For any ( ϕ, ψ ) ∈ Φ$_{c}$ , one can write which is a continuous convex combination of a continuum of pairs of The nontrivial case to check is when the LHS in (A.4) is positive. Consider the case where s + t < ϕ ( y$_{0}$ ) ≤ s + t + 1 and − ( s + t ) < ψ ( y$_{1}$ ) ≤ − ( s + t − 1) for some nonnegative integer t and s ∈ [0 , 1]. Then, and so the LHS in (A.4) is 1. Also, it follows from (A.1) that for ( y$_{0}$, y$_{1}$ ) ∈$_{R}$×$_{R}$ s.t. s + t ≤ ϕ ( y$_{0}$ ) < s + t +1 and − ( s + t ) < ψ ( y$_{1}$ ) , and thus (A.4) is satisfied in this case from the following: Consider another case where s + t ≤ ϕ ( y$_{0}$ ) < s + t + 1 and − ( s + t − 1) < ψ ( y$_{1}$ ) ≤ − ( s + t − 2) for 57 1 { y$_{1}$ − y$_{0}$ < δ } + λ (1 − 1$_{C}$ ( y$_{0}$, y$_{1}$ )) ≥ 1 . 0 < ϕ ( y$_{0}$ ) + ψ ( y$_{1}$ ) ≤ 1 { y$_{1}$ − y$_{0}$ < δ } + λ (1 − 1$_{C}$ ( y$_{0}$, y$_{1}$ )) , ∞ ∑ k =0 $^{(}$1$_{A}$ + $_{k}$( ϕ,s $_{)}$( y$_{0}$ ) − 1$_{A}$ − k ( ϕ,s $_{)}$( y$_{0}$ ) ) = t + 1 , ∞ ∑ k =0 $^{(}$1$_{A}$ + $_{k}$( ψ,s $_{)}$( y$_{1}$ ) − 1$_{A}$ − k ( ψ,s $_{)}$( y$_{1}$ ) ) = − t, ( ∞ ∑ k =0 $^{(}$1$_{A}$ + $_{k}$( ϕ,s ) − 1$_{A}$ − k ( ϕ,s ) $^{)}$, ∞ ∑ k =0 $^{(}$1$_{A}$ + $_{k}$( ψ,s ) − 1$_{A}$ − k ( ψ,s ) ) ) s ∈ [0 , 1] . ∞ ∑ k =0 $^{(}$1$_{A}$ + $_{k}$( ϕ,s $_{)}$( y$_{0}$ ) − 1$_{A}$ − k ( ϕ,s $_{)}$( y$_{0}$ ) $^{)}$+ ∞ ∑ k =0 $^{(}$1$_{A}$ + $_{k}$( ψ,s $_{)}$( y$_{1}$ ) − 1$_{A}$ − k ( ψ,s $_{)}$( y$_{1}$ ) (A.4) ≤ 1 { y$_{1}$ − y$_{0}$ < δ } + λ (1 − 1$_{C}$ ( y$_{0}$, y$_{1}$ )) . ( ϕ, ψ ) = ( ϕ$_{+}$ + ϕ$_{−}$, ψ$_{+}$ + ψ$_{−}$ ) = ∫ 1 0 ( ∞ ∑ k =0 $^{(}$1$_{A}$ + $_{k}$( ϕ,s ) − 1$_{A}$ − k ( ϕ,s ) $^{)}$ds, ∞ ∑ k =0 $^{(}$1$_{A}$ + $_{k}$( ψ,s ) − 1$_{A}$ − k ( ψ,s ) ) ) ds, To see if ( ∑ ∞ k =0 $^{(}$1$_{A}$ + $_{k}$d ( ϕ,s ) − 1$_{A}$ − k ( ϕ,s ) $^{)}$, ∑ ∞ k =0 $^{(}$1$_{A}$ + $_{k}$( ψ,s ) − 1$_{A}$ − k ( ψ,s ) )) ∈ Φ$_{c}$ , check the following: for any s ∈ [0 , 1] and λ = ∞ ,