Functional Analysis
Anteprima
ESTRATTO DOCUMENTO
Contents
Introduction 4
Chapter 1. Normed Vector Spaces 5
1. Vector Spaces 5
2. Normed Vector Spaces 8
3. Continuity and Convergence in Normed Spaces 11
4. Topology in Normed Spaces 12
5. Banach Spaces 14
p
Chapter 2. Lebesgue Integration and L spaces 21
1. A Reminder on Riemann’s Integral 21
2. The Lebesgue Measure and Integrable Functions 22
p
3. The L Spaces 29
Chapter 3. Linear Operators 32
1. Basic Definitions 32
2. Three Cornerstones of Functional Analysis 36
Chapter 4. Dual Spaces and Weak Topologies 42
1. Linear Functionals and Dual Spaces 42
2. The Weak Topology 45
3. Sobolev Spaces and Distributions 48
Chapter 5. Hilbert Spaces 52
1. Basic Definitions 52
2. Orthogonality 54
3. Fourier Series in Hilbert Spaces 57
4. Dual Spaces and Adjoint Operators 60
Chapter 6. Spectral Theory for Bounded Linear Operators 69
Chapter 7. Some Applications 73
3
Introduction
Roughly speaking, “Functional Analysis” is Linear Algebra and Analysis in infinite dimen
sional vector spaces. Its development started about 100 years ago from the observation that
problems in quite different fields of mathematics enjoy similar features and properties. By
omitting unnecessary details FA provides an effective and unified approach to these prob
lems. The advantage of this abstract approach is that it concentrates on the essential facts,
so that these facts become clearly visible since the attention is not disturbed by unimportant
details. In this respect this abstract method is the simplest and most economical method
for treating certain mathematical systems. Moreover, it helps to find relations between fields
which have at a first glance no contact with each other.
In the abstract approach, one usually starts from a set of elements and operations satisfying
certain axioms. The theory then consists of logical consequences resulting from the axioms
and are derived as theorems. These general theorems can then be applied to various special
sets satisfying those axioms.
In this connection the concept of “space” is used in a wide and general sense. By choosing
different sets of axioms one obtains different types of abstract spaces. This idea goes back to
(1906, definition of metric space) and is justified by its great success.
Maurice Fréchet 4
CHAPTER 1
Normed Vector Spaces
In the spirit of the Introduction, we start by pursuing the following
3 n
k·k) k·k),
Aim 1.1. Generalize the concept of the euclidian space (R , (or, more generally, (R ,
1
≥
n 1), where n ∈ ∀ ≤ ≤
:= x = (x , . . . , x )x 1 i n ,
R R
1 n i
n 1
X 2
2 n
kxk x  ∈
:= = length of the vector x .
R
i
i=1
n
In the first place, is a
R 1. Vector Spaces
6 ∅
Definition 1.2. Let K be a field (e.g. K = or X = a set and consider two mappings
R C),
× → 7→
+ : X X X, (x, y) x + y (addition)
· × → 7→ ·
: K X X, (α, y) α x (scalar multiplication)
∀ ∈ ∀ ∈
such that x, y, z X, α, β K
(i) x + y = y + x (commutativity),
(ii) (x + y) + z = x + (y + z) (associativity),
∃ ∈
(iii) 0 X such that 0 + x = x (∃ zero element),
∃ − ∈
(iv) x X such that x + (−x) = 0 (∃ inverse element),
· · ·
(v) (α + β) x = α x + β x (distributivity),
· · ·
(vi) α (x + y) = α x + α y (distributivity),
· · · ·
(vii) (α β) x = α (β x) (associativity),
·
(viii) 1 x = x (1 = unit element of K).
·,
Then X (more precisely (X, K, +) is called a vector space (over the field K).
• ⇐⇒
Remarks 1.3. (i)–(iv) (X, +) is an abelian group.
• If X is a Kvector space, then ∈
x X = vectors,
∈
α K = scalars.
• We will only consider vector spaces over (i.e., real vector spaces) and (i.e., complex
R C
vector spaces) and use the notation to indicate either of them.
K
Let’s consider some n
•
Examples 1.4. X = with the operations
K
(x , . . . , x ) + (y , . . . , y ) := (x + y , . . . , x + y )
1 n 1 n 1 1 n n
· · ·
α (x , . . . , x ) := (α x , . . . , α x )
1 n 1 n
is a space.
Kvector
1 Here A := B or B =: A indicates that A = B by definition
5
6 1. NORMED VECTOR SPACES
• 6 ∅
Let S = be a set, then {f →
F (S) : = : S K}
= set of all functions on S
Kvalued
equipped with the operations
(f + g)(s) := f (s) + g(s),
· ·
(α f )(s) := α f (s)
n
is a space. Note that = F ({1, 2, . . . , n}).
Kvector K
• The set X = C([a, b], = C[a, b] of all continuous, functions on [a, b] is a
K) Kvalued
space with respect to the addidion and scalar multiplication defined in the
Kvector
previous example.
⊂
Remark 1.5. C[a, b] F ([a, b]), i.e. C[a, b] is a subspace of F ([a, b]) in the following sense.
⊂
Definition 1.6. Let X be a vector space and let Y X. If
· · ∈ ∀ ∈ ∀ ∈
(1.1) α x + β y Y x, y Y, α, β K
 {z }
=linear combination
(i.e., Y is closed under addition and scalar multiplication), then Y is called (linear) subspace
of X. •
Remarks 1.7. A subspace of a vector space is always a vector space.
• Condition (1.1) is equivalent to
· ∈ ∈ ∀ ∈ ∀ ∈
α x Y and x + y Y x, y Y, α K.
3 3
• {(x, ∈ ⊂
Examples 1.8. Y = 0, z)x, z is a subspace of .
R} R R
• ∈ {f → 
If s S, then : S f (s ) = 0} is a subspace of F (S).
K
0 0
• P {p ⊂
:= : p is a polynomial} C(R) is a subspace of C(R).
• An arbitrary intersection of subspaces of a vector space X is again a subspace. However,
the union of two subspaces in general is not a subspace anymore.
• Let 1
Z 2
→ f ⊂
Y := f : [0, 1] : f is Riemann integrable and (s) ds < +∞ F ([0, 1]).
R 0
Then Y is a subspace since
∈ ⇒ · ∈ ·
– f Y α f Y , i.e. α f is Riemann integrable and
X,
1 1
Z Z
2 2 2
α · α · f
f (s) ds = (s) ds < +∞ X,
0 0
∈ ⇒ ∈
– f, g Y f + g Y , i.e., f + g is Rintegrable and
X,
1
Z 2
f (s) + g(s) ds < +∞.
0
Here the last statement follows from Schwarz’s inequality
2 1 1
1 Z Z
Z 2 2
≤ f · g(s)
· (s) ds ds
f (s) g(s) ds
0 0 0
which implies
1 1
Z Z
2 2 2
f ≤ f · f g(s)
(s) + g(s) ds (s) + 2 (s)g(s) + ds < +∞.
0 0 1. VECTOR SPACES 7
∈
Translating a subspace Y of a vector space X by some fixed vector x X we obtain
0
{x ∈ }
x + Y := + y : y Y = affine subspace.
0 0
∀ ⊂
If Y is maximal (i.e. subspaces Z X satisfying Y Z it follows Z = X), then x + Y is
$ 0
called hyperplane. 2
•
Examples 1.9. Every line passing through the origin is a subspace of , every line is
R
2
a hyperplane in .
R 3
• Every line or plane passing through the origin is a subspace of , every line is an affine
R
3 3
subspace of , planes are exactly the hyperplanes in .
R R
Next we study how vector spaces can be “generated”. ∈
Definition 1.10. Let X be a vector space, then x , . . . x X are called
1 n
• linearly independent if
n
X ⇒
· =0 α = α = . . . = α = 0;
α x 1 2 n
k k
k=1 {z }

= linear combination nk=1
P ·
• α x = 0.
linearly dependent if there exist α , . . . , α not all = 0, such that
1 n k k
• ⊂
More generally a set V X is called linearly independent if every nonempty finite subset
of V is linearly independent 3
• ∈
Examples 1.11. (1, 0, 0), (1, 1, 0), (1, 1, 1) are linearly independent while (1, 1),
R
2
∈
(−1, 1), (1, 1) are linear dependent.
R 2
k
• {p ∈ } ⊂
Let p (s) = s , then : k C[0, 1] is linearly independent. In fact, any linear
N 0
k k ∈
combination of the p , k being = 0 is a polynomial having infinitely many zeros
N
0
k
and hence is zero. ∈
Definition 1.12. If X ia a vector space and x , . . . , x X, then
1 n
n
X ∈
·
} α , . . . , α
α x
span{x , . . . , x := K
1 n
1 n k k
k=1 ⊂
is called the subspace generated (or spanned) by x , . . . , x . If V X (possibly infinite), then
1 n
we define its span as the set of all (finite!) linear combinations of elements in V , i.e.,
n
X · ∈ ∈ ∈
span V := α x x , . . . , x V, α , . . . , α n .
K, N
1 n 1 n
k k
k=1 ⊂
In other words, for a subset V X of a vector space X, span V is the smallest subspace of
X containing V . 3
• −1), {(x, −x) ∈ ⊂
Examples 1.13. span{(1, 0, (0, 1, 0)} = y, : x, y .
R} R
k
• } {p ≤
If p (s) = s , then span{p , . . . , p = : p is a polynomial of degree n}, while
0 n
k ∈ } {p
span{p : k = : p is a polynomial}.
N 0
k }
Definition 1.14. If X = span{x , . . . , x for n linearly independent vectors x , . . . , x , then
1 n 1 n
{x }
we say that X has dimension dim(X) := n and call , . . . , x a (Hamel) basis of X. If in
1 n
X there exists an infinite linearly independent subset, then X is called infinite dimensional
and we write dim(X) = +∞.
• {x } {y }
Remarks 1.15. If , . . . , x and , . . . , y are two bases of X, then necessarily
1 n 1 m
m = n, i.e., the concept of dimension is well defined.
• 6 {0}
Using Zorn’s lemma one can show that every vector space = has a basis.
2 ∪ {0}.
Here and in the sequel :=
N N
0
8 1. NORMED VECTOR SPACES
•
Examples 1.16. is a onedimensional space.
K Kvector
• considered as a space has dimension 2. For example = span{1, i} and 1
C Rvector C
and i are linearly independent over R.
• considered as a space is infinite dimensional.
