∈ D), then we have
x+th =
s−t
s
x+
t
s
(x+sh)
and, consequently,
f(x+th) ≤
s−t
s
f(x)+
t
s
f(x+sh).
Thus,
1
t
(f(x+th)− f(x)) ≤
1
s
(f(x+sh)− f(x)).
If we consider t > 0 sufficiently small, then
−
f(x−2th)− f(x)
2t
≤
f(x+2th)− f(x)
2t
, (2.3)
because
2 f(x) = 2 f
x−2th+x+2th
2
≤ f(x−2th)+ f(x+2th). (2.4)
37
Therefore (see (2.3)), we have
−d+
f(x)(−h) ≤ d+
f(x)(h).
Especially, the considered limit exists.
It remains to prove the convexity of the functional. If λ > 0, then
d+
f(x)(λh) = λ lim
t→0+
f(x+tλh)− f(x)
λt
= λd+
f(x)(h).
For h,k ∈ X, it holds (see (2.4))
d+
f(x)(h+k) = lim
t→0+
f(x+(h+k)t)− f(x)
t
≤ lim
t→0+
f(x+2th)− f(x)
2t
+
f(x+2tk)− f(x)
2t
= d+
f(x)(h)+d+
f(x)(k).
Remark 2.15. In the proof of the previous theorem, we have obtained the inequality
−d+
f(x)(−h) ≤ d+
f(x)(h).
It is easy to show that Dh f(x) exists for all h ∈ X if and only if
−d+
f(x)(−h) = d+
f(x)(h), h ∈ X.
Theorem 2.16. Let f be a convex function on an open convex set D ⊆ X, let it be
continuous at x ∈ D, and let it have the derivative Dh f(x) linearly in the direction of
all h ∈ X. Then, f has the Gâteaux derivative d f(x) at x.
Proof. By Theorem 2.9, there exist K > 0 and δ > 0 such that
|f(u)− f(v)| ≤ K u−v
for all u,v ∈ B(x,δ) ⊆ D. We consider h ∈ X. Let λ > 0 be such that x+λh ∈ B(x,δ).
Then,
|f(x+λh)− f(x)| ≤ K λh = Kλ h .
Therefore,
|d+
f(x)(h)| ≤ K h
and, consequently, the derivative d+
f(x)(h) = Dh f(x) is continuous.
2.3 Tangent functional
We know that the function f : X → R given by x → x is continuous and convex.
We show that, under certain assumptions, the weak derivative of this function f is the
so-called tangent functional. We recall the well-known corollary of the Hahn–Banach
theorem which says that, for any non-zero element x ∈ X, there exists g ∈ X such that
g = 1 and g(x) = x . This functional is called the tangent functional at x.
38
Theorem 2.17. Let the function f : x → x have the Gâteaux derivative at x = 0, x ∈
X. Then, the Gâteaux derivative d f(x) ∈ X has the norm d f(x) = 1 and d f(x)(x) =
x . At the same time, if g ∈ X satisfies g = 1 and g(x) = x , then g = d f(x).
Proof. For h ∈ X and sufficiently small t, from the estimation
1
t
[ x+th − x ] ≤
1
|t|
x+th−x = h ,
it follows that d f(x) ≤ 1. Since
d f(x)
x
x
= lim
t→0
1
t
x+t
x
x
− x
= lim
t→0
1
t
1+
t
x
x − x
= lim
t→0
x
t
1+
t
x
−1 = 1,
we have d f(x) = 1 and d f(x)(x) = x .
Let g ∈ X be such that g = 1 and g(x) = x . We consider h ∈ X and we define
ε(t) = df(x)(h)−
1
t
[ x+th − x ]
for sufficiently small t = 0. Obviously, ε(t) → 0 as t → 0. For the considered t, we
have
g(x+th) ≤ g · x+th = x +td f(x)(h)−tε(t) = g(x)+td f(x)(h)−tε(t).
Therefore,
tg(h) ≤ td f(x)(h)−tε(t)
which gives (as t → 0+
)
g(h) ≤ d f(x)(h).
Since this inequality is valid for all h ∈ X (also for −h), it is enough to consider the
linearity of g and df(x). Therefore, g = d f(x).
The previous theorem relates to the geometry of a Banach space. According to this
theorem, at all non-zero point in which the norm has the Gâteaux derivative, the space
is “smooth” in the sense that there exists just one tangent functional. We discuss this
topic in the next chapter.
39
3. Strictly and uniformly
convex spaces
In this chapter, we consider a real Banach space X.
3.1 Strictly convex space
Definition 3.1. An extreme point of a convex set C ⊆ X is a point x ∈ C for which
x = (a+b)/2, where a,b ∈ C, implies a = b. The set of all extreme points of the set C
is denoted by ext C.
Remark 3.2. Definition 3.1 says that x ∈ C is an extreme point if there does not exist a
non-degenerated line segment in C having the center x. It is seen that x is an extreme
point of C if and only if the set C {x} is convex.
