Tutorial 'll①

From Lecture 18 we knowhow to construct LYX.IR ) for an integral pair( X

,b) as a set of equivalence classes of Cauchy sequences ( fn Into

where each fn :X→ IR is continuous and the metic used to define

Cauchy - ness is the one derived from the LZ - norm

Hfllz -_ { 1×142 )"

we know LYX.IR ) is a Banach space and (CtsHR )

,H -Hz ) → (LYX, 1121,11 - Ha )

as normed spaces , but [ space remainsmysterious : what are these vectors in

[(x, IR ) lctslx.IR) ? Are they just some formal objects or do they have

some deeper significance ? To examine this question we return to the topic of Tutorial 10 .

Example48-2_ Consider X=[Oil] and the sequence of functions fn :X → IR

given for n> 4 by

x

\ - - - - . . . - . - - - - - - - - - - - -

y

ii÷:i; tnex-fmca-i.mn ,

'

Ii 't:"i 1 Yztyncx EliLi.

'I- Yu 'k 'K'Hn l

(non - continuous but integrable

1 x >'12

The sequence ( fu )F4 convenes pointwise to f- (x ) - { yz x=yz

O XL'Iz

fat ctscx.IR)

we proved (fn )F=4 is Cauchy in ( Cb(X.IR ) .dz ) but does not converge . Hence

T-gectslx.IR )I ( (fnliiyJELCX.IR ) ) (XR)

( fnyii.ir (g) 5=4

dffn , g) → O

fu - , g .

②

We will proveLYX , IR ) is a Hilbert space 4

> '

- L4x.IR )xL4x, IR)-s IR

is continuous in each variable and for g, HE Cts (X, IR )

< g , h>= 1×9 h

.

Hence for g Ects (X, IR)

,Lyx , IR ) lots

(X, IR )

Lg , I > = Lg , limn→ of n> i --

f= nlim, Lg , fn> ↳-

- nlihf, fxgfn

= Linn.( f igfnklx - '

ht 'M tf+ "is ]

= fig = fxgf

LYX, R )a f×gf

✓f

Cts ( x, IR) •i

r.

i

.

.

" L 9, fn> = f×gfn

i i

•

o

° ^

,

'

\ tis .

'

< 9,f ,> = ↳ g f ,

claim we should think of I as"

being" f .

Moreover any Riemann

integrable function can be represented in this way by a vector

in L2 ( X,IR ) .

③

There are several things to check before you should believe this

00

(1) Does I depend on the choice of approximation (fn ) n -- 4 of f- bycontinuous functions ?

(2) How to find Cfn )F=4 fora general Riemann integrable f ?

(3) Is the function

{ Riemann integrable f }→ LYX , IR)

injective ? Surjective ?

Thekey is :

since this u is unique , we denote

u by Oz and call this the representingelement for R .

theorem (Riesz representation theorem ) Let (H ,47 ) be a Hilbert space .

If 7 : H→ IF is continuous and linear there exists a unique

fedor *H with.

2 =L- , u>.

Strategy . Let (X,Sx ) = ( co, D , fo

'

).

. Let f :X→ 112 be Riemann integrable

• Question : I E LYX,IR ) " representing

" f.

. H = ( L2( X, IR ) , C , > ),7 : H→ IF gives Oz C- ECHR ) .

Lemme Given f integrable consider the function

2 : Cts ( X, IR )→ IR

Z ( g ) = fo'

gfdx .

is continuous and linear with respect to dz , hence

ME Cts (X,R )"

.

Proof Linearity is clear . For continuity observe that

I fgfd .- f g ' foul = I f ( g - g ' ) fat(

z (g) 4cg ' ) E f l g - g 'll f ldx H G -g 't 't He

Holder"E H g - g

' Hall f- theinequality

Hence 2 is bounded 11211 s y fth. D

[ { J! If I2)"Z

Let c : Cfs ( X , IR )→ LYX , IR) be the inclusion.

Since 2 : Cts (x, R) → IR

is continuous and linear there is a unique continuous linear Z'

making

Z'

↳ ( x , IR) - - - - → IR

t .Ctslx , as ,

↳

commute - Nole

,gnwnhnuous

'

Z' ( limn→ a 9N ) = him 219N) = him,

f! gnfdx

By the Riesz representation theorem there is a unique representingelement I

'

for Z'

.

That is,I C-E (X , IR ) and C-, I > = Z

'.

Equivalently

§,

idCg, I > = f! gfdx . V-gectslx.IR )

Lemmon HE Ha - H f Ha - L f'

HH.

"'

II I

limllfnllz { Riemann integrable }→ L2 (X , IR)n→N

u n't

f d

• It is not injective , Kev (I) = { almost everywhere zero functions } .• I is not surjective either !

€

{ Lebesgue integrable } -77 L2 (X, IR )

1a. e. equality