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!



Convergence of $\sup_{l \leq \rho \leq u} | \frac{1}{\sqrt{n}} \sum_{i=[\rho n]}^n \sum_{k=1}^i i^{-1} x_i y_k |$ in probability to zero
$\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\Var}{\operatorname{\mathsf Var}} 
\renewcommand{\P}{\operatorname{\mathsf P}}$


Looking at the calculation of the second moment in the statement of the question, it appears that it was tacitly assumed there that $\E x\_1^2+\E y\_1^2+\E x\_1^2y\_1^2 <\infty$, which let us assume as well. Then let us show somewhat more than was asked: for any sequence $(b\_n)$ of real numbers such that $b\_n/\ln n\to\infty$, 
\begin{equation\*} \tag{0}
 \frac1{b\_n}\,\max\_{1\le r\le n} |S\_r| \stackrel{P}{\longrightarrow} 0 
\end{equation\*}
as $n\to\infty$, 
where
\begin{equation\*}\tag{1}
 S\_r:=\sum\_{i = 1}^r x\_i Y\_i=T\_r+U\_r,\quad T\_r:=\sum\_{i = 1}^r \frac{x\_i}i\,y\_i
 ,\quad U\_r:=\sum\_{i = 1}^r \frac{x\_i}i\,Z\_i, \quad Z\_i:=\sum\_{k=1}^{i-1} y\_k. 
\end{equation\*}
We have 
\begin{equation\*}\tag{2}
 \max\_{1\le r\le n}\E|T\_r|\ll\sum\_{i = 1}^n \frac1i\ll\ln n=o(b\_n). 
\end{equation\*}
Also, by [Doob's inequality](https://en.wikipedia.org/wiki/Doob%27s_martingale_inequality),  

\begin{equation\*}
 \E\max\_{1\le r\le n}(T\_r-\E T\_r)^2\le2\Var T\_n\ll\sum\_{i = 1}^n \frac1{i^2}\ll1=o(b\_n^2). 
\end{equation\*}
So, by (2) and Markov's inequality, 
\begin{equation\*} \tag{3}
 \frac1{b\_n}\,\max\_{1\le r\le n} |T\_r| \stackrel{P}{\longrightarrow} 0. 
\end{equation\*}


Next, 
\begin{multline\*}
 \E U\_n^2=\sum\_{1\le k<i\le n}\sum\_{1\le \ell<j\le n}\frac1{ij}\E x\_i x\_j y\_k y\_\ell
 =\sum\_{1\le k<i\le n}\frac1{i^2}\E x\_i^2 y\_k^2\ll\sum\_{1\le k\le n}\frac1k\ll\ln n. 
\end{multline\*}
The key point of this answer is that $(U\_r)$ is a martingale. Hence, again by [Doob's inequality](https://en.wikipedia.org/wiki/Doob%27s_martingale_inequality) we have  

\begin{equation\*}
 \E\max\_{1\le r\le n}U\_r^2\le2\E U\_n^2\ll\ln n=o(b\_n^2). 
\end{equation\*} 
So, again by Markov's inequality, 
\begin{equation\*} \tag{4}
 \frac1{b\_n}\,\max\_{1\le r\le n} |U\_r| \stackrel{P}{\longrightarrow} 0. 
\end{equation\*}


Now (0) follows from (1), (3), (4). 


