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Normed Linear Space

INTRODUCTION

Functional analysis is a promising part of analysis. What is analysis that is more functional now provides a common language for all areas of mathematics involving the concept of continuity. No serious investigation in the theory of functions, differential equations or mathematical physics, in numerical methods, mathematical economics or control theory, or in numerous other fields take place or could take place without extensive use of the language and results of functional analysis.

Another important feature of functional analysis is the general abstract approach to the study of problems of analysis, which makes it possible to combine and subject to a single investigation problems that at first sight appear quite diverse.

In our project, we discuss about Normed Linear Space and Banach Space.

Normed Linear Space

Introduction:

The concept elaborated in the form of definitions, examples and properties of a normed linear space and the functions defined on such spaces are discussed in this chapter. The notion of the norm of a vector is a generalization of the notion of length and it obeys some simple and natural geometric principles.

Some definitions and examples:

Definition

Let L be a non-empty set and we assume that each pair of elements x and y in L can be combined by a process called addition and each scalar ?, and each element x in L can be combined by a process called scalar multiplication.

These operations of addition and scalar multiplication satisfy the following conditions:

(1) x + y = y + x;

(2) x + (y + z) = (x + y) + z;

(3) there exists in L a unique element, denoted by 0 and called the zero element, or the origin, such that x + 0 = x for every x;

(4) to each element x in L there corresponds a unique element in L denoted by –x and called the negative of x, such that x + (–x) = 0;

(5) ?(x + y) = ? x + ? y;

(6) (? + ?)x = ? x + ? x;

(7) (? ?)x = ?(? x);

(8) 1.x = x.

The algebraic system L defined by those operations and axioms is called a linear space.

Remark

A linear space is often called a vector space, and its elements are called vectors.

Definition

A non-empty subset M of a linear space L is called a linear subspace of L in x + y is in M whenever x and y are and if ?x is in M (for any scalar ?) whenever x is.

Sum and Direct Sum of Subspaces:

Given linear subspaces M1, M2,……,Mk of a linear space L, the set

M1 + M2 + ……..+Mk = {x1 + x2 +………+ xk :xi Mi; for i = 1,2,…..,k} is a linear subspace of L, it is the smallest linear subspace of L containing M1 M2 …. Mk .

M1 + M2 +…………….+Mk is called a direct sum iff every one of its elements can be expressed in one and only on way as x1 + x2 +……………+ xk , where xi Mi for

i =1,2,…….,k. That is, if x = x1 + x2 +……….+ xk and y = y1 + y2 +….….+ yk, (where xi, yi Mi for i = 1,2,….,k) implies xi = yi for every i = 1,2,….k, then the sum M1 + M2 +……..+ Mk is direct. We indicate this by writing

L = M1 M2 … Mk

Criterion for direct sum:

The sum M1 + M2 +………+ Mn is direct if and only if Mi ? Mj = {0} holds for every i ? j, where i = 1,2,…..,and j = 1,2,…….,k.

In particular, M1 + M2 is a direct sum M1 ? M2 = {0}.

Example

Let F be a field. Therefore, FN is a vector space. Fn is the direct sum of the n subspaces, Mi = {(x1,…..,xn) Fn, xj = 0 for every j ? i}, where i = 1,2,…,n. These subspaces are referred to as the axes of co-ordinates.

Definition

Let M be a subspace of a linear space L, and let the coset of an element x in L be defined by x + M = {x + m: m M}.Then the distinct cosets form a partition of L; and if addition and scalar multiplication are defined by

(x + M) + (y + M) = (x + y) +M

and ?(x + M) = ?x + M,

then these cosets constitute a linear space denoted by L/M and called the quotient space of L with respect to M. The origin in L/M is the coset 0 + M = M, and the negative of x + M is (–x) + M.

Definition

L be a linear space over (where will be either or ). A norm on L is a mapping of into having the following properties:

1. ||x|| ? 0 always and ||x|| = 0 x = 0 (positive definiteness)

2. ||?x|| = |?| ||x|| (homogeneity)

3. ||x + y|| ? ||x|| + ||y|| (triangle inequality)

Here x, y are any elements of L and ? is any scalar in .

Definition

A linear space equipped with a norm is called a normed linear space.

Example

If X = and ||x|| = ||x1|| + ||x2| +……………+ ||xn||, then show that

(X, ||.||) is a normed linear space.

Proof:

X = is a linear space with respect to the following operations of addition and scalar multiplication: x + y = (x1 + y1, x2 + y2,……….,xn + yn)

Where x = (x1, x2,…..,xn) X

y = (y1, y2,…..,yn) X

and ?x = (?x1, ?x2,…….,?xn), where x=(x1, x2,…..,xn) X and ? is real.

Verification of the norm axioms:

1.||x|| ? 0 as it is the sum of non-negative numbers.

2. Let x = 0, then xi = 0, i = 1,2,……n.

Thus ||x|| = 0, ||x2|| = 0,………,||xn|| = 0

Therefore ||x|| = 0.

Conversely, let ||x|| = 0.Then, each ||xi|| (i = 1,2,….,n) is zero. By the properties of norms on Xi , i = 1,2,……..,n, each xi , is zero. i.e., x = (0,0,0,……,0) = 0, zero vector of X.

Now,

||?x|| = || (?x1, ?x2,……,?xn) ||

=||?x1|| + ||?x2|| +……+ ||?xn||

=|?| ||x1|| + |?| ||x2|| +…+ |?| ||xn||

=|?| (||x1|| +……+ ||xn||)

=|?| ||x||

and ||x + y|| = ||x1 + y1, x2 + y2,………..,xn + yn||

= ||x1 + y1|| + ||x2 + y2|| +……+ ||xn + yn||

? (||x1|| + ||y1||) + (||x2|| + ||y2||) +……..+ (||xn|| + ||yn||)

= (||x1|| + ||x2|| +……. + ||xn||) + (||y1|| + ||y2|| +……+ ||yn||)

= ||x|| + ||y||

||x + y|| ? ||x|| + ||y|| for x, y X.

