Certain relations between the Fourier and Hilbert transforms

Certain relations between the Fourier and Hilberttransforms

Elijah Liflyand

Bar-Ilan University

March, 2012

Elijah Liflyand (Bar-Ilan University) Fourier and Hilbert transforms March, 2012 1 / 24

The Fourier transform of a function of bounded variation

Let us start with the next result:

L 1993; Fridli, 2001

We define the following T -transform of a functiong : R = [0,∞)→ C:

Tg(t) =

∫ t/2

g(t+ s)− g(t− s)s

where the integral is understood in the improper (principal value)sense, that is, as lim

δ→0+

∫δ .

Here and in what follows we use the notation “. ” and “& ” asabbreviations for “≤ C ” and “≥ C ”, with C being an absolutepositive constant.

L 1993; Fridli, 2001

Tg(t) =

∫ t/2

δ→0+

∫δ .

L 1993; Fridli, 2001

Tg(t) =

∫ t/2

δ→0+

∫δ .

Theorem 1. Let f : R+ → C be locally absolutely continuous, of boundedvariation and lim

t→∞f(t) = 0. Let also Tf ′ ∈ L1(R+).

Then the cosine Fourier transform of f

Fc(x) =

∫ ∞0

f(t) cosxt dt

is Lebesgue integrable on R+,

‖Fc‖L1(R+) . ‖f ′‖L1(R+) + ‖Tf ′‖L1(R+),

t→∞f(t) = 0. Let also Tf ′ ∈ L1(R+).

Fc(x) =

∫ ∞0

f(t) cosxt dt

‖Fc‖L1(R+) . ‖f ′‖L1(R+) + ‖Tf ′‖L1(R+),

t→∞f(t) = 0. Let also Tf ′ ∈ L1(R+).

Fc(x) =

∫ ∞0

f(t) cosxt dt

‖Fc‖L1(R+) . ‖f ′‖L1(R+) + ‖Tf ′‖L1(R+),

and for the sine Fourier transform, we have, with x > 0,

Fs(x) =

∫ ∞0

f(t) sinxt dt =1

xf( π

)+ F (x),

‖F‖L1(R+) . ‖f ′‖L1(R+) + ‖Tf ′‖L1(R+).

Fs(x) =

∫ ∞0

f(t) sinxt dt =1

xf( π

)+ F (x),

‖F‖L1(R+) . ‖f ′‖L1(R+) + ‖Tf ′‖L1(R+).

Fs(x) =

∫ ∞0

f(t) sinxt dt =1

xf( π

)+ F (x),

‖F‖L1(R+) . ‖f ′‖L1(R+) + ‖Tf ′‖L1(R+).

The Hilbert transform

Let us now turn to the Hilbert transform of an integrable function g

Hg(x) =1

t− xdt,

where the integral is also understood in the improper (principal value)sense, now as lim

δ→0+

∫|t−x|>δ .

It is not necessarily integrable,

and when it is, we say that g is in the (real) Hardy space H1(R).

If g ∈ H1(R), then it satisfies the cancelation property (has meanzero)

∫Rg(t) dt = 0.

It was apparently first mentioned by Kober.

Hg(x) =1

t− xdt,

δ→0+

∫|t−x|>δ .

∫Rg(t) dt = 0.

Hg(x) =1

t− xdt,

δ→0+

∫|t−x|>δ .

∫Rg(t) dt = 0.

Hg(x) =1

t− xdt,

δ→0+

∫|t−x|>δ .

∫Rg(t) dt = 0.

Hg(x) =1

t− xdt,

δ→0+

∫|t−x|>δ .

∫Rg(t) dt = 0.

Hg(x) =1

t− xdt,

δ→0+

∫|t−x|>δ .

∫Rg(t) dt = 0.

An odd function always has mean zero.

However, not every odd integrable function belongs to H1(R),

a counterexample is explicitly given by L-Tikhonov.

When in the definition of the Hilbert transform the function g is odd,we will denote this transform by H0,

and it is equal to

H0g(x) =2

∫ ∞0

t2 − x2dt.

