Almost-Everywhere Convergence and Polynomials

Denote by $\Gamma$ the set of pointwise good sequences. Those are sequences of real numbers $(a_k)$ such that for any measure preserving flow $(U_t)_{t\in \mathbb R}$ on a probability space and for any $f\in L^\infty$, the averages $\frac{1}{n} \sum_{k=1}^{n} f(U_{a_k}x) $ converge almost everywhere. We prove the following two results. [1.] If $f: (0,\infty)\to\mathbb R$ is continuous and if $\big(f(ku+v)\big)_{k\geq 1}\in\Gamma$ for all $u, v>0$, then $f$ is a polynomial on some subinterval $J\subset (0,\infty)$ of positive length. [2.] If $f: [0,\infty)\to\mathbb R$ is real analytic and if $\big(f(ku)\big)_{k\geq 1}\in\Gamma$ for all $u>0$, then $f$ is a polynomial on the whole domain $[0,\infty)$. These results can be viewed as converses of Bourgain's polynomial ergodic theorem which claims that every polynomial sequence lies in $\Gamma$.


Introduction
For 1 ≤ p ≤ ∞, a sequence of real numbers (a k ) k≥1 is said to be p-good pointwise if for any measure preserving flow (U t ) t∈R on a probability space (X, B, µ) and for any f ∈ L p (X, µ), the averages 1 n n k=1 f (U a k x) converge almost everywhere.
Denote by Γ p the set of p-good pointwise sequences and set Γ = Γ ∞ . J. Bourgain proved in a series of papers (see [?] for the proof and further references) that polynomial sequences lie in Γ p , for all p > 1. Bourgain, in fact, formulated his result for Z d actions instead of R actions. To see the idea how we can translate the Z d results to results on a flow, consider the example f (x) = √ 2x 2 +x. The transformation T u,v = U √ 2u+v is a Z 2 action, Date: November 10, 2009.
1 and U f (n) = T n 2 ,n , hence Bourgain's result applied to T gives the result for the flow U t . For a simpler proof (in the case p = 2) of Bourgain's result, and for references see [?]. It is a famous open problem whether or not (k 2 ) ∈ Γ 1 .
In the paper we prove the following two theorems. (Each can be viewed as a converse of J. Bourgain's result mentioned above). Denote Theorem A. Let f : R + → R be a real analytic function such that the set is uncountable. Then f (x) must be a polynomial.
Then there exists a subinterval J ⊂ R + (of positive length) such that f restricted to J is a polynomial.
We remark that, in general, the conclusion of Theorem B cannot be that f (x) agrees with a polynomial on a halfline, say. Indeed, we have satisfies the conditions of Theorem A with U = V = R + .
The above proposition shows that f (x) satisfying the conditions of Theorem B (even with U = V = R + ) may be a periodic non-constant function, so definitely not a polynomial on large subintervals of R + .
The claim of Proposition 1.1 follows from the results in [?]. Note that our proof of Theorem A depends on the fact that f is analytic at 0. (See Remark 2.1 in the next section).
The following result (first proved in [?]) plays a central role in the proofs of both theorems.
Lemma 1.2 ([?]). Any sequence (a k ) ∈ Γ must be linearly dependent over the field Q of rational constants.
For a simpler proof of this lemma (avoiding the use of Bourgain's entropy method) we refer the reader to [?].

Proof of Theorem A
Proof of Theorem A. In view of Lemma 1.2, for every x ∈ U , there are an integer n(x) ≥ 1 and an n(x)-tuple of rationals Since the set of possible pairs n(x), q(x) is countable while the set of x ∈ U is uncountable, there is a pair n, q corresponding to an uncountable set U ′ ⊂ U of x: Since f (x) is analytic, the identity q k k r c r x r one deduces that, for all integers r ≥ 0, either c r = 0 or n k=1 q k k r = 0. It follows that n k=1 q k k r = 0, for all r in the set K = {r ∈ Z + : c r = 0}.
We observe that lim r→+∞ n k=1 q k k r = +∞ (the last term q n n r = n r is dominant in the sum). We conclude that the set K = {r ∈ Z + : c r = 0} is finite, and f (x) is a polynomial.
Remark 2.1. We don't know whether Theorem A holds if the domain of f is assumed to be (0, ∞) rather than [0, ∞). The recurrence relation (2.1) still holds in this setting. The conclusion of Theorem can be derived under the assumptions that f is analytic on (0, ∞) and that either 0 or ∞ is an isolated singularity of the analytic extension of f .