Thus, (X, ||.||) is a normed linear space.

Example

X = , ||x|| = max (|x1|, |x2|……,|xn|).

If x = (x1,….…,xn) and (y = y1,…….,yn ) then x + y = (x1 + y1,…….,xn + yn). We observe that |xi + yi| ? |xi| + |yi| ? ||x|| + ||y|| holds for every i = 1,2,……..,n since ||x|| + ||y|| is independent of i, we conclude that the maximum value of |xi + yi| taken over all i = 1,2,..……,n is ? ||x|| + ||y||.

Therefore ||x + y|| ? ||x|| + ||y||.

Example

X be the set of all bounded sequence of elements of . It becomes a vector space over under the definitions

x + y = (x1 + x2 + …..,xn + yn), ?x = (?x1,…….,?xn,……),

Where x = (x1,…..,xn), and y = (y1,…..,yn,….)

Proof:

Let ||x|| = |xn|. This defines a norm on X, n

|xi + yi| ? |xi| + |yi| ? ||x|| + ||y|| holds for every i = 1,2,…..,n

Since ||x|| + ||y|| is independent of i, we conclude that the supremum value of

|xi + yi| taken over all i = 1,2,….,n is ? ||x|| + ||y||.

Therefore ||x + y|| ? ||x|| + ||y||.

Topological Isomorphism, Equivalent Norms:

Definition

Two normed linear spaces N, N? over the same field is called topologically isomorphic iff there exists a bijective linear operator T: N?N? which is a homomorphism of the associated topological spaces. Such a mapping is called a topological isomorphism.

Theorem

Two normed linear spaces N, N? over the same field are topologically isomorphic if and only if there exists a surjective linear operator T: N?N? are constants m >0, M >0, such that

m||x|| ? ||T(x) || ? M ||x|| holds for all x X

Proof:

Necessity:

Suppose N, N? are topologically isomorphic. So there exists a bijective linear mapping T: N?N? such that T and T-1 are continuous on N and N? respectively. Hence T and T-1 are bounded. That means there exist constants M > 0, K > 0 such that ||T(x)|| ? M||x|| and ||T-1(y)|| ? K||y|| hold for all x N and y N? respectively. Writing x = T-1(y) we get ||x|| ? K ||T(x)||, whence m||x|| ? ||T(x)|| follows, where m = . Here x runs through N as y runs through N? because T-1 is bijective.

Hence m||x| |? ||T(x)|| ? M ||x|| holds for all x X

Sufficiency:

The hypothesis m||x|| ? ||T(x)|| implies ker T = {0}; So T is injective. T is surjective by hypothesis; so T-1 exists on y and is bijective and linear. The hypothesis ||T(x)|| ? M ||x|| for all x X, implies that T is continuous on X. Writing y = T(x) the hypothesis m||x|| ? ||T(x)|| for all x X, implies

||T-1(y)|| ? K||y|| for all y N? where K = , for y runs through N? as x runs through N, because T is bijective.

Hence T is a topological isomorphism of N onto N?.

Definition

Two norms on a linear space L are called equivalent iff they induce the same topology on L.

Theorem

Two norms ||.||1, ||.||2 on a linear space X are equivalent if and only if there exist constants m > 0, M > 0 such that

m||x||1 ? ||x||2 ? M ||x||1 holds for all x X

Proof:

The normed linear space obtained by equipping X with the norm ||.||1 be denoted by X1 and that obtained by equipping X with the norm ||.||2 be denoted by X2, respectively. Let T(x) = x for every x X; T bijective and linear.

Consider T as linear operator from X1 and X2; the hypothesis implies

m||x||1 ? ||T(x)||2 ? M||x||1 for all x X. This is precisely the condition that T and T-1 be continuous. So T is a homomorphism of X1 and X2. Since T is the identity mapping on X, we conclude from the theorem: every continuous function is sequentially continuous, that A X is an open set in the topological space X2 if and only if A is an open set in the topological space X1.That means, the norms ||.||1, ||.||2 induce the same topology on X; so they are equivalent.

Example

The Euclidean norm ||x||e = ?|x1|2+|x2|2+……+|xn|2, the maximum norm

||x||m = max (|x1| + |x2| +…….+ |xn|)

and the zigzag norm ||x||z = |x1| + |x2| +……+ |xn| on are mutually equivalent.

From the definition of these norms we get the inequalities

i) ||x||m ? ||x||z ? n||x||m

ii) ||x||m ? ||x||e ? ?n ||x||m

(i)implies that the maximum norm and the zigzag norm are equivalent.

(ii) implies that the maximum norm and the Euclidean norm are equivalent. The equivalence of norms is a transitive relation; so the zigzag norm and the Euclidean norm are equivalent.

Theorem

Any two norms on a finite dimensional vector space X are equivalent.

Proof:

The proof is by induction on n, the dimension of X. It suffices to prove that any norm ||.|| on an n-dimensional vector space X is equivalent to the maximum norm. Choose a basis {e1,e2,……,en}of X and let x = ?1e1 + ?2e2 +……..+ ?nen be any vector in X. For n = 1we have x = ?1e1; ||x||m = |?1|,

while ||x|| = ||?1e1|| = |?1| ||e1|| = ||e1|| ||x|| (in the given norm on X). So the relations m ||x||m ? ||x|| ? M ||x||m hold with m = M = ||e1|| and equality all through.

Therefore ||.|| is equivalent to ||.||m .

