Georgian Mathematical Journal Volume 8 (2001), Number 4, 727–732

THE REARRANGEMENT INEQUALITY FOR THE ERGODIC MAXIMAL FUNCTION

L. EPHREMIDZE

Abstract. The equivalence of the decreasing rearrangement of the ergodic maximal function and the maximal function of the decreasing rearrangement is proved. Exact constants are obtained in the corresponding inequalities.

2000 Mathematics Subject Classification: 28D05, 26D15.

Key words and phrases: Measure-preserving ergodic transformation, ergodic maximal function, decreasing rearrangement.

Let (X, 𝕊, μ) be a σ-finite measure space and T : X → X be a measurepreserving ergodic transformation. For a measurable function f the ergodic maximal function is defined as

Mf(x) = sup_{N} \frac{1}{N} ∑_{k=0}^{N−1} |f(T^{k}x)|, x ∈ X.

The decreasing rearrangement of f is the function f^{∗} defined on [0, ∞) by

f^{∗}(t) = inf {λ : μ(|f| > λ) ≤ t}  (1)

and its maximal function is denoted by f^{∗∗}:

f^{∗∗}(t) = \frac{1}{t} ∫_{0}^{t} f^{∗}(τ)dτ, t > 0.

The equivalence of (Mf)^{∗} and f^{∗∗}, i.e., the validity of inequalities

cf^{∗∗}(t) ≤ (Mf)^{∗}(t) ≤ Cf^{∗∗}(t)

with constants c and C independent of f and t (these inequalities sometimes are called rearrangement inequalities) was proved by several authors when M stands for Hardy–Littlewood maximal operator (see [8], [5] for the one-dimen-sional case and [1] for higher dimensions). This fact is very useful in the proofs of many theorems on the related topics (see [2]).

In the present paper, we prove analogous inequalities for the ergodic maximal operator (see (2) below). The constants \frac{1}{2} and 1 in these inequalities are exact and the corresponding examples are constructed.

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Theorem. Let f ∈ L(X). Then

\frac{1}{2} f^{∗∗}(t) ≤ (Mf)^{∗}(t) ≤ f^{∗∗}(t)  (2)

when 0 < t < μ(X).

Remark. If μ(X) < ∞ and t ≥ μ(X), then (Mf)^{∗}(t) = 0. Thus the second inequality in (2) is valid for each t > 0, while the first inequality fails to hold whenever t ≥ μ(X) unless f is identically zero.

In the proof of the theorem we can take function f nonnegative since all functions considered depend only on the modulus of f. We shall also assume that the measure space (X, 𝕊, μ) is nonatomic. The case when the space has atoms can easily be reduced to the nonatomic case by “putting” suitable measurable sets into the atoms, keeping the values of f inside the atoms unchanged and defining T correspondingly. This process does not change the distribution functions λ ⟼ μ(f > λ) and λ ⟼ μ(Mf > λ), λ > 0. Consequently f^{∗}(t) and (Mf)^{∗}(t) keep the same values for each t > 0.

The following notation will be used: f^{+} = max(f, 0), f^{−} = max(−f, 0). S_{n}(f)(x) = ∑_{k=0}^{n} f(T^{k}x) and A_{n}(f)(x) = \frac{1}{n+1}S_{n}(f)(x). 1_{E} stands for the characteristic function of E. {f > 0} or (f > 0) means {x ∈ X : f(x) > 0}.

Since a weak-type estimate for the ergodic maximal operator has a simple form

μ(Mf > λ) ≤ \frac{1}{λ} ∫_{(Mf>λ)} f dμ,  (3)

where f ∈ L(X), λ > 0 (see, e.g., [7]), the second inequality in (2) can be proved easily and it is given below for the sake of completeness.

Proof of the inequality (Mf)^{∗}(t) ≤ f^{∗∗}(t), t > 0. Since \frac{1}{μ(E)} ∫_{E} f dμ ≤ \frac{1}{t} ∫_{0}^{t} f^{∗}(τ)dτ for each measurable E with μ(E) = t and f^{∗∗}(t) is a decreasing function (see, e.g., [2]), we have

f^{∗∗}(t) ≥ sup_{μ(E)≥t} \frac{1}{μ(E)} ∫_{E} f dμ.  (4)

Consider the nontrivial case when (Mf)^{∗}(t) > 0. It follows from definition (1) that

0 < λ < (Mf)^{∗}(t) ⟹ μ(Mf > λ) > t.  (5)

Because of (3) we have

λ ≤ \frac{1}{μ(Mf > λ)} ∫_{(Mf>λ)} f dμ, λ > 0.  (6)

It follows from (5) and (4) that

sup_{0<λ<(Mf)^{∗}(t)} \frac{1}{μ(Mf > λ)} ∫_{(Mf>λ)} f dμ ≤ f^{∗∗}(t).

