We assume the following:
$\bullet$ $U$ be a separable real Hilbert space.
$\bullet$ $\{W_t\}_{0\leq t\leq T}$ be a $U$ valued Q-Wiener process.
$\bullet$ $Q:U\to U$ be a non-negative definite trace class operator with eigen sequence $\{e_j,\lambda_j\}$.
It is know that for a sub-interval $[\xi,v]$ of $[0,T]$, since $\frac{\langle W_t,e_j\rangle_U}{\sqrt{\lambda_j}}$ is Brownian motion we have
$$E\Big[\Big(\tfrac{\langle W_v-W_\xi,e_j\rangle_U}{\sqrt{\lambda_j}}\Big)^2\Big]=v-\xi \text{ and } E\Big[\Big(\tfrac{\langle W_v-W_\xi,e_j\rangle_U}{\sqrt{\lambda_j}}\Big)^4\Big]=3(v-\xi)^2.$$ such properties was utilize for the following results $$E\big[\Vert W_v-W_\xi\Vert_U^2\big]=(v-\xi)\text{tr}Q$$ and $$E\big[\Vert W_v-W_\xi\Vert_U^4\big]=(v-\xi)^2\Big(2\displaystyle\sum_{j=1}^\infty \lambda_j^2 + (\text{tr}Q)^2\Big).$$
Let $P=\{[\xi_i,v_i]\}_{i=1}^{n}$ be any partition of $[0,T]$, Since for $[\xi',v']\neq [\xi,v]\in P$, $W_{v'}-W_{\xi'}$ and $W_{v}-W_{\xi}$ independent we have
\begin{align*}E\Big[\Big(\displaystyle\sum_{i=1}^n & \Vert W_v-W_\xi \Vert^3\Big)^2\Big] \\ &=E\Big[\displaystyle\sum_{i=1}^n\Vert W_v-W_\xi \Vert^6+\displaystyle\sum_{[\xi',v']\neq [\xi,v]} \Vert W_{v'}-W_{\xi'} \Vert^3_U \Vert W_{v}-W_{\xi} \Vert^3_U\Big] \\ &=\displaystyle\sum_{i=1}^n E[\Vert W_v-W_\xi \Vert^6\big]+\displaystyle\sum_{[\xi',v']\neq [\xi,v]} E\big[\Vert W_{v'}-W_{\xi'} \Vert_U^3\big]E\big[\Vert W_{v}-W_{\xi} \Vert^3_U\big] \\ \end{align*}
I was badly got stuck on this following calculation and couldnt find a close form of $E[\Vert W_v-W_\xi \Vert^6\big]$ and $E[\Vert W_v-W_\xi \Vert^3\big]$ or any bounds in terms of the difference $v-\xi$. I'm starting to think that this expectation can be unbounded.
Any tips or insights would be a great help.