Definition 3.3. A Banach space is called strictly convex if all point of the unit sphere
∂B(0,1) is an extreme point of the closed unit ball B[0,1], i.e., if ext B[0,1] = ∂B(0,1).
Strictly convex spaces are also called rotund.
Remark 3.4. In strictly convex spaces, any line segment cannot lie on any sphere.
Therefore, the following implication is valid. If C is a convex set in the strictly convex
space X, x ∈ X, a,b ∈ C, and if x−a = x−b = dist(x,C), then a = b. Indeed, for
λ ∈ [0,1], it suffices to consider that
x−λa−(1−λ)b ≤ λx−λa + (1−λ)x−(1−λ)b = dist(x,C).
Remark 3.5. Since extreme points of the closed unit ball B[0,1] have to lie on the unit
sphere ∂B(0,1), we can say that X is strictly convex if and only if
x+y < 2, x,y ∈ B[0,1], x = y.
Let the identity
x+y 2
= 2 x 2
+2 y 2
(3.1)
be valid for points x,y of a Hilbert space. Since, in any Hilbert space, the identity
x+y 2
+ x−y 2
= 2 x 2
+2 y 2
is valid, we see that x = y. Geometrically, (3.1) says that, in the parallelogram with the
sides x and y, the second diagonal x −y is missing. Now, we consider points x,y of X
for which (3.1) is valid. Since
0 = 2 x 2
+2 y 2
− x+y 2
≥ 2 x 2
+2 y 2
−( x + y )2
= 2 x 2
+2 y 2
− x 2
−2 x · y − y 2
= ( x − y )2
≥ 0,
we get x = y , but not x = y necessarily. This observation motivates the following
theorem.
40
Theorem 3.6. For the Banach space X, the following statements are equivalent:
i) X is strictly convex;
ii) if x+y = x + y , x = 0,y = 0, then x = λy for some λ > 0;
iii) if x,y ∈ X and x+y 2
= 2 x 2
+2 y 2
, then x = y,
Proof. We begin with the implication i) ⇒ ii). Let X be strictly convex and let x,y
be non-zero points of X satisfying the equality x+y = x + y . For example, let
x ≤ y . We have
x
x
+
y
y
≥
x
x
+
y
x
−
y
x
−
y
y
=
1
x
( x + y )− y
1
x
−
1
y
= 2.
Therefore, we obtain (see Remark 3.5)
x
x
=
y
y
,
i.e.,
x =
x
y
y.
Let us consider the implication ii) ⇒ iii). Let the implication in ii) be true and let
x+y 2
= 2 x 2
+2 y 2
.
Hence (see the text before Theorem 3.6), x = y . We do not consider the trivial
case x = 0,y = 0. In the non-zero case, we obtain
x+y 2
= 4 x 2
.
Therefore,
x+y = 2 x = x + y .
Based on the assumption, x = λy for some λ > 0. Considering x = y , we obtain
that λ = 1. The implication has been proved.
In the last part of the proof, we can use, e.g., Remark 3.4 or Remark 3.5. If
x = y =
1
2
(x+y) = 1,
then
x+y 2
= 4 = 2 x 2
+2 y 2
.
41
3.2 Uniformly convex space
If points x,y of a strictly convex space have a constant distance and they are on the
unit sphere ∂B(0,1), then the midpoint of the line segment between x and y is in the
open unit ball. But, in a space whose dimension is infinite, it is not clear, whether this
midpoint can be arbitrarily close to the sphere ∂B(0,1). This motivates the following
definition.
Definition 3.7. A Banach space is called uniformly convex if, for all ε ∈ (0,2], there
exists δ > 0 such that, if x,y are in the unit ball B[0,1] and x−y ≥ ε, then
1
2
(x+y) ≤ 1−δ.
The basic characteristics of uniformly convex spaces are mentioned in the following
theorem.
Theorem 3.8. For the Banach space X, the following statements are equivalent:
i) X is uniformly convex;
ii) for any ε ∈ (0,2], there exists δ > 0 such that, if x,y are on the unit sphere
∂B(0,1) and x−y ≥ ε, then
1
2
(x+y) ≤ 1−δ;
iii) if {xn}∞
n=1 ,{yn}∞
n=1 ⊆ ∂B(0,1) satisfy
lim
n→∞
xn +yn
2
= 1,
then xn −yn → 0 as n → ∞.
Proof. If X is uniformly convex, then ii) is true. The implication ii) ⇒ iii) is trivial as
well.
If X is not uniformly convex, then there exist ε > 0 and sequences
{xn}∞
n=1 ,{yn}∞
n=1 ⊆ B[0,1]
such that
xn −yn ≥ ε, n ∈ N,
and
1−
1
n
≤
1
2
xn +yn ≤
1
2
( xn + yn ) ≤ 1, n ∈ N.
Therefore,
xn + yn → 2 as n → ∞.