Linear operators on Normed Linear Spaces:

Definition

Let X and Y be two normed linear spaces. Then a mapping T from X into Y is called a linear operator. T is called a linear operator if the following conditions are satisfied:

a) T(x + y) = Tx + Ty, for all x, y X

b) T(?x) = ?Tx, for all x X and real ?.

In linear algebra the term “linear operator” signifies a linear transformation of a linear space into itself.

Theorem

If a linear operator T: X?Y is continuous at only one point x0 X, then it is continuous at every point.

Proof:

Given any point x X and ? > 0, we have to find ? > 0 such that

||T(u) T(x)|| < ?, whenever ||u x|| < ?. Since T is continuous at x0 , there

exists ? > 0 such that ||T(u) T(x0)|| < ?, whenever ||u x0|| < ?. Replacing

u by u x + x0 and using the linearity of T we get ||T(u) T(x)|| < ?, whenever

||u x|| < ?.

Definition

A linear operator T: X?Y is called bounded iff there exists a constant

M > 0 such that ||T(x)|| ? M||x|| holds for all x X. Thus T: X?Y is bounded iff the set { : x ? 0, x X} of non-negative real numbers is bounded above.

Theorem

A linear operator T: X?Y is continuous on X if and only if it is bounded.

Proof:

Necessity:

Corresponding to ? = 1 there exists ? > 0 such that ||T(x)|| < 1 whenever ||x|| < ?, because T is continuous at 0 X.

Claim

||T(x)||

t x.

Then || || = < ?

Hence ||T( )|| = ||T(x)|| < 1

Hence||T(x)|| ? ||x||.

This proves that T is bounded.

Sufficiency:

Suppose T is bounded, say ||T(x)|| ? M||x|| holds for all x X, where M > 0 is a constant.

Given ? > 0, let ? = .

Then ||T(x) T(x0)|| = ||T(x x0)|| ? M||x x0|| < ? whenever ||x x0||<?.

Therefore T is continuous at x0.

So T is continuous on X.

Theorem

Let N and N? be normed linear spaces and T is a linear operator on N into N?. Then the following statements are equivalent:

1) T is continuous

2) T is continuous at the origin

3) T is bounded

4) TS1 is a bounded subset of N?.

Proof:

We have to prove that (1) (2), (2) (3), (3) (4) which means that all are equivalent.

a)(1) (2)

Suppose T is continuous, which means that T is continuous at every point of N. Thus (1) (2) is obvious. Now suppose that T is continuous at the origin, that is for xn N with xn ? 0, Txn ? 0.This implies that if xn ? x then

T(xn–x) = Txn–Tx?0.That is, for xn ? x in N, Txn ? Tx in N?.

Then T is continuous and (2) (1).

b)(2) (3)

Suppose T is continuous at the origin. Then xn ? 0 implies Txn ? 0.Assume further that T is not bounded. This implies that for each natural number n we can find a vector xn such that

||Txn|| > n||xn|| or ||Txn|| > 1

or ||T ( xn)|| > 1.

By choosing yn = xn / n||xn||, we see that yn ? 0 but Tyn ? 0.This indicates the first assumption.

Hence, T is bounded whenever T is continuous at the origin, i.e. (2) (3).

To prove the converse, we assume that T is bounded, i.e. there exists K > 0 such that ||Tx|| ? K||x||.

From here, it follows immediately that is xn ? 0, then Txn ? 0.Thus (3) (2).

c)(3) (4):

Suppose T is bounded, there exists K > 0 such that

||Tx|| ? K||x||.From here it is clear that if ||x|| ? 1, then ||Tx|| ? K. This means that the image of closed unit sphere S1 under T is a bounded subset of N?.

Thus (3) (4).

For the converse, assume that T is bounded subset of N?. This means that TS1 is contained in a closed sphere of radius K centered at the ‘0’ of N?. If x=0, then Tx = 0 and ||Tx|| ? K||x||, and the result is proved .

In case x ? 0, and so ||T( )|| ? K

or ||Tx|| ? K||x||.

This shows that (4) (3).

Hence the proof.

Banach Spaces

Introduction:

Banach space is a normed linear space which is also, in a special way, a complete metric space. Most of our work in this chapter centers around two fundamental theorems relating to continuous linear transformations. The Hahn-Banach theorem guarantees that a Banach space is richly supplied with continuous linear functional. The open mapping theorem enables us to give a satisfactory description of the projections on a Banach space. This chapter is strongly oriented toward the algebraic and topological aspects of the matters at hand.

Definition examples of Banach space:

Definition

A normed linear space is a linear space N in which to each vector x there corresponds a real number, denoted by ||x|| and called the norm of x, in such a manner that

1) ||x|| ? 0, and ||x|| = 0 x = 0;

2) ||x + y|| ? ||x|| + ||y||

3) ||?x|| = |?| ||x||

The non-negative real number ||x|| is to be through of as the length of the vector x. If we regard ||x|| as a real function defined on N, this function is called the norm on N. A normed linear space is a metric space with respect to the induced metric defined by

d (x , y) = ||x – y||.

A Banach space is a normed linear space which is complete as a metric space.

Example l? is a Banach space under the supremum norm.

Proof:

Suppose <xn> is a Cauchy sequence in l?.

Here each xn l? let xn = (x1(n),x2(n),x3(n),……). Given ? > 0, which we assume to be < 0, there exists N (depending on ?) such that ||xm – xn|| < whenever m >N, n > N. Therefore |xk(m) – xk(n)| < holds for every k ? 1, provided m>N, n>N, because

||xm – xn|| = sup |xk(m)– xk(n)|,

so for every fixed but arbitrary value of k, the sequence (xk(1),xk(2),xk(3),……) is a Cauchy sequence in ; so it converges to a limit in (because is complete), which we denote by xk. So we get an infinite sequence x = (x1, x2, x3,…….). Since |xk(m) – xk(n)| < holds for all m > N, n > N and k ? 1, keeping n and k fixed and taking limits as m ? ? we get

|xk – xk(n)| ? for all n > N and k ? 1.

