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Functional analysis Distant Learning. Week 3.

Functional analysis

Lesson 9.

April 21, 2020


Review

In the WEIGHTED L2 SPACE we applied G-S orthogonalization:

{1, x , . . . , xn . . . } −→ {ϕ0, ϕ1, . . . , ϕn . . . } ON polynomials.

E.g. Legendre-, Chebishev-, Hermite-polynomials. Do you know?

Questions.

I Why are these systems of orthogonal polynomials important?

I What can we use the ON polynomials for?


Review


{1, x , . . . , xn . . . }

−→ {ϕ0, ϕ1, . . . , ϕn . . . } ON polynomials.


Questions.




Review


{1, x , . . . , xn . . . } −→

{ϕ0, ϕ1, . . . , ϕn . . . } ON polynomials.


Questions.




Review


{1, x , . . . , xn . . . } −→ {ϕ0, ϕ1, . . . , ϕn . . . } ON polynomials.


Questions.




A detour

Theorem. (Classical Fourier theorem.)

Assume f : [−π, π]→ IR satisfies the Dirichlet conditions. ??

Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

k=1

(ak cos(kx) + bk sin(kx)) , with

ak =1π

∫ π

−πf (x) cos(kx) dx , bk =

1π

∫ π

−πf (x) sin(kx) dx .

Corollary. The trig. system is complete in L2[−π, π].

−→ Moreover, the coefficients are known.


A detourTheorem. (Classical Fourier theorem.)


Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

k=1


ak =1π

∫ π


1π

∫ π






Assume f : [−π, π]→ IR satisfies the Dirichlet conditions.

??

Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

k=1


ak =1π

∫ π


1π

∫ π







Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

k=1


ak =1π

∫ π


1π

∫ π







Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

k=1

(ak cos(kx) + bk sin(kx)) ,

with

ak =1π

∫ π


1π

∫ π







Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

k=1


ak =1π

∫ π


1π

∫ π







Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

k=1


ak =1π

∫ π

−πf (x) cos(kx) dx ,

bk =1π

∫ π







Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

k=1


ak =1π

∫ π


1π

∫ π







Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

k=1


ak =1π

∫ π


1π

∫ π


Corollary.

The trig. system is complete in L2[−π, π].





Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

k=1


ak =1π

∫ π


1π

∫ π


Corollary. The trig. system

is complete in L2[−π, π].





Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

k=1


ak =1π

∫ π


1π

∫ π





General Fourier series


In a Hilbert space.

(H, 〈·, ·〉) is a Hilbert space. (Can you recall the definition?)

Let (ϕk , ) ⊂ H be an ON system.

Theorem. Assume, that for some f ∈ H we have

f =∞∑

k=1

ckϕk .

Then ck = 〈f , ϕk 〉. I.e. the coefficients can be recovered from f .

Remark. If (ϕn) ⊂ H is complete, then every f ∈ H: ∃(cn)

f =∞∑

n=1

cnϕn.


In a Hilbert space.




f =∞∑

k=1

ckϕk .

Then ck = 〈f , ϕk 〉.

I.e. the coefficients can be recovered from f .


f =∞∑

n=1

cnϕn.


In a Hilbert space.




f =∞∑

k=1

ckϕk .



f =∞∑

n=1

cnϕn.


In a Hilbert space.




f =∞∑

k=1

ckϕk .


Remark. If (ϕn) ⊂ H is complete, then every f ∈ H:

∃(cn)

f =∞∑

n=1

cnϕn.


In a Hilbert space.




f =∞∑

k=1

ckϕk .



f =∞∑

n=1

cnϕn.


Proof.

Let us define sn :=n∑

k=1

ckϕk .

Then by the Thm.’s assumption

limn→∞

‖f − sn‖ = 0.

It follows, that for all ϕj , j ≤ n

limn→∞〈f − sn, ϕj〉 = 0. (Why?) =⇒ 〈f , ϕj〉 = lim

n→∞〈sn, ϕj〉

If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

k=1

ckϕk , ϕj

⟩= ??? =

n∑k=1

ck 〈ϕk , ϕj〉 = cj .