If it is integrable, we will denote the corresponding Hardy space byH1

0 (R).

and it is equal to

H0g(x) =2

∫ ∞0

t2 − x2dt.

0 (R).

and it is equal to

H0g(x) =2

∫ ∞0

t2 − x2dt.

0 (R).

and it is equal to

H0g(x) =2

∫ ∞0

t2 − x2dt.

0 (R).

and it is equal to

H0g(x) =2

∫ ∞0

t2 − x2dt.

0 (R).

and it is equal to

H0g(x) =2

∫ ∞0

t2 − x2dt.

0 (R).

H0g(x) = Tg(x) + Γ(x),

where Γ is such that

∫ ∞0|Γ(x)| dx .

∫ ∞0|g(t)| dt,

the estimates in Theorem 1 can be treated as ‖f ′‖H10 (R+).

The space of integrable functions g with integrable Tg, or justH1

0 (R+), is one of the widest spaces the belonging of the derivative f ′

to which ensures the integrability of the cosine Fourier transform of f.

However, the possibility of existence (or non-existence) of a widerspace of such type is of considerable interest.

H0g(x) = Tg(x) + Γ(x),

∫ ∞0|Γ(x)| dx .

∫ ∞0|g(t)| dt,

H0g(x) = Tg(x) + Γ(x),

∫ ∞0|Γ(x)| dx .

∫ ∞0|g(t)| dt,

H0g(x) = Tg(x) + Γ(x),

∫ ∞0|Γ(x)| dx .

∫ ∞0|g(t)| dt,

H0g(x) = Tg(x) + Γ(x),

∫ ∞0|Γ(x)| dx .

∫ ∞0|g(t)| dt,

Let us show that such a space does exist, moreover, it is the widestpossible, at least provides a necessary and sufficient condition for theintegrability of the cosine Fourier transform.

In fact, it has in essence been introduced (for different purposes) byJohnson-Warner in 2010 as

Q = {g : g ∈ L1(R),

|g(x)||x|

dx <∞}.

With the obvious norm ‖g‖L1(R) +∫R|g(x)||x| dx it is a Banach

space and ideal in L1(R).

What we will actually use is the space Q0 of the odd functions from Q

Q0 = {g : g ∈ L1(R), g(−t) = −g(t),

∫ ∞0

|gs(x)|x

dx <∞};

such functions naturally have mean zero.

Q = {g : g ∈ L1(R),

|g(x)||x|

dx <∞}.

Q0 = {g : g ∈ L1(R), g(−t) = −g(t),

∫ ∞0

|gs(x)|x

dx <∞};

Q = {g : g ∈ L1(R),

|g(x)||x|

dx <∞}.

Q0 = {g : g ∈ L1(R), g(−t) = −g(t),

∫ ∞0

|gs(x)|x

dx <∞};

Q = {g : g ∈ L1(R),

|g(x)||x|

dx <∞}.

Q0 = {g : g ∈ L1(R), g(−t) = −g(t),

∫ ∞0

|gs(x)|x

dx <∞};

t→∞f(t) = 0. Then the cosine Fourier transform of f is

Lebesgue integrable on R+ if and only if f ′ ∈ Q0.

The situation is more delicate with the sine Fourier transform, wherea sort of asymptotic relation can be obtained.

In what follows we shall denote Tg(x) = gs(x)x .

Theorem 3. Let f : R+ → C be locally absolutely continuous, ofbounded variation and lim

t→∞f(t) = 0. Then for any x > 0

Fs(x) =

∫ ∞0

f(t) sinxt dt =1

xf( π

)+H0Tf ′(x) +G(x),

‖G‖L1(R+) . ‖f ′‖L1(R+).

Fs(x) =

∫ ∞0

f(t) sinxt dt =1

xf( π

)+H0Tf ′(x) +G(x),

‖G‖L1(R+) . ‖f ′‖L1(R+).

Fs(x) =

∫ ∞0

f(t) sinxt dt =1

xf( π

)+H0Tf ′(x) +G(x),

‖G‖L1(R+) . ‖f ′‖L1(R+).

