Che materia stai cercando?

Anteprima

### ESTRATTO DOCUMENTO

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

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

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 (Picard-Lindelöf 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 well-defined, 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

K-vector

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

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 s-axis 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 s-axis. 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 =

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 non-negative 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 s-axis

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 well-defined.

∗ 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

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” non-measurable 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

non-measurable subset, i.e. there exist very many non-measurable sets. Moreover we mention

that non-measurable 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 so-called Lebesgue measure λ has much better

properties than λ .

Proposition 2.10. (a) Σ is a σ-algebra, i.e., it satisfies

∅ ∈

(i) Σ, c n

∈ \ ∈

(ii) D Σ implies that also its complement D := D Σ,

## 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 σ-algebra 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 σ-algebra 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 non-zero 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 non-negative 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 non-negative 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, non-negative, 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 non-negative 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 non-negative 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 complex-valued 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 real-valued 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

## 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 real-valued 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 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 so-called 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

semi-norm) 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 well-defined, 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 1-dimensional

## 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.

## 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.

## 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 n-matrices, 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 .

## 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 .

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 left-shift operator

T (x ) := (x )

n n+1

n∈N n∈N

l kT k

is linear and bounded satisfying = 1. Moreover, the right-shift 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 well-defined 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

## 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

1. BASIC DEFINITIONS 35

Definition 3.8. Let X be a normed space over the field then the Banach space

0 L(X, kϕk

X := := sup ϕ(x)

K), 0

X kxk ≤1

## X

is called the (topological) dual space of X and ϕ : X linear functional on X.

## K

The next chapter will be dedicated to the study of dual spaces.

Let X and Y be normed spaces and D X be a normed subspace of X. If Y is complete,

then every bounded operator T : D Y can be uniquely extended to a bounded operator

→ kT k k k. ⊆

T̄ : D Y preserving the norm, i.e. = T̄ In particular, if D X is dense then T can

∈ L(X,

be uniquely extended to some T̄ Y ).

∈ L(X, ∈ L(Y,

If X, Y and Z are normed spaces and T Y ), S Z) we can consider their product

◦ → ∀ ∈

ST := S T : X Z, (ST )x := S(T x) x X.

∈ L(X,

Then it is clear that ST Z) and

kST k ≤ kSk · kT k. n n

L(X) kT k ≤ kT k

In particular, if X = Y = Z we conclude that is an algebra and for all

n N. ∈ ∈ L(X).

Example 3.9. Let X be a Banach space, t and A Then we can (as for scalars)

## R

define the exponential +∞ k k

t A

tA

e := ,

k!

k=0 n k k

t A

## L(X).

where the series converges in In fact, if T := denotes the partial sum of the

n k!

k=0

exponential series, then for n > m we obtain

n k k

|t| · kAk

## X

kT − k ≤ → → ∞.

T 0 as n, m

n m k!

k=m+1

⊂ L(X) L(X)

Hence (T ) is a Cauchy sequence and since X, hence also is complete it

n n∈N

## L(X).

converges in Moreover, we obtain |t|·kAk

tA

ke k ≤ e .

We can introduce different notions of convergence on spaces of bounded operators. The first

3

## L(X,

one, based on the operator norm in Y ), is called uniform convergence.

⊂ L(X,

Definition 3.10. If X and Y are normed spaces, (T ) Y ) is a sequence of opera-

n n∈N

∈ L(X,

tors and T Y ) such that kT − k

lim T = 0,

n L(X,Y )

n→+∞ →

then we say that (T ) converges uniformly to T as n +∞.

n n∈N ∈ L(X)

Example 3.11. If X is a Banach space and A then the sequence (T ) defined by

n n∈N

the partial sums of the exponential series n k k

t A

X ,

T :=

n k!

k=0

tA ∈ L(X),

converges uniformly to T = e cf. page 35.

In the applications we frequently need the following less restrictive notion of convergence.

3 {x ∈ kxk ≤

Uniform means uniformly on the unit ball U = X : 1}.

## X X

36 3. LINEAR OPERATORS ⊂ L(X,

Definition 3.12. If X and Y are normed spaces, (T ) Y ) is a sequence of opera-

n n∈N

∈ L(X,

tors and T Y ) such that kT − ∀ ∈

lim x T xk = 0 x X

n Y

n→+∞ →

we say that (T ) converges strongly to T as n +∞.

n n∈N

It is clear that uniform convergence implies strong convergence and that for finite dimensional

X the two notions are equivalent. In infinite dimensions this is not true anymore.

1 n

Example 3.13. On X := l consider the left shift T (see page 34) and define T := T ,

n

l l

∈ ∈

n Then for x = (x ) X we obtain

N. k k∈N T x = (x )

n k+n k∈N

and hence +∞

## X

kT |x | → → ∀ ∈

xk = 0 as n +∞ x X.

n k

k=n+1

Therefore (T ) converges strongly to T = 0, however this convergence is not uniform since

n n∈N

kT k ∈

= 1 for all n N.

n 2. Three Cornerstones of Functional Analysis

The following three theorems, known as Uniform Boundedness Principle, Open Mapping The-

orem and Closed Graph Theorem, are, together with the Hahn–Banach theorem, cf. page 43,

considered the four cornerstones of (linear) functional analysis. While Hahn–Banach holds in

arbitrary normed spaces, the results in this section are based on the following result which

heavily depends on completeness. 6 {0}

Theorem 3.14 (Baire’s Category Theorem). If the Banach space X = can be written as

a countable union [

X = A

k

k∈N

of closed sets A X, then at least one A contains a non-empty open subset.

