Bounded Power Series on the Real Line

We investigate power series that converge to a bounded function on the real line. First, we establish relations between coefficients of a power series and boundedness of the resulting function; in particular, we show that boundedness can be prevented by certain Tur\'an inequalities and, in the case of real coefficients, by certain sign patterns. Second, we show that the set of bounded power series naturally supports three topologies and that these topologies are inequivalent and incomplete. In each case, we determine the topological completion. Third, we study the algebra of bounded power series, revealing the key role of the backward shift operator.


Introduction
Relating the coefficients of a power series to the values of the corresponding function has a long history [16].In 1885, Cauchy proved that a power series bounded on the complex plane must be constant, and thus all but one coefficients must be equal to zero [4]; the result is now known as Liouville's theorem.In 1917, Schur characterized the coefficients of power series bounded in the interior of the unit circle [7,14,16].In 1940, Boas studied power series of exponential type bounded on the real line [2,Ch. 7,8].Despite the popularity of some power series bounded on the real line but not of exponential type, like the Gaussian exp(−πx 2 ) or the Uppuri-Carpenter generating function exp(1 − exp(x)) [1,6,17], this general case remains largely unexplored.
We say that a power series (1) is bounded on the real line if it converges for every x ∈ R and if the corresponding function f : R → C is bounded.Boundedness on a line is a remarkable property: infinitely many monomials have to balance each other out, uniformly, on an unbounded set.In contrast to convergence, boundedness is not a tail property: it is not preserved by changing finitely many coefficients.Indeed, dividing (1) by x n for some n ≥ 1 gives (2) implying that the tail (a k ) k>n determines a 1 , . . ., a n (the coefficient a 0 is arbitrary).On the opposite, tails are not uniquely determined by heads.Indeed, for every polynomial p(x) = a 0 + a 1 x + • • • + a n x n the power series is bounded on the real line, implying that every finite complex sequence is the head of a bounded power series, and thus that tails are not uniquely determined.While boundedness is not a tail property itself, it entails some tail properties.In Section 2 we related boundedness to certain Turán inequalities [10,15], namely (4) where N ≥ 1 and χ > 0 are constants.The constant χ controls the logarithmic concavity of the sequence (a n ) n≥0 .For simplicity, we call a power series real (complex ) if its coefficients are real (complex).This is the main result of Section 2: x n be a real power series, bounded on the real line, and satisfying (4) for some N ≥ 1 and 0 < χ < 1.Let and let L(n) denote the supremum of the integers L ≥ 0 for which the tail (a k ) k≥n contains L consecutive elements with the same sign.Then The proof of Theorem 1.1 relies on a discrete Legendre transform, see Remark 2.3.This theorem is useful to show that certain power series are not bounded: Corollary 1.2.A real power series satisfying (4) for some N ≥ 1 and 0 < χ ≤ 1  2 is unbounded on the real line.
The coefficient sequence of a real bounded power series must change sign infinitely many times.If Turán inequalities hold, something stronger is true: Proof.The first statement follows from (5) since L(0) ≤ 2ψ(χ) < ∞.The second statement follows from replacing f (x) by f (−x).□ The techniques developed in Section 2 might help deducing inequalities for combinatorial sequences from the boundedness of their generating function.In contrast to Corollary 1.3, we conjecture that exp(1−exp(x)) contains arbitrarily long subsequences of consecutive coefficients with the same sign; see the related Wilf's Conjecture [1,6] and Remark 2.9 for details.
Let B denote the set of complex power series bounded on the real line.Equation (3) suggests that the set B is dense in the set of complex power series with respect to some topology.Section 3 discusses three natural topologies on B: Theorem 1.4.Given a power series bounded on the real line f The following facts are true: ), the set of bounded continuous complex-valued functions on the real line; (iii) The set B endowed with uniform convergence on compact subsets of C is (isometric to) a dense subset of the set entire functions.Moreover, the three topologies are inequivalent.
