03_AJMS_279_20_20210128_V2.pdf

www.ajms.com 14
ISSN 2581-3463
INTRODUCTION AND RESULTS
Introduction
Let X be a linear vector space.Alinear operator from X into the space R is called a real linear functional
on X . Similarly for X a normed linear space a bounded linear operator from X into R is called a
continuous linear functional on X .
Results
The Hahn–Banach theorem is basically defined for R and sometimes holds for a complex linear functional
on X when X is a complex space while a complex linear functional on X is obtained when X is a
complex space and R is replaced by R.
Theorem 1.2.1 (Hahn–Banach Theorem):[1]
Let X be a real vector space, M a subspace of X , and P
a real function defined on X satisfying the following conditions:
1. P x y P x p y
( ) ( ) ( )
.
2. P x p x
( ) ( )

x y X
, and positive real α .
Further, suppose that f is a linear functional on M such that f x p x

x M . Then, there exists a
linear functional F defined on X for which F x f x

x M and F x p x x X
. In other
words, there exists an extension F of f having the property of f .
Theorem 1.2.2 (Topological Hahn–Banach Theorem):[2]
Let X be a normed space, M a subspace of
X , and f a bounded linear functional on M .
1. F x f x x M
.
2. F f
= .
RESEARCH ARTICLE
On Analytic Review of Hahn–Banach Extension Results with Some Generalizations
Chigozie Emmanuel Eziokwu
Department of Mathematics, Michael Okpara University of Agriculture, Umudike, Abia State, Nigeria
Received: 15-09-2020; Revised: 30-10-2020; Accepted: 15-11-2020
ABSTRACT
The useful Hahn–Banach theorem in functional analysis has significantly been in use for many years
ago. At this point in time, we discover that its domain and range of existence can be extended point
wisely so as to secure a wider range of extendibility. In achieving this, we initially reviewed the existing
traditional Hahn–Banach extension theorem, before we carefully and successfully used it to generate the
finite extension form as in main results of section three.
Key words: Extensions, linear functional, lower bounds, normed space, upper bounds, vector space
2010 Mathematics Subject Classification: 46BXX, 54C20
Address for correspondence:
Chigozie Emmanuel Eziokwu
E-mail: okereemm@yahoo.com

Eziokwu: Hahn Banach extension results with generalizations
AJMS/Oct-Dec-2020/Vol 4/Issue 4 15
In other words, there exists an extension F of f which is also bounded linear and preserves the norm.
The proof of Theorem 1.2.1 depends on the following lemma:
Lemma1.2.1:[3]
Let X beavectorspaceand M itspropersubspace.For x X M
0 ,let N M x

0

. Furthermore, suppose that f is a linear functional on M and p a functional on X satisfying the
conditions in theorem 1.2.1 such that f x p x x M
. Then, there exists a linear functional F
defined on N such that F x f x x M
and F x p x x N
.
In short, this lemma tells us that Theorem 1.2.1 is valid for the subspace generated or spanned byM x0

 .
Consequences of the Extension Form of the Hahn–Banach Theorem
The proofs of the following important results mainly depend on the proof of Lemma 1.2.1.
Theorem 1.2.3:[4]
Let w be a nonzero vector in a normed space X then there exists a continuous linear
functional F , defined on the entire space X such that 1
=
F
‖ ‖ and ( ) =
F w w
‖ ‖
.
Theorem 1.2.4:[5]
If X is a normed space such that F w F X

0 *, then w = 0.
Theorem 1.2.5:[6]
Let X be a normed space and M its closed subspace. Further assuming that
w X M w X w M

but . Then, there exists F X
∈ * such that F m
( ) = 0 for all m M
∈ , and
F w
( ) =1.
Theorem 1.2.6:[7]
Let X be a normed space, M its subspace and w X
∈ such that d w m

inf 0. It
may be observed that this condition is satisfied if m is closed and w X M