R Qvector 3
• ∈
The vectors e := (1, 0, 0), e := (0, 1, 0) and e := (0, 0, 1) are linearly independent
R
1 2 3
3 3
}.
and = span{e , e , e Hence is 3dimensional.
R R
1 2 3
n
• More generally, dim(K ) = n.
• S S
If denotes the number of elements of the set S (where = +∞ if S is infinite), then
S.
dim F (S) = k
• P {p ∈ } ⊂
As we have seen above, = : n C[0, 1] (p (s) = s ) is linearly independent
N
n 0 k
and hence C[0, 1] is infinite dimensional. n
As we will see next among the finite dimensional spaces is fundamental. To this
Kvector K
end we need the following
Definition 1.17. Two vector spaces X and X over the same field are (algebraically)
K
1 2
→ ∀ ∈ ∀ ∈
isomorphic if there exists a bijective map T : X X such that α , α x , x X
K,
1 2 1 2 1 2
· · · ·
T (α x + α x ) = α T (x ) + α T (x ) (i.e., T is linear )
1 1 2 2 1 1 2 2
Now one can show: n
Proposition 1.18. Every ndimensional space is isomorphic to .
Kvector K
2. Normed Vector Spaces n
Up to know we generalized only the algebraic structure of . Now we add to the previous
R
considerations the concept of “length” of a vector. This will then allow us to measure the
distance between two vectors.
Definition 1.19. If X is a space, a map
Kvector
k · k →
: X := [0, +∞)
R
+
∀ ∈ ∀ ∈
is called a norm on X if x, y X, α K
kxk ⇐⇒
(i) = 0 x = 0 (positive definiteness);
kx ≤ kxk kyk
(ii) + yk + (triangle inequality);
kα · α · kxk
(iii) xk = (positive homogeneity).
k · k
X equipped with a norm is called a normed (vector) space
•  ·   · )
Examples 1.20. The absolute value is a norm on hence (K, is a one
K,
dimensional normed vector space.
n
• k · k
More generally, (K , ) where for x = (x , . . . , x ) we define
2 1 n
1
n
2
X 2
kxk x 
:=
2 k
k=1
is a normed space. Here the properties (i) and (iii) of a norm are trivially satisfied, while
the triangle inequality in (ii) follows from Schwarz’s inequality
n n n
2
X
X X 2 2
· ≤ x  · y 
x y
k k k k
k=1 k=1 k=1
2. NORMED VECTOR SPACES 9
• k · k
The 2norm introduced in the previous example is the natural (euclidian) general
2
ization of length from 3 to n dimensions. However, this is by far not the only example
n
of a norm on and also
K kxk x ,
:= max
∞ k
1≤k≤n
n 1
X p
p ≤
kxk kx  , (1 p < +∞)
:=
p k
k=1
n
define norms on . The only nontrivial part is to show the triangle inequality for the
K 3
k · k ≤
pnorm , 1 p < +∞, which follows from Minkowski’s inequality
p 1 1 1
n n n
p p p
X X X
p p
p ≤ x  y 
x  +
+ y k k
k k
k=1 k=1 k=1
If in the last example we (formally) consider n = +∞ we arrive at various sequence spaces:
Example 1.21. Let
∈ ∀ ∈
X := x = (x , x , x , . . .) : x k = space of all sequences.
K N Kvalued
1 2 3 k
Since X = F (N) it is clear that X is an infinite dimensional vector space with respect to
coordinatewise operations. On X we define the (possibly infinitevalued) maps
k · k → ∪ {+∞}, kxk x ,
: X := sup
R
∞ ∞
+ k
k∈N
+∞ 1
X p
p ≤
k · k → ∪ {+∞}, kxk x  , (1 p < +∞).
: X :=
R
p + p k
k=1
Considering only elements with finite norm we obtain
( bounded sequences if p = +∞,
p {x ∈ kxk
l := X : < +∞} =
p ≤
psummable sequences if 1 p < +∞.
p k · k
By Minkowski’s inequality on page 9 for n = +∞ it follows that (l , ) is a normed space
p
∀ ≤ ≤
1 p +∞. Moreover, ∞
1 p ∀ ∞.
l l l 1 < p <
$ $
1 p p
∈ ⇐⇒ ∈ ⇐⇒
For example ( ) l p > 1, while (1) l p = +∞.
n∈N n∈N
n ≤
Example 1.22. Let X := C[a, b], then for 1 p < +∞
kf k f
:= sup (s),
∞ s∈[a,b]
b 1
Z p
p
kf k f ≤
:= (s) ds , (1 p < +∞)
p a ≤ ∞
define norms on X. This is trivial for p = +∞. For 1 p < condition (i) of the norm
follows from b
Z h(s) ⇒ ∀ ∈
∈ ds = 0 h(s) = 0 s [a, b].
h C[a, b], a
Finally, the triangle inequality follows from Minkowski’s inequality
1 1 1
b b b
Z Z Z
p p p
p p p
f ≤ f g(s)
(s) + g(s) ds (s) ds + ds
a a a
3 This inequality also holds for n = +∞ if the two series on the righthandside converge.
10 1. NORMED VECTOR SPACES
As we have seen the same vector space can be equipped with different norms, which might
have quite different properties (see below). To relate two normed spaces we make the following
k · k k · k
Definition 1.23. Two normed vector spaces (X, ) and (Y, ) are called topologically
X Y
→
isomorphic if there exists a bijective linear map T : X Y and constants c , c > 0 such
1 2
that · kxk ≤ kT ≤ · kxk ∀ ∈
c xk c x X.
1 2
X Y X
'
In this case we write X Y .
In finite dimensions we have
k · k
Proposition 1.24. If (X, ) is a ndimensional normed space over then
K,
X n
'
X ,
K
n
where can be equipped with an arbitrary norm.
K
If in the previous definition we can choose c = c = 1 we arrive at the following
1 2
k · k k · k
Definition 1.25. Two normed vector spaces (X, ) and (Y, ) are called isometrically
X Y
→
isomorphic if there exists a bijective linear map T : X Y such that
kT kxk ∀ ∈
(∗) xk = x X.
Y X
Moreover, a map satisfying (∗) is called an isometry.
Starting from one or more normed space(s), next we will construct new related ones. The first
construction is very simple. k · k
Definition 1.26. A subspace Y of a normed space (X, ) with the induced norm
X
kyk kyk ∈
:= , y Y
Y X
is called a normed subspace of X.
{p →  ≤
Example 1.27. Let Y := : [0, 1] p is a real polynomial of degree 5} then Y is a
R
k · k
normed subspace of (C[0, 1], ) if equipped with the norm
∞ kpk p(s).
:= sup
Y s∈[0,1]
Next we consider products of normed spaces.
k · k k · k
Proposition 1.28. Let (X , ), . . . , (X , ) be normed spaces and define
1 n
X X
n
1
× · · · × {(x ∈ ∀ ≤ ≤
X := X X := , . . . , x ) : x X 1 k n}.
1 n 1 n k k
Then each of the following maps
k · k → kxk kx k
: X , := max ,
R
∞ ∞
+ X
k k
1≤k≤n
n 1
pX
X p ≤
k · k → kxk kx k , (1 p < +∞).
: X , :=
R
p + p k k
k=1
k · k ≤ ≤
defines a norm on X, i.e., (X, ) is a normed space for all 1 p +∞.
p
3. CONTINUITY AND CONVERGENCE IN NORMED SPACES 11
3. Continuity and Convergence in Normed Spaces
With the concept of “length” (=norm) of a vector we can measure distances between two
elements x, y in a normed space X:
kx − yk = distance between x and y.
Using this fact it is very easy to generalize concepts like “continuity”, “convergence” etc. from
to normed spaces.
R → k · k
Definition 1.29. Consider the map F : X Y between two normed spaces (X, ) and
X
k · k
(Y, ). Then F is said to be
Y
• ∈ ∀ ∃ kF −
continuous in x X, if ε > 0 δ = δ(x , ε) > 0 such that (x ) F (x)k < ε
0 0 0 Y
∀ ∈ kx −
x X satisfying xk < δ;
0 X
• ∈
continuous if it is continuous in every x X;
0
• ∀ ∃ kF −
uniformly continuous if ε > 0 δ = δ(ε) > 0 such that (x ) F (x )k < ε
0 1 Y
∀ ∈ kx − k
x , x X satisfying x < δ.
0 1 0 1 X
• ⇒
Examples 1.30. It is clear that “uniform continuity” “continuity”, while the con
→
verse even in a onedimensional normed space is not true. For example F : R R,
2
F (x) = x is continuous but not uniformly continuous.
n m n m
• → k · k
Let F : be linear where we equip and with the supnorm .
K K K K ∞
4 t
×
Then F can be represented by a m nmatrix A = (a ) and for x = (x , . . . , x ) ,
ij m×n 1 n
01 0 t
x = (x , . . . , x ) we obtain
0 n
kF − kF −
(x) F (x )k = (x x )k
∞ ∞
0 0
n
0
P · −
a (x x )
1j j j
j=1
..
=
.
n
0
P · −
a (x x )
mj j j
j=1 ∞
n
X 0
· −
= max a (x x )
ij j j
1≤i≤m j=1
n
X 0
≤ a  · (x −
max max x )
ij j j
1≤i≤m 1≤j≤m
j=1
0
· kx − k
= q x ∞
nj=1
P a 
for q := max = maximal row sum of A. Hence, given ε > 0 we obtain
1≤i≤m ij
ε kx − k
and x < δ
for δ := 0
q+1 ε
kF − ≤ · kx − k ·
(x) F (x )k q x < q < ε.
∞ ∞
0 0 q +1
This shows that F is uniformly continuous.
Remark 1.31. Since every finite dimensional normed space X is topologically isomorphic to
n k · k) k · k → k · k
some (K , (see above, page 10), every linear map F : (X, ) (Y, ) with finite
X Y
dimensional X is uniformly continuous.
Next we consider sequences in normed spaces.
4 t
Here (. . .) denotes the transposed vector.
12 1. NORMED VECTOR SPACES
⊂ k · k ∈
Definition 1.32. We say that the sequence (x ) (X, ) converges to x X if
n 0
X
n∈N
kx − k
lim x = 0.
0 n X
n→+∞
Remarks 1.33. As in the scalar case
• the limit of a sequence, if it exists, is unique;
• sequences can be used to characterize continuity:
lim F (x ) = F (x )
n 0
k · k → k · k
F : (X, ) (Y, )
X Y n→+∞
⇐⇒
∈ ∀ ⊂
is continuous in x X (x ) X with lim x = x .
