# 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

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!

$$\newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\Var}{\operatorname{\mathsf Var}} \renewcommand{\P}{\operatorname{\mathsf P}}$$
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,
$$\begin{equation*} \E\max_{1\le r\le n}(T_r-\E T_r)^2\le2\Var T_n\ll\sum_{i = 1}^n \frac1{i^2}\ll1=o(b_n^2). \end{equation*}$$ 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, $$\begin{multline*} \E U_n^2=\sum_{1\le k The key point of this answer is that $$(U_r)$$ is a martingale. Hence, again by Doob's inequality we have
$$\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*}$$