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Let $m,M>0$ be such that $m\le c_k\le M$ for all $k$. If $|z|<1$, then $|c_kz^k|\le M|z|^k\to 0$. Then we have $$\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^n c_k z^k=0$$

But I cannot show for $|z|=1$. If $c_k=c$ is a constant sequence. Then $\frac{1}{n}\sum\limits_{k=1}^n c_k z^k=c\frac{z(z^n-1)}{n(z-1)}\to0$ as $z\ne1$ and $|z|=1$.

Can anyone help with any idea or hint for the problem? Thanks for your help in advance.

Nafula12
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1 Answers1

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This is false. Take $z=-1, c_k=2$ for $k$ even and $c_k=1$ for $k$ odd. Then $\frac 1 {2n} \sum\limits_{k=1}^{2n}c_kz^{k}=\frac 1 2$ for each $n$.

Kavi Rama Murthy
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  • Maybe we can impose some extra conditions to make it true, like $z^k$ is dense in $S^1$? – Ricky Sep 09 '23 at 07:32
  • @Ricky $z^k$ being dense might not be enough, e.g. $z=e^i$ and $c_k=2$ if Re$(z^k)\geq0$ and $c_k=1$ otherwise, then the number of $c_k=2$ should grow linearly in $k$, making $\lim\frac1n\sum Re(c_kz^k)>0$. In fact, for any $|z|=1$ we could probably use this idea to find bounded $c_k$ such that $\lim\frac1n\sum Re(c_kz^k)>0$, so boundedness seems not that useful in determining $\lim\frac1n\sum Re(c_kz^k)$. I think this is because boundedness does not tell us whether $c_k$ are biased towards some $z^k$. – Mengchun Zhang Sep 21 '23 at 16:29