E. Liflyand Fourier Transform Versus Hilbert Transform

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 where the integral is understood in the improper (principal value) sense, as lim R . δ→0+ |t−x|>δ It is not necessarily integrable, 1 and when it is, we say that g is in the (real) Hardy space H (R). 1 If g ∈ H (R), then it satisﬁes the cancelation property (has mean zero)

Z g(t) dt = 0. R It was apparently ﬁrst mentioned by Kober.

The Hilbert transform

The Hilbert transform of an integrable function g:

1 Z g(t) Hg(x) = dt, π R x − t

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 2 / 19 It is not necessarily integrable, 1 and when it is, we say that g is in the (real) Hardy space H (R). 1 If g ∈ H (R), then it satisﬁes the cancelation property (has mean zero)

Z g(t) dt = 0. R It was apparently ﬁrst mentioned by Kober.

The Hilbert transform

The Hilbert transform of an integrable function g:

1 Z g(t) Hg(x) = dt, π R x − t where the integral is understood in the improper (principal value) sense, as lim R . δ→0+ |t−x|>δ

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 2 / 19 1 and when it is, we say that g is in the (real) Hardy space H (R). 1 If g ∈ H (R), then it satisﬁes the cancelation property (has mean zero)

Z g(t) dt = 0. R It was apparently ﬁrst mentioned by Kober.

The Hilbert transform

The Hilbert transform of an integrable function g:

1 Z g(t) Hg(x) = dt, π R x − t where the integral is understood in the improper (principal value) sense, as lim R . δ→0+ |t−x|>δ It is not necessarily integrable,

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 2 / 19 1 If g ∈ H (R), then it satisﬁes the cancelation property (has mean zero)

Z g(t) dt = 0. R It was apparently ﬁrst mentioned by Kober.

The Hilbert transform

The Hilbert transform of an integrable function g:

1 Z g(t) Hg(x) = dt, π R x − t where the integral is understood in the improper (principal value) sense, as lim R . δ→0+ |t−x|>δ It is not necessarily integrable, 1 and when it is, we say that g is in the (real) Hardy space H (R).

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 2 / 19 It was apparently ﬁrst mentioned by Kober.

The Hilbert transform

The Hilbert transform of an integrable function g:

Z g(t) dt = 0. R

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 2 / 19 The Hilbert transform

The Hilbert transform of an integrable function g:

Z g(t) dt = 0. R It was apparently ﬁrst mentioned by Kober.

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 2 / 19 1 However, not every odd integrable function belongs to H (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

2 Z ∞ tg(t) Hog(x) = 2 2 dt. π 0 x − t If this transform is integrable, we will denote the corresponding Hardy 1 space by H0 (R).

The Hilbert transform: continuation

An odd function always has mean zero.

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 3 / 19 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

2 Z ∞ tg(t) Hog(x) = 2 2 dt. π 0 x − t If this transform is integrable, we will denote the corresponding Hardy 1 space by H0 (R).

The Hilbert transform: continuation

An odd function always has mean zero. 1 However, not every odd integrable function belongs to H (R),

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 3 / 19 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

2 Z ∞ tg(t) Hog(x) = 2 2 dt. π 0 x − t If this transform is integrable, we will denote the corresponding Hardy 1 space by H0 (R).

The Hilbert transform: continuation

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 3 / 19 and it is equal to

2 Z ∞ tg(t) Hog(x) = 2 2 dt. π 0 x − t If this transform is integrable, we will denote the corresponding Hardy 1 space by H0 (R).

The Hilbert transform: continuation

An odd function always has mean zero. 1 However, not every odd integrable function belongs to H (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,

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 3 / 19 If this transform is integrable, we will denote the corresponding Hardy 1 space by H0 (R).

The Hilbert transform: continuation

2 Z ∞ tg(t) Hog(x) = 2 2 dt. π 0 x − t

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 3 / 19 The Hilbert transform: continuation

2 Z ∞ tg(t) Hog(x) = 2 2 dt. π 0 x − t If this transform is integrable, we will denote the corresponding Hardy 1 space by H0 (R).

