Existence and non-existence of ergodic Hilbert transform ...

Existence and non-existence of ergodic Hilbert transform foradmissible processes

Dogan Comez

Department Colloquium, NDSU

March 29, 2016

Set upHistory.

Admissible processes.Local eHt for admissible processes.

Sketch of the proofeHt along moving averages sequences

Set up

Dirichlet problem in the upper half plane:

Given f (x), find u(x , y), harmonic in theupper half plane and agrees with f (x) on the real line; that is,

∂2u∂x2 + ∂2u

∂x2 = 0, and u(x , 0) = f (x).

Solution: u(x , y) = yπ

∫∞−∞

(x−s)2+y2 ds.

Harmonic conjugate of u(x , y) in the upper half plane is

v(x , y) =1

∫ ∞−∞

f (x − s)s

s2 + y2ds.

Notice: limy→0 v(x , y) = 1π

∫∞−∞

f (x−s)s

Hilbert transform of f :

Hf (x) =1

πp.v.

∫ ∞−∞

f (x − s)

sds = lim

ε→0

∫ε<|s|<1/ε

f (x − s)

Besicovitch (early 1930’s): Hf exists a.e. for all integrable f on R.

Dogan Comez Existence and non-existence of ergodic Hilbert transform for admissible processes

Set upHistory.

Set up

Dirichlet problem in the upper half plane: Given f (x), find u(x , y), harmonic in theupper half plane and agrees with f (x) on the real line; that is,

∂2u∂x2 + ∂2u

∂x2 = 0, and u(x , 0) = f (x).

∫∞−∞

(x−s)2+y2 ds.

v(x , y) =1

∫ ∞−∞

f (x − s)s

s2 + y2ds.

∫∞−∞

f (x−s)s

Hf (x) =1

πp.v.

∫ ∞−∞

f (x − s)

sds = lim

ε→0

∫ε<|s|<1/ε

f (x − s)

Set upHistory.

Set up

∂2u∂x2 + ∂2u

∂x2 = 0, and u(x , 0) = f (x).

∫∞−∞

(x−s)2+y2 ds.

v(x , y) =1

∫ ∞−∞

f (x − s)s

s2 + y2ds.

∫∞−∞

f (x−s)s

Hf (x) =1

πp.v.

∫ ∞−∞

f (x − s)

sds = lim

ε→0

∫ε<|s|<1/ε

f (x − s)

Set upHistory.

Set up

∂2u∂x2 + ∂2u

∂x2 = 0, and u(x , 0) = f (x).

∫∞−∞

(x−s)2+y2 ds.

v(x , y) =1

∫ ∞−∞

f (x − s)s

s2 + y2ds.

∫∞−∞

f (x−s)s

Hf (x) =1

πp.v.

∫ ∞−∞

f (x − s)

sds = lim

ε→0

∫ε<|s|<1/ε

f (x − s)

Set upHistory.

Set up

∂2u∂x2 + ∂2u

∂x2 = 0, and u(x , 0) = f (x).

∫∞−∞

(x−s)2+y2 ds.

v(x , y) =1

∫ ∞−∞

f (x − s)s

s2 + y2ds.

∫∞−∞

f (x−s)s

Hf (x) =1

πp.v.

∫ ∞−∞

f (x − s)

sds = lim

ε→0

∫ε<|s|<1/ε

f (x − s)

Set upHistory.

Set up

∂2u∂x2 + ∂2u

∂x2 = 0, and u(x , 0) = f (x).

∫∞−∞

(x−s)2+y2 ds.

v(x , y) =1

∫ ∞−∞

f (x − s)s

s2 + y2ds.

∫∞−∞

f (x−s)s

Hf (x) =1

πp.v.

∫ ∞−∞

f (x − s)

sds = lim

ε→0

∫ε<|s|<1/ε

f (x − s)

Set upHistory.

Ergodic setting

Observe: Lebesgue measure is translation invariant; hence, the map Us(x) = x − s

preserves Lebesgue measure, namely, m(U−1s E) = m(E).