Proof.


k=1

ckϕk .


limn→∞

‖f − sn‖ = 0.


limn→∞〈f − sn, ϕj〉 = 0.

(Why?) =⇒ 〈f , ϕj〉 = limn→∞〈sn, ϕj〉

If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

k=1

ckϕk , ϕj

⟩= ??? =

n∑k=1



Proof.


k=1

ckϕk .


limn→∞

‖f − sn‖ = 0.


limn→∞〈f − sn, ϕj〉 = 0. (Why?)

=⇒ 〈f , ϕj〉 = limn→∞〈sn, ϕj〉

If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

k=1

ckϕk , ϕj

⟩= ??? =

n∑k=1



Proof.


k=1

ckϕk .


limn→∞

‖f − sn‖ = 0.




If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

k=1

ckϕk , ϕj

⟩= ??? =

n∑k=1



Proof.


k=1

ckϕk .


limn→∞

‖f − sn‖ = 0.




If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

k=1

ckϕk , ϕj

⟩=

??? =n∑

k=1



Proof.


k=1

ckϕk .


limn→∞

‖f − sn‖ = 0.




If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

k=1

ckϕk , ϕj

⟩= ???

=n∑

k=1



Proof.


k=1

ckϕk .


limn→∞

‖f − sn‖ = 0.




If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

k=1

ckϕk , ϕj

⟩= ??? =

n∑k=1

ck 〈ϕk , ϕj〉

= cj .


Proof.


k=1

ckϕk .


limn→∞

‖f − sn‖ = 0.




If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

k=1

ckϕk , ϕj

⟩= ??? =

n∑k=1



Fourier series expansion

Let (ϕn) ⊂ H be a complete ON system. For any f ∈ H we define

I FOURIER COEFFICIENTS of f with respect to (ϕn) as

〈f , ϕn〉 , n = 1,2, . . .

I FOURIER SERIES EXPANSION of f with respect to (ϕn) as

∞∑n=1

〈f , ϕn〉 ϕn.

Notation. f ∼∞∑

n=1

cn ϕn, with cn = 〈f , ϕn〉.

It is a formal definition yet. Why?





〈f , ϕn〉 , n = 1,2, . . .


∞∑n=1

〈f , ϕn〉 ϕn.


n=1


It is a formal definition yet.

Why?





〈f , ϕn〉 , n = 1,2, . . .


∞∑n=1

〈f , ϕn〉 ϕn.


n=1


It is a formal definition yet. Why?


Sum of the Fourier series

Theorem. If (ϕn) is a complete ON system, then

f =∞∑

n=1

〈f , ϕn〉 ϕn.

I.e. the sum of the Fourer series gives back the original function.

Analogy. V is a finite dim. vector space. v1, . . . , vn ∈ V is a basis, if

I these vectors are linearly independent,

I ∀v ∈ V can be written as v =n∑

k=1

ck vk (i.e. a generator system).

In infinite dimensional Hilbert space basis ≡ complete ON system




f =∞∑

n=1

〈f , ϕn〉 ϕn.


Analogy.

V is a finite dim. vector space. v1, . . . , vn ∈ V is a basis, if



k=1






f =∞∑

n=1

〈f , ϕn〉 ϕn.


Analogy. V is a finite dim. vector space.

v1, . . . , vn ∈ V is a basis, if



k=1






f =∞∑

n=1

〈f , ϕn〉 ϕn.





k=1






f =∞∑

n=1

〈f , ϕn〉 ϕn.





k=1


In infinite dimensional Hilbert space

basis ≡ complete ON system




f =∞∑

n=1

〈f , ϕn〉 ϕn.





k=1




Parseval equality

Try to recall ”the original” one

Theorem. Let f ∈ H.

1. (ϕn) ⊂ H is an ON system. Then

∞∑n=1

c2n ≤ ‖f‖2, cn = 〈f , ϕn〉 ϕn.

2. (ϕn) is ON and complete ⇐⇒∞∑

n=1

c2n = ‖f‖2.

The latter identity is called PARSEVAL EQUALITY.


