Fundamental Theorems For Normed And Banach Spaces

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10 Οκτ 2013 (πριν από 3 χρόνια και 10 μήνες)

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1

Chapter 4

Fundamental Theorems For Normed And Banach Spaces


4.1 Zorn’s Lemma


4.1
-
1 Definition
.

A partially ordered set is a set M with a partial order relation


satisfying the following:



(PO1) a


a for all a

M.


(PO
2
)
If
a


b

and b


a, then a = b
.

(PO3) If a


b and b


c, then a


c.

4.1
-
2

Definition
.

An upper bound

of a subset W of a partially ordered set M is
u

M s
uch that x


u for all x

M.
A maximal element

of M is m

M such that
if m


x, then m = x. A subset C of M is called
a chain or

a

totally ordered set

if for any a, b

C, a


b or b


a ( or both ).

4.1
-
3

Zorn’s Lemma
.

Let M be a non empty partially orde
red set. If every
chain C


M has an upper bound, then M has at least one maximal element.

4.1
-
4 Theorem.

( Total Orthonormal Set )

In every Hilbert space H


{0}
there exists a total orthonormal set.

Proof.

Let M be the set of all orthonormal subsets of
H. Since H


{0}, then
there is x

H and

x


0. Hence {
} is an orthonormal subset of H.

Hence

M



.
Also note that (M,

) defines a partially ordered set ( set inclusion). Let
C


M be any chain, then X =

is an upper bound of C. By Zorn’s Lemma
M has a maximal element F. Suppose that F is not total in H. Then by Theorem
3.6
-
2
,

F




{0} and so
there is z

F


and z


0. Then F
1

= F


{
} is
orthonormal and F is a proper subse
t of F
1

which is a contradiction of
maximality of F. Therefore, the orthonormal se F is total in H




4.2 Hahn
-
Banach Theorem


4.
2
-
1 Definition
.

A sublinear functional is a real valued function p on a vector
space X which is a) subadditive, p(x + y)


p(x)

+ p(y) for all x, y

X.
b) positive homogeneous, p(

x ) =

p(x) for all

x

X and all




0.

4.
2
-
2

Hahn
-
Banach Theorem
.

( Extension of Linear Functionals )

Let X be a real vector space and p
is
a sublinear functional on X. Furthermo
re,
let f

be a linear functional which is defined on a subspace Z of X and satisfies
f(x)


p(x) for all x

Z. Then f has a linear extension

from Z to X satisfying
(x)




p(x)

for all x

X.


Proof.

(a) Let E be the set of all linear extensions g of f that satisfy g(x)


p(x)
for all x

D(g). Since f

E, then E



. Define on E a partial ordering by g


h
if and only if h is
an extension of g.
For any chain C


E, define

by
(x) =
g(x) if x

D(g) and
g

C. Then
is a linear functional and D(
) =


2

which is a vector space

s
ince C is a chain

(why?).
It is clear that g



for all
g

C
.
Then

is an upper bound of C. Hence by Zorn’s Lemma E has a
maximal element say,
. By t
he definition of E this is a linear extension of f
which satisfies
(x)


p(x) for all x

D(
).

b) We show that D(
) = X
. Suppose that D(
)


X. Choos
e
y
1

X
\
D(
) and
let Y
1

be the subspace spanned by D(
) and y
1
. Since 0

D(
), then y
1


0.
Any x

Y
1
can be written uniquely as x =
y +

y
1
, y

D(
). To s
ee this
suppose that x =
y +

y
1
and x =


+

y
1
. Then (


-


)

y
1

=
-

y. However,
y
1

D(
) and
-

y

D(
), then


-



= 0, so
that


=


and
= y. This
proves the uniqueness.

Define g
1

: Y
1


R by g
1
(y +

y
1
) =
(y) +

c, where
c is any real constant.
It is easy to see that g
1

is linear ( How?). If


= 0 then
g
1
(y) =
(y) for all y

D(
). Hence g
1

is a proper extension of
. However,
g
1
(x)


p(x) for all x

= y +

y
1

D(g
1
), [see
its

proof below]
..………..
……(1)
Hence

g
1

E which is contradicts

the maxima
l
it
y of
. Therefore, D(
) = X.
(c) Finally we prove (1).
Let

y
,

z


D(
).
For

any
fixed y
1

X,
(y)
-
(z) =
(y
-

z)





p(y
-

z)


=


p(y +

y
1

-

y
1
-

z)



p(y + y
1
) + p(
-

y
1
-

z). Then
-
p(
-
y
1
-

z)

-
(z)


p(y + y
1
)
-
(y). Hence u = sup{
-
p(
-
y
1
-

z)
-
(z): z

D(
)}



inf{

p(y + y
1
)
-
(y): y

D(
)} = w. Then there is a real number c such that
u




c



w
. So that
-
p(
-
y
1
-

z)
-
(z)



c for all z

D(
) .………………..(2),
and c


p(y + y
1
)
-
(y) for all y

D(
) ..……………..………………..(
3
)

To prove (1) we have three steps.
First step, when



<‰
.

P
ut z =


-
1
y in (2),
then multiply both sides by
-


to get,


p(
-
y
1
-



-
1
y) +
(y)



-

c. Then
,

for

x = y +

y
1

g
1
(
x
)

=

(y) +

c



-

p(
-
y
1
-



-
1
y) = p(y +

y
1
) = p(x) since p
is sublinear. Thus (1) is true
when



< 0.
Second

step, when


=



Then x =
y

D(
) and g
1
(x) =
(y)



p(
y
)

= p(x).

Thus (1) is true when


= 0.

Third

step, when


>



By (3), but replace y by


-
1
y, and then multiply both sides
by


to get


c



p(


-
1
y + y
1
)
-

(


-
1
y)

= p(x)
-

(y). Hence g
1
(x) =
(y) +

c


p(x).

