FUNCTIONAL ANALYSIS BANACH SPACE HANH-BANACH THEOREM MAPPING THEOREM

2. Normed Linear Space:-
A Normed Space is a vector space N endowed
with a function(called Norm)
)_(: fieldscalarFN 
Which satisfies the conditions
NyxFxx
yxyx
xx
x




,...4
.3
00.2
0.1

MANIKANTA SATYALA FUNTIONAL ANALYSIS

3. Properties of Normed Linear Space:-
xx
xx
nn
n
n








yxyx 
xxxx nn 
yxyx
yy
xx
nn
n
n






1.
2.
3.
4.
(i.e.,) Norm is continuous function
MANIKANTA SATYALA FUNTIONAL ANALYSIS

4. Properties of Normed Linear Space:-
Norm induces a metric on a Normed linear space N
𝑑 ∶ 𝑁 × 𝑁 → 𝑅 𝑏𝑦 𝑑 𝑥, 𝑦 = 𝑥 − 𝑦 , 𝑥, 𝑦 ∈ 𝑁.
𝐶𝑙𝑒𝑎𝑟𝑙𝑦 𝑑 𝑥, 𝑦 ≥ 0 𝑎𝑛𝑑 𝑑 𝑥, 𝑦 = 0 ⇔ 𝑥 − 𝑦 = 0 ⇔ 𝑥 = 𝑦
𝑑 𝑥, 𝑦 = 𝑥 − 𝑦 = −( 𝑦 − 𝑥) = 𝑦 − 𝑥 = 𝑑 𝑦, 𝑥 ,
𝑑 𝑥, 𝑧 = 𝑥 − 𝑧 𝑓𝑜𝑟 𝑧 ∈ 𝑁.
𝑑 𝑥, 𝑧 = 𝑥 − 𝑧 = 𝑥 − 𝑦 + 𝑦 − 𝑧
≤ ||𝑥 − 𝑦|| + ||𝑦 − 𝑧|| = 𝑑(𝑥, 𝑦) + 𝑑(𝑦, 𝑧)
Note: -
A metric 𝑑 induced by a norm on a normed space 𝑁 satisfies
1.𝑑 𝑥 + 𝑎, 𝑦 + 𝑎 = 𝑑 𝑥, 𝑦
2.𝑑(𝛼𝑥, 𝛼𝑦) = 𝛼 𝑑(𝑥, 𝑦)
MANIKANTA SATYALA FUNTIONAL ANALYSIS

5. Properties of Normed Linear Space:-
 The metric induced by norm i.e. d(x, y) = ||x − y||,
x, y ∈ N has translational invariant property and
absolute homogeneity property..
 A normed linear space N is said to be complete if
every Cauchy sequence in N is convergent to an
element of N
 If L be a subspace of a normed linear space N then
its closure L is also a subspace of N
MANIKANTA SATYALA FUNTIONAL ANALYSIS

6. A complete Normed Linear Space is called a
Banach Space.
A Banach Space is a normed linear space which
is complete with respect to a metric induced by
a norm
Banach Space :-
Remark :-
A Normed Linear Space N is said to be Complete
if every Cauchy sequence in N is convergent to
an element of N
MANIKANTA SATYALA FUNTIONAL ANALYSIS

7. Banach Space :-Examples
1. R is a Banach space
Define by : 𝑅 → 𝑅 𝑏𝑦 𝑥 = 𝑥 .
2. Euclideanspace Rn and unitary space Cn.
They are Banach spaces with norm defined by
𝑥 : 𝜉𝑗
2𝑛
𝑗=1
1/2
= 𝜉1
2 + ⋯+ 𝜉 𝑛
2.
3. Let 1 ≤ 𝑝 ≤ ∞.
𝑳 𝒑
𝒏 denotes the space of all numbers
𝑥 𝑝 = (𝑥1,𝑥2,… , 𝑥 𝑛) 𝑝 = ( 𝑥𝑖
𝑝𝑛
𝑖=1 )1/𝑝
Remark: 𝑥 = 𝑥𝑖
𝑝𝑛
𝑖=1
1/𝑝
look at a closed unit ball in R2
with different norms.
MANIKANTA SATYALA FUNTIONAL ANALYSIS

8. Banach Space
𝒀𝒐𝒖𝒏𝒈’𝒔 𝒊𝒏𝒆𝒒𝒖𝒂𝒍𝒊𝒕𝒚
Let 𝑝 > 1 and set 𝑞 =
𝑝
𝑝 − 1
. For every 𝑎, 𝑏 ∈ 𝐶:
𝑎𝑏 ≤
𝑎 𝑝
𝑝
+
𝑏 𝑞
𝑞
𝑃𝑟𝑜𝑜𝑓:
Note that
1
𝑝
+
1
𝑞
= 1.Since –log is a convexfuction thenfor every 𝛼, 𝛽 > 0:
−log
𝛼
𝑝
+
𝛽
𝑞
≤ −
1
𝑝
log 𝛼 −
1
𝑞
log 𝛽 = − log 𝛼
1
𝑝 𝛽
1
𝑞 .
It follows that:
𝛼
𝑝
+
𝛽
𝑞
≥ 𝛼
1
𝑝 𝛽
1
𝑞.
Setting 𝛼 = 𝑎 𝑝
and β = 𝑏 𝑞
we reoverthe desiredresult
MANIKANTA SATYALA FUNTIONAL ANALYSIS

