The phrase "a then for" is not correct and does not make sense in written English.
It appears to be a combination of words that lacks clarity and proper grammatical structure.
Example: "I need a then for my project" does not convey a clear meaning.
Alternatives: "a time for" or "a reason for".
If (A,Bin operatorname{CB}(X)) and (xin A), then for each (epsilon>0), there is (bin B) such that d a, b)< H A, B +epsilon.
For example, if we choose (K=frac{1}{2} A), then for arbitrary positive definite matrix Λ, (28) can always be guaranteed.
If (A,B in operatorname{CB}(X) ) and (a in A), then, for each (varepsilon> 0), there exists (b in B ) such that (d a,b) leq H A,B) + varepsilon).
Moreover, if there exist an element x 0 ∈ Q and an operator S 0 ∈ L ( X, Y ) satisfying (11) from Theorem A then for δ = ∥ x 0 ∥ X, ∥ A x 0 ∥ Y = Ω ( δ, A ; Q ) = E δ ( O, A ; Q ) = E δ ( L, A ; Q ).
The idea is to show that if π 1 and π 2 are admissible projections on A, then for some.
If | f ′ | p / p − 1 is preinvex on A then, for every a, b ∈ A with θ ( a, b ) ≠ 0 the following inequality holds (11).
We show here an example: if the medium value is A = ϵ A ∗, then for ϵ A = ϵ A ∗ − ε an editor at x = 0 will disagree with the article and change it by Δ = ϵ A ∗ μ A, so the new medium value would be A = 1 − ϵ A ∗. Afterwards an agent at x = 1 can restore the article to its previous state and avoid consensus.
Let X be a Banach space with norm ∥ ⋅ ∥, and let − A : D ( A ) ⊂ X → X be the infinitesimal generator of an analytic semigroup S ( t ) ( t ≥ 0 ) of a uniformly bounded linear operator in X, that is, there exists M ≥ 1 such that ∥ S ( t ) ∥ ≤ M for all t ≥ 0. Without loss of generality, let 0 ∈ ρ ( A ). Then for any α > 0, we can define A − α by A − α = 1 Γ ∫ 0 ∞ t α − 1 S ( t ) d t.
Theorem 5 Suppose that f ∈ CM [ a, ∞ ), then for any δ ≧ 0, the sequence { f ( a + n δ ) } n = 0 ∞ is completely monotonic. This result was generalized in [28] as follows.
A pain, then, for everyone.
A victory, then, for common sense?