k k

This theorem, which can be generalized to “complete metric spaces” has, besides the above

mentioned theorems, many surprising applications in analysis. Here we just give the following

Example 3.15. An infinite dimensional Banach space does never possess a countable Hamel

basis. {e ∈

By contradiction assume that X possesses a countable Hamel basis : k

Proof. N}.

k

} ∈

Then define X := span{e , . . . , e which is closed for every k Moreover,

## N.

1

k k [

X = X

k

k∈N

∈ ∈

and hence there exists k such that X contains an open set, i.e., there exists x X

N k k k

and ε > 0 such that ∈ kx − k ⊂

x X : x < ε X .

k k

Since X is a subspace this implies X X which is a contradiction since by assumption

k k

dim(X) = +∞.

4

Theorem 3.16 (Uniform Boundedness Principle ). Let (T ) be a sequence of bounded

n n∈N

linear operators T : X Y from a Banach space X into a normed space Y . Then

n

kT ∈ ⇒ kT k

sup xk < +∞ for all x X sup < +∞.

n n

Y L(X,Y )

n∈N n∈N

4 Banach, Steinhaus (1927)

2. THREE CORNERSTONES OF FUNCTIONAL ANALYSIS 37

This theorem states that (under the hypothesis stated) pointwise (i.e. strong) boundedness

of an operator family implies its uniform boundedness. Clearly, we can invert the statement

and obtain kT k ⇒ ∃ ∈ kT

sup = +∞ x X such that sup xk = +∞.

n n X

L(X,Y )

n∈N n∈N

As we will see next, this result (and its immediate consequences) has applications in classical

analysis. k · k) ∈

Example 3.17. Let X := (C[−π, π], and define for n the functions

## N

·

sin (n + 1/2) s 6

k (s) := , s = 0

n sin(s/2)

which can be continuously extended for s = 0. Then

π

Z · ∈

k (s) f (s) ds, f X

T (f ) := n

n −π 0

∈ L(X,

defines a linear functional T = X . By an exercise on page 40 it follows that

n π

kT k = ds

k (s)

n n

−π

π ·

sin (n + 1/2) s

·

=2 ds

sin(s/2)

0

·

sin (n + 1/2) s

π

## Z

≥ · ds

4 s

0 (n+1/2)·π sin(r)

## Z

·

=4 dr

r

0

n−1 (k+1)·π sin(r)

≥ ·

4 dr

r

k·π

k=1

n−1 (k+1)·π

1

≥ · · dr

sin(r)

4 ·

k π k·π

k=1

n−1

8 1

## X

· → →

= +∞ as n +∞,

π k

k=1

≤ ≥ ·

where we used the estimate sin(r) r for t 0 and the substitution r = (n + 1/2) s. Since

X is complete, by the uniform boundedness principle there exists f X such that

sup T f = +∞.

n

n∈N

Remark 3.18. The previous example is related to Fourier Series which we will study from

an abstract (i.e. functional analytic) point of view in Chapter 5. More precisely, one can show

1 ·

that T f = s (0) where s (t) denotes the n-th partial sum of the (classical) Fourier

n n n

2π ∈

expansion of the function f C[−π, π]. Hence, using the uniform boundedness principle, we

proved the existence of a continuous function f whose Fourier series diverges at t = 0.

From the Uniform Boundedness Theorem it is easy to deduce the following →

Corollary 3.19. Let (T ) be a sequence of bounded linear operators T : X Y from a

n n

n∈N

Banach space X into a normed space Y . If ∀ ∈

lim T x =: T x converges x X,

n

n→+∞

38 3. LINEAR OPERATORS

∈ L(X, kT k ≤ kT k.

then T Y ) and lim inf n

n∈N

It il clear that the operator T in the previous result is linear, however if X is not complete it

has not to be bounded as is shown in the following example.

k · k

Example 3.20. Let X = (Φ, ) be the space of all finite sequences equipped with the

∞ →

sup-norm, cf. page 13. On X, which is not complete, define T : X X by

n

· · ·

T (x , x , . . . , x , x , . . .) := (1 x , 2 x , . . . , n x , 0, 0, . . .).

n 1 2 n n+1 1 2 n

∈ L(X) kT k ·

Then T (with = n) and lim T (x ) = (k x ) =: T (x ) converges

n n n k k∈N k k∈N k k∈N

n→+∞

∈ ke k kT k

for all x = (x ) Φ. However, since for e := (δ ) we have = 1 and e =

∞ ∞

n n n

k k∈N nk k∈N

n for all n the operator T is unbounded.

