Fourier transform versus Hilbert transform

Elijah Liflyand

Bar-Ilan University

June, 2013

Elijah Liflyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 1 / 19

The Hilbert transform

The Hilbert transform of an integrable function g:

Hg(x) = 1

x− tdt,

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

x− tdt,

δ→0+

∫|t−x|>δ .

∫Rg(t) dt = 0.

Hg(x) = 1

x− tdt,

δ→0+

∫|t−x|>δ .

∫Rg(t) dt = 0.

Hg(x) = 1

x− tdt,

δ→0+

∫|t−x|>δ .

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.

Hg(x) = 1

x− tdt,

δ→0+

∫|t−x|>δ .

∫Rg(t) dt = 0.

Hg(x) = 1

x− tdt,

δ→0+

∫|t−x|>δ .

∫Rg(t) dt = 0.

The Hilbert transform: continuation

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; see also”Functional Analysis” by Stein-Shakarchi.

When in the definition of the Hilbert transform the function g is odd,we will denote this transform by Ho,and it is equal to

Hog(x) =2

∫ ∞0

x2 − t2dt.

If this transform is integrable, we will denote the corresponding Hardyspace by H1

0 (R).

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

a counterexample is explicitly given by L-Tikhonov; see also”Functional Analysis” by Stein-Shakarchi.

Hog(x) =2

∫ ∞0

x2 − t2dt.

0 (R).

Hog(x) =2

∫ ∞0

x2 − t2dt.

0 (R).

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

and it is equal to

Hog(x) =2

∫ ∞0

x2 − t2dt.

0 (R).

Hog(x) =2

∫ ∞0

x2 − t2dt.

0 (R).

Hog(x) =2

∫ ∞0

x2 − t2dt.

0 (R).

Correspondingly, when in the definition of the Hilbert transform thefunction g is even,

we will denote this transform by He,

and it is equal to

Heg(x) =2

∫ ∞0

x2 − t2dt.

Less important but sometimes works.

and it is equal to

Heg(x) =2

∫ ∞0

x2 − t2dt.

and it is equal to

Heg(x) =2

∫ ∞0

x2 − t2dt.

and it is equal to

Heg(x) =2

∫ ∞0

x2 − t2dt.

1. An old problem: re-expansion

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) <∞.

∞∑k=1

|ak| ln(k + 1) <∞.

∞∑k=1

|ak| ln(k + 1) <∞.

Similar problem for Fourier transforms

We study a similar problem for Fourier transforms defined on R+ = [0,∞).

Let∫∞0 |Fc(x)| dx <∞,

and hence

f(t) =1

∫ ∞0

Fc(x) cos tx dx,

or, alternatively,

∫ ∞0|Fs(x)| dx <∞

and hence

f(t) =1

∫ ∞0

Fs(x) sin tx dx.

Let∫∞0 |Fc(x)| dx <∞,

and hence

f(t) =1

∫ ∞0

Fc(x) cos tx dx,

or, alternatively,

∫ ∞0|Fs(x)| dx <∞

and hence

f(t) =1

∫ ∞0

Fs(x) sin tx dx.

Let∫∞0 |Fc(x)| dx <∞,

and hence

f(t) =1

∫ ∞0

Fc(x) cos tx dx,

or, alternatively,

∫ ∞0|Fs(x)| dx <∞

and hence

f(t) =1

∫ ∞0

Fs(x) sin tx dx.

Let∫∞0 |Fc(x)| dx <∞,

and hence

f(t) =1

∫ ∞0

Fc(x) cos tx dx,

or, alternatively,

∫ ∞0|Fs(x)| dx <∞

and hence

f(t) =1

∫ ∞0

Fs(x) sin tx dx.

Let∫∞0 |Fc(x)| dx <∞,

and hence

f(t) =1

∫ ∞0

Fc(x) cos tx dx,

or, alternatively,

∫ ∞0|Fs(x)| dx <∞

and hence

f(t) =1

∫ ∞0

Fs(x) sin tx dx.