Fs(x) =

∫ ∞0

f(t) sinxt dt =1

xf( π

)+H0Tf ′(x) +G(x),

‖G‖L1(R+) . ‖f ′‖L1(R+).

Fs(x) =

∫ ∞0

f(t) sinxt dt =1

xf( π

)+H0Tf ′(x) +G(x),

‖G‖L1(R+) . ‖f ′‖L1(R+).

A Hardy type space

This theorem makes it natural to consider a Hardy type space H1Q(R+)

which consists of Q0 functions g with integrable H0Tg.Corollary. Let a function f satisfy the assumptions of Theorem 3 andsuch that f ′ ∈ H1

Q(R+). Then

Fs(x) =1

xf( π

)+G(x),

‖G‖L1(R+) . ‖f ′‖H1Q(R+).

Technically, this is an obvious corollary of Theorem 3.

A Hardy type space

which consists of Q0 functions g with integrable H0Tg.

Corollary. Let a function f satisfy the assumptions of Theorem 3 andsuch that f ′ ∈ H1

Q(R+). Then

Fs(x) =1

xf( π

)+G(x),

‖G‖L1(R+) . ‖f ′‖H1Q(R+).

A Hardy type space

Q(R+). Then

Fs(x) =1

xf( π

)+G(x),

‖G‖L1(R+) . ‖f ′‖H1Q(R+).

A Hardy type space

Q(R+). Then

Fs(x) =1

xf( π

)+G(x),

‖G‖L1(R+) . ‖f ′‖H1Q(R+).

A Hardy type space

Q(R+). Then

Fs(x) =1

xf( π

)+G(x),

‖G‖L1(R+) . ‖f ′‖H1Q(R+).

About the proofs

Proof of Theorem 2.

The assumptions of the theorem give a possibility to integrate byparts.

This yields

Fc(x) = −1

∫ ∞0

f ′(t) sinxt dt = −1

xf ′s(x).

Integrating both sides over R+ completes the proof.

About the proofs

Proof of Theorem 2.

This yields

Fc(x) = −1

∫ ∞0

xf ′s(x).

About the proofs

Proof of Theorem 2.

This yields

Fc(x) = −1

∫ ∞0

xf ′s(x).

About the proofs

Proof of Theorem 2.

This yields

Fc(x) = −1

∫ ∞0

xf ′s(x).

About the proofs

Proof of Theorem 3.

The main ingredient:

I = I(x) =

∫ ∞π2x

f(t) sinxt dt.

We prove that

I(x) = H0Tf ′(x) +G(x),

‖G‖L1(R+) . ‖f ′‖L1(R+).

About the proofs

Proof of Theorem 3.

I = I(x) =

∫ ∞π2x

f(t) sinxt dt.

We prove that

I(x) = H0Tf ′(x) +G(x),

‖G‖L1(R+) . ‖f ′‖L1(R+).

About the proofs

Proof of Theorem 3.

I = I(x) =

∫ ∞π2x

f(t) sinxt dt.

We prove that

I(x) = H0Tf ′(x) +G(x),

‖G‖L1(R+) . ‖f ′‖L1(R+).

About the proofs

Proof of Theorem 3.

I = I(x) =

∫ ∞π2x

f(t) sinxt dt.

We prove that

I(x) = H0Tf ′(x) +G(x),

‖G‖L1(R+) . ‖f ′‖L1(R+).

At first sight, Theorem 2 does not seem to be a result at all, at mosta technical reformulation of the definition.

This could be so but not after the appearance of the analysis of Q byJohnson-Warner.

Indeed, the well-known extension of Hardy’s inequality

|g(x)||x|

dx . ‖g‖H1(R)

implies

H1(R) ⊆ Q ⊆ L10(R),

where the latter is the subspace of g in L1(R) which satisfy thecancelation property.

|g(x)||x|

dx . ‖g‖H1(R)

implies

H1(R) ⊆ Q ⊆ L10(R),

|g(x)||x|

dx . ‖g‖H1(R)

implies

H1(R) ⊆ Q ⊆ L10(R),

|g(x)||x|

dx . ‖g‖H1(R)

implies

H1(R) ⊆ Q ⊆ L10(R),

|g(x)||x|

dx . ‖g‖H1(R)

implies

H1(R) ⊆ Q ⊆ L10(R),

Discussion

It is worth noting that this embedding immediately proves the cosinepart from Theorem 1.