## N,

Also the next result follows quite easily from the Uniform Boundedness Theorem. ⊂

Corollary 3.21 (Banach–Steinhaus Theorem). Let X and Y be Banach spaces, (T )

n n∈N

L(X, ∈ L(X, ⊆

Y ), T Y ) and S X be a dense subset of X. Then

 kT k

sup < +∞ and

n

n∈N

∀ ∈ ⇐⇒

lim T x = T x x X

n

n→+∞ ∀ ∈

lim T x = T x x S.

n

n→+∞

We give an application to numerical integration. 0

k · k ∈ L(C[a,

Example 3.22. On X := (C[a, b], ) we consider ϕ X = b], defined by

∞ b

## Z

ϕ(f ) := f (s) ds.

a 0

Our aim is to approximate ϕ by a sequence (ϕ ) X given by

n n∈N

n

X nk

n · ),

f (s

ϕ (f ) = ω

n k

k=0

nk n

∈ ∈

where the nodes s [a, b] and the weights ω Since polynomials are easy to integrate

## R.

k

it is reasonable to ask that ∀

(∗) lim ϕ (p) = ϕ(p) polynomial p.

n

n→+∞

Now observe that n

X n |

kϕ k |ω

=

n k

k=0

P ⊂

(see an exercise on page 40) and that the set X of all polynomials is dense in X (by

Weierstraß’s theorem). Hence, assuming (∗), it follows by the previous result that

n

b

Z X n

∀f ∈ ⇐⇒ |ω |

lim ϕ (f ) = f (s) ds C[a, b] sup < +∞.

n k

n→+∞ n∈N

a k=0

1

n ≥ kϕ k → − → ∞.

If all ω 0, then by the same exercise and (∗) we obtain = ϕ ( ) b a as n

n n

k

Thus if all weights are non-negative, (∗) already implies

b

Z ∀f ∈

f (s) ds C[a, b].

lim ϕ (f ) =

n

n→+∞ a

Another main result in functional analysis is the following

∈ L(X,

Theorem 3.23 (Open Mapping Theorem). Let T Y ) where both X and Y are Banach

spaces. If T is surjective, then it is open, i.e. it maps open sets of X onto open sets of Y .

2. THREE CORNERSTONES OF FUNCTIONAL ANALYSIS 39

If in the above result T is also injective then it is invertible and the fact that T is open

characterizes exactly the continuity of its inverse. Since for linear operators boundedness and

continuity are equivalent this proves the following

∈ L(X,

Corollary 3.24. Let T Y ) be (algebraically) invertible where both X and Y are

−1 ∈ L(Y,

Banach spaces. Then T X).

This result gives conditions implying that the inverse of an invertible continuous function

is continuous again, which is not true in general. We note that similar conclusions hold for

invertible continuous functions with compact domains or real functions defined on intervals.

k·k k·k k·k

Example 3.25. Let and be two norms on a vector space X such that X = (X, )

1 2 1 1

k · k →

and X = (X, ) are both complete. If the identity operator Id : X X , Id x := x,

2 2 1 2

∈ kxk ≤ · kxk

x X is continuous (i.e. there exists c > 0 such that c ), then the previous

2 1

k · k ' k · k

corollary implies (cf. page 19).

1 2 k · k k · k

Now consider on X = C[0, 1] the norms and . Then it is easy to see that Id :

1 −1

k · k → k · k k · k → k · k

(X, ) (X, ) is bounded while its inverse Id = Id : (X, ) (X, ) is

∞ ∞

1 1

k · k

not. This shows that again that (C[0, 1], ) cannot be complete.

1

Many operators of practical importance are unbounded. For example differential operators on

infinite dimensional normed spaces are in general unbounded. However, they have a property

which is closely related to continuity. To introduce it we first need the following

⊂ →

Definition 3.26. For a linear operator T : D(T ) X Y we define its graph by

G(T ∈ ⊂ ×

) := (x, T x) : x D(T ) X Y. ⊂ →

Definition 3.27. If X and Y are normed spaces then a linear operator T : D(T ) X Y

G(T ×

is called closed if its graph ) is closed in X Y .

×

Since convergence in X Y is equivalent to convergence in each of the coordinates X and Y ,

⊂ →

an operator T : D(T ) X Y is closed iff )

3 → ∈

D(T ) x x X &

n ⇒ ∈

x D(T ) & T x = y.

→ ∈ →

T x y Y (n +∞)

n

Hence in the definition of closed operator we assume that (x ) and (T x ) both converge

n n

n∈N n∈N

and then have to conclude that they converge in the “right” way. On the other hand, for

∈ L(X,

T Y ) convergence of (x ) already implies convergence of (T x ) .

n n

n∈N n∈N 0

1

k · k {f ∈

Example 3.28. Consider X = (C[0, 1], ) and define on D(T ) := C [0, 1] : f (1) =

0} the linear operator 0

⊂ →

T : D(T ) X X, T f := f . 0

⊂ → →

We show that T is closed. To this end let (f ) D(T ) such that f f and T f = f g

n n n

n∈N n

1

→ ∈

in X as n +∞. Since convergence in X means uniform convergence this implies f C [0, 1],

0 0

f (1) = 0, i.e., f D(T ), and T f = f = g. →

Theorem 3.29 (Closed Graph Theorem). A closed linear operator T : X Y between

Banach spaces X and Y is bounded. ⊂ →

Example 3.30. Assume that T : D(T ) X Y is a closed (not necessarily bounded)

−1 ∈ L(Y,

operator for Banach spaces X and Y . If T is (algebraically) invertible, then T X).

−1 −1

This follows from the previous result applied to T observing that the graphs of T and T

are isometrically isomorphic. More precisely, if we put y := T x for x D(T ) then

−1 −1

G(T ∈ ∈ S G(T

) = (x, T x) : x D(T ) = (T y, y) : y Y = )

−1

S × → × S(x, G(T

where : X Y Y X is given by y) = (y, x). In particular, ) is closed since

−1

G(T →

) is closed, and therefore by the closed graph theorem T : X Y is bounded.