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 3 / 19 we will denote this transform by He,

and it is equal to

2 Z ∞ xg(t) Heg(x) = 2 2 dt. π 0 x − t Less important but sometimes works.

The Hilbert transform: continuation

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 4 / 19 and it is equal to

2 Z ∞ xg(t) Heg(x) = 2 2 dt. π 0 x − t Less important but sometimes works.

The Hilbert transform: continuation

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

we will denote this transform by He,

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 4 / 19 Less important but sometimes works.

The Hilbert transform: continuation

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

we will denote this transform by He,

and it is equal to

2 Z ∞ xg(t) Heg(x) = 2 2 dt. π 0 x − t

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 4 / 19 The Hilbert transform: continuation

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

we will denote this transform by He,

and it is equal to

2 Z ∞ xg(t) Heg(x) = 2 2 dt. π 0 x − t Less important but sometimes works.

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 4 / 19 ∞ Let {ak}k=0 be the sequence of the Fourier coeﬃcients of the absolutely convergent sine (cosine) Fourier series of a function P f : 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:

∞ X |ak| ln(k + 1) < ∞. k=1

1. An old problem: re-expansion

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 5 / 19 The obtained condition is quite simple and is the same in both cases:

∞ X |ak| ln(k + 1) < ∞. k=1

1. An old problem: re-expansion

In 50-s (Kahane, Izumi-Tsuchikura, Boas, etc.), the following problem in Fourier Analysis attracted much attention: ∞ Let {ak}k=0 be the sequence of the Fourier coeﬃcients of the absolutely convergent sine (cosine) Fourier series of a function P f : 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?

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 5 / 19 1. An old problem: re-expansion

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

∞ X |ak| ln(k + 1) < ∞. k=1

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 5 / 19 R ∞ Let 0 |Fc(x)| dx < ∞, and hence

1 Z ∞ f(t) = Fc(x) cos tx dx, π 0 or, alternatively,

Z ∞ |Fs(x)| dx < ∞ 0 and hence

1 Z ∞ f(t) = Fs(x) sin tx dx. π 0

Similar problem for Fourier transforms

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 6 / 19 and hence

1 Z ∞ f(t) = Fc(x) cos tx dx, π 0 or, alternatively,

Z ∞ |Fs(x)| dx < ∞ 0 and hence

1 Z ∞ f(t) = Fs(x) sin tx dx. π 0

Similar problem for Fourier transforms

We study a similar problem for Fourier transforms deﬁned on R+ = [0, ∞). R ∞ Let 0 |Fc(x)| dx < ∞,

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 6 / 19 or, alternatively,

Z ∞ |Fs(x)| dx < ∞ 0 and hence

1 Z ∞ f(t) = Fs(x) sin tx dx. π 0

Similar problem for Fourier transforms

We study a similar problem for Fourier transforms deﬁned on R+ = [0, ∞). R ∞ Let 0 |Fc(x)| dx < ∞, and hence

1 Z ∞ f(t) = Fc(x) cos tx dx, π 0

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 6 / 19 and hence

1 Z ∞ f(t) = Fs(x) sin tx dx. π 0

Similar problem for Fourier transforms

We study a similar problem for Fourier transforms deﬁned on R+ = [0, ∞). R ∞ Let 0 |Fc(x)| dx < ∞, and hence

1 Z ∞ f(t) = Fc(x) cos tx dx, π 0 or, alternatively,

Z ∞ |Fs(x)| dx < ∞ 0

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 6 / 19 Similar problem for Fourier transforms

We study a similar problem for Fourier transforms deﬁned on R+ = [0, ∞). R ∞ Let 0 |Fc(x)| dx < ∞, and hence

1 Z ∞ f(t) = Fc(x) cos tx dx, π 0 or, alternatively,

Z ∞ |Fs(x)| dx < ∞ 0 and hence

1 Z ∞ f(t) = Fs(x) sin tx dx. π 0

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 6 / 19 under which (additional) conditions on Fs we get the integrability of Fc?

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

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

Similar problem for Fourier transforms: continuation

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 7 / 19 Theorem 1.1. In order that the re-expansion Fs of f with the integrable cosine Fourier transform Fc be integrable, it is necessary and suﬃcient that its Hilbert transform HFc(x) be integrable.

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

Similar problem for Fourier transforms: continuation

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

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 7 / 19 Similarly, in order that the re-expansion Fc of f with the integrable sine Fourier transform Fs be integrable, it is necessary and suﬃcient that its Hilbert transform HFs(x) be integrable.