Therefore ||x – xn|| = sup |xk – xk(n)| ? < ? for all n > N. So <xn> converges to x in the supreme norm. It remains to show that x belongs to l?. Choose n >N and fix it. Then there exists Mn > 0 such that

|xk(n)| ? Mn for all k ? 1, because xn l?.

So |xk| ? ||xk – xk(n)| + |xk(n)| ? ? + Mn < 1 + Mn .

Example

For every fixed p ? 1, is a Banach space.

Proof:

Suppose <xn> is a Cauchy sequence in . Here each xn is an sequence of elements of , say xn = (x1(n),x2(n),x3(n),………),of elements of such that the series is convergent. Given ? > 0 there exists N (depending on ?) such that ||xm – xn|| < whenever m > N, n > N. Therefore |xk(m) – xk(n)| < for all k ? 1, whenever m > N, n > N, because

|xk(m) – xk(n)| ? – | ? ( – = ||xm – xn||.

So for every fixed but arbitrary value of k ? 1, the sequence (xk(1),xk(2),xk(3),……) is a Cauchy sequence in so it converges to a limit in (because is complete); we denote it by xk. We thus get in infinite sequence x = (x1, x2, x3,………).

For every r we have

– ? ||xm – xn|| < whenever m > N, n > N.

Keeping r, n fixed and taking limits as m ? ? we get

( – ? .

This is true for all r; so letting r ? ? we get

( – ? whenever n > N.

Therefore ||x – xn|| ? < ? for all n > N. So <xn> converges to x in the – norm. It remains to prove that x There exists M > 0 such that ||xn|| ? M holds for all n, because every r , we have ( ||x|| ? M.

Letting n ? ? we get ( ? M.

This is true for every r . Therefore ( ? M.

Hence x .

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},………….(1)

then N/M is a normed linear space. Further, if N is a Banach space then so is N/M.

Proof :

We first verify that (1) defines a norm in the required sense. It is obvious that ||x + M|| ? 0; and since M is closed, it is easy to see that ||x + M|| = 0 there exists a sequence {mk} in M such that

||x + mk|| ? 0 x is in M x + M = M = the zero element of N/M.

Next, we have ||(x + M) + (y + M)||

= ||(x + y) + M||

= inf {||x + y + m||: m M}

= inf {||x + y + m + m?||: m and m? M}

= inf {||(x + m) + (y + m?)||: m and m? M}

? inf {||x + m|| + ||y + m?||: m and m? M}

= inf{||x + m||: m M} + inf{||y + m?||: m? M}

=||x + M|| + ||y + M||.

Now, ||?(x + M)||

= inf {||?(x + m)||: m M}

= inf {|?| ||x + m)||: m M}

= |?| inf {||x + m||: m M}

= |?| ||x + M||.

Finally, we assume that N is complete, and we show that N/M is also complete.

Let {xn + M} be a Cauchy sequence in N/M. In order to show that {xn + M} is convergent, it is sufficient to show that it has a convergent subsequence.

Now, we suppose that {xn} is a Cauchy sequence in (X, ||.||) which is also convergent in it. Then every subsequence of it will be convergent in X.

Conversely, we assume that {xnk} is a subsequence of {xn} which converges to

S X, i.e. ||xnk S|| ? 0 as nk ? ?.

Then we have,

||xn – S|| = ||xn – xnk + xnk – S|| ? ||xn – xnk|| + ||xnk – S||?0 as n??.

Therefore, {xn} is convergent.

Then we choose a subsequence in the following manner:

Let y1 x1 + M and we choose a y2 x2 + M such that ||y – y2|| < .

Next we choose a vector y3 x3 + M such that ||y2 – y3|| < .

Continuing this process, we obtain a subsequence {yn} in N such that

||yn – yn+1|| < .

If m < n, then

||ym – yn|| = ||(ym – ym+1) + (ym+1 – ym+2) +………+ (yn-1 – yn)||

? ||ym – ym+1|| + ||ym+1 – ym+2|| + …….+ ||yn-1 – yn||

<

< ( + +…….+ )

= ? 0 as m ? ?,

So {yn} is a Cauchy sequence in N. Since N is complete, i.e. N is a Banach space, there exists a vector y in N such that yn ? y.

It now follows from || (xn+ M) – (y+ M) || ? || yn – y|| that {xn+ M} ? y + M in N/M and so N/M is complete, i.e. N/M is a Banach space.

Problem Let N be a non-zero normed linear space and prove that N is a Banach space {x: ||x|| = 1} is complete.

Proof:

Let N? = {x: ||x|| = 1}

First we will prove that N is a Banach space N? is complete.

Let <xn> be any Cauchy sequence in N?.

Since N? N, therefore <xn> is also a Cauchy sequence in N and converges to a vector x, since N is complete.

That means, xn ? x ||xn|| ? ||x||……………(1)

Since <xn> in N?, n N ||x|| = 1

Therefore ||xn|| ? 1…………(2)

By (1) and (2) we get ||x|| = 1

which implies x N?.

<xn> be any Cauchy sequence in N? and converges in N?.

Hence N? is complete.

Now we will prove that N? is complete N is complete.

Let <xn>be any Cauchy sequence in N.

Therefore for every ?>0 there exists a natural number K such that m, n > K ,

||xm – xn|| < ? .

Now |||xm ||xn||| ? ||xm xn|| < .

Therefore <||xn||> is also a Cauchy sequence in .

Since is complete, <||xn||> converges to a real number say r.

