ON CONDITIONS FOR THE STRONG LAW OF LARGE NUMBERS IN GENERAL BANACH SPACES

We give Chung-Teicher type conditions for the SLLN in general Banach spaces under the assumption that the weak law of large numbers holds. An example is provided showing that these conditions can hold when some earlier known conditions fail.

Let (Ꮾ, ) be a real, separable Banach space.A strongly measurable mapping X from a probability space (Ω, Ᏺ, ᏼ) into Ꮾ is said to be a random element.
If E X < ∞, then the expectation EX is defined by the Bochner integral.
The aim of our paper is to give conditions for the SLLN in Banach spaces which can be applied in more general cases than those of Choi and Sung [5], Sung [15], Adler, Rosalsky, and Taylor [1] and Kuczmaszewska and Szynal [12].We present also an example showing that these conditions can be applied when some earlier known conditions fail.We make use of the following lemmas.
Theorem 3. Let {X n , n ≥ 1} be a sequence of independent Ꮾ-valued random elements.Suppose that in the case (a) for some p, or in the case (b) for some p, and (C) is satisfied for some sequence {a n , n ≥ 1} of positive numbers such that if and only if where Proof.Suppose that (4) holds.
Let r ≥ 1.We note that where we put r = p in the case (a) and r = 1 in the case (b).Hence {X n , n ≥ 1} and {X n , n ≥ 1} are equivalent.Therefore, by (4), we have Write Define where Now we prove that E S * n /n → 0, n → ∞.Using (9) and Lemma 1, we get Hence in the case (a) by (B), we get while in the case (b) by (B 1 ) Therefore, in the case (a) and (b) we have Thus we conclude, from S * n /n P → 0, n → ∞, and (14), that Now we are going to prove that Taking into account the identity we see that Now, put Then by Chebyshev's inequality, equation ( 9) and Lemma 1, we have Hence we see that in the case (a) under the assumptions (C) and (D) we have 13  15 Similarly, in the case (b) under the assumptions (C) and (D 1 ) we lead to Therefore, by the Borel-Cantelli lemma, we state that Now, note that in the case (a) the assumption (B) implies Similarly, in the case (b), by B 1 , we get Therefore, in the case (a) and (b), we obtain Using the assumption (C), we get which proves in the end that Now we see that Y n,i i−1 j=1 Y n,j , 2 ≤ i ≤ n is a martingale difference for fixed n.Therefore, in the case (a), after using Chebyshev's inequality, (9), and Lemma 1, we get Similarly, in the case (b), we have Now using the Borel-Cantelli lemma we obtain Thus by (15) and (18) we see that so we have But {X n , n ≥ 1} and {X n , n ≥ 1} are equivalent, so that which completes to proof of Theorem 3.
Before giving an example showing that the presented conditions under which WLLN is equivalent to the SLLN can be applied when some earlier known ones fail we quote the following results in this subject.
Theorem 4 [13].Let {X n , n ≥ 1} be a sequence of independent Ꮾ-valued random variables such that (a) X j /j → 0 a.s., j → ∞, (b) for some p ∈ [1,2] and some r ∈ (0, ∞) Then Theorem 5 [13].Let {X n , n ≥ 1} be a sequence of independent Ꮾ-valued random variables such that (a) |X j | ≤ M j /LLj for some constant M < ∞, where LLj = log(log(j ∨ e e )), and Theorem 6 [15].Let {X n , n ≥ 1} be a sequence of independent B-valued random variables, and let {a n } and {b n } be constants that 0 < b n .Suppose that and let e n denotes the element having 1 for its nth coordinate and 0 in the other coordinates.
But we can show that the assumptions of Theorem 3 are fulfiled as Now it is enough to see that (4) holds.Taking into account that Using Chebyshev's inequality and the fact that l 2 is a space of the type 2, we have which completes the proof that Corollary 8.If (B) and (B 1 ) are replaced by the condition and in the case (b) additionally Proof.It is enough to show that the condition implies Indeed in the case (a) we have while in the case (b), for some sequence {a n , n ≥ 1} of positive numbers with To prove the above-given assertion it is enough to use in the case (b) of the Theorem 3 the function φ(x) = x 2 .
Corollary 10.Let {X n , n ≥ 1} be a sequence of independent random elements in a Banach space of Ᏻ α , 0 < α ≤ 1 [12].Suppose that in the case (a) the conditions (A), (B), (C) and (D) are satisfied with p = 1 + α, or in the case (b) the conditions (A 1 ), (B 1 ), (C), and (D 1 ) are satisfied with p = 1 + α.Then Proof.It is enough to show that (4) holds.Indeed in the case (a), we get and in the case (b), we have But{X n , n ≥ 1} and {X n , n ≥ 1} are equivalent, therefore we have (4) which completes the proof of Corollary 10.Now we give a generalization of Theorem 3 replacing the condition (A) or (A 1 ) by less restrictive ones.
Lemma 11.If x j , 1 ≤ j ≤ n, are real numbers, S n = n j=1 x j and where c k is a generic designation for a finite linear combination (coefficients indepen- Using this lemma we can prove the following result. Theorem 12. Let {X n , n ≥ 1} be a sequence of independent Ꮾ-valued random elements.Suppose that in the case (a) for some p, ∞ Proof.Using Lemma 11 we get where Now we show that 2 −kn To prove (66) it is enough to see that in the case (a) while in the case (b) By (61), we see that 2 −n 2 n i=1 Y 2 n ,i is a root of a kth degree polynomial in which the leading coefficient is unity and the remaining coefficients (cf.( 65) and ( 66)) converge almost surely to zero.Therefore, the conclusion follows from the well-known relations between the roots and coefficients of a polynomial.Similarly, as in the proof of Theorem 3 we can complete the proof of Theorem 12.
Remark 13.The presented results extend also to random elements theorems of [18] and [11].