Let $(z_n)_{n \in \mathbb{N}}$ be a sequence of complex numbers of the unit circle. Suppose that for all integers $k\geq 1$ :
\begin{equation*} \frac{1}{n} \sum_{l=0}^{n} \overline{z_{k+l}}z_l \underset{n \to \infty}{\longrightarrow} 0 \end{equation*}
We want to prove that :
\begin{equation*} \frac{1}{n} \sum_{l=0}^{n} z_l \underset{n\to \infty}{\longrightarrow} 0 \end{equation*}
I have computed :
\begin{equation*} \left| \frac{1}{n} \sum_{l=0}^{n} z_l \right|^2 = \frac{1}{n^2} \sum_{k=0}^n \sum_{l=0}^k \overline{z_{k-l}} z_l \end{equation*}
And it thus suffices to show that :
\begin{equation*} \sum_{l=0}^k \overline{z_{k-l}} z_l= o(k) \quad \text{as $k$ approaches $\infty$} \end{equation*}
But I am having difficulty proving it... We would like the $-$ in the summation $\overline{z_{k-l}}$ to be a $+$ i.e. $ \overline{z_{k+l}}$... Does anyone have an idea?