. Then, there exists
F X
∈ * with 1
=
F
‖ ‖ , ( ) 0
≠
F w , and F m
( ) = 0 for all m M
∈ .
Theorem 1.2.7:[8,9]
If X * is separable, then X is itself separable.
PROOF OF HAHN–BANACH RESULTS
Proof of Lemma 1.2.1[1,9]
due to Siddiqi
This will help us in developing the proof of theorem 1.2.1 of the Hahn–Banach Theorem. Since
( ) ( )
≤
f x p x for x M
∈ and f is linear, we have arbitrary 1 1
, .
∈
y y M
f y y f y f y P y y
( , ) ( ) ( ) ( , )
1 2 1 2 1 2

or
f y f y p y x y x
( ) ( ) ( , )
1 2 1 0 2 0

p y x p y x
( ) ( )
1 0 2 0
by condition (1) of Theorem 1.2.1.

p y x f y p y x f y
2 0 2 1 0 1
 (2.1.1)
Suppose y1 is kept fixed and y2 is allowed to vary over M , then equation (2.1.1) implies that the set of
real numbers p y x f y y M
1 0 2 2

has upper bounds and hence the least upper bound. Let

sup p y x f y y M
1 0 2 2 . If we keep y2 fixed and y1 is allowed to vary over M , equation

(2.1.1) implies that the set of numbers p y x f y y M
2 0 1 1

has lower bounds and hence the
greatest lower bound.
Let

inf p y x f y y M
1 0 1 1 . As it is well known that between any two real numbers, there is
an always a third real numbers. Let y be a real number such that

(2.1.2)
It may be observed that if
, then
. Therefore, for M , we have
p y x f y p y x f y
( ) ( ) ( ) ( )

0 0
(2.1.3)
From the definition of N , it is clear that every element x in N can be written as
x y x
0 (2.1.4)
Where x M
0 ∈ or x XM
0 ∈ , λ is a uniquely determined real number and γ a uniquely determined vector
in M . We now define a real-valued function on N as follows:
F x F y x f y y

0 (2.1.5)
We shall now verify that[8]
the well-defined function satisfies the desired conditions, i.e.,
i. F is linear,
ii. F x f x x M
,
iii. F x p x x N
.
iv. F is linear: For
z z N z y x z y x
1 2 1 1 1 0 2 2 2 0
, ,

F z z F y x y x
1 2 1 1 0 2 2 0

F y y x f y y
1 2 1 2 0 1 2 1 2

f y f y
1 2 1 2

as f is linear: Or
F z z f y f y
1 2 1 1 2 2

Similarly, we can show that F z F z z N

and for real µ .
2. If x M
∈ , then γ must be zero in equation (2.1.4) and then equation (2.1.5) gives F x f x

Here, we consider three cases.[9]
(See equation 2.1.4)
Case 1, 0: We have seen that F x f x
and as f x p x
, we get that.
F x p x

Case 2, 0: From equation (2.1.3), we have.

p y x f y
0 (2.1.6)

Since N is a subspace, y N
/ replacing y by y N
/ in equation (2.1.6), we have

p y x f f
y
/ 0
or

p y x f y
1
0 /
By condition (2) of theorem 1.2.1
p y x p y x
1 1
0 0

For 0 and f y f y
/

1
as f is linear. Therefore,

p y x f
0 or
f y p y x

0 . Thus, from equations (2.1.4) and (2.1.5), we have F x p x x N
.
Case 3, 0: From equation (2.1.3), we have.