0 n n 0
n∈N
n→+∞
n k kn
• ⊂
If X = equipped with an arbitrary norm an (x ) X, x = (x , . . . , x ), then
K k k∈N k 1
k 01
lim x = x ,
1
k→+∞
..
⇐⇒
lim x = x .
0
k
k→+∞ kn 0
lim x = x .
n
k→+∞
Hence convergence in norm is equivalent to convergence in each coordinate.
Next we consider series in normed spaces.
+∞
P ⊂ k · k
x for elements (x ) (X, ) is said to converge
Definition 1.34. The series n X
n∈N
k
k=1
∈
to the sum x X if n
X
−
lim x
x = 0.
k
n→+∞ X
k=1
 {z }
partial sum
4. Topology in Normed Spaces
In real analysis the concepts of open, closed and compact sets are of fundamental importance.
Using the notion of convergence of sequences it is easy to generalize them to normed spaces.
k · k ⊂
Definition 1.35. Let (X, ) be a normed space and A X a subset. Then
X
• ⊂ ∈
A is closed if (x ) A, lim x = x in X implies x A;
n n 0 0
n∈N n→+∞
• \
A is open if X A is closed;
• ∈ ⊂ \ {x }
x X is an accumulation point of A if there exists (x ) A such that
0 n 0
n∈N
lim x = x ;
n 0
n→+∞
• ∪ {x ∈
Ā := A X : x is an accumulation point of A} is called the closure of A in X.
0 0
• ⇐⇒ ∀ ∈ ∃ {x ∈ kx−x k ⊂
Remarks 1.36. A is open x A ε > 0 such that X : < ε} A,
0 0
∈
i.e., each x A is an interior point of A.
0
• Ā is the smallest closed set containing A. By the following proposition it can be repre
sented as \
Ā = C.
A⊆C
C closed
•
Proposition 1.37. Arbitrary unions of open sets are open, arbitrary intersections of
closed sets are closed;
• finite unions of closed sets are closed, finite intersections of open sets are open;
•
Examples 1.38. In general, arbitrary unions of closed sets are not closed anymore. For
1
1 − ∪
, 1 ], then each A is closed while A = (0, 1)
example if X = and A := [
R n n n
n∈N
n n
is not closed anymore. 4. TOPOLOGY IN NORMED SPACES 13
• In general, arbitrary intersections of open sets are not open anymore. For example if
1 1 ∩ {0}
X = and A := (− , ), then each A is open while A = is not open
R n n n
n∈N
n n
anymore.
Definition 1.39. A subset A of a normed space X is called dense, if Ā = X.
∈ ∈
Hence A is dense in X means that any x X can be approximated by elements y A as
∈ kx −
closely as we like, i.e., for every ε > 0 there exists y A such that yk < ε.
•
Examples 1.40. is dense in
Q R.
• Consider the vector space
∈ ∃ ∈ ∀ ∈
Φ : = (x ) F (N) : k such that x = 0 k
N N
0
k k∈N k
= space of all finite sequences.
Kvalued
Then p ≤ ≤
– Φ l for all 1 p +∞;
$ p ⇐⇒ ≤
– Φ is dense in l 1 p < +∞.
k·k
∞
– Φ = c , where
0
∈
c := (x ) F (N) : lim x = 0 = space of all null sequences.
0 k k∈N k
k→+∞
Here the first assertion is trivial. The third assertion and the first part of the second
p
∈
assertion follows by truncation of a given sequence x l while the second part is clear
∞
∈
by the fact that for e := (1, 1, 1, 1, . . .) l we have
ke − ≥ ∀ ∈
xk 1 x Φ.
∞
• P
By Weierstrass’s approximation theorem the space of all real polynomials is dense in
k · k
(C([a, b], ).
R), ∞
Definition 1.41. A normed vector space is said to be separable, if it contains a countable
dense subset S. •
Examples 1.42. Every finite dimensional normed space X is separable. (Consider the
set of all rational linear combinations of a finite set of vectors forming a basis of X.)
p
• ≤ ≤ ⇐⇒ ≤ {(x ∈
X := l , 1 p +∞ is separable 1 p < +∞. In fact Φ := ) Φ :
k k∈N
Q
≤
all x are rational} is dense in X for 1 p < +∞.
k
• k · k
(C[a, b], ) is separable. (Consider polynomials wit rational coefficients.)
∞ ⊂ k · k
Definition 1.43. A set A (X, ) is called bounded if
X kxk
sup < +∞.
X
x∈A
Next we look at compactness. ⊂
Definition 1.44. Let X be a normed space and A X a subset.
• (A ) (here J denotes an arbitrary index set) is called an open cover of A if each A is
i i
i∈J
open and [
⊂
A A .
i
i∈J
• A is compact if every open cover (A ) has a finite subcover, i.e., if there exists a finite
i i∈J
S
⊂ ⊂
I J such that A A .
i
i∈I
• A is relatively compact if the closure Ā is compact.
The next result relates some of the previous concepts.
k · k ⊂
Proposition 1.45. Let (X, ) be a normed space and A X.
X
14 1. NORMED VECTOR SPACES
• ⇐⇒
A is compact every sequence in A has a subsequence converging in A (i.e., A is
⇐⇒
sequentially compact) every infinite subset in A has an accumulation point in A
(Bolzano–Weierstraß property).
• If A is compact, then it is bounded and closed; the converse holds if and only if dim(X) <
+∞ (Theorem of HeineBorel).
• ⊂
If A is compact and C A, then C is relatively compact. In particular, every closed
subset of a compact set is compact. {x ∈ kxk ≤
Example 1.46. In a normed space X let U := X : 1} be the closed unit ball.
Then U is always bounded and closed. However, U is compact if and only if dim X < +∞.
5
p
≤ ≤ k · k {e ∈ ⊂
For example, for 1 p +∞ take X = (l , ). Then A := : k U where
N}
p k
ke − k ∀ 6
e := (δ ) . However, A has no accumulation points since e = 1 k = l.
n∈N
k nk k l
Next we consider continuous maps on compact domains ⊂ →
Proposition 1.47. Let X, Y be two normed spaces, A X be compact and let F : A Y
be continuous. Then
• {f ∈
F (A) := (x) : x A} is compact in Y ;
• if Y = then F achieves its minimum and its maximum (in particular F is bounded).
R, n
⊂
Example 1.48. By the previous proposition it follows that if K is compact
K
kf k f
:= sup (s) < +∞
∞ s∈K
defines a norm on C(K). Compact sets in C(K) are characterized by the following result.
⊂ k · k
Theorem 1.49 (Arzela–Ascoli Theorem). A set A (C(K), ) is relatively compact if
∞
and only if
{f ∈ ∈
(i) (s) : f A} is bounded in for every s K (i.e. A is pointwise bounded), and
K
(ii) A is equicontinuous.
⊂ ∈
Here a set A C(K) is called equicontinuous, if for every s K and every ε > 0 there exists
∈
δ > 0 such that for all f A ∈ s −
− < ε if r K satisfies r < δ.
f (s) f (r)
Remark 1.50. The concept of open sets (and consequently closed and compact sets) can be
defined in the much more general context of “topological spaces”.
5. Banach Spaces
 · )  · )
Note that both (Q, and (R, are normed spaces, however the latter is much more
important. The fundamental difference between these spaces is their completeness: while R
is complete is not. The completeness of can be described in various ways (e.g. using the
Q R
order structure), in a normed space we use the following concept.
⊂ k · k
Definition 1.51. A sequence (x ) (X, ) is called Cauchy sequence if for every
n X
n∈N
ε > 0 there exists n = n (ε) such that
0 0 kx − k ∀ ≥
x < ε n, m n .
n m 0
− → →
Hence (x ) is Cauchy, if x x 0 as n, m +∞. It is easy to show that every
n n m
n∈N
convergent sequence in a normed space is Cauchy, however the converse in general does not
hold.
5 6
Here δ denotes the Kronecker delta: δ = 1 if n = k, δ = 0 if n = k.
nk nk nk
5. BANACH SPACES 15
Example 1.52. Let X = C[−1, 1] be equipped with the 1norm
1
Z
kf k f
= (s) ds.
1 −1
⊂
Consider the sequence (f ) X defined by
n n∈N ( 1 ∈
s if s (0, 1]
n
f (s) :=
n ∈
0 if s [−1, 0].
This sequence is Cauchy, since for n > m 1
1 1
1 1+
1+
Z s
s n m
1 1
kf − k −
−
f = s s ds =
n m
n m 1 1 1
1 + 1+
0 0
n m
1 1
− → →
= 0 as m, n +∞.
1 1
1 + 1+
n m
Clearly, the pointwise limit of f is
n ( ∈
1 if s (0, 1]
f (s) := lim f (s) =
n ∈
0 if s [−1, 0].
n→+∞
∈
which defines a function f / X because of the discontinuity in s = 0.
The previous example justifies the following fundamental definition
Definition 1.53. Let X be a normed space. Then
• X is said to be complete if every Cauchy sequence in X converges in X;
• X is called Banach space if it is complete.
•
Examples 1.54. Every finitedimensional normed space over is complete.
K
• k · k ≤ ≤
(Φ, ), 1 p +∞ is never complete. For example
p 1 1 1 1 ∈
x := , , , . . . , 0, . . . Φ
n n
1 2 3 2
2 2 2 k · k
defines a Cauchy sequence (x ) in (Φ, ) converging to the infinite sequence
n p
n∈N
1 p
∈ \
( ) l Φ.
n∈N
n
2
• k · k
Since uniform limits of continuous functions are continuous, (C[a, b], ) is complete,
∞
hence a Banach space. n
• ⊂ kk
More generally, if K is compact, then (C(K), ) is a Banach space.
R ∞
p
• ≤ ≤
For each 1 p +∞ the sequence space l is a Banach space.
• ⊂ k · k
A subspace Y X of a Banach space (X, ) is complete if and only if it is closed in
X
X.
Proposition 1.55. Let X be a normed space. Then there exists a Banach space X̃ and an
→ {T ∈ ⊂
isometry T : X X̃ with dense range T (X) = (x) : x X} X̃. The space X̃ is unique
up to isometric isomorphy and is called the completion of X.
•
Examples 1.56. The completion of is i.e., =
Q R, Q̃ R.
• k · k
Using similar arguments as in Example 1.52 one can show that X = (C[a, b], ) (cf.
p
≤
Example 1.22) for 1 p < +∞ is not complete. It is clear that its completion must
contain discontinuous functions. One might guess that
∼
k · k
X̃ = C[a, b], p
coincides with the space of all Riemann integrable functions on [a, b] with finite pnorm.
