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Consider the following two properties of a real divergent sequence $\{c_n\}$.

$(1) \;$ $\displaystyle\lim_{n\to\infty}\frac{c_n}{n}=M$ for some $M\in\mathbb{R}$.

$(2) \;$ There exist two real convergent sequences $\{a_n\}$, $\{b_n\}$ such that $c_n=\sum_{k=0}^{n}a_kb_{n-k}$.

Are $(1)$ and $(2)$ equivalent?

I know how to prove $(2) \implies (1)$, but I don't know whether $(1)$ implies $(2)$.

jdhejw
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    The title does not match what you ask – Fedor Petrov Oct 15 '23 at 11:42
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    A test case. $c_n = (-1)^n$. Find $a_n, b_n$. – GEdgar Oct 15 '23 at 13:09
  • @GeraldEdgar coefficients of the power series $(1+x)^{1 /2}$ do converge to 0 – Fedor Petrov Oct 15 '23 at 13:23
  • So: assume $\lim c_n/n = 0$. Let $f(x) = \sum_{n=0}^\infty c_n x^n$ which has radius of convergence (at least) $1$. Then do the Taylor coefficients of $f(x)^{1/2}$ converge to zero? Is there an example of this where $f(x)$ has no single-valued square-root on the unit disk? – GEdgar Oct 15 '23 at 13:29
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    @GeraldEdgar this power series may easily have 0 inside the unit didk which makes taking a square root quite problematic – Fedor Petrov Oct 15 '23 at 16:26
  • The power series really don't seem that useful since it's hard to tell when the coefficients of $f$ converge (or maybe I'm missing something). –  Oct 15 '23 at 22:00
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    A reformulation is: Consider the bilinear map $F: c\times c \to c$ ($c$ = convergent sequences), $F(a,b)_n=(1/n)(a*b)_n$. Is $F$ surjective? –  Oct 16 '23 at 14:35
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    I suspect it was not a good idea migrating the question here. – Pietro Majer Oct 20 '23 at 13:04

1 Answers1

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This question was migrated from MathOverflow because it can be answered on Mathematics Stack Exchange.

That is technically correct, of course, but that is true about every question (because both sites are frequented by pretty much the same people and the snobs that make it a matter of principle to post on MO but not on MSE are very few if they exist at all).

The power series really don't seem that useful since it's hard to tell when the coefficients of $f$ converge (or maybe I'm missing something).

Actually, the power series is useful to the extent that it solves the problem in no time combined with another famous principle, which is "If you don't know how to build something, build it at random".

So, let $f(z)=\sum_ka_kz^k$, $g(z)=\sum_k b_k z^k$, $h(z)=\sum_k c_kz^k$, $|z|<1$. If $a,b$ are bounded, then $$ \|f(r\cdot)\|_{L^2(\mathbb T)}^2=\sum_k |a_k|^2r^{2k}=O(1/(1-r)) $$ as $r\to 1$ and similarly for $g$. So we must have $$ \|h\|_{L^1(\mathbb T)}=O(1/(1-r)) $$ by Cauchy.

Now take a random sequence $c_k=\pm k^{p}$ with $p\in(\frac 12,1))$, say (so $\lim_{n\to\infty}\frac{c_n}n=0$). Then the expectation of the $L^1$-norm of the corresponding random $h(r\cdot)$ on the unit circumference is comparable to the expectation of its $L^2$-norm (or any other $L^p$ norm) by the usual probabilistic nonsense about sums of independent Rademacher random variables. But the $L^2$-norm is deterministic and its square equals $\sum_k k^{2p}r^{2k}\approx (1-r)^{-1-2p}$, which is too much. It formally remains to show that we can choose signs so that the corresponding equivalences hold for a whole sequence of $r$'s going to $1$, but that is easy if we notice that for every fixed $r$, only a finite number of the coefficients really matter in the norm computation and that for every fixed finite number of indices, the contribution of the corresponding $c_k$ is bounded and, therefore, negligible when comparing the norms to $(1-r)$ to a negative power for $r$ close to $1$.

This raises several interesting questions however, the simplest of which being what happens if we assume that $c_n=O(\sqrt n)$ rather than $c_n=O(n)$, so don't hurry to close the discussion :-).

fedja
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