Consequently, if we let λ in (6) tend to (Mf)^{∗}(t) from the left, we get the second inequality in (2).

For the proof of the first inequality in (2) we need

Lemma. Let g : X → ℕ_{0} = {0, 1, 2, … } and g ∈ L(X). Then

μ(Mg ≥ 1) = min ( ∫_{X} g dμ, μ(X)).

Proof. That μ(Mg ≥ 1) = μ(X) whenever ∫_{X} g dμ ≥ μ(X) follows from the Individual Ergodic Theorem:

lim_{n→∞} A_{n}(g)(x) = \frac{1}{μ(X)} ∫_{X} g dμ  (7)

for a.a. x ∈ X (see, e.g., [7]). Thus it is sufficient to consider the case where

∫_{X} g dμ < μ(X).  (8)

We shall use the filling scheme method (see [6], [7] or [3]) truncating the function g at level 1. Let

g_{0} = g and g_{n+1} = 1_{(g_{n}≥1)} + (g_{n} − 1)^{+} ∘ T.  (9)

Observe that g_{n} takes only nonnegative integer values and

g_{n} = 1_{(g_{n}≥1)} + (g_{n} − 1)^{+}, n = 0, 1, ….  (10)

If we consider another sequence

h_{0} = g − 1 and h_{n+1} = −h_{n}^{−} + h_{n}^{+} ∘ T,

then, as it can easily be checked by induction,

h_{n} = g_{n} − 1, n = 0, 1, ….  (11)

That

lim_{n→∞} ∫_{X} h_{n}^{+} dμ = lim_{n→∞} ∫_{X} (g_{n} − 1)^{+} dμ = 0  (12)

is proved in [3] (see (19) therein). At the same time, since T is measurepreserving and (10) holds, we obtain

∫_{X} g_{n+1}d μ = ∫_{X} 1_{{g_{n}≥1}} dμ + ∫_{X} (g_{n} − 1)^{+} ∘ T dμ =

= ∫_{X} 1_{{g_{n}≥1}} dμ + ∫_{X} (g_{n} − 1)^{+} dμ = ∫_{X} g_{n} dμ,

n = 0, 1, . . .. Thus, for each n ≥ 0, we have

∫_{X} g_{n} dμ = ∫_{X} g dμ.  (13)

We also use the equality of sets

{x : max_{0≤m≤n} S_{m}(h_{0})(x) ≥ 0} = (h_{n} ≥ 0),  (14)

n = 0, 1, …, which is proved in [4] (see Lemma 2; see also Lemma 1.1 in [3], where the basic idea of the proof is given). Since

lim_{n→∞} \frac{1}{n + 1} ∑_{k=0}^{n} g(T^{k}x) = lim_{n→∞} A_{n}(g)(x) < 1

for a.a. x (see (7), (8)), we have

(Mg ≥ 1) = {x : A_{n}(g)(x) ≥ 1 for some n ≥ 0}

= ∞ ; ⋃_{n=0}{x : max_{0≤m≤n} A_{m}(g)(x) ≥ 1} = ∞ ; ⋃_{n=0}{x : max_{0≤m≤n} S_{m}(h_{0})(x) ≥ 0}

= ∞ ; ⋃_{n=0}(h_{n} ≥ 0) = ∞ ; ⋃_{n=0}(g_{n} ≥ 1)

(the first equality holds if we neglect the sets of measure 0 and all other equalities are exact; (see (11), (14)). Thus

μ(Mg ≥ 1) = lim_{n→∞} μ(g_{n} ≥ 1)  (15)

(that (g_{n} ≥ 1) = (h_{n} ≥ 0), n = 0, 1, …, is an increasing sequence of sets follows from definition (9) and also from (14)).

It follows from (13) and (10) that

∫_{X} g dμ = ∫_{X} g_{n} dμ = ∫_{X} (1_{{g_{n}≥1}} + (g_{n} − 1)^{+}) dμ = μ(g_{n} ≥ 1) + ∫_{X} (g_{n} − 1)^{+} dμ.