However, this space is not complete either. In Chapter 2 below we will introduce a more
sophisticated type of integration in order to obtain the explicit representation
∼ p
k · k
C[a, b], = L [a, b],
p
16 1. NORMED VECTOR SPACES
6
the space of all measurable, pintegrable functions.
The following examples are important for the study of differential equations.
n
• ⊂ ∈
More Examples 1.57. For an open set Ω and k we consider the (vector)
R N
0
space of all real, ktimes continuously differentiable functions on Ω, i.e.,
α
( )
D f exists and can be continuously
k →
C (Ω) := f : Ω .
R ∀ α ≤
extended to Ω̄ k
Here α
∂ f
α
D f := α α
· · · n
1 ∂
∂ n
1 n k
α ∈
where := α + . . . + α for a multiindex (α , . . . , α ) . Then C (Ω) equipped
N
1 n 1 n 0
with the norm α
kf k kD k
:= max f ∞
k,∞ 0≤α≤k
is complete, hence a Banach space.
∞
• Let X := C [a, b], the (vector) space of all infinitely differential functions on [a, b]. Then
∈ ≤
for all k 1 p < +∞
N, 1
k
b
Z p
p
X i
kf k ds
D f (s)
:= ,
k,p a i=0
i
where D denotes the ith derivative, defines a norm on X for which X is not complete.
k · k
In this way we obtain a whole scale of normed spaces (X, ) and since
k,p
kf k ≥ kf k ∀ ∈
f X
k+1,p k,p
it follows that ∼ ∼
∞ ∞
k · k ⊂ k · k
C [a, b], C [a, b],
k+1,p k,p
with continuous inclusion.
The previous example can be generalized to higher dimensions.
n
• ⊂
For an open set Ω we consider the (vector) space of all infinitely differentiable
R
functions on Ω having compact support, i.e.
supp f is compact and
∞ →
C (Ω) := f : Ω .
R α n
c ∀ ∈
D f exists α N
0
{s ∈ 6 ∈
Here the support of f is given by supp f := Ω : f (s) = 0}. Then for all k and
N
0
≤
1 p < +∞ 1
Z p
X p
α
kf k D
:= f (s) ds ,
k,p Ω α≤k k · k
defines a norm on X for which X is not complete. The completions of (X, )
k,p
are called Sobolev spaces, and are very important in the study of partial differential
equations. We will come back to these spaces later, see Definition 4.23.
We close this chapter by the following important result which has numerous applications in
applied mathematics. First we need a → ≤
Definition 1.58. Let X be a normed space and F : X X. If there exists 0 q < 1 such
that kF − ≤ · kx − ∀ ∈
(x) F (y)k q yk x, y X,
then F is called a contraction.
6 in the sense of Lebesgue 5. BANACH SPACES 17
It is easy to see that every contraction is uniformly continuous.
7 →
Theorem 1.59 (Banach Fixed Point Theorem ). Let X be a Banach space and F : X X
∗ ∗
∈
be a contraction. Then there exists a unique fixed point x X of F , i.e., a unique x such
that ∗ ∗
F (x ) = x .
∈
Moreover, if we choose an arbitrary x X and define (recursively) the sequence (x ) by
0 n n∈N
x := F (x ),
n+1 n
∗
n
then lim x = lim F (x ) = x and
n 0
n→+∞ n→+∞ n
q
∗
kx − k ≤ · kx −
x F (x )k.
n 0 0
−
1 q +∞ 1
n
P q =
Using the triangle inequality and the geometric series
Sketch of Proof: n=0 1−q
∗
one shows that (x ) is a Cauchy sequence. Since X is complete, lim x =: x exists and
n n
n∈N n→+∞
from the continuity of F one obtains
∗ ∗
F (x ) = F lim x = lim F (x ) = lim x = x .
n n n+1
n→+∞ n→+∞ n→+∞
∗ ∗
Finally, if we suppose that x and y are two different fixed points of F , then from
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
kx − k kF − ≤ · kx − k kx − k
y = (x ) F (y )k q y < y
it follows a contradiction.
Note that the previous proof is “constructive”, i.e., not only gives existence of the fixed point
but also a procedure how to construct it. This fact is used in various numerical methods in
the form of iteration schemes.
Frequently in applications the following corollary is very useful. k
→
Corollary 1.60. If X is a Banach space and F : X X is such that F is a contraction
∈
for some k then F has a unique fixed point.
N, ∗ ∗
k
By Banach’s fixed point theorem G := F has a unique fixed point x = Gx =
Proof. ∗
n ∈
lim G (x ) for arbitrary x X. Hence, choosing x := F (x ) we obtain
0 0 0
n→+∞
∗ ∗ ∗ ∗ ∗ ∗
n n·k+1 n n
x = lim G F (x ) = lim F (x ) = lim F G x = F lim G (x ) = F (x ),
n→+∞ n→+∞ n→+∞ n→+∞
∗
i.e., x is a fixed point of T . Since every fixed point of T is also a fixed point of G, this fixed
point is unique.
As already mentioned above, the previous two results have various applications.
n
∈
Example 1.61 (Linear systems). For a matrix A = (a ) and b we consider the
K
ij n×n
n n
→
map F : defined by
K K F (x) := Ax + b.
n
P a 
If max =: q < 1 (row sum criterion), then by the calculations from the
1≤i≤n ij
j=1 n k · k
example on page 11 it follows that F is a contraction on (K , ). Hence the linear system
∞
x = Ax + b
∗ ∗ k
has a unique solution x . This solution is given by x = lim F (x ) for any initial value
0
n→+∞
n
∈
x , i.e., can be obtained as the limit of the iteration scheme
K
0 ∈
x := Ax + b, m .
N
m+1 m 0
7 also called Contraction Mapping Principle
18 1. NORMED VECTOR SPACES ×
We note that there are various ways to transform a linear system Cx = d for a given n n
n
∈
matrix C and a given d into the fixed point form x = Ax + b.
K × →
Example 1.62 (PicardLindelöf Theorem). For a continuous function f : and
R R R
∈
u we consider the Cauchy problem
R
0 ( 0
u (t) = f u(t), t ,
(CP) u(t ) = u .
0 0
Substituting in (CP) t by s and integrating from s = t to s = t we obtain the Volterra
0
8
integral equation t
Z
f u(s), s ds,
u(t) = u +
0 t
0
which is equivalent (proof!) to (CP). Now we choose δ > 0 and consider on the Banach space
− k · k →
X := (C[t δ, t + δ], ) the operator F : X X defined by
∞
0 0 t
Z
f u(s), s ds.
F (u) (t) := u +
0 t
0
∈ ∈
Since f is continuous, F is welldefined, i.e., F (u) X for all u X. Moreover, u is a solution
⇐⇒
of (CP) F (u) = u.
In order to apply Banach’s fixed point theorem, we assume that there exists a Lipschitz
constant L > 0 such that ≤ · u − ∀ ∈ ∀ ∈ −
− L v u, v t [t δ, t + δ].
f (u, t) f (v, t) R, 0 0
∈
Using this we conclude for u(·), v(·) X t
Z
− −
F (u) (t) F (v) (t) = ds
f u(s), s f v(s), s
t
0
t
Z · −
≤ L u(s) v(s) ds
t
0
≤ · t −  · ku −
L t vk .
∞
0
Hence t
Z
2 2
− −
F (u) (t) F (v) (t) = f [F (u)](s), s f [F (v)](s), s ds
t
0
t
Z
≤ · −
L [F (u)](s) [F (v)](s) ds
t
0  {z }
≤L·s−t ·ku−vk ∞
0
t
Z
2
≤ · s −  · ku −
L t ds vk ∞
0
t
0 2
t − 
t
0
2 · · ku −
= L vk .
∞
2
Proceeding in this way one can show by induction that
k k
· t − 
L t
0
k k
− ≤ · ku − ∀ ∈ ∀ ∈ − ∀u, ∈
F (u) (t) F (v) (t) vk k t [t δ, t + δ], v X.
N,
∞ 0 0
k!
This implies that k k
·
L δ
k k
− ≤ · ku − ∀ ∈ ∀u, ∈
F (u) F (v) vk k v X.
N,
∞
k!
8 Integral equations had a huge effect on the development and promotion of FA.
5. BANACH SPACES 19
Since k k
·
L δ
lim =0
k!
k→+∞
k
this shows that F for k sufficiently large is a contraction and hence F has a unique fixed
∈ −
point u which gives the unique solution u(t), t [t δ, t + δ] of (CP). Since δ > 0 can be
0 0
∈
chosen arbitrarily large, the solution exists for all times t R.
Banach’s fixed point theorem does not depend on the linear (i.e. vector space) structure of
the underlying space and is valid in the more general context of “complete metric spaces”.
We will not proceed in this direction but at least give a “local” version in Banach spaces.
⊂ k·k →
Theorem 1.63. Let S X be a closed subset of a Banach space (X, ) and let F : S S
X
k ∈
be a map such that G := F is a contraction for some k Then F has a unique fixed point
N.
∗ ∈
x S given by ∗ n
x = lim G (x )
0
n→+∞
∈
for any x S.
0 x 2x
• {1, }
Exercises 1.64. Show that the set e , e is linearly independent in C[0, 1].
• {f ∈ {f ∈
Which of the sets X := X : f (0) = 1}, X := X : f (1) = 0},
1 2
{f ∈ ≥
X := X : f 0} are subspaces or affine subspaces of C[0, 1]?
3
• ⊂
Let X be a space and Y X be a subset of X. Show that the following
Kvector
assertions are equivalent:
 Y is a hyperplane,
∈ → 6
 there exists x X and a linear functional ϕ : X ϕ = 0, such that Y =
K,
0 {x ∈
x + ker(ϕ), where ker(ϕ) := X : ϕ(x) = 0} denotes the kernel of ϕ.
0
• f
Does p(f ) := (0) define a norm on C[0, 1]?
• k · k
Let (X, ) be a normed space. Show that
X ≤ kx − ∀ ∈
kxk − kyk yk x, y X
k · k →
and conclude that the norm : X is always continuous.
