By Volkodavov V.F., Radionova I.N., Bushkov S.V.

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**Example text**

S. ) on {ω : t < τ (ω)}. We take up the proof of this theorem after discussing its hypotheses and assertions. 5) exists if h(s) is completely measurable and, for t < τ (ω), t 0 ˜ |h(s)|d A t s + 0 2 ˜ |h(s)| dm s < ∞, where A s = lim n→∞ ∞ k=0 A s∧ k+1 2n −A s∧ k 2n . ˜ Both these conditions are satisfied because h(s) is continuous in s and is Fs consistent, while m t + A t < ∞. We point out that by a stochastic integral we always understand a continuous (for all ω) process. 5) are therefore meaningful.

It may be assumed with no loss of generality that the original sequence has this property. Moreover, we set π0 = 0; then, as is well know, in probability for r > 0, t ∈ [0, 1], lim n→∞ tn j+1 ≤t (1 − πr )(h(tnj+1 ) − h(tnj ) 2 = (1 − πr )m t . 21) Therefore, there exists a subsequence along which the last equality, understood in the sense of pointwise convergence, is true for all r > 0, t ∈ I almost surely. To simplify the notation, we assume that this subsequence also coincides with the original sequence.

Ref. [42]). 2. If a sequence vn converges strongly in V to v, then the sequence A(vn ) converges weakly to A(v) in V ∗ . Proof. Because of assumption (A4 ), for every subsequence {µ} of natural numbers, the sequence A(vµ ) is bounded in V ∗ , and therefore there exists a subsequence {η} of the sequence {µ} along which A(vη ) converges weakly to some A∞ ∈ V ∗ . We will now show that A∞ = A(v). Let u be an arbitrary element of V . By assumption (A2 ) 2 (u − vη )(A(u) − A(vη )) − K |u − vη |H ≤ 0.