# (1)The ontinuous wavelet transform and its disretization 1

## Full text

(1)

The ontinuous wavelet

transform and its

disretization

1. The ontinuous transform

Supp ose that 2L 2

(R). We dene the familyof funtions a;b

by

a;b

(x)= 1

p

jaj

x b

a

; a6=0; b2R: (8.1)

Often one only onsiders the ase where a>0 and the normalization 1

p

jaj

isusedsothatk a;b

k

L 2

(R)

=k k

L 2

(R)

. Othernormalizations an,ofourse,

b eusedas well. Then wean dene

(W f)(a;b)= D

f;

a;b E

= Z

R f(x)

a;b

(x)dx:

(8.2)

It islear that jW f(a;b)jk k

L 2

(R) kfk

L 2

(R) .

Wehave thefollowing result:

Theorem8.1. Assume 2L 2

(R)nf0g issuh that

C def

= Z

R j

^

(!)j 2

j!j

d ! <1:

Then

Z

R Z

R

W f(a;b)W g(a;b) 1

a 2

for all f and g2L 2

(R).

65

(2)

This theorem says that in a weak sense, we have (provided, of ourse,

that 6=0)

f(x)= 1

C Z

R Z

R

W f(a;b) a;b

(x) 1

a 2

Pro of. Firstwenote that

d

a;b

(!)= p

jaje i2 b!

^

(a!);

sothat byPlanherel's theorem (2.4)we have

W f(a;b)= p

jaj Z

R

^

f(!)e i2 ! b

^

(a!)d! = p

jaj

\

^

f()

^

(a)( b):

Thuswe get,again using Planherel's theorem,

(8.3) Z

R Z

R

W f(a;b)W g(a;b) 1

a 2

=jaj Z

R Z

R

\

^

f()

^

(a)( b)

\

^ g()

^

(a)( b)db

!

1

a 2

da

= Z

R Z

R

^

f(!)^g(!)j

^

(a!)j 2

d !

1

jaj da

= Z

R

^

f(!)g (!)^ Z

R j

^

(a!j 2

1

jaj da

d!:

A simple hangeof variables nowshowsthat

Z

R j

^

(a!j 2

1

jaj

da=C ;

and thenthelaim followsfrom equation (8.3).

If one do es not want to use negative as well as p ositive values for the

dilation athenone getsalmostthesameresult,provided

Z

0

1 j

^

(!)j 2

j!j

d! = Z

1

0 j

^

(!)j 2

j!j d !

def

= : (8.4)

This isofoursethe asewhen isreal-valued.

Corollary 8.2. Assume 2L 2

(R)nf0gissuh that (8.4)holdswith <

1. Then

Z

R Z

1

0

W f(a;b)W g(a;b) 1

a 2

for all f and g2L 2

(R).

Thisresult saysthat in a weak sense we have

f = 1

Z

R Z

1

0

W f(a;b) a;b

(x) 1

a 2

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2. Frames of wavelets

The ontinuous wavelet transform is not neessarily a very pratial to ol

sine it is notlear towhatextent theintegralsan atually b eomputed.

If one uses disretizations, one question to ask is when the inner pro duts

hf;

m;n

i, where where

m;n

(x) = a m

2

(a

m

x nb

), really haraterize

the funtion f. If thesequene (

m;n )

m;n2Z

is an orthonormalbasis, there

arenoproblems,butif issomequitegeneralfuntion,thereisnoreasonto

exp et thattob ethease. Butitturns outtob ep ossible togiverelatively

simple onditions for this sequene to b e a frame, see Denition 3.6. We

have thefollowing result:

Theorem8.3. Assume a

and b

>0 and that 2L 2

(R) is suh that

0<essinf

! 2R X

m2Z j

^

(a m

!)j 2

esssup

! 2R X

m2Z j

^

(a m

!)j 2

<1;

(8.5)

and

X

k 2Z

k 6=0 s

k

b

k

b

<essinf

! 2R X

m2Z j

^

(a m

!)j 2

; (8.6)

where

(s) def

= esssup

! 2R X

m2Z j

^

(a m

!)jj

^

(a m

!+s )j:

Then the sequene (

m;n )

m;n2Z where

m;n

(x) = a m

2

(a

m

x nb

) is a

frame in L 2

(R) with frame bounds

A= 1

b

0

B

B

essinf

! 2R X

m2Z j

^

(a m

!)j 2

X

k 2Z

k 6=0 s

k

b

k

b

1

C

C

A

;

B = 1

b

0

B

B

esssup

! 2R X

m2Z j

^

(a m

!)j 2

+ X

k 2Z

k 6=0 s

k

b

k

b

1

C

C

A :

Pro of. A simple alulation shows that

[

m;n

(!)=a m

2

^

(a m

!)e i 2 a

m

bn!