Parseval equality Try to recall ”the original” one



∞∑n=1

c2n ≤ ‖f‖2, cn = 〈f , ϕn〉 ϕn.


n=1

c2n = ‖f‖2.




Theorem.

Let f ∈ H.


∞∑n=1

c2n ≤ ‖f‖2, cn = 〈f , ϕn〉 ϕn.


n=1

c2n = ‖f‖2.






∞∑n=1

c2n ≤ ‖f‖2, cn = 〈f , ϕn〉 ϕn.


n=1

c2n = ‖f‖2.





1. (ϕn) ⊂ H is an ON system.

Then

∞∑n=1

c2n ≤ ‖f‖2, cn = 〈f , ϕn〉 ϕn.


n=1

c2n = ‖f‖2.






∞∑n=1

c2n ≤ ‖f‖2,

cn = 〈f , ϕn〉 ϕn.


n=1

c2n = ‖f‖2.






∞∑n=1

c2n ≤ ‖f‖2, cn = 〈f , ϕn〉 ϕn.


n=1

c2n = ‖f‖2.






∞∑n=1

c2n ≤ ‖f‖2, cn = 〈f , ϕn〉 ϕn.

2. (ϕn) is ON and complete

⇐⇒∞∑

n=1

c2n = ‖f‖2.






∞∑n=1

c2n ≤ ‖f‖2, cn = 〈f , ϕn〉 ϕn.

2. (ϕn) is ON and complete ⇐⇒

∞∑n=1

c2n = ‖f‖2.






∞∑n=1

c2n ≤ ‖f‖2, cn = 〈f , ϕn〉 ϕn.


n=1

c2n = ‖f‖2.



∞∑n=1

c2n ≤ ‖f‖2

cn = 〈f , ϕn〉 ϕn.,

Proof. 1. Let us define sn :=n∑

k=1

ckϕk . Geometrically it is try to

finish...the projection of f onto span{ϕ1, ..., ϕn}. Thus (f − sn)⊥sn.

Then we can use the Pythagorean theorem:

‖f‖2 = ‖f − sn‖2 + ‖sn‖2 =⇒ ‖sn‖2 ≤ ‖f‖2 ∀n.

By orthogonality ‖sn‖2 =n∑

k=1

c2k . Finally, with n→∞

√.


∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,


k=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2 =⇒ ‖sn‖2 ≤ ‖f‖2 ∀n.


k=1


√.


∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,

Proof.

1. Let us define sn :=n∑

k=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2 =⇒ ‖sn‖2 ≤ ‖f‖2 ∀n.


k=1


√.


∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,


k=1

ckϕk .

Geometrically it is try to



‖f‖2 = ‖f − sn‖2 + ‖sn‖2 =⇒ ‖sn‖2 ≤ ‖f‖2 ∀n.


k=1


√.


∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,


k=1

ckϕk . Geometrically it is

try to



‖f‖2 = ‖f − sn‖2 + ‖sn‖2 =⇒ ‖sn‖2 ≤ ‖f‖2 ∀n.


k=1


√.


∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,


k=1


finish...

the projection of f onto span{ϕ1, ..., ϕn}. Thus (f − sn)⊥sn.


‖f‖2 = ‖f − sn‖2 + ‖sn‖2 =⇒ ‖sn‖2 ≤ ‖f‖2 ∀n.


k=1


√.


∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,


k=1


finish...the projection of f onto span{ϕ1, ..., ϕn}.

Thus (f − sn)⊥sn.


‖f‖2 = ‖f − sn‖2 + ‖sn‖2 =⇒ ‖sn‖2 ≤ ‖f‖2 ∀n.


k=1


√.


∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,


k=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2 =⇒ ‖sn‖2 ≤ ‖f‖2 ∀n.


k=1


√.


∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,


k=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2

=⇒ ‖sn‖2 ≤ ‖f‖2 ∀n.


k=1


√.


∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,


k=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2 =⇒ ‖sn‖2 ≤ ‖f‖2 ∀n.


k=1


√.


∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,


k=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2 =⇒ ‖sn‖2 ≤ ‖f‖2 ∀n.

