Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The following question came up while I was working through an example:

Does there exist an $\ell^1$ sequence of complex numbers $a_n$, not all zero, such that $\sum_n a_n n^{-p} = 0$ for all $p \in [0,1]$?

share|improve this question
You mean, aside from the zero sequence? –  Gerry Myerson Mar 8 '13 at 11:20
Thanks for point that out, Gerry. –  Andre Mar 8 '13 at 11:24

1 Answer 1

up vote 8 down vote accepted

Let $U\subset \Bbb C$ be the set of complex number with positive real part and $f\colon U\to\Bbb C$ given by $f(z):=\sum_{n=1}^{+\infty}a_n\exp(-z\log n)$. Since there is uniform convergence on compact subsets of $\Bbb C$, $f$ is holomorphic. We have $f(z)=0$ if $z\in (0,1]$, hence $f$ vanishes identically on the connected set $U$. In particular, the initial assumption is valid for any $p>0$.

Then we can prove by induction that $a_n=0$ for all $n$.

We have $|a_1|=\left|\sum_{n\geqslant 2}a_n\exp(-p\log n)\right|\leqslant \exp(-p\log 2)\lVert a\rVert_{\ell^1}$ for each $p$, hence $a_1=0$.

Assume that for $n\geqslant 2$, $a_0=\dots=a_{n-1}=0$, then $$|a_n| \leqslant \exp\left(-p\log \frac{n+1}{n}\right)\lVert a\rVert_{\ell^1},$$ giving what we want.

Note that we can relax the initial assumption "for all $p\in [0,1]$" by for "all $p$ in an non-discrete subset of the unit interval".

share|improve this answer
That's a very nice solution. Thank you. –  Andre Mar 9 '13 at 6:45
You are welcome. (and I thank you for the interesting problem) In which context did it appear? –  Davide Giraudo Mar 9 '13 at 8:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.