Definition of banach spaces

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Banach Spaces
Definition 1.1 (Norms and Normed Spaces). Let be a vector space over the field of complex
scalars. Then is a normed linear space if for every there is a real number ‖ ‖, called
norm of such that:
1) ‖ ‖ ,
2) ‖ ‖ if and only if ,
3) ‖ ‖ | |‖ ‖ for every scalar , and
4) ‖ ‖ ‖ ‖ ‖ ‖
Definition 1.2 (Convergent and Cauchy sequences). Let be a vector space, and let * + be a
sequence of elements of .
1) * + converges to if ‖ ‖ , i.e., if
, , , ‖ ‖
In this case, we write or .
2) * + is Cauchy is if
, , , ‖ ‖
Definition 1.3 (Banach Spaces). It is easy to show that any convergent sequence in a normed
linear space is a Cauchy sequence. However, it may or may not be true in an arbitrary normed
linear space that all Cauchy sequences are convergent. A normed linear space which does have
property that all Cauchy sequences are convergent is said to be complete. A complete normed
linear space is called a Banach space.
Definition 1.4 A Banach space is a normed vector space in which each Cauchy sequence
* + converges to some .
Example 1.5. The following are all Banach spaces under the given norms. Here can be in the
range .
1. ( ) { ∫ | ( )| }, ‖ ‖ .∫ | ( )| /
2. ( ) * +, ‖ ‖ | ( )|
3. ( ) * + ‖ ‖ | ( )|
4. ( ) { ( ) | | ( ) } ‖ ‖ | ( )|
5. , - * +, ‖ ‖ | ( )|, -
6. ( ) ** + | ∑ | | +, ‖* + ‖ ∑ | |
7. ( ) ** + | ∑ | | +, ‖* + ‖ (∑ | | )
Proof: (1)
First step we have to prove (1) is satisfies properties of norm.
1) ‖ ‖ .∫ | ( )| / trivially
2) ‖ ‖ .∫ | ( )| / ( )
( )
.∫ | ( )| / ∫ | ( )|
| ( )|
( )
( )
( ) | ( )|

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∫ | ( )|
∫ | ( )|
.∫ | ( )| /
3) ‖ ‖ | |‖ ‖ ,
‖ ‖ (∫ | ( )| )
(∫ | | | ( )| )
| | (∫ | ( )| )
| |‖ ‖
4) ‖ ‖ ‖ ‖ ‖ ‖ , ( )
‖ ‖ (∫ | ( ) ( )| )
.∫ | ( )| / .∫ | ( )| / (Minkowski’s Inequality)
‖ ‖ ‖ ‖
Second step, we need to prove (1) is completeness.
Use Definition 1.4:
Proof: Let * + be a Cauchy sequence in ( ),i.e.
, , such that:
‖ ‖ , for all .
Need to show: there are exists a function ( ) such that:
‖ ‖ for
Idea: use that for ,
| ( ) ( )| .∫ | ( ) ( )| /
‖ ‖
Thus:
, , such that: | ( ) ( )| , for all . (*)
This shows that the sequence of complex numbers * ( )+ is a Cauchy sequence in .
Since is a Banach space, * ( )+ convergent. Put:
( ) ( )
We need to show that:
‖ ‖ as

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By (*), by letting :
, , such that: | ( ) ( )| , for all .
So, ‖ ‖ .∫ | ( ) ( )| / . This shows that ‖ ‖ for .
( ) is complete and hence ( ) is a Banach space.
Proof: (2)
First step we have to prove (2) is satisfies properties of norm.
1) ‖ ‖ | ( )| trivially
2) ‖ ‖ | ( )| ( )
( )
| ( )| | ( )|
| ( )|
( )
( )
( ) | ( )|
| ( )|
| ( )|
3) ‖ ‖ | |‖ ‖ ,
‖ ‖ | ( )| | | | ( )| | |‖ ‖
4) ‖ ‖ ‖ ‖ ‖ ‖ , ( )
| ( ) ( )| | ( )| | ( )|
Proof: Work first on | ( ) ( )| (without “ ”).
| ( ) ( )| | ( )| | ( )|
| ( )| | ( )| (This holds for all )
So, | ( ) ( )| | ( )| | ( )|
Second step, we need to prove (2) is completeness.
Proof: Use Definition 1.4:
Let * + be a Cauchy sequence in ( ),i.e.
, , such that:
‖ ‖ , for all .
Need to show: there are exists a function ( ) such that:
‖ ‖ for
Idea: use that for ,
| ( ) ( )| | ( ) ( )|

