taking on the both sides of equality above and in view of the weak lower semi-continuity of norm, it yields that (3.34).
In view of the continuity of φ on ([0,infty)), we have varphi z lemax_{zin[0,z_{0}]}varphi z) quad text{for all } z>0, which implies (2.4).
In view of the continuity of (f(x)) and Assumption 1, we have (0<lambda_{1}leq+infty) and (0leqlambda_{2}<+infty). Suppose Assumption 1 holds.
In view of the Hölder continuity of u, i.e., (3.48), Arzelà-Ascoli theorem asserts ((u(cdot,t))_{t>1}) is relatively compact in (C^{0}(bar{Omega})).
In view of the continuity of φ for (lambda>1), from (10) and (11) there follows the existence of a point that satisfies (8) and (9).
In view of the absolute continuity of the Lebesgue integral and properties (c) and (f), for any (epsilon>0), we can choose a number (k'inmathbf{N}) large enough such that |l|_{L^{2} omega_{t^backslashomega_{t'})}< sqrt{epsilon}, qquad| tilde{u}|_{L^{2} omega_{t^backslashomega_{t'})}< sqrt{ epsilon}.
In view of the continuity of the eigenvalues, Lemma 26, we may now assume that (mu <mu ^0(theta _i,c_k,d_k)) for all (kin mathbb {N}).
(2) Let (uin K), in view of the nonnegativeness and continuity of functions (G t,s)), (a(t)), (f t, u(t))), (I_{k} u)) and (lambda>0), we conclude that (T K to K) is continuous.
Proof The operator T : K → X is continuous in view of the nonnegativity and continuity of the functions G ( t, s ) and f ( t, u ( t ), D 0 + β u ( t ) ).
Further (mathcal{S}_{1}) is continuous in view of the continuity of f.