Thus (1) is true when


> 0.

Therefore
, (1) is true for all

,
that is g
1
(x)


p(x) for all x

= y +

y
1

D(g
1
). This completes the pr
oof




H. W. 5
-
10. H.W.* 8, 10
.


3

4.
3

Hahn
-
Banach Theorem
For Complex Vector Spaces And Normed Spaces


4.
3
-
1
Hahn
-
Banach Theorem
.

(
Generalized

)

Let

X


be a real
or
a
complex
vector space and p is a
subadditive real
-
valued
functional on X
, and for any
sc
alar

, p(

x) = |

|p(x)

……………………………………………………...(1)


Furthermore, let f be a linear functional which is defined on a subspace Z of X
and satisfies
|
f(x)
|



p(x) for all x

Z
…………………………………………(
2
)

Then f has a linear extension
from Z t
o X satisfying
|
(x)
|

p(x) for all x

X.
Proof.

Left to the reader.


4.
3
-
2

Hahn
-
Banach Theorem
.

( Normed Space )
Let f be a bounded linear
functiona
l on a subspace Z of a normed space X. Then there exists a bounded
linear functional

on
X which is an extension of f to
X
and has the same
norm,


|
|
|
|
X

= || f ||

Z
,

where


||
||
X

=
sup
{

|
(x)| :
x

X
, || x || = 1

},

|| f ||

Z

=
sup
{ |

f
(x)

| : x

Z
, || x || = 1}
, and || f ||

Z

= 0 in the trivial case Z = {0}
.

Proof.

If Z = {0}, then

f = 0 and the extension is

= 0.
Let Z


{0}. Then for
any x

Z we have | f(x) |



|| f ||
Z

|| x ||. Define p : X



R by p(x) = || f ||
Z

|| x ||.
Then | f(x) |


p(x) for all x

Z. Furthermore, for a
ny x, y

X and any


R we
have, (1) p( x + y ) = || f ||
Z

|| x + y ||



|| f ||
Z

( || x || + || y || ) = p( x ) + p( y )
and (2) p(

x )

= || f ||
Z

||

x || = |


| p( x )
. Hence by Theorem 4.3
-
1, there
exists a linear functional

on X which is an extension of f and satisfies



|
(x)|


p(x) = || f ||
Z

|| x || for all x

X
. Then ||
||
X


|| f ||
Z
.

However,
|| f ||
Z
=

||

||
Z


||
||
X
. Hence ||
||
X

= || f ||

Z




4.
3
-
3

Theorem
.

( Bounded Linear Functional )
Let X be a normed space and
let x
0



0 be any element of X. Then there exists a bounded linear functional

on X such that ||
|| = 1 and
(

x
0

) = || x
0

||.
Proof.

Consider the subspace Z = {

x
0

:


is
a
scala
r }. Define
f : Z



k by
f(x) = f(

x
0
) =


|| x
0

||, where k is the scalar field ( real or complex ). It is eas
y
to see that f is linear (how?
)
. Also
f is bounded, where |f(x)| = |f(

x
0
)| = ||

x
0
||

=
|| x ||

for all x

X. Hence || f || = 1. By Theorem 4.3
-
2, f has a linear extension

from Z to X of norm ||
|| = || f || = 1. Moreover,
(

x
0

) = f(

x
0

) = || x
0

||





4.
3
-
4

Corolllary.


For any x in a normed space

X
we have
,

|
| x || = sup{

:
f

X

/

,
f



0
}
. Hence if x
0

is s
uch that
f
(

x
0

) =
0 for all f

X

/
, then

x
0

= 0
.
Proof.

By Theorem 4.3
-
3, there exists

X

/

such that ||
|| = 1 and
(

x ) =
|| x || where x is a nonzero element in X.

So that
sup{

: f

X

/

, f


0 }


4


= || x ||. However, |
f(x) | ≤ || f || || x ||, then sup{

: f

X

/

, f


0 } ≤
|| x ||. Therefore, sup{

: f

X

/

, f


0 }
= || x ||. If x
0

is such that f(

x
0

) = 0
for a
ll f

X

/
, then || x
0

|| = sup{

: f

X

/

, f


0 } = . Hence x
0

= 0



H. W. 1, 2, 4, 5, 8, 11, 15. H.W.* 4
.



4.
5

Adjoint operator


4.
5
-
1
Definition
.
Let
T : X



Y

be a
bounded linear operator, where X and
Y are normed

space
s. Then the adjoint operator T

×
: Y

/



X

/

of T is defined by

(T

×
g)(x) = g(Tx) = f(x), where X

/


and Y

/

are the dual spaces of X and Y,
respectively, and x

X,

f

X

/

and g

Y

/
.


4.
5
-
2

Theorem.

The adjoint operator
of T in the above definition is l
inear and
bounded with || T

×

|| = || T ||.

Proof.

For any x

X, g
1
,

g
2

Y

/

and any scalar


we have, (T

×

(

g
1
+ g
2
))(x) =
(

g
1
+ g
2
)(Tx)

=


g
1
(Tx)


+ g
2
(Tx)


=

(T

×
g
1
)(x) + (T

×
g
2
)(x). Hence
T

×
(

g
1
+ g
2
) =

T

×
g
1

+ T

×
g
2
. Therefor
e, T is linear.

Since g

Y

/

and T are
bounded then, | f(x) | = | g(Tx) |
≤ || g || || Tx || ≤ || g || || T || || x || and so || f || ≤
|| g || || T ||.
B
y the definition of T

×
, T

×
g = f,

and so

|| T

×
g || = || f || ≤ || g || || T ||.
Hence || T

×
||


= sup{|| T

×
g || : g

Y

/
, || g || = 1} ≤ || T ||……………………(1)

This me
ans that T

×

is bounded.
By Theorem 4.3
-
3
for any
x
0



0
in

X

t
he
re

exists
g
0

Y

/

such that ||

g
0

|| = 1 and
g
0
(

T
x
0

) = ||
T
x
0

||.