Section 4 concerns the algebraic structure of B. Taking n = 1 in (2) shows that B is invariant under backward shift σ ((a n ) n≥0 ) = (a n+1 ) n≥0 .Note that the backward shift plays a central role in the theory of power series in the unit circle [5,11].The following theorem collects the main findings of Section 4: Roughly speaking, the backward shift σ on B is almost invertible: it is one-to-one on a dense subset and for every small perturbation λ ̸ = 0 the map σ − λ is one-to-one on the whole set.
By definition, the elements of σ k (B) are exactly the bounded power series that can be extended k times to the left to a bounded power series.Such extension is unique for the exception of the constant term, which does not affect boundedness but does affect further extensibility.The strict inclusions (6) state that for every k ≥ 0 there is a bounded power series that can be extended k times but not k + 1 times.Since σ does not preserve products, the fact that the set σ k (B) is a subring is not trivial.Equation ( 7) relates left-extensibility of a power series to the behavior at x = ∞ of the corresponding function.If O(x k ) is replaced by o(x k ), then (7) turns into the notion of Peano differentiability [8].In particular, with the exception of f = 0, a power series f (x) ∈ σ ∞ (B) is infinitely Peano differentiable (but not analytic) at x = ∞.

Boundedness and Coefficients
In this section we are interested in asymptotic properties of the coefficient sequence that are entailed by the boundedness of the resulting function.We begin by reviewing some elementary facts about convergent power series.
Lemma 2.1.Let f (x) = n≥0 a n x n be a power series convergent on the real line and not a polynomial.For every fixed x ∈ R, let arg max n≥0 |a n x n | denote the minimum index realizing the maximum max n≥0 |a n x n |.Then: Proof.Convergence implies that for every fixed x ∈ R the sequence (|a n x n |) n≥0 has a maximum.Moreover, for every fixed n < m such that a n , a m ̸ = 0 we have x n a m x m = 0. Since there are infinitely many non-zero coefficients, the statement follows.□ Let f (x) = n≥0 a n x n be a power series with non-zero coefficients and convergent on the real line.Then the sequence ( a n a −1 n+1 ) n≥0 converges to +∞.If (4) holds for some constants N ≥ 1 and 0 < χ < 1, then the sequence ( a n a −1 n+1 ) n≥0 is strictly monotone and, furthermore, for every m ≥ N (9) In (9) the maximizer is unique; this represents the generic case.If x = |a m−1 a −1 m | for some m ≥ N there are two maximizers.As a particular case of (9), in Lemma 2.2 we will take x equal to the geometric mean of the endpoints.Lemma 2.2.Let f (x) = n≥0 a n x n be a power series convergent on the real line and satisfying (4) for some constants N ≥ 1 and 0 < χ < 1. Fix m ≥ N and let x = a m−1 a −1 m+1 1 2 .Then for every p ≥ q ≥ 0 the following inequalities hold: (10) a m+p x m+p a m+q x m+q ≤ χ .
Proof.Let us prove the first inequality in (10).Multiple applications (4) give Multiplying both sides by .
The second inequality in (10) can be obtained in a similar way.□ Remark 2.3.The result of Lemma 2.2 can be interpreted as a discrete Taylor-like inequality.Letting g(n) = − log(|a n |) and taking q = 0 in (10) gives Define the discrete derivative of a sequence g as (∆g)(n) = g(n + 1) − g(n).Then the quantity 1 2 (g(m + 1) − g(m − 1)) is the average of ∆g(m) and ∆g(m − 1), while log(χ −1 ) is bounded by ∆ 2 g(n − 1).Let x = exp(s).Then |a n x n | = exp(ns − g(n)) and the maximum in ( 9) can be interpreted as a discrete Legendre transform.
Proof of Theorem 1.1.Let f (x) = n≥0 a n x n be a real power series, convergent on the real line, and satisfying (4) for some constants N ≥ 1 and 0 < χ < 1.