9. Banach Space
𝑯ö𝒍𝒅𝒆𝒓 𝒊𝒏𝒆𝒒𝒖𝒂𝒍𝒊𝒕𝒚:
Let 𝑝 > 1 and set 𝑞 =
𝑝
𝑝 − 1
. For every 𝑥, 𝑦 ∈ Cn
𝑥𝑖 𝑦𝑖
𝑛
𝑖=1
≤ 𝑥 𝑝 𝑦 𝑞 .
𝑃𝑟𝑜𝑜𝑓:
Using Young′
s inequality termby term:
𝑥𝑖 𝑦𝑖
𝑛
𝑖=1
𝑥 𝑝 𝑦 𝑞
=
𝑥𝑖
𝑥 𝑝
𝑦𝑖
𝑦 𝑞
𝑛
𝑖=1
≤
1
𝑝
𝑥𝑖
𝑥 𝑝
𝑝𝑛
𝑖=1
+
1
𝑞
𝑦𝑖
𝑦 𝑞
𝑞𝑛
𝑖=1
=
1
𝑝
+
1
𝑞
= 1.
MANIKANTA SATYALA FUNTIONAL ANALYSIS

10. Banach Space
𝑴𝒊𝒏𝒌𝒐𝒘𝒔𝒌𝒊 𝒊𝒏𝒆𝒒𝒖𝒂𝒍𝒊𝒕𝒚:
Let 𝑝 > 1 . For every 𝑥, 𝑦 ∈ Cn
𝑥 + 𝑦 𝑝 ≤ 𝑥 𝑝 + 𝑦 𝑞 .
𝑃𝑟𝑜𝑜𝑓:
For every 𝑖 ∈ 1,2,…, 𝑛 ,it follows from the triangleinequality that
𝑥𝑖 + 𝑦𝑖
𝑝 = 𝑥𝑖 + 𝑦𝑖 𝑥𝑖 + 𝑦𝑖
𝑝−1 ≤ 𝑥𝑖 𝑥𝑖 + 𝑦𝑖
𝑝−1 + 𝑦𝑖 𝑥𝑖 + 𝑦𝑖
𝑝−1
Summing over 𝑖,and using Holder′ s inequality:
𝑥 + 𝑦 𝑝
𝑝
≤ 𝑥 𝑝 + 𝑦 𝑝 𝑥𝑖 + 𝑦𝑖
𝑞 𝑝−1
𝑛
𝑖=1
1
𝑞
,
where 𝑞 =
𝑝
𝑝 − 1
. Noting that 𝑞 𝑝 − 1 = 𝑝 and
1
𝑞
=
𝑝 − 1
𝑝
:
𝑥 + 𝑦 𝑝
𝑝
≤ 𝑥 𝑝 + 𝑦 𝑝 𝑥 + 𝑦 𝑝
𝑝−1
,
which completesthe proof.
MANIKANTA SATYALA FUNTIONAL ANALYSIS

11. Banach Space
Schwarz inequality :-
MANIKANTA SATYALA FUNTIONAL ANALYSIS

12. Banach Space
MANIKANTA SATYALA FUNTIONAL ANALYSIS

13. Banach Space :-
Theorem: -
Let 𝑀 be a closed linear subspace of a normed linear space 𝑁. If
the norm of a coset 𝑥 + 𝑀 in the Quotient space 𝑁/𝑀 is
defined by 𝑥 + 𝑀 = inf 𝑥 + 𝑚 /𝑚 ∈ 𝑀 then 𝑁/𝑀 is a
normed linear space. Further if 𝑁 is complete then 𝑁/𝑀 is
complete
Remark: 𝑁/𝑀 is Banach if 𝑁 is Banach
MANIKANTA SATYALA FUNTIONAL ANALYSIS

14. Banach Space :-
Proof: -
First we have to show that 𝑁/𝑀 is Linear Space:-
Let 𝑥 + 𝑀, 𝑦 + 𝑀 ∈ 𝑁/𝑀
⟹ 𝑥 + 𝑀 + 𝑦 + 𝑀 = 𝑥 + 𝑦 + 𝑀
⟹ 𝑎 𝑥 + 𝑀 = 𝑎𝑥 + 𝑀 ∀𝑥, 𝑦 ∈ 𝑁 & 𝑎 ∈ 𝐹
Clearly 𝑁/𝑀is a linear space
Define : 𝑁/𝑀 → 𝐹 by 𝑥 + 𝑀 = inf 𝑥 + 𝑚 /𝑚 ∈ 𝑀
Now we show that 𝑁/𝑀 is Normed linear space:-
⟹ Since inf 𝑥+ 𝑚 /𝑚 ∈ 𝑀 ≥ 0.
We have 𝑥 + 𝑀 ≥ 0
⟹ Suppose 𝑥 + 𝑀 = 0 ⟺ inf 𝑥 + 𝑚 /𝑚 ∈ 𝑀 = 0
Since 𝑀 is closed, there certainly exists a sequence < 𝑚 𝑘 > such that
𝑚 𝑘 ⟶ 𝑚 ∀ 𝑘
⟹ 𝑥 + 𝑚 𝑘 ⟶ 𝑥 + 𝑚
⟺ 𝑥 + 𝑀 ∈ 𝑀 ⟺ 𝑥 ∈ 𝑀
⟺ 𝑥 + 𝑀 = 𝑀 = 𝑧𝑒𝑟𝑜 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑁/𝑀.
MANIKANTA SATYALA FUNTIONAL ANALYSIS

