Suppose $X_1^{(d)}, X_2^{(d)}, \cdots, X_n^{(d)} \stackrel{\text{i.i.d.}}{\sim} \mathbf{F}_d$ are $d$-dimensional random variables such that there exists a function $g_d(\cdot,\cdot)$ taking values in $[-2,+2]$, and satisfying $g_d(X_1^{(d)},X_2^{(d)}) \to 0$ in probability as $d \to \infty$. Also, $\mathbb{E}[g_d(X_i^{(d)}, X_j^{(d)})] = 0$, for all $i,j = 1,2,\cdots,n ~(i \neq j)$.
Define the following random variable: $$S_{n,d} = \frac{1}{\binom{n}{2}} \sum_{1 \leqslant i \neq j \leqslant n} g_d(X_i^{(d)},X_j^{(d)})$$
Can we comment on the probability convergence of $S_{n,d}$ as $n \to \infty$ and $d \to \infty$ simultaneously?
NOTE: $g_d(\cdot,\cdot)$ is a function which doesn't vary over $d$. e.g. $g_d(a^{(d)},b^{(d)})= \frac{||a^{(d)}+b^{(d)}||_2}{||a^{(d)}||_2+||b^{(d)}||_2}$.
The problem is that I am facing difficulty understanding whether the different $g_d(X_i^{(d)},X_j^{(d)})$'s are independent or not. If they are independent, it would have been a straightforward application of WLLN which is not the case here. However, we have an additional information that each summand goes to $0$ in probability as $d \to \infty$. I have a hunch that $S_{n,d} \to 0$ in probability as $n \to \infty$ and $d \to \infty$. Any help will be appreciated.
We can use the Hoeffding's decomposition: let $f_{d}(x,y)=g(x,y)+g(y,x)$, $g_{d,1}=\mathbb E\left[f_d\left(X_1^{(d)},x\right)\right]$ and $$ g_{d,2}(x,y)=f_d(x,y)-g_{d,1}(x)-g_{d,1}(y). $$ Then \begin{align} S_{n,d}&=\frac{1}{\binom{n}{2}}\sum_{1\leqslant i<j\leqslant n}f_d\left(X_i^{(d)},X_j^{(d)}\right)\\&= \frac{1}{\binom{n}{2}}\sum_{1\leqslant i<j\leqslant n}g_{d,1}\left(X_i^{(d)} \right)+\frac{1}{\binom{n}{2}}\sum_{1\leqslant i<j\leqslant n}g_{d,1}\left(X_j^{(d)} \right)+\frac{1}{\binom{n}{2}}\sum_{1\leqslant i<j\leqslant n}g_{d,2}\left(X_i^{(d)},X_j^{(d)}\right)\\ &=\frac{2}{n}\sum_{k=1}^ng_{d,1}\left(X_k^{(d)}\right)+\frac{1}{\binom{n}{2}}\sum_{1\leqslant i<j\leqslant n}g_{d,2}\left(X_i^{(d)},X_j^{(d)}\right) \end{align} and taking the $\mathbb L^2$-norms gives $$ \mathbb E\left[S_{n,d}^2\right]\leqslant \frac{4}{n^2}\mathbb E\left[\left(\sum_{k=1}^ng_{d,1}\left(X_k^{(d)}\right)\right)^2\right] +\frac{2}{\binom{n}{2}^2}\mathbb E\left[\left(\sum_{1\leqslant i<j\leqslant n}g_{d,2}\left(X_i^{(d)},X_j^{(d)}\right)\right)^2\right]. $$ By independence and the fact that the random variables $g_{d,1}(X_k^{(d)})$ are centered and identically distributed, we get that $$ \mathbb E\left[S_{n,d}^2\right]\leqslant \frac{4}{n}\mathbb E\left[\left( g_{d,1}\left(X_1^{(d)}\right)\right)^2\right] +\frac{2}{\binom{n}{2}^2}\mathbb E\left[\left(\sum_{1\leqslant i<j\leqslant n}g_{d,2}\left(X_i^{(d)},X_j^{(d)}\right)\right)^2\right]. $$ For the last term, notice that $$ \mathbb E\left[ g_{d,2}\left(X_i^{(d)},X_j^{(d)}\right)g_{d,2}\left(X_k^{(d)},X_\ell^{(d)}\right)\right]=0 $$if $(i,j)\neq (k,\ell)$ hence $$ \mathbb E\left[S_{n,d}^2\right]\leqslant \frac{4}{n}\mathbb E\left[\left( g_{d,1}\left(X_1^{(d)}\right)\right)^2\right] +\frac{2}{\binom{n}{2}}\mathbb E\left[\left( g_{d,2}\left(X_1^{(d)},X_2^{(d)}\right)\right)^2\right]. $$ Under the assumptions of the questions, it turns out that (by a uniform integrability argument) $$ \lim_{d\to\infty}\sup_{n\geqslant 1}\mathbb E\left[S_{n,d}^2\right]=0=\lim_{n\to\infty}\sup_{d\geqslant 1}\mathbb E\left[S_{n,d}^2\right] $$