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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 $p \in [0,1]$?

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  • $\begingroup$ You mean, aside from the zero sequence? $\endgroup$ Mar 8, 2013 at 11:20

1 Answer 1

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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".

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  • $\begingroup$ That's a very nice solution. Thank you. $\endgroup$
    – Andre
    Mar 9, 2013 at 6:45
  • $\begingroup$ You are welcome. (and I thank you for the interesting problem) In which context did it appear? $\endgroup$ Mar 9, 2013 at 8:56

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