I am dealing with a sequence $\{(x_i,y_i)\}$ of **zero-mean** random variables. For simplicity we can assume that the sequence is i.i.d. Define $Y_i := i^{-1} \sum_{k=1}^i y_k$. I would like show that
\begin{equation} \tag{1} \label{eq:1}
\sup_{l \leq \rho \leq u} \bigg| \frac{1}{\sqrt{n}} \sum_{i = [\rho n]}^n x_i Y_i \bigg| \stackrel{p}{\longrightarrow} 0
\end{equation}
where $0 < l < u < 1$ are constants.

The proof would be easy if there were no supremum involved. We can show $$ \mathbb{E} \left[ \left( \sum_{i=[\rho n]}^n x_i Y_i \right)^2 \right] = O(\log(n / [\rho n])) = O(1) $$ and so $\frac{1}{\sqrt{n}} \sum_{i = [\rho n]}^n x_i Y_i \stackrel{p}{\longrightarrow} 0$ by Markov's inequality.

I cannot show \eqref{eq:1} when the supremum gets involved. We have \begin{align*} \mathbb{P} \left( \sup_{l \leq \rho \leq u} \bigg| \frac{1}{\sqrt{n}} \sum_{i = [\rho n]}^n x_i Y_i \bigg| > \epsilon \right) &\leq \sum_{R = [ln]}^{[un]} \mathbb{P} \left( \bigg| \frac{1}{\sqrt{n}} \sum_{i = R}^n x_i Y_i \bigg| > \epsilon \right) \\ &\leq \epsilon^{-2} n^{-1} \sum_{R = [ln]}^{[un]} \log(n/R) \end{align*} and we can only conclude that the rightmost term is bounded. It seems that the first inequality uses a bound that is too large.

So can anyone suggest if it's possible to establish \eqref{eq:1}? Maybe using some other inequalities?

Thanks!