By orthogonality ‖sn‖2 =

n∑k=1


√.


∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,


k=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2 =⇒ ‖sn‖2 ≤ ‖f‖2 ∀n.


k=1

c2k .

Finally, with n→∞√

.


∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,


k=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2 =⇒ ‖sn‖2 ≤ ‖f‖2 ∀n.


k=1


√.

Functional analysis Distant Learning. Week 3.∞∑

n=1

c2n = ‖f‖2

⇐⇒ (ϕn) is complete,

2. To verify a proposition with ⇐⇒ inside has to parts.

Part A. Assume (ϕn) is ON. From the previous slide:

‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)

From the completeness of (ϕn) follows, that f =∞∑

n=1

cnϕn, thus

limn→∞

‖f − sn‖2 = 0. From (1) we get ‖f‖2 = limn→∞

‖sn‖2 =∞∑

k=1

c2k .

Part B. Assuming ‖f‖2 =∞∑

k=1

c2k ∀f , prove (ϕn) is COMPLETE. Do it

Yourself. HW.


n=1

c2n = ‖f‖2 ⇐⇒ (ϕn) is complete,



‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)


n=1

cnϕn, thus

limn→∞


‖sn‖2 =∞∑

k=1

c2k .


k=1


Yourself. HW.


n=1



Part A. Assume (ϕn) is ON.

From the previous slide:

‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)


n=1

cnϕn, thus

limn→∞


‖sn‖2 =∞∑

k=1

c2k .


k=1


Yourself. HW.


n=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)


n=1

cnϕn, thus

limn→∞


‖sn‖2 =∞∑

k=1

c2k .


k=1


Yourself. HW.


n=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)


n=1

cnϕn, thus

limn→∞

‖f − sn‖2 = 0.

From (1) we get ‖f‖2 = limn→∞

‖sn‖2 =∞∑

k=1

c2k .


k=1


Yourself. HW.


n=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)


n=1

cnϕn, thus

limn→∞


‖sn‖2 =

∞∑k=1

c2k .


k=1


Yourself. HW.


n=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)


n=1

cnϕn, thus

limn→∞


‖sn‖2 =∞∑

k=1

c2k .


k=1


Yourself. HW.


n=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)


n=1

cnϕn, thus

limn→∞


‖sn‖2 =∞∑

k=1

c2k .

Part B.

Assuming ‖f‖2 =∞∑

k=1


Yourself. HW.


n=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)


n=1

cnϕn, thus

limn→∞


‖sn‖2 =∞∑

k=1

c2k .

Part B. Assuming ‖f‖2 =

∞∑k=1


Yourself. HW.


n=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)


n=1

cnϕn, thus

limn→∞


‖sn‖2 =∞∑

k=1

c2k .


k=1

c2k ∀f ,

prove (ϕn) is COMPLETE. Do it

Yourself. HW.


n=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)


n=1

cnϕn, thus

limn→∞


‖sn‖2 =∞∑

k=1

c2k .


k=1

c2k ∀f , prove (ϕn) is COMPLETE.

Do it

Yourself. HW.


n=1




‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)


n=1

cnϕn, thus

limn→∞


‖sn‖2 =∞∑

k=1

c2k .


k=1


Yourself. HW.


Generalized Parseval equality.

Theorem. Let (ϕn) be a complete ON system in L2(R)-ben.

f ,g ∈ L2(R) are arbitrary functions.Then

〈f ,g〉 =∞∑

k=1

ck dk ,

where c = (ck ) and d = (dk ) are the Fourier coefficients of f and g.

This relation can be also written as:

〈f ,g〉L2 = 〈c,d〉`2 .




f ,g ∈ L2(R) are arbitrary functions.

Then

〈f ,g〉 =∞∑

k=1

ck dk ,



〈f ,g〉L2 = 〈c,d〉`2 .





〈f ,g〉 =∞∑

k=1

ck dk ,



〈f ,g〉L2 = 〈c,d〉`2 .





〈f ,g〉 =∞∑

k=1

ck dk ,



〈f ,g〉L2 =

〈c,d〉`2 .





