Sentence examples for continuity of ψ from inspiring English sources

By the continuity of ψ , ξ e and since K is continuous with compact values in X × Λ, thus, by Proposition 23 in Section 1 of Chapter 3 [20], we can deduce that g ( x, z, γ, μ ) is continuous with respect to ( x, z, γ, μ ).

This shows that for any bounded subset B ⊆ X, the family { Ψ ( ⋅, u ) : u ∈ B } is equicontinuous at each λ 0 ∈ R. Hence it follows from the continuity of Ψ ( λ 0, ⋅ ) that Ψ is continuous on R × X, on observing the following relation: ∥ Ψ ( λ, u ) − Ψ ( λ 0, v ) ∥ L q ( x ) ≤ ∥ Ψ ( λ, u ) − Ψ ( λ 0, u ) ∥ L q ( x ) + ∥ Ψ ( λ 0, u ) − Ψ ( λ 0, v ) ∥ L q ( x ).

by continuity of ψ(3).

Therefore, Ψ ′ is compact by the weakly continuity of Ψ ′ since E is a Hilbert space.

This implies that ψ ( x n i ) ⇀ ψ ( z ) due to the weak continuity of ψ.

Using the continuity of ψ and lower semi-continuity of ϕ, we get ψ , (17).

On taking n → ∞ in (2.9), using (2.7) and (2.8), the continuity of μ and lower semi-continuity of ψ, we get that μ ≤ μ ( 1 2 ) − ψ.

Letting n → ∞ in (2.1) and using the continuity of μ and lower semi-continuity of ψ, we obtain that μ ( r ) ≤ μ ( r ) − ψ ( 0, 2 r ).

Having in mind the continuity of φ and the lower semi-continuity of ψ, we obtain φ ε ≤ φ ε − lim inf k → + ∞ Ω ( x n k, x m k, x m k ) ≤ φ ε − ψ ε, which is impossible, since ε > 0. It follows that lim m, n → + ∞ Ω ( x n, x m, x m ) = 0, m > n.

By supposition, we can replace x by u and y by v in (4.1), and by the lower semi-continuity of ψ, we obtain d ( v, u ) = d ( T v, S u ) ≤ ψ ( d ( R v, R u ), d ( R v, T v ), d ( R u, S u ), d ( R v, S u ), d ( u, T v ) ) ≤ ψ ( d ( v, u ), 0, 0, d ( v, u ), d ( u, v ) ) < d ( u, v ), a contradiction.

Letting n → + ∞ in relation (6), the continuity of φ and the lower semi-continuity of ψ imply φ r ≤ φ r − lim inf n → + ∞ ψ ( Ω ( g x n, g x n + 1, g x n + 1 ) + Ω ( g y n, g y n + 1, g y n + 1 ) ) ≤ φ r − ψ r, which forces φ r = 0, that is, r = 0. Since Ω takes nonnegative values, lim n → + ∞ Ω ( g x n, g x n + 1, g x n + 1 ) = 0 and lim n → + ∞ Ω ( g y n, g y n + 1, g y n + 1 ) = 0.