Again <xn/||xn||> is also a Cauchy sequence in N?.

Since N? is complete, <xn / ||xn||> converges to a vector say x0 in N?

? x0 =

Since ||xn|| ? r, therefore xn ? rx0 N.

Hence N is complete, i.e. N is a Banach space.

Problem

(a)Let p >1 and q be defined by the relationship

+ = 1. For a ? 0, b ? 0 show that

? + .

Proof:

The inequality is clearly true if a = b or b = 0.So we can assume that a > 0 and b > 0.For a fixed ? with 0 < ? < 1 we consider the function

f(t) = t ? – ?t + ? – 1 for t ? 0……..

Then f?(t) = ?(t?-1–1) is positive for 0 < t < 1 because ? – 1< 0,and negative for

t >1. Also f(1) = 0. By Langrange’s mean value theorem, we have any t >1,

f(t) = f(t) – f(1) = t – 1f?(?), for some ? between 1 and t. For 0 < t < 1, we have,

–f(t) = f(1) – f(t) = (1 – t)f?(?), for some ? between t and 1.At t = 1 we have

f(t) = 0; therefore we conclude that f(t) ? 0 for all t > 0

Setting ? = and t = in we get

f ( ) = )1/p . + – 1 ? 0 .

Multiplying both sides with b, we get,

– . a + b ( – 1) ? 0

? + because = 1–

b)Prove Hölder’s inequality:

for any n-tuples (x1,x2,……..,xn) and (y1,y2,……..,yn) of real or complex numbers, we have

? |p (

where p>1 and q is defined by the relation

= 1.

Proof:

Let ? = ( and ? = Hölder’s inequality is clearly true if ? = 0 and ? = 0, for ? = 0 implies x1 = x2 =……..= xn = 0.

Let ai = |xi|p/?p, bi = |yi|q/?q for i = 1,2,………..,n.

By the relation (a)

. for every i = 1,2,………,n, we get

because = 1 and

= 1.

Hence ? ? ? which is a Hölder’s inequality.

c)Prove Minkowski’s inequality:

For any n-tuples (x1,x2,……,xn) and (y1,y2,…….,yn) of real or complex numbers, we have

( , where p > 1.

Proof:

When all xi and all yi are zero, the two sides are equal. So we assume that the xi, yi are not all zero.

Then |xi| + |yi|)p > 0.

Raising both sides of |xi + yi| ? |xi| + |yi| to the p-th power and then summing up from i = 1and i = n, we get

xi + yi|p ? |xi| + |yi|}p .

Extracting p-th root of the both sides we get

( |xi| + |yi|p ? { (|xi| + |yi|)p}1/p……………….(1)

Writing (|xi + yi|)p = |xi| (|xi| + |yi|)p-1 + |yi| (|xi| + |yi|)p-1

and summing up from i = 1 to i = n, we have

(|xi| + |yi|)p = {|xi|(|xi| + |yi|)p-1}+ {|yi|(|xi| + |yi|)p-1} …………………….(2)

Applying Hölder’s inequality to the first sum on the right side of (2), we get

{|xi|(|xi| + |yi|)p-1} ? { |xi|p}1/p[{(|xi| + |yi|)p-1}q]1/q

= { |xi|p}1/p[(|xi| + |yi|)p]1/q…………….(3),

because (p-1)q = p.

There is a similar result for the second sum on the right side of (2), viz

{|yi|(|xi| + |yi|)p-1} ? { |yi|p}1/p[ (|xi| + |yi|)p]1/q…..…(4)

Adding (3) & (4), then using (2) , we get

(|xi| + |yi|)p ? [{ |xi|p}1/p+{ |yi|p}1/p][ (|xi| + |yi|)p]1/q

Dividing both sides by [ (|xi| + |yi|)p]1/q, which is > 0 , we get

[ (|xi| + |yi|)p]1/p ? ( |xi|p}1/p+{ |yi|p}1/p, because 1 –

Now the inequality (1) implies that

|xi + yi|p)1/p ? ( |xi|p}1/p + { |yi|p}1/p

which is Minkowski’s inequality.

Definition

Let N and N? be normed linear spaces with the same scalar and let T be a linear transformation of N into N?. When T is continuous as a mapping of the metric space N into the metric space N?, then T is continuous if and only if

xn ? x in N T(xn)?T(x) in N?.

Theorem

Let N and N? be normed linear spaces and T a linear transformation of N into N?. Then the following conditions on T are all equivalent to one another:

1) T is continuous;

2) T is continuous at the origin, in the sense that xn?0 T(xn)?0;

3) There exists a real number K ? 0 with the property that ||T(x)|| ? K||x|| for every x N;

4) If S = {x = ||x|| ? 1} is the closed unit sphere in N, then the image T(S)is a bounded set in N?.

Proof :

(1) (2). If T is continuous, then since T(0) = 0 it is certainly continuous at the origin. On the other hand, if T is continuous at the origin, then

xn ? x xn – x?0

T(xn – x) ? 0 T(xn) – T(x) ? 0

T(xn) ? T(x),

so T is continuous.

(2) (3).It is obvious that (3) (2), for if such a K exists then xn?0 clearly implies that T(xn) ? 0.To show that (2) (3), we assume that there is no such K. If follows from xn such that ||T(xn)|| > n||xn||, or equivalently, such that

||T(xn/n||xn||)|| > 1.

If we now put

yn = xn / n||xn||,

then it is easy to see that yn ? 0 but T(yn)?0,so T is not continuous at the origin.