p y x f y
0 (2.1.7)
Replacing γ by
/ in equation (2.1.1), we have

p
y
x f
y

0
or

p
y
x f
y
f y

0
1
As f is linear, i.e.,

p
y
x f y

0
1
(2.1.8)
Multiplying (2.1.8) by λ, we have

p
y
x f y
0
(The inequality in equation (2.1.8) is reversed as λ is negative),

p
y
y x F x
0 (2.1.9)
Since
1
0

, by condition (2) of Theorem 1.2.1, we have
P y x p y x

1 1
0 0

(2.1.10)
and so

1
0
p y x F x
or
F x p x x N

Proof of Theorem 1.2.1.[2,9]
due to Siddiqi
Let S be the set of all linear functionals F such that F x f x x M
( ) ( )
and F x p x x X
( ) ( )
.
That is to say,S is the set of all functionals F extending f and ( ) ( )
≤
F x p x over X . S is non-empty as
not only does F belong to it but there are other functionals also which belong to it by virtue of Lemma
1.2.1, we introduce a relation in S as follows.
For F F S
1 2
, ∈ , we say that F1 is in relation to F2 and we write F F
1 2
if DF DF
1 2
⊂ and
F DF F
2 1 1
/ = (let DF1 and DF2 denote, respectively, the domain of F1 and F F DF
2 2 1
: / denotes the
restriction of F2 on the domain of F1. S is a partially ordered set. The relation $$ is reflexive as F F
1 1
.
is transitive, because for F F F F
1 2 2 3

, , we have
DF DF DF DF F DF F
1 2 2 3 2 1 1

, /
. and F DF F
3 2 2
/ = , which implies that
DF DF
1 3
⊂ and F DF F
3 1 1
/ = . $$ is anti-symmetric. For F F
1 2
;
DF DF
F DF F
1 2
2 1 1

For F F
2 1
;
DF DF
F DF F
2 1
1 2 2

Therefore, we have F F
1 2
= .
We now[5]
show that every totally ordered subset of S has an upper bound in S . Let T F

be a
totally ordered subset of S . Let us consider a functional, say F defined over DFσ
σ
 . If x DF

 ,
there must be some σ such that x DF
, and we define F x F x

. F is well defined and its
domain Fσ
σ
 is a subspace of X . DFσ
σ
 is a subspace: Let x y DF
,

 . This implies that x DF
1
and y DF
2
. Since T is totally ordered, either DF DF

1 2
or DF DF

2 1
. Let DF DF

1 2
. Then,
DFσ , x DF
1
which implies that real
σ µ
∈ ∀
x DF . This shows that DFσ is a subspace. F is well
defined: Suppose x DF
. Then, by the definition of F , we have F x F x
( ) ( )
and F x F x
( ) ( )
. By
the total ordering of T either Fσ extends Fυ or vice-versa and so F x F x

( ) ( )
which shows that F is
well defined. It is clear from the definition that F is linear, F x f x
( ) ( )
= for x D M
and
( ) ( )
≤ ∀ ∈
F x p x x DF . Thus, for each F F
; i.e., is an upper bound of T . By Zorn’s lemma, there
exists a maximal element F̂ in S ; i.e., ˆ
i
F is a linear extension of ˆ( ) ( )
≤
F x p x and ˆ

F F for every
F S
∈ . The theorem will be proved if we show that ˆ =
F
D X . We know that ˆ ⊂
F
D X . Suppose there is
an element x X
∈ such that ˆ
0 ∉ =
F
x D X . By lemma 1.1.1, there exists F̂ such that F̂ is linear,
ˆ
ˆ
( ) ( )
= ∀ ∈ F
F x F x x D , and ˆ( ) ( )
≤
F x p x for x D x
F

0 is also an extension of f . This implies that
F̂ is not maximal element for S which is a contradiction. Hence, D X
F = .

Proof of Theorem 1.2.2[3.9]
due to Siddiqi
Since f is bounded and linear, we have f x f x x

 , . If we define p x f x
then p x
( )
satisfies the conditions of theorem 1.1.1. By theorem 1.2.1, there exists F extending f which is linear
and ( ) ( )
≤ =
F x p x f x
‖ ‖‖ ‖which implies that F is bounded and
F x F x f

1sup

On the other hand,[9]
for x M
∈ , f x F
 . Hence, =
f F
‖ ‖‖ ‖
.
Remark 2.2.1: The Hahn–Banach theorem is also valid for normed spaces defined over the complex
field.
Consequences of the Extension Form of the Hahn–Banach Theorem
The proofs of the following important results mainly depend on theorem 1.2.2.
due to Siddiqi
Let M w m m w R