Similar problem for Fourier transforms: continuation

Under which (additional) conditions on Fc we get the integrability ofFs,or, in the alternative case,

under which (additional) conditions on Fs we get the integrability ofFc?

Theorem 1.1. In order that the re-expansion Fs of f with theintegrable cosine Fourier transform Fc be integrable, it is necessaryand sufficient that its Hilbert transform HFc(x) be integrable.

Similarly, in order that the re-expansion Fc of f with the integrablesine Fourier transform Fs be integrable, it is necessary and sufficientthat its Hilbert transform HFs(x) be integrable.

About the proof

In fact, we proved

Fs(x) = HFc(x)

Fc(x) = −HFs(x)

by means of

Theorem A. If|f(t)|1 + |t|

is integrable on R, then the (C, 1) means

∫ ∞−∞

f(x+ t)

t− sinNt

converge to the Hilbert transform Hf(x) almost everywhere asN →∞.

About the proof

In fact, we proved

Fs(x) = HFc(x)

Fc(x) = −HFs(x)

by means of

∫ ∞−∞

f(x+ t)

t− sinNt

About the proof

In fact, we proved

Fs(x) = HFc(x)

Fc(x) = −HFs(x)

by means of

∫ ∞−∞

f(x+ t)

t− sinNt

About the proof: continuation

These formulas are known (see, e.g., the monograph by King) for,say, square integrable functions.

But that proof is equivalent to (Carleson’s solution of) Lusin’sconjecture.

In our L1 setting this is by no means applicable. And, indeed, ourproof is different and rests on the less restrictive Theorem A.

This corresponds to what E.M. Dyn’kin wrote in his well-knownsurvey on singular integrals: ”In fact, the theory of singular integralsis a technical subject where ideas cannot be separated from thetechniques.”

Analyzing the proofs for series, one can see that in fact the results aresimilar to ours,

that is, can also be given in terms of the (discrete) Hilbert transform.

In that case the above condition with logarithm is simply a sufficientcondition for the summability of the discrete Hilbert transform.

A similar analog for functions cannot be the only sufficient conditionfor the integrability of the Hilbert transform.

Indeed, known counter-examples stand up to the multiplication bylogarithm.

We give sufficient conditions for the integrability of the Hilberttransforms in the spirit of that for sequences supplied by someadditional properties, in addition to few known.

2. A Hardy-Littlewood theorem

The following result is due to Hardy and Littlewood:If a (periodic) function f and its conjugate f are both of boundedvariation, their Fourier series converge absolutely.

We generalize the Hardy-Littlewood theorem (joint work with U.Stadtmuller) to functions on the real axis, hence the absoluteconvergence of the Fourier series should be replaced by theintegrability of the Fourier transform.

Since a function f of bounded variation may be not integrable, itsHilbert transform, a usual substitute for the conjugate function, maynot exist. One has to use the modified Hilbert transform

f(x) = (P.V.)1

∫Rf(t)

x− t+

1 + t2

well adjusted for bounded functions.

As a singular integral, it behaves like the usual Hilbert transform; theadditional term in the integral makes it to be well defined near infinity.

f(x) = (P.V.)1

∫Rf(t)

x− t+

1 + t2

f(x) = (P.V.)1

∫Rf(t)

x− t+

1 + t2

f(x) = (P.V.)1

∫Rf(t)

x− t+

1 + t2

A generalized Hardy-Littlewood theorem

Theorem 2.1. Let f be a function of bounded variation and vanish atinfinity: lim

|t|→∞f(t) = 0. If its conjugate f is also of bounded variation,

then the Fourier transforms of both functions are integrable on R.

Thew proof contains two main ingredients:

one of them isLemma. Under the assumptions of the theorem, we have at almostevery x

dxf(x) = Hf ′(x).

Interesting that the initial Hardy-Littlewood theorem is a partial caseof this extension, when the function is taken to be of compactsupport.

dxf(x) = Hf ′(x).

