Let $X_i$ be iid zero mean, unit variance random vectors taking values in $\mathbb R^{d(n)}$ where ${d(n)}$ is a sequence (dependent on a positive integer $n$) that grows such that ${d(n)}/n \rightarrow \gamma >0$ as $n\rightarrow \infty$ for some positive constant $\gamma \in (0,1)$. Let $\{w_j\}^m$ be a set of $m$ orthogonal vectors in $\mathbb R^{d(n)}$ with $\|w_j\|_2 = 1$. Consider the following quantity:
\begin{equation} Y_n = \max_j \left \langle w_j, \frac{1}{\sqrt n} \sum_i^n X_i\right \rangle \end{equation}
Find the limiting distribution of $Y_n$.
What I've tried
Rearranging:
\begin{equation} Y_n = \max_j \left \langle w_j, {\sqrt n} \bar X\right \rangle \end{equation}
where $\bar X$ is the average of the $n$ vectors $X_i$. Notice that $\sqrt n \bar X = \mathcal N \left(0, I_d\right) + o_p(1)$ so that for each $w_j$, the term inside the max is distributed as $\mathcal N \left(0, w_j^TI_dw_j\right) + o_p(1)$ which is simply $\mathcal N \left(0, 1\right) + o_p(1)$. This shows that $Y_n$ is asymptotically the max of $m$ normal variables, but they are clearly not independent so I'm not sure how to analyze it.
Another way to think about this is that $Y_n$ is asymptotically the maximum magnitude of the projection of a gaussian vector onto the subspace formed by the $\{w_j\}^m$.
Any hints would be appreciated.
EDIT: Since $m$ is fixed, then as $n \rightarrow \infty$, the measure of the subspace spanned by $\{w_j\}^m$ decreases. I will see if I can use the fact that in high dimensions a normal vector will be orthogonal to a given vector with high probability; a union bound with this might be sufficient to show that $Y_n \rightarrow 0$.