[ / , ]
and f M R
: ⇒ such that ( ) λ
=
f m w
‖ ‖
.
f is linear
1 2 1 2
[ ( ) ( ) ]
λ λ
+ = +
f m m w
‖ ‖
where m w
1 1
and m w
2 2
or
f m m w w w f m f m
1 2 1 2 1 2 1 2

Similarly, f m f m R

. f is bounded f m w m

and so f m k m
where
0 1

k and
f w w m w

if then
, 1
By theorem 1.2.2,
f f m w m
m M
m
m m

1
1 1
1
sup sup sup

Since f , defined on M , is linear and bounded (and hence continuous) and satisfies the conditions
f w w
and f =1; by Theorem 1.2.2, there exists a continuous linear functional F over X
extending f such that F =1 and F w w
.
due to Siddiqi
Suppose 0
≠
w but F w
0 for all F X
∈ *. Since 0
≠
w , by theorem 1.2.1., by theorem 1.2.3, there
exists a functional F X
∈ * such that 1
=
F
‖ ‖ and ( ) =
F w w
‖ ‖
. This shows that F w
0 which
contradiction is. Hence, if F w F X

0 *, then w must be zero.
due to Siddiqi
Let w XM
∈ and d w m
m M

inf
. Since M is a closed subspace and M d
, 0. Suppose N is the subspace
spanned by w and M ; i.e.,n N
∈ if and only if

N w m R m M

, , (2.1.11)
Define a functional on N as follows:
F n

F is linear and bounded: f n n
1 2 1 2

, where n w m
1 1

and n w m
2 2

. Hence,
f n n f n f n
1 2 1 2

. Similarly, f n f

for real µ . Thus, f is linear. To show that f is
bounded, we need to show that there exists K 0 such that f n n N
n
. We have
n m w
m
w
m
w

Since
m M
and d inf w m
m M

, we see that
m
w d

. Hence, n or
n d
/ By
definition, f n n d

/ or f n k
where k
d

1
0. Thus, f is bounded. N w
= implies that
1 and therefore, f w
1. N m M
implies that 1 and therefore, from the definition of f ,
f m
0. Thus, f is bounded linear and satisfies the conditions f w
1 and f m
0. Hence, by
theorem 1.2.2, there exists F defined over X such that F is an extension of f and F is bounded linear,
i.e., F X F w

*, 1 and F m m M

0 .
due to Siddiqi
Let N be the subspace spanned by M and (see equation (2.1.11)). Define f on N as f n d
,
proceeding exactly as in the proof of theorem 1.2.5, we can show that f is linear and bounded on N ,
f n d n

, f w d
0, and f m
0 for all m M
∈ since f n n
, we have
1
≤
f
(2.1.12)
For arbitrary 0, by the definition of d , there must exist an m M
∈ such that w m d
Let
z
w m
w m

. Then, z
w m
w m

1 and f z f w m d w m
( ) ( ) /
. By definition, f n d
( ) ;
n w m
, then 1; and so f w m d
( )
;
f z
d
d
( )

(2.1.13)
By theorem 1.2.2. f sup f m
m
=
=1
( ) . Since z =1, equation (2.1.13) implies that f z
d
d
( )

. Since 0
is arbitrary, we have
f ≥1(2.1.14)
From equations (2.1.12) and (2.1.14) have f =1. Thus, f is bounded and linear,
f m m M f w
( ) ; ( )

0 0 and f =1. By theorem 1.2.2, there exists F X
∈ * such that ( ) 0
≠
F w ;
F m
( ) = 0 for all m M
∈ and f =1.

due to Siddiqi
Let fn
be a sequence in the surface of the unit sphere S of
X S F X F
* *
/