The initial proof is essentially more complicated.

It is doubtful that Q (or Q0) may be defined in terms of g itselfrather than its Fourier transform, therefore it is of interest to findcertain proper subspaces of Q0 wider than H1 belonging to which iseasily verifiable.

Back to Theorem 3, let us analyze I. On the one hand, we have

∫ ∞0|I(x)| dx =

∫ ∞0|H0Tf ′(x)| dx+O(‖f ′‖H1

0 (R+)).

On the other hand, it is proved by L in 1993 that

∫ ∞0|I(x)| dx = O(‖f ′‖H1

0 (R+)).

Discussion

∫ ∞0|I(x)| dx =

∫ ∞0|H0Tf ′(x)| dx+O(‖f ′‖H1

0 (R+)).

∫ ∞0|I(x)| dx = O(‖f ′‖H1

0 (R+)).

Discussion

∫ ∞0|I(x)| dx =

∫ ∞0|H0Tf ′(x)| dx+O(‖f ′‖H1

0 (R+)).

∫ ∞0|I(x)| dx = O(‖f ′‖H1

0 (R+)).

Discussion

∫ ∞0|I(x)| dx =

∫ ∞0|H0Tf ′(x)| dx+O(‖f ′‖H1

0 (R+)).

∫ ∞0|I(x)| dx = O(‖f ′‖H1

0 (R+)).

Discussion

∫ ∞0|I(x)| dx =

∫ ∞0|H0Tf ′(x)| dx+O(‖f ′‖H1

0 (R+)).

∫ ∞0|I(x)| dx = O(‖f ′‖H1

0 (R+)).

Discussion

This leads to

Proposition. If g is an integrable odd function, then

‖H0Tg‖L1(R+) . ‖g‖H10 (R+).

The above proof of this proposition looks ”artificial”. A direct proof,preferable simple enough will be very desirable.

In any case, this implies an updated chain of embeddings

H10 (R+) ⊆ H1

Q(R+) ⊆ Q0 ⊆ L10(R+).

It is very interesting to figure out which of these embeddings areproper. Correspondingly, intermediate spaces are of interest, boththeoretical and practical.

Discussion

This leads to

‖H0Tg‖L1(R+) . ‖g‖H10 (R+).

H10 (R+) ⊆ H1

Q(R+) ⊆ Q0 ⊆ L10(R+).

Discussion

This leads to

‖H0Tg‖L1(R+) . ‖g‖H10 (R+).

H10 (R+) ⊆ H1

Q(R+) ⊆ Q0 ⊆ L10(R+).

Discussion

This leads to

‖H0Tg‖L1(R+) . ‖g‖H10 (R+).

H10 (R+) ⊆ H1

Q(R+) ⊆ Q0 ⊆ L10(R+).

Discussion

This leads to

‖H0Tg‖L1(R+) . ‖g‖H10 (R+).

H10 (R+) ⊆ H1

Q(R+) ⊆ Q0 ⊆ L10(R+).

An old problem

In 50-s (Kahane, Izumi-Tsuchikura, Boas, etc.), the following problem inFourier Analysis attracted much attention:

Let {ak}∞k=0 be the sequence of the Fourier coefficients of theabsolutely convergent sine (cosine) Fourier series of a functionf : T = [−π, π)→ C, that is

∑|ak| <∞. Under which conditions on

{ak} the re-expansion of f(t) (f(t)− f(0), respectively) in the cosine(sine) Fourier series will also be absolutely convergent?

The obtained condition is quite simple and is the same in both cases:

∞∑k=1

|ak| ln(k + 1) <∞.

An old problem

∞∑k=1

|ak| ln(k + 1) <∞.

An old problem

∞∑k=1

|ak| ln(k + 1) <∞.

Certain relations between the Fourier and Hilbert transforms

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