Similar problem for Fourier transforms: continuation

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

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

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 7 / 19 Similar problem for Fourier transforms: continuation

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

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

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

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 7 / 19 and

Fc(x) = −HFs(x) by means of |f(t)| Theorem A. If is integrable on , then the (C, 1) means 1 + |t| R

1 Z ∞ 1 sin Nt − f(x + t) − 2 dt π −∞ t Nt converge to the Hilbert transform Hf(x) almost everywhere as N → ∞.

About the proof

In fact, we proved

Fs(x) = HFc(x)

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 8 / 19 by means of |f(t)| Theorem A. If is integrable on , then the (C, 1) means 1 + |t| R

1 Z ∞ 1 sin Nt − f(x + t) − 2 dt π −∞ t Nt converge to the Hilbert transform Hf(x) almost everywhere as N → ∞.

About the proof

In fact, we proved

Fs(x) = HFc(x) and

Fc(x) = −HFs(x)

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 8 / 19 About the proof

In fact, we proved

Fs(x) = HFc(x) and

Fc(x) = −HFs(x) by means of |f(t)| Theorem A. If is integrable on , then the (C, 1) means 1 + |t| R

1 Z ∞ 1 sin Nt − f(x + t) − 2 dt π −∞ t Nt converge to the Hilbert transform Hf(x) almost everywhere as N → ∞.

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 8 / 19 But that proof is equivalent to (Carleson’s solution of) Lusin’s conjecture.

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

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

About the proof: continuation

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 9 / 19 In our L1 setting this is by no means applicable. And, indeed, our proof is diﬀerent and rests on the less restrictive Theorem A.

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’s conjecture.

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 9 / 19 This corresponds to what E.M. Dyn’kin wrote in his well-known survey on singular integrals: ”In fact, the theory of singular integrals is a technical subject where ideas cannot be separated from the techniques.”

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’s conjecture.

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 9 / 19 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’s conjecture.

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 9 / 19 that is, can also be given in terms of the (discrete) Hilbert transform.

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

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

Indeed, known counter-examples stand up to the multiplication by logarithm. We give suﬃcient conditions for the integrability of the Hilbert transforms in the spirit of that for sequences supplied by some additional properties, in addition to few known.

About the proof: continuation

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 10 / 19 In that case the above condition with logarithm is simply a suﬃcient condition for the summability of the discrete Hilbert transform.

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

About the proof: continuation

Analyzing the proofs for series, one can see that in fact the results are similar to ours, that is, can also be given in terms of the (discrete) Hilbert transform.

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 10 / 19 A similar analog for functions cannot be the only suﬃcient condition for the integrability of the Hilbert transform.

About the proof: continuation

Analyzing the proofs for series, one can see that in fact the results are similar 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 suﬃcient condition for the summability of the discrete Hilbert transform.

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 10 / 19 Indeed, known counter-examples stand up to the multiplication by logarithm. We give suﬃcient conditions for the integrability of the Hilbert transforms in the spirit of that for sequences supplied by some additional properties, in addition to few known.

About the proof: continuation

Analyzing the proofs for series, one can see that in fact the results are similar 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 suﬃcient condition for the summability of the discrete Hilbert transform.

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 10 / 19 We give suﬃcient conditions for the integrability of the Hilbert transforms in the spirit of that for sequences supplied by some additional properties, in addition to few known.

About the proof: continuation

Analyzing the proofs for series, one can see that in fact the results are similar 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 suﬃcient condition for the summability of the discrete Hilbert transform.

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

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 10 / 19 About the proof: continuation

Analyzing the proofs for series, one can see that in fact the results are similar 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 suﬃcient condition for the summability of the discrete Hilbert transform.

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

Elijah Liflyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 10 / 19 We generalize the Hardy-Littlewood theorem (joint work withU. Stadtmüller) to functions on the real axis, hence the absolute convergence of the Fourier series should be replaced by the integrability of the Fourier transform. Since a function f of bounded variation may be not integrable, its Hilbert transform, a usual substitute for the conjugate function, may not exist. One has to use the modified Hilbert transform

1 Z 1 t fe(x) = (P.V.) f(t) + 2 dt π R x − t 1 + t well adjusted for bounded functions. As a singular integral, it behaves like the usual Hilbert transform; the additional term in the integral makes it to be well deﬁned near inﬁnity.