Hardy type inequalities

The second ingredient (and one of the main tools in what follows): thewell-known extension of Hardy’s inequality:∫

|g(x)||x|

dx . ‖g‖H1(R)

This inequality shows that for the Fourier transform of a functionfrom the Hardy space we have more than just the Riemann-Lebesguelemma for an integrable function and its Fourier transform.

There holdsProposition. Let

∫R|g(x)||x| dx be finite. Then g is the derivative of a

function of bounded variation f which is locally absolutelycontinuous, lim

|t|→∞f(t) = 0, and its Fourier transform is integrable.

The obtained estimates lead to refinement of Hardy’s inequality incertain particular cases.

All above is true in the discrete case, for sequences.

|g(x)||x|

dx . ‖g‖H1(R)

|g(x)||x|

dx . ‖g‖H1(R)

|g(x)||x|

dx . ‖g‖H1(R)

3. The Fourier transform of a function of boundedvariation

The possibility of existence (or non-existence) of the widest space forthe integrability of the Fourier transform of a function of boundedvariation is of considerable interest.

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

Q = {g : g ∈ L1(R),∫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).The space Qo of the odd functions from Q is of special importance:

Qo = {g : g ∈ L1(R), g(−t) = −g(t),∫ ∞0

|gs(x)|x

dx <∞};

such functions naturally have mean zero.

The possibility of existence (or non-existence) of the widest space forthe integrability of the Fourier transform of a function of boundedvariation is of considerable interest.In fact, it has in essence been introduced (for different purposes) byJohnson-Warner in 2010 as

Q = {g : g ∈ L1(R),∫R

|g(x)||x|

dx <∞}.

Qo = {g : g ∈ L1(R), g(−t) = −g(t),∫ ∞0

|gs(x)|x

dx <∞};

Q = {g : g ∈ L1(R),∫R

|g(x)||x|

dx <∞}.

space and ideal in L1(R).

The space Qo of the odd functions from Q is of special importance:

Qo = {g : g ∈ L1(R), g(−t) = −g(t),∫ ∞0

|gs(x)|x

dx <∞};

Q = {g : g ∈ L1(R),∫R

|g(x)||x|

dx <∞}.

Qo = {g : g ∈ L1(R), g(−t) = −g(t),∫ ∞0

|gs(x)|x

dx <∞};

such functions naturally have mean zero.Elijah Liflyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 14 / 19

The Fourier transform of a function of bounded variation:continuation

An even counterpart of Qo is

Qe = {g : g ∈ L1(R), g(−t) = g(t),

∫ ∞0

|gc(x)|x

dx <∞}.

This makes sense only if∫∞0 g(t) dt = 0.

Theorem 3.1. Let f : R+ → C be locally absolutely continuous, ofbounded variation and lim

t→∞f(t) = 0.

a) Then the cosine Fourier transform of f is Lebesgue integrable onR+ if and only if f ′ ∈ Q.b) The sine Fourier transform of f is Lebesgue integrable on R+ ifand only if f ′ ∈ Q.

In fact, in a) and b) the conditions are given in terms of differentsubspaces of Q, namely Qo and Qe, respectively.

Qe = {g : g ∈ L1(R), g(−t) = g(t),

∫ ∞0

|gc(x)|x

dx <∞}.

t→∞f(t) = 0.

Qe = {g : g ∈ L1(R), g(−t) = g(t),

∫ ∞0

|gc(x)|x

dx <∞}.

t→∞f(t) = 0.

Qe = {g : g ∈ L1(R), g(−t) = g(t),

∫ ∞0

|gc(x)|x

dx <∞}.

t→∞f(t) = 0.

The situation is more delicate with the sine Fourier transform,

where a sort of asymptotic relation can be obtained.

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

t→∞f(t) = 0. Then for any x > 0

Fs(x) =

∫ ∞0

f(t) sinxt dt =1

xf( π2x

)−HoTf ′(x) +G(x),

‖G‖L1(R+) . ‖f ′‖L1(R+).

Fs(x) =

∫ ∞0

f(t) sinxt dt =1

xf( π2x

)−HoTf ′(x) +G(x),

‖G‖L1(R+) . ‖f ′‖L1(R+).