R
• {x ∈ kxk ≤
Show that the closed unit ball U := X : 1} in a normed space is always
X
9
closed and convex . 1
2 p p
• → x y
For 0 < p < 1 consider the map f : , f (x, y) := + . Explain why
p
R R
p + p
2 ≤
this does not define a norm on . (Hint: Look at f (x, y) 1 and use the previous
R p
exercise.) 2 2
• {x ∈ kxk ≤ ≤
Draw the unit spheres S := : = 1} in for 1 p +∞. What kind of
R R
p p
kxk kxk
relation between and for p < p does this picture imply?
p p 1 2
1 2
• k · k k · k
Two norms and on a vector space X are called equivalent if there exist
1 2
c , c > 0 such that
1 2 · kxk ≤ kxk ≤ · kxk ∀ ∈
c c x X.
1 1 2 2 1
k · k ' k · k
In this case we write . Show that all norms on a finite dimensional vector
1 2
space are equivalent.
• k · k k · k
True or false? Let X be a vector space equipped with two norms and . Define
1 2
k · k
X := (X, ) for k = 1, 2. Then
k k ' ⇐⇒ k · k ' k · k
X X .
1 2 1 2
• k · k k · k
Show that the norms and are not equivalent on C[0, 1]. Which relation holds
∞
1
between this two norms?
9 ∈ ⇒ · − · ∈ ∀ ∈
A set A in a vector space is called convex if x, y A α x + (1 α) y A α (0, 1).
20 1. NORMED VECTOR SPACES
∞
• {s ⊂
Show that X := l is not separable. (Hint: for a set S = , s , s , . . .} X, where
1 2 3
k k k ∈
s = (s , s , s , . . .) consider x = (x , x , x , . . .) X defined by
1 2 3
k 1 2 3 ( kk kk
s  ≤
s + 1 if 1,
x :=
k 0 else. n n
• →
Find conditions implying that a linear map A = (a ) : is a contraction on
K K
ij n×n
n n
k · k k · k
(K , ) or (K , ). Use these conditions in order to solve linear equations of the
1 2
form Ax + b = x as in the example on page 17.
n n n
• →
As before, let A = (a ) : be a linear map where is equipped with an
K K K
ij n×n k
∈ ⇐⇒ λ
arbitrary norm. Show that there exists k such that A is a contraction < 1
N
for all eigenvalues λ of A.
• Apply in an appropriate way Banach’s fixed point theorem to the function F (s) :=
1
√ 3
3 − −
1 + s to show that f (s) := s s 1 has a unique zero in [1, 2]. Approximate this zero
−2
with an error < 10 . Explain why one can not argue using
3 −
– F (s) = s 1 on R;
2 1 on [1 + δ, +∞) for some δ > 0 sufficiently small.
– F (s) =
3 2 −1
s
• Using in an appropriate way Banach’s fixed point theorem, calculate the solution of the
following Cauchy problem: ( 0
u (t) = u(t),
(CP) u(0) = 1.
• Using in an appropriate way Banach’s fixed point theorem, prove that the equation
s
Z · · ∈
u(s) = 1 + s r u(r) dr, s [0, 1]
0
∈
has a unique solution u(·) C[0, 1]. Moreover, find a power series representation of u(s).
• ∈
Using in an appropriate way Banach’s fixed point theorem, show that for every g C[a, b]
∈
there exists a unique solution u C[a, b] of the following equations
(a) Volterra integral equation
t
Z
− · · ∈
u(t) µ k(t, s) u(s) ds = g(t), t [a, b],
a
∈ × ∈
where k C([a, b] [a, b]) and µ C.
(b) Fredholm integral equation
b
Z
− · · ∈
u(t) µ k(t, s) u(s) ds = g(t), t [a, b],
a
∈ × ∈
where k C([a, b] [a, b]) and µ sufficiently small.
C
• Compare the solvability of Volterra’s and Fredholm’s integral equations in the previous
n
exercise with the solvability of the following linear systems in X = , where x =
C
t t
∈ ∈ ∈ ×
(x , . . . , x ) X, b = (b , . . . , b ) X, µ and A = (a ) is a n n matrix
C
1 n 1 n ij n×n
i
X
− · · ≤ ≤
(a) x µ a x = b , 1 i n;
i ij j i
j=1
n
X
− · · ≤ ≤
(b) x µ a x = b , 1 i n;
i ij j i
j=1
≤ ≤
satisfying a = 0, 1 i n in case (a).
ii CHAPTER 2 p
Lebesgue Integration and L spaces
1. A Reminder on Riemann’s Integral
→
If f : [a, b] is bounded and P : a = s < s < . . . < s = b is a partition of [a, b] we
R 0 1 m
∈ ∈ ≤ ≤
define m := inf{f (s) : s [s , s ]} and M := sup{f (s) : s [s , s ]} for 1 i m. Then
i i−1 i i i−1 i
the expressions m
X · −
s(P, f ) : = m (s s ) (lower sum),
i i i−1
k=1
m
X · −
S(P, f ) : = M (s s ) (upper sum)
i i i−1
k=1
approximate the area between the graph of f and the saxis from below and from above,
respectively. f (s)
f (s) M
M S(f, P )
s(f, P ) m
m s
s
s = b
a = s s s s s a = s s s s s s = b
5
0 1 2 3 4 0 1 2 3 4 5
Lower sum s(P, f ) and upper sum S(P, f ) (m = 5).
Figure 1.
Clearly, the best approximation from below would be the biggest possible one, while the best
from above the smallest possible one and coincidence of the two would give the area between
the graph of f and the saxis. However, these best approximations usually do not exist and
we are thus led to the following 1 →
Definition 2.1 (Riemann Integral ). A bounded function f : [a, b] is called Riemann
R
integrable if
sup{s(P, f ) : P partition of [a, b]} = inf{S(P, f ) : P partition of [a, b]} =: A
In this case we set b
Z f (s) ds := A (Riemann integral).
a
While for many purposes the Riemann integral is sufficient, for our needs it is too restrictive.
For example the space of Riemann integrable functions is not closed with respect to pointwise
convergence.
1 More precisely, Riemann–Darboux integral 21 p
22 2. LEBESGUE INTEGRATION AND L SPACES ⊂
Example 2.2. Since is countable there exists a sequence (r ) such that [0, 1]∩Q =
Q Q
k k∈N
{r ∈
: k Now define
N}.
k ( ∈ {r },
1 if s , . . . , r
1 k
f (s) :=
k 0 else.
Then all f are Riemann integrable (having integral = 0) and converge pointwise, i.e.,
k ( ∈ ∩
1 if s [0, 1] Q,
lim f (s) =: f (s) =
k 0 else.
k→+∞
However, the limit function f is not Riemann integrable. In fact, since is dense in the
Q R,
{r ∈
set : k is dense in [0, 1] and hence for every partition P of [0, 1] it follows
N}
k s(P, f ) = 0, S(P, f ) = 1.
Summing up, the Riemann integral is simple to define, however has drawbacks which make
it of few use in advanced mathematics. In the sequel we will construct a “better” theory of
integration which then will be used to define certain very important (complete) spaces of
integrable functions.
2. The Lebesgue Measure and Integrable Functions
While in Riemann’s approach to integration the domain of a function f is partitioned, in
Lebesgue’s approach the range of the function will be divided. To explain the basic idea
→ ⊆
suppose that f : D [0, +∞) is a nonnegative function defined on some domain D R.
Then every partition 0 = t < t < . . . < t
0 1 m 2
induces a partition of of the domain D, i.e., we can decompose
(
m ∈ ≤ ≤ ≤ −
D := s D : t f (s) < t if 0 i m 1
˙ i i i+1
[
D = D where
i ∈ ≤
D := s D : t f (s) .
i=0 m m
This partition can be used to approximate the area between the graph of f and the saxis
from below by the expression (cf. Figure 2)
m m
X X
· ·
t λ(D ) = t λ(D ),
i i i i
i=0 i=1
where λ(D ) denotes the “measure” of the set D . If D is just an interval, then clearly λ(D )
i i i i
is just its length. However, in general the sets D will be much more complicated and it is by
i
no means clear how to define λ(D ). Hence, before we pursue this idea to construct a “better”
i
integral we need to define the measure of a set.
2.1. The Outer Measure. For simplicity we considered above domains in but now we
R n
⊂
turn to the general case and pose the problem to define the “measure” of a set D . In
R
ni=1
×
case D = B = [a , b ] is a box things are easy and we define its measure as
i i n
Y −
l(B) := (b a ).
i i
i=1
For an arbitrary set we make the following
2 ˙
∪”
Here “ denotes the disjoint union of sets.
2. THE LEBESGUE MEASURE AND INTEGRABLE FUNCTIONS 23
f (s)
t
2 m
X ·
t λ(D )
i i
i=1
t
1 s
t 0 D 2
D 0 D
1
m
˙
∪
Decomposition D = D of the domain (m = 2).
Figure 2. i
i=0 n
⊆
Definition 2.3. The Lebesgue Outer Measure of a set D is
R
+∞ +∞
X [
∗ ⊆
λ (D) := inf l(B ) : D B ,
i i
i=1 i=1
where each B denotes a box.
i ∗
≥
Note that l(B ) 0 and hence the sum of the series appearing in the definition of λ (D) is
i ∗
independent of the order of summation, i.e., λ (D) is welldefined.
∗ n
P(R
The function λ is defined on the whole power set ) and takes values in [0, +∞] =:
∪ {+∞},
[0, +∞) i.e., ∗ n
P(R → ∪ {+∞}.
λ : ) [0, +∞] =: [0, +∞)
∗ n
⊂
We remark that λ (B) = l(B) for every box B . Some further properties of the outer
R
measure are collected in the following
∗
Proposition 2.4. (i) λ (∅) = 0.
∗ ∗
n
⊂ ⊂ ≤
(ii) If D D then λ (D ) λ (D ) (i.e., the outer measure is monotone).
R
1 2 1 2
n
⊂ ∈
(iii) For every sequence D , k one has
R N
k
+∞ +∞
[ X
∗ ∗
≤
λ D λ (D ) (countable subadditivity)
k k
k=1 k=1 {s ∈
Example 2.5. We show that every countable set D = : k has outer measure
N}
k
∗ n
⊂
λ (D) = 0. To this end choose ε > 0 and a box B such that
R
k ∗ ε
∈ ≤
s B and λ (B ) .
k k k k
2
+∞
S
⊂
Then clearly D B and by monotony and subadditivity we conclude
k
k=1 +∞ +∞ +∞
[ X X
∗ ∗ ∗ ε
≤ ≤ ≤
λ (D) λ B λ (B ) = ε.
k k k
2
k=1 k=1 k=1
∗
Since we can choose ε > 0 arbitrarily small this implies λ (D) = 0 as claimed.