Let f
0
= T

×
g
0
.

Then || Tx
0

||

= g
0
(

Tx
0

) = f
0
(

x
0

) ≤ || f
0

|| || x
0

|| =
|| T

×
g
0

|| || x
0

|| ≤

|| T

×
|| || g
0

||
|| x
0
||

=


|| T

×
|| || x
0
||. This implies that || T ||

≤ || T

×

||. By this and (1) we have,
|| T

×
|| = || T ||



4.
5
-
3

Theorem.

If S, T : X



Y
and W : Y



Z
are bounded linear operators,
where X
, Y

and
Z

are normed spaces.
The
n


(1)
( S +
T

)

×

=
S
×

+ T

×

and (

T)
×

=

T
×

for any scalar

.

(2)
( WT )

×

= T
×
W
×
.

(3) If T

-
1

BL(Y, X), then

(T

×
)
-
1

BL(X
/
, Y
/
) and (T

×
)
-
1

= (T

-
1
)

×
.

Proof.

Left to t
he reader.


4.
5
-
4

Theorem.

Let H
1

and H
2
be two Hilbert spaces, T

BL(H
1
, H
2
).
The
n

there exist A
1

:
H
/
1



H
1

and A
2

: H
/
2


H
2

such that T* = A
1
T
×
A
2
-
1

where T*
and T
×

are the Hilbert adjoint and the adjoint
operators
of T, respectively
, and
bot
h A
1
and A
2

are bijective, isometric and conjugate linear.


Proof.
L
et T : H
1


H
2

be a bounded linear operator and T
×

: H
/
2


H
/
1

be its
adjoint. Then T
×
g = f where g(Tx) = f(x), f

H
/
1
, g

H
/
2

and x

H
1
. Then by

5

Theorem 3.8
-
1 there exist a unique x
0

H
1

and

a unique

y
0

H
2

such that
f(x) =
< x, x
0
>, || f || = || x
0

||, g(y) = < y, y
0
> and || g || = || y
0

||. By noting that x
0

and
y
0

are uniquely determined by f and g, respectively we can define A
1
:

H
/
1



H
1

and A
2

: H
/
2


H
2

by A
1
f = x
0

and A
2
g = y
0
. Then ||
A
1
f || = || x
0

|| = || f ||. Hence
A
1

is isometric and so it is one to one. Let h

H
1
. Then f : H
1


k defined by
f(x) = < x, h > for all x

H
1

is a bounded linear operator ( k is the scalar field).
Then A
1
f = h, hence A
1

is onto. Similarly, for

A
2
. To show
that A
1

is conjugate
linear, let

f
1
, f
2


H
/
1

and


any scalar. Then there exist x
1
, x
2

H
1

such that
f
1
(x) = < x, x
1

> and f
2
(x) = < x, x
2

> for all x

H
1
. Then (

f
1

+ f
2
)(x) =

f
1
(x) +
f
2
(x) =

< x, x
1

> + < x, x
2

> = < x,
x
1
+

x
2

>
. Hence A
1
(

f
1

+ f
2
) =
x
1
+

x
2

=
A
1
f
1

+ A
1
f
2
. Therefore, A
1

is conjugate linear. Similarly for A
2
.
For any
y
0

H
2

there exists g

H
/
2

such that A
2
g = y
0
,
so A
2
-
1
y
0

= g. Then (A
1
T
×
A
2
-
1
)(y
0
)
= (A
1
T
×
)(
g
)
=

A
1
f
= x
0
. Also, < Tx, y
0

> = g(Tx) = f(x) = < x, x
0

> =
< x, (A
1
T
×
A
2
-
1
)(y
0
) >. However, < Tx, y
0

> = < x, T*y
0

> and T* is unique,
hence T* = A
1
T
×
A
2
-
1



H. W. 1
-
5, 8,9


H.W.* 8
.



4.
6 Reflexive Spaces


4.
6
-
1

Lemma.

For every fixed x in
a normed space X, the functional g
x

defined on
X
/

by
g
x
(f) = f(x)


(
for all f

X
/

)

is a bounded linear functional so
that
g
x

X
//

and has the norm ||
g
x

||
= || x ||.

Proof.

Left to the reader.


4.
6
-
2

Lemma.
Let X be a normed space. Then the canonical mapp
ing C:X

X
//

defined by C(x) =
g
x
, where
g
x
(f) = f(x)

(
for all f

X
/

)

is an isomorphism from
X onto R(C), the range of C.

Proof.


For any
f

X
/
, x, y

X and any scalar

, C(

x + y)(f) =
g


x + y
(f) =
f(

x + y) =

f(x) + f(y) =

g
x
(f)

+
g
y
(f)


=

C(x)(f) + C(y)(f).
Hence C(

x + y) =

C(
x) + C(y), and so
C

is linear. Note that
for all f

X
/
,

g

x


y

(f) = f(x

y)
= f(x)


f(y) =
g
x
(f)



g
y
(f)

= (
g
x

g
y
)(
f)
. Hence
g

x


y

=
g
x

g
y
.
By Lemma 4.6
-
1,


|| C(x)


C(y)

|| =

||
g
x


g
y
|| =


||
g

x


y

||

= || x


y

||. This
means that C is isometric and so it is one to one. Therefore, C is an
isomorphism from X onto R(C)



4.
6
-
3

Definition.
A normed space X is said to be reflexive if R(C) = X

//
,
where C as in Lemma 4.6
-
2.


No
tes.

(1) By Lemma 4.6
-
2, the normed space X is isomorphic to a subspace
of X
//
. Hence X is embeddable in X
//
, and in this case C is called the canonical
embedding of X into X
//
.

