and then φ φ ( t ) := ∑ ∑ k = 1 ∞ ∞ ξ ξ k ψ ψ k ( t ) is easily seen to be a one-dimensional Brownian motion (or Brownian bridge, if the boundary values for φ φ k are set up that way). So, in this case, it is a random continuous function (not belonging to H 1 [ 0 , 1 ] , however). For instance, if ( φ φ k ) is the Haar basis, then this is Lévy's construction of Brownian motion, see, e.g., Section 3 of Peres (2001).