Since xn ≤ 1, yn ≤ 1, n ∈ N, we have
lim
n→∞
xn = lim
n→∞
yn = 1.
Especially, without loss of generality, we can assume that xn · yn > 0 for all n ∈ N.
For
un =
xn
xn
, vn =
yn
yn
, n ∈ N,
42
we obtain un = 1 = vn , n ∈ N, and
| un +vn − xn +yn | ≤ un −xn + vn −yn , n ∈ N.
The right side converges to 0 as n → ∞, because
un −xn =
xn
xn
−xn = xn ·
1
xn
−1 = 1− xn , n ∈ N,
and, analogously,
vn −yn = 1− yn , n ∈ N.
With respect to
xn +yn
2
→ 1 as n → ∞,
we get
un +vn
2
→ 1 as n → ∞. (3.2)
We have two sequences {un}∞
n=1 ,{vn}∞
n=1 ⊆ ∂B(0,1) for which (3.2) is valid and, at
the same time, for which
liminf
n→∞
un −vn ≥ ε > 0,
because
0 < ε ≤ xn −yn ≤ xn −un
→0
+ un −vn + vn −yn
→0
, n ∈ N.
Remark 3.9. Many other equivalences can be mentioned in Theorems 3.6 and 3.8.
The statements in Theorem 3.6 are also equivalent to:
iv) if p ∈ (1,∞) and x,y ∈ X, x = y, then
x+y
2
p
< x p
+ y p
;
v) if
x−y = x−z + z−y ,
then there exists λ ∈ [0,1] such that z = λx+(1−λ)y.
Similarly, the statements in Theorem 3.8 are equivalent to:
iv) for any ε > 0, there exists δ > 0 such that, if x < 1+δ, y < 1+δ, and if
1
2
(x+y) ≥ 1,
then x−y < ε.
Remark 3.10. Obviously, any uniformly convex space is strictly convex. In spaces
whose dimension is finite, these notions are same. It follows from the compactness of
the closed unit ball B[0,1] in spaces with finite dimensions and the continuity of the
function
(x,y) →
x+y
2
.
43
Remark 3.11. We consider X = C[0,1] with the norm
f = max{|f(t)|; t ∈ [0,1]}+


1
0
f2


1
2
, f ∈ X.
One can show that this space X is strictly convex, but it is not uniformly convex.
Example 3.12. Any Hilbert space H is uniformly convex. It is enough to consider that,
for x,y ∈ H, x ≤ 1, y ≤ 1, and x−y ≥ ε, we have
x+y
2
2
+
x−y
2
2
=
x 2
2
+
y 2
2
and, consequently,
x+y
2
2
≤
1
2
+
1
2
−
ε
2
2
,
i.e.,
x+y
2
≤ 1−
ε2
4
< 1.
Example 3.13. The spaces
Zp = Lp
(Ω), · p
are strictly convex for all p ∈ (1,∞) and any measurable set Ω ⊆ R. This fact is possible
to easily show taking into account the case, when the Minkowski inequality becomes
the equality. But, we prove the stronger result that the space Zp is uniformly convex
for all p > 1. It is enough to prove that, for all ε > 0, there exists δ > 0 such that, if
u,v ∈ Lp
(Ω), u p = v p = 1, and if
u+v
2
p
p
> 1−δ,
then
u−v
2
p
p
≤ 2εp
.
We consider an arbitrary number ε > 0. For simplicity, we denote
s =
1
2
(u+v), t =
1
2
(u−v),
where u = s+t, v = s−t. We put
S = {ω ∈ Ω; |t(ω)| ≤ ε|s(ω)|},
S0 = {ω ∈ Ω; t(ω) = 0},
S+ = {ω ∈ Ω; 0 < |t(ω)| ≤ ε|s(ω)|},
T = {ω ∈ Ω; |t(ω)| > ε|s(ω)|}.
Evidently,
S
|t(ω)|p
dω ≤ εp
Ω
|s(ω)|p
dω ≤ εp
. (3.3)
44
It is well-known that the function λ → |λ|p
is strictly convex and continuous on R.
Therefore,
|λ +1|p +|λ −1|p
2
> |λ|p
=
λ +1
2
+
λ −1
2
p
, λ ∈ R,
and there exists γ > 0 such that
1
2
[|λ +1|p
+|λ −1|p
]−|λ|p
≥ γ, λ ∈ −
1
ε
,
1
ε
. (3.4)
In (3.4), we consider
λ =
s(ω)
t(ω)
, ω ∈ Ω S0.
For ω ∈ T, we obtain
1
2
[|s(ω)+t(ω)|p
+|s(ω)−t(ω)|p
] ≥ γ|t(ω)|p
+|s(ω)|p
and, for ω ∈ S = S0 ∪S+, it holds
1
2
[|s(ω)+t(ω)|p
+|s(ω)−t(ω)|p
] ≥ |s(ω)|p
.