(2) If the normed space
X
is reflexive, then
it is isomorphic with X
//
.


6

4.
6
-
4

Theorem.
If a normed space X is reflexive, then it is complete.

Proof.


Since
X is reflexive, then X is isomorphic to
X
/
/
. However, X
//

is a
Banach space, then X is a Banach space




4.
6
-
5

Theorem.
Every finite dimensional normed space is reflexive.

Proof.


Since
dim
X is
finite, then X
/

= X*. By Theorem 2.9
-
3 dimX* = dimX
,

then

X
//

= X**
.
However,

C : X


X** i
s a
n isomorphism, so that C : X


X
//

is an isomorphism. Hence

X is
reflexive



Note.

Since

=


( 1 < p <


), then

is reflexive.


4.
6
-
6

Theorem.
Every Hilbert space is reflexive.

Proof.


Left to the reader
.


4.
6
-
7

Lem
m
a
.
Let Y be a proper closed subspace of a normed space X. Let
x
0

X
-
Y be arbitrary an
d δ =
inf { ||
-

x
0

|| :

Y}, the distance from x
0

to Y.
Then there exists

X
/


such that

||
|| = 1,
(y) = 0 for all
y

Y and
(

x
0
) = δ.

Proof.


Consider the subspace Z = { y +

x
0

: y

Y,


is a scalar } of X, and
define the

functional f on Z by f(z) = f(y +

x
0
) =

δ.

Then f is linear (how?)

Since Y is closed,


then

δ > 0 and so f ≠

0. Note
that for all y

Y
,

f(y) =
f(y +
0
x
0
) =
0 and
f(x
0
)=δ

(let y = 0 and


=1). Now we show that f is
bounded
. If


= 0, then f(z) = 0. If




0, then
-
y

Y

for all y

Y

and

| f(z)|
= | f(y +

x
0
) |
= |

|
δ

= |

|

inf{||
-

x
0

|| :

Y}


|

| ||
-
y
-

x
0

|| = ||

y +

x
0

|| = || z ||.
Hence f is bounded and || f ||


1 ……………………………………………..(1)

By the definition of infimum there is a sequence (y
n
) of elements in Y such

that
|| y
n

-

x
0

||


δ

as n



. Then || f ||



=



1 as n



.
By this an (1) we have || f || = 1. By Hahn
-
Banach Theorem 4.3
-
2, there exists

X
/

such that ||
||

= 1,
(y) = 0 for all y

Y and
(

x
0
) = δ




4.6
-
8

Theorem.
If the dual space X
/

of a normed space X is separable, then X
itself is separable.

Remark.
A separable normed space X with a nonseparabl
e dual space X
/


can’t
be reflexive.

Proof.


Suppose that X
/


is
non
separable
but

X is separable
and

reflexive. Then
X
//


isomorphic to X which implies the separability of X
//

. Then by Theorem
4.6
-
8, X
/


is separable which is a contradiction.
Therefore,
X can’t be reflexive


Example.


is not reflexive.

Proof.
Since

is separable but

=

is not separable, then

is not
refl
exive


H. W. 1, 5, 7
-
9 H.W.* 8
.


7

4.
7 Category Theorem. Uniform Boundedness Theorem



4.
7
-
1
Definition.
A subset M of a metric space X is said to be

(a) rare ( or nowhere dense ) in X if
has no interior
point (

= φ

),
(b) meager ( or of first category ) in X if M is the union of countably many sets
each of which is rare in X.
(c) nonmeager ( or of the second category ) in

X if M is not meager in X.


4.
7
-
2

Bair Category Theorem.

If a metric space X ≠ φ is complete then it is
nonmeager in it self. Hence if X ≠ φ is complete and X =
, A
k

is closed
for all k then at least one A
k

contains a nonemp
ty open subset.

Proof.

Suppose that the complete metric space X ≠ φ is meager in itself. Then
X =

with each M
k

is rare ( nowhere dense ) in X. Then

= φ, that
means

does
not contain a nonempty open set. But X does ( for example X
itself ), so that
≠ X. Hence (~
) = X
-


is
nonempty and open
, so
we can choose p
1

(~
)

and an open ball about it, say B
1

= B(p
1
;

ε
1

)



(~
) and ε
1

<
½

. Since M
2

is nowhere dense in X, then

does not
contain a nonempty open set. Hence B(p
1
;

½
ε
1

) is not a subset of

which
implies that (~
)


B(p
1
;

½
ε
1

) is
nonempty and open
, so we can choose
p
2

(~
)


B(p
1
;

½
ε
1

) and an open ball about it, say B
2

= B(p
2
;

ε
2

)



(~
)


B(p
1
;

½
ε
1

) and
ε
2

<
½

ε
1
. So by induction we obtain a sequence of
balls B
k

= B(p
k
;

ε
k

), ε
k

< 2
-
k

such that B
k+1


B(p
k
;

½
ε
k

)



B
k
. Since ε
k

< 2
-
k
,
then



and so it converges. Hence (p
k
) is a
C
auchy
sequence in t
he complete metric space X so it converges, say lim p
k

= p

X.
Also for all m and all n > m we have, B
n


B(p
m
;

½
ε
m

) so that d(p
m
, p)


d(p
m
, p
n

) + d(p
n
, p) <
½

ε
m

+ d(p
n
, p)


½

ε
m

as n




Hence p

B
m

for all m.
Since B
m


(~
), then p
M
m
for all m, so that p

= X. This
contradicts p

X. Therefore, X must be nonmeager in it self



4.
7
-
3

Uniform Boundedness Theorem.

Let (T
n

) be a sequence of bound
ed
linear operators T
n

: X



Y from a Banach space X into a normed space Y
such that ( || T
n

|| ) is bounded for every x

X, say || T
n
x || ≤ c
x

..………….(1)
where c
x

is a real number. Then the sequence of the norms || T
n

|| is bounded,
that is, there is c such that || T
n

|| ≤ c …………………………………….(2)


8

Proof.