15. Banach Space :-
⟹ 𝑥 + 𝑀 + 𝑦 + 𝑀
= 𝑥 + 𝑦 + 𝑀
= inf 𝑥 + 𝑦 + 𝑚 /𝑚 ∈ 𝑀
= inf 𝑥 + 𝑦 + 𝑚1 + 𝑚2 /𝑚 = 𝑚1 𝑚2 ∈ 𝑀 .
≤ inf 𝑥 + 𝑚1 𝑦 + 𝑚2 / 𝑚1, 𝑚2 ∈ 𝑀
≤ inf 𝑥 + 𝑚1 / 𝑚1 ∈ 𝑀 + inf 𝑦 + 𝑚2 / 𝑚2 ∈ 𝑀
= 𝑥 + 𝑀 + 𝑦 + 𝑀
⟹ 𝛼 𝑥 + 𝑀 = 𝛼𝑥 + 𝑀
= inf 𝛼𝑥 + 𝑚 /𝑚 ∈ 𝑀
= inf 𝛼 𝑥 + 𝑚/𝛼 /𝑚 ∈ 𝑀
= inf 𝛼 𝑥 +
𝑚
𝛼
/
𝑚
𝛼
∈ 𝑀
= 𝛼 inf 𝑥 + 𝑚1 / 𝑚1 =
𝑚
𝛼
∈ 𝑀
= 𝛼 𝑥 + 𝑀
∴ 𝑁/𝑀 is a normed linear space
MANIKANTA SATYALA FUNTIONAL ANALYSIS

16. Banach Space :-
So it is enough to show that 𝑁/𝑀 is Complete.
⟹ Let 𝑥 𝑛 + 𝑀 be a Cauchy sequence in 𝑁/𝑀.
Choose 𝑦1 ∈ 𝑥1 + 𝑀. & 𝑦2 ∈ 𝑥2 + 𝑀 ∋ 𝑦1 − 𝑦2 <
1
2
Now choose 𝑦3 ∈ 𝑥3 + 𝑀 ∋ 𝑦2 − 𝑦3 <
1
22.
On continue this processes, we get 𝑦𝑛 − 𝑦𝑛+1 <
1
2 𝑛
Suppose 𝑚 < 𝑛 in 𝑁. Then,Consider
𝑦 𝑚 − 𝑦𝑛 = 𝑦 𝑚 − 𝑦 𝑚 +1 + 𝑦 𝑚 +1 − 𝑦 𝑚 +2 + 𝑦 𝑚 +2 + ⋯ + 𝑦 𝑛 −1 − 𝑦𝑛
≤ 𝑦 𝑚 − 𝑦 𝑚 +1 + 𝑦 𝑚 +1 − 𝑦 𝑚 +2 + ⋯ + 𝑦 𝑛−1 − 𝑦𝑛
≤
1
2 𝑚 +
1
2 𝑚 +1 +
1
2 𝑚 +2 + ⋯ +
1
2 𝑛 −1
<
1
2 𝑛−1
⟹ 𝑦𝑛 Converges.
𝑖. 𝑒. , 𝑦𝑛 ⟶ 𝑦(𝑠𝑎𝑦) Since 𝑁 is Complete
Consider 𝑥 𝑛 + 𝑀 − 𝑦 + 𝑀 = 𝑥 𝑛 − 𝑦 + 𝑀 ⟶ 0
⟹ 𝑥 𝑛 + 𝑀 ⟶ 𝑦 + 𝑀
⟹ 𝑁/𝑀 is complete
MANIKANTA SATYALA FUNTIONAL ANALYSIS

17. Banach Space :-
Bounded Linear : -Transformation
Let 𝑁, 𝑁 be two normed linear spaces. A linear transformation
𝑇: 𝑁 → 𝑁 is said to be bounded if 𝑇𝑥 ≤ 𝑀 𝑥 ∀𝑥 ∈ 𝑁 & 𝑀 > 0.
Continuous Linear Transformation: -
A transformation 𝑇 is said to be Continuous if it is continuous as a
mapping of Metric space 𝑁 into a Metric space 𝑁 .
(i.e,) 𝑥 𝑛 → 𝑥 ⟹ 𝑇𝑥 𝑛 → 𝑇𝑥
MANIKANTA SATYALA FUNTIONAL ANALYSIS