We will prove that if lim n→∞ L(n) ≤ ϑ(χ) or if lim n→∞ L(n) > 2ψ(χ), then the function f : R → R is unbounded on the real line.We will do so by exhibiting a sequence (x m ) m≥N and a constant C > 0 such that lim m→∞ x m = +∞ and (11) |f for every m sufficiently large.By Lemma 2.1, this implies that f is unbounded.Note that for every fixed M ≥ 0 we have Therefore, in obtaining (11) we can always discard a finite number of terms.In particular, without loss of generality, we can assume N = 0 in (4).Define By definition of ϑ(χ) and ψ(χ) it follows that Θ < 1 and Ψ < 1 2 respectively.Suppose that lim n→∞ L(n) ≤ ϑ(χ), that is, L(n) ≤ ϑ(χ) for every n sufficiently large.In light of (12), without loss of generality we can assume L(n) ≤ ϑ(χ) for every n ≥ 0, that is, every subsequence of consecutive coefficients with the same sign has length at most ϑ(χ).If coefficients are eventually alternating signs, that is, a n a n−1 < 0 for every n sufficiently large, then the coefficients of f (−x) have eventually all the same sign, thus f (−x) is unbounded, and therefore f (x) is unbounded.Otherwise, there are infinitely many m ≥ 0 such that a m and a m+1 have the same sign and, without loss of generality, we assume them positive.Fix such m and let Let p, n ≥ 1 be any integers such that a m+p > 0 and a m+p+1 , . . ., a m+p+n < 0. By hypothesis n ≤ ϑ(χ).From the first inequality of ( 10) we obtain It follows that every sequence of consecutive negative terms appearing after the index m + 1 is dominated by the positive term preceding it.Therefore Similarly, from the second inequality of ( 10) we obtain Since lim m→∞ x m = +∞ and Θ < 1, by Lemma 2.1 it follows that f is unbounded.Now suppose that lim n→∞ L(n) > 2ϑ(χ), that is, L(n) ≥ 2ϑ(χ) + 1 eventually.Then there are infinitely many m ≥ 0 such that the coefficients a m−ϑ(χ) , . . ., a m+ϑ(χ) have the same sign and, without loss of generality, we can suppose them positive.Fix such m and let From the first inequality of ( 10) we obtain Similarly, from the second inequality of (10) we obtain Since lim m→∞ x m = +∞ and Ψ < 1 2 , by Lemma 2.1 it follows that f is unbounded.□ We conclude this session with some open problems.As discussed in the introduction, bounded power series on the real line cannot have arbitrary convexity.Definition 2.4.Let χ denote the supremum of the constants χ such that if a complex power series satisfies (4) for some N ≥ 1, then it is unbounded on the real line.Let χR denote the supremum of the constants χ such that if a real power series satisfies (4) for some N ≥ 1, then it is unbounded on the real line.
Example 2.5.The coefficients of the power series associated to sin(x) + cos(x) satisfy |a n | = 1 n! , thus (4) with N = 1 and χ = 1.Corollary 1.2 in the introduction gives χR ≥ 1  2 .Its proof relies on the inequalities ϑ(χ) ≥ 2 and ψ(χ) ≤ 1, which can be seen numerically to hold for χ ≤ 0.67522.The argument of the second part of the proof of Theorem 1.1 generalizes without modifications to the case of complex coefficients, showing that if ψ(χ) = 0 then f is unbounded on the real line.We see that ψ(χ) = 0 holds for χ ≤ 0.207875.We are unable to close the gaps 0.67522 ≤ χR ≤ 1 and 0.207875 ≤ χ ≤ 1.
Conjecture 2.6.The constants χ and χR are equal to each other.
The techniques developed in this section might help deducing asymptotic inequalities of combinatorial sequences from the boundedness of their generating function.For example, let exp(1 − exp(x)) = n≥0 bn n! x n .The quantity b n , known as complementary Bell number or Uppuri-Carpenter number [17], is equal to the difference between the number of partitions of {1, . . ., n} with an even number of blocks and those with an odd number of blocks.We make the following conjecture: ≤ 1 follows from the elementary fact that the power series is convergent on the real line.

