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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
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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
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I think the nice added conjecture goes to the core of the problem, nevertheless it has to be modified because it doesn't hold as it is.

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 $a\in\mathbb{N}:$

$$\left|\ f(3^{2a})\right| \geq 3^{a^2} -\sum_{n\neq a} 3^{n(2a-n)}\geq 3^{a^2}\left( 1-2\sum_{k>0 }3^{-k^2}\right)\ge \frac{3^{a^2}}{4}.$$

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