1
such that F F Fn
1 2
, , ,

is a dense subset of S . By theorem 1.2.2,
( )
1
=
= =
v
F sup F v
‖
‖
‖ ‖
and so for 0, there exists v X
∈ such that 1
=
v
‖ ‖ and
( ) ( )
1−∈ ≤
F F v
‖ ‖
(2.1.15)
Putting
1
2
in equation (2.1.15), there exists v X
∈ such that 1
=
v
‖ ‖ and ( )
1
2
≤
F F v
‖ ‖ .
Let vn
be a sequence such that 1
=
n
v
‖ ‖ ; ( )
1
2
≤
n n n
F F v
‖ ‖ ; and M be a subspace spanned by vn
.
Then, M is separable by its construction. In other to prove that X is separable, we show that M X
=
suppose ≠
X M ; then, there exists w X w M

; by theorem 1.2.2, there exists F X
∈ *such that 1
=
F
‖ ‖
( ) 0
≠
F w (2.1.16)
and F m m M

0 . In particular, F v n
n

0 , where
1
2
F F v F v F v F v F v F v F v
n n n n n n n n n n n

Since vn =1 and
F v n
n

0
We have
1
2
F F F
n n
(2.1.17)
We can choose Fn
such that
lim
n
n
F F

0(2.1.18)
Because Fn
is a dense subset of S . This implies from equation (2.1.17) that F n
n
0 .
Thus, using equations (2.1.16), (2.1.18), we have
I F F F F F F F F F F F
n n n n n n

2
or
1 0
= =
F ,
which is contradiction. Hence, our assumption is false and X M
= .
MAIN RESULTS ON THE GENERALIZED HAHN–BANACH THEOREM
Theorem 3.1: Let X be a real vector space, M − a subspace of X , and Pi a sequence of real function s
defined on X satisfying the following conditions:
i. P x p x
i i
i
n
i i
i
n

1 1

ii. P x p x
i i i i i i

For each x X
i ∈ and αi all positive.
Further, suppose that fi is a sequence of linear functional on M such that
f x p x x M
i i
i
n
i i
i
n
i

1 1
Then, there exists sequence of linear functional i
F defined on X for which
F x f x x M
i i
i
n
i i
i
n
i

1 1
and
F x p x x X
i i
i
n
i i
i
n
i

1 1
In other words, there exists sequence of extensions Fi of F having the property of Fi .
Proof: The statement and proof of the following Lemma will be very significant in the proof of the
Generalized Hahn–Banach theorem.
Lemma: Let X be a vector space and µ its proper subspace. For each x X M
i , let N m xi

 .
Furthermore, suppose that fi is a sequence of linear functionals on M and pi sequence of functionals
on X satisfying the conditions of theorem 3.1 such that
f x p x x M
i i
i
n
i i
i
n
i

1 1
Then, there exists a sequence of linear functional Fi defined on N such that
F x f x x M
i i
i
n
i i
i
n
i

1 1
,
and
F x p x x N
i i
i
n
i i
i
n
i

1 1
,
This Lemma implies that theorem 3.1 is valid for the subspace generated or spanned by M xi
.
Proof: Since
f x p x
i i
i
n
i i
i
n

1 1
For x M
i ∈ and fi
i
n

1
are linear, we have for arbitrary y P
i ∈
f y f y p y
i i
i
n
i i
i
n
i i
i
n

1 1 1
or
( ) ( ) ( ) ( )
0 1 0 1 1 0
1 1
= = +
= = ≠ ≠
∆ ≤ ∆ + ≤ + + − −
∑ ∑
∑ ∑
n n
n n
i i i i i i i i i i
i i i even i odd
f y p y x p y x p y x
By condition 1 of theorem 1.2.1., thus by regrouping the terms of i1
on one side and those of γi on the
other side, we have


p y x f y p y x f y
i i i i i i i i
i
n
i
n
1 0 1 0 1
1
1
(3.1)
Suppose y s
i
′ are kept fixed and y s
i

1 are allowed to vary over M , then equation (3.1) implies that the
set of real number p y x f y y M
i i i i i