18. Banach Space :-
𝑻𝒉𝒆𝒐𝒓𝒆𝒎 ∶ 𝑳𝒆𝒕 𝑻 𝒃𝒆 𝒂 𝒍𝒊𝒏𝒆𝒂𝒓 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒕𝒊𝒐𝒏 𝒇𝒓𝒐𝒎 𝑵 𝒊𝒏𝒕𝒐 𝑵.
𝑻𝒉𝒆𝒏 𝒕𝒉𝒆 𝒇𝒐𝒍𝒍𝒐𝒘𝒊𝒏𝒈 𝒂𝒓𝒆 𝒆𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕
I. 𝑻 𝒊𝒔 𝑪𝒐𝒏𝒕𝒊𝒏𝒖𝒐𝒖𝒔
II. 𝑻 𝒊𝒔 𝒄𝒐𝒏𝒕𝒊𝒏𝒏𝒖𝒐𝒖𝒔 𝒂𝒕 𝒕𝒉𝒆 𝑶𝒓𝒊𝒈𝒊𝒏 𝒊. 𝒆., 𝒙 𝒏 → 𝟎 ⟹ 𝑻𝒙 𝒏 ⟶ 𝟎
III. 𝒕𝒉𝒆𝒓𝒆 𝒆𝒙𝒊𝒕𝒔 𝒂 𝒓𝒆𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝑴 ≥ 𝟎 𝒔𝒖𝒄𝒉 𝒕𝒉𝒂𝒕 𝑻𝒙 ≤
𝑴 𝒙 𝒊. 𝒆., 𝑻 𝒊𝒔 𝒃𝒐𝒖𝒏𝒅𝒆𝒅.
IV. 𝑰𝒇 𝑺 =
𝒙 𝒙 ≤ 𝟏 𝒊𝒔 𝒂 𝒄𝒍𝒐𝒔𝒆𝒅 𝒖𝒏𝒊𝒕 𝒃𝒂𝒍𝒍 𝒕𝒉𝒆𝒏 𝒕𝒉𝒆 𝒊𝒎𝒂𝒈𝒆 𝑻 𝑺 𝒊𝒔 𝒃𝒐𝒖𝒏𝒅𝒆𝒅
Proof: for practice
MANIKANTA SATYALA FUNTIONAL ANALYSIS

19. Banach Space :-
𝑹𝒆𝒎𝒂𝒓𝒌
∶ 𝑭𝒐𝒓 𝒕𝒉𝒆 𝒃𝒐𝒖𝒏𝒅𝒆𝒅 𝒍𝒊𝒏𝒆𝒂𝒓 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒕𝒊𝒐𝒏 𝑻 𝒇𝒓𝒐𝒎 𝑵 𝒊𝒏𝒕𝒐 𝑵.
𝑻𝒉𝒆 𝒇𝒐𝒍𝒍𝒐𝒘𝒊𝒏𝒈 𝒂𝒓𝒆 𝒆𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕
I. 𝑻 = 𝐒𝐮𝐩{ 𝑻𝒙 / 𝒙 ≤ 𝟏}
II. 𝑻 = 𝐒𝐮𝐩{ 𝑻𝒙 / 𝒙 = 𝟏}
III. 𝑻 = 𝐒𝐮𝐩
𝑻𝒙
𝒙
/𝒙 ≠ 𝟎, 𝒙 ∈ 𝑵
IV. 𝑻 = 𝐈𝐧𝐟 𝑴 > 0/ 𝑻 ≤ 𝑴 𝒙 , ∀𝒙 ∈ 𝑵
Proof: for practice
MANIKANTA SATYALA FUNTIONAL ANALYSIS

20. Banach Space :-
𝑻𝒉𝒆𝒐𝒓𝒆𝒎: −
𝑰𝒇 𝑵, 𝑵 𝒂𝒓𝒆 𝒏𝒐𝒓𝒎𝒆𝒅 𝒍𝒊𝒏𝒆𝒂𝒓 𝒔𝒑𝒂𝒄𝒆𝒔 𝒕𝒉𝒆𝒏 𝑩 𝑵, 𝑵 𝒐𝒇
𝒂𝒍𝒍 𝒄𝒐𝒏𝒕𝒊𝒏𝒖𝒐𝒖𝒔 𝒍𝒊𝒏𝒆𝒂𝒓 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝑵 𝒊𝒏𝒕𝒐 𝑵
𝒊𝒔 𝒂 𝒏𝒐𝒓𝒎𝒆𝒅 𝒍𝒊𝒏𝒆𝒂𝒓 𝒔𝒑𝒂𝒄𝒆 𝒘𝒊𝒕𝒉 𝒓𝒆𝒔𝒑𝒆𝒄𝒕 𝒕𝒐 𝒕𝒉𝒆 𝒑𝒐𝒊𝒏𝒕 𝒘𝒊𝒔𝒆
𝒍𝒊𝒏𝒆𝒂𝒓 𝒐𝒑𝒆𝒓𝒂𝒕𝒊𝒐𝒏𝒔 𝒂𝒏𝒅 𝒃𝒆 𝒏𝒐𝒓𝒎
∥ 𝑻 ∥= 𝐬𝐮𝐩 ∥ 𝑻 𝒙 ∥: ∥ 𝒙 ≤ 𝟏
𝒇𝒖𝒓𝒕𝒉𝒆𝒓 𝒊𝒇 𝑵 𝒊𝒔 𝒃𝒂𝒏𝒂𝒄𝒉 𝒔𝒑𝒂𝒄𝒆 𝒕𝒉𝒆𝒏 𝑩(𝑵, 𝑵) 𝒊𝒔 𝒂𝒍𝒔𝒐 𝒃𝒂𝒏𝒂𝒄𝒉 𝒔𝒑𝒂𝒄𝒆 .
MANIKANTA SATYALA FUNTIONAL ANALYSIS