(3) (4). Since a non-empty subset of a normed linear space is bounded it is contained in a closed sphere centered on the origin, it is evident that (3) (4); for if ||x|| ? 1, then ||T(x)|| ? K. To show that (4) (3), we assume that T(S) is contained in a closed sphere of radius K centered on the origin. If x = 0, then

T(x) = 0 and clearly ||T(x)|| ? K||x||; and if x ? 0, then x||x|| S, and therefore ||T(x/||x||)|| ? K, so origin we have ||T(x)|| ? K ||x||.

If the linear transformation T in this theorem satisfies condition (3), so that there exists a real number K ? 0 with the property that

||T(x)|| ? K||x||

for every x, then K is called a bound for T, and such a T is often referred to as a bounded linear transformation. According to our theorem, T is bounded it is continuous, so these two adjectives can be used interchangeably .We now assume that T is continuous, so that it satisfies condition (4), and we define its norm by

||T|| = sup {||T(x)||:||x|| ? 1}…………………(1)

When N ? {0}, this formula can clearly be written in the equivalent form

||T|| = sup {||T(x)||:||x|| = 1}…………………(2)

Theorem

If N and N? are normed linear spaces, then the set B(N,N?) of all continuous linear transformations of N into N? is itself a normed linear space with respect to the pointwise linear operations and the norm defined by

||T|| = = sup {||T(x)||:||x|| ? 1}.

Further, if N? is a Banach space, then B(N,N?) is also a Banach space.

Proof :

First, we prove that B(N,N?)is a normed linear space.

Given T B(N,N?) we had defined ||T|| = and observed that

||T(x)|| ? ||T|| holds for all x N with ||x|| = 1.

1. Clearly ||T|| ? 0 and ||T|| = 0 if T = 0 (the zero operator).

Suppose ||T|| = 0.Then ||T(x)|| = 0 for all x; hence T(x) = 0 for all x.

2. ||?T|| = |?| ||T||.

This is true if ? = 0. Let ? ? 0; for any x N with ||x|| = 1, we have

||(?T)(x)|| = ||?T(x)|| = |?| ||T(x)|| ? |?| ||T||.

Taking supremum over all x with ||x|| = 1 we get ||?T|| ? |?| ||T||.On the other hand, for all x with

||x|| = 1, we have ||T(x)|| = = ||(?T)(x)|| ? ||?T||

Therefore ||T|| ? ||?T||, whence |?| ||T|| ? ||?T||

Altogether we have ||?T|| = |?| ||T||.

3. ||T + S|| ? ||T|| + ||S|| for all T, S B(N,N?).

For any x with ||x|| = 1 we have

||(T + S)(x)|| = ||T(x) + S(x)|| ? ||T(x)|| + ||S(x)|| ? ||T|| + ||S||.

Taking supremum over all x with ||x|| = 1 we get

||T + S|| ? ||T|| + ||S||.

Now, we prove that B(N,N?) is complete when N? is. Let {Tn} be a Cauchy sequence in B(N,N?).If x is an arbitrary vector in N, then

||Tm(x) – Tn(x)|| = ||(Tm – Tn)(x)|| ? ||Tm – Tn|| ||x||

shows that {Tn(x)} is a Cauchy sequence in N?; and since N? is complete, there exists a vector in N? we denote it by T(x) such that Tn(x)?T(x).This defines a mapping T of N into N?, and by the joint continuity of addition and scalar multiplication, T is easily seen to be a linear transformation. To conclude the proof, we have only to show that T is continuous and that Tn ?T with respect to the norm on B(N,N?).

By the inequality: |||x|| – ||y||| ? ||x – y||, the norms of the terms of a Cauchy sequence in a normed linear space form a bounded set of numbers, so

||T(x)|| = ||lim Tn(x)|| = lim||Tn(x)|| ? sup(||Tn||||x||) = (sup ||Tn||)||x||

shows that T has a bound and is therefore continuous.

It remains to be proved that ||Tn – T||?0.

Let ? > 0 be given, and let n0 be a positive integer such that

m, n ? n0 ||Tm – Tn||< ?, If ||x|| ? 1 and m, n ? n0, then

||Tm(x) – Tn(x)|| = ||Tm – Tn(x)|| ? ||Tm – Tn|| ||x|| ? ||Tm – Tn|| < ?.

We now hold m fixed and allow n to approach as ?, and we see that

||Tm(x) – Tn(x)||?||Tm(x) – T(x)||,from which we conclude that ||Tm¬(x) – T(x)|| ? ? for all m ? n0 and all x such that ||x|| ? 1.This shows that ||Tm – T|| ? ? for all m ? m0, and the proof is complete.

Hahn-Banach Theorem:

Theorem

(Hahn-Banach Theorem for Vector Spaces over )

Suppose X is a vector space over , M be a proper linear subspace of X; P: X? be a function satisfying the conditions:

P(x + y) ? P(x) + P(y), P(?x) = ?(Px) for all ? ? 0 and x, y X.

Suppose f: M? is a given linear functional on M. Then there exists a linear functional f: X? such that g(x) = f(x) holds for all x M and

–p(–x) ? g(x) ? p(x) holds for all x X.

Now here we make two observations,

(i)P(0) = 0; (ii)if P(–x) = P(x) then P(x) ? 0 for all x X.

In this situation the chain of inequalities

–P(–x) ? g(x) ? P(x) is equivalent to the inequality

|g(x)| ? P(x)

Taking ? = 0 in P(?x) = ?P(x) we get (i). So 0 = P(x+(–x) ) ? P(x)+P(–x) = 2P(x); this proves (ii).

Proof :

Given x0 M we show how to extend f to the linear subspace

M0 = M + L(x0), so that the extended linear functional has the desired properties.

