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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

Questions.

Review

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

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

Questions.

Review

{1, x , . . . , xn . . . } −→

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

Questions.

Review

Questions.

Review

Questions.

Review

Questions.

Review

Questions.

Review

Questions.

A detour

Theorem. (Classical Fourier theorem.)

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

Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

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

ak =1π

∫ π

−πf (x) cos(kx) dx , bk =

∫ 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+∞∑

ak =1π

∫ π

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

Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

ak =1π

∫ π

Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

ak =1π

∫ π

Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

ak =1π

∫ π

Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

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

ak =1π

∫ π

Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

ak =1π

∫ π

Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

ak =1π

∫ π

−πf (x) cos(kx) dx ,

bk =1π

∫ π

Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

ak =1π

∫ π

Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

ak =1π

∫ π

Corollary.

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

Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

ak =1π

∫ π

Corollary. The trig. system

is complete in L2[−π, π].

Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

ak =1π

∫ π

Then ∀x ∈ [−π, π]:

f (x) =a0

2+∞∑

ak =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 =∞∑

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 =∞∑

cnϕn.

In a Hilbert space.

f =∞∑

ckϕk .

f =∞∑

cnϕn.

In a Hilbert space.

f =∞∑

ckϕk .

f =∞∑

cnϕn.

In a Hilbert space.

f =∞∑

ckϕk .

f =∞∑

cnϕn.

In a Hilbert space.

f =∞∑

ckϕk .

Then ck = 〈f , ϕk 〉.

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

f =∞∑

cnϕn.

In a Hilbert space.

f =∞∑

ckϕk .

f =∞∑

cnϕn.

In a Hilbert space.

f =∞∑

ckϕk .

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

∃(cn)

f =∞∑

cnϕn.

In a Hilbert space.

f =∞∑

ckϕk .

f =∞∑

cnϕn.

In a Hilbert space.

f =∞∑

ckϕk .

f =∞∑

cnϕn.

Proof.

Let us define sn :=n∑

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∑

ckϕk , ϕj

⟩= ??? =

n∑k=1

ck 〈ϕk , ϕj〉 = cj .

Proof.

ckϕk .

limn→∞

‖f − sn‖ = 0.

If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

ckϕk , ϕj

⟩= ??? =

n∑k=1

Proof.

ckϕk .

limn→∞

‖f − sn‖ = 0.

If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

ckϕk , ϕj

⟩= ??? =

n∑k=1

Proof.

ckϕk .

limn→∞

‖f − sn‖ = 0.

If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

ckϕk , ϕj

⟩= ??? =

n∑k=1

Proof.

ckϕk .

limn→∞

‖f − sn‖ = 0.

limn→∞〈f − sn, ϕj〉 = 0.

(Why?) =⇒ 〈f , ϕj〉 = limn→∞〈sn, ϕj〉

If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

ckϕk , ϕj

⟩= ??? =

n∑k=1

Proof.

ckϕk .

limn→∞

‖f − sn‖ = 0.

limn→∞〈f − sn, ϕj〉 = 0. (Why?)

=⇒ 〈f , ϕj〉 = limn→∞〈sn, ϕj〉

If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

ckϕk , ϕj

⟩= ??? =

n∑k=1

Proof.

ckϕk .

limn→∞

‖f − sn‖ = 0.

If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

ckϕk , ϕj

⟩= ??? =

n∑k=1

Proof.

ckϕk .

limn→∞

‖f − sn‖ = 0.

If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

ckϕk , ϕj

⟩= ??? =

n∑k=1

Proof.

ckϕk .

limn→∞

‖f − sn‖ = 0.

If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

ckϕk , ϕj

⟩= ??? =

n∑k=1

Proof.

ckϕk .

limn→∞

‖f − sn‖ = 0.

If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

ckϕk , ϕj

??? =n∑

Proof.

ckϕk .

limn→∞

‖f − sn‖ = 0.

If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

ckϕk , ϕj

⟩= ???

Proof.

ckϕk .

limn→∞

‖f − sn‖ = 0.

If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

ckϕk , ϕj

⟩= ??? =

n∑k=1

ck 〈ϕk , ϕj〉

= cj .

Proof.

ckϕk .

limn→∞

‖f − sn‖ = 0.

If n ≥ j , then

〈sn, ϕj〉 =

⟨n∑

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 ∼∞∑

cn ϕn, with cn = 〈f , ϕn〉.

