MTH641 Midterm Short Notes

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.

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.

 c x  D(T ) -{0}

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

We call the value as T norm c  sup xD(T ), x0 as

c  T

If D(T ) {0}, T  0

c  T

Tx  sup xD(T ), x0  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 T  sup Tx xD(T ) x 1 The norm defined on T satisfies (N1) to (N4).

Proof: Part (a)

Tx  c x Tx x

Tx x

Tx x We have to prove c  T  sup xD(T ), x0  sup Tx xD(T ) x 1

Let x  a; set y= x

a , x  0, sup xD(T ), x0

y  sup Tx xD(T ) x 1

 x  1

a T  sup Tx

xD(T ), a x0 as T is linear so, we take constant a inside the norm T  sup T  1 x    sup Ty as 1  y

xD(T ),  a  yD(T ), a

x0 y 1 Here variable is y which can be any other. Part (a) of lemma is proved.

Part (b):

N1: T  0 is obvious. T  sup xD(T ), x0  sup Tx xD(T ), x 1

N2: T  0

T  0  T=0,

 Tx  0,  x  D(T )  T  0

N3: T  sup xD(T ), x 1 Tx  sup  x 1 Tx   sup Tx   T x 1 as sup Tx  T x 1

N4: T1  T2  T1  T2

T1  T2  sup xD(T ) x 1 (T1  T2 )x

 sup T1x  T2 x x 1  sup  T1x x 1  T2 x 

 sup T1x x 1  sup T2 x x 1  T1  T2

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

MODULE NO. 62

Tx x

Tx x

Tx x