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We may choose sequences { m k }, { n k } such that corresponding to m k, natural number n k is smallest satisfying (3.13).
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Then one may choose a sequence such that and define (37).
Since ∥ φ ∥ ∞ = 1, we may choose a sequence { z k } k ∈ N ⊂ D such that | φ ( z k ) | → 1 as k → ∞.
For (f(x) in L^{p_{i}}(Omega)), we may choose a sequence ({f_{n}}subset L^{2}(Omega)) such that (f_{n}(x) rightarrow f(x)) in (L^{p_{i}}(Omega)), as (n rightarrowinfty).
end{aligned} (25)Now, since (u) is bounded and Lipschitz continuous in (,overline{!Q}), in view of the Ascoli Arzela theorem, we may choose a sequence (tau _jrightarrow infty ) and a bounded function (zin mathrm{Lip}(mathbb{T }^n times (-infty,+infty ))) so that begin{aligned} lim _{jrightarrow infty } u x,t+tau _j)=z x,t) quad text{ locally } text{ uniformly } text{ on } ; mathbb{R }^{n+1}.
As one solution, the researcher may choose to sequence additional orthologous genes of appropriate evolutionary rate.
As one solution, the researcher may choose to sequence additional orthologous genes of appropriate evolutionary rate for the taxa in the study.
Second, we choose sequences and satisfy (4.8).
We may choose an increasing sequence ({j_k}_{kin mathbb {N}}) of natural numbers so that as (krightarrow infty ), the sequence ({(r_{j_k,0},r_{j_k,1},ldots,r_{j_k,n+1})}) converges to a point ((r_0,ldotsdots,r_{n+1} in mathbb {R}^{n+2}).
We may choose the diagonal sequence { x n k + 1 ( n 0 + k ) ( t ) } which converges everywhere in ( 0, 1 ) and it is easy to verify that { x n k + 1 ( n 0 + k ) ( t ) } converges uniformly on any interval [ c, d ] ⊆ ( 0, 1 ).
We may choose a distinct nondecreasing sequence { λ n } n = 1 ∞ ⊂ Λ such that lim n → ∞ λ n = λ ˜ μ.
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com