For every
k

N, let A
k


X be the set of all x such that || T
n
x || ≤ k for
all n. To

show that A
k

is closed, let x

, then there is a sequence (x
j

) of
elements in A
k

such that lim x
j

= x. Then for any fixed n, we have || T
n
x
j

|| ≤ k,
hence by the continuity of T
n

and the continuity of the norm we have, || T
n
x |
| ≤
k and so x

A
k
. Therefore, A
k

is closed. Since for every x

X there exists c
x

such that || T
n
x || ≤ c
x

for all n. Then each x

X belongs to some A
k
.
Hence X =
, however, X is complete, then by Bairs Category Theorem there exists

A
k0

contains an open ball, say B
0

= B(
x
0
;

r

)


A
k0


=
. Now let x

X be
arbitrary and nonzero

and

l
et z = x
0

+ γx, γ =
≠ 0 ………………(3)

Then || z
-

x
0
|| = || γx

|| =

< r. Hen
ce z

B
0

and so z


A
k0
. Then by the
definition of
A
k0

we have, || T
n
z || ≤ k
0

for all n. Since x
0

B
0



A
k0
, then
|| T
n
x
0
|| ≤ k
0

for all n. By (3), x =
(z
-

x
0
). Then for all n, || T
n
x || =
|| T
n
(z
-

x
0
) || ≤
( || T
n
z || +

|| T
n
x
0

|| ) ≤
(2k
0
) =
(2k
0
). Hence
|| T
n

|| ≤
. Therefore, || T
n

|| ≤ c for all n, where c
=



4.7
-
4 Space of Polynomials.

The normed space X of all polynomials is not
complete under the norm defined by || x || =
, where α
0
, α
1
, α
2
, … are
the coefficients of x.

Proof.

We construct a seque
nce of bounded linear
functional
s on X which
satisfies (1) but not (2). Let x be a nonzero polynomial of degree N
x
, then
x(t) =
, where α
j

= 0 for j > N
x
. Let ( f
n

) be a sequence of functionals
that are defined on X by

f
n
(0) = 0, f
n
(x) = α
0

+ α
1

+ α
2

… + α
n
-
1
. It is left to
the reader to show that for any
fixed
n, f
n

is linear. Since |α
j

|


= || x ||,
then | f
n
(x) | ≤ n|| x ||. Hence f
n

is bounded. Furthermore, for each fixed x

X,


|
f
n
(x) | ≤ (N
x

+ 1)

= c
x
, (since x is a polynomial of degree N
x

that has
N
x

+ 1 coefficients ). Therefore, (| f
n
(x) |) satisfies (1). Finally, we show that
(f
n
)
does not
satisf
y

(2). Choose a polynomial x defined by x(t) =
. Then
|| x || = 1 and f
n
(x) = 1 + 1 + … + 1 = n = n || x ||. Hence || f
n

|| ≥

= n.
Hence the sequence (|| f
n
||) is unbounded and so (2) is not satisfied. Therefore,
by Uniform Boundedness Theorem,
X is not complete





H. W. 1, 3, 9, 5, 11, 13. H.W.*1 4
.



9

4.
8 Strong and Weak Convergence



4.
8
-
1
Definition.
A s
equence ( x
n

) in a normed space X is said to be strongly
convergent (or convergent in the norm) if there is x

X such that || x
n



x |
|


0
as n


∞. This is written as lim x
n

= x or x
n



x. x is called the strong limit of
( x
n

) and we say that ( x
n

) converges strongly to x.


4.
8
-
2

Definition.
A s
equence ( x
n

) in a normed space X is said to be weakly
convergent (x
n
x) i
f there is x

X such that for every f

X

/
, lim f(x
n

) = f(x).
x is called the weak limit of ( x
n

) and we say that ( x
n

) converges weakly to x.


4.
8
-
3

Lemma.
Let ( x
n

) be a

weakly convergent
s
equence in a normed space
X, say x
n
x. Then
(a) The weak limit x of
( x
n

)
is unique.
(b) Every subsequence of
( x
n

)
converges weakly

to x.
(c) The sequence
(
||
x
n

||
)
is bounded.

Proof.
The proof of
(a) and (b) are left to the

reader.

(c) Since x
n
x, then f(x
n

)


f(x)

for all f

X

/
, so that the sequence of
numbers

( f(x
n

) ) is bounded. That is, there is c
f


a constant depending on f
such that | f

(x
n
) |
≤ c
f

for all n. Using the canonical mapping C: X



X

//
, we
can define g
n

X

//

by g
n
(f) = f(x
n

) for all f


X

/
. Then for all n, | g
n
(f) | = | f(x
n
)

|
≤ c
f

, that is the sequence (| g
n
(f) |) is bounded for all f

X
/
. However, X
/

is
complete, Then by unifor
m bounded Theorem 4.7
-
3, ( || g
n

|| ) is bounded, and
by Lemma 4.6
-
1, || g
n

|| = || x
n

||. Hence ( || x
n

|| ) is bounded



4.
8
-
4

Theorem.
Let ( x
n

) be a

s
equence in a normed space X. Then

(a) Strong convergence implies weak convergence with the same limit.
(b) The converse of (a) is not generally true.

(c) If dim X < ∞, then weak convergence implies strong convergence.

Proof.
The proof of
(a)
is

left to the

reader.