0 has lower bounds and hence greatest lower bound by
Remark 1.1.
Let
R p y x f y y M
i i i i i

inf :
0
From equation (3.1), it is clear that
. As it is well known that between any two real numbers, there
is always a third real number. Let p be a real number such that

(3.2)
It may be observed that if
, then
. Therefore, for all y M
∈ , we have
f p y x f y p y x f y
i i i i i i i i
i
n
i
n

0 0
1
1
(3.3)
From the definition of N , it is clear that every element xi in N can be written as
x y x
i i
0 (3.4)
Where x M
0 ∈ or x X M
0 , λ is uniquely determined real number and γ is uniquely determined
vector in M . We now define a sequence of real valued functions on N as follows
F x F y x f y
i i i i i i

0 (3.5)
Where y is given by equation (3.2) and x is as in equation (3.4). We shall now verify that the well-
defined sequence of functions F x
i i
satisfies the desired conditions, i.e.,
i. Fi
i
n

1
is linear
ii. f x F x x M
i i
i
n
i i
i
n
i

1 1
iii. F x p x x N
i i
i
n
i i
i
n
i

1 1
1. ∑ i
F is linear
For
z z z N z y x z y x z y x
n n n n
1 2 1 1 1 0 2 2 2 0 0
, , , , , , ,

F y x y x y x
n n
1 1 1 0 2 2 0 0


= f y y y x
n n
1 1 2 1 2 0

f y y y
i n n
1 2 1 2

f y f y f y
n n n
1 1 2 2 1 2

as is linear
fi

or
F z z z f y f y f y
i n n n n
1 2 1 1 1 2 2 2

F z F z F z
n n
1 1 2 2
Similarly,
2. f z F Z
i
n
i
n

1
1
for each z N
∈ and for real α
3. If x M
1 ∈ , then λi must be zero in equation (3.4).
Case 1: i 0:} We have seen that
F x f x
i i
i
n
i i
i
n

1 1
and as
f x p x
i i
i
n
i i
i
n

1 1
we get that
F x P x
i i
i
n
i
i
n
i

1 1
Case 2: 0: From equation (3.2), we have

p y x f y
i i i
i
n
0
1
(3.6)
since N is a subspace, y N
i replacing y by y / λ in equation (3.6), we have

p
y
x f
y
i
i
i
i
i
i
i
n
0
1
or

p y x f
y
i i i i
i
i
i
n
1
1
0
1
By condition (2) of theorem (2.1),
p y x p y x
i
i
i
i
i i
i
n
i
n
1 1
0 0
1
1

For 0 and f
y
x
f y
i
i
i i
i i

1

as fi is linear.
Therefore,

i i i i
i
n
i i
i
n
p y x f y

0
1
1
or
f y y p y x
i i i i i i
i
n
i
n

0
1
1

Thus, from equations (3.4) and (3.5), we have
p y x f y
i i i i i
i
n
i
n

0
1
1
(3.7)
Replacing γi by
yi
λ
in equation (3.7), we have
p
y
x
x f
y
x
i
i
i
i
i
i
i
i
n
i
n

0
1
1

or
p
y
x
x f
y
x
f y
i
i
i
i i
i
i
i i i

0
1

i
n
i
n
i
n
1
1
1
As fi is linear,
p
y
x
x f y
i
i
i
i i i
i
n
i
n

0
1
1
1
(3.8)
Multiplying (3.8) by λ, we have

i i
i
i
i i i i
i
n
i
n
p
y
x
x f y

0
1
1
(The inequality in (3.8) is reversed as λ is negative) or

i i
i
i i i i
i
n
i
n
p y x F x
1
0
1
1
Since
1
0
i
, by theorem 2.1, we have
p y x p y x
i
i
i i
i
i i
i
n
i