〈f ,g〉 =∞∑

k=1

ck dk ,



〈f ,g〉L2 = 〈c,d〉`2 .


Special case: H = L2(R)

Corollary. For any f ∈ L2(R) it is possible to assign (cn) ∈ `2, usingany (ϕn) complete ON system.

The other direction is the following important Thm.

Theorem. (Riesz-Fisher thm.) Let (dk ) ∈ `2, i.e.∞∑

k=1

d2k <∞.

Then ∃f ∈ L2(R), such that ‖f‖2 =∞∑

k=1

d2k , and it’s Fourier coefficients

are dk .

Proof. (Hint) A ”candidate” is f :=∞∑

k=1

dkϕk . It is OK. Finish the proof.






k=1

d2k <∞.


k=1


are dk .

Proof. (Hint)

A ”candidate” is f :=∞∑

k=1







k=1

d2k <∞.


k=1


are dk .


k=1

dkϕk .

It is OK. Finish the proof.






k=1

d2k <∞.


k=1


are dk .


k=1

dkϕk . It is OK.

Finish the proof.






k=1

d2k <∞.


k=1


are dk .


k=1



L2 and `2

Corollary. L2(R) es `2 are isometrically isomorphic.

The linear isometry is based an any (ϕn) complete ON system,

using the Fourier coefficients: f ←→ (cn).

PLEASE STOP FOR A WHILE, AND UNDERSTAND THIS POINT.

L2(R) and `2 are the ”same”.


L2 and `2



using the Fourier coefficients:

f ←→ (cn).




L2 and `2



using the Fourier coefficients: f ←→

(cn).




L2 and `2



using the Fourier coefficients: f ←→ (cn).




Example. H = L2[−1,1]

In L2[−1,1] a complete ON system are the Legendre polynomials.

We have seen some elements of (Pn(x)):

P0(x) =1√2, P1(x) =

√32

x , P2(x) = it was a HW . . . ..

Then every f ∈ L2[−1,1] can be written as

f (x) =∞∑

n=0

cnPn(x), with cn =

∫ 1

−1f (x)Pn(x)dx .

Thus every f ∈ L2[−1,1] can be approximated by a polynomial of

degree n with KNOWN coefficients. Can you recall sg. similar?





P0(x) =1√2,

P1(x) =

√32

x , P2(x) = it was a HW . . . ..


f (x) =∞∑

n=0

cnPn(x), with cn =

∫ 1

−1f (x)Pn(x)dx .







P0(x) =1√2, P1(x) =

√32

x , P2(x) =

it was a HW . . . ..


f (x) =∞∑

n=0

cnPn(x), with cn =

∫ 1

−1f (x)Pn(x)dx .







P0(x) =1√2, P1(x) =

√32

x , P2(x) = it was a HW . . . ..


f (x) =∞∑

n=0

cnPn(x), with cn =

∫ 1

−1f (x)Pn(x)dx .







P0(x) =1√2, P1(x) =

√32

x , P2(x) = it was a HW . . . ..


f (x) =∞∑

n=0

cnPn(x), with cn =

∫ 1

−1f (x)Pn(x)dx .


degree n with KNOWN coefficients.

Can you recall sg. similar?





P0(x) =1√2, P1(x) =

√32

x , P2(x) = it was a HW . . . ..


f (x) =∞∑

n=0

cnPn(x), with cn =

∫ 1

−1f (x)Pn(x)dx .




An example in H = L2[0,1]

This example gives an ON system in L2[0,1]-ben.

They are called Haar-functions.

They are not polynomials, but this is the simplest wavelet family .

( More details on that can be found in the in the book.)

They are defined in blocks.

Hn,k with n = 0,1,2, . . . k = 1, ...,2n.

For all indices Hn,k : [0,1]→ IR.






(

More details on that can be found in the in the book.)


Hn,k with n = 0,1,2, . . . k = 1, ...,2n.









Hn,k with n = 0,1,2, . . . k = 1, ...,2n.









Hn,k

with n = 0,1,2, . . . k = 1, ...,2n.