Hilbert transform of f can be rewritten as:

Hf (x) = limε→0

∫ε<|s|<1/ε

f (x − s)

sds = lim

ε→0

∫ε<|s|<1/ε

f (Usx)

More generally, if τ = Ttt∈R is a group of invertible m.p.t’s on (X ,Σ, µ), then theergodic Hilbert transform (eHt) of f is defined as

Hτ f (x) = limε→0

∫ε<|s|<1/ε

f (Tsx)

Discrete version: If T : X → X is an i.m.p.t. (i.e., τ = T nn∈Z), the (discrete)ergodic Hilbert transform of f is

HT f (x) = limn→∞

∑0<|k|≤n

f (T kx)

Set upHistory.

Ergodic setting

Hf (x) = limε→0

∫ε<|s|<1/ε

f (x − s)

sds = lim

ε→0

∫ε<|s|<1/ε

f (Usx)

∫ε<|s|<1/ε

f (Tsx)

∑0<|k|≤n

f (T kx)

Set upHistory.

Ergodic setting

Hf (x) = limε→0

∫ε<|s|<1/ε

f (x − s)

sds = lim

ε→0

∫ε<|s|<1/ε

f (Usx)

∫ε<|s|<1/ε

f (Tsx)

∑0<|k|≤n

f (T kx)

Set upHistory.

Ergodic setting

Hf (x) = limε→0

∫ε<|s|<1/ε

f (x − s)

sds = lim

ε→0

∫ε<|s|<1/ε

f (Usx)

∫ε<|s|<1/ε

f (Tsx)

∑0<|k|≤n

f (T kx)

Set upHistory.

Ergodic theorems

Let (X ,Σ, µ) be a probability space. A m.pt. T : X → X is called ergodic ifT−1E = E implies µ(E) = 0 or µ(E) = 1.

Crown jewel of ergodic theory:

Birkhoff’s Ergodic Theorem. (1931)

If T : X → X is m.p.t., then, limn→∞1n

∑n−1k=0 f (T kx) = f ∗(x) exists a.e. for all

f ∈ L1. If, furthermore, T is ergodic, then f ∗ =∫f .

Continuous parameter version:

Local Ergodic Theorem. (Wiener, 1939)

If τ is a m.p. flow and f ∈ L1, then limt→0+1t

∫ t0 f (Tsx)ds = f (x) a.e.

Question: Analogous results for the (discrete or local) eHt?

Set upHistory.

Ergodic theorems

Set upHistory.

Ergodic theorems

Set upHistory.

Ergodic theorems

Set upHistory.

Existence of eHt

Recall:

The (discrete/local) eHt of f ∈ Lp(X ) is

∑0<|k|≤n

f (T kx)

kor HT f (x) = lim

ε→0

∫ε<|s|≤1/ε

f (Tsx)

M. Cotlar (1955)

If τ is an i.m.p. flow (or T is an i.m.p.t.) on X , then Hτ f (x) exists a.e. for all f ∈ L1.

K. Petersen (1983, 1985) - Another proof of eHt (both local and discrete).

R. Sato (1989) - eHt in the operator setting (discrete).

L.M. Fernadez-Cabrera, F. Martin-Reyes and J.L. Torrea (1995) - connectionbetween ergodic everages and eHt (discrete).

D. Comez (2006) - eHt for admissible processes (discrete).

Set upHistory.

Existence of eHt

Recall: The (discrete/local) eHt of f ∈ Lp(X ) is

∑0<|k|≤n

f (T kx)

kor HT f (x) = lim

ε→0

∫ε<|s|≤1/ε

f (Tsx)

M. Cotlar (1955)

Set upHistory.

Existence of eHt

∑0<|k|≤n

f (T kx)

kor HT f (x) = lim

ε→0

∫ε<|s|≤1/ε

f (Tsx)

M. Cotlar (1955)

Set upHistory.

Existence of eHt

∑0<|k|≤n

f (T kx)

kor HT f (x) = lim

ε→0

∫ε<|s|≤1/ε

f (Tsx)

M. Cotlar (1955)

Set upHistory.

Existence of eHt

∑0<|k|≤n

f (T kx)

kor HT f (x) = lim

ε→0

∫ε<|s|≤1/ε

f (Tsx)

M. Cotlar (1955)

Set upHistory.

