In this case, for each n ∈ N , there exist integers k n 0 between 1 and N such that Therefore, it can be verified that The last inequality implies the existence of numbers η$_{n}$ ∈ [ − k n $_{0}$, k n $_{0}$] satisfying Suppose that g ( s ) = ( g$_{1}$ ( s ) , g$_{2}$ ( s ) , . . . , g$_{m}$ ( s )) , where each g$_{i}$, 1 ≤ i ≤ m, is a real-valued function. In accordance with (4.6), for each n ∈ N , there is an integer j$_{n}$, 1 ≤ j$_{n}$ ≤ m, with where τ$_{n}$ = τ$_{n}$ + η$_{n}$, n ∈ N . Since the function g is uniformly continuous, there exists a positive number Δ ≤ 1 , which does not depend on the sequences { t$_{n}$ } and { τ$_{n}$ } , such that both of the inequalities and are valid for s ∈ [ − Δ , Δ] . Thus, we have for s ∈ [ − Δ , Δ] that For each n ∈ N , one can find numbers s n $_{1}$, s n $_{2}$, . . . , s n m ∈ [ − Δ , Δ] such that 7 ∥ ∫ Δ − Δ [ g ( t$_{n}$ + τ$_{n}$ + u ) − g ( τ$_{n}$ + u )] du ∥ = 2Δ ( m ∑ i =1 [ g$_{i}$ ( t$_{n}$ + τ$_{n}$ + s n i ) − g$_{i}$ ( τ$_{n}$ + s n i )] 2 ) 1 / $^{2}$. (4.8) | g$_{j}$$_{n}$ ( t$_{n}$ + τ$_{n}$ + s ) − g$_{j}$$_{n}$ ( τ$_{n}$ + s ) | ≥ | g$_{j}$$_{n}$ ( t$_{n}$ + τ$_{n}$ ) − g$_{j}$$_{n}$ ( τ$_{n}$ ) | − | g$_{j}$$_{n}$ ( t$_{n}$ + τ$_{n}$ ) − g$_{j}$$_{n}$ ( t$_{n}$ + τ$_{n}$ + s ) | − | g$_{j}$$_{n}$ ( τ$_{n}$ ) − g$_{j}$$_{n}$ ( τ$_{n}$ + s ) | ≥ ϵ$_{0}$ 2 N m . (4.7) ‖ g ( τ$_{n}$ ) − g ( τ$_{n}$ + s ) ‖ ≤ ϵ$_{0}$ 4 N m ‖ g ( t$_{n}$ + τ$_{n}$ ) − g ( t$_{n}$ + τ$_{n}$ + s ) ‖ ≤ ϵ$_{0}$ 4 N m | g$_{j}$$_{n}$ ( t$_{n}$ + τ$_{n}$ ) − g$_{j}$$_{n}$ ( τ$_{n}$ ) | ≥ ϵ$_{0}$ N m , ‖ g ( t$_{n}$ + τ$_{n}$ + η$_{n}$ ) − g ( τ$_{n}$ + η$_{n}$ ) ‖ ≥ ϵ$_{0}$ N , n ∈ N . (4.6) sup s ∈ [ − k n $_{0}$,k n $_{0}$] ‖ g ( t$_{n}$ + τ$_{n}$ + s ) − g ( τ$_{n}$ + s ) ‖ ≥ ϵ$_{0}$ N , n ∈ N . ρ$_{k}$ n $_{0}$( g$_{t}$$_{n}$$_{+}$$_{τ}$$_{n}$, g$_{τ}$$_{n}$ ) ≥ 2 k n 0 ϵ$_{0}$ 2 N ≥ ϵ$_{0}$ N .