Exact(2)
Since, for each and, the function is continuous on, from Lemma 2.5(1), we can see that and are l.s.c., hence (1) is true.
(1) Since, for each and, the function is continuous on, from Lemma 2.5(1), we can see that and are l.s.c., hence (1) is true. .
Similar(58)
Hence is continuous on, and from we conclude.
A function (f Irightarrowmathbb{R}) is n-exponentially convex on I if it is n-exponentially convex in the Jensen sense and continuous on I. From Definition 4 it follows that 1-exponentially convex functions in the Jensen sense are exactly nonnegative functions.
Thus, in view of the fact that x ( t ) ≡ x ( t 0 − r ) for t ∈ ( − ∞, t 0 − r ], we obtain that x ( t ) is uniformly continuous on R. From (2.1), for any ϵ > 0, there exists l = l > 0 such that every interval [ α, α + l ], α ∈ R, contains δ for which | ϵ ( δ, t ) | ≤ 1 2 η ϵ for all t ∈ R. (2.16).
Since is uniformly norm-to-norm continuous on bounded sets, from (3.22) we have (3.24).
Since is uniformly norm-to-norm continuous on bounded sets, from (3.38), we have.
Since the functions are uniformly continuous on, it follows from the above estimates that is an equicontinuous set.
by (2.41), and the function is increasing and continuous on, we deduce from (2.45) and (2.46) that.
In the case where is continuous on, it follows from the Stone-Weierstrass theorem that can be approximated over an arbitrarily large compact set.
Since J is uniformly norm-to-norm continuous on bounded sets, from (2.3), we have lim n → ∞ J x n - J u n = 0. (2.4).
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