Assume that $Y=(Y_t, t \in \mathbb{N})$ is a strictly stationary $\alpha$-mixing process, and let $G_h(x)=G(x/h)=\int_{x/h}^{+\infty}K(u)du$, where $K$ is a Kernel function(symetric density in this case), and $h$ is the sequence of bandwidths, that satisfy certain properties, which can be found in the assumptions of Chen's work, that I will provide a link for, shortly. Moreover, assume that Y is a geometric $\alpha$-mixing process (i.e, there exist $C, \rho>0$ such that $\alpha(k)\leq \rho^k$, for all $k\geq 1$), and let $\nu_p$ be the $(1-p)$-th order quantile of Y_t.
I am trying to show that, for $i=0, 1$ and $j=0, 1$, \begin{align*} \Bigg|\sum_{k=1}^{n-1}\left(1-\frac{k}{n}\right)\Big[\text{Cov}\Big(Y_{1}^{i}G_{h}&(\nu_{p}-Y_{1}), Y_{k+1}^{j}G_{h}(\nu_{p}-Y_{k+1})\Big) \\ &-\text{Cov}\left(Y_{1}^{i}I\left(Y_{1}>\nu_{p}\right), Y_{k+1}^{j}I\left(Y_{k+1}>\nu_{p}\right)\right)\Big]\Bigg| = o\left(h\right). \end{align*}
This lemma is in Chen's work, and he doesn't prove it. He refers to another article of his for the proof of i=j=0, and says the other cases are similar. You can find it here, under Lemma 4. http://www.public.iastate.edu/~songchen/shortfall.pdf
The proof for i=j=0 is done in two steps. The first is to conclude that $$\Bigg|\text{Cov}\Big(G_{h}(\nu_{p}-Y_{1}), G_{h}(\nu_{p}-Y_{k+1})\Big)-\text{Cov}\left(I\left(Y_{1}>\nu_{p}\right), I\left(Y_{k+1}>\nu_{p}\right)\right)\Bigg| \leq Ch^2,$$
which can be found in Cai and Roussas http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.47.7779&rep=rep1&type=pdf under lemma 2.2 (i). The proof provides interesting insight.
The second step is to notice that, under the assumption of an $\alpha$-mixing process, we get that $$\sup\limits_{(x,y) \in \mathbb{R}^2} \left|F_k(x,y)-F(x)F(y)\right| \leq \alpha(k)$$ Now, since $$\text{Cov}\left(I\left(Y_{1}>\nu_{p}\right), I\left(Y_{k+1}>\nu_{p}\right)\right) = F_k(\nu_p,\nu_p)-F^2(\nu_p) \leq \sup\limits_{(x,y) \in \mathbb{R}^2} \left|F_k(x,y)-F(x)F(y)\right| \leq \alpha(k),$$
and using the same idea as is done in Cai and Roussas, from a change of variables and Hoeffding's identity we get
\begin{align*} \text{Cov}&\Big(G_{h}(\nu_{p}-Y_{1}), G_{h}(\nu_{p}-Y_{k+1})\Big) \\ &= \int_{\mathbb{R}^2}\left[ F_k\left(\nu_p-hr, \nu_p-hs\right)-F\left(\nu_p-hr\right)F\left(\nu_p-hs\right) \right] K\left(r\right)K\left(s\right) dr ds \\ &\leq \sup\limits_{(x,y) \in \mathbb{R}^2} \left|F_k(x,y)-F(x)F(y)\right| \int_{\mathbb{R}^2} K\left(r\right)K\left(s\right) dr ds \\ &= \sup\limits_{(x,y) \in \mathbb{R}^2} \left|F_k(x,y)-F(x)F(y)\right| \leq \alpha(k) \end{align*} we get that $$\Bigg|\text{Cov}\Big(G_{h}(\nu_{p}-Y_{1}), G_{h}(\nu_{p}-Y_{k+1})\Big)-\text{Cov}\left(I\left(Y_{1}>\nu_{p}\right), I\left(Y_{k+1}>\nu_{p}\right)\right)\Bigg| \leq 2\alpha(k).$$
Let us write $ \gamma_h^{i,j}(k) = \text{Cov}\left(Y_{1}^i G_{h}\left(\nu_{p}-Y_{1}\right), Y_{k+1}^j G_{h}\left(\nu_{p}-Y_{k+1}\right)\right)$ and $\gamma^{i,j}(k) = \text{Cov}\left(Y_{1}^i I\left(Y_{1}>\nu_{p}\right), Y_{k+1}^j I\left(Y_{k+1}>\nu_{p}\right)\right)$.
These two bounds we provided for $\Bigg|\gamma_h^{0,0}(k)-\gamma^{0,0}(k)\Bigg|$ allow us write
$$|\gamma_h^{0,0}(k)-\gamma^{0,0}(k)|= |\gamma_h^{0,0}(k)-\gamma^{0,0}(k)|^{2/3} |\gamma_h^{0,0}(k)-\gamma^{0,0}(k)|^{1/3}\leq Ch^{4/3}\alpha^{1/3}(k),$$
which allows us to conclude
$$\left|\sum_{k=1}^{n-1}\left(1-\frac{k}{n}\right) \left[\gamma_h^{0,0}\left(k\right)-\gamma^{0,0}\left(k\right)\right]\right| \leq \sum_{k=1}^{n-1} \left|\left[ \gamma_h^{0,0}\left(k\right)-\gamma^{0,0}\left(k\right)\right]\right| \leq Ch^{4/3}\sum_{k=1}^\infty \alpha^{1/3}(k) = o\left(h\right),$$ which concludes the proof for the case i=j=0.
In his article, Chen says that for the other cases of i and j, the proof is similar, but I can not conclude. I have already shown that, for i=j=1, $$\Bigg|\gamma_h^{1,1}(k)-\gamma^{1,1}(k)\Bigg|\leq Ch^2.$$ However, I am not being able to show anything of the sort $\Bigg|\gamma_h^{1,1}(k)-\gamma^{1,1}(k)\Bigg| \leq C\alpha(k)$. This would allow me to conclude.
Thank you very much for reading and for any help you can provide.