(b) Let (e
n

) be an orthonormal sequence in a Hilbert space H. By Riez’s
representation Theorem, any
f

H

/

has a representation f(x) = < x, z >. Hence
f(e
n
) = <e
n
, z >. By Bessel’s inequality
≤ || z ||

2
. Hence

converges. So that |<e
n
, z >|
2

0

as n

∞. This implies that f(e
n
) = <e
n
, z>

0 =
f(0) as n


∞. Since f

H
/

was arbitrary, then

e
n
0. However, (e
n
) does not
converge strongly, because for all m ≠ n, || e
m

-

e
n

||
2

= <e
m
, e
m

> + <e
n
, e
n
> = 2



(c) Suppose that x
n
x and dim X = k. Let {

e
1
, e
2
, …
, e
k

} be any basis of X
and, say, x
n

=

and x =
. By assumption, f(x
n

)


f(x)

for all
f

X

/
, in particular f
j
(x
n

)


f
j
(x)

for all j = 1, 2, …, k, where f
j
(e
j
) = 1 and
f
j
(e
m
) = 0 ( m ≠ j ). Hence f
j
(x
n

) = α
j

(n)


and f
j
(x) = α
j

and so α
j

(n)



α
j
. Then

10

|| x
n



x || = ||
|| ≤

|| e
j
||


0 as n


∞. Therefore,
(x
n

)


converges strongly to x



Examples

4.
8
-
5

Hilbert Space.

In a Hilbert spa
ce H, x
n
x if and only if
<x
n
, z >


< x, z > for all z in H.
Proof.
Left to the reader.

4.
8
-
6

Space
.

In
, where 1 < p < ∞, x
n
x if and only if:
(A) The sequence ( || x
n

|| ) is bounded.

(B) For every fixed j we have

j

(n)



j

as n


∞; here, x
n

= (

j

(n)
) and x = (

j
).
Proof.
Left to the reader.


4.
8
-
7

Lemma ( Weak Convergence ).

In a normed space X we have,


x
n
x if and only if:

(A) The sequence ( || x
n

|| ) is bounded.

(B) For every element f of a total subset M of X
/

we have, f(x
n
)


f(x)

as n

∞.
Proof.
If
x
n
x, then (A) follows from Lemma 4.8
-
3, and (B) follows from
the definition of weak convergence. Conversely, suppose that (A) and (B) hold.

By (A),
|| x
n

|| ≤ c for all n and || x || ≤ c for sufficiently large c. Since M is total
in
X
/
, then for every f

X
/

there is a sequence (f
i
) in span M such that f
i


f
(since span M is dense in X

/
).Then for every


> 0 there is j such that || f
i
-

f || <
. Moreover, since f
i

span M, then by (B) there is N such that |

f
i
(x
n
)
-

f
i
(x) |
<

for all n > N. Then for all n > N we have, | f(x
n
)
-

f(x) | ≤ | f(x
n
)
-

f
i
(x
n
) |
+ | f
i
(x
n
)
-

f
i
(x) | + | f

i
(x)
-

f(x) | < || f

-

f
i

|| || x
n

|| +

+ || f
i
-

f || || x || <
c
+

+
c =

. Since f

X
/


was arbitrary, then x
n
x



H. W. 1
-
10 H.W.* 4, 5.



4.
9 Convergence of sequences of operators and functionals


4.
9
-
1
Definition.
Let X and Y be normed spaces.
A s
equence ( T
n

) of
operators in B(X, Y) is said to be

(1) Uniformly convergent if ( T
n

) converges in the norm of B(X, Y), that is
there is T :X

Y such that || T
n



T ||


0 as n


∞.
(2) Strongly convergent if ( T
n
x) converges strongly in Y for every x

X, that is
there is T :X

Y such that || T
n
x



Tx ||


0 as n




for every x

X.

(3) Weakly convergent if ( T
n
x) converges weakly in Y for eve
ry x

X, that is
there is T :X

Y such that || f(T
n
x)



f(Tx) ||


0 as n




for every x

X and
for every f

Y

/
.



11

Remark.
Uniform operator convergence implies strongly operator
convergence implies weakly operator convergence.

Proof.
Left to the reader.

Exa
mples

4.
9
-
2

Space
.

The strong operator convergence need not imply the uniform
operator convergence.

Proof.

Let ( T
n

) be a
s
equence of operators, where T
n
:

is defined by

T
n
x = ( 0, 0, ….., 0,

n+1
,

n+2
, ……. ), ( note: the number of zeros is n ) where x
= (

j
)

. It is easy to see that T
n

is linear

for any n. Also, for any n, since
||T
n
x|| =



= || x ||, then T
n

is bounded and || T
n

|| ≤ 1.
Moreover, if we choose x
0

= ( 0, 0, ….., 0,

n+1
,

n+2
, ……. ) we get, T
n
x
0

= x
0

and so || T
n

|| ≥

= 1. Therefore, || T
n

|| = 1 and so || T
n



0 || =
|| T
n

|| = 1,
which implies that ( T
n
) is not uniformly convergent

to the zero operator
. But, it
is clear that for all x


we have, T
n
x

( 0, 0, …..) as n


∞. Hence

( T
n
) is
strongly operator convergent to 0




4.
9
-
3

Space
.

The weak operator convergence need not imply the strong
operator convergence.

Proof.

Let ( T
n

) be a
s
equence of operators, where T
n
:

is defined by
T
n
x = ( 0, 0,..., 0,

1
,

2
, ... ), where the number of zeros is n and x = (

j
)

.
It is easy to see that T
n

is linear

and bounded for any n (how?)
Since

is a
Hilbert space, then every linear functional f on

has a Riesz representation
f(x) = < x, z > =
, where z = (

j
)

. But f(T
n
x) = <T
n
x, z > =

=
. By Cauchy
-
Schwarz inequality, | f(T
n
x) |

2

= | <T
n
x, z > |

2



0 as n


∞. Therefore, f(T
n
x)


0 = f(0x) for all x

. Thus
(
T
n
) is weakly o
perator convergent to 0. But, (
T
n
) is not strongly operator
convergent to 0, because for x
= (1
, 0, 0,...), we have ||

T
m
x

T
n
x

|| =

2 for all
m ≠ n



Note.
Since the real and the complex fields are finite dimensional, then by
Theorem 4.8
-
4 (c), the weak and the strong convergence of any sequence of
bounded linear
operators

are the same.

