The basic idea of $L^2$ constructions of Brownian motions is as follows. Let $\{\xi_k\}$ be a sequence of iid Gaussian random variables. Let $\{\phi_k\}$ be a complete orthonormal basis system in $L^2([0,1])$. The idea is to show that the limit
$$ \lim_{N\to \infty} B^N(t) $$
is a Brownian motion, where
$$ B^N(t) = \sum^N_{k=1} \xi_k \, \langle \mathbb{1}_{[0,t]}, \phi_k \rangle. $$
It can be shown that the aximos of the Brownian motions always hold except for the continuity axiom. In literature, in order to prove the continuity axiom, one has to spefically choose the basis $\{\phi_k\}$, for example, the Haar and Schauder basis.
My question: is it so that the continuity actually hold for all choices of system $\{\phi_k\}$? If not, is there an example of $\{\phi_k\}$ such that the continuity does not hold?
The Kolmogorov continuity theorem tells us that if $X$ is a stochastic process satisfying $\mathbb{E}[|X_t-X_s|^{\gamma}] \le c|t-s|^{1+\varepsilon}$ for some positive constants $\gamma,c,\varepsilon$ and all $t,s \in [0,1]$, then there exists a continuous process $\tilde X$ that is indistinguishable from $X$, i.e. $\mathbb{P}(\tilde X_t = X_t \text{ for all }t \in [0,1]) = 1$. Since $\mathbb{E}[|X_t-X_s|^{\gamma}] \le c|t-s|^{1+\varepsilon}$ is implied by $X_t - X_s \sim N(0,t-s)$, $B(t)$ satisfying all of the axioms except continuity implies $B(t)$ is continuous up to indistinguishability. Therefore, regardless of the choice of $\{\phi_k\}$ used in the construction, the limit is continuous up to indistinguishability.