Existence of eHt

∑0<|k|≤n

f (T kx)

kor HT f (x) = lim

ε→0

∫ε<|s|≤1/ε

f (Tsx)

M. Cotlar (1955)

Set upHistory.

Existence of eHt

∑0<|k|≤n

f (T kx)

kor HT f (x) = lim

ε→0

∫ε<|s|≤1/ε

f (Tsx)

M. Cotlar (1955)

Set upHistory.

Results on the modulated eHt

Some other eHt results:

C. Demeter, M. Lacey, T. Tao & C. Thiele (2007)- Return time theorem for theeHt (discrete).

M. Lacey & E. Terwilleger (2008)- Wiener-Wintner type theorem for the eHt(local and discrete).

Corollary: limN→∞∑

0<|k|≤Nak f (T k x)

kexists a.e., where ak is a sequence

induced by a trigonometric polynomial.

Comez (2006) - There is a bounded Besicovitch sequence ak such that thelimit fails to exist.

A. Akhmedov & D. Comez, (2014)- Let Mα = a :∑n−n |ak | = O( nα−1

logα n),

1 < α ≤ 2. If a ∈ Mα, then it is universally good for the eHt in L1.

There is a subclass Bα of bounded Besicovitch sequences s.t. if a ∈ Bα, then it isuniversally good for the eHt in L2.

Question. Does eHt exist along moving averages sequences? We will return!

Set upHistory.

logα n),

Set upHistory.

logα n),

Set upHistory.

logα n),

Set upHistory.

logα n),

Set upHistory.

logα n),

Set upHistory.

logα n),

Set upHistory.

logα n),

Question. Does eHt exist along moving averages sequences?

We will return!

Set upHistory.

logα n),

Set upHistory.

Admissible processes

F = ftt∈R ⊂ Lp is called a τ -admissible process on R ifTs ft ≤ fs+t and T−s f−t ≤ f−s−t for t, s ≥ 0.

Discrete version: F = fnn∈Z is T -admissible on Z ifTfk ≤ fk+1 and T−1f−k ≤ f−k−1 for k ∈ Z+.

An admissible process F = ftt∈R is called

positive if ft ≥ 0 for all t ∈ R,strongly bounded if γF := supt∈R ‖ft‖p <∞,symmetric if T2t f−t = ft for all t ∈ R+.

D. Comez (2006) - If T : X → X is an i.m.p.t and F = fnn∈Z ⊂ L1 is a strongly

bounded symmetric T -admissible process, then limn∑

1≤|k|≤nfk (x)k

exists a.e.

Question. How about extending the local eHt result to the admissible processes?

Set upHistory.

1≤|k|≤nfk (x)k

exists a.e.

Set upHistory.

1≤|k|≤nfk (x)k

exists a.e.

Set upHistory.

1≤|k|≤nfk (x)k

exists a.e.

Set upHistory.

1≤|k|≤nfk (x)k

exists a.e.

Set upHistory.

LeHt for admissible processes-set up

Recall: Local eHt of f isHf (x) = limε→0

∫ε≤|t|≤1/ε

f (Ts x)s

Local ergodic Hilbert transform of a symmetric admissible process F = ftt∈R :

HF (x) = limε→0

∫ε≤|t|≤1/ε

fs (x)s

Let F = ft be a positive, symmetric, strongly bounded τ -admissible process. Then:

M.A. Akcoglu & U. Krengel (1981) - ∃ a maximal τ -additive process g Ttsuch that g Tt(x) ≤ ft(x) a.e. for all t ∈ R. (Hence, we can assume a givenadmissible F a positive process.)

∃ a family of non-negative integrable functions ut s.t. ft = Ttu|t| for all t ∈ R∃ δ ∈ L1 s.t. ft ≤ Ttδ, ∀ t ∈ R, and ‖δ‖1 = supt∈R ‖ft‖1 (exact dominant of F ).

Set upHistory.

∫ε≤|t|≤1/ε

f (Ts x)s

HF (x) = limε→0

∫ε≤|t|≤1/ε

fs (x)s

Set upHistory.