21. Banach Space :-
𝑷𝒓𝒐𝒐𝒇: −
𝑭𝒐𝒓 𝑻, 𝑺 ∈ 𝜷 𝑵, 𝑵 , 𝒂 𝒊𝒔 𝒂 𝒔𝒄𝒂𝒍𝒂𝒓
𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝑻 + 𝒃 = 𝑻𝒙 + 𝒃𝒙 & 𝒂𝑻 𝒙 = 𝒂 𝑻𝒙 ∀𝒙 ∈ 𝑵
𝑪𝒍𝒆𝒂𝒓𝒍𝒚, 𝜷 𝑵, 𝑵 𝒊𝒔 𝒂 𝒍𝒊𝒏𝒆𝒂𝒓 𝒔𝒑𝒂𝒄𝒆
𝑶𝒃𝒔𝒆𝒓𝒗𝒆 𝒕𝒉𝒂𝒕 : 𝜷 𝑵, 𝑵 → 𝑭
1. 𝐒𝐢𝐧𝐜𝐞 𝑻𝒙 ≥ 𝟎, 𝐒𝐮𝐩 𝑻𝒙 ≥ 𝟎 ⇒ 𝑻 ≥ 𝟎
2. 𝐒𝐮𝐩𝐩𝐨𝐬𝐞 𝑻 = 𝟎 ⇔ 𝐒𝐮𝐩{ 𝑻𝒙 / 𝒙 ≤ 𝟏} = 𝟎
⇔ 𝑻𝒙 = 𝟎
⇔ 𝑻𝒙 = 𝟎
⇔ 𝑻 = 𝟎
MANIKANTA SATYALA FUNTIONAL ANALYSIS

22. Banach Space :-
3. 𝑻 + 𝑺 = 𝐒𝐮𝐩{ 𝑻 + 𝑺 (𝒙) / 𝒙 ≤ 𝟏}
= 𝐒𝐮𝐩{ 𝑻𝒙 + 𝑺𝒙 / 𝒙 ≤ 𝟏}
= 𝐒𝐮𝐩{ 𝑻𝒙 + 𝑺𝒙 / 𝒙 ≤ 𝟏}
≤ 𝐒𝐮𝐩{ 𝑻𝒙 / 𝒙 ≤ 𝟏} + 𝐒𝐮𝐩{ 𝑺𝒙 / 𝒙 ≤ 𝟏}
= 𝑻 + 𝑺
∴ 𝑻 + 𝑺 ≤ 𝑻 + 𝑺 .
4. (𝜶𝑻) = 𝐒𝐮𝐩{ 𝜶𝑻 𝒙 / 𝒙 ≤ 𝟏}
= 𝐒𝐮𝐩{ 𝜶 𝑻𝒙 / 𝒙 ≤ 𝟏}
= 𝐒𝐮𝐩{ 𝜶 𝑻𝒙 / 𝒙 ≤ 𝟏}
= 𝜶 𝐒𝐮𝐩{ 𝑻𝒙 / 𝒙 ≤ 𝟏}
= 𝜶 𝑻 .
𝑻𝒉𝒖𝒔 𝜷 𝑵, 𝑵 𝒊𝒔 𝒂 𝒏𝒐𝒓𝒎𝒆𝒅 𝒍𝒊𝒏𝒆𝒂𝒓𝒔𝒑𝒂𝒄𝒆.
𝑵𝒐𝒘 𝒊𝒕 𝒊𝒔 𝒆𝒏𝒐𝒖𝒈𝒉 𝒕𝒐 𝒔𝒐 𝒕𝒉𝒂𝒕 𝑵 𝒊𝒔 𝑪𝒐𝒎𝒑𝒍𝒆𝒕𝒆.( 𝒑𝒓𝒐𝒐𝒇 ∶ −𝒑𝒓𝒂𝒄𝒕𝒊𝒄𝒆 )
MANIKANTA SATYALA FUNTIONAL ANALYSIS

