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.
Enter the password to open this PDF file:
File name:

File size:

Title:

Author:

Subject:

Keywords:

Creation Date:

Modification Date:

Creator:

PDF Producer:

PDF Version:

Page Count:

Preparing document for printing…
0%
Σχόλια 0
Συνδεθείτε για να κοινοποιήσετε σχόλιο