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Let $X_1,\dots ,X_n$ be independent random variables taking values in a normed vector space, each symmetric in law: $X_i$ and $-X_i$ have the same probability distribution. With partial sums $S_k={\sum }_{i=1}^k{X}_i$, the Lévy maximal inequality isThe statement requires no finite moments. Symmetry is stronger than having zero expected value and is the key reflection hypothesis.

For a proof, let $\tau$ be the first crossing index. On $\{\tau =k\}$, the future sum $R=X_{k+1}+\cdots +X_n$ is independent of the past and symmetric. For any vector $s$ with $\Vert s\Vert >t$, the two vectors $s+r$ and $s-r$ cannot both have norm at most $t$: their average is $s$, contradicting the triangle inequality. Symmetry therefore gives conditional terminal escape probability at least $1/2$. Sum over the disjoint first-crossing events to obtain the inequality. This controls every intermediate partial sum using a single terminal tail.

Letbe 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 $X_1,\dots ,X_k$, the random variable $R_k$ is independent and symmetric. Sinceat least one of the two norms on the right exceeds $t$. Symmetry of $R_k$ consequently giveson $A_k$. The events $A_k$ are disjoint, so summation proves the Lévy maximal inequality

For the Gaussian series, put $S_m={\sum }_{j=1}^m{g}_j{u}_j$. Apply the inequality to the symmetric independent increments from $n+1$ through $m$, followed by Markov inequality in squared norm:Here the cross terms vanish by orthogonality of independent centered Hilbert-space random variables. Letting $m\to \infty$ and then $n\to \infty$ showsThe convergence criterion in the question now shows that ${\sum }_j{g}_j{u}_j$ converges almost surely in $V$.