Hn,k with n = 0,1,2, . . . k = 1, ...,2n.



Haar functions

For n = 0 there are two functions: H0,0 and H0,1.

H0,0(x) = 1.

H0,1(x) =

1 if 0 ≤ x < 1/2

−1 if 1/2 ≤ x ≤ 1

x ∈ [0,1]This is the so called mother wavelet

Easy to check, that ‖H0,0‖ = ‖H0,1‖ = 1 and H0,0⊥H0,1. DO IT.


Haar functions


H0,0(x) = 1.

H0,1(x) =

1 if 0 ≤ x < 1/2

−1 if 1/2 ≤ x ≤ 1

x ∈ [0,1]

This is the so called mother wavelet



Haar functions


H0,0(x) = 1.

H0,1(x) =

1 if 0 ≤ x < 1/2

−1 if 1/2 ≤ x ≤ 1




Haar functions


H0,0(x) = 1.

H0,1(x) =

1 if 0 ≤ x < 1/2

−1 if 1/2 ≤ x ≤ 1


Easy to check, that ‖H0,0‖ = ‖H0,1‖ = 1 and

H0,0⊥H0,1. DO IT.


Haar functions


H0,0(x) = 1.

H0,1(x) =

1 if 0 ≤ x < 1/2

−1 if 1/2 ≤ x ≤ 1


Easy to check, that ‖H0,0‖ = ‖H0,1‖ = 1 and H0,0⊥H0,1.

DO IT.


Haar functions


H0,0(x) = 1.

H0,1(x) =

1 if 0 ≤ x < 1/2

−1 if 1/2 ≤ x ≤ 1




Haar functions, nth block.

For n ≥ 1 divide [0,1] into 2n equal parts with pointsk2n . Let’s define:

Hn,k (x) =

√2n if

k − 12n ≤ x <

k − 1/22n

−√

2n ifk − 1/2

2n ≤ x <k2n

0 otherwise

, n ≥ 1, 1 ≤ k ≤ 2n.

The nonzero part is the ”mother wavelet”, squished and stretched.

Easy to check, that ‖Hn,k‖ = 1 and Hn,k⊥Hn,j for j 6= k . DO IT.



For n ≥ 1 divide [0,1] into 2n equal parts with pointsk2n .

Let’s define:

Hn,k (x) =

√2n if

k − 12n ≤ x <

k − 1/22n

−√

2n ifk − 1/2

2n ≤ x <k2n

0 otherwise

, n ≥ 1, 1 ≤ k ≤ 2n.






Hn,k (x) =

√2n if

k − 12n ≤ x <

k − 1/22n

−√

2n ifk − 1/2

2n ≤ x <k2n

0 otherwise

, n ≥ 1, 1 ≤ k ≤ 2n.






Hn,k (x) =

√2n if

k − 12n ≤ x <

k − 1/22n

−√

2n ifk − 1/2

2n ≤ x <k2n

0 otherwise

, n ≥ 1, 1 ≤ k ≤ 2n.


Easy to check, that ‖Hn,k‖ = 1 and

Hn,k⊥Hn,j for j 6= k . DO IT.




Hn,k (x) =

√2n if

k − 12n ≤ x <

k − 1/22n

−√

2n ifk − 1/2

2n ≤ x <k2n

0 otherwise

, n ≥ 1, 1 ≤ k ≤ 2n.


Easy to check, that ‖Hn,k‖ = 1 and Hn,k⊥Hn,j for j 6= k .

DO IT.




Hn,k (x) =

√2n if

k − 12n ≤ x <

k − 1/22n

−√

2n ifk − 1/2

2n ≤ x <k2n

0 otherwise

, n ≥ 1, 1 ≤ k ≤ 2n.




E.g. Haar functions H2,k

As an example, hereare the graphs of the

H2,k

Haar functions fork = 1,2,3,4.

Remark. This ON system is complete. (Not trivial to prove. )

Pázmány Péter Catholic Universityusers.itk.ppke.hu/~vago/funkanal_9_20_online.pdf · Functional...

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