Proof of Theorem B
Proof of Theorem B. In view of Lemma 1.2, for every u ∈ U and v ∈ V there is an integer n(u, v) ≥ 1 and an n(u, v)-tuple of rationals For every integer n ≥ 1 and an n-tuple of rationals q ∈ Q n , denote Since U ×V is a countable union of its closed subsets K(n, q), by the Baire category theorem there is a choice of n and q = (q 1 , q 2 , . . . , q n ) ∈ Q n , with not all q k = 0, such that the set K(n, q) contains a non-empty interior, say the set U ′ × V ′ where U ′ ⊂ U and V ′ ⊂ V are non-empty open subintervals of (0, ∞). We conclude that Definition 3.1. Let U ⊂ R be an open set, and let X ⊂ U be a finite subset, card(X) ≥ 1. Let f : U → R, g : X → R and h : X → R be three functions such that f is continuous and g is injective. The provided that |t| and |s| are small enough.
Example 3.2. Let U ′ , V ′ , n ≥ 1 and q = (q 1 , q 2 , . . . , q n ) ∈ Q n be such as described in the paragraph preceding (3.1). Pick u 0 ∈ U ′ , v 0 ∈ V ′ and let X be the set X = {x k | 1 ≤ k ≤ n} where x k = ku 0 + v 0 . Define g, h : X → R as follows: g(x k ) = k, h(x k ) = q k . With these choices, it follows from (3.1) that the quintuple (f, g, h, X, R + ) is balanced. Indeed, by setting u = u 0 + t and v = v 0 + s, we obtain Now the claim of Theorem B follows from the following result.
Proposition 3.3. Let (f, g, h, X, U ) be a balanced quintuple in the sense of Definition 3.1, with card(X) = n ≥ 1. Let x 0 ∈ X be such that h(x 0 ) = 0.
Then f (x) is a polynomial of degree ≤ n − 2 in a neighborhood of x 0 .
By definition, the zero constant is a polynomial of degree −1.

Proof of Proposition 3.3; smooth case
Proof of Proposition 3.3; smooth case. First we provide proof under the additional assumption that f ∈ C ∞ (U ). Without loss of generality, we may assume that 0 / ∈ h(X). The proof is by induction on n = card(X). If n = 1, then by setting t = 0 we get f (x 0 + s) = 0, for all |s| small enough, i. e. f (x) vanishes in a neighborhood of x 0 . The case n = 1 is validated.
Next we assume that n ≥ 1, that X = {x 0 , x 1 , . . . , x n } and that the claim of proposition is validated if card(X) ≤ n. Observe that for an arbitrary real constant c a quintuple (f, g, h, X, U ) is balanced if and only if a quintuple (f, g + c, h, X, U ) is. This is because f (x + s + t g(x)) = f (x + s − ct + (g(x) + c)t), and the pair (s, t) is close to (0, 0) (in the R 2 metric) if and only if (s − ct, t) is.
By the induction hypothesis, for any x ′ ∈ X ′ , the derivative f ′ is a polynomial of degree ≤ n − 2 in some neighborhood of x ′ . It follows that f (x) is a polynomial of degree ≤ n − 1 = (n + 1) − 2 in a neighborhood of x 0 . This completes the proof of Proposition 3.3 under the added assumption that f ∈ C ∞ (R + ).

Proof of Proposition 3.3; continuous case
Proof of Proposition 3.3; continuous case. Since (f, g, h, X, U ) is a balanced quintuple, there exists an ǫ > 0 such that (3.2) holds provided that |s|, |t| < ǫ. In the preceding section we proved that, under the additional condition that f ∈ C ∞ (R + ), there exists a neighborhood W of a point x 0 such that f | W is a polynomial of degree ≤ n − 2.
Our proof provides slightly more: This neighborhood W depends only on ǫ, g, h, X and U but not on f ∈ C ∞ (R + ). The observation will be used in what follows. Now we move to the general case of f ∈ C(R + ) (rather than f ∈ C ∞ (R + )).
Fix any function φ ∈ C ∞ (R) which satisfies R φ(t) dt = 1 and vanishes outside the interval [−1, 1]. For integers m ≥ 1, denote The sequence of kernels φ m (x) is known to converge the δ-function at 0 in the sense that Note that all , and dist(x, ∂U ) stands for the distance between x and the boundary of U ⊂ R.
Let V be a neighborhood of the set X such that its closure V is compact and is contained in U . Then the pointwise convergence in (5.1) is uniform on V , and in fact f m ∈ C ∞ (V ) holds for all m large enough. It is also clear that for m large enough, (f m , g, h, X, V ) forms a balanced quintuple because (f, g, h, X, V ) does; moreover, there exists ǫ ′ > 0 such that x∈X h(x) f m (x + s + tg(x)) = 0 holds simultaneously for all large m (say, m > m 0 ) and s, t ∈ (−ǫ ′ , ǫ ′ ). It follows that there exists a neighborhood W ⊂ V of a point x 0 such that each function f m is a polynomial of degree ≤ n − 2 in it. (Here we use the observation made in the second paragraph of this section). In view of the uniform convergence (5.1), f | W is also a polynomial of degree ≤ n − 2, completing the proof of Proposition 3.3.

Concluding remarks
The following is a slightly more general version of Theorem A.
Theorem A ′ . Let (r k ) k≥1 be a sequence of distinct positive numbers, let f : R + → R be a real analytic function such that the set is uncountable. Then f (x) must be a polynomial.
The proof of Theorem A ′ is very similar to the proof of Theorem A.