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Consequently,, and so is continuous at, or equivalently,.
Since and, so is continuous for all and we have exponentially convexity of the function.
Moreover, the maps and are continuous, so is continuous.
Since we have, so is continuous for all ; then by (3.6) and Proposition 1.2 we have that is exponentially convex.
So, is continuous on since is arbitrary.
and so is Lipschitz continuous with constant.
Let us define the function F : X → ℝ by F ( x ) = T x - x + T 2 x - T x + T 2 x - x. Since by Lemma 4.7 the function T is continuous and so is F. It is easy to see that F is bounded form below and not identically +∞.
It is easy to see that the function (numapsto I^{0} u; nu)) is sublinear, continuous and so is the support function of a nonempty, convex and (omega^ -compact set (partial I (u)subset X^) defined by partial I (u)=bigl{ u^in X^:bigllangle u^,nubigrrangle _{X}leq I^{0}(u; nu) mbomega^ -compactin Xbigr}.
Furthermore, the cost-effectiveness of sequencing is rapidly increasing and so is the accuracy due to continuous technical improvements, including some bench-top sequencing platforms [ 36].
Hence, ∥ T x − T y ∥ < ϵ and so T is continuous on K.
