Let
$$
A_k=\{\lVert S_j\rVert\leq t\text{ for }j<k, \lVert S_k\rVert>t\}
$$
be the event that the partial sums first cross level $t$ at time $k$. On $A_k$, write $R_k=S_n-S_k$. Conditional on $X_1,\ldots,X_k$, the random variable $R_k$ is independent and symmetric. Since
$$
2\lVert S_k\rVert
\leq\lVert S_k+R_k\rVert+\lVert S_k-R_k\rVert,
$$
at least one of the two norms on the right exceeds $t$. Symmetry of $R_k$ consequently gives
$$
\mathbb P(\lVert S_n\rVert>t\mid X_1,\ldots,X_k)\geq\frac12
$$
on $A_k$. The events $A_k$ are disjoint, so summation proves the <Lévy maximal inequality>
$$
\mathbb P\left(\max_{1\leq k\leq n}\lVert S_k\rVert>t\right)
\leq2\mathbb P(\lVert S_n\rVert>t).
$$

For the Gaussian series, put $S_m=\sum_{j=1}^mg_ju_j$. Apply the inequality to the symmetric independent increments from $n+1$ through $m$, followed by <Markov inequality> in squared norm:
$$
\begin{aligned}
\mathbb P\left(\max_{n<k\leq m}\lVert S_k-S_n\rVert_V>\varepsilon\right)
&\leq2\mathbb P(\lVert S_m-S_n\rVert_V>\varepsilon)\\
&\leq\frac2{\varepsilon^2}\mathbb E\lVert S_m-S_n\rVert_V^2
=\frac2{\varepsilon^2}\sum_{j=n+1}^m\lVert u_j\rVert_V^2.
\end{aligned}
$$
Here the cross terms vanish by <orthogonality of independent centered Hilbert-space random variables>. Letting $m\to\infty$ and then $n\to\infty$ shows
$$
\mathbb P\left(\sup_{k\geq n}\lVert S_k-S_n\rVert_V>\varepsilon\right)\longrightarrow0.
$$
The convergence criterion in the question now shows that $\sum_jg_ju_j$ converges almost surely in $V$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 217 / 2 / Solution

If $X_i$ are independent centered square-integrable random variables in a Hilbert space, then $\mathbb E\langle X_i,X_j\rangle=0$ for $i\ne j$, and hence
$$
\mathbb E\left\lVert\sum_iX_i\right\rVert^2
=\sum_i\mathbb E\lVert X_i\rVert^2.
$$

= Independent random variable
{synonym}


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