Thus,
1 =
Ω
1
2
[|s(ω)+t(ω)|p
+|s(ω)−t(ω)|p
] dω
≥
Ω
|s(ω)|p
dω +
T
γ|t(ω)|p
dω.
(3.5)
If
Ω
|s(ω)|p
dω > 1−δ,
then, using (3.5), we obtain
T
|t(ω)|p
dω ≤
δ
γ
. (3.6)
The choice δ = γεp
, (3.3), and (3.6) give
Ω
|t(ω)|p
dω =
T
|t(ω)|p
dω +
S
|t(ω)|p
dω ≤ εp
+εp
= 2εp
.
3.3 Projection
Now, we focus on projections in uniformly convex Banach spaces. Let X be a uniformly
convex Banach space.
Theorem 3.14. Let C(= /0) be a closed convex subset of X. Then, for any y ∈ X, there
exists just one c ∈ C such that y−c = dist(y,C).
45
Proof. The uniqueness is presented in Remark 3.4. Hence, it suffices to prove the
existence of an element x ∈C with a minimal norm, because, without loss of generality,
we can assume that y = 0. We put
d = inf{ c ; c ∈ C}.
If d = 0, then we see that 0 ∈ C (the set C is closed). Therefore, without loss of
generality, let d = 1. In this case, there exists a sequence {xn}∞
n=1 ⊆ C such that
lim
n→∞
xn = 1.
If {xn}∞
n=1 is Cauchy, then there exists the limit
x = lim
n→∞
xn, where x ∈ C, x = 1.
The convexity of C guarantees that
1
2
(xn +xk) ≥ 1, n,k ∈ N.
We consider an arbitrary number ε > 0. For the given ε, there exists n0 ∈ N such that
xn < 1 +ε for all n ≥ n0, n ∈ N. Therefore (using the uniform convexity), we have
xn −xk < ξ(ε) for all n,k ≥ n0, n,k ∈ N, where ξ(ε) → 0 as ε → 0+
. The theorem
is proved.
Definition 3.15. Let C be a closed convex subset of X. For any x ∈ X, the uniquely
determined PC(x) ∈ C satisfying x−PC(x) = dist(x,C) is called the projection (of x
on C).
Theorem 3.16. Let C be a closed convex subset of X. The projection PC is continuous.
Proof. For simplicity, we consider dist(0,C) = 1. Let {xn}∞
n=1 ⊆ X be a sequence such
that xn → 0 as n → ∞. We want to show that PC(xn) → PC(0) as n → ∞. Since
| xn −PC(xn) −1| = |dist(xn,C)−dist(0,C)|
≤ xn −0 = xn → 0 as n → ∞,
we obtain that xn −PC(xn) → 1 as n → ∞. From
1 ← | xn −PC(xn) − xn | ≤ PC(xn) ≤ xn −PC(xn) + xn → 1
as n → ∞, we have PC(xn) → 1 as n → ∞. The convexity of C gives
1 ≤
1
2
(PC(0)+PC(xn)) ≤
1
2
PC(0) +
1
2
PC(xn) → 1 as n → ∞.
We use iii) from Theorem 3.8 for the sequences
{PC(0)}∞
n=1 ,
PC(xn)
PC(xn)
∞
n=1
.
Thus, PC(xn)−PC(0) → 0 as n → ∞. It suffices to consider that
1
2
(PC(0)+PC(xn)) −
1
2
PC(0)+
PC(xn)
PC(xn)
≤
1
2
PC(xn)−
PC(xn)
PC(xn)
→ 0 as n → ∞.
46
Remark 3.17. The projection on closed convex sets does not need to be linear (even in
Hilbert spaces). It is known that the projection on (closed) subspaces of Hilbert spaces
is linear. But, it is not true in uniformly convex spaces.
For a better understanding of Theorem 3.20 below, we mention the following result.
Theorem 3.18. Any uniformly convex space is reflexive.
Remark 3.19. Strictly convex spaces do not need to be reflexive (see the example in
Remark 3.11).
Now, without a proof, we mention a generalization of Theorem 3.14.
Theorem 3.20. Let C be a closed convex subset of a strictly convex reflexive Banach
space Y and y ∈ Y. Then, there exists just one c ∈ C such that y−c = dist(y,C).
Remark 3.21. We add that there exist strictly convex reflexive spaces, which are not
uniformly convex.
Now, we generalize the concept of the projection to the concept of the so-called
metric projections in the following definition.
Definition 3.22. Let M be a subset of a Banach space Y. For x ∈ Y, we denote
PM(x) = {m ∈ M; x−m = dist(x,M)}.
The set M is called:
• proximinal if PM(x) = /0 for all x ∈ Y;
• semi-Chebyshev if PM(x) has at most one element for all x ∈ Y;
• Chebyshev if PM(x) has just one element for all x ∈ Y.
We repeat that, in strictly convex spaces, any closed convex set is semi-Chebyshev
and that any closed convex subset of a strictly convex reflexive Banach space is Chebyshev
(see Remark 3.4 and Theorem 3.20). We add that any compact set is proximinal.