Fs(x) =

∫ ∞0

f(t) sinxt dt =1

xf( π2x

)−HoTf ′(x) +G(x),

‖G‖L1(R+) . ‖f ′‖L1(R+).

Fs(x) =

∫ ∞0

f(t) sinxt dt =1

xf( π2x

)−HoTf ′(x) +G(x),

‖G‖L1(R+) . ‖f ′‖L1(R+).

Fs(x) =

∫ ∞0

f(t) sinxt dt =1

xf( π2x

)−HoTf ′(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 Qo functions g with integrable HoTg.Corollary. Let a function f satisfy the assumptions of Theorem 3.2and such that f ′ ∈ H1

Q(R+). Then

Fs(x) =1

xf( π2x

)+G(x),

‖G‖L1(R+) . ‖f ′‖H1Q(R+).

Technically, this is an obvious corollary of Theorem 3.2.

A Hardy type space

which consists of Qo functions g with integrable HoTg.

Corollary. Let a function f satisfy the assumptions of Theorem 3.2and such that f ′ ∈ H1

Q(R+). Then

Fs(x) =1

xf( π2x

)+G(x),

‖G‖L1(R+) . ‖f ′‖H1Q(R+).

A Hardy type space

Q(R+). Then

Fs(x) =1

xf( π2x

)+G(x),

‖G‖L1(R+) . ‖f ′‖H1Q(R+).

A Hardy type space

Q(R+). Then

Fs(x) =1

xf( π2x

)+G(x),

‖G‖L1(R+) . ‖f ′‖H1Q(R+).

A Hardy type space

Q(R+). Then

Fs(x) =1

xf( π2x

)+G(x),

‖G‖L1(R+) . ‖f ′‖H1Q(R+).

At first sight, Theorem 3.1 does not seem to be a result at all, atmost a technical reformulation of the definition.

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

Indeed, 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 (mean zero).

|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 doubtful that Q (or Qo) may be defined in terms of g itselfrather than its Fourier transform,

therefore it is of interest to find certain proper subspaces of Qo widerthan H1 belonging to which is easier to be checked.Proposition. If g is an integrable odd function, then

‖HoTg‖L1(R+) . ‖g‖H10 (R+).

A direct, molecular proof exists. In any case, this implies an updatedchain of embeddings

H10 (R+) ⊆ H1

Q(R+) ⊆ Qo ⊆ L10(R+).

It is very interesting to figure out which of these embeddings areproper. At least, H1

0 (R+) is strictly smaller than H1Q.

Correspondingly, intermediate spaces are of interest, both theoreticaland practical.

Discussion

It is doubtful that Q (or Qo) may be defined in terms of g itselfrather than its Fourier transform,therefore it is of interest to find certain proper subspaces of Qo widerthan H1 belonging to which is easier to be checked.

Proposition. If g is an integrable odd function, then

‖HoTg‖L1(R+) . ‖g‖H10 (R+).

H10 (R+) ⊆ H1

Q(R+) ⊆ Qo ⊆ L10(R+).

Discussion

It is doubtful that Q (or Qo) may be defined in terms of g itselfrather than its Fourier transform,therefore it is of interest to find certain proper subspaces of Qo widerthan H1 belonging to which is easier to be checked.Proposition. If g is an integrable odd function, then

‖HoTg‖L1(R+) . ‖g‖H10 (R+).

H10 (R+) ⊆ H1

Q(R+) ⊆ Qo ⊆ L10(R+).

Discussion

‖HoTg‖L1(R+) . ‖g‖H10 (R+).

H10 (R+) ⊆ H1

Q(R+) ⊆ Qo ⊆ L10(R+).

Discussion

‖HoTg‖L1(R+) . ‖g‖H10 (R+).

H10 (R+) ⊆ H1

Q(R+) ⊆ Qo ⊆ L10(R+).

Discussion

‖HoTg‖L1(R+) . ‖g‖H10 (R+).

H10 (R+) ⊆ H1

Q(R+) ⊆ Qo ⊆ L10(R+).

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