Hence, taking into account (15) and (12), we get

μ(Mg ≥ 1) = ∫_{X} g dμ.

Proof of the inequality \frac{1}{2}f^{∗∗}(t) ≤ (Mf)^{∗}(t), 0 < t < μ(X). Fix t ∈ (0, μ(X)) and assume f^{∗∗}(t) = λ_{0}. We shall show that

μ(Mf ≥ \frac{1}{2} λ_{0}) > t.  (16)

The first inequality in (2) follows from (16) by virtue of definition (1). Let E ∈ 𝕊 be a measurable set with

μ(E) = t  (17)

such that

\frac{1}{μ(E)} ∫_{E} f dμ = \frac{1}{t} ∫_{0}^{t} f^{∗}(τ)dτ = λ_{0}.  (18)

Since we assume that the space is nonatomic, such E exists (see, e.g., [2], Lemma 2.2.5). Define the function g as follows

g = ∑_{m=0}^{∞} \frac{λ_{0}}{2} m1_{(\{ \frac{λ_{0}}{2} m≤f< \frac{λ_{0}}{2} (m+1)\}∩E)}.

Observe that g ≤ f, \frac{2}{λ_{0}} g takes only nonnegative integer values and f(x)−g(x) < \frac{λ_{0}}{2} for each x ∈ E. We have

∫_{E} g dμ > ∫_{E} f dμ − \frac{λ_{0}}{2} μ(E) = \frac{λ_{0}}{2} μ(E)

(see (18)). Thus

∫_{X} \frac{2}{λ_{0}} g dμ > μ(E)

and because of Lemma we have

μ(Mg ≥ \frac{λ_{0}}{2} ) = μ(M(\frac{2}{λ_{0}} g) ≥ 1) = min (\frac{2}{λ_{0}} ∫_{X} g dμ, μ(X))
> min(μ(E), μ(X)) = t

(see (17)). Since Mf ≥ Mg, we have proved (16).

At the end of the paper we shall show that the constants \frac{1}{2} and 1 are exact in the inequalities in (2) and cannot be improved. This is clear for 1 since it may happen that (Mf)^{∗}(t) and f^{∗∗}(t) are equal (e.g., for constant functions). A simple example below shows that the equality

\frac{1}{2} f^{∗∗}(t) = (Mf)^{∗}(t)

can hold for t such that f^{∗∗}(t) does not vanish.

Example. Let T̃ be a (Lebesgue) measure-preserving ergodic transformation of [0; \frac{1}{2}) and define T by the equalities T (x) = x + \frac{1}{2} when x ∈ [0; \frac{1}{2}) and T (x) = T̃(x − \frac{1}{2}) when x ∈ [\frac{1}{2}; 1). Then T is a measure-preserving ergidic transformation of [0; 1). If f = 1_{[ \frac{1}{2} ;1)}, then Mf(x) = \frac{1}{2} when x ∈ [0; \frac{1}{2}) and Mf(x) = 1 when x ∈ [\frac{1}{2}; 1). Thus (Mf)^{∗}(\frac{1}{2}) = \frac{1}{2}, while f^{∗∗}(\frac{1}{2}) = 1.

References

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2. C. Bennett and R. Sharpley, Interpolation of operators. Academic Press, Boston etc., 1988.

3. L. Ephremidze, On the distribution function of the majorant of ergodic means. Studia Math. 103(1992), 1–15.

4. L. Ephremidze, On the uniqueness of the ergodic maximal function. Fund. Math. (to appear).

5. C. Herz, The Hardy–Littlewood maximal theorem. Symposium on harmonic analysis, 1–27, University of Warwick, 1968.

6. J. Neveu, The filling scheme and the Chacon–Ornstein theorem. Israel J. Math. 33(1979), 368–377.

7. K. Petersen, Ergodic theory. Cambridge University Press, Cambridge etc., 1983.

8. F. Riesz, Sur un theoreme de maximum de MM. Hardy et Littlewood. J. London Math. Soc. 7(1932), 10–13.

(Received 17.08.2001)

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

Current address: Institute of Mathematis Academy of Sciences of the Czech Republic Žitná 25, 115 67 Prague 1 Czech Republic E-mail: lasha@math.cas.cz