It is a formal definition yet. Why?

〈f , ϕn〉 , n = 1,2, . . .

∞∑n=1

〈f , ϕn〉 ϕn.

〈f , ϕn〉 , n = 1,2, . . .

∞∑n=1

〈f , ϕn〉 ϕn.

〈f , ϕn〉 , n = 1,2, . . .

∞∑n=1

〈f , ϕn〉 ϕn.

〈f , ϕn〉 , n = 1,2, . . .

∞∑n=1

〈f , ϕn〉 ϕn.

〈f , ϕn〉 , n = 1,2, . . .

∞∑n=1

〈f , ϕn〉 ϕn.

It is a formal definition yet.

〈f , ϕn〉 , n = 1,2, . . .

∞∑n=1

〈f , ϕn〉 ϕn.

Sum of the Fourier series

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

f =∞∑

〈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∑

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

In infinite dimensional Hilbert space basis ≡ complete ON system

f =∞∑

〈f , ϕn〉 ϕn.

f =∞∑

〈f , ϕn〉 ϕn.

f =∞∑

〈f , ϕn〉 ϕn.

Analogy.

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

f =∞∑

〈f , ϕn〉 ϕn.

Analogy. V is a finite dim. vector space.

v1, . . . , vn ∈ V is a basis, if

f =∞∑

〈f , ϕn〉 ϕn.

f =∞∑

〈f , ϕn〉 ϕn.

f =∞∑

〈f , ϕn〉 ϕn.

In infinite dimensional Hilbert space

basis ≡ complete ON system

f =∞∑

〈f , ϕn〉 ϕn.

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 ⇐⇒∞∑

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.

c2n = ‖f‖2.

Theorem.

Let f ∈ H.

∞∑n=1

c2n ≤ ‖f‖2, cn = 〈f , ϕn〉 ϕn.

c2n = ‖f‖2.

∞∑n=1

c2n ≤ ‖f‖2, cn = 〈f , ϕn〉 ϕn.

c2n = ‖f‖2.

1. (ϕn) ⊂ H is an ON system.

∞∑n=1

c2n ≤ ‖f‖2, cn = 〈f , ϕn〉 ϕn.

c2n = ‖f‖2.

∞∑n=1

c2n ≤ ‖f‖2,

cn = 〈f , ϕn〉 ϕn.

c2n = ‖f‖2.

∞∑n=1

c2n ≤ ‖f‖2, cn = 〈f , ϕn〉 ϕn.

c2n = ‖f‖2.

∞∑n=1

c2n ≤ ‖f‖2, cn = 〈f , ϕn〉 ϕn.

2. (ϕn) is ON and complete

⇐⇒∞∑

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.

c2n = ‖f‖2.

∞∑n=1

c2n ≤ ‖f‖2, cn = 〈f , ϕn〉 ϕn.

c2n = ‖f‖2.

∞∑n=1

c2n ≤ ‖f‖2

cn = 〈f , ϕn〉 ϕn.,

Proof. 1. Let us define sn :=n∑

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∑

c2k . Finally, with n→∞

∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,

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

∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,

Proof.

1. Let us define sn :=n∑

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

∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,

ckϕk .

Geometrically it is try to

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

∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,

ckϕk . Geometrically it is

try to

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

∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,

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.

∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,

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.

∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,

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

∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,

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

∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,

‖f‖2 = ‖f − sn‖2 + ‖sn‖2

=⇒ ‖sn‖2 ≤ ‖f‖2 ∀n.

∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,

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

∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,

‖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.,

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

Finally, with n→∞√

∞∑n=1

c2n ≤ ‖f‖2 cn = 〈f , ϕn〉 ϕn.,

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

Functional analysis Distant Learning. Week 3.∞∑

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 =∞∑

cnϕn, thus

limn→∞

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

‖sn‖2 =∞∑

Part B. Assuming ‖f‖2 =∞∑

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

Yourself. HW.

c2n = ‖f‖2 ⇐⇒ (ϕn) is complete,

‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)

cnϕn, thus

limn→∞

‖sn‖2 =∞∑

Yourself. HW.

‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)

cnϕn, thus

limn→∞

‖sn‖2 =∞∑

Yourself. HW.

Part A. Assume (ϕn) is ON.