Remark 2.6. Not every set having outer measure zero is countable. For example the Cantor
∗
set C is an example of an uncountable set having outer measure λ (C) = 0.
p
24 2. LEBESGUE INTEGRATION AND L SPACES
One would expect in Proposition 2.4.(iii) to have equality if the sets D are pairwise disjoint.
k
However, in general ∗ ∗ ∗
n
⊆ ∩ ∅ 6⇒ ∪
(2.1) D , D , D D = λ (D D ) = λ (D ) + λ (D )
R
1 2 1 2 1 2 1 2
∗ ∗
i.e., λ is not even finitely additive. Surprisingly, this is not a defect of the function λ but of
∗
n
P(R
its domain ) which is too big. If one restricts λ to a smaller domain Σ, things become
much nicer, and as we well see below one gets even countable additivity.
n n
⊂ ⊂
Definition 2.7. A set D is called (Lebesgue) measurable if for every subset S
R R
one has ∗ ∗ ∗
∩ \
(2.2) λ (S) = λ (S D) + λ (S D). ∗
Moreover, if D is measurable we define its Lebesgue measure λ(D) := λ (D).
n
⊂
We mention that one can show that D is measurable iff for every ε > 0 there exist a
R ∗
n n
⊂ ⊂ ⊂ ⊂ \
closed set C and an open set O such that C D O and λ (O C) < ε.
R R n
⊂
Returning to the previous definition, we note that for arbitrary sets S, D one has
R
∩ ∪ \
S = (S D) (S D) and hence by subadditivity
∗ ∗ ∗
≤ ∩ \
(2.3) λ (S) λ (S D) + λ (S D) ≥
always holds. This shows that a set is already measurable if in (2.2) we only have “ ”.
The (vague) motivation for the definition of measurable sets (going back to C. Caratheodory)
∗
is that the sets we want to single out as measurable should be such that λ will become additive
on them. ∗
n
⊂
Example 2.8. We show that every set D of outer measure λ (D) = 0 is measurable.
R n
⊂
By monotony of the outer measure we have for every set S R
∗ ∗
≤ ∩ ≤
0 λ (S D) λ (D) = 0
∗ ∗ ∗
∩ ≥ \ ⊇ \
hence λ (S D) = 0. Thus we have only to show that λ (S) λ (S D). Since S S D
∗
this follows again by the monotonicity of λ . n
Example 2.9. One can show that every open and every closed subset of is measurable.
R
It is not easy to “construct” nonmeasurable sets and one needs the axiom of choice to do so.
∗
n
⊂
However, one can show that every set D having outer measure λ (D) > 0 contains a
R
nonmeasurable subset, i.e. there exist very many nonmeasurable sets. Moreover we mention
that nonmeasurable sets have very weird properties. As an example we refer to the Banach
Tarski paradox which also gives an example for the failing implication in (2.1).
Next we study the set n n
{D ⊂ ⊂ P(R
Σ := : D is measurable} )
R
and the restriction →
λ : Σ [0, +∞]
∗
of the outer measure λ . As we will see next the socalled Lebesgue measure λ has much better
∗
properties than λ .
Proposition 2.10. (a) Σ is a σalgebra, i.e., it satisfies
∅ ∈
(i) Σ, c n
∈ \ ∈
(ii) D Σ implies that also its complement D := D Σ,
R
S
∈ ∈ ∈
(iii) D Σ for all k implies D Σ.
N
k k
k∈N
(b) λ is a measure, i.e., it satisfies
(i) λ(∅) = 0, S P
∈ ∈
(ii) if all D Σ, k are pairwise disjoint then λ D = λ(D ).
N
k k k
k∈N k∈N
2. THE LEBESGUE MEASURE AND INTEGRABLE FUNCTIONS 25
Having defined the Lebesgue measure λ on the σalgebra of all Lebesgue measurable sets we
return to our original problem of constructing a more general theory of integration.
2.2. Integrable Functions. We first need to define the class of functions for which Lebesgue’s
approach sketched above possibly works. 3
n
⊆ → is called (Lebesgue)
Definition 2.11. Let D be a measurable set. A function f : D R
R
∈
measurable if for all t R {s ∈ ∈
D : f (s) < t} Σ →
Since Σ is a σalgebra it is not difficult to show that f : D is measurable if and only of
R
one of the following properties is satisfied:
{s ∈ ∈ ∈
(i) D : f (s) > t} Σ for all t R;
{s ∈ ≤ ∈ ∈
(ii) D : f (s) t} Σ for all t R;
{s ∈ ≥ ∈ ∈
(iii) D : f (s) t} Σ for all t R;
{s ∈ ≤ ∈ ∈
(iv) D : r f (s) < t} Σ for all r, t R.
Using this characterizations of measurability one can verify the following result.
∈ →
Proposition 2.12. Let D Σ and f, g : D Then
R.
−f f ,
(i) If f is measurable then also , f and f are measurable. Here we define
−
+ −f
f (x)+f (x) (x)+f (x)
≥ ≥
f (x) := max{f (x), 0} = 0, f (x) := max{−f (x), 0} = 0.
−
+ 2 2
·
(ii) If f and g are measurable then also f + g and f g are measurable.
(iii) If f is continuous, then it is measurable. n
⊂
Next we introduce an important class of measurable functions. Here for a set D we
R
define its characteristic function as ( ∈
1 if s D,
1 (s) :=
D 0 else.
1
Using the above characterizations it is clear that is measurable if and only if D is mea
D
surable.
Example 2.13. Since D = is countable it is also measurable (use Examples 2.5 and 2.8)
Q 1
and hence the Dirichlet function is a measurable.
Q n →
Definition 2.14 (Simple function). A function ϕ : is called simple if it assumes
R R
only a finite number of values.
If α , . . . , α are the distinct nonzero values of the simple function ϕ then we can write
1 l l 1
X ·
(2.4) ϕ(s) = α (s),
D
k k
k=1
n
{s ∈ }.
where D := : ϕ(s) = α It is clear that ϕ is measurable if and only if all D , . . . , D
R 1
k k l
are measurable.
For nonnegative simple functions its simple to define the integral.
n →
Definition 2.15. If ϕ : [0, +∞] is a measurable simple function of the form (2.4) and
R 4
∈
if D Σ, we define the (Lebesgue) integral of ϕ in D as
l
Z X · ∩ ∈
ϕ dλ := α λ(D D) [0, +∞].
k k
D k=1
3 ∪ {−∞,
Here we define := +∞}.
R R
4 ∩ · ∩
In case α = +∞ and λ(D D) = 0 we define α λ(D D) := 0.
k k k k p
26 2. LEBESGUE INTEGRATION AND L SPACES
Next, using the order in we will extend Lebesgue’s integral to arbitrary nonnegative mea
R
surable functions. ∈ →
Definition 2.16. If D Σ and f : D [0, +∞] is measurable we define the (Lebesgue)
integral of f in D as
Z
Z ≤ ∀ ∈
f dλ := sup ϕ dλ ϕ simple, nonnegative, measurable & ϕ(s) f (s) s D .
D D
Note that for simple functions we have defined the integral in two ways, however these assign
the same value and hence all is well defined.
After these preparations we can finally give the following
⊆
Definition 2.17 (Lebesgue integral). Let D be measurable.
R.
→ ∈
(i) A measurable nonnegative function f : D [0, +∞] on a domain D Σ is called
R
(Lebesgue) integrable if f dλ < +∞.
D ∈
→ on a domain D Σ is called (Lebesgue) integrable
(ii) A measurable function f : D R
if the nonnegative functions f and f (cf. Proposition 2.12.(i)) are integrable. In this
−
+
case one defines the (Lebesgue) integral of f in D as
Z Z Z
−
f dλ := f dλ f dλ.
−
+
D D D
1
Example 2.18. The Dirichlet function (see Example 2.13) is Lebesgue integrable on
Q
D = [0, 1] with Lebesgue integral = 0. Note that we verified in Example 2.2 that this function
is not Riemann integrable.
We close this section by noting that by decomposing a complexvalued function in its real
∈
and imaginary part it is easy to extend the above definitions. More precisely, if D Σ and
→
f : D we define the realvalued functions
C ¯ ¯
1 1
→ − →
Re f := (f + f ) : D Im f := (f f ) : D
R, R.
2 2i
·
and call f = Re f + i Im f
(i) measurable if Re f and Im f are both measurable,
(ii) integrable if Re f and Im f are both integrable.
Moreover, in the latter case we define
Z Z Z
·
f dλ := (Re f ) dλ + i (Im f ) dλ
D D D
2.3. Properties of the Lebesgue Integral. In this subsection we will collect some of the
principal results on the Lebesgue integral. We start with the following important facts.
∈ →
Proposition 2.19. Let D Σ and f, g : D be integrable. then
C
∈ · ·
(i) for all α, β also α f + β g is integrable and
C
Z Z Z
· · · ·
(α f + β g) dλ = α f dλ + β g dλ (i.e., the integral is linear),
D D D
≤
(ii) if f and g are realvalued and f g then
Z Z
≤
f dλ g dλ (i.e., the integral is monotone),
D D
f 
(iii) also is integrable and
Z Z
≤ f 
f dλ dλ (triangle inequality),
D D
2. THE LEBESGUE MEASURE AND INTEGRABLE FUNCTIONS 27
∪ ∈ ∩
(iv) if D = D D with D , D Σ and λ(D D ) = 0 then
1 2 1 2 1 2
Z
Z
Z f dλ.
f dλ +
f dλ = D
D
D 2
1
In the sequel the following notation which will be quite handy.
Definition 2.20. We say that a property P holds almost everywhere (a.e.) in a measurable
n
⊆
set D if
R
{s ∈
λ D : P is wrong} = 0 1
Examples 2.21. (i) Since λ(Q) = 0 the Dirichlet function (s) = 0 a.e.
Q
6
(ii) f = g a.e. means λ({s : f (s) = g(s)}) = 0.
→ → 6→
(iii) f f a.e. as k +∞ means λ({s : f (s) f (s)}) = 0.
k k
Using this notation we will now compare the Riemann and Lebesgue integral. 5
→
Proposition 2.22. A bounded function f : [a, b] is Riemann integrable iff it is contin
R
uous a.e. In this case it is also Lebesgue integrable and both integrals coincide, i.e.
b
Z Z
f (s) ds = f dλ.
a [a,b]
By this result rules to calculate Riemann integrals carry over to to Lebesgue integrals. Note
also that in the definition of the Lebesgue integral neither the domain D nor the function f
had to be bounded and thus the definition of the Lebesgue integral also includes improper in
tegrals. However, not every function which is improperly Riemann integrable is also Lebesgue
integrable, see Figure 3. In this example the Riemann integral is given by the converging
f (s)
1 1 1
3 5 s
16
−
1 14
−
1
− 2
Improperly Riemann but not Lebesgue integrable function.
