GEORGIAN MATHEMATICAL JOURNAL: Vol. 5, No. 2, 1998, 101-106

ON THE INTEGRABILITY OF THE ERGODIC HILBERT

TRANSFORM FOR A CLASS OF FUNCTIONS WITH EQUAL ABSOLUTE VALUES

L. EPHREMIDZE

Abstract. It is proved that for an arbitrary non-atomic finite measure space with a measure-preserving ergodic transformation there exists an integrable function f such that the ergodic Hilbert transform of any function equal in absolute values to f is non-integrable.

Let T be a one-to-one measure-preserving ergodic transformation of a finite measure space (X, 𝕊, μ). For a real integrable function f, f ∈ L(X), the ergodic Hilbert transform is defined by

H(f)(x) =\begin{matrix}∞ \\ ∑ & ′ \\ k=−∞ \end{matrix} \frac{f(T^{k}x)}{k} , x ∈ X

(the mark ′ indicates that the 0-th term in the sum is omitted). It exists for almost all x ∈ X (in the sense of measure μ) and, furthemore, the operator H satisfies a weak-type (1, 1) inequality

μ(|H(f)| > λ) ≤ \frac{c}{λ} ∥f∥_{1}, λ > 0, f ∈ L(X),  (1)

where c is an absolute constant (see, e.g., [2]).

It is well known that the condition f ∈ L lg L implies the integrability of H(f) (see[1]) but, in general, it is rather difficult to give by means of any integral classes an exact description of the set of functions f for which H(f) ∈ L(X). Therefore the theorem below seems to be interesting for investigation of this set.

Theorem. Let (X, 𝕊, μ) be a non-atomic finite measure space, and let T be a one-to-one measure-preserving ergodic transformation on X. Then

101 there exists f ∈ L(X) such that for any function g with the same absolute values, |g| = |f|, we have

1991 Mathematics Subject Classification. 28D05.

Key words and phrases. Ergodic Hilbert transform.

1072-947X/98/0300-0101$15.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1998 Plenum Publishing Corporation

H(g) ∉ L(X).

Analogous questions as a general problem of “kernel and hull” were formulated by Prof. O. Tsereteli (see [3],[4]). He also proved the following theorem on the conjugate operator: for any integrable function f on the unit circle there exists a function g with the same absolute values such that g̃ is integrable (see [5]).

The proof of the following lemma is simple and we give it for the sake of completeness.

Lemma. Let T be a one-to-one measure-preserving ergodic transformation of a non-atomic finite measure space (X, 𝕊, μ). Then there exists a set A of positive measure, μ(A) > 0, such that

μ(X\ ∪^{n}_{k=0} T^{k}A) > 0  (2)

for all positive integers n.

Proof. Let us construct successively a sequence of imbedded measurable sets (A_{n})_{n=0}^{∞}. Let A_{0} be a set of positive not full measure, i.e., μ(A_{0}) > 0 and the inequality

μ(X\ ∪^{n}_{k=0} T^{k}A_{n}) > 0  (3)

holds for n = 0. Assume now that A_{0} ⊃ A_{1} ⊃ ⋯ ⊃ A_{n} have already been chosen, μ(A_{n}) > 0, and inequality (3) holds for A_{n}. Choose A_{n}^{0} ⊂

X\ ∪^{n}_{k=0} T^{k}A_{n} such that

0 < μ(A_{n}^{0}) < \frac{μ(A_{0})}{2^{n+2}} ,

and assume A_{n+1} = A_{n}\T^{−(n+1)}A_{n}^{0}. Then

A_{n}^{0} ⊂ X\( ∪^{n+1}_{k=0} T^{k}A_{n+1}).  (4)

Let

A = ∩^{∞}_{n=0}A_{n}.

ON THE INTEGRABILITY OF THE ERGODIC HILBERT TRANSFORM 103

Obviously,

μ(A) = μ(A_{0}) − ∑_{n=0}^{∞} μ(A_{n}\A_{n+1}) ≥

≥ μ(A_{0}) − ∑_{n=0}^{∞} μ(T^{−(n+1)}A_{n}^{0}) = μ(A_{0}) − ∑_{n=0}^{∞} μ(A_{n}^{0}) ≥

≥ μ(A_{0}) − ∑_{n=0}^{∞} \frac{μ(A_{0})}{2^{n+2}} = \frac{μ(A_{0})}{2} > 0,

and (2) also holds for each positive integer n because of (4).

Proof of the Theorem. Let A be a set of positive measure with property (2), and let us represent the space X as a skyscraper construction with the base A, i.e., assume

X = ∪^{∞}_{n=0} ∪^{n}_{k=0} T^{k}B_{n},

where ∪_{n=0}^{∞}B_{n} = A and T^{i}B_{n} ∩ T^{j}B_{m} = ∅ when n ≠ m, 0 ≤ i ≤ n,0 ≤ j ≤ m. Because of (2), there exists arbitrarily large n for which μ(B_{n}) > 0.

