MTH641 Final Term Handouts Searchable

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Final Handsout · MTH 641 · FUNCTIONAL ANALYSIS · MODULE #60TO 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) · Zain Ahsan · MODULE No. 60 · LINEAR OPERATORS · >» Bounded Linear Operator

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

to iscuss

Bounded Linear Oprator.

g

Bounded Linear Operator (Definition):

Let X and Y be normed spaces and 7: D(T) > ¥ a linear operator, where D(T)c X.

The operator T 1s said to be bounded if there 1s a real number c such that for all x € D(T). yo

[Fx < eel

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 ts fix.

xe

D(T)-{0}

IZ. <c¢c

I

The smallest possible value of c is supremum of left hand side. Then the value of c 1s called

c= sup — as

xelhiT }, |x| x#0

iz

We call the value as T norm

c=|T|

If D(T)={0},

|T|=0

Tx||

[|= sup

xe Pit }, |x| x#O [rx] <r hl

This 1s 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 1.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

7] = sup |/7x|

xvyED(T)

IF!

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

Proof: Part (a) · Fx] · sel] · MTH 641 Funtional Analysis - by ABU SULTAN

“int.

e=|T|= sup

sup |?

xeD(T) xreD{T) xzx0 [q=!

We have to prove

su

Z|

Ir

p sup |x xeD(T) |x| xeDi(T} x40 |x]

Let ||x|=a; set y=—

, x#0,

a

_ lel _ · p= Ht · ir|= sup Eh · pm.

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

l

7" = sup r(+ | = sup l7y||

as—=y

xeEDI(T), if veD(T), it

xa)

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

Part (b): · Iz · Iz

sup sup reD(T ren(T x0

NI:

7 >Q is obvious.

N2: |7|>0 > T=,

T||=0

Tx=0,

V xED(T) > T=0

N3: j 7||= sup | Tx||=sup|al|[7| =|e|sup|[7x]=|a\ |r| as sup|Tx||= ||

xeD(T), Ie! |x|

IF

I=!

N4: T, +T,|| <|Z,||+|Z|

T,+T7,||= sup

|(7, +7,)a

er

= sup||7,x+ T,x| = Sup (| Lx

| + Ir)

[fH

Ft

= sup |Z,x||+ sup|[7,2| =|[7,]+ [Z|

bE!

IF

First we define a 7 x7 norm and then prove the four properties of norm.