From the previous slide:

‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)

cnϕn, thus

limn→∞

‖sn‖2 =∞∑

Yourself. HW.

‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)

cnϕn, thus

limn→∞

‖sn‖2 =∞∑

Yourself. HW.

‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)

cnϕn, thus

limn→∞

‖sn‖2 =∞∑

Yourself. HW.

‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)

cnϕn, thus

limn→∞

‖f − sn‖2 = 0.

From (1) we get ‖f‖2 = limn→∞

‖sn‖2 =∞∑

Yourself. HW.

‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)

cnϕn, thus

limn→∞

‖sn‖2 =

∞∑k=1

Yourself. HW.

‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)

cnϕn, thus

limn→∞

‖sn‖2 =∞∑

Yourself. HW.

‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)

cnϕn, thus

limn→∞

‖sn‖2 =∞∑

Part B.

Assuming ‖f‖2 =∞∑

Yourself. HW.

‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)

cnϕn, thus

limn→∞

‖sn‖2 =∞∑

Part B. Assuming ‖f‖2 =

∞∑k=1

Yourself. HW.

‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)

cnϕn, thus

limn→∞

‖sn‖2 =∞∑

c2k ∀f ,

prove (ϕn) is COMPLETE. Do it

Yourself. HW.

‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)

cnϕn, thus

limn→∞

‖sn‖2 =∞∑

c2k ∀f , prove (ϕn) is COMPLETE.

Yourself. HW.

‖f‖2 = ‖f − sn‖2 + ‖sn‖2. (1)

cnϕn, thus

limn→∞

‖sn‖2 =∞∑

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〉 =∞∑

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〉 =∞∑

ck dk ,

〈f ,g〉L2 = 〈c,d〉`2 .

f ,g ∈ L2(R) are arbitrary functions.

〈f ,g〉 =∞∑

ck dk ,

〈f ,g〉L2 = 〈c,d〉`2 .

〈f ,g〉 =∞∑

ck dk ,

〈f ,g〉L2 = 〈c,d〉`2 .

〈f ,g〉 =∞∑

ck dk ,

〈f ,g〉L2 = 〈c,d〉`2 .

〈f ,g〉 =∞∑

ck dk ,

〈f ,g〉L2 =

〈c,d〉`2 .

〈f ,g〉 =∞∑

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.∞∑

d2k <∞.

Then ∃f ∈ L2(R), such that ‖f‖2 =∞∑

d2k , and it’s Fourier coefficients

are dk .

Proof. (Hint) A ”candidate” is f :=∞∑

dkϕk . It is OK. Finish the proof.

d2k <∞.

are dk .

d2k <∞.

are dk .

d2k <∞.

are dk .

d2k <∞.

are dk .

Proof. (Hint)

A ”candidate” is f :=∞∑

d2k <∞.

are dk .

dkϕk .

It is OK. Finish the proof.

d2k <∞.

are dk .

dkϕk . It is OK.

Finish the proof.

d2k <∞.

are dk .

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 ←→

L2 and `2

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) =

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

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

f (x) =∞∑

cnPn(x), with cn =

−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) =

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

f (x) =∞∑

cnPn(x), with cn =

−1f (x)Pn(x)dx .

P0(x) =1√2,

P1(x) =

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

f (x) =∞∑

cnPn(x), with cn =

−1f (x)Pn(x)dx .

P0(x) =1√2, P1(x) =

x , P2(x) =

it was a HW . . . ..

f (x) =∞∑

cnPn(x), with cn =

−1f (x)Pn(x)dx .

P0(x) =1√2, P1(x) =

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

f (x) =∞∑

cnPn(x), with cn =

−1f (x)Pn(x)dx .

P0(x) =1√2, P1(x) =

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

f (x) =∞∑

cnPn(x), with cn =

−1f (x)Pn(x)dx .

P0(x) =1√2, P1(x) =

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

f (x) =∞∑

cnPn(x), with cn =

−1f (x)Pn(x)dx .

P0(x) =1√2, P1(x) =

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

f (x) =∞∑

cnPn(x), with cn =

−1f (x)Pn(x)dx .

degree n with KNOWN coefficients.

Can you recall sg. similar?

P0(x) =1√2, P1(x) =

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

f (x) =∞∑

cnPn(x), with cn =

−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.

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

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

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

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

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

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.

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

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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