2. A Hardy-Littlewood theorem

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 11 / 19 Since a function f of bounded variation may be not integrable, its Hilbert transform, a usual substitute for the conjugate function, may not exist. One has to use the modiﬁed Hilbert transform

2. A Hardy-Littlewood theorem

Elijah Liflyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 11 / 19 As a singular integral, it behaves like the usual Hilbert transform; the additional term in the integral makes it to be well defined near infinity.

2. A Hardy-Littlewood theorem

The following result is due to Hardy and Littlewood: If a (periodic) function f and its conjugate fe are both of bounded variation, their Fourier series converge absolutely. We generalize the Hardy-Littlewood theorem (joint work withU. Stadtm¨uller) to functions on the real axis, hence the absolute convergence of the Fourier series should be replaced by the integrability of the Fourier transform. Since a function f of bounded variation may be not integrable, its Hilbert transform, a usual substitute for the conjugate function, may not exist. One has to use the modiﬁed Hilbert transform

1 Z 1 t fe(x) = (P.V.) f(t) + 2 dt π R x − t 1 + t well adjusted for bounded functions.

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 11 / 19 2. A Hardy-Littlewood theorem

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 11 / 19 Thew proof contains two main ingredients: one of them is Lemma. Under the assumptions of the theorem, we have at almost every x

d fe(x) = Hf 0(x). dx Interesting that the initial Hardy-Littlewood theorem is a partial case of this extension, when the function is taken to be of compact support.

A generalized Hardy-Littlewood theorem

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 12 / 19 one of them is Lemma. Under the assumptions of the theorem, we have at almost every x

d fe(x) = Hf 0(x). dx Interesting that the initial Hardy-Littlewood theorem is a partial case of this extension, when the function is taken to be of compact support.

A generalized Hardy-Littlewood theorem

Theorem 2.1. Let f be a function of bounded variation and vanish at inﬁnity: lim f(t) = 0. If its conjugate fe is also of bounded variation, |t|→∞ then the Fourier transforms of both functions are integrable on R. Thew proof contains two main ingredients:

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 12 / 19 Interesting that the initial Hardy-Littlewood theorem is a partial case of this extension, when the function is taken to be of compact support.

A generalized Hardy-Littlewood theorem

d fe(x) = Hf 0(x). dx

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 12 / 19 A generalized Hardy-Littlewood theorem

d fe(x) = Hf 0(x). dx Interesting that the initial Hardy-Littlewood theorem is a partial case of this extension, when the function is taken to be of compact support.

Elijah Liflyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 12 / 19 There holds Proposition. Let R |gb(x)| dx be finite. Then g is the derivative of a R |x| function of bounded variation f which is locally absolutely continuous, lim f(t) = 0, and its Fourier transform is integrable. |t|→∞ The obtained estimates lead to refinement of Hardy’s inequality in certain particular cases. All above is true in the discrete case, for sequences.

Hardy type inequalities

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 13 / 19 The obtained estimates lead to reﬁnement of Hardy’s inequality in certain particular cases. All above is true in the discrete case, for sequences.

Hardy type inequalities

The second ingredient (and one of the main tools in what follows): the well-known extension of Hardy’s inequality: Z |gb(x)| dx . kgkH1(R) R |x| This inequality shows that for the Fourier transform of a function from the Hardy space we have more than just the Riemann-Lebesgue lemma for an integrable function and its Fourier transform. There holds Proposition. Let R |gb(x)| dx be ﬁnite. Then g is the derivative of a R |x| function of bounded variation f which is locally absolutely continuous, lim f(t) = 0, and its Fourier transform is integrable. |t|→∞

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 13 / 19 All above is true in the discrete case, for sequences.

Hardy type inequalities

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 13 / 19 Hardy type inequalities

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 13 / 19 In fact, it has in essence been introduced (for diﬀerent purposes) by Johnson-Warner in 2010 as

Z 1 |g(x)| Q = {g : g ∈ L (R), b dx < ∞}. R |x| R |g(x)| With the obvious norm kgk 1 + b dx it is a Banach L (R) R |x| 1 space and ideal in L (R). The space Qo of the odd functions from Q is of special importance:

Z ∞ 1 |gbs(x)| Qo = {g : g ∈ L (R), g(−t) = −g(t), dx < ∞}; 0 x such functions naturally have mean zero.