The Topology of Bounded Power Series
Let B denote the set of complex power series bounded on the real line.This section investigates topological aspects of B. Note that if a power series f (x) converges for every x ∈ R, then it converges for every z ∈ C. Therefore, there is a natural bijection between the set B and the set of entire functions that bounded on the real line.On the other hand, there is a natural bijection between the set B and the set of complex sequences (a n ) n≥0 satisfying There are three natural notions of convergence on B: first, ℓ 1 -convergence of coefficient sequences; second, uniform convergence of bounded functions on R; third, uniform convergence of entire functions on compact subsets of C. Theorem 1.4 states that these topologies are inequivalent and incomplete, and describes their topological completions.
Proof.If f (x) = n≥0 a n x n ∈ ker σ k , then the sequence (a n ) n is eventually 0, that is, the function f is a bounded polynomial, and thus a constant.It follows that f (x) ∈ ker σ.This proves ker σ k ⊆ ker σ.The converse ker σ ⊆ ker σ k is trivial.□ The map σ : B → B is not surjective, but it is not far from being surjective: Proof.Let k ≥ 0. Identify the function sin(x) with its power series.We claim that σ k sin(x) ∈ σ k (B) \ σ k+1 (B).Clearly we have σ k sin(x) ∈ σ k (B).For the sake of contradiction, assume that σ k sin(x) ∈ σ k+1 (B).Then there is g(x) ∈ B such that σ k sin(x) = σ k (σg(x)) and, by Lemma 4.1, there is a ∈ C such that sin(x) − a = g(x) − g(0) x .
This is impossible since the right hand side converges to 0 as x → ∞.In particular, the set σ k (B) is a subring of B.
Proof.We have f (x) ∈ σ k (B) if and only if there are a 0 , . . ., a k−1 ∈ C such that which is the same as ( 14) with c j = −a k−j .It remains to prove that σ k (B) is a ring.Since σ is linear, the set σ k (B) is closed under sums.From ( 14) it follows that σ k (B)σ h (B) ⊆ σ min(k+1,h+1) (B).
for every k, h ≥ 1. Taking k = h shows that σ k (B) is closed under products.□ We will need the following algebraic result, which we state in general: Lemma 4.4.Let (B, +) an abelian group and σ an endomorphism satisfying ker σ k = ker σ for every k ≥ 1. Define σ ∞ (B) = k≥0 σ k (B).Then the restriction of σ to σ ∞ (B) is one-to-one.

Conjecture 2 . 8 . 2 n ≥ 1 .
The coefficient sequence of the power series exp(1 − exp(x)) contains arbitrarily long subsequences of consecutive coefficients with the same sign.Remark 2.9.Wilf's Conjecture states that b n ̸ = 0 for n > 2 [1, 6].Since the function exp(1 − exp(x)) is bounded on the real line, if both Wilf's Conjecture and Conjecture 2.8 are true, then by Corollary 1.3 we have lim sup n→∞ nb n+1 b n−1 (n + 1)b In contrast, note that the inequality lim inf n→∞ nb n+1 b n−1 (n+1)b 2 n |x| → ∞ for every n ≥ 1 or, equivalently, for n = 1.The set B has the cardinality of the continuum: to see this, note that |C| ≤ |B| ≤ C N and |C| = C N = |R|.

Proposition 4 . 3 .
It remains to prove that σ ∞ (B) is ℓ 1 -dense in B. Let m ∈ N and λ > 0. Since the function x k (x m exp(−λx 2 )) is bounded for every k ≥ 0, we have x m exp(−λx 2 ) ∈ σ ∞ (B).By the proof of Theorem 1.4 the set {z m exp(−λz 2 )} λ>0,m∈N is dense in ℓ 1 (N, C), thus in B. This shows that σ ∞ (B) is dense in B. It follows that every σ k (B) is dense in B.□The following result characterizes the elements of σ k (B): Let f ∈ B and k ≥ 1.Then f ∈ σ k (B) if and only if there are c 1 , . . ., c k−1 ∈ C such that(14) f 1 x = c 1 x + c 2 x 2 + • • • + c k−1 x k−1 + O(x k ) as x → 0.