40 3. LINEAR OPERATORS

• →

Exercises 3.31. Show that a linear operator T : X Y between two vector spaces is

{x ∈ {0}.

injective iff its kernel (or null space) ker(T ) := X : T x = 0} =

• Prove Proposition 3.3. 1 3 2

• k · k

Calculate the norm of the operator T = on X = (R , ).

2

3 1

• →

If X is a finite dimensional vector space and T : X X is linear, then

⇐⇒ ⇐⇒

T is injective T is surjective T is bijective.

Show that this is not true anymore if dim(X) = +∞, even if X is a Banach space and

p

∈ L(X).

T (Hint: Consider the left- and right-shift T and T on l , cf. page 34.)

r

l

• ∈ L(X) kT k ⇐⇒

True or false? For T one always has = 1 T is an isometry?

• k·k →

Let X = (C[0, 1], ). Verify that T : X X is linear and bounded and then calculate

kT k, where s

## R

– (T f )(s) := f (r) dr;

0 2 ·

– (T f )(s) := (1 + r ) f (r);

2

– (T f )(s) := f (r ).

Which of these operators is invertible? Which is an isometry?

• k · k →

On X := (C[a, b], ) consider ϕ : X defined by

## R

∞ n

X ·

ϕ(f ) = ω f (s ),

k k

k=0

∈ ∈

where s [a, b] are pairwise different and ω for all k = 0, . . . , n.

## R

k k

0

∈ L(X,

– Show that ϕ = X .

n

## P

kϕk |ω |.

– Show that = k

k=0 1

5

kϕk − − ≥

– Show that = b a if ϕ( ) = b a and ω 0 for all k = 0, . . . , n.

k

• ·) ∈ × k · k

For k(·, C [a, b] [a, b] define on X := (C[a, b], ) Fredholm’s integral operator

by b

Z · ∈ ∈

(T f )(t) := k(t, s) f (s) ds, f X, t [a, b].

a

∈ L(X)

Show that T and verify that b

## Z

kT k = sup k(t, s) ds

a≤t≤b a k(t ,s)

0 ∈

where t [a, b] is chosen such that

(Hint: To show equality, consider f (s) := 0

n 1

|k(t ,s)|+

0 n

b

b R

R |k(t

|k(t, , s)| ds.)

sup s)| ds = 0

a≤t≤b a a

• ⊂ →

Let T : D(T ) X Y be a closed linear operator between two normed spaces X and

Y . Show that the kernel {x ∈ ⊆

ker(T ) := D(T ) : T x = 0} X

is always closed.

• ⊆ →

Let T : D(T ) X Y be a bounded linear operator between two normed spaces X

and Y . Show that

⊆ ⇒

– D(T ) X closed T is closed.

⇒ ⊆

– Y complete and T closed D(T ) X closed.

• Let X be a vector space and X , X two subspaces of X such that X is given by the

1 2

⊕ ∈

(algebraic) direct sum X = X X , i.e. every x X can be uniquely decomposed as

1 2

x = x + x , where x X , k = 1, 2.

1 2 k k 2

– Show that P : X X , P x := x is a projection, i.e. P = P .

1 1 1 1 1

1

1 1

5 ∀ ∈

Here denotes the constant function (s) = 1 s [a, b].

2. THREE CORNERSTONES OF FUNCTIONAL ANALYSIS 41

– Show that P := Id P is also a projection.

2 1 {x ∈

– Determine the kernels ker P = X : P x = 0}, k = 1, 2.

k k

{P ∈

– Determine the ranges rg(P ) := x : x X}, k = 1, 2.

k k

• →

Let X and X be kernel and range of a projection P : X X.

1 2 ⊕

– Show that X is given by the (algebraic) direct sum X = X X .

1 2

– If X is a Banach space, then P is bounded iff X and X are both closed in X.

1 2

× →

(Hint: Consider the operator T : X X X, T (x , x ) := x + x .)

1 2 1 2 1 2

CHAPTER 4

Dual Spaces and Weak Topologies

0 L(X,

In this chapter we will study the (topological) dual space X = of a normed vector

## K)

space X. 1. Linear Functionals and Dual Spaces

A linear functional on a vector space X over the field is a linear map

## K

ϕ : X K.

The set of all linear functionals on X is a vector space, called the algebraic dual space and

denoted by L(X, If X is a normed space one is usually only interested in the bounded (i.e.

K). 0

∈ L(X, ⊂

continuous) linear functionals ϕ =: X L(X, cf. Definition 3.8 on page 35.