23. Banach Space :-
𝑻𝒉𝒆𝒐𝒓𝒆𝒎:−𝑳𝒆𝒕 𝑴 𝒃𝒆 𝒂 𝒍𝒊𝒏𝒆𝒂𝒓 𝒔𝒖𝒃𝒔𝒑𝒂𝒄𝒆 𝒐𝒇 𝒂 𝒏𝒐𝒓𝒎𝒆𝒅 𝒍𝒊𝒏𝒆𝒂𝒓 𝒔𝒑𝒂𝒄𝒆 𝑵,𝒂𝒏𝒅
𝒍𝒆𝒕 𝒇 𝒃𝒆 𝒂 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒐𝒏 𝑴 𝑻𝒉𝒆𝒏 𝒇 𝒄𝒂𝒏 𝒃𝒆 𝒆𝒙𝒕𝒆𝒏𝒅𝒆𝒅 𝒕𝒐 𝒂 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏
𝒇 𝟎 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒐𝒏 𝒕𝒉𝒆 𝒘𝒉𝒐𝒍𝒆 𝒔𝒑𝒂𝒄𝒆 𝑵 𝒔𝒖𝒄𝒉 𝒕𝒉𝒂𝒕 𝒇 𝟎 = 𝒇 .
𝑷𝒓𝒐𝒐𝒇:−𝑾𝒆 𝒑𝒓𝒐𝒗𝒆 𝒕𝒉𝒊𝒔 𝒃𝒚 𝒛𝒐𝒓𝒏′
𝒔𝒍𝒆𝒎𝒎𝒂
𝑳𝒆𝒕 𝑷 = {(𝒇 𝝀,𝑴 𝝀)/𝒇 𝝀 𝒊𝒔 𝒂𝒏 𝒆𝒙𝒕𝒆𝒏𝒕𝒊𝒐𝒏 𝒐𝒇 𝒇 𝒐𝒏 𝑴 𝝀 ⊇ 𝑴, 𝒇 𝝀 = 𝒇 }
𝑫𝒆𝒇𝒊𝒏𝒆 ≤ 𝒐𝒏 𝑷 𝒂𝒔 𝒇𝒐𝒍𝒍𝒐𝒘𝒔.
𝒇 𝝀, 𝑴 𝝀 ≤ 𝒇 𝝁, 𝑴 𝝁 ⇔ 𝑴 𝝀 ⊂ 𝑴 𝝁
𝒇 𝝀 𝒙 = 𝒇 𝝁 𝒙 ∀ 𝒙 ∈ 𝑴 𝝀. 𝑪𝒍𝒆𝒂𝒓𝒍𝒚 𝑷 ≠ ∅ 𝒔𝒊𝒏𝒄𝒆 𝒇, 𝑴 ∈ 𝑷, 𝒂𝒍𝒔𝒐 𝑷,≤ 𝒊𝒔 𝒂 𝒄𝒐𝒔𝒆𝒕
𝑳𝒆𝒕 𝑸 = 𝒇𝒊,𝑴𝒊 𝒃𝒆 𝒂 𝒄𝒉𝒂𝒊𝒏 𝒊𝒏 𝑷.
𝑪𝒍𝒆𝒂𝒓𝒍𝒚 ∅, 𝑴𝒊 𝒊𝒔 𝒂𝒏 𝒖𝒑𝒑𝒆𝒓𝒃𝒐𝒖𝒏𝒅 𝒇𝒐𝒓 𝑸. 𝑾𝒉𝒆𝒓𝒆 ∅ 𝒙 = 𝒇𝒊 𝒙 ,∀ 𝒙 ∈ 𝑴𝒊
𝑨𝒍𝒔𝒐 𝒐𝒃𝒔𝒆𝒓𝒗𝒆 𝒕𝒉𝒂𝒕 𝑴𝒊
𝒊
𝒊𝒔 𝒂 𝒔𝒖𝒃𝒔𝒑𝒂𝒄𝒆 𝒐𝒇 𝑵
MANIKANTA SATYALA FUNTIONAL ANALYSIS

24. Banach Space :-
∅ 𝒊𝒔 𝒘𝒆𝒍𝒍 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 − 𝑳𝒆𝒕 𝒙 ∈ 𝑴𝒊 ⟹ 𝒙 ∈ 𝑴𝒊 𝒇𝒐𝒓 𝒔𝒐𝒎𝒆 𝒊
⟹ 𝒙 ∈ 𝑴 𝒇𝒐𝒓 𝒔𝒐𝒎𝒆
𝒊𝒇 𝑴𝒊 𝑴 𝒕𝒉𝒆𝒏 ⟹
∅ 𝒙 𝒇𝒊 𝒙
∅ 𝒙 𝒇 𝒙
⟹ 𝒇𝒊 𝒙 𝒇 𝒙 ⟹ ∅ 𝒊𝒔 𝒘𝒆𝒍𝒍 𝒅𝒆𝒇𝒊𝒏𝒆𝒅
𝑯𝒆𝒏𝒄𝒆 𝒃𝒚 𝒆𝒓𝒐 𝒔 𝒍𝒆𝒎𝒎𝒂 𝑻𝒉𝒆𝒓𝒆 𝒊𝒔 𝒂 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝑭 𝑯 ∋ 𝑭 𝑯 ∈ 𝑷
𝑪𝒍𝒂𝒊𝒎 ∶-𝑯 𝑵
𝑺𝒖𝒑𝒑𝒐𝒔𝒆 𝑯 ≠ 𝑵
𝑯𝒆𝒏𝒄𝒆 𝒃𝒚 𝒂𝒃𝒐𝒗𝒆 𝒍𝒆𝒎𝒎𝒂 𝒂 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝑭 𝟎 𝒐𝒏 𝑯 𝟎 ∋ 𝑭 𝟎 𝒇
⟹ 𝑭 𝟎 𝑯 𝟎 ∈ 𝑷 𝒕𝒐 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒂𝒍𝒊𝒕𝒚 𝒐𝒇 𝑭 𝑯
𝑯𝒂𝒏𝒄𝒆 𝑯 𝑵
MANIKANTA SATYALA FUNTIONAL ANALYSIS

