We firstly prove that there exists, independent of and, such that.
Firstly, if there exists some external attackers, who want to make cluster paralysis, they have to follow this protocol.
Firstly, if there exists other subsequence of { x n } which is denoted by { x n t } such that x n t ⇀ q ∈ Ω with q ≠ p, then, by Opial's condition, lim inf t → ∞ ∥ x n t − q ∥ < lim inf t → ∞ ∥ x n t − p ∥ < lim inf t → ∞ ∥ x n t − q ∥.
In view of assumption (e′1), firstly we assume that g ( X ) is complete, then there exists u ∈ X such that lim n → ∞ g ( x n ) = g ( u ).
Then for the expansion mapping T, there exists a unique fixed point in X. Proof Firstly, T is injective.
Firstly, let us consider ω ∈ F − 1 ( B 0 ) and hence there exists x 0 ∈ B 0 such that x 0 ∈ ( T − 1 ( x 0 ).
Firstly, there implicitly exists a constraint from the QPSK constellation construction that eliminates nearby lattice points that do not belong to ( {xi}^{n_{mathrm{T}}} ), due to the quantization operation ( {Q}_{xi^{n_{mathrm{T}}}}left{.right} ).
Firstly, we claim that there exists a point (t_{1}in[0,T]) such that d_{1}leq x(t_{1} leq d_{2}.
Firstly, we show that there exists (rho>0) such that (inf_{uin{mathbf{S}}_{rho}}varphi u)>0).
Firstly, we show that there exists (underline{t}>0) independent of k such that (3.15) and (3.16) are satisfied for all (tgequnderline{t}).
Firstly, we show that there exists a simple O(n -time algorithm if the size of the alphabet is un -timed, where n is the length of the output string.