## K) K),

0

We recall that by Proposition 3.7 the space X equipped with the norm

kϕk ϕ(x)

:= sup

0

X kxk ≤1

## X

is always complete, i.e. a Banach space, even if X is not.

n n

• ∈

Examples 4.1. If X = , then every y = (y , . . . , y ) defines a linear functional

## K K

1 n

ϕ given by

y n

X · ∀ ∈

ϕ (x) = x y , x = (x , . . . , x ) X.

y 1 n

k k

k=1 0

Choosing an arbitrary norm on X, ϕ will be always bounded and hence ϕ X for

y y

0

n

∈ ∈

all y . On the other hand, every ϕ X can be represented uniquely as ϕ = ϕ

K y

n ni=1

for y = (ϕ(e ), . . . , ϕ (e )) , where e = (δ ) denotes the k-th canonical basis

## K

1 n n k ik 0

n n n

vector of . This shows that as vector spaces and (K ) coincide

K K 1 1

n k·k

Now consider X := (K , ) for some 1 < p < +∞. Then for q > 1 satisfying + = 1

p p p q

Hölder’s Inequality 1

1 n

n n

p q

## X X X

p q

· |y |

|x · | ≤ |x |

y k

k k k k=1

k=1 k=1

implies ≤ kyk · kxk ∀ ∈

ϕ (x) x X.

y q p kϕ k ≤ kyk

This shows (as already mentioned above) that ϕ is bounded and . In order

y y q

n n

∈ ∈

to show equality, for given y = (y , . . . , y ) we define x = (x , . . . , x ) by

## K K

1 n 1 n

q

( |y | 6

k if y = 0,

k

y

x := k

k 0 if y = 0.

k

1 1

Then using that + = 1 it follows

p q |ϕ kyk · kxk

(x)| =

y q p

and therefore kϕ k kyk ∀

(∗) = 1 < p < +∞.

0

y q

(X )

p 42

1. LINEAR FUNCTIONALS AND DUAL SPACES 43

Similar arguments show that (∗) also holds for p = 1, q = +∞ and p = +∞, q = 1 and

therefore we obtain that 0

n n

k · k ' k · k ∀ ≤ ≤

(K , ) (K , ) 1 p +∞

p q

isometrically.

• k · k

On the space X := (Φ, ) of all finite sequences (cf. page 13) consider the linear

functional ϕ : X defined by

K +∞

X ∀ ∈

ϕ(x) = x x = (x ) Φ.

k k k∈N

k=0

Since x Φ is a finite sequence, the above series is finite and hence converges for all

∈ ∈

x X. However, if we define e := (1, . . . , 1, 0, 0 . . .) Φ having the first n-entries = 1,

n

ke k ∈

then = 1 and ϕ(e ) = n for all n i.e. ϕ is not bounded.

## N,

n n

• k · k →

On X = (C[a, b], ) consider the linear functional ϕ : X given by

∞ b

## Z

ϕ(f ) = f (s) ds.

a 1

|ϕ(f ≤ − · kf k ∈ kϕk ≤ −

Then )| (b a) for all f X and hence (b a). Now if f = , then

− |ϕ(f ≤ − · kf k − kϕk −

(b a) = )| (b a) = (b a) and therefore = (b a).

• k · k ∈

As before we take X = (C[a, b], ) and consider for some s [a, b] the linear

∞ 0

functional δ : X given by

## K

s 0 δ (f ) = f (s ).

s 0

0 1

|δ |f ≤ kf k ∈ kδ k ≤

Then (f )| = (s )| for all f X and hence 1. Now if f = , then

s 0 s

0 0

|δ ≤ · kf k kδ k

1 = (f )| 1 = 1 and therefore = 1. This particular functional is usually

s s

0 0

called the Dirac measure in s .

0

Having seen examples of bounded linear functionals on several normed spaces, one can ask if

6

there exist always non-trivial (i.e. = 0) bounded linear functionals on general normed spaces

6 {0}.

= The answer is affirmative as shows the following result which forms the fourth corner

stone of functional analysis. Its proof is based on Zorn’s Lemma. 0

Theorem 4.2 (Hahn–Banach Theorem). Every bounded linear functional ϕ Y defined

0 0

⊂ k · k ∈

on a subspace Y X of a normed space (X, ) can be extended to a functional ϕ X

## X

kϕk kϕ k

having the same norm = .

0 0

0

X Y 6 ∈ }

To see that this result guarantees the existence of 0 = ϕ X consider Y = span{x =

0

{α · ∈ 6 ∈

x : α for some 0 = x X. Then

## K}

0 0

→ · kx k ∀ · ∈

ϕ : Y ϕ (y) := α y = α x Y

## K,

0 0 0 0

defines a linear functional on Y . Since

|ϕ · kx k kyk ∀ · ∈

(y)| = α = y = α x Y

0 0 0

kϕ k

ϕ is bounded on Y with norm = 1. Hence, by Hahn–Banach, there exists an extension

0 0

0

∈ kϕk

ϕ X of ϕ having norm = 1.

0

A very useful application of the Hahn–Banach theorem is the following

k · k span(S) = X iff the

Corollary 4.3. Let S be a subset of the normed space (X, ). Then

## X

zero functional is the only bounded linear functional vanishing on all of S.

44 4. DUAL SPACES AND WEAK TOPOLOGIES

0

⊂ ∈

If span(S) X is dense and ϕ X vanishes on S, then by linearity and continuity

Proof.

ϕ vanishes on all of X. On the other hand, if span(S) is not dense in X, then there exists

∈ ∈

x X such that x / span(S) and therefore

0 0 kx − k

inf x = δ > 0.

0

x∈span S

} ⊕

Next define on Y := span{x span S the linear functional

0 · · ∀ ∈

ϕ (α x + x) := α δ, x span S.

0 0

6 ∈ · ∈ 6

Then ϕ (x ) = δ = 0 and ϕ (x) = 0 for all x S. Moreover, for all y := α x + x Y , α = 0

0 0 0 0

and x span S we have ∈span(S)

}| {

z

−1

−1

kyk −α · · − |α| · · −x ≥ |α| ·

= x x = x δ = ϕ (y) .