Figure 3.
series +∞
+∞
Z X k+1 1
f (s) ds = (−1) = ln(2)
k
0 k=1 f 
On the other hand, f is not Lebesgue integrable. Otherwise, by Proposition 2.19.(iii) also
would be Lebesgue integrable, yielding a contradiction since
+∞
Z 1
X
f  dλ = = +∞.
k
,
[0 +∞) k=1
Our motivation to construct the Lebesgue integral were the bad properties of the Riemann
integral concerning, e.g., limits, see Example 2.2. We will now study how Lebesgue’s integrals
behaves in this respect.
5 “iff” means “if and only if”. p
28 2. LEBESGUE INTEGRATION AND L SPACES ∈
Theorem 2.23 (Lebesgue’s dominated convergence theorem). Let D Σ be measurable and
→ ∈ → →
let f : D k be a sequence of integrable functions such that f f a.e. as k +∞.
R, N,
k k
→
If there exists an integrable function g : D such that
R
f  ≤ ∈
(2.5) g for all k a.e. in D,
N
k
then also f is integrable and Z Z
Z f −  f dλ = f dλ.
f dλ = 0 and lim
lim k
k k→+∞
k→+∞ D D
D
We show that condition (2.5) cannot be omitted.
→
Example 2.24. For α > 0 define f : [0, 1] as in Figure 4
R
k
α
h = k
k f (s)
k 1
α
s s
2
1 1
k k The sequence f .
Figure 4. k
∈
Then lim f (s) = 0 =: f (s) for all s [0, 1] hence
k
k→+∞ Z f dλ = 0.
[0,1]
On the other hand, 1
Z Z h k α−1
f dλ = f (s) ds = = k .
k k k
[0,1] 0
If we now choose
• α = 1 then Z 6 ∈
f dλ = 1 = 0 for all k N;
k
[0,1]
• α > 1 then Z → →
f dλ +∞ as k +∞;
k
[0,1]
• 0 < α < 1, then Z → →
f dλ 0 as k +∞.
k
[0,1] 1
≤ ∈
This follows also from Theorem 2.23 since in this case f (s) g(s) := for all k N,
k α
α·s
∈
s [0, 1] and g is integrable on [0, 1]. ∈
Theorem 2.25 (Lebesgue’s monotone convergence theorem). Let D Σ be measurable and
→ ∈
let f : D [0, +∞), k be an increasing sequence of integrable functions such that
N,
k
→ →
f f as k +∞. Then f is measurable and
k Z Z
lim f dλ = f dλ.
k
k→+∞ D D
p
3. THE L SPACES 29
Example 2.26. We consider again the sequence (f ) of simple functions from Example 2.2
k k∈N ≥ ≥
converging to the Dirichlet function f . Since every f is measurable and f f 0 for
k k+1 k
∈
all k by Lebesgue’s monotone convergence theorem f is measurable and
N, Z
Z f dλ.
f dλ = 0 =
lim k
k→+∞ [0,1]
[0,1] 2
We close this section with a result which allows to calculate integrals in via iterated
R
integrals in R. × → ⊆
Theorem 2.27 (Fubini–Tonelli ). Let f : X Y be a measurable function where X, Y
R
are measurable. Then f is integrable iff at least one of the iterated integrals
R Z Z
Z Z
f f
(x, y) dλ(y) dλ(x), (x, y) dλ(x) dλ(y)
X Y Y X
is finite, and in this case Z Z
Z Z Z
f (x, y) dλ(x, y) = f (x, y) dλ(y) dλ(x) = f (x, y) dλ(x) dλ(y).
X×Y X Y Y X 2
Here “λ(x), λ(y)” and “λ(x, y)” denote the Lebesgue measures on and , respectively.
R R
p
3. The L Spaces p
The aim of this section is to define the socalled Lebesgue spaces L (Ω) for a measurable set
n
⊆ ≤ ≤
Ω and 1 p +∞. To this end we first introduce the spaces
R
n o
p
→ f  ≤
f : Ω : f is measurable and is integrable if 1 p < +∞,
K
p
L (Ω) := n o
→ ≥ ≤
f : Ω : there exists M 0 such that f M a.e. if p = +∞.
K
p
L
Next we equip (Ω) with a (semi)norm. To this end we define
1
p
p
R f  ≤
dλ if 1 p < +∞,
p
L →
N : (Ω) [0, +∞), N (f ) := Ω
p p ≥ f  ≤
inf{M 0 : M a.e.} if p = +∞.
The following inequalities which we already encountered in Chapter 1 are important in order
to proceed. 1
1
6
≤ ≤ ≤ ≤
Theorem 2.28. Let 1 p +∞ and take 1 q +∞ such that + = 1.
p q
p q 1
∈ L ∈ L · ∈ L
(i) If f (Ω) and g (Ω) then f g (Ω) and
· ≤ ·
N (f g) N (f ) N (g) (Hölder’s inequality)
1 p q
2 1
∈ L · ∈ L
(ii) If f, g (Ω) then f g (Ω) and
· ≤ ·
N (f g) N (f ) N (g) (Schwarz’s inequality)
1 2 2
p p
∈ L ∈ L
(iii) If f, g (Ω) then f + g (Ω) and
≤
N (f + g) N (f ) + N (g) (Minkowski’s inequality)
1 p p
6 1
Here we define := 0
+∞ p
30 2. LEBESGUE INTEGRATION AND L SPACES
p
L
Using Minkowski’s inequality one easily verifies that (Ω) is a vector space. Moreover, it
follows that N fulfils conditions (ii) and (iii) of Definition 1.1.19 for a norm (i.e. it is a
p
seminorm) but not condition (i) since 6⇒
N (f ) = 0 f = 0.
p p
L ∼
The idea to resolve this problem is to introduce on (Ω) an equivalence relation and then
∼
consider the equivalence classes modulo . More precisely, we define
∼ ⇐⇒
f g f = g a.e.
and
p
L (Ω)
p p
∈ L
L (Ω) := = [f ] : f (Ω) .
∼
p
{g ∈ L ∼
Here [f ] := (Ω) : f g} denotes an equivalence class which are the elements of the
p
L
p (Ω)
quotient space L (Ω) = / . This construction just formalizes the fact that we identify
∼ p
two functions which are equal a.e. Now L (Ω) is a vector space for the operations
· · · · ∈
α [f ] + β [g] := [α f + β g], α, β K
and [f ] := N (f )
p
p
p
defines a norm on L (Ω). One can verify that these definitions are welldefined, i.e., indepen
dent of the special choice of the representant of an equivalence class. Moreover, the inequalities
1 1
p ≤ ≤
in Theorem 2.2.28 carry over to L (Ω), i.e., if 1 p, q +∞ satisfy + = 1 then
p q
p q 1
∈ ∈ ⇒ · ∈ kf · ≤ kf k · kgk
f L (Ω), g L (Ω) f g L (Ω) & gk (Hölder);
1 p q
2 1
∈ ⇒ · ∈ kf · ≤ kf k · kgk
f, g L (Ω) f g L (Ω) & gk (Schwarz);
1 2 2
p 1
∈ ⇒ ∈ kf ≤ kf k kgk
f, g L (Ω) f + g L (Ω) & + gk + (Minkowski).
1 p p
In abuse of notation it is common to identify [f ] with f and work with it as if it was a
function. This turns out to be very convenient, and after some practice one almost forgets
about the fact that these objects are not really functions but sets of equivalent functions. The
p
∈
most annoying consequence of this sloppiness is that for f L (Ω) the value f (s) of f at a
∈
single point s Ω is meaningless. In fact, if we change f in a single point s we remain in the
same equivalence class since the measure of the set λ({s}) = 0.
We are now able to state the main result of this section.
n p
⊆ ≤ ≤ k · k
Theorem 2.29. If Ω is measurable and 1 p +∞, then (L (Ω), ) is a Banach
R p
∞ p ≤
space. Moreover, C (Ω) (cf. page 16) is dense dense in L (Ω) for all 1 p < +∞. Finally,
c
if λ(Ω) < +∞ then ∞ p 1
⊂ ⊂ ⊂ ⊂
L (Ω) . . . L (Ω) . . . L (Ω)
and there exist positive constants c such that
p
· kf k ≤ ≤ · kf k ≤ ≤ · kf k
c . . . c . . . c
∞ ∞
1 1 p p p
Note the “opposite” behavior to the sequence spaces l in Example 1.1.21. where we saw that
∞
1 p
⊂ ⊂ ⊂ ⊂
l . . . l . . . l .
From the previous result we easily obtain
≤
Corollary 2.30. For all 1 p < +∞ ∼ p
k · k
C[a, b], = L [a, b].
p ∗
n
• ⊂
Exercises 2.31. A set N is called null set if λ (N ) = 0. Show that
R
– every subset of a null set is a null set;
– the countable union of null sets is a null set;
p
3. THE L SPACES 31
– the union and the intersection of a measurable set and a null set is measurable.
∗ ∗
n
• ∈ ⊆ \ \ ∈
Let D Σ and S such that λ (S D) = λ (D S) = 0. Show that also S Σ.
R
• Show that a countable intersection of measurable sets is measurable.
2
• −
Show that every line in is a null set. More generally, every at most n 1dimensional
R
n
affine subspace of is a null set.
R
• ∈ →
Let D Σ. Show that a function f : D is measurable if and only if it satisfies one
R
of the conditions (i)(iv) following Definition 2.11. n
• → ⊆
Show that every continuous function f : D defined on a measurable set D is
R R
measurable. n
• → ⊆
Consider f, g : D for a measurable set D . Show that if f is measurable and
R R
g = f almost everywhere, then also g is measurable.
• ∈ ∈
Show that if f, g C[a, b], then f = g a.e. implies f (s) = g(s) for all s [a, b].
1
• Show that the Dirichlet function is integrable on every measurable set D with
Q
Lebesgue integral = 0 R R
• →
Let f, g : D be integrable. Show that f = g a.e. implies f dλ = g dλ.
C D D
R
• →
Let f : D [0, +∞) be integrable with f dλ = 0. Show that f = 0 a.e.