3. The Fourier transform of a function of bounded variation

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 14 / 19 R |g(x)| With the obvious norm kgk 1 + b dx it is a Banach L (R) R |x| 1 space and ideal in L (R). The space Qo of the odd functions from Q is of special importance:

Z ∞ 1 |gbs(x)| Qo = {g : g ∈ L (R), g(−t) = −g(t), dx < ∞}; 0 x such functions naturally have mean zero.

3. The Fourier transform of a function of bounded variation

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

Z 1 |g(x)| Q = {g : g ∈ L (R), b dx < ∞}. R |x|

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 14 / 19 The space Qo of the odd functions from Q is of special importance:

Z ∞ 1 |gbs(x)| Qo = {g : g ∈ L (R), g(−t) = −g(t), dx < ∞}; 0 x such functions naturally have mean zero.

3. The Fourier transform of a function of bounded variation

Z 1 |g(x)| Q = {g : g ∈ L (R), b dx < ∞}. R |x| R |g(x)| With the obvious norm kgk 1 + b dx it is a Banach L (R) R |x| 1 space and ideal in L (R).

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 14 / 19 3. The Fourier transform of a function of bounded variation

Z ∞ 1 |gbs(x)| Qo = {g : g ∈ L (R), g(−t) = −g(t), dx < ∞}; 0 x

Elijah Liﬂyandsuch (Bar-Ilan functions University) naturallyFourier have and mean Hilbert transforms zero. June, 2013 14 / 19 R ∞ This makes sense only if 0 g(t) dt = 0. Theorem 3.1. Let f : R+ → C be locally absolutely continuous, of bounded variation and lim f(t) = 0. t→∞ a) Then the cosine Fourier transform of f is Lebesgue integrable on 0 R+ if and only if f ∈ Q. b) The sine Fourier transform of f is Lebesgue integrable on R+ if and only if f 0 ∈ Q. In fact, in a) and b) the conditions are given in terms of diﬀerent subspaces of Q, namely Qo and Qe, respectively.

The Fourier transform of a function of bounded variation: continuation

An even counterpart of Qo is

Z ∞ 1 |gbc(x)| Qe = {g : g ∈ L (R), g(−t) = g(t), dx < ∞}. 0 x

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 15 / 19 Theorem 3.1. Let f : R+ → C be locally absolutely continuous, of bounded variation and lim f(t) = 0. t→∞ a) Then the cosine Fourier transform of f is Lebesgue integrable on 0 R+ if and only if f ∈ Q. b) The sine Fourier transform of f is Lebesgue integrable on R+ if and only if f 0 ∈ Q. In fact, in a) and b) the conditions are given in terms of diﬀerent subspaces of Q, namely Qo and Qe, respectively.

The Fourier transform of a function of bounded variation: continuation

An even counterpart of Qo is

Z ∞ 1 |gbc(x)| Qe = {g : g ∈ L (R), g(−t) = g(t), dx < ∞}. 0 x R ∞ This makes sense only if 0 g(t) dt = 0.

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 15 / 19 In fact, in a) and b) the conditions are given in terms of diﬀerent subspaces of Q, namely Qo and Qe, respectively.

The Fourier transform of a function of bounded variation: continuation

An even counterpart of Qo is

Z ∞ 1 |gbc(x)| Qe = {g : g ∈ L (R), g(−t) = g(t), dx < ∞}. 0 x R ∞ This makes sense only if 0 g(t) dt = 0. Theorem 3.1. Let f : R+ → C be locally absolutely continuous, of bounded variation and lim f(t) = 0. t→∞ a) Then the cosine Fourier transform of f is Lebesgue integrable on 0 R+ if and only if f ∈ Q. b) The sine Fourier transform of f is Lebesgue integrable on R+ if and only if f 0 ∈ Q.