3.4 Smooth space
Now, let us define smooth spaces explicitly.
Definition 3.23. The space X is called smooth at x ∈ ∂B(0,1) if there exists just one
functional ϕ ∈ X such that ϕ = 1, ϕ(x) = 1. We say that X is smooth if it is smooth
at any point of the unit sphere ∂B(0,1).
We emphasize that, according to the above mentioned corollary of the Hahn–Banach
theorem, the tangent functional exists at any point x ∈ ∂B(0,1). Concerning
Definition 3.23, we point out the uniqueness.
The following reinforcement of Theorem 2.17 is valid.
Theorem 3.24 (Šmuljan). The space X is smooth at x ∈ ∂B(0,1) if and only if the
function f : t → t has the Gâteaux derivative at x.
The most important connection between Chapters 2 and 3 is presented in the following
theorem.
47
Theorem 3.25 (Klee). If X is strictly convex, then X is smooth. If X is smooth, then
X is strictly convex.
Proof. We assume that X is not smooth at x ∈ ∂B(0,1). Thus, there exist ϕ,ψ ∈ X
such that
ϕ = ψ, ϕ = ψ = 1 = ϕ(x) = ψ(x).
Since
ϕ +ψ
2
= 1,
the space X cannot be strictly convex.
If X is not strictly convex, then there exist x,y ∈ ∂B(0,1) such that x = y and
x+y
2
∈ ∂B(0,1).
By the recalled corollary of the Hahn–Banach theorem, there exists a functional ϕ such
that
ϕ = 1, ϕ
x+y
2
= 1.
Since
1 = ϕ
x+y
2
=
1
2
ϕ(x)+
1
2
ϕ(y) ≤
1
2
+
1
2
= 1,
we see
ϕ(x) = ϕ(y) = 1.
For the elements of the unit sphere in X given by x and y (denoted by fx and fy), we
have
fx(ϕ) = fy(ϕ) = 1.
Since fx = fy, the space X cannot be smooth at ϕ.
Corollary 3.26. Let X be reflexive. The space X is smooth if and only if X is strictly
convex; and X is strictly convex if and only if X is smooth.
Remark 3.27. Note that there exist smooth spaces, whose dual spaces are not strictly
convex. Similarly, there exist strictly convex spaces, whose dual spaces are not smooth.
We repeat that smooth Banach spaces are the spaces whose norms have the Gâteaux
derivative at any point of the unit sphere. The Banach spaces, whose norms have the
Fréchet derivative uniformly on the unit sphere, are called uniformly smooth spaces.
See the following definition.
Definition 3.28. The space X is called uniformly smooth if there exists the limit
lim
τ→0
x+τy − x
τ
uniformly for x,y ∈ ∂B(0,1).
We end this chapter with the analogy of Corollary 3.26. We add that uniformly
smooth spaces are reflexive.
Theorem 3.29. The space X is uniformly smooth if and only if X is uniformly convex;
and X is uniformly convex if and only if X is uniformly smooth.
48
4. Fixed point theorems
In this chapter, we consider a real Banach space X.
Definition 4.1. Let Y be a metric space. A point y is called a fixed point of a map
f : D ⊆ Y → Y if f(y) = y.
At first, we recall the most important fixed point theorem.
Theorem 4.2 (Banach). If Y is a complete metric space and f : Y →Y is a contraction,
then there exists just one fixed point of f.
Remark 4.3. Theorem 4.2 is well-known, e.g., from the theory of ODEs.
Definition 4.4. We say that a subset D of X has the fixed point property if any continuous
map f : D → D has a fixed point.
Remark 4.5. For example, for (R,|·|), it is seen that any closed interval [a,b] has the
fixed point property.
4.1 Topological degree
Let f : G → X be a map, where G is an open set in X. Our goal is to define the number
deg(f,G, p), i.e., the so-called topological degree of f, which means “the number of
solutions of the equation f(x) = p on G”. This number depends on f and on p continuously.
Thus, for small perturbations of f, the number deg(f,G, p) has to be constant
in a sufficiently small neighbourhood of p.
Let us consider the Euclidean space Rn
.
Definition 4.6. To any (f,G, p), where G is from the system of all bounded open
subsets of Rn
, f : G → Rn
is a continuous map, and p ∈ Rn
f(∂G), we assign the
integer deg(f,G, p) satisfying the following conditions:
i) if f is the identity on G and p ∈ G, then deg(f,G, p) = 1;
ii) if G1,G2 ⊆ G are open sets satisfying G1 ∩ G2 = /0 and p /∈ f G (G1 ∪G2) ,
then
deg(f,G, p) = deg(f,G1, p)+deg(f,G2, p);
iii) if H : [0,1]×G → Rn
is a continuous map, f0(x) = H(0,x), f1(x) = H(1,x), and
if H(t,x) = p for t ∈ [0,1], x ∈ ∂G, then
deg(f0,G, p) = deg(f1,G, p);
iv) if deg(f,G, p) = 0, then there exists x ∈ G such that f(x) = p.