0 0 0

α α

Clearly, this estimate also holds for α = 0 and hence

≤ kyk ∀y ∈ ⇒ kϕ k ≤

Y 1.

ϕ (y) 0

0 6

Now by Hahn–Banach there exists a bounded extension ϕ = 0 of ϕ to all of X which vanishes

0

on S. This completes the proof.

More Examples 4.4. Using Hölder’s inequality for series, by similar arguments as in

0

n n

k · k ' k · k ≤ ≤

the proof of (K , ) (K , ) for all 1 p +∞ (see page 43) one can show

p q

that 0

p q 1

1

' ∀ ≤ + = 1,

(l ) l (isometrically) 1 p < +∞, p q

∞ 0 1

(l ) l .

%

∞ 0 1

6

To verify that (l ) = l we consider

c : = x = (x ) : lim x converges

k k∈N k

k→+∞

= space of all convergent sequences.

Then c l is a normed subspace and

ϕ : c ϕ (x) = lim x

## K,

0 0 k

k→+∞

kϕ k

defines a linear functional satisfying = 1. Hence, by Hahn–Banach there exists a

0

∞ 1

bounded extension of ϕ to all of l . However, there is no y = (y ) l such that

0 k k∈N

+∞

X · ∈

ϕ (x) := x y = lim x for all x c.

y k k k

k→+∞

k=0 p

• By Theorem 2.2.29 the space of all p-integrable functions L (a, b) is a Banach space for

1

1

1

p q

≤ ≤ ∈ ∈ ≤ ≤

all 1 p +∞. If f L (a, b) and g L (a, b) for 1 p +∞ and + = 1,

p q

1

· ∈

then by Hölder’s inequality (see Theorem 2.2.28) we conclude f g L (a, b). Therefore

p →

ϕ : L (a, b) given by

## K

g b

Z ·

ϕ (f ) := f (s) g(s) ds

g a kϕ k kgk ∈

defines a bounded linear functional and = . Conversely, for every ϕ

0

p

g q

(L (a,b))

0

p q

≤ ∈

(L (a, b)) , 1 p < +∞ there exists g L (a, b) such that ϕ = ϕ . In this way we arrive

g

1 1

Recall that := 0

+∞ 2. THE WEAK TOPOLOGY 45

at the following representations of the dual spaces which are in complete analogy to the

p

l case. 0

p q 1 1

' ∀ ≤

L (a, b) L (a, b) (isometrically) 1 p < +∞, + = 1,

p q

0

∞ 1

L (a, b) L (a, b).

% 1 1 1

1 the relation + = 1 implies q = and hence we conclude

Remark 4.5. Note that for p = 2 p q 2 0

0 0

n n 2 2 2 2

k · k ' k · k ' '

(K , ) (K , ), (l ) l , and L (a, b) L (a, b)

2 2

isometrically, i.e., these spaces are self-dual. We will return to this fact in Chapter 5 on Hilbert

spaces. 0

Since the dual space X of a normed space X is a Banach space, we may look at the dual

0 00 ∈

space of X which we denote by X and call the bidual space of X. Note that every x X

0 →

gives rise to a functional f : X by means of

## K

x 0

∀ ∈

f (ϕ) := ϕ(x), ϕ X .

x

Since |f |ϕ(x)| ≤ kϕk · kxk

(ϕ)| = 0

x X

## X

00

∈ kf k ≤ kxk

we conclude f X and . Using Hahn–Banach’s extension theorem one can

00

x

x X

## X

kf k kxk

show that indeed = (use the exercise on page 47). This shows that the natural

00

x X

## X

embedding 00

J : X X , J(x) = f ,

x

00

of X in X is an isometry. 1

1 + = 1)

Examples 4.6. By the above examples of dual spaces we conclude (for p q

00 0

n n n

k · k ' k · k k · k ∀ ≤ ≤

(K , ) (K , ) = (K , ) 1 p +∞,

p q p

00 0

p q p

' ∀

(l ) (l ) = l , 1 < p < +∞,

00 ∞ 0

1 1

'

(l ) (l ) l ,

%

00 0

p q p

' ∀

L (a, b) L (a, b) = L (a, b) 1 < p < +∞,

00 0

1 1

'

L (a, b) L (a, b) L (a, b).

% 00 '

These examples show that in many (but not all) cases we obtain X X. This motivates the

following 00 '

Definition 4.7. A Banach space X is called reflexive if X X via the natural embedding.

Examples 4.8. All finite dimensional spaces are reflexive.

p p

• l and L (a, b) are reflexive for all 1 < p < +∞.

∞ ∞

1 1

• l , L (a, b), l and L (a, b) are not reflexive.

• C[a, b] is not reflexive. 2. The Weak Topology

The norm topology on a (infinite dimensional) normed space is for many purposes to strong.