D R
• → f 
Let f : D be integrable. Show that f = 0 a.e. iff dλ = 0.
C D
R
• → ⊃ ∈
Let f : D be integrable such that f dλ = 0 for every D E Σ. Show that
C E
f = 0 a.e. p
• N N {f ∈ L
Show that [f ] = f + where = (Ω) : f = 0 a.e.}.
2
• Let α > 0. Study the almost everywhere convergence in L (R) of the sequence (f )
k k∈N
( 1
α ≤ ,
k if 0 x < k
f (x) =
k 0 otherwise.
1
• → →
Show that f f and f g in L (R) implies f = g almost everywhere.
k k
• Let ( ∈
sin(x) if x Q,
f (x) = ∈ \
cos(x) if x R Q.
R
Explain why f is measurable and calculate f dλ in the sense of Lebesgue.
[0,π] p
p
• ≤ ≤ (R).
(R) and L
Show that for 1 p < p +∞ there is no inclusion between L 2
1
1 2 1 1
1 1
· ·
(Hint: Consider the functions f (s) := (s) and g (s) := (s).)
α α
(0,1) (1,+∞)
α α
s s
CHAPTER 3
Linear Operators
In this chapter we will be concerned with linear maps between two normed spaces X and Y
(over the same field of scalars K). 1. Basic Definitions
Definition 3.1. Let X and Y be vector spaces over the same field and let D be a subspace
K
1
⊂ →
of X. A map T : D X Y such that
· · · · ∀ ∈ ∀ ∈
T (α x + α x ) = α T x + α T x α , α x , x D
K,
1 1 2 2 1 1 2 2 1 2 1 2
is called linear operator with domain D = D(T ). If D(T ) = X, and only then, we write
→
T : X Y .
Clearly, linearity immediately implies that
n
n
X
X · ∀ ∈ ∈ ≤ ≤ ∈
· α T x α x D, 1 k n and n
=
T α x K, N.
k k k k
k k k=1
k=1
Linear operators between finite dimensional spaces can (after choosing two bases in X and
×
Y ) always be represented as m nmatrices, where n = dim(X) and m = dim(Y ). Moreover,
→
if dim(X) < +∞ then every linear operator T : X Y is (uniformly) continuous. As we will
see in the sequel, in infinite dimensions things are far more complicated.
The above definition of a linear operator is of purely algebraic nature. However, the main
ingredient for a rich and powerful “operator theory” is the topological structure of the un
derlying spaces. →
Definition 3.2. A linear operator T : X Y between two normed spaces X and Y is said
≥
to be bounded if there exists L 0 such that
kT ≤ · kxk ∀ ∈
xk L x X.
Y X
Note that boundedness of a linear operator not means boundedness on all of X (the only
6 {0}
linear operator which is bounded on X = is T = 0), but that it is bounded on the unit
{x ∈ kxk ≤
ball U := X : 1} (or on any bounded subset of X).
X
As we will see next, for linear operators boundedness and continuity are equivalent.
→
Proposition 3.3. Let T : X Y be a linear operator between two normed vector spaces
k · k k · k
(X, ) and (Y, ). Then the following are equivalent.
X Y
(a) T is bounded.
(b) T is uniformly continuous.
(c) T is continuous in x = 0.
0 ∈
(d) T is continuous in some x X.
0
1 In case T is linear it is common to use the notation T x instead of T (x).
32
1. BASIC DEFINITIONS 33
It is clear that for two vector spaces the set
{T → 
L(X, Y ) := : X Y T is linear}
is again a vector space under the operations · ·
(T + S)x := T x + Sx, (α T )(x) := α (T x),
∈ ∈ ∈
where T, S L(X, Y ), α x X. If X and Y are normed vector spaces then we can make
K,
2
also
L(X, ∈ 
Y ) := T L(X, Y ) T is bounded
into a normed space as follows:
k · k k · k
Proposition 3.4. If (X, ) and (Y, ) are normed vector spaces then
X Y
kT k ≥ kT ≤ · kxk ∀ ∈
(1.1) = inf L 0 : xk L x X .
Y X
L(X,
defines a norm on Y ).
One can show that the operator norm can also be characterized as
kT xk
Y
kT k kT
(1.2) = sup = sup xk .
Y
kxk X kxk
x6 =0 =1
X
From this representations in particular it follows that the “inf” in (1.1) is attained, i.e., is a
“min”. This implies that for bounded T
kT ≤ kT k · kxk ∀ ∈
xk x X
Y X
kT k
and is the best (i.e. smallest) possible constant for which such an estimate holds.
→
Remark 3.5. To show boundedness of a linear operator T : X Y between two normed
k · k k · k ≥
spaces (X, ) and (Y, ) it suffices to find some L 0 such that
X Y kT ≤ · kxk ∀ ∈
xk L x X.
Y X kT k
If one is also interested in calculating the operator norm one has to find the best possible
kT k
constant L. To do so one first has to find a candidate L for by carefully estimating
kT ≤ kxk kT k ≤ ∈ kx k
xk L for = 1. This shows L. If then one finds some x X, = 1
0 0
Y X X
kT k kT k
such that x = L it follows = L.
0 Y
Unfortunately, the sup in (1.2) in general is not attained and only in special cases one finds
⊂ kx k ≤
such x̄. In these cases one has to construct a sequence (x ) X, 1 such that
n n X
n∈N
kT k → → kT k
x L as n +∞. This again proves that = L.
n X •
Examples 3.6. Linear operators between finite dimensional normed spaces are always
bounded. n m
• k · k → k · k
Consider the matrix A = (a ) : (K , ) (K , ). Then from the compu
∞ ∞
ij m×n
tations in Example 1.30 it follows that
n
X
kAk a 
= max (row sum norm).
ij
1≤i≤m j=1
• If in the previous example we change the norms and consider the matrix A = (a ) :
ij m×n
n m
k · k → k · k
(K , ) (K , ), then
1 1 m
X
kAk a 
= max (column sum norm).
ij
1≤j≤n i=1
2 L(X).
If X = Y we simply write
34 3. LINEAR OPERATORS
p
• ≤ ≤
Let X := l for some 1 p +∞, then the leftshift operator
T (x ) := (x )
n n+1
n∈N n∈N
l kT k
is linear and bounded satisfying = 1. Moreover, the rightshift operator
l
T (x ) := (x )
r n n−1
n∈N n∈N kT k
where x := 0 is linear and isometric, hence bounded with norm = 1.
0 r
• ∈ →
Let X := C[a, b] and q X. Then the multiplication operator M : X X defined by
q
· ∈
(M f )(s) := q(s) f (s), s [a, b]
q
kM k kqk
is linear and bounded and = .
∞
q
2
• → ·) ∈
On X := L (a, b) consider the integral operator T : X X with kernel k(·,
2
×
L (a, b) (a, b) defined by b
Z · ∈ ∈
(T f )(t) := k(t, s) f (s) ds, f X, t (a, b).
a
Clearly this operator is welldefined and linear. To show that its bounded we will use
∈
Schwarz inequality (see Theorem 2.2.28) and conclude for t (a, b)
b
Z ·
≤ ds
f (s)
k(t, s)
(T f )(t) a 1 12
b b
Z
Z
2
≤ ·
ds
k(t, s) ds
f (s)
a a
kk(t, ·)k · kf k
= ,
2 2
hence 2 22 22
≤ kk(t, ·)k · kf k .
(T f )(t)
Integrating from a to b with respect to t this implies
b
b Z
Z 2 22 22
≤ kk(t, ·)k · kf k
(T f )(t) dt dt
a
a
and by taking the square root we finally conclude
kT k ≤ kk(·, ·)k · kf k ∀ ∈
f f X.
2 2
2
L (a,b)×(a,b)
Hence, T is bounded and kT k ≤ kk(·, ·)k .
2
L (a,b)×(a,b)
1
• k · k ⊂
Let X := (C[0, 1], ) and let D := C [0, 1] X. Then the differentiation operator
∞
0 n
→ ∈
T : D X, T p := p is linear but not bounded. In fact, if for n we define p (s) := s ,
N n
∈ kp k kT k · kp k → →
s [0, 1], then = 1 while p = n = n +∞ as n +∞.
∞ ∞
n n n−1
L(X,
As we have seen for normed spaces X and Y the space Y ) is again a normed space and
⊂ L(X,
one can ask under which conditions it is complete. If (T ) Y ) is a Cauchy sequence
n n∈N
L(X, ∀ ∈
in Y ) it is easy to see that (T x) is Cauchy in Y x X and in order to define T
n n∈N
as a pointwise limit T x := lim T x
n
n→+∞
we need completeness of Y . This explains the hypothesis in our next result.
k · k k · k
Proposition 3.7. If (X, ) and (Y, ) are normed spaces and Y is complete, then
X Y
L(X, L(X)
Y ) is a Banach space. In particular is a Banach space if X is.
A particularly important case is Y = which clearly is complete.
K
DESCRIZIONE DISPENSA
Normed Vector Spaces: Basic denitions, continuity, convergence and compactness in normed spaces, separable spaces, Banach spaces, Banach fixed point theorem. Integration theory: Riemann integral, Lebesgue measure and Lebesgue integral, Lpspaces. Linear Operators: Basic denitions, bounded linear operators, uniform boundedness theorem, BanachSteinhaus theorem, open mapping theorem, closed graph theorem. Dual Spaces and Weak Topologies: Linear functionals and dual spaces, HahnBanach theorem, weak topology, reexive spaces, best approximation problem. Hilbert Spaces: Basic Denitions, Riesz representation theorem, orthogonality, orthogonal projection theorem, GramSchmidt orthonormalization, Fourier series in Hilbert spaces, dual Spaces and adjoint operators, normal, selfadjoint and unitary operators, Fourier transform. Partial differential equations: Distribution theory, Sobolev spaces, weak derivatives, LaxMilgram theorem. Differential calculus in Banach spaces: Gateaux and Frechet derivatives. Spectral Theory for Bounded Linear Operators: Spectrum and resolvent, Neumann series. Linear system theory: Controllability, stability, observability of linear systems.
I contenuti di questa pagina costituiscono rielaborazioni personali del Publisher Atreyu di informazioni apprese con la frequenza delle lezioni di Functional analysis in applied mathematics and engineering e studio autonomo di eventuali libri di riferimento in preparazione dell'esame finale o della tesi. Non devono intendersi come materiale ufficiale dell'università L'Aquila  Univaq o del prof Engel Klaus Jochen Otto.
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