4.
9
-
4

De
finition.
Let ( f
n

) be a

s
equence of bounded linear functionals on a
normed space X, then
(1) strongly convergence of ( f
n
) means that there is f

X
/

such that || f
n



f ||


0
as n


∞. This written as f
n



f

12

(2) weak* convergence of ( f
n
) means that there is f

X
/

such that f
n
(x)


f(x)
as n




for every x

X. This written as f
n


f


Remark.
Let

X and Y be normed

spaces, and

( T
n

) be a sequence of elements
in B(X, Y).

(1)
If
( T
n

) converges uniformly to T, then T

B(X, Y).


(2)

Strongly or weakly convergent of
( T
n

) to some T :X

Y need not imply
that T

B(X, Y).

Proof.
The easy
proof of (1) is left to the reader.

To verify (2), consider the
subspace X = {x = (

j
)

: x has finitely many nonzero} of
. Let
T
n
:X

X
be defined by T
n
x = (

1
, 2

2
, 3

3
, .....

, n

n
,

n+1
,

n+2
, ……. ) a
nd
T
:X

X be
defined by Tx = (j

j
). It is left to the reader to show that for any fixed n, T
n

is
linear and bounded. Moreover, T is linear but is not bounded, then T

B(X).
Finally, it is clear that
( T
n

) converges strongly to T, where for all x

X,

||
T
n
x


Tx ||


0 as n







4.
9
-
5

Lemma.

Let

X
be a Banach space,
Y
a
normed

space and

( T
n

) be a
sequence of elements in B(X, Y) that converges strongly to a limit T :X

Y,
then T

B(X, Y).

Proof.
It

is left to the reader

to prove that T is linear. Since
T
n
x


Tx for all
x

X,
then by Lemma 1.4
-
2(a), the sequence (
T
n
x) is bounded
.

However, X is
complete, then by U. B. Theorem 4.7
-
3, there is c > 0 such that ||
T
n

|| ≤ c for
all n. Hence, ||
T
n
x || ≤ ||
T
n

|| || x || ≤ c || x ||. Then by using the continuity of the
norm we have,

|| Tx || = lim ||
T
n
x || ≤

c || x ||. Hence T is bounded and then
T

B(X, Y)



4.
9
-
6

Theorem.

Let

X
and Y be Banach spaces and

( T
n

) be a sequence of
elements in B(X, Y). Then,
( T
n

) converges strongly if and only if:

(A) The sequence ( || T
n

|| ) is b
ounded.

(B) The sequence (T
n
x) is Cauchy in Y for every x in a total subset M of X.

Proof.

If
T
n
x


Tx for all x

X,
then (
T
n
x) is bounded and so by U. B.
Theorem 4.7
-
3, ( || T
n
||

) is bounded. (B)
follows from the definition of strongly
operator convergent

and the fact that every convergent sequence is Cauchy.

Conversely,
suppose that (A) and (B) hold, then there is c > 0 such that ||
T
n
|| ≤ c
for all n.

Since M is total in X, then span M is dense in X. Hence for every
x

X and every


> 0 there is y

span M such that || x


y || <
. By (B), the
sequence (
T
n
y) is Cauchy. Hence there is k

N such that || T
n
y


T
m
y |
| <

for
m, n > k. Therefore, for
all
m, n > k
and any fixed x

X
we have,



|| T
n
x


T
m
x || ≤ || T
n
x


T
n
y || + || T
n
y


T
m
y || + || T
m
y


T
m
x ||
< || T
n
|| || x


y ||
+

+ || T
m
|| || y


x ||
< c
+

+

c

=

. Hence (
T
n
x) is Cauchy in Y for
all x

X. However, Y is complete, then
(
T
n
x) converges in Y; that is
( T
n

)
converges strongly



13


4.
9
-
7

Corollary.

A sequence
(
f
n

) of boun
ded linear functional on a Banach
space X is weak* convergent, the limit being a bounded linear functional on X
if and only if:

(A) The sequence ( || f
n

|| ) is bounded.

(B) The sequence (f
n
x) is Cauchy for every x in a total subset M of X.


H. W. 1
-
4 H
.W.* 4.




4.
12 Open Mapping Theorem


4.
12
-
1
Definition.
Let X and Y

be metric spaces.

Then T : D(T)


Y with
D(T)


X

is called an open mapping if the image of any open set in D(T) is an
open set in Y.


4.
12
-
2

Lemma.
A bounded linear operator T from a
Banach space X onto a
Banach space Y has the propert
y

that the image T(B)

of

the

open
unit ball
B = B(0; 1)



X

contains an open ball a bout 0

Y.


4.
12
-
3

Open Mapping Theorem, Bounded Inverse Theorem.
A bounded
linear operator from a Banach space X

onto a Banach space Y
is an open
mapping. Hence if T is bijective, then T

-
1

is continuous and thus is bounded.

Proof.

Let A be an open set in X. Let y = Tx be an element in T(A).