Here M0 = M + L(x0) = {u + ?x0: u M, ? } is the subspace generated by the set M {x0} which is a strict superset of M; so M0 is a strictly larger linear subspace of X than M, we have for all x, y M.

f(u) + f(v) = f(u + v) ? P(u – x0) + (v + x0) ? P(u – x0) + P(v + x0);

hence f(u) – P(u – x0) ? P(v + x0) – f(v)……………(i)

Let ? be the supremum of the left hand side of (i) as u ranges over M, keeping

v M fixed. Then

? ? f(u) – P(u – x0) for every u M.

So f(u) – ? ? P(u – x0) for all u M…………..(ii)

From (i) we get ? ? P(v + x0) – f(v). So

f(v) + ? ? P(v + x0) for all v M……………(iii)

Define f0:M0? by f0(u + ?x0) = f(u) + ?? for u M.

Claim 1:

f0 is a linear functional on M0. For u, u? M0 and ?, ?? we have

f0((u + ?x0)+(u? + ??x0))=f0((u + u?)+(? + ??)x0)

=f(u + u?)+(? + ??)?, because u + u? M

=f(u) + f(u?) + ?? + ??? , because f is linear

=f0(u + ?x0) + f0(u? + ??x0); and for any ?

We have,

f0 (?(u + ?x0)) = f0((?u + (??)x0)

=f(?u) + (??)?, because ?u M

=?f(u) + ?(??)

=?(f(u) + ??)

=?f0(u + ?x0).

f0(x) = f(x) holds for every x M; so f0 is an extension of f to M0.

Claim 2:

–P(–x) ? f0(x) ? P(x) for all x M0.

We take any ? >0 and replace u by ?-1u in (ii) and v by ?-1v in (iii); using the homogeneity of f and of P we get

?-1f(u ) – ? ? ?-1P(u – ?x0) and ?-1f(v)? ? ?-1P(v + ?x0).

Hence f(u) – ?? ? P(u – ?x0) and f(v) + ?? ? P(v + ?x0) hold for any u, v M.

So f0(x) ? P(x) holds for all x M0, because every element of M0 has the form

x = u ± ?x0, with ? > 0. Replacing x by –x we get

f0(–x) ? P(–x) so that

–P(–x) ? –f0(–x)

But –f0(–x) = f0(x).

Hence –P(–x) ? f0(x) ? P(x) for all x M0.

To prove the existence of an extension of f with domain X having the desired properties, we apply Zorn’s lemma(Let P be a non-empty partially ordered set with the property that every totally ordered subset of P has an upper bound in P. Then P contains at least one maximal element.) to the family of all pairs (N, g)where N is a extension of f to a linear functional on N having the required properties. Partial order on this family is defined as follows:

(N1, g1) ? (N2, g2) N1 N2 and g2(x) = g1(x) holds for all x N1.

Let {(Ni, gi)}i=I be a totally ordered subfamily; then is a subspace of X containing M, and the correspondence g : defined by g(x) = gi(x) if x Ni, is M fact a linear functional on having the required properties.

Zorn’s lemma guarantees the existence of a maximal element (N, g) in the family. N must be the whole of X, for otherwise there would exist a further extension of g to a strictly larger subspace of X, which also has the required properties. Therefore N = X and the linear functional g: X? is an extension of f having the required properties.

Remark

A function P: X? satisfying the conditions:

P(x + y) ? P(x) + P(y), P(?x) = ?P(x) for all ? ? 0 and x, y X is called a sublinear functional. In most applications P satisfies the stronger condition P(?x) = |?|P(x) for all ? and x X. Then P is called a seminorm.

Theorem (Hahn-Banach Theorem for Vector Space over )

Suppose X is a vector space over M be a proper vector subspace of X and p : X be a seminorm. Suppose f : M is a given linear functional such that |f(x)| p(x) holds for all x M . Then there exists a linear fumctional ? : X? such that ?(x) = f(x) holds for all x M and |?(x)| ? P(x) holds for all x X.

Proof :

First we observe that P(0) = 0 and P(x) ? 0 for all x X. Split f(x) up into and imaginary parts, f(x) = g(x) + ih(x).We regard M and X as a vector spaces over and,for the moment, indicate this by writing and . Then g and h are real valued linear functional on .

Further g(ix) + ih(ix) = f(ix) = if(x) = – h(x) + ig(x) holds for every x M. Therefore equating real parts, we get g(ix) = h(x) for all x in M.

Also g(x) ? |f(x)| ? P(x) holds for all x M. By the Hahn-Banach theorem or vector spaces over , g has an extension to a linear functional ?:X? such that ?(x) ? P(x) for all x X. Now put ?(x) = ?(x) – i?(ix) for all x X. For any x M we have

?(x) = ?(x) – i?(ix) = g(x) – ig(ix) = g(x) + ih(x) = f(x)

So ? is an extension of f to X. It remains to prove that

(i)? is a linear functional on the complex vector space X

(ii)|?(x)| ? P(x) holds for all x X.

The proof of (i) uses the property g(ix) = – h(x) for all x in M, and the linearity of ? on the real vector space . For any a, b in we have

?((a + ib)x) = ?(ax + ibx) – i?(iax – bx)

= a?(x) + b?(ix) – ia?(ix) + ib?(x)

= (a + ib)(?(x) – i?(ix))

= (a + ib)?(x)

Clearly, ?(x + y) = ?(x) + ?(y).

The proof of (i) is complete.

(ii)|?(x)| ? P(x) is true when ?(x) = 0.When ?(x) ? 0, let ? be the argument of the number ?(x) , so that ?(x) = |?(x)| ei?. Then

|?(x)| = ?(x) e-i? = ? (e-i?x); therefore ?(e-i?x) is equal to its real part, which is here positive and is equal to ?(e-i?x) but

?(e-i?x) ? P(e-i?x) = P(x),because |e-i?| = 1.

So (ii) is proved.

Theorem

(Hahn-Banach Theorem for Normed Linear Spaces)

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 functional f0 defined on the whole space N such that

||f0|| = ||f||.