:

Thuswegetforf 2L 2

(R),byrstusingPlanherel'stheorem forfuntions

in L 2

(R),thenwriting theintgeral overRasa sumof integrals,thenusing

(4)

Planherel'stheoremforp erio difuntions,thenexpandingthepro dutand

again writingthesum of integralsas one integral,

X

m;n2Z jhf;

m;n ij

2

= X

m;n2Z j D

^

f; [

m;n E

j 2

= X

m;n2Z

Z

R

^

f(!)a m

2

^

(a m

!)e i2 a

m

b

n!

d!

2

= X

m2Z a

m

X

n2Z

Z

1

a m

b

0 e

i2 a m

b

n!

X

j2Z

^

f !+ j

a m

b

^

a m

!+ j

b

d!

2

(byParseval'sequality applied tothe 1

a m

b

p erio di funtion

X

j2Z

^

f !+ j

a m

b

^

a m

!+ j

b

)

= 1

b

X

m2Z Z 1

a m

b

0

X

j2Z

^

f !+ j

a m

b

^

a m

!+ j

b

2

d !

= 1

b

X

m2Z Z 1

a m

b

0

X

j2Z

^

f !+ j

a m

b

^

a m

!+ j

b

X

k 2Z

^

f !+ j+k

a m

b

^

a m

!+ j+k

b

d!

= 1

b

X

m2Z X

k 2Z Z

R

^

f(!)

^

f !+ k

a m

b

^

(a m

!)

^

a m

!+ k

b

d!

= 1

b

Z

R j

^

f(!)j 2

X

m2Z j

^

(a m

!)j 2

!

d !

+ 1

b

X

m2Z X

k 2Z

k 6=0 Z

R

^

f(!)

^

f !+ k

a m

b

^

(a m

!)

^

a m

!+ k

b

d!

We have to get some estimates for the seond term, and we get by using

theCauhy-Shwarzinequality, ahangeofvariables, andthentheCauhy-

Shwarzinequality in the sumoverm:

1

b

X

m2Z X

k 2Z

k 6=0 Z

R

^

f(!)

^

f !+ k

a m

b

^

(a m

!)

^

a m

!+ k

b

d!

(5)

1

b

X

k 2Z

k 6=0 X

m2Z Z

R j

^

f(!)j 2

j

^

(a m

!)jj

^

(a m

!+ k

b

)jd!

1

2

Z

R j

^

f(!+ k

a m

b

)j 2

j

^

(a m

!)jj

^

(a m

!+ k

b )jd!

1

2

= 1

b

X

k 2Z

k 6=0 X

m2Z Z

R j

^

f(!)j 2

j

^

(a m

!)jj

^

(a m

!+ k

b

)jd!

1

2

Z

R j

^

f()j 2

j

^

(a m

k

b )jj

^

(a m

)jd

1

2

1

b

X

k 2Z

k 6=0 Z

R j

^

f(!)j 2

X

m2Z j

^

(a m

!)jj

^

(a m

!+ k

b

)j

!1

2

Z

R j

^

f()j 2

X

m2Z j

^

(a m

)jj

^

(a m

k

b

)j

! 1

2

1

b

kfk

2

L 2

(R) X

k 2Z

k 6=0 s

k

b

k

b

:

When we ombine the results we have dedued ab ove, we get the desired

onlusion.

It is lear that for (8.5)to hold we must have a

6=1, but it turns out

that one an quite easily get suÆient onditions for the assumptions of

Theorem8.3to hold.

Corollary 8.4. Assume that 2L 2

(R;R) isreal-valued,notthezerofun-

tion,

^

is ontinuous and satises

sup

! 2Rnf0g j

^

(!)j(1+j!j

)

j!j

<1;

for some onstants > +1 > 1. Then the assumptions of Theorem 8.3

hold, provided a

has beenhosensuÆientlyloseto1 andthenb

>0 has

beenhosento besuÆiently small.