1 1
0 0
1
1
n
n
and so

i
i
i i i i i
i
n
i
n
p y x F x
1
0
1
1
or
F x p x x N
i
i
n
i i i
i
n
i

1 1
and hence, the proof.
Now, having established the proof of the above stated lemma, we then make its use in the proof of
theorem 3.1 earlier stated. Hence: Let S be the set of sequence of all linear functional Fi such that
F x f x x M
i i
i
n
i i
i
n
i

1 1
and
F x p x x X
i i i
i
n
i
i
n
i

1
1

That is to say that S is the set of sequences of all functional Fi extending fi and F x p x
i i i
i
n
i
i
n

1
1
over X ; S is a non-empty as not only does Fi belong to it but there are other functional also which
belong to it by virtue of theorem (1.2.1), we introduce a relation which is as follows.
For F F S
i i
,
1 , we say that Fi is in relation to Fi+1 and we write F F
i i
1. If DF DF
i i
1 and
F DF F
i i i

1 / , let DFi and DFi+1 denote, respectively, the domain of Fi and Fi+1 . F DF
i i
+1 / on the
partially ordered set. The relation $$ is reflexive as F F
i i
1:� is transitive because for
F F F F
i i i i

1 1 2
; , we have DF DF
i i
1; DF DF
i i

1 2 . F DF F
i i i

1 / ; and F DF F
i i i

2 1 1
/ , which
implies that DF DF
i ⊂ 3 and F DF F
i i i

2 / : $$ is anti-symmetric forF F
i i
1.
DF DF
i i
1
F DF F
i i i

1 /
for
F F
DF DF
F DF F
i i
i i
i i i

1
1
1 1
/
Therefore, we have F F
i i
1. We now show that every totally ordered subject of S has an upper bound
in S Let T F

1
be a sequence of totally ordered subset of S . Let us consider a sequences of functional
sayF defined over DF

1
1
.
If x DF

1
1
, there must be some σi such that x DF
i i
and we defined F x F x
i i i
i

. F is well
defined and its domain DF i
i

is a subspace of X . DF i
i

is a subspace. Let x x x DF
n i
i
1 2
, , ,

.
This implies that x DF
i i
i

and x DF
i i
i

1 1

.
Since T is totally ordered, either DF DF
i i

1
or DF DF
i i

1
. Let DF DF
i i

1
. Then, x DF i

1
and
so
x x DF
i i i

1 1

or
x x DF
i i i
i

1

Let x DF
i i
implies that

x DF i
i

real . This shows that DF i
i

is a subspace. F is well
defined: Suppose x DF
i i
and x DF
i i
. Then, by the definition of Fi , we have F x F x
i i i
i

and
F x F x
i i i
i

.Bythetotalorderingof T ,either F i
σ andextend F i
ϑ orvice-versaandso F x F x
i i
i i

which shows that Fi is well defined. It is clear from the definition that Fi is linear,
F x f x x D M
i i
i
n
i i
i
n
i f

1 1
and
F x p x x D
i i
i
n
i i
i
n
i f

1 1

Thus, for each F T F F
i i i

, , i.e., is an upper bound of T . By Zorn’s lemma, there exists a maximal
element Fi in S , i.e., Fi is a linear extension of
f F x p x
i i i
i
n
i i
i
n
.

1 1
and
F F F S
i
i
n
i
i
n
i

1 1
for every
The theorem will be proved if we show that D X
Fi
= . We know that D X
Fi
⊂ . Suppose there is an
element x X
∈ such that x DFi
0 ∉ . By the above lemma 3.1, there exists Fi such that Fi is linear,
F x F x x D
i i i i i Fi

and
F x p x x D V x
i i
i
n
i i
i
n
i F

1 1
0
for
Is also an extension of f . This implies that F is not maximal element S which is a contradiction. Hence,
D X
fi
= . Hence, the proof.
Theorem 3.2 (on the generalized form of the topological Hahn–Banach theorem): Let x be a normed
space M − a subspace of X and fi − a sequence of bounded linear functional of M , then there exist a
sequence of bounded functional Fi on x such that
F x f x x M
i i
i
n
i i
i
n
i