Proof:

Let us define P(x) = ||f|| ||x||.This function P is a seminorm and the hypotheses of theorem 3.4/3.5 are satisfied. Hence there exists an extension of f to a linear functional on N such that

|f0(x)| ? ||f|| ||x|| for all x N. This implies ||f0|| ? ||f||.

Since f0 is an extension of f, the set { : x ? 0 in N}

Contains the set { : x ? 0 in M}

Therefore ||f0|| ? ||f|| .

Hence ||f0|| = ||f|| .

Example Let M be a closed linear subspace of a normed linear space N, and let x0 be a vector not in M. If d is the distance from x0 to M, show that there exists a linear, functional f on N such that f(M) = 0,f(x0) = d and ||f|| = 1.

Proof:

Let d = inf {||x0 u||: u M}.We observe that d>0 because M is closed and x0 M. Let us consider the subspace

M0 = M + L(x0) = {u + ?x0: u M, ? } .

The representation of an element x M0 in the form x = u + ?x0 is unique, because

x = u + ?x0, x = u? + ??x0 imply (?? – ?)x0 = u – u? M ?? –?=0(for,otherwise,x0 would belong to M) ?? = ? u?= u. We define f0:M0? by setting f0(x) = ?d if x = u + ?x0.

Because of the uniqueness the representation of x M0 as u + ?x0, f0 is a well-defined mapping. f0 is clearly linear and has properties:

f0(M) = 0 because ? = 0 x M. Also f0(x0) = d

Claim ||f0|| = 1.

We have, taking x = 0 in the definition of d, the inequality

f0(x0) = d ? ||x0|| ? 1.

Hence ||f0|| ? 1.

To prove the reverse inequality, given any 0 < ? < 1 we let d ?= . Then d? > d; so by definition of d there exists u M such that d ? ||x0 – u|| < d?.

For x = x0 – x = – u + x0 belonging to M0, we have f0(x) = d;

Therefore ? = 1 – ?. so ||f0|| ? 1 – ?.

Since ? > 0 may be as small as we please, it follows

that ||f0|| ? 1

Hence ||f0|| = 1.

By the Hahn-Banach Theorem f0 has an extension to a linear functional f on X such that f(y) = f0(y) for every x M0 and ||f|| = ||f0||.

Hence f(M) = f0(M) = 0, f(x0) = f0(x0) = d and ||f|| = 1.

Definition

A mapping T: X?Y is called open if the image of every open set in the domain is an open set in the codomain.

The Open Mapping Theorem:

Theorem

(The Open Mapping Theorem)

If B and B? are Banach spaces and if T is a continuous linear transformation of B onto B?, then T is an open mapping.

Proof:

In order to prove the main theorem, first we have to prove the following lemma which makes our main theorem easy to prove.

Lemma:

If B and B? are Banach spaces and if T is a continuous linear transformation of B onto B?, then the images of each open sphere centered on the origin in B contains an open sphere centered on the origin in B?.

Proof of the lemma:

We denote by Sr and Sr? the open spheres with radius r centered on the origin in B and B?. If is obvious that

T(Sr) = T(rS1) = rT(S1),

so it suffices to show that T(S1) continuous some Sr?. We begin by proving that contains some Sr?. Since T is onto, we see that B? = .B? is complete. So Baire’s theorem : If a complete metric space is the union of a sequence of its subsets, then the closure of at least one set in the sequence must have non-empty interior, implies that some has an interior point y0,which may be assumed to lie in T( ).The mapping y ? y – y0 is a homomorphism of B? onto itself, so – y0 has the origin as an interior point. Since y0 is in T( ),we have T( ) – y0 T( ); and from this we obtain

– y0 = , which shows that the origin is an interior point of . Multiplication by any non-zero scalar is a homomorphism of B? onto itself, so = 2n0 and it follows from this that the origin is also an interior point of , so Se? for some positive number ?.

We conclude the proof by showing that Se? , which is clearly equivalent to S?/3? Let y be a vector in B? such that ||y||<?. Since y is in , there exists a vector x1 in B such that ||x1|| < 1 and ||y – y1|| < , where y1 = T(x1).

We now observe that S??/2 , so there exists a vector x2 in B such that ||x2|| < ½ and ||(y – y1) – y2|| < , where y2 = T(x2).

Continuing in this way, we obtain a sequence {xn} in B such that ||xn|| < ½n 1 and ||y (y1 + y2 + ……….+ yn)|| ,where yn = T(xn).

If we put Sn = x1 + x2 + ………..+ xn, then it follows from ||xn||<½n – 1 that {Sn}is a Cauchy sequence in B for which

||Sn|| ? ||x1|| + ||x2|| +………..+ ||xn|| < 1 + ½ +……..+ ½n–1 < 2

B is complete, so there exists a vector x in B such that Sn? x; and ||x||=||lim Sn||= lim ||Sn|| ? 2 < 3 shows that x is in S2.All that remains is to notice that the continuity of T yields

T(x) =T (lim Sn) = lim (Sn) = lim(y1 + y2 +……….+ yn)

from which we see that y is in T(S?3).

Proof of the main theorem:

We must show that if G is an open set in B, then T(G) is also an open set in B?. If y is a point in T(G),it suffices to produce an open sphere centered in y and contained in T(G).Let x be a point in G such that T(x) = y.Since G is open, x is the center of an open sphere which can be written in the form x + Sr contained in G. Our lemma now implies that T(Sr) contains some Sr?

It is clear that y + Sr? is an open sphere centered on y, and the fact that it is contained in T(G) follows at once from

y + Sr? y + T(Sr) = T(x) + T(Sr?) = T(x + Sr) T(G).

The proof is complete.