∫ε≤|t|≤1/ε

f (Ts x)s

HF (x) = limε→0

∫ε≤|t|≤1/ε

fs (x)s

Set upHistory.

∫ε≤|t|≤1/ε

f (Ts x)s

HF (x) = limε→0

∫ε≤|t|≤1/ε

fs (x)s

∃ a family of non-negative integrable functions ut s.t. ft = Ttu|t| for all t ∈ R

∃ δ ∈ L1 s.t. ft ≤ Ttδ, ∀ t ∈ R, and ‖δ‖1 = supt∈R ‖ft‖1 (exact dominant of F ).

Set upHistory.

∫ε≤|t|≤1/ε

f (Ts x)s

HF (x) = limε→0

∫ε≤|t|≤1/ε

fs (x)s

Set upHistory.

LeHt for admissible processes-results

Let τ = Ttt∈R be an i.m.p. flow and F = ft ⊂ L1 be a symmetric, stronglybounded τ -admissible process with exact dominant δ.

Theorem.1 (Maximal Inequality)

For any λ > 0, there exists a constant C such that,

µx : supq>0|∫q≤|s|≤1/q

sds| > λ ≤

λ‖δ‖1.

Theorem.2 (Local eHT for admissible processes)

HF (x) = limq→0

∫q≤|s|≤1/q

sds exists a.e.

Set upHistory.

sds| > λ ≤

λ‖δ‖1.

HF (x) = limq→0

∫q≤|s|≤1/q

sds exists a.e.

Set upHistory.

sds| > λ ≤

λ‖δ‖1.

HF (x) = limq→0

∫q≤|s|≤1/q

sds exists a.e.

Set upHistory.

One-sided leHt

One-sided (ergodic) Hilbert transform.

Recall Hilbert transform: limq→0

∫q<|s|<1/q

f (s+t)s

ds exists a.e. for all f ∈ L1.

One-sided variant: limq→0

∫q<|s|≤a

f (s+t)s

ds exists a.e. for all f ∈ L1, a > 0.

Ergodic version:

(∗) Huf (x) := limq→0

∫q<|s|≤a

f (Tsx)

sds exists a.e. for all f ∈ L1, a > 0.

Hence: existence of the local eHt and (*) implies that

(∗∗) H l f (x) := limq→0

∫a≤|s|<1/q

f (Tsx)

Set upHistory.

One-sided leHt

∫q<|s|<1/q

f (s+t)s

∫q<|s|≤a

f (s+t)s

Ergodic version:

(∗) Huf (x) := limq→0

∫q<|s|≤a

f (Tsx)

(∗∗) H l f (x) := limq→0

∫a≤|s|<1/q

f (Tsx)

Set upHistory.

One-sided leHt

∫q<|s|<1/q

f (s+t)s

∫q<|s|≤a

f (s+t)s

Ergodic version:

(∗) Huf (x) := limq→0

∫q<|s|≤a

f (Tsx)

(∗∗) H l f (x) := limq→0

∫a≤|s|<1/q

f (Tsx)

Set upHistory.

One-sided leHt

∫q<|s|<1/q

f (s+t)s

∫q<|s|≤a

f (s+t)s

Ergodic version:

(∗) Huf (x) := limq→0

∫q<|s|≤a

f (Tsx)

(∗∗) H l f (x) := limq→0

∫a≤|s|<1/q

f (Tsx)

Set upHistory.

One-sided leHt

∫q<|s|<1/q

f (s+t)s

∫q<|s|≤a

f (s+t)s

Ergodic version:

(∗) Huf (x) := limq→0

∫q<|s|≤a

f (Tsx)

(∗∗) H l f (x) := limq→0

∫a≤|s|<1/q

f (Tsx)

Set upHistory.

Sketch of proof

Sketch of proof of Theorem.2 (assuming Theorem.1):

We can assume that ft ≥ 0 for each t. Recall ft = Ttu|t|, and ut ≥ 0, for allt ∈ R.

Let G1 = Ttu0t∈R. Then G1t =

∫ t0 Tsu0ds ≤ Ft for all t ∈ R.

