Let $c_n\ge0$ be a sequence. Sufficient conditions of $(c_n)$ such that $\lim 1/n\sum\limits_{k=1}^n c_k z^k=0$ for $|z|=1,z\ne 1$

Question

Let $(c_n)$ be a sequence of non-negative reals which is bounded below and above i.e. $m\le c_n\le M$ for some $m,M>0$. But this is not enough to say about the limit of $\frac{1}{n}\sum\limits_{k=1}^n c_kz^k$ to be zero (this post). Suppose I assume $\lim c_n$ exists or $(c_n)$ is Cesaro summable, can we conclude anything? Observe that $(z^n)$ is Cesaro summmable with cesaro sum is equal to $0$. Can we say something about the Cesaro summability of the product sequence $(c_nz^n)$? This post shows that the product of Cesaro summable sequence may not be Cesaro summable.

Can anyone help me with some idea, suggestions? Thanks for your help in advance.

Answer 1

We need the sequence $(|c_{k+1}-c_k|)_k$ to have $0$ Cesaro sum. More specifically,

Let $S_n=\sum_{k=1}^nc_kz^k$ where $c_k\in\mathbb C$ and $$\frac1n\sum_{k=1}^n|c_{k+1}-c_k|\to0 \tag{$\star$},$$ then $\frac1nS_n\to0$ for all $|z|\leq1$ with $z\neq1$.

First, some remarks:

• $(c_k)$ being Cesaro summable is not enough, consider $(c_k)=(1,2,1,2,...)$ and $z=-1$.

• $(c_k)$ is convergent$\;\Rightarrow(\star)$, but the reverse is false, consider $c_k=\sqrt k$.

• $(\star) \nRightarrow (c_k)$ is Cesaro summable or bounded , consider $c_k=\sqrt k$.

• $|c_{k+1}-c_k|\to0\Rightarrow(\star)$, but reverse is false, consider $c_k=1$ if $k$ is a square and $c_k=2$ otherwise.

• $(c_k)$ is Cesaro summable and $\big|\frac{c_{k+1}}{c_k}\big|\to1\Rightarrow(\star)$, but reverse is false, consider the same example in the above line.

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Proof sketch:

Let $S_n(z)=\sum_{k=1}^nc_kz^k$ and it follows \begin{align*} S_n(z)&=c_nz^n+c_{n-1}z^{n-1}+\cdots+c_1z\\[.6em] &=z^n(c_n-c_{n-1})+\big(z^n+z^{n-1}\big)(c_{n-1}-c_{n-2})+\cdots+\big(z^n+z^{n-1}+\cdots+z^2\big)(c_2-c_1)\\ &\ \ \ \ +\big(z^n+\cdots+z\big)c_1\\[.5em] &=z^n(c_n-c_{n-1})+\frac{1-z^2}{1-z}z^{n-1}(c_{n-1}-c_{n-2})+\cdots+\frac{1-z^{n-1}}{1-z}z^2(c_2-c_1)+\frac{1-z^n}{1-z}zc_1. \end{align*} Since $|z|\leq1$ we have $|1-z^k|\leq2$ for all $k$ and therefore $$\frac1n|S_n(z)|\leq\frac2{1-z}\left(\frac{c_1}n+\frac1n\sum_{k=1}^{n-1}|c_{k+1}-c_k|\right).$$ As $n\to\infty$, the right hand side tends to $0$ due to $z\neq1$ and the condition $(|c_{k+1}-c_k|)$ has $0$ Cesaro sum.

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For a sanity check, note that without $(c_k)$ we have $\frac1n\sum z^k\to0$ since the 'directions' of all $z^k$ cancel each other out. With the weight $(c_n)$, the directions may not cancel out since some directions can have larger weights than others. Hence, to ensure $\frac1n\sum c_kz^k\to0$ we need $(c_k)$ to be sufficiently 'uniform', in the sense that the weights for all directions are roughly the same in average. This loosely justifies the condition $|c_{k+1}-c_k|$ having $0$ Cesaro sum.

Also, boundedness of $c_k$ seems rather irrelevant here, since $c_k$ being bounded doesn't tell us anything about whether $c_k$ are biased towards some directions. (Unless you want to control the magnitude of the weights)