Since A is open,
then A contains an open ball with center x. Hence A
-
x con
tains an open ball with
center 0; let the radius of the ball be r, and set k =
. Then k(A
-
x)

contains the open
unit ball B(0; 1). Then by Lemma 4.12.2, T(k(A
-
x)) = k[T(A)


T(x)] contains a ball a
bout 0

Y
, and so does T(A)


T(x
). Hence T(A) cotains an open ball a bout Tx = y.
But y was arbitrary in T(A), then T(A) is open. Therefore, T is an open mapping.
Finally, if T is bijective, then T

-
1

exists and
linear
by Theorem 2.6
-
10. However, T is
open then
by Theorem 1.3
-
4
T is cont
inuous, and by Theorem 2.7
-
9, T is bounded




H. W. 1, 2, 5
-
7 H.W.* 6.













14

4.
13 Closed Linear Operators. Closed Graph Theorem


4.
13
-
1
Definition.
Let X and
Y be normed spaces and
T : D(T)


Y
be a
linear operator with

D(T)


X
. Then T is called a

closed linear operator if its
graph G(T) = { (x, y) : x

D(T), y = Tx } is closed in the normed space X

Y,
where the two algebraic operations of the vector space X

Y are defined as
(x
1
, y
1
) + (x
2
, y
2
) = (x
1
+ x
2
, y
1
+ y
2
);


(x, y) = (

x,

y) (


is
scalar) and the
norm on X

Y is defined by || (x, y) || = || x || + || y ||.


4.
13
-
2

Closed Graph Theorem.
Let X and
Y be
Banach

spaces and
let
T

:

D(T)



Y
be a closed linear operator with

D(T)


X
. If D(T) is closed in
X, then T is bounded
.

P
roof.

Let

(z
n
) be any Cauchy sequence in X

Y, where z
n

= (x
n
, y
n
). Then for
every


> 0 there is k

N such that for all m > n > k, || x
n


x
m

|| + || y
n


y
m

|| =
|| ( x
n


x
m
, y
n


y
m
)|| = || z
n


z
m
|| <


……………………………………...(1)
Hence (x
n
) and (y
n
) are
Cauchy sequences in X and Y, respectively. However,
X and Y are complete, then there are x

X and y

Y such that x
n

x and y
n


y.
Let m

∞ in (1) to get, || z
n


(x, y)

|| = || x
n


x || + || y
n


y || <



for all n > k.
This means that (z
n
) converges to z =

(x, y)

X

Y. Therefore, X

Y is
complete. Since G(T) is closed in X

Y and D(T) is closed in X, then G(T) and
D(T) are complete. Now c
onsider the mapping P
:

G
(T)



D(T), that defined
by P((x, Tx)) = x.

It is left to the reader to show that P is linear. Since, || P((x,
Tx)) || = || x || for all x

X, then P is bounded. It is easy to see that P is bijective
(how?) Hence


P
-
1

:

D
(T)



G(T)
exists and is defined by P
-
1
(x) = (x, Tx).
Since G(T) and D(T) are complete, then by open mapping Theorem 4.12
-
2, P
-
1

is bounded, say || (x, Tx) || = || P
-
1
(x) || ≤ b || x || for some b > 0 and all x

D
(T)
.
Hence || Tx || ≤ || Tx || + || x || = || (x, Tx)

|| ≤ b || x ||. Therefore, T is bounded.


4.
13
-
3 Theorem.
Let T:D(T)

Y
be a linear operator with

D(T)

X

and

X and
Y
be
normed

spaces
. Then, T is closed if and only if it has the following property:
if x
n

x, where x
n

D
(T)

for all n and Tx
n

y, then

x

D
(
T)

and Tx = y………(2)


Proof.

Suppose that T is closed, then G(T) is closed. Let x
n

x, where x
n

D
(T)

for all n and Tx
n

y, then (x
n
, Tx
n
)

G(T) and (x
n
, Tx
n
)

(x, y). Hence (x,
y)

G(T). This means that x

D
(T)

and Tx = y. Conversely, suppose that the
property (2
) holds. Let (x, y)

. Then there is ( (x
n
, Tx
n
) ) a sequence of
elements in G(T) such that (x
n
, Tx
n
)

(x, y). Hence x
n

x, x
n

D
(T)

for all n
and Tx
n

y. Then by assumption [ the property (2) ], x

D
(T)

and Tx = y; that
is (x, y)

G(T
). Therefore, G(T) is closed, and so T is closed.


4.
13
-
4 Example Defferential operator.
Let
X = C[0, 1] and
T:D(T)

X

be such
that Tx = x

/
, where x

/

is the derivative of x and D(T) is a subspace of functions
x

X which have a continuous derivative. Then

T is not bounded, but is closed.


15

Proof.

By 2.7
-
5, T is linear but not bounded. Let x
n

D
(T)

for all n such that
x
n

x and Tx
n
= x
n

/


y. By 1.5
-
5, the convergence in the normed space C[0,
1] is uniform convergence. Then

=

= lim

= lim

x
n
(t)
-

x
n
(0) = x(t)
-

x(0); that is x(t) = x(0)


+
. Then x

D
(T)

and Tx

= x

/
= y. Then by Theorem 4.13
-
3, T is closed.


Note.
D
(T)

in Example 4.13
-
4, is

not closed , because if it is closed then by
Theorem 4.13
-
2, T is bounded, because T is closed. Then we have a
contradiction with T is not bounded. Therefore, D
(T)

in Example 4.13
-
4, is not
closed.

Remark.
(1) Closedness does not imply boundedness of line
ar operator. [ see
Example 4.13
-
4 above. ]

(2) Boundedness does not imply closedness of linear operator.

Proof.

(2) Let T
:

D(T)



D(T)

be the identity operator on
D(T)
, where
D(T)

is a proper dense subspace of a normed space X. It is clear that T is line
ar and
bounded. But T is not closed. This follows from Theorem 4.13
-
3 if we take an
x

X


D
(T)

and a sequence (x
n
) in
D(T)

which converges to x.


4.
13
-
5 Lemma.
Let T

:

D(T)



Y
be a bounded linear operator with

D(T)



X
,

where

X and
Y
are

normed

spaces
.

Then,

(a) If D(T) is a closed subset of X, then T is closed.

(b) If T is closed and Y is complete, then D(T) is a closed subset of X.

Proof.

Left to the reader.



H. W.
3
-
6, 11
-
13. H. W*. 6, 12.