Fix ε > 0, and pick t0 > 0 such that ‖ut0‖1 > ‖δ‖1 − ε, and let G2 = Tt ft0.Then G2

t =∫ t

0 Tsut0ds ≤ Ft for all t ≥ t0 and γG2 = ‖ut0‖1 > γF − ε.

Define the process G = gt by

Ttu0 when 0 ≤ t < t0

Ttut0 when t0 ≤ t.

Then: G is τ -admissible, Gt ≤ Ft for all t ∈ R, and γG > γF − ε.

Set upHistory.

Sketch of proof

t =∫ t

Set upHistory.

Sketch of proof

t =∫ t

Set upHistory.

Sketch of proof

t =∫ t

0 Tsut0ds ≤ Ft for all t ≥ t0 and

γG2 = ‖ut0‖1 > γF − ε.

Set upHistory.

Sketch of proof

t =∫ t

Set upHistory.

Sketch of proof

t =∫ t

Set upHistory.

Sketch of proof

t =∫ t

Set upHistory.

Sketch of proof

Define a new process Z = zt by zt(x) = ft(x)− Ttg(x) for all t ∈ R.

Then: 0 < zt(x) ≤ Tt(δ − g)(x), and ‖δ − g‖1 < ε.

Furthermore, Z = zt is a strongly bounded, symmetric τ -admissible process withexact dominant δ − g such that γZ = ‖δ − g‖ < ε.

Also, HqF (x) = HqZ(x) +∫q≤|s|≤t0

u0(Ts x)s

ds +∫t0≤|s|≤1/q

ut0 (Ts x)

Since the last two limits exist a.e. by (*) and (**), it remains to prove thatlimq→0 HqZ(x) exists a.e.

Set upHistory.

Sketch of proof

Define a new process Z = zt by zt(x) = ft(x)− Ttg(x) for all t ∈ R.Then: 0 < zt(x) ≤ Tt(δ − g)(x), and ‖δ − g‖1 < ε.

u0(Ts x)s

ds +∫t0≤|s|≤1/q

ut0 (Ts x)

Set upHistory.

Sketch of proof

u0(Ts x)s

ds +∫t0≤|s|≤1/q

ut0 (Ts x)

Set upHistory.

Sketch of proof

u0(Ts x)s

ds +∫t0≤|s|≤1/q

ut0 (Ts x)

Set upHistory.

Sketch of proof

u0(Ts x)s

ds +∫t0≤|s|≤1/q

ut0 (Ts x)

Set upHistory.

Sketch of proof

Let E = x : lim supq HqF (x)− lim infq HqF (x) > λ, then

E ⊂ x : supq|HqZ(x)| >

From the maximal inequality,

µ(E) ≤C

λ‖δ − gε‖1 < ε.

Since ε > 0 is arbitrary, so µ(E) = 0.

Thus HF (x) = limq→0 HqF (x) exists a.e.

Set upHistory.

Sketch of proof

µ(E) ≤C

λ‖δ − gε‖1 < ε.

Set upHistory.

Sketch of proof

µ(E) ≤C

λ‖δ − gε‖1 < ε.

Set upHistory.

Moving averages

Moving averages sequences.

A sequence w = (vn, rn) in Z× Z+, where rn > 0, satisfies the cone condition (CC)if ∃ C such that ∀ s ∈ R+, |Ωα(s)| ≤ Cs, where

Ωα = (t, s) : ∃n, |t − vn| ≤ α(s − rn), and Ωα(s) = x ∈ R : (x , s) ∈ Ωα.Examples: (n, n) and (22n ,

√22n ) satisfy CC, but (n,

√n) does not.

M.A. Akcoglu and A. delJunco (1975): limn1√n

∑√nk=0 f (T n+kx) fails to exist a.e.

Notice: (n,√n) does not satisfy cone condition.

A. Bellow, R. Jones and J. Rosenblatt (1990): (vn, rn) satisfy CC is necessaryand sufficient for the moving average 1

∑rnk=0 f (T vn+kx) to converge a.e.

S. Ferrando (1995): extended it to superadditive processes setting.

Question. How about the analogous results for the eHt?

