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Suppose $(z_i)$ is a sequence of complex numbers such that $|z_i|\to 0$ strictly decreasing. If $(a_i)$ is a sequence of complex numbers that has the property that for any $n\in\mathbb{N}$

$$ \sum_{i}a_iz_i^{n}=0 $$ does this imply that $a_i=0$ for any $i$?

Edit: For the "finite dimensional" case, when we have $n$ distinct $(z_i)$, then $(a_i)$ must be $0$. This amounts to solving a homogeneous system of $n$ equations with $n$ unknowns, which only has the trivial solution in the case of distinct $(z_i)$. I am really curious what happens in the infinite dimensional case. My intuition tells me the same must be true, but I don't have a proof for it.

Edit 2: Very interesting, looking were this question originated, the fact that all $a_i=0$ when $(a_i)\in l_1$ is "expected". I was hoping to get a counterexample otherwise. However, if a non-trivial sequence $a_i$ exists (at least for some sequences $z_i$), I would "expect" to be able to choose it in $l_2$. Looking at Davide and Julien answers below, it seems $(a_i)\in l_1$ is an essential assumption in their argument.

Theo
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  • For the sake of clarity: am I right that the series is not assumed absolutely convergent? –  Dec 21 '12 at 06:44
  • No, it is not. All it is known is that $(z_i)$ converge to $0$, decreasing. So I cannot assume a priori anything about the convergence of $\sum a_i z_i$. – Theo Dec 21 '12 at 20:06
  • My guess is that a heavy-tailed sequence such as $z_i=1/\log(i+1)$ could yield a counterexample, but I could not come up with an argument for existence of $a_i$. –  Dec 23 '12 at 21:34
  • Does the trivial case of $z_i = \frac{1}{2^i}$ and $a_i = 1$ (if $i$ odd) and $a_i = -2$ (if $i$ even) (My index starts at $i=1$) show a counter example? – picakhu Dec 24 '12 at 18:23
  • @picakhu I don't think so. If I understand it correctly, your example works only for $n=1$. – Theo Dec 24 '12 at 18:42
  • The sum would be $1/2 - 2/4 + 1/8 - 2/16 + ...$ I think I misunderstood your question, although it would be good for you to clarify where I misread it. – picakhu Dec 24 '12 at 19:05
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    @picakhu Your example shows that $\sum a_i z_i=0$, but not $\sum a_i z_i^n$ when $n\in\mathbb{N}$, $n\geq 2$. – Theo Dec 24 '12 at 21:29
  • @PavelM: Do you know to prove it under the extra assumption that the series absolutely convergent? – Lior B-S Dec 25 '12 at 13:26
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    The following special cases already looks interesting: the case $z_i = \frac{i}$ is equivalent to showing that if a Dirichlet series vanishes on the integers then it is identically zero. – Ofir Dec 25 '12 at 22:15
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    If we know that $(a_i)\in \ell^1$, can we conclude? Just for curiosity, if we define $T:\ell^1\rightarrow\ell^{\infty}$ by $Ta=(\sum_i a_iz_i,\sum_i a_iz_i^2,...,\sum_i a_iz^n,...)$,where $a=(a_1,a_2,...,a_n,...)$, can we conclude that the only solution of the equation $Ta=0$ is $a=0$ – Tomás Dec 26 '12 at 19:25

3 Answers3

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Under the assumption that $\sum_{i=0}^\infty |a_i| < \infty$, we can use power series. This assumption can be weakened: it is enough to assume that there exists $n \in \mathbb{N}$ such that $(a_i z_i^n) \in \ell^1$ (since we can work with $a'_i = a_i z_i^n$).

For $|\lambda| < |z_0|^{-1}$, define $$ f(\lambda) = \sum_{i=0}^\infty \frac{a_i}{1 - \lambda z_i} = \sum_{n=0}^\infty \sum_{i=0}^\infty (\lambda z_i)^na_i =0. $$ If $k = \min\{i \in \mathbb{N} \mid a_i \neq 0\}$ exists, then $f(\lambda) \sim \dfrac{a_k}{1-\lambda z_k}$ when $\lambda \to z_k^{-1}$, wich is absurd.