Let us choose successively an increasing sequence of positive integers m_{n}, for which μ(B_{m_{n}}) > 0, and decreasing to 0 sequences δ_{n} and ε_{n}, n = 1, 2, …, which satisfy the inequalities

\begin{matrix}δ_{1} < \frac{1}{6} , ∑_{n=1}^{∞} ε_{n} < 1, \\ δ_{n} · max_{1≤i≤n−1} (\frac{ε_{i}}{μ(B_{m_{i}})}) < \frac{1}{4} , \\ ε_{n} · log(δ_{n} · m_{n}) > 3, \\ \frac{ε_{n}}{μ(B_{m_{n}})} · c ∑_{k=n+1}^{∞} ε_{k} < 1, n = 1, 2, …, \end{matrix}  (5)

where c is the constant appearing in inequality (1). After a suitable choice of ε_{n}, δ_{n−1} and m_{n−1} one can take δ_{n} small enough, m_{n} large enough and ε_{n+1} small enough again to satisfy the second, the third, and the fourth inequalities, respectively.

To simplify our expressions let us use the following notations: \overline{n} = [\frac{m_{n}}{2} ] ([a] is the nearest integer less than or equal to a) and β_{n} = μ(B_{m_{n}}). Define f_{n}, n = 1, 2, …, by the equalities

f_{n}(x) = \begin{cases}\frac{ε_{n}}{β_{n}} , & when x ∈ T \overline{n}B_{m_{n}} \\ 0, & when x ∈ X\T \overline{n}B_{m_{n}} \end{cases} .

Note that ∥f_{n}∥ = ε_{n}. Thus the function f,

f = ∑_{n=1}^{∞} f_{n},

is integrable and let us show that it has the property required in the theorem. Let g be any function equal in absolute value to f. Assume

g = ∑_{n=1}^{∞} g_{n},

where |g_{n}| = f_{n}, n = 1, 2, . . .. Partial sums of this series will be denoted by h_{n},

h_{n} = ∑_{k=1}^{n} g_{k}.

The relations below are assumed to hold for almost all x, in particular, for those x for which the Hilbert transform of the corresponding functions exists.

Temporarily, suppose n to be fixed. Let us show that the inequality

|H(h_{n})(T^{i}x) − H(h_{n})(T^{−i}x)| ≥ \frac{2ε_{n}}{iβ_{n}} − \frac{12ε_{n}}{m_{n}β_{n}} − 8δ_{n}∥h_{n−1}∥_{∞}, (6)

x ∈ T \overline{n}B_{m_{n}}, holds for each i, 0 < i < δ_{n}m_{n}. Indeed, because of the linearity of the Hilbert transform,

|H(h_{n})(T^{i}x) − H(h_{n})(T^{−i}x)| ≥

≥ |H(g_{n})(T^{i}x) − H(g_{n})(T^{−i}x)| − |H(h_{n−1})(T^{i}x) − H(h_{n−1})(T^{−i}x)| = = S_{1} − S_{2}.

Taking into account that T^{k}x ∈ T \overline{n}B_{m_{n}} implies T^{k+j}x ∉ T \overline{n}B_{m_{n}}, j = 1, 2, …, m_{n}, we have

S_{1} = | − \frac{g_{n}(x)}{i} +\begin{matrix}∑ & ′ \\ k≠−i \end{matrix} \frac{g_{n}(T^{k+i}x)}{k} − \frac{g_{n}(x)}{i} − \begin{matrix}∑ & ′ \\ k≠i \end{matrix} \frac{g_{n}(T^{k−i}x)}{k}| =

= | − \frac{2g_{n}(x)}{i} + ∑_{|k|>m_{n}/2} (\frac{g_{n}(T^{k+i}x)}{k} − \frac{g_{n}(T^{k+i}x)}{k + 2i})| ≥

≥ \frac{2f_{n}(x)}{i} − ∑_{|k|>m_{n}/2} f_{n}(T^{k+i}x)(\frac{1}{k} − \frac{1}{k + 2i}) ≥

≥ \frac{2f_{n}(x)}{i} − 2f_{n}(x) ∑_{k=0}^{∞} (\frac{1}{m_{n}/2 + km_{n}} − \frac{1}{m_{n}/2 + km_{n} + 2i}) ≥

≥ \frac{2f_{n}(x)}{i} − \frac{4if_{n}(x)}{m_{n}^{2}} ∑_{k=0}^{∞} \frac{1}{(1/2 + k)^{2}} ≥ \frac{2f_{n}(x)}{i} − \frac{12δ_{n}f_{n}(x)}{m_{n}} =