This map is called the topological degree in Rn
.
Remark 4.7. In Rn
, the map from the previous definition exists and it is determined by
the conditions i)–iv) uniquely.
49
Before Remark 4.9 mentioned below, we recall the notion of the so-called homo-
topy.
Definition 4.8. Let Z and Y be metric spaces and let f,g: Z → Y be continuous maps.
We say that f is homotopic with g, if there exists a continuous map H : [0,1] × Z →
Y such that H(0,x) = f(x) and H(1,x) = g(x) for x ∈ Z. The map H is called the
homotopy.
Remark 4.9. Now, we comment the conditions i)–iv) from Definition 4.6.
The condition i) says that the equation idx = p has one solution x = p.
The condition ii) says that, if the equation f(x) = p has just n1 solutions on G1, just
n2 solutions on G2, and no solution on G (G1 ∪ G2), then this equation has n1 + n2
solutions on G.
The condition iii) expresses the invariance of the topological degree with respect to
homotopies.
The condition iv) says when the equation f(x) = p has solutions on G.
Remark 4.10. Now, using a simple example, we explain why we cannot consider also
p ∈ f(∂G) in Definition 4.6. Let us consider the function f(t) = t on G = (0,1) ⊆ R.
If p ∈ (−∞,0) ∪ (1,∞), then deg(f,G, p) = 0, because the equation f(x) = p has no
solution from the interval (0,1) for this p. Next, deg(f,G, p) = 1 for p ∈ (0,1). In any
neighbourhood of p = 0 or p = 1, the topological degree deg(f,G, p) takes the both
values 0 and 1. Hence, for p ∈ f(∂G) = {0,1}, deg(f,G, p) cannot be defined if the
topological degree depends on p continuously.
Remark 4.11. In spaces whose dimension is infinite, one can construct the theory of the
topological degree as well. It is called the Leray–Shauder degree and it is introduced
for maps of the type I −T, where T is a completely continuous operator.
Theorem 4.12. The topological degree in Rn
has the following properties:
1. if f,g: G → Rn
are continuous maps, f = g on ∂G, and p ∈ Rn
f(∂G), then
deg(f,G, p) = deg(g,G, p);
2. the map deg(f,G,−) is constant on any connected component of the open set
Rn
f (∂G).
4.2 Brouwer and 1. and 2. Shauder theorem
Now, we use the topological degree in the Euclidean space Rn
.
Theorem 4.13 (Brouwer). The closed unit ball B[0,1] ⊆ Rn
has the fixed point pro-
perty.
Proof. By contradiction, we consider a continuous map f : B[0,1] → B[0,1] with the
property that f(x) = x for all x ∈ B[0,1]. The map
H(t,x) = x−t f(x), x ∈ Rn
, t ∈ [0,1],
is a homotopy. We show that H(t,x) = 0 for t ∈ [0,1] and x = 1. For t = 1, it follows
from the assumption. If t ∈ [0,1), then we have the inequality t f(x) ≤ t < 1 for
x = 1. Therefore, x = t f(x). We denote
g0(x) = H(0,x) = x, g1(x) = H(1,x) = x− f(x), x ∈ B(0,1).
50
From the condition iii) in Definition 4.6, we obtain that
deg(g0,B(0,1),0) = deg(g1,B(0,1),0).
But, g0 is the identity, which gives deg(g0,B(0,1),0) = 1. Thus,
deg(g1,B(0,1),0) = 1
and the condition iv) from Definition 4.6 gives the existence of x ∈ B(0,1) for which
g1(x) = 0. Of course, this is a contradiction.
Remark 4.14. From Theorem 4.13, it follows that any compact convex subset of a
Banach space with a finite dimension has the fixed point property.
Naturally, we obtain the question, whether the Brouwer theorem, i.e., Theorem 4.13,
is valid also for the closed unit balls in spaces whose dimension is infinite. But, it is
enough to consider, e.g., the map
h: x = {x1,x2, ...} → 1− x 2
,x1,x2, ... ,
which does not have any fixed point on the closed unit ball of the space l2
.
For spaces whose dimension is infinite, we have the following result.
Theorem 4.15 (Shauder). Let K be a (non-empty) compact convex subset of X and let
f : K → K be a continuous map. Then, there exists x ∈ K such that f(x) = x.
Proof. We consider ε > 0. There exist x1,...,xn ∈ K such that
K ⊆
n
j=1
B(xj,ε).
We define
ϕj(x) = max 0,ε − x−xj , x ∈ K, j ∈ {1,...,n}.
The functions ϕj are non-negative on K and the function
n
∑
j=1
ϕj
is positive on K. Thus, we can define the function ϕ on K by
ϕ : x →
n
∑
j=1
ϕj(x)xj
n
∑
j=1
ϕj(x)
−1
.
