Sentence examples for continuity of the involved from inspiring English sources

S without the sequential weak continuity of the involved mappings.

By the continuity of the involved operators, we have ∥ v 1 ∥ 2 m, p, R + n ≤ C ( ∥ u 1 ∥ 0, p, R + n + ∥ f 1 ∥ 0, p, R + n ).

Remark 4.2 Theorem 4.1 improves the results of Aalam et al. [[18], Theorem 3.1], Rao et al. [[12], Theorem 3.1] and Chauhan and Kumar [[20], Theorem 3.1] as Theorem 4.1 never requires conditions on completeness (or closedness) of the underlying space (or subspace), continuity of the involved mappings and containment amongst range sets of the involved mappings.

Thus, the continuity of the involved contractive mapping cannot be derived from the contractivity condition (as in the Banach theorem): in fact, Nieto and Rodríguez-López replaced such condition by the regularity of the partially ordered ambient metric space (which is a condition about the behavior of nondecreasing convergent sequences).

Jungck and Rhoades [22] weakened the notion of compatibility by introducing the notion of weakly compatible mappings (extended by Singh and Jain [23] to probabilistic metric space) and proved common fixed-point theorems without assuming continuity of the involved mappings in metric spaces.

Closure of the two sets simply follows from the continuity of the mappings involved.

In a recent study [24], the Hölder continuity assumption is relaxed to C p continuity of the functions involved in the Riemann-Liouville fractional differential equation.

Meanwhile, a certain topological continuity of the mapping involved in the considered variational inequality must be required, such as continuity and semicontinuity.

In this section, we apply the variational characterization of the metric projection in Hilbert spaces to study the solvability of general variational inequalities without the continuity of the mappings involved in the considered general variational inequalities.

Then, because of the continuity of the functions involved and (2.9), it follows that there exists a t1 such that 0 < t1 ≤ 1 and v ( t 1 ) = w ( t 1 ), v ( t ) < w ( t ), 0 < t < t 1. (2.11).

Then, there exists a t1> t0 such that v t1) > w t1) and t0< t1 ≤ T. Since v t0) ≤ w t0) we encounter with two cases in view of the continuity of functions involved: (i) If v t0) < w t0), then one can find a τ0 such that v τ0) = w τ0) and t0< τ0< t0<