For example, in the infinite dimensional case the unit ball is never compact which, however, is

a property of great theoretical and practical interest. Hence we now define a weaker concept

of convergence which will produce a much bigger class of compact sets. ∈

Definition 4.9. A sequence (x ) in a normed space X is said to converge weakly to x X

n n∈N

if 0

→ ∀ ∈

ϕ(x ) ϕ(x) ϕ X .

n

46 4. DUAL SPACES AND WEAK TOPOLOGIES

In this case we write w

→ →

x as n +∞, or w-lim x = x.

x n

n n→+∞

In order to distinguish weak convergence from convergence (in norm), for the latter one uses

also the term strong convergence. Moreover, sometimes it is convenient to use the notation

s

→ → kx − → →

x as n +∞, for strong convergence which just means that xk 0 as n +∞.

x n

n

Before giving examples me make some

Remarks 4.10. The weak limit, if it exists, is unique by Hahn–Banach’s theorem.

s

• x then

If x n − ≤ kϕk · kx − → →

ϕ(x ) ϕ(x) xk 0 as n +∞.

n n

w

→ →

x as n +∞, i.e., strong convergence always implies weak convergence.

implies x n

The contrary is not true as we will see in a following example.

w

→ → ∈

• x as n +∞, then (x ) is bounded. In fact, identifying x X with

If (x ) n n

n n∈N

n∈N

0 00

→ →

f : X via the natural embedding of X in X , ϕ(x ) = f (ϕ) ϕ(x) as

## K

x n x

n n

0

→ (ϕ) is bounded for all ϕ X and hence the uniform

n +∞ implies that f

x n n∈N

boundedness principle (see page 36) implies

kf k kx k

sup = sup < +∞.

x n

n

n∈N n∈N

Examples 4.11. In finite dimensional normed spaces weak and strong convergence are

equivalent. 0

p

• ∈ ∈

Consider X = l for 1 < p < +∞. Then e := (δ ) X and for ϕ X there exists

n nk k∈N

1 1

q

y = (y ) l , + = 1, such that ϕ = ϕ . Hence,

y

k k∈N p q +∞

X · → →

ϕ(e ) = ϕ (e ) = δ y = y 0 as n +∞

n y n n

nk k

k=1 w

since q < +∞ and therefore y is a sequence converging to 0. This shows that e 0 as

n

n +∞. On the other hand, since strong convergence implies weak convergence, the

only candidate for the strong limit of (e ) is 0. However,

n n∈N

ke − ke k ∀ ∈

0k = = 1 n N

n n

and hence (e ) does not converge strongly.

n n∈N

Next we give some explicit characterizations of weak convergence.

p

Examples 4.12. On can show that in X = l , 1 < p < +∞ one has

( kx k

sup < +∞ and

w n p

nk 0

→ → ⇐⇒

x = (x ) as n +∞

x = (x )

n 0

k∈N k∈N

k nk 0

→ → ∀ ∈

x x as n +∞ k N.

k

p

Hence weak convergence in l is characterized by boundedness and convergence in each

coordinate.

• k · k

On can show that in X = (C[a, b], ) one has

( kf k

sup < +∞ and

w n

n∈N

→ → ⇐⇒

f x = f as n +∞

n 0 → → ∀ ∈

f (s) f (s) as n +∞ s [a, b].

n

Hence weak convergence in C[a, b] is characterized by uniform boundedness and pointwise

convergence in each s [a, b]. Since the pointwise limit of a sequence of continuous

functions in general is not continuous anymore, it is easy to find a sequence in C[a, b]

which is weakly but not strongly convergent.

2. THE WEAK TOPOLOGY 47

The next result, which has various important applications, relates weak convergence and

reflexivity.

Theorem 4.13 (Eberlein–Šmulian). A Banach space X is reflexive if and only if every

(norm) bounded sequence of X contains a subsequence converging weakly to some x X.

We give the following sketch of a possible application to

k · k ∈

Best Approximation. Let (X, ) be a normed space and suppose that a fixed x X is to

## X

be approximated by a y C where C is a fixed subset of X. Let

0 kx −

δ := dist(x, C) := inf yk = distance between x and C.

y∈C ∈

Clearly, δ depends on both x and C. If there exists y C such that

0

kx − k

y = δ,

0

i.e., if y C has minimum distance from the given x, then y is called best approximation

0 0

to x out of C. k · k ⊂

Corollary 4.14. Let (X, ) be a reflexive Banach space and C X be a closed convex

## X

∈ ∈

subset of X. Then for each x X there exists a best approximation y C to x. In other

0

words, the minimizing problem kx − kx − k

inf yk = y

0

y∈C

has a (possibly non-unique) solution y .

0

Let (y ) C be a sequence such that

Sketch of the Proof. n n∈N

kx − k → kx − →

y δ = inf yk as n +∞.

n y∈C

Then (y ) is bounded and hence by the previous result there exists a subsequence (y )

n n

n∈N k∈N

k

w

∈ → →

and y X such that y y as k +∞. Now one can show (using Hahn-Banach’s

0 n 0

k w

∈ − → − →

theorem) that convexity of C implies that y C. Moreover, x y x y as k +∞

0 n 0

k

0

∈ L(X kϕ k

implies that the sequence T := ϕ , converges to T := ϕ . Since =

## K)

x−y 0 x−y z

k n 0

k

kzk ∈

for all z X, Corollary 3.19 implies

≤ kx − k ≤ kx − k kx − k

δ y lim inf y = lim y = δ

0 n n

k k

k→+∞ k→+∞

kx − k

this implies y = δ and hence y is a solution of the best approximation problem.

0 0

We will return to the problem of best approximation in Chapter 5, cf. Proposition 5.6.