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‖ ‖
Thus:
, , such that: | ( ) ( )| , for all . (*)
This shows that the sequence of complex numbers * ( )+ is a Cauchy sequence in .
Since is a Banach space, * ( )+ convergent. Put:
( ) ( )
We need to show that:
‖ ‖ as
By (*), by letting :
, , such that: | ( ) ( )| , for all .
So, ‖ ‖ | ( ) ( )| . This shows that ‖ ‖ for .
( ) is complete and hence ( ) is a Banach space.
Proof: (3)
First step we have to prove (3) is satisfies properties of norm.
1) ‖ ‖ | ( )| trivially
2) ‖ ‖ | ( )| ( )
( )
| ( )| | ( )|
( )
( )
( ) | ( )|
| ( )|
3) ‖ ‖ | |‖ ‖ ,
‖ ‖ | ( )| | | | ( )| | |‖ ‖
4) ‖ ‖ ‖ ‖ ‖ ‖ , ( )
| ( ) ( )| | ( )| | ( )|
Proof: Work first on | ( ) ( )| (without “ ”).
| ( ) ( )| | ( )| | ( )|
| ( )| | ( )| (This holds for all )
So, | ( ) ( )| | ( )| | ( )|
Second step, we need to prove (3) is completeness.
Proof: Use Definition 1.4:
Let * + be a Cauchy sequence in ( ),i.e.
, , such that:

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‖ ‖ , for all .
Need to show: there are exists a function ( ) such that:
‖ ‖ for
Idea: use that for ,
| ( ) ( )| | ( ) ( )|
‖ ‖
Thus:
, , such that: | ( ) ( )| , for all . (*)
This shows that the sequence of complex numbers * ( )+ is a Cauchy sequence in .
Since is a Banach space, * ( )+ convergent. Put:
( ) ( )
We need to show that:
‖ ‖ as
By (*), by letting :
, , such that: | ( ) ( )| , for all .
So, ‖ ‖ | ( ) ( )| . This shows that ‖ ‖ for .
( ) is complete and hence ( ) is a Banach space.
Proof: (4)
First step we have to prove (4) is satisfies properties of norm.
1) ‖ ‖ | ( )| trivially
2) ‖ ‖ | ( )| ( )
( )
| ( )| | ( )|
( )
( )
( ) | ( )|
| ( )|
3) ‖ ‖ | |‖ ‖ ,
‖ ‖ | ( )| | | | ( )| | |‖ ‖
4) ‖ ‖ ‖ ‖ ‖ ‖ , ( )
| ( ) ( )| | ( )| | ( )|
Proof: Work first on | ( ) ( )| (without “ ”).
| ( ) ( )| | ( )| | ( )|
| ( )| | ( )| (This holds for all )

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So, | ( ) ( )| | ( )| | ( )|
Second step, we need to prove (4) is completeness.
Proof: Use Definition 1.4:
Let * + be a Cauchy sequence in ( ),i.e.
, , such that:
‖ ‖ , for all .
Need to show: there are exists a function ( ) such that:
‖ ‖ for
Idea: use that for ,
| ( ) ( )| | ( ) ( )|
‖ ‖
Thus:
, , such that: | ( ) ( )| , for all . (*)
This shows that the sequence of complex numbers * ( )+ is a Cauchy sequence in .
Since is a Banach space, * ( )+ convergent. Put:
( ) ( )
We need to show that:
‖ ‖ as
By (*), by letting :
, , such that: | ( ) ( )| , for all .
So, ‖ ‖ | ( ) ( )| . This shows that ‖ ‖ for .
( ) is complete and hence ( ) is a Banach space.
Proof: (5)
First step we have to prove (5) is satisfies properties of norm.
1) ‖ ‖ | ( )|, - trivially
2) ‖ ‖ | ( )|, - ( )
( )
| ( )|, - | ( )|
( )
( )
( ) | ( )|
| ( )|, -
3) ‖ ‖ | |‖ ‖ ,