Siméon
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  • While this does not completely answer my question, it is enough to settle the problem from which the question originated. My sequences $(z_i)$ converge very fast to $0$, so in particular are in $l_1$. In this situation it is impossible to find a non-trivial, bounded (let alone in $l_2$ as I require) sequence $(a_i)$. By your observation, $(a_i z_i)$ will be in $l_1$, thus $(a_i)$ is necessarily trivial. Thanks guys, this was very helpful! – Theo Dec 28 '12 at 04:36
  • Just one more comment. I would be still very much interested what happens in the general case. For example, if $z_i=1/\log(i+1)$ as Pavel M suggested in the comments, can one find a sequence $(a_i)$ which is in $l_2$? Perhaps I should formulate this as a different question. – Theo Dec 28 '12 at 04:47
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It's a partial answer to @Tomas' comment which ask what happens if we assume the sequence $\{a_j\}$ in $\ell^1$, and it would be too long for a comment.

I interpret the hypothesis $\sum_{j=0}^{+\infty}a_jz_j^n=0$ for all $n$ as "the sequence of partial sums converges and its limit is $0$". In particular, it implies that $\{a_j\}$ is convergent to $0$, hence $\{|a_j|\}$ is in particular a bounded sequence, say by $M$. We show by induction that $a_j=0$ for all $j$. First, we have for all integer $p$ that $$|a_0z_0^p|\leqslant\sum_{j\geqslant 1}|a_j||z_j|^p$$ so $$|a_0|\leqslant \sum_{j\geqslant 1}|a_j|\left(\frac{|z_j|}{|z_0|}\right)^p.$$ Let $b_{j,p}:=|a_j|\left(\frac{|z_j|}{|z_0|}\right)^p$. We have $|b_{j,p}|\leqslant |a_j|\left(\frac{|z_j|}{|z_0|}\right)^p$, and as $\{a_j\}\in\ell^1$, we can conclude by monotone convergence theorem that $a_0=0$.

Now assume that $a_0=\dots=a_n=0$. Then $$|a_{n+1}|\leqslant \sum_{j\geqslant n+2}|a_j|\left(\frac{|z_j|}{|z_{n+1}|}\right)^p,$$ and an other application of monotone convergence theorem yields $a_{n+1}=0$.

Davide Giraudo
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    I think you dont need monotone convergence theorem. In the second inequality you can use the fact that $$\Big(\frac{|z_j|}{|z_0|}\Big)^p\leq \Big(\frac{|z_1|}{|z_0|}\Big)^p,\ \forall\ i\geq 1$$ and by hypothesis $\frac{|z_1|}{|z_0|}<1$, hence, by making $p\rightarrow\infty$ you can conclude. Am I right? – Tomás Dec 27 '12 at 12:37
  • @Tomás I think yes. So the case $\{a_j\}\in\ell^1$ seems to be too simple with respect to the complexity of the problem. – Davide Giraudo Dec 27 '12 at 13:21
  • Maybe it is possible to show that $(a_i)\in\ell^1$. – Tomás Dec 27 '12 at 13:25
  • @DavideGiraudo Just an observation: when I say $\sum a_i z_i^n=0$, for all $n$, I actually mean $n>0$. So the conclusion in your second sentence follows from the added assumption that $(a_i)\in l_1$, but not from the original hypothesis alone. – Theo Dec 27 '12 at 14:30
  • @Theo Ok. (in France, $\Bbb N$ usually mean the set of integers greater or equal than $0$) – Davide Giraudo Dec 27 '12 at 16:52
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Suppose that the statement is not true, and let $N$ be the smallest index such that $a_N \neq 0$. Then $-a_N z_N^n= \sum_{i > N} a_i z_i^n \neq 0$ for all $n$. Then $0 < |a_N| \leq |\sum_{i \geq N} a_i \frac{z_i^n}{z_N^n}| \leq \sum_{i > N}|a_i|(|z_i|/|z_N|)^n$. Taking $n \to \infty$ in this strict inequality would yield $0 < 0$, a contradiction.

Rookatu
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    I don't think you can interchange the limits to conclude the RHS converges to $0$. Or am I missing something? – Theo Dec 19 '12 at 17:33
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    Indeed the RHS could even be infinite for all n. –  Dec 20 '12 at 07:51