= \frac{2ε_{n}}{iβ_{n}} − \frac{12δ_{n}ε_{n}}{m_{n}β_{n}}

and

S_{2} = | ∑_{|k|>(1−δ_{n})m_{n}} (\frac{h_{n−1}(T^{k+i}x)}{k} − \frac{h_{n−1}(T^{k+i}x)}{k + 2i})| ≤

≤ 2∥h_{n−1}∥_{∞} ∑_{k>(1−δ_{n})m_{n}} (\frac{1}{k} − \frac{1}{k + 2i}) ≤ 4i∥h_{n−1}∥_{∞} ∑_{k>(1−δ_{n})m_{n}} \frac{1}{k^{2}} ≤

≤ 4δ_{n}m_{n}∥h_{n−1}∥_{∞} \frac{1}{(1 − δ_{n})m_{n} − 1} ≤ 8δ_{n}∥h_{n−1}∥_{∞}.

Thus (6) holds and therefore the inequality

max (|H(h_{n})(T^{i}x)|, |H(h_{n})(T^{−i}x)|) ≥
≥ \frac{ε_{n}}{iβ_{n}} − \frac{6δ_{n}ε_{n}}{m_{n}β_{n}} − 4δ_{n}∥h_{n−1}∥_{∞} ≡ η_{n}(i)  (7)

is true.

Denote by the same letter η_{n} the function on X defined by the equalities: η_{n}(T^{i}x) = η_{n}(|i|) whenever x ∈ T \overline{n}B_{m_{n}} and |H(h_{n})(T^{i}x)| ≥ η_{n}(|i|), 0 < |i| < δ_{n}m_{n}; η_{n}(x) = 0 otherwise.

Obviously,

|H(h_{n})| ≥ η_{n}.

Let E_{n} = ∪_{\overline{n}−δ_{n}m_{n}<i<\overline{n}+δ_{n}m_{n}}T^{i}B_{m_{n}} and

M_{n} = {x ∈ X : |H( ∑_{k=n+1}^{∞} g_{k})| > 1}.

Because of inequality (1),

μ(M_{n}) ≤ c · ∥ ∑_{k=n+1}^{∞} g_{k}∥ = c · ∑_{k=n+1}^{∞} ε_{k}.

Taking into account (7), we have

∫_{E_{n}} |H(g)|dμ ≥ ∫_{E_{n}} | |H(h_{n})| − |H( ∑_{k=n+1}^{∞} g_{k})| |dμ ≥

≥ ∫_{E_{n}\M_{n}} (|H(h_{n})| − 1)dμ ≥ ∫_{E_{n}\M_{n}} η_{n}dμ − μ(E_{n}) ≥

≥ ∫_{E_{n}} η_{n}dμ − ∥η_{n}∥_{∞}μ(M_{n}) − μ(E_{n}) ≥

≥ ∑_{i=1}^{[δ_{n}m_{n}]} η_{n}(i)β_{n} − \frac{ε_{n}}{β_{n}} μ(M_{n}) − μ(E_{n}) ≥

≥ ε_{n} log(δ_{n}m_{n}) − \frac{6δ_{n}ε_{n}[δ_{n}m_{n}]}{m_{n}} − 4δ_{n}∥h_{n−1}∥_{∞}μ(E_{n}) −

− \frac{ε_{n}}{β_{n}} c ∑_{k=n+1}^{∞} ε_{k} − μ(E_{n}).

Thus, by virtue of (5),

∫_{E_{n}} |H(g)|dμ ≥ 1 − 2μ(E_{n}),

and since E_{n}, n = 1, 2, …, are pairwise disjoint, we can conclude that

∥H(g)∥ = ∞.

References

1. M. Cotlar, A unified theory of Hilbert transforms and ergodic theorems. Rev. Mat. Cuyana 1(1955), No. 2, 105–167.

2. K. Petersen, Ergodic Theory. Cambridge Univ. Press., 1983.

3. O. D. Tsereteli, On some problems arising in the theory of the Fourier series. (Russian) Seminar Inst. Prikl. Mat. Tbilis. Univ. Annot. Dokladov 6(1972), 33–36.

4. O. D. Tsereteli, Metric properties of conjugate functions. J. of Soviet Math. 7(1977), 317–343.

5. O. D. Tsereteli, On the integrability of conjugate functions. (Russian) Mat. Zametki 4(1968), 461–465.

(Received 21.12.1995)

Author’s address: A. Razmadze Mathematical Institute Georgian Academy of Sciences 1, M. Aleksidze St., Tbilisi 380093 Georgia