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‖ ‖ | ( )|, - | | | ( )|, - | |‖ ‖
4) ‖ ‖ ‖ ‖ ‖ ‖ , ( )
| ( ) ( )| | ( )| | ( )|
Proof: Work first on | ( ) ( )| (without “ ”).
| ( ) ( )| | ( )| | ( )|
| ( )| | ( )| (This holds for all )
So, | ( ) ( )| | ( )| | ( )|
Second step, we need to prove (5) is completeness.
Proof: Use Definition 1.4:
Let * + be a Cauchy sequence in , -, i.e.
, , such that:
‖ ‖ , for all .
Need to show: there are exists a function , -, such that:
‖ ‖ for
Idea: use that for , -,
| ( ) ( )| | ( ) ( )|, -
‖ ‖
Thus:
, , such that: | ( ) ( )| , for all . (*)
This shows that the sequence of complex numbers * ( )+ is a Cauchy sequence in .
Since is a Banach space, * ( )+ convergent. Put:
( ) ( )
We need to show that:
‖ ‖ as
By (*), by letting :
, , such that: | ( ) ( )| , for all .
So, ‖ ‖ | ( ) ( )|, - . This shows that ‖ ‖ for .
, - is complete and hence , - is a Banach space.
Proof (6):
First step we have to prove (6) is satisfies properties of norm.
1) ‖* + ‖ ∑ | | trivially
2) ‖* + ‖ ∑ | |
( )
∑ | | | |

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( )
| |
∑| |
3) ‖ ‖ | |‖ ‖ ,
‖ ‖ ∑| | | | ∑| | | |‖ ‖
4) ‖ ‖ ‖ ‖ ‖ ‖ , ( )
‖ ‖ ∑| |
∑| | ∑| |
‖ ‖ ‖ ‖
Second step, we need to prove (6) is completeness.
Proof: Use Definition 1.4:
Let * + be a Cauchy sequence in ( ), i.e.
, , such that:
‖ ‖ , for all .
Need to show: there are exists a function ( ), such that:
‖ ‖ for
Idea: use that for
| ( ) ( )| ∑ | ( ) ( )|
‖ ‖
Thus:
, , such that: | ( ) ( )| , for all . (*)
This shows that the sequence of complex numbers * + is a Cauchy sequence in .
Since is a Banach space, * + convergent. Put:
We need to show that:
‖ ‖ as
By (*), by letting :
, , such that: | ( )| , for all .
So, ‖ ‖ ∑ | | . This shows that ‖ ‖ for .
( ) is complete and hence ( )is a Banach space.

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Remark 1.5
a) To be more precise, the elements of ( ) are equivalence classes of functions that are
equal almost everywhere, i.e., if a.e. then we identify and as elements of
( )
b) If ( ) then | ( )| | ( )|. Thus we regard ( ) as being a subspace
of ( ) more precisely, each element ( ) determines an equivalence class of
functions that are equal to it a.e., and it is this equivalence class that belongs to ( )
c) The ( ) norm on ( ) is called the uniform norm. Convergence with respect to this
norm coincides with the definition of uniform convergence.
d) Sometimes the notation ( ) is used to denote the space that we call ( ) or ( )
may denote the space of all possible continuous function on . In this latter case, ‖ ‖
would not be a norm on ( ), since ‖ ‖ would not be finite for all ( ).
Definition 1.6 (Closed Subspace). Let be a Banach space.
a) A vector is a limit point of a set if there exist vectors then converge
to . In a particular, every element of is a limit point of
b) A subset of a Banach space is closed if it contains all its limit points. In another
words, is closed if whenever * + is a sequence of elements of and
then must be an element of
If is a closed subspace of Banach space , then it is itself a Banach space under the norm of
. Conversely, if subspace of and is a Banach space under the norm of , then is a
closed a subspace of
Example 1.7 ( ) and ( ) are closed subspaces of ( ) under the norm. The
following are also subspaces of ( ), but they are not a Banach spaces under the norm,
because they are not closed subspace of ( ):
( ) * ( ) +
( ) * ( ) ( ) +
( ) * ( ) ( ) ( ) +.
The space ( ) is called the Scwartz space.
Definition 1.8 (Sequence spaces). There are also many useful Banach spaces whose elements are
sequences of complex numbers. Be careful to distinguish between an element of such a space,
which is a sequence of number, and a sequence of elements of a such a space. Which would be a
sequence of sequences of numbers.
The elements of a sequence space may be indexed by any countable index set , typically the
natural number * + or the integer * +. If a sequence is indexed
by the natural numbers, then a typical sequence has the form ( ) ( ).
While if they are indexed by the integer, then a typical sequence has the form ( )
( ). In either case, the are complex numbers.
The following are Banach spaces whose elements are sequences of complex numbers:
( ) *( ) ∑ | | + ‖( ) ‖ (∑ | | ) ,
( ) {( ) | | } ‖( ) ‖ | | ,
Example 1.9 (Finite dimensional Banach spaces). The finite-dimensional complex Ecludian-
space is Banach space. There are infinitely many possible choices of norms for . In
particular, each of the following determines a norm for

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| | (| | | | | | )
| | *| | | | | |+
The norm | | is the usual Euclidian norm on .