Set upHistory.

Moving averages

Ωα = (t, s) : ∃n, |t − vn| ≤ α(s − rn), and Ωα(s) = x ∈ R : (x , s) ∈ Ωα.

Examples: (n, n) and (22n ,√

22n ) satisfy CC, but (n,√n) does not.

Set upHistory.

Moving averages

√n) does not.

Set upHistory.

Moving averages

√n) does not.

Set upHistory.

Moving averages

√n) does not.

Set upHistory.

Moving averages

√n) does not.

Set upHistory.

Moving averages

√n) does not.

Set upHistory.

Moving averages

√n) does not.

Set upHistory.

Moving eHt

Let T : X → X be an i.m.p.t. and w = (vn, rn). Define the moving eHt of f by

HTw f = limn

∑1≤|k|≤rn

f (T vn+k x)k

, if exists.

As usual, one needs the maximal inequality and the a.e. existence of HTw f for all f in a

dense subset of Lp .

Maximal inequality is OK:For continuous parameter additive processes (Ferrando, Jones and Reinhold, 1995).For (discrete) admissible processes (Comez, 2015).

Convergence on a dense subset of Lp?

Set upHistory.

Moving eHt

HTw f = limn

∑1≤|k|≤rn

f (T vn+k x)k

, if exists.

Set upHistory.

Moving eHt

HTw f = limn

∑1≤|k|≤rn

f (T vn+k x)k

, if exists.

Set upHistory.

Moving eHt

HTw f = limn

∑1≤|k|≤rn

f (T vn+k x)k

, if exists.

Set upHistory.

Example

Example.Let (X ,F , µ,T ), where X = 0, 1, 2, F = P(X ), µ(i) = 1/3, 0 ≤ i ≤ 2,

andT (0) = 1, T (1) = 2, T (2) = 0.

Let w = (n, n). w satisfies the cone condition and vn = n = rn →∞.

Consider two subsequences: w1 consists of those that are multiples of 3, w2 consistsof those that are of the form n = 3k + 1.

Let f : X → R be such that f (0) = f (2) = 0 and f (1) = 1.

Hf (1) =∑k 6=0

f (T k1)

Hf (0) =∑k 6=0

f (T k0)

k= (1−

2) + (

5) + (

8) + · · · 6= 0.

Set upHistory.

Example

Example.Let (X ,F , µ,T ), where X = 0, 1, 2, F = P(X ), µ(i) = 1/3, 0 ≤ i ≤ 2, andT (0) = 1, T (1) = 2, T (2) = 0.

Hf (1) =∑k 6=0

f (T k1)

Hf (0) =∑k 6=0

f (T k0)

k= (1−

2) + (

5) + (

8) + · · · 6= 0.

Set upHistory.

Example

Let w = (n, n).

w satisfies the cone condition and vn = n = rn →∞.

Hf (1) =∑k 6=0

f (T k1)

Hf (0) =∑k 6=0

f (T k0)

k= (1−

2) + (

5) + (

8) + · · · 6= 0.

Set upHistory.

Example

Hf (1) =∑k 6=0

f (T k1)

Hf (0) =∑k 6=0

f (T k0)

k= (1−

2) + (

5) + (

8) + · · · 6= 0.

Set upHistory.

Example

Consider two subsequences:

w1 consists of those that are multiples of 3, w2 consistsof those that are of the form n = 3k + 1.

Hf (1) =∑k 6=0

f (T k1)

Hf (0) =∑k 6=0

f (T k0)

k= (1−

2) + (

5) + (

8) + · · · 6= 0.

Set upHistory.

Example

Consider two subsequences: w1 consists of those that are multiples of 3,

w2 consistsof those that are of the form n = 3k + 1.

Hf (1) =∑k 6=0

f (T k1)

Hf (0) =∑k 6=0

f (T k0)

k= (1−

2) + (

5) + (

8) + · · · 6= 0.

Set upHistory.

Example

Hf (1) =∑k 6=0

f (T k1)

Hf (0) =∑k 6=0

f (T k0)

k= (1−

2) + (

5) + (

8) + · · · 6= 0.