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 15 / 19 The Fourier transform of a function of bounded variation: continuation

An even counterpart of Qo is

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 15 / 19 where a sort of asymptotic relation can be obtained. gbs(x) In what follows we shall denote Tg(x) = x . Theorem 3.2. Let f : R+ → C be locally absolutely continuous, of bounded variation and lim f(t) = 0. Then for any x > 0 t→∞

Z ∞ 1 π Fs(x) = f(t) sin xt dt = f − HoTf 0 (x) + G(x), 0 x 2x where

0 kGk 1 kf k 1 . L (R+) . L (R+)

The Fourier transform of a function of bounded variation: continuation

The situation is more delicate with the sine Fourier transform,

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 16 / 19 gbs(x) In what follows we shall denote Tg(x) = x . Theorem 3.2. Let f : R+ → C be locally absolutely continuous, of bounded variation and lim f(t) = 0. Then for any x > 0 t→∞

Z ∞ 1 π Fs(x) = f(t) sin xt dt = f − HoTf 0 (x) + G(x), 0 x 2x where

0 kGk 1 kf k 1 . L (R+) . L (R+)

The Fourier transform of a function of bounded variation: continuation

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 16 / 19 Theorem 3.2. Let f : R+ → C be locally absolutely continuous, of bounded variation and lim f(t) = 0. Then for any x > 0 t→∞

Z ∞ 1 π Fs(x) = f(t) sin xt dt = f − HoTf 0 (x) + G(x), 0 x 2x where

0 kGk 1 kf k 1 . L (R+) . L (R+)

The Fourier transform of a function of bounded variation: continuation

The situation is more delicate with the sine Fourier transform, where a sort of asymptotic relation can be obtained. gbs(x) In what follows we shall denote Tg(x) = x .

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 16 / 19 where

0 kGk 1 kf k 1 . L (R+) . L (R+)

The Fourier transform of a function of bounded variation: continuation

The situation is more delicate with the sine Fourier transform, where a sort of asymptotic relation can be obtained. gbs(x) In what follows we shall denote Tg(x) = x . Theorem 3.2. Let f : R+ → C be locally absolutely continuous, of bounded variation and lim f(t) = 0. Then for any x > 0 t→∞

Z ∞ 1 π Fs(x) = f(t) sin xt dt = f − HoTf 0 (x) + G(x), 0 x 2x

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 16 / 19 The Fourier transform of a function of bounded variation: continuation

Z ∞ 1 π Fs(x) = f(t) sin xt dt = f − HoTf 0 (x) + G(x), 0 x 2x where

0 kGk 1 kf k 1 . L (R+) . L (R+)

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 16 / 19 which consists of Qo functions g with integrable HoTg. Corollary. Let a function f satisfy the assumptions of Theorem 3.2 0 1 and such that f ∈ HQ(R+). Then

1 π F (x) = f + G(x), s x 2x where

0 kGk 1 kf k 1 . L (R+) . HQ(R+) Technically, this is an obvious corollary of Theorem 3.2.

A Hardy type space

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 17 / 19 Corollary. Let a function f satisfy the assumptions of Theorem 3.2 0 1 and such that f ∈ HQ(R+). Then

1 π F (x) = f + G(x), s x 2x where

0 kGk 1 kf k 1 . L (R+) . HQ(R+) Technically, this is an obvious corollary of Theorem 3.2.

A Hardy type space

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

which consists of Qo functions g with integrable HoTg.

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 17 / 19 where

0 kGk 1 kf k 1 . L (R+) . HQ(R+) Technically, this is an obvious corollary of Theorem 3.2.

A Hardy type space

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

which consists of Qo functions g with integrable HoTg. Corollary. Let a function f satisfy the assumptions of Theorem 3.2 0 1 and such that f ∈ HQ(R+). Then

1 π F (x) = f + G(x), s x 2x

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 17 / 19 Technically, this is an obvious corollary of Theorem 3.2.

A Hardy type space

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

which consists of Qo functions g with integrable HoTg. Corollary. Let a function f satisfy the assumptions of Theorem 3.2 0 1 and such that f ∈ HQ(R+). Then

1 π F (x) = f + G(x), s x 2x where

0 kGk 1 kf k 1 . L (R+) . HQ(R+)

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 17 / 19 A Hardy type space

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

which consists of Qo functions g with integrable HoTg. Corollary. Let a function f satisfy the assumptions of Theorem 3.2 0 1 and such that f ∈ HQ(R+). Then

1 π F (x) = f + G(x), s x 2x where

0 kGk 1 kf k 1 . L (R+) . HQ(R+) Technically, this is an obvious corollary of Theorem 3.2.