Obviously, ϕ is a continuous function on K which maps K to the set
Kε = conv{x1,...,xn} ⊆ K,
where conv{x1,...,xn} is the convex hull of x1,...,xn. We have
ϕ(x)−x ≤ ε, x ∈ K.
51
The composition ϕ ◦ f maps Kε into Kε . According to Theorem 4.13 (see Remark 4.14),
it has a fixed point xε ∈ Kε . Since
xε − f(xε ) ≤ xε −ϕ(f(xε )) + ϕ(f(xε ))− f(xε )
= ϕ(f(xε ))− f(xε ) ≤ ε,
we have
inf{ x− f(x) ; x ∈ K} = 0.
Considering that f is a continuous map on a compact set, we know that there exists
x ∈ K such that f(x) = x.
Definition 4.16. A map f : D ⊆ X → X is called compact if it is continuous and if it
maps any bounded subset of D into a set whose closure is a compact set.
Remark 4.17. We repeat that a linear map f : X → X, which maps bounded sets into
sets with compact closures, is continuous.
Theorem 4.18 (Shauder). Let K be a closed bounded convex subset of X and let
f : K → K be compact. Then, f has a fixed point.
52
5. Integration in Banach
spaces
In this chapter, we consider a real Banach space X.
5.1 Preliminaries and basic definitions
At first, we recall basic definitions.
Definition 5.1. A system S of subsets of a given set Ω is called σ-algebra if
a) Ω ∈ S;
b) A ∈ S ⇒ Ω A ∈ S;
c) An ∈ S,n ∈ N ⇒
∞
n=1
An ∈ S.
Then, (Ω,S) is called a measure space.
Definition 5.2. Let S by a system of subsets of a set Ω. A non-negative map µ : S →
[0,+∞] is called a measure if
a) S is σ-algebra;
b) µ(/0) = 0;
c) for any sequence {An}∞
n=1 of pairwise disjoint sets from S, it holds
µ
∞
n=1
An =
∞
∑
n=1
µ(An).
We say that a measure is finite if µ(Ω) < ∞. A measure is called complete if the
implication
A,B ⊆ Ω, A ⊆ B, B ∈ S, µ(B) = 0 ⇒ A ∈ S
is valid.
Definition 5.3. Let Y be an arbitrary set. We define the function χA : A ⊆ Y → R by
χA(y) =
1, for y ∈ A;
0, for y /∈ A.
Let a measurable space (Ω,S) be equipped with measure µ, whereas µ is probability
complete measure.
Definition 5.4. Function f : Ω → X is called:
53
• simple if there exist x1,...,xn ∈ X and E1,...,En ∈ S such that
n
∑
i=1
xiχEi ≡ f;
• measurable if there exists sequence { fn}∞
n=1 of simple functions such that
lim
n→∞
fn(ω) = f(ω)
for all ω ∈ Ω up to a set with zero measure;
• weakly measurable if ϕ ◦ f is measurable function for all ϕ ∈ X .
Remark 5.5. It is easy to show that f1 + f2 and λ f2 are measurable if f1, f2 are measurable
and λ ∈ R. (analogously weakly measurable)
Theorem 5.6 (Pettis). Function f : Ω → X is measurable if and only if f is weakly
measurable and there exists set E ∈ S with measure µ(E) = 0 such that f (Ω E) is
separably subset of X. Especially for separably Banach space X terms measurable and
weakly measurable merge.
Definition 5.7. If xn for n ∈ N be elements of Banach space X, we say that series
∞
∑
i=1
xi
• converges, if there exists
lim
n→∞
n
∑
i=1
xi;
• converges absolutely if
∞
∑
i=1
|xi| < +∞;
• converges unconditionally to x ∈ X, if
n
∑
i=1
xp(i)
for all permutation p.
Remark 5.8. Due to the completeness of X, any absolutely convergent series is also
unconditionally convergent. If the dimension of X is finite, then any unconditionally
convergent series converges also absolutely. In spaces whose dimension is infinite, this
implication is not valid. We consider
xn = 0,...,0,
1
n
,0, ... ∈ C0, n ∈ N,
where C0 is the space of all sequences of real numbers which converge to zero with the
norm
x = max
n∈N
|xn|.
54
Definition 5.9. Let F : S → X be a set of functions. We say that F is additive or
σ-additive vector measure if F(/0) = 0 and
F
n
En = ∑
n
F(En)
for all finite or countable sequence of disjoint sets En ∈ S, respectively.
Remark 5.10. Convergence in X in Definition 5.9 has the meaning of convergence of
series (in case of σ-additivity of measure). This convergence is necessary unconditional
(union is commutative).
Definition 5.11. We say that vector measure F : S → X is absolutely continuous with
respect to measure µ, if for each ε > 0 exists δ > 0 such that F(E) < ε if µ(E) < δ.
Theorem 5.12 (Pettis). Let F be σ-additive vector measure. Then, F is absolutely
continuous with respect to the measure µ if and only if F(E) = 0 if µ(E) = 0.