25. Banach Space :-
𝑪𝑶𝑵𝑺𝑬𝑸𝑼𝑬𝑵𝑪𝑬𝑺 𝑶𝑭 𝑯𝑨𝑯𝑵 − 𝑩𝑨𝑵𝑨𝑪𝑯 𝑻𝑯𝑬𝑶𝑹𝑬𝑴
1. 𝐋𝐞𝐭 𝐍 𝐛𝐞 𝐚 𝐧𝐨𝐫𝐦𝐞𝐝 𝐥𝐢𝐧𝐞𝐚𝐫 𝐬𝐩𝐚𝐜𝐞 𝐚𝐧𝐝 𝟎 ≠ 𝒙 𝟎 ∈ 𝐍 𝐭𝐡𝐞𝐧
𝐚 𝐟𝐮𝐧𝐭𝐢𝐨𝐧𝐚𝐥 𝒇 ∈ 𝐍 ∋ 𝐟 𝒙 𝟎 = 𝒙 𝟎
Remark:- 𝐍 𝒔𝒆𝒑𝒆𝒓𝒂𝒕𝒆𝒔 𝒗𝒆𝒄𝒕𝒐𝒓𝒔 𝒊𝒏 𝐍 𝐢. 𝐞. , 𝒙 ≠ 𝒚 ⇒ 𝒇(𝒙) ≠ 𝒇(𝒚)
2. 𝐋𝐞𝐭 𝐌 𝐛𝐞 𝐭𝐡𝐞 𝐜𝐥𝐨𝐬𝐞𝐝 𝐥𝐢𝐧𝐞𝐫 𝐨𝐟 𝐚 𝐧𝐨𝐫𝐦𝐞𝐝 𝐥𝐢𝐧𝐞𝐫 𝐬𝐩𝐚𝐜𝐞 𝐍 𝐚𝐧𝐝
𝒙 𝟎 ∉ 𝐌. 𝐓𝐡𝐞𝐧 𝐚 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐚𝐥 𝒇 ∈ 𝐍 ∋ 𝐟 𝐌 = 𝟎, 𝒇 𝒙 𝟎 ≠ 𝟎.
Definition:-
𝑨 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒕𝒊𝒐𝒏 𝑻 𝒊𝒔 𝒄𝒂𝒍𝒍𝒆𝒅 𝒂𝒏 𝒊𝒔𝒐𝒎𝒆𝒕𝒓𝒚 𝒊𝒇 𝑻𝒙 − 𝑻𝒚 𝒙 − 𝒚
𝐍𝐚𝐭𝐮𝐫𝐚𝐥 𝐞𝐦𝐛𝐞𝐝𝐝𝐢𝐧 𝐨𝐟 𝐍 𝐭𝐨 𝐍
𝑳𝒆𝒕 𝑵 𝒃𝒆 𝒂 𝒏𝒐𝒓𝒎𝒆𝒅 𝒍𝒊𝒏𝒆𝒓 𝒔𝒑𝒂𝒄𝒆,∀ 𝒙 ∈ 𝑵 𝒊𝒏𝒅𝒖𝒄𝒆𝒔 𝒂 𝒇𝒖𝒏𝒕𝒊𝒐𝒏𝒂𝒍 𝑭 𝒙 𝒐𝒏
𝑵 𝒃𝒚 𝑭 𝒙 𝒇 = 𝒇 𝒙 ∀ 𝒇 ∈ 𝑵 ∋ 𝑭 𝒙 = 𝒙 .
𝒇𝒖𝒓𝒕𝒉𝒆𝒓 𝒂 𝒎𝒂𝒑𝒑𝒊𝒏𝒈 𝑱:𝑵 → 𝑵 ∋ 𝑱 𝒙 = 𝑭 𝒙∀ 𝒙 ∈ 𝑵
𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒐𝒏 𝒊𝒔𝒐𝒎𝒆𝒕𝒓𝒊 𝒊𝒔𝒐𝒎𝒐𝒓𝒑𝒉𝒊𝒔𝒎 𝒏𝒐𝒇 𝑵 𝒊𝒏𝒕𝒐 𝑵
MANIKANTA SATYALA FUNTIONAL ANALYSIS

26. Banach Space :-
OPEN MAPPING THEOREM
𝑻𝒉𝒆𝒐𝒓𝒆𝒎:−
𝑳𝒆𝒕 𝑩,𝑩′ 𝒃𝒆 𝑩𝒂𝒏𝒂𝒄𝒉 𝒔𝒑𝒂𝒄𝒆𝒔 𝒂𝒏𝒅 𝑻 𝒃𝒆 𝒄𝒐𝒏𝒕𝒊𝒏𝒖𝒐𝒖𝒔 𝒍𝒊𝒏𝒆𝒂𝒓 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒕𝒊𝒐𝒏 𝒕𝒉𝒆𝒏
𝑻 𝒊𝒔 𝒐𝒑𝒆𝒏 𝒎𝒂𝒑𝒑𝒊𝒏𝒈
𝑷𝒓𝒐𝒐𝒇:−
𝑳𝒆𝒕 𝑻 𝑩 → 𝑩 𝒊𝒔 𝒂 𝒄𝒐𝒏𝒕𝒊𝒏𝒖𝒐𝒖𝒔 𝒍𝒊𝒏𝒆𝒂𝒓 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒕𝒊𝒐𝒏
𝒔𝒖𝒑𝒑𝒐𝒔𝒆 𝒊𝒔 𝒂𝒏 𝑶𝒑𝒆𝒏𝒔𝒆𝒕 𝒊𝒏 𝑩
𝑪𝒍𝒂𝒊𝒎 − 𝑻 𝒊𝒔 𝒐𝒑𝒆𝒏𝒊𝒏 𝑩
𝑳𝒆𝒕 𝒚 ∈ 𝑻 ⇒ 𝒚 𝑻𝒙 𝒇𝒐𝒓 𝒙 ∈
𝑺𝒊𝒏𝒄𝒆 𝒊𝒔 𝒐𝒑𝒆𝒏 𝒂𝒏 𝒐𝒑𝒆𝒏𝒔𝒑𝒉𝒆𝒓𝒆 𝑺 𝒙 𝒓 𝒔𝒖𝒄𝒉 𝒕𝒉𝒂𝒕 𝒙 ∈ 𝑺 𝒙 𝒓 ⊂
⇒ 𝒙 ∈ 𝒙 𝑺 𝒓 ⊂ ⇒ 𝑻𝒙 ∈ 𝑻 𝒙 𝑺 𝒓 ⊂ 𝑻 − − − − − − 𝟏
𝒃𝒖𝒕 𝒃𝒚 𝒌𝒏𝒐𝒘𝒏 𝒍𝒆𝒎𝒎𝒂 𝒂𝒏 𝒐𝒑𝒆𝒏 𝒔𝒑𝒉𝒆𝒓𝒆 𝑺 ⊂ 𝑻 𝑺 𝒓
𝑵𝒐𝒘 𝑪𝒐𝒏𝒔𝒊𝒅𝒆𝒓 𝒚 𝑺 ⊂ 𝒚 𝑻 𝑺 𝒓 ⇒ 𝒚 𝑺 ⊂ 𝑻𝒙 𝑻 𝑺 𝒓
⇒ 𝒚 𝑺 ⊂ 𝑻 𝒙 𝑺 𝒓 ⊂ 𝑻 𝑩𝒚 𝟏
⇒ 𝑺 𝒚 ⊂ 𝑻 𝒂𝒏𝒅 𝒚 ∈ 𝑺 𝒚
⇒ 𝑻 𝒊𝒔 𝒐𝒑𝒆𝒏 ⇒ 𝑻 𝒊𝒔 𝒂𝒏 𝒐𝒑𝒆𝒏 𝒎𝒂𝒑𝒑𝒊𝒏𝒈
MANIKANTA SATYALA FUNTIONAL ANALYSIS

