Theorem 5.1 The function φ ( t ) is unpredictable. Proof. First of all, we will show that ρ ( φ$_{i}$$_{n}$ , φ ) → 0 as n → ∞ , where ρ is the metric defined by equation (2.1). Let us fix an arbitrary natural number n. The functions φ ( i$_{n}$ + s ) and φ ( s ) satisfy the equation It is worth noting that for each r ∈ $^{Z}$both of the points ζ$_{r}$ and ζ$_{i}$$_{n}$$_{+}$$_{r}$ − i$_{n}$ belong to the interval ( r, r +1) . Moreover, ‖ ν ( i$_{n}$ + s ) − ν ( s ) ‖ = ‖ d$_{1}$ − d$_{2}$ ‖ for s ∈ ⋃ r = −∞ ̂ ( ζ$_{r}$, ζ$_{i}$$_{n}$$_{+}$$_{r}$ − i$_{n}$ ] , and ‖ ν ( i$_{n}$ + s ) − ν ( s ) ‖ = 0 , otherwise. Since for each r = − 2 k$_{n}$, − 2 k$_{n}$ + 1 , . . . , k$_{n}$ − 1 the distance between the points ζ$_{r}$ and ζ$_{i}$$_{n}$$_{+}$$_{r}$ − i$_{n}$ are at most δ$_{n}$, one can verify for each s ∈ [ − k$_{n}$, k$_{n}$ ] that Hence, we have The last inequality implies that ρ ( φ$_{i}$$_{n}$ , φ ) → 0 as n → ∞ , i.e., φ ( t ) is a Poisson function. Now, let us show the existence of a positive number ϵ$_{0}$ satisfying ϵ$_{0}$ → 0 as ϵ$_{0}$ → 0 such that ρ ( φ$_{i}$$_{n}$$_{+}$$_{j}$$_{n}$, φ$_{j}$$_{n}$ ) ≥ ϵ$_{0}$ for each n ∈ $^{N}$. For a fixed natural number n, using the equations and 9 φ ( j$_{n}$ + s ) = e − $^{γs}$φ ( j$_{n}$ ) + s ∫ 0 e − γ ( s − u $^{)}$ν ( j$_{n}$ + u ) du, φ ( i$_{n}$ + j$_{n}$ + s ) = e − $^{γs}$φ ( i$_{n}$ + j$_{n}$ ) + s ∫ 0 e − γ ( s − u $^{)}$ν ( i$_{n}$ + j$_{n}$ + u ) du ρ ( φ$_{i}$$_{n}$ , φ ) = ∞ ∑ k =1 2 − $^{k}$ρ$_{k}$ ( φ$_{i}$$_{n}$ , φ ) ≤ k$_{n}$ ∑ k =1 2 − $^{k}$ρ$_{k}$ ( φ$_{i}$$_{n}$ , φ ) + 2 − k$_{n}$ ≤ 1 n . ‖ φ ( i$_{n}$ + s ) − φ ( s ) ‖ ≤ − 2 k$_{n}$ ∫ −∞ e − γ ( s − u $^{)}$‖ ν ( i$_{n}$ + u ) − ν ( u ) ‖ du + k$_{n}$ − 1 ∑ r = − 2 k$_{n}$ ∣ ζ$_{r}$ ∫ ζ$_{i}$$_{n}$$_{+}$$_{r}$ − i$_{n}$ ‖ ν ( i$_{n}$ + u ) − ν ( u ) ‖ du ∣ ≤ ‖ d$_{1}$ − d$_{2}$ ‖ γ e − γk$_{n}$ + 3 k$_{n}$δ$_{n}$ ‖ d$_{1}$ − d$_{2}$ ‖ ≤ 1 2 n . φ ( i$_{n}$ + s ) − φ ( s ) = s ∫ −∞ e − γ ( s − u $^{)}$( ν ( i$_{n}$ + u ) − ν ( u )) du. ( a, b ] if a < b and ̂ ( a, b ] = ( b, a ] if a > b.