Definition 5.13. Let F be vector measure. If
|F(Ω)| = sup
n
∑
k=1
F(Ak) ; Ak ∈ S are pairwise disjoint,
n
k=1
Ak = Ω < +∞,
then we say that F has a bounded variation.
5.2 Bochner integral
Definition 5.14. A function f : Ω → X is called Bochner integrable if f is measurable
and there exist simple functions fn for n ∈ N such that
lim
n→∞
Ω
f − fn dµ = 0.
It can be shown that if
lim
n→∞
B
f − fn dµ = 0 = lim
n→∞
B
f −gn dµ
for measurable function f and simple functions fn,gn for n ∈ N and B ∈ S, then there
exist limits and
lim
n→∞
B
fn dµ = lim
n→∞
B
gn dµ ∈ X,
and we put for simple function
ϕ ≡
p
∑
i=1
x1χEi
just
B
ϕ dµ =
p
∑
i=1
xiµ (Ei ∩B).
The previous one guarantees correctness of the following definition.
55
Definition 5.15. Let f be Bochner integrable function, B ∈ S and { fn}∞
n=1 be sequence
of simple functions satisfying
lim
n→∞
B
f − fn dµ = 0.
The limit
lim
n→∞
B
fn dµ ∈ X
is called Bochner integral f over B and it is denoted as
B
f dµ.
Theorem 5.16 (Bochner). Let f : Ω → X be measurable. Then, f is integrable in the
Bochner sense if and only if
Ω
f dµ < +∞.
As L1
X , respectively L1
X (Ω,S,µ) is denoted the space of all Bochner integrable func-
tions.
Theorem 5.17. For f ∈ L1
X and E ∈ S, the inequality
E
f dµ ≤
E
f dµ
is valid.
Theorem 5.18. After identification functions differing on a set with zero measure, L1
X
is Banach space with the norm
f L1
X
=
Ω
f dµ.
Definition 5.19. Indefinite Bochner integral is F : S → X defined by
F(E) =
E
f dµ,
where E ∈ S.
Theorem 5.20. Let function f : Ω → X be Bochner integrable. Indefinite Bochner
integral is σ-additive vector measure, which is absolutely continuous with respect to
µ. If En ∈ S are pairwise disjoint, the series ∑F(En) converges absolutely.
5.3 Gelfand and Pettis integral
Lemma 5.21. Let f : Ω → X. If ϕ ◦ f ∈ L1
X for all ϕ ∈ X , then for each E ∈ S exists
LE ∈ X such that
LE(ϕ) =
E
ϕ ◦ f dµ
for all ϕ ∈ X .
56
Definition 5.22. Let f : Ω → X. If ϕ ◦ f ∈ L1
X for all ϕ ∈ X , we say that f is weakly
integrable function. Element LE ∈ X, which existence is guaranteed by Lemma 5.21,
is called Gelfand integral of f over E. Similarly as for Bochner integral is being introduced
indefinite Gelfand integral, which E ∈ S assigns values LE ∈ X .
In addition, if LE ∈ X for all E ∈ S, i.e., there exists pE ∈ X such that LE(ϕ) =
ϕ(pE) for all ϕ ∈ X , we say that f is Pettis integrable. Element pE ∈ X is called Pettis
integral of f over E. Indefinite Pettis integral assigns for E ∈ S values pE ∈ X.
Remark 5.23. If X is reflexive space, Gelfand integral merges with Pettis integral.
Theorem 5.24. Let function f : Ω → X be weakly integrable. Then, the following
conditions are equivalent:
i) f is Pettis integrable;
ii) indefinite Gelfand integral of f is σ-additive vector measure;
iii) indefinite Gelfand integral of f is absolutely continuous with respect to µ.
Remark 5.25. If function f is Bochner integrable, f is also Pettis integrable. More
specifically, the relationship between “strong” and “weak” integral describes following
theorem.
Theorem 5.26. Let f : Ω → X be measurable and Pettis integrable function. Then, f
is Bochner integrable if and only if indefinite Pettis integral of f is vector measure of
bounded variation.
Theorem 5.27. Let K be compact subset of Banach space X and µ be probability
complete measure on K. If function f : K → X is continuous, then there exists Pettis
integral of f over K.
Definition 5.28. Let function f : [0,1] → X be given. Let
D = {0 = x0 < x1 < ··· < xn = 1}
be partition of interval [0,1]. We denote
D(f,D) =
n
∑
i=1
sup{ f(s)− f(t) ;s,t ∈ [xi−1,xi]}·(xi −xi−1).
We say that f is Darboux integrable, if for all ε > 0 exists δ > 0 such that D(f,D) < ε
if norm of partition D is smaller than δ.
Theorem 5.29. Function f : [0,1] → X is Darboux integrable if and only if f is bounded
function, which is continuous on [0,1] up to the set with zero Lebesgue measure.
Theorem 5.30. Any function f : [0,1] → X integrable in the Darboux sense is integrable
in the Bochner sense.
57