Maximize the trace of the convariance of a bounded random vector Given a random variable $X$, satifying $P(0\leq X \leq 1)=1$, and $\mathsf{E}[X^2] = \alpha$. We know its maximum variance $\text{Var}(X) = \alpha(1-\alpha)$ achived by a binary random variable $P(X =x) = \begin{cases} &1-\alpha, &x=0 \\ &\alpha, &x=1 \end{cases}$.
My problem is given a random vector $\boldsymbol{X}$, and $\text{supp}\boldsymbol{X} = [0,1]^n$, but with a second-moment constraint $\mathsf{E}\|\boldsymbol{X}\|^2_2={{\alpha}}$. With a linear transformation $\boldsymbol{H} (\boldsymbol{H} \succ \boldsymbol{0})$, I want to know whether the maximum trace of the covaraince matrix $\text{cov}(\boldsymbol{H}\boldsymbol{X})=\mathsf{E}[(\boldsymbol{H}\boldsymbol{X}-\mathsf{E}[\boldsymbol{H}\boldsymbol{X}])(\boldsymbol{H}\boldsymbol{X}-\mathsf{E}[\boldsymbol{H}\boldsymbol{X}])^\text{T}]$ can still be achived by a discrete random vector whose support $\text{supp}\boldsymbol{X}=\{0,1\}^n$.
The trace can be expanded as $\text{tr}(\text{cov}(\boldsymbol{H}\boldsymbol{X}))=\sum_{k=1}^{n} h_{i,k}^2 \mathsf{E}{\bigl(X_k-\mathsf{E}{X_k}\bigr)^2} 
  + \sum_{k=1}^{n} \sum_{\substack{\ell=1\\\ell\neq k}}^{n}
  2h_{i,k} h_{i,\ell}  \bigl( \mathsf{E}{X_{k} X_{\ell}} -
  \mathsf{E}{X_{k}}\mathsf{E}{X_{\ell}}\bigr)$. If we use the similar method as in the random variable case, we can maximize the first term of RHS of above equation, be the change of second term of RHS cannot be determined.
Edit: This is a follwoing-up problem from the following one, but with different moment constraints:
https://mathoverflow.net/questions/421113/the-maximum-trace-of-a-covariance-can-be-achieved-by-a-discrete-random-vector
 A: Using linearity of expectation and some properties of the trace which I don't recall the name (mainly $\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$) we can write
\begin{align*}
\mathrm{Tr}(\mathrm{cov}(HX)) &= \mathrm{Tr}(\mathbb E[(HX−\mathbb E[HX])(HX−\mathbb E[HX)^T])\\
&= \mathbb E[\mathrm{Tr}((HX−\mathbb E[HX])(HX−\mathbb E[HX])^T)]\\
&= \mathbb E[\mathrm{Tr}((HX−\mathbb E[HX])^T(HX−\mathbb E[HX]))]\\
&= \mathbb E[(HX−\mathbb E[HX])^T(HX−\mathbb E[HX])]\\
&= \mathbb E[(X−\mathbb E[X])^TH^TH(X−\mathbb E[X])]\\
&\leq \mathbb E[\|X-\mathbb E[X]\|^2_2] \lambda_{\max}(H^TH)
\end{align*}
Where the last line is because for any vector $u$ and any square matrix $A$, $u^TAu\leq \lambda_{\max}(A)\|u\|^2$ which is attained with equality if and only if $u$ is in the eigen-space of $A$ with corresponding eigen-value $\lambda_{\max}$, this means we want $X-\mathbb E[X]$ to be in this space.
In the case where $V_{\max}$ the eigen-space of eigen-value $\lambda_{\max}$ is such that $[0,1]^n\cap V_{\max}$ is not $\{0\}$, let $v=\mathrm{argmax}_{v\in V_{\max}\cap [0,1]^n} \|v\|_2$, and let us suppose that $X=Y v$ for some random variable $Y\in[0,1]$, then $\mathbb E[\|X\|_2^2]=\|v\|_2^2 \mathbb E[Y^2]$ and so $\mathbb E[Y^2]=\frac{\alpha}{\|v\|_2^2}$ and $\mathbb E[(X-\mathbb E[X])^T(X-\mathbb E[X])]=\|v\|_2^2 \mathrm{Var}[Y]\leq \alpha\cdot\left(1-\frac{\alpha}{\|v\|_2^2}\right)$ by your argument which means that in this scenario we can achieve $\mathrm{Tr}(\mathrm{Cov}(HX))=\alpha\cdot\left(1-\frac{\alpha}{\max_{v\in V_{\max}\cap[0,1]^n} \|v\|_2^2} \right)\cdot\lambda_{\max}(H)^2$.
I would expect the expression to get a lot messier in the case where $V_{\max}\cap[0,1]^n=\{0\}$, the reason why this is not very satisfying is that we are mixing $\|\cdot\|_1$ and $\| \cdot\|_2$ in the definition of the problem.
