# Criteria for boundedness of power series

Consider a power series $\sum_{n=0}^{\infty} a_n x^n$ that is convergent for all real x, thus defining a function $f: \mathbb{R} \to \mathbb{R}$. Can one give necessary and sufficient criteria the sequence of the coefficients $(a_n)$ has to meet in order for $f$ to be bounded on $\mathbb{R}$? (Let's disregard the trivial case that $a_0$ is the only non-zero coefficient and let's call a sequence "function-bounded" if the power series is bounded.) Criteria for boundedness seem to be far more difficult to obtain than the usual criteria for convergence of a power series, here some remarks:

a) A necessary condition for $\sum_n a_n x^n$ to be bounded is that there are infinitely many non-zero coefficients which change sign infinitely many times.

b) The boundedness of $f$ is an "unstable" property of the sequence of coefficients: any non-zero change in any finite subset (except $a_0$) will destroy boundedness. Thus the linear subspace of all function-bounded sequences is rather "sparse" in the vector space of all sequences representing convergent power series.

c) On the other hand, the linear subspace of all function-bounded sequences contains at least all power series of functions that can be written as $\cos \circ h$ with $h$ an entire, real-analytic function, and the algebraic span of these functions. One could conjecture that this space is already the space of all bounded functions that can be represented as power series[EDIT: seems to be refuted, cf. comment below]. And perhaps this could be a starting point for deducing the criteria.

EDIT (conjecture added): Is is true, that every power series $\sum_{n=0}^{\infty} a_n x^n$ that is convergent for all real $x$ can be modified only by changing the signs of the terms to a convergent power series $\sum_{n=0}^{\infty} \epsilon_n a_n x^n, \quad \epsilon_n \in \{\pm1\}$ that is bounded for all real $x$?
Example: One can modifify the signs of the power series of the exponential function $\sum_{n=0}^{\infty} x^n/n!$ pretty easily to a bounded power series by $\epsilon_n = +1$ for $n = 0 or 1 \mod 4$ and $\epsilon_n = -1$ for $n = 2 or 3 \mod 4$, yielding the function $\sin(x) + \cos(x)$. (One can modify the signs pretty easily a bit further such that the power series is not only bounded on the real axis, but also on the imaginary axis - but this is not the question here). I have neither succeeded in finding a counterexample nor in prooving this conjecture.

EDIT2: Thanks for the nice counterexample. I would like to improve the conjecture as follows: Define a power series $\sum_{n=0}^{\infty} a_n x^n$ as nondominant, if for all $x \in \mathbb{R}$ the absolute value of every term $a_kx^k$ is smaller or equal than the sum of the absolute values of all the other terms: $|a_kx^k| \le \sum_{n \neq k} |a_n x^n|$. The improved conjecture is: Is is true, that every nondominant power series $\sum_{n=0}^{\infty} a_n x^n$ that is convergent for all real $x$ can be modified only by changing the signs of the terms to a convergent power series $\sum_{n=0}^{\infty} \epsilon_n a_n x^n, \quad \epsilon_n \in \{\pm1\}$ that is bounded for all real $x$?

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Even your first derivative can be unbounded, as $\cos x^2 .$ So, for an easier problem, can you characterize those functions for which every derivative is also bounded? The comparison need not be helpful, but I am thinking of Hardy fields, which are usually defined in one direction only. –  Will Jagy Jun 4 '10 at 21:11
Great Question. –  Charlie Frohman Jun 5 '10 at 1:41
Without attempting at being deep enough, I hardly believe that a nice criterion can exist by means of the Taylor coefficients. The Weierstrass products (en.wikipedia.org/wiki/Weierstrass_factorization_theorem) look more natural in the context. –  Wadim Zudilin Jun 5 '10 at 8:24
Many thanks for that remark. I had thought a bit about it before asking the question the way I did, but I had not continued thinking along these lines, because I'm mainly interested in a "real analysis" solution, and it is obviously not true that f is bounded if and only if all zeroes (in Weierstrass factorization) are real. But I will rethink about approaching the problem via Weierstrass. –  Andreas Rüdinger Jun 27 '10 at 10:47
This is an open problem, according to the Open Problem Garden, though I am not aware of recent progress on the problem : garden.irmacs.sfu.ca/?q=op/… –  Malik Younsi Jun 27 '10 at 18:44

Consider for instance an entire function $f(x):=\sum_{n=0}^\infty a_n x^n$ with
$$|a_n|=3^{-n^2}.$$
then $f$ is unbounded on $\mathbb{R}$ since we have, for all $a\in\mathbb{N}:$
$$\left|\ f(3^{2a})\right| \geq 3^{a^2} -\sum_{n\neq a} 3^{n(2a-n)}\geq 3^{a^2}\left( 1-2\sum_{k>0 }3^{-k^2}\right)\ge \frac{3^{a^2}}{4}.$$