0 0

n n n n

• k · k ' k · k k · k ' k · k

Exercises 4.15. Show that (K , ) (K , ) and (K , ) (K , )

∞ ∞

1 1

isometrically.

• k · k ∞. →

Let X = (C[0, 1], ) for k = 1, Show that ϕ : X K,

k k k

1

Z ·

ϕ(f ) := s f (s) ds

0 kϕk ∞.

defines a linear bounded functional and calculate for k = 1,

k

• k · k ∈

Let (X, ) be a normed space and x , x X be two linearly independent vectors.

0 1

X 0

∈ kϕk kx k.

Show that there exists ϕ X , = 1 such that ϕ(x ) = 0 and ϕ(x ) =

0 1 1

• Why a normed, not complete space X can never be reflexive?

• →

Let X be a normed space and ϕ : X be a linear functional. Show that ϕ is bounded

K 6

iff ker(ϕ) is closed in X. Conclude that ϕ = 0 is unbounded iff ker(ϕ) is dense in X.

48 4. DUAL SPACES AND WEAK TOPOLOGIES

3. Sobolev Spaces and Distributions

Many problems arising naturally in differential equations call for a generalized definition of

functions and derivatives. In this paragraph we give a brief introduction to this topic defining

the concepts of weak derivative, distribution and Sobolev spaces.

Let us illustrate the basic ideas in the simplest possible context: Let I = (a, b) be an open

interval and f : I a continuously differentiable function. Then for every C function

## R

ϕ : I with compact support (cf. page 16) one obtains using integration by parts taking

## R

into account that ϕ(a) = ϕ(b) = 0 Z

Z 0

0 ·

· − f (s) ϕ (s).

f (s) ϕ(s) ds = I

I ∞

It is not difficult to show that considering all possible test functions ϕ C (a, b) this equation

c

0 →

uniquely determines f , i.e., if g : I is continuous such that

## Z

Z 0 ∞

· ∀ ∈

· − f (s) ϕ (s) ϕ C (a, b),

g(s) ϕ(s) ds =

(∗) c

## I

I 0 ∈

then necessarily g = f . The idea is now that equation (∗) might still have a solution g

1 1

L (a, b) even if one drops the differentiability assumption on f and considers just f L (a, b).

1

By the following result also in this case (∗) uniquely determines g L (a, b).

n

⊂ →

Lemma 4.16 (du Bois-Reymond lemma). If Ω is open and g : Ω is integrable

## R R

such that Z ∞

· ∀ ∈

g ϕ dλ = 0 ϕ C (Ω),

c

Ω 1

then g = 0 almost everywhere, i.e. g = 0 considered as a function in L (Ω).

1 1

∈ ∈

If now for f L (a, b) there exists a g L (a, b) such that (∗) holds then we call f weakly

0

differentiable and g = f the weak derivative of f . This approach enables us to differentiate

many functions which are not differentiable in a classical sense. Note, however, that the weak

derivative can only be defined for the whole function f and not just in a single point as the

classical derivative. → |s|.

Example 4.17. Let I = (−1, 1) and consider f : I defined by f (s) := Then for

## R

every ϕ C (−1, 1) we have

c 1 0 1

## Z Z Z

0 0 0

− · · − ·

f (s) ϕ (s) ds = s ϕ (s) ds s ϕ (s) ds

−1 −1 0

0 1

## Z Z

−1 · ·

= ϕ(s) ds + 1 ϕ(s) ds

−1 0

1

Z ·

= g(s) ϕ(s) ds

−1

where (

−1 if s < 0,

g(s) = +1 if s > 0.

0

| · |

This shows that f = is weakly differentiable and f = g.

Using ideas from the previous sections one can generalize things even further. We start by

n

considering for an open set Ω the set

R ∞

D(Ω) := C (Ω)

c

PAGINE

72

PESO

787.35 KB

AUTORE

PUBBLICATO

+1 anno fa

### DESCRIZIONE DISPENSA

Normed Vector Spaces: Basic de nitions, continuity, convergence and compactness in normed spaces, separable spaces, Banach spaces, Banach fixed point theorem. Integration theory: Riemann integral, Lebesgue measure and Lebesgue integral, Lp-spaces. Linear Operators: Basic de nitions, bounded linear operators, uniform boundedness theorem, Banach-Steinhaus theorem, open mapping theorem, closed graph theorem. Dual Spaces and Weak Topologies: Linear functionals and dual spaces, Hahn-Banach theorem, weak topology, reexive spaces, best approximation problem. Hilbert Spaces: Basic De nitions, Riesz representation theorem, orthogonality, orthogonal projection theorem, Gram-Schmidt orthonormalization, Fourier series in Hilbert spaces, dual Spaces and adjoint operators, normal, self-adjoint and unitary operators, Fourier transform. Partial differential equations: Distribution theory, Sobolev spaces, weak derivatives, Lax-Milgram 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.

DETTAGLI
Corso di laurea: Corso di laurea magistrale in ingegneria matematica
SSD:
Università: L'Aquila - Univaq
A.A.: 2011-2012

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.

Acquista con carta o conto PayPal

Scarica il file tutte le volte che vuoi

Paga con un conto PayPal per usufruire della garanzia Soddisfatto o rimborsato

Recensioni
Ti è piaciuto questo appunto? Valutalo!

Dispensa

Dispensa

Dispensa

Dispensa