Then the sequence { x n } generated by (2.1) converges strongly to x = Π F x 0, where Π F is the generalized projection of E onto F. Proof For each n = 1, 2, 3, …, N, define G n : C × C → R by G n ( x, y ) = f n ( x, y ) + 〈 A n x, y − x 〉.
Then the operator A has a fixed point in E. Proof For any x ∈ E, since A : E ⟶ E, we have A x ∈ E. Since E is a Banach space with lattice structure, there exists u 0 ∈ E such that inf { A x, x } = u 0. That is, u 0 ≤ A x, u 0 ≤ x. (2.8).
Then I n ′ ( u ) v = ( u, v ) 0 − ∫ 0 σ ( T ) f n ( t, ( u + ) σ ( t ) ) v σ ( t ) Δ t, u, v ∈ E. Lemma 3.5 Assume that (H1) and (H2) hold, then there exists n 0 > 0 such that the functional I n satisfies the P.S. condition in E for n > n 0. Proof For given n, let { u m } ⊆ E be the P.S. sequence of I n, that is, { I n ( u m ) } is bounded and I n ′ ( u m ) → 0 as m → ∞.
Notice that the proof for necessity e-conversion depends on possibility i-conversion, and the proof for possibility i-conversion on necessity e-conversion.
Hence, c is an eigenvalue with corresponding eigenfunction x, i.e., c = sup E ∈ E. This completes the proof for Case (b).
In particular for almost every λ ∈ [ 1, 2 ] there is a bounded sequence { v j } ⊂ E satisfying I λ ( v j ) → c λ and I λ ′ ( v j ) → 0. Proof For the v ∈ E obtained in Lemma 2.2, we have I λ ( v ) ≤ 0. It follows from (H1) that I λ ( v ) ≤ I ( v ) ≤ 0, ∀ λ ∈ [ 1, 2 ].
Proof: Please refer to Appendix E for the proof of this theorem.
Proof For ∀ u ∈ E, T u ( t ) = ∫ 0 1 G ( t, s ) f ( s, u ( s ), C D 0 + σ u ( s ) ) d s, by Lemma 3.1 and Lemma 3.2, we have T : E → E. Now we separate the proof into three steps.
Theorem 2: The above algorithm yields T i, j (L, M ) = T i, j q (X, Π s (X ) ) and E i (y, L, M ) = E i q (y, X, Π s (X ) ). Proof: The proof for this theorem is very similar to the proof of theorem 1 for Viterbi training and therefore omitted.
Then f ( X ) is an increasing function for X > e. Proof Since f ′ ( X ) = ( log X − 1 ) / ( log X ) 2, we have f ′ ( X ) > 0 for X > e.
I don't have proof for now".