1 1
F f
i
i
n
i
i
n

1 1
Proof 3.2: Since fi is bounded and linear, we have
f x f x x
i
i
n
i i
i
n
i
i
n
i

1 1 1
If we have defined p x f x
i i
i
n
i
i
n
i
i
n

1 1 1
then p x
i i
i
n

1
satisfies the conditions of the theorem (3.1) and
by this theorem, there exists Fi
i
n

1
extending fi
i
n

1
which is linear and F x p x x X
i i
i
n
i i
i
n
i

1 1
, we
have

F x F x
i i
i
n
i i
i
n
1 1
as Fi is linear and so by the above relation
F x p x f x f x p x
i i
i
n
i i
i
n
i i i i i i
i
n

1 1 1
Thus,
F x p x f x
i i
i
n
i i
i
n
i i

1 1
Which implies that Fi is bounded and
F F x f
i x i i i
i

1
sup
(3.9)

On the other hand, for x M
∈ ,
f x F x x
i i
i
n
i i
i
n
i
i
n

1 1 1
and so
f f x
i
i
n
x

1
1
sup
f f x F
i x
i
n
i i
i
n
i
i
n

1
1 1 1
sup
(3.10)
Hence, by (3.9) and (3.10), we have
f F
i
i
n
i
i
n

1 1
Proof of theorem 3.3: LetM w m m R
i i i i

: and F M R
i : → such that
f m w
i i
i
n
i i
i
n

1 1

fi is a linear since
f m m w
i i j
i
n
i j i
i
n

1 1

where m w
i i i
and m w
j j i
or
f m m w w f m
i i j
i
n
i i
i
n
i
j i
n
i i i i
i
n
i
j i

1 1
1
1 1
1

1
1
n
i j
f m
We now state the rest of the generalized results without their proofs as they directly follow.
Theorem 3.4: Let wi be a sequence of non-zero vectors in a normed space X . Then, there exists a
sequence of continuous linear functional Fi defined on the entire space X such that Fi =1 and
F w w
i i
i
n
i
i
n

1 1
Theorem 3.5: If X is a normed space such that F w F X
i i
i
n
i

1
0 *. Then, wi
i
n

1
0.
Theorem 3.6: Let X be a normed space and M its closed subspace. Further assume that w XM
i ∈ .
Then, there exists F X
i ∈ * such that F m
i i
0 for all m M
i ∈ and F w
i i
1.
Theorem 3.7: Let X be a normed space, m its subspace and w X
i ∈ such that d w m
i i
m M
i

inf 0.
Theorem 3.8: If xi
i
*
=1
 is separable, then Xi
i=1
 is itself separable.
REFERENCES
1. Banach S. Theories Operations Linear. New York: North Western University Dover Publication Inc; 1932.
2. Betherians S. Introduction to Hilbert Space. New York: Oxford University Press; 1963.
3. Davie A, Gamelin TA. Theorem on polynomial star approxiation. Proc Am Math Soc 1989;106:351-6.
4. Day MM. Normed Linear Spaces. 3rd
ed. New York: Springer; 1973.
5. Edwards RE. Functional Analysis. New York: Holt, Rinehart and Winston; 1965.
6. Avner F. Foundations of Modern Analysis. New York: North Western University Dover Publications Inc.; 1970.
7. Luisternik LA, Sobolev VJ. Elements of Functional Analysis. 3rd
ed. New Delhi: Hindustan Publishing Company; 1974.
8. Riesz F, Nagy BS. Functional Analysis. New York: Ungar; 1955.
9. Siddiqi AH. Applied Functional Analysis: Numerical Methods. Wave Let Methods and Image Processing. Boca Raton,
Florida: CRC Press; 2004.

03_AJMS_279_20_20210128_V2.pdf

More Related Content

Similar to 03_AJMS_279_20_20210128_V2.pdf (20)

More from BRNSS Publication Hub (20)

Recently uploaded (20)

03_AJMS_279_20_20210128_V2.pdf