27. Banach Space :-
𝑻𝒉𝒆𝒐𝒓𝒆𝒎 ∶ −
𝑳𝒆𝒕 𝑩 𝑩 𝒃𝒆 𝑩𝒂𝒏𝒂𝒄𝒉 𝒔𝒑𝒂𝒄𝒆𝒔 𝑻 𝒃𝒆 𝒂 𝑪𝒐𝒏𝒕𝒊𝒏𝒖𝒐𝒖𝒔 𝟏 − 𝟏 𝒐𝒏𝒕𝒐
𝑳𝒊𝒏𝒆𝒂𝒓 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒕𝒊𝒐𝒏 𝒇𝒓𝒐𝒎 𝑩 𝒐𝒏𝒕𝒐 𝑩 𝒕𝒉𝒆𝒏 𝑻 𝒊𝒔 𝑯𝒐𝒎𝒆𝒐𝒎𝒐𝒓𝒑𝒉𝒊𝒄
𝑰𝒏𝒑𝒂𝒓𝒕𝒊𝒄𝒖𝒍𝒂𝒓 𝑻 𝟏
𝒊𝒔 𝑪𝒐𝒏𝒕𝒊𝒏𝒖𝒐𝒖𝒔
𝑷𝒓𝒐𝒐𝒇 𝒇𝒐𝒓 𝒑𝒓𝒂𝒄𝒕𝒊𝒄𝒆
𝑷𝒓𝒐 𝒆𝒄𝒕𝒊𝒐𝒏 𝒊𝒏 𝑩𝒂𝒏𝒂𝒄𝒉 𝒔𝒑𝒂𝒄𝒆:−
𝑨 𝒎𝒂𝒑𝒑𝒊𝒏𝒈 𝑬: 𝑩 → 𝑩 𝒊𝒔 𝒄𝒂𝒍𝒍𝒆𝒅 𝒑𝒓𝒐 𝒆𝒄𝒕𝒊𝒐𝒏 𝒊𝒇 𝟏 𝑬 𝟐
= 𝑬 𝒊𝒅𝒆𝒎𝒑𝒐𝒕𝒆𝒏𝒕
𝟐 𝑬 𝒊𝒔 𝒄𝒐𝒏𝒕𝒊𝒏𝒖𝒐𝒖𝒔
𝑹𝒆𝒎𝒂𝒓𝒌:− 𝟏 𝑨 𝒑𝒓𝒐 𝒆𝒄𝒕𝒊𝒐𝒏 𝑬 𝒅𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒆𝒔 𝒂 𝒑𝒂𝒊𝒓 𝒐𝒇 𝒔𝒖𝒃𝒔𝒑𝒂𝒄𝒆𝒔 𝑴 𝒂𝒏𝒅 𝑵 ∋
𝑩 = 𝑴⨁𝑵. 𝒘𝒉𝒆𝒓𝒆 𝑴 = {𝑬(𝒙)/𝒙 ∈ 𝑩} 𝒂𝒏𝒅 𝑵 = {𝒙 ∈ 𝑩/𝑬(𝒙) = 𝟎}
𝟐 𝑨𝒏𝒚 𝒓𝒆𝒑𝒓𝒆𝒔𝒆𝒏𝒕𝒂𝒕𝒊𝒐𝒏 = 𝒙 + 𝒚, 𝒙 ∈ 𝑴, 𝒚 ∈ 𝑵 𝒊𝒔 𝒖𝒏𝒊𝒒𝒖𝒆.
MANIKANTA SATYALA FUNTIONAL ANALYSIS