Set upHistory.

Example

Hf (1) =∑k 6=0

f (T k1)

Hf (0) =∑k 6=0

f (T k0)

k= (1−

2) + (

5) + (

8) + · · · 6= 0.

Set upHistory.

Example

Hf (1) =∑k 6=0

f (T k1)

Hf (0) =∑k 6=0

f (T k0)

k= (1−

2) + (

5) + (

8) + · · · 6= 0.

Set upHistory.

Example

Along w1,

Hw1 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T k0)

k→ Hf (0) 6= 0.

Along w2,

Hw2 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T 1+k0)

0<|k|≤rn

f (T k1)

k→ Hf (1) = 0.

Hence limn H(vn,rn)f (0) does not exist!

Question. Is it possible to have an affirmative answer under some additionalconditions?

Set upHistory.

Example

Along w1,

Hw1 f (0) =∑

0<|k|≤rn

f (T vn+k0)

0<|k|≤rn

f (T k0)

k→ Hf (0) 6= 0.

Along w2,

Hw2 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T 1+k0)

0<|k|≤rn

f (T k1)

k→ Hf (1) = 0.

Set upHistory.

Example

Along w1,

Hw1 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T k0)

→ Hf (0) 6= 0.

Along w2,

Hw2 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T 1+k0)

0<|k|≤rn

f (T k1)

k→ Hf (1) = 0.

Set upHistory.

Example

Along w1,

Hw1 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T k0)

k→ Hf (0) 6= 0.

Along w2,

Hw2 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T 1+k0)

0<|k|≤rn

f (T k1)

k→ Hf (1) = 0.

Set upHistory.

Example

Along w1,

Hw1 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T k0)

k→ Hf (0) 6= 0.

Along w2,

Hw2 f (0) =∑

0<|k|≤rn

f (T vn+k0)

0<|k|≤rn

f (T 1+k0)

0<|k|≤rn

f (T k1)

k→ Hf (1) = 0.

Set upHistory.

Example

Along w1,

Hw1 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T k0)

k→ Hf (0) 6= 0.

Along w2,

Hw2 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T 1+k0)

0<|k|≤rn

f (T k1)

k→ Hf (1) = 0.

Set upHistory.

Example

Along w1,

Hw1 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T k0)

k→ Hf (0) 6= 0.

Along w2,

Hw2 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T 1+k0)

0<|k|≤rn

f (T k1)

→ Hf (1) = 0.

Set upHistory.

Example

Along w1,

Hw1 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T k0)

k→ Hf (0) 6= 0.

Along w2,

Hw2 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T 1+k0)

0<|k|≤rn

f (T k1)

k→ Hf (1) = 0.

Set upHistory.

Example

Along w1,

Hw1 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T k0)

k→ Hf (0) 6= 0.

Along w2,

Hw2 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T 1+k0)

0<|k|≤rn

f (T k1)

k→ Hf (1) = 0.

Set upHistory.

Example

Along w1,

Hw1 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T k0)

k→ Hf (0) 6= 0.

Along w2,

Hw2 f (0) =∑

0<|k|≤rn

f (T vn+k0)

∑0<|k|≤rn

f (T 1+k0)

0<|k|≤rn

f (T k1)

k→ Hf (1) = 0.

Set upHistory.

Some special dynamical systems

Since, moving eHt of an integrable function fails to exist for arbitrary i.m.p. dynamicalsystem; positive results are possible if one considers:

restricted classes of functions, or

some special classes of dynamical systems, or

further restrictions of the sequence (vn, rn).

T has Lebesgue spectrum if L02(X ) = ⊕jHj , where Hj = spanT k fj for some fj ∈ L0

such that < fj ,Tk fj >= 0 if k 6= 0.

Theorem (Comez, 2016) Let (vn, rn) satisfy CC. If T has Lebesgue spectrum, then,for all f ∈ L0

2, limn H(vn,rn)f (x) exists a.e.

Set upHistory.

THANK YOU!

Existence and non-existence of ergodic Hilbert transform ...

Documents

Transcript of Existence and non-existence of ergodic Hilbert transform ...