MTH641 Midterm Notes 2 Modules 60-113

MTH 641 · FUNCTIONAL ANALYSIS · MODULE # 60 To 113 · (FINAL TERM SYLLABUS)

Don’t look for someone who can solve your problems,

Instead go and stand in front of the mirror,

Look straight into your eyes,

And you will see the best person who can solve your problems!

Always trust yourself.

(BY ABU SULTAN) · Module No. 60 · Linear Operators ·  Bounded Linear Operator

Norms spaces are generalization of distances. By using Norm spaces we are going to discuss

Bounded Linear Operator.

Bounded Linear Operator (Definition):

Let X and Y be normed spaces and T D T Y a linear operator, where D T X

The operator T is said to be bounded if there is a real number c such that for all x D T

Tx c x

If this condition satisfied then we call T to be a bounded linear operator. Bounded

function mean range is bounded but here bounded set is mapping over a bounded set so we

call this a bounded linear operator. c is fix.

( )-{0} Tx c x D T x

The smallest possible value of c is supremum of left hand side. Then the value of c is called

sup

x D T x Tx c x as

We call the value as T norm

c T

If {0}, D T T

sup

x D T x Tx c T x

Tx T x

This is the formula that we use for bounded linear operator.

Module No. 61 · Bounded Linear Operators ·  Lemma (Norm)

First we define the norm (equivalent definition) and then prove that the norm defined on T

satisfies all four properties of Norm i.e. (N1) to (N4).

Lemma (Statement):

Let T be a bounded linear operator as defined before then an alternate formula for the norm of

T is

sup

x D T x T Tx

The norm defined on T satisfies (N1) to (N4).

Proof: Part (a)

Tx c x sup sup

x D T x D T x x Tx c T Tx x

We have to prove

sup sup

x D T x D T x x Tx Tx x

Let ; set y= , x 0, x x a a

x y a

sup

x D T x Tx T a

as T is linear so, we take constant a inside the norm

sup sup

x D T y D T x y T T x Ty a as 1 y a 

Here variable is y which can be any other. Part (a) of lemma is proved.

Part (b):

sup sup

x D T x D T x x Tx T Tx x

N1: T  is obvious.

N2: T=0, T 

0, T Tx x D T T

N3: sup sup sup

x D T x x x T Tx Tx Tx T as sup

x Tx T

N4: T T T T

sup (

sup sup

sup sup x D T x

x x

x x T T T T x

T x T x T x T x

T x T x T T

First we define a T T norm and then prove the four properties of norm.