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 17 / 19 This could be so but not after the appearance of the analysis of Q by Johnson-Warner. Indeed, Hardy’s inequality

Z |gb(x)| dx . kgkH1(R) R |x| implies

1 1 H (R) ⊆ Q ⊆ L0(R), 1 where the latter is the subspace of g in L (R) which satisfy the cancelation property (mean zero).

The Fourier transform of a function of bounded variation: continuation

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 18 / 19 Indeed, Hardy’s inequality

Z |gb(x)| dx . kgkH1(R) R |x| implies

1 1 H (R) ⊆ Q ⊆ L0(R), 1 where the latter is the subspace of g in L (R) which satisfy the cancelation property (mean zero).

The Fourier transform of a function of bounded variation: continuation

At ﬁrst sight, Theorem 3.1 does not seem to be a result at all, at most a technical reformulation of the deﬁnition. This could be so but not after the appearance of the analysis of Q by Johnson-Warner.

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 18 / 19 implies

1 1 H (R) ⊆ Q ⊆ L0(R), 1 where the latter is the subspace of g in L (R) which satisfy the cancelation property (mean zero).

The Fourier transform of a function of bounded variation: continuation

Z |gb(x)| dx . kgkH1(R) R |x|

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 18 / 19 1 where the latter is the subspace of g in L (R) which satisfy the cancelation property (mean zero).

The Fourier transform of a function of bounded variation: continuation

Z |gb(x)| dx . kgkH1(R) R |x| implies

1 1 H (R) ⊆ Q ⊆ L0(R),

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 18 / 19 The Fourier transform of a function of bounded variation: continuation

Z |gb(x)| dx . kgkH1(R) R |x| implies

1 1 H (R) ⊆ Q ⊆ L0(R), 1 where the latter is the subspace of g in L (R) which satisfy the cancelation property (mean zero).

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 18 / 19 therefore it is of interest to ﬁnd certain proper subspaces of Qo wider than H1 belonging to which is easier to be checked. Proposition. If g is an integrable odd function, then

kHoTgk 1 kgk 1 . L (R+) . H0 (R+) A direct, molecular proof exists. In any case, this implies an updated chain of embeddings

1 1 1 H0 (R+) ⊆ HQ(R+) ⊆ Qo ⊆ L0(R+). It is very interesting to ﬁgure out which of these embeddings are 1 1 proper. At least, H0 (R+) is strictly smaller than HQ. Correspondingly, intermediate spaces are of interest, both theoretical and practical.

Discussion

It is doubtful that Q (or Qo) may be deﬁned in terms of g itself rather than its Fourier transform,

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 19 / 19 Proposition. If g is an integrable odd function, then

kHoTgk 1 kgk 1 . L (R+) . H0 (R+) A direct, molecular proof exists. In any case, this implies an updated chain of embeddings

Discussion

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

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 19 / 19 A direct, molecular proof exists. In any case, this implies an updated chain of embeddings

Discussion

kHoTgk 1 kgk 1 . L (R+) . H0 (R+)

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 19 / 19 It is very interesting to ﬁgure out which of these embeddings are 1 1 proper. At least, H0 (R+) is strictly smaller than HQ. Correspondingly, intermediate spaces are of interest, both theoretical and practical.

Discussion

kHoTgk 1 kgk 1 . L (R+) . H0 (R+) A direct, molecular proof exists. In any case, this implies an updated chain of embeddings

1 1 1 H0 (R+) ⊆ HQ(R+) ⊆ Qo ⊆ L0(R+).

Elijah Liﬂyand (Bar-Ilan University) Fourier and Hilbert transforms June, 2013 19 / 19 Correspondingly, intermediate spaces are of interest, both theoretical and practical.

Discussion

kHoTgk 1 kgk 1 . L (R+) . H0 (R+) A direct, molecular proof exists. In any case, this implies an updated chain of embeddings

1 1 1 H0 (R+) ⊆ HQ(R+) ⊆ Qo ⊆ L0(R+). It is very interesting to ﬁgure out which of these embeddings are 1 1 proper. At least, H0 (R+) is strictly smaller than HQ.

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

kHoTgk 1 kgk 1 . L (R+) . H0 (R+) A direct, molecular proof exists. In any case, this implies an updated chain of embeddings

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