Pro of. Sine isreal-valuedwehave

^

( !)=

^

(!)and sine

^

is ontin-

uousandnotidentiallyzerotherearep ositivenumb ers!

0

,Æand sothat

(6)

j

^

(!)jÆ when!

0

j!j!

0

+. Itfollows thatif 0<ja

1j<

!

0 then

essinf

! 2R X

m2Z j

^

(a m

!)j 2

Æ 2

>0;

b eause if for example 1 < a

< 1+

!

0

and a m

j!j < !

0

then a m+1

j!j

a

!

0

<!

0

+ and thereforethere isforevery !6=0 anindex m

suhthat

!

0 ja

m

!j!

0

+ and we havethe rstinequality in (8.5).

Nextwederive someuseful inequalities. Clearly

jxj

1+jxj

jx+sj

1+jx+sj

jxj

jx+sj

;

and if jsj>2jxjthen jx+sjjsj jxjjsj 1

2 jsj=j

s

2

jso thatweget

jxj

1+jxj

jx+sj

1+jx+sj

jxj

j s

2 j

; jsj2jxj:

(8.7)

In thesamewaywe getthefollowing rudeestimate

(8.8) jxj

1+jxj

jx+sj

1+jx+sj

1

jxj

1

jx+sj

1

jxj

j s

2 j

=

1

jxj 1

2

j s

2 j

+1

2

1

jxj +1

2

j s

2 j

1

2

1

jxj 1

2

j s

2 j

+1

2

;

jsj2jxj2:

b eause 1>0. On theother hand we haveforthe samereason

(8.9) jxj

1+jxj

jx+sj

1+jx+sj

1

jxj

=

1

jxj 1

2

jxj +1

2

1

jxj 1

2

j s

2 j

+1

2

; jsj2jxj:

Ifa

>0anda

6=1and we maywithoutlossof generalityassumethat

a

>1b eause inthesums involving a

we mayreplaem by mwhih is

thesameasreplaing a

by

1

a .

It followsfrom theassumptions that there isaonstant C suh that

j

^

(!)jC j!j

1+j!j

; !2R: (8.10)

Sine

^

(0) = 0 we have P

m2Z j

^

(a m

!)j 2

= 0 if ! = 0. Let ! 6= 0 and

let m

0

b e suh that a m

0

j!j 1 but a m

0 +1

j!j > 1, i.e., a m

1

j!j> 1 where

(7)

m

1

=m

0

+1. Then

X

m2Z

^

(a m

!)

2

C 2

m0

X

m= 1 ja

m

!j 2

1+ja m

!j 2

+C 2

1

X

m=m

1 ja

m

!j 2

1+ja m

!j 2

C 2

m

0

X

m= 1 ja

m

!j 2

+C 2

1

X

m=m1 ja

m

!j 2( )

=C 2

ja m

0

!j 2

1 ( 1

a

)

2 +C

2 ja

m

1

!j 2( )

1 a

2(

) C

2 1

1 a 2

+C 2

1

1 a 2( )

:

by the formula for the sum of a geometri series and b eause ja m

0

!j 1,

ja m

1

!j 1, 2 > 0 ja 2( ) < 0. This gives the seond inequality in

(8.5).

Ifnowjsj2thenwehaveby(8.7){(8.10)andthefatthata m0

j!j1

and a m

1

j!j>1that

X

m2Z

^

(a m

!)

^

(a m

!+s)

C 2

m

0

X

m= 1 ja

m

!j

s

2

+

+C 2

1

X

m=m1 ja

m

!j 1

2

s

2

+1

2

=C 2

ja m

0

!j

1 a

s

2

+

+C 2

ja m

1

!j 1

2

1 a 1

2

s

2

+1

2

:

FromthisweseethatthereisaonstantC

1

=C 2

2 +1

2

1

1 a

+

1

1 a 1

2

!

suh that

(s)C

1 jsj

+1

2

; jsj2:

Then

X

k 2Z

k 6=0 s

k

b

k

b

2C

1 b

+1

2

1

X

k =1 k

+1

2

;

when 0<b

1

2

. Thus we see that(8.6)holds provided

2C

1 b

+1

2

1

X

k =1 k

+1

2

<Æ 2

;

whih isp ossible if b

issuÆiently small.

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