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A compact condition of sequence of distribution functions
Nonlinear Analysis 63 (2005) e2347 – e2351 www.elsevier.com/locate/na
A compact condition of sequence of distribution functions Suyuan Cao Institute of Applied Maths, Guizhou University, Guiyang, Guizhou 550003, China
Abstract In this note, we establish a condition to give a sequence of distribution functions being compact. 䉷 2005 Elsevier Ltd. All rights reserved. Keywords: Compact condition; Distribution functions
1. Introduction In the probability theorem, Helly’s selection theorem is a famous one which is stated as follows. Theorem (Helly). For every sequence {Fn } of distribution functions, there exists a subsequence {Fnk } and a nondecreasing left-continuous function F such that lim Fnk = F (x) (k → ∞) at continuous points x of F . However, F in Helly’s theorem above may not be a distribution function. In this case, we say that the {Fn } is weakly compact. If F is a distribution function, then {Fn } is called compact. Naturally, we shall ask what condition is added to the family {Fn } such that it is compact. In this paper, we suggest such a condition. 0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.03.006
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S. Cao / Nonlinear Analysis 63 (2005) e2347 – e2351
Definition. Let {Fn } be a sequence of distribution functions. ∀ > 0, if there exists a absolute constant A > 0, A is only correlated to , such that |x|>A
dF (x) < ,
then we call the sequence {Fn } to be an equi-convergence. We shall prove the following main theorem in the paper. Main theorem. Suppose that the sequence {Fn } of distribution functions is equiconvergence, then the sequence is compact.
2. Some lemmas Before the proof of main theorem, we firstly introduce some lemmas. They are famous theorems in probability theory or analysis and thus their proofs are omitted. Lemma 1. Suppose that the sequence {Fn } of distribution functions is equi-convergence. {n (t)} is a sequence of characteristics functions corresponding to {Fn }, then {n (t)} is bounded uniformly and equi-continuous for t ∈ (−∞, +∞). Proof. Because {n (t)} is a characteristic function of {Fn }, n (0) = 1, and for every n, t ∈ (−∞, +∞), |(t)| 1. It shows that {n (t)} is bounded uniformly. On the other hand, for every n, ∀t1 , t2 ∈ (−∞, +∞), according to the definition of characteristic function, we have +∞ |n (t1 ) − n (t2 )| = (eixt 1 − eixt 2 ) dFn (x) −∞ +∞ |eix(t1 −t2 ) − 1| dFn (x) −∞ +A = |eix(t1 −t2 ) − 1| dFn (x) + |eix(t1 −t2 ) − 1| dFn (x). −A
|x|>A
Obviously for ∀ > 0, |eix(t1 −t2 ) − 1|2, there exists A > 0, not be correlated with any n, such that |eix(t1 −t2 ) | dFn (x)2 · = . 4 2 |x|>A Now fix A, consider the integral |x|>A
|eix(t1 −t2 ) − 1| dFn (x).
S. Cao / Nonlinear Analysis 63 (2005) e2347 – e2351
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In this case, due to |x|A (bounded) and eixt(t1 −t2 ) is continuous on variable t, there exists > 0, such that (|t1 − t2 | < ), |eix(t1 −t2 ) − 1| < 2 where is not correlated with any n and n (t). Therefore dFn (x) = |eix(t1 −t2 ) − 1| dFn (x) < 2 2 |x| A |x|>A i.e |n (t1 ) − n (t2 )| <
(|t1 − t2 | < ).
In other words, ∀ > 0, there exist > 0, for each n and n (t), when |t1 − t2 | < , we always have |n (t1 ) − n (t2 )| < . This is equi-continuity.
Definition. (t) is a continuous function with realvariable t ∈ (−∞, +∞), for each natural number m, t1 , t2 , . . . , tm are real numbers, 1 , 2 , . . . , m are complex numbers arbitrarily given. If we always have m m
(tj − tk )j k 0
k=1 k=1
hold, then we say the (t) is a positive definite, where k is the conjugate of k . Lemma 2 (Bochner). Suppose that (t) is a continuous function, and (0)=1. A sufficient and necessary condition that (t) is characteristic function is that (t) should be positive definite Lemma 3 (Levy–Cramer). Suppose that {Fn (x)} and {n (x)} are sequences of distribution functions and its (see [1]) characteristic functions, respectively. F (x) and (x) are the distribution function and its characteristic function. A sufficient and necessary condition lim Fn (x) = F (x) (n → ∞) at continuous points x of F (x), is lim n (x) = (x)
(n → ∞, x ∈ (−∞. + ∞))
and the convergence of {n (x)} is uniform in each definite close interval [T1 , T2 ] (see [1]). Lemma 4 (Ascoli–Arzela). Suppose that A = {f (x)} is a family of continuous functions defined on a compact metric space. If the family of functions is bounded uniformly and is equi-continuous, then we can select a sequence {fn } in the family, such that {fn (x)} is convergent uniformly. Now we turn to the proof of main theorem.
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S. Cao / Nonlinear Analysis 63 (2005) e2347 – e2351
3. The proof of main theorem First of all, we know from Lemma 1 that sequence B = {n (t)} satisfies the conditions of Lemma 4 on each close interval, therefore given the close interval I1 =[−1, 1]. We can select (1) (1) a subsequence B1 = {n (t)} in {n (t)}, such that B1 = {n (t)} converges uniformly on (2) I1 = [−1, 1]. For B1 and I2 = [−2, 2] ⊃ I1 , we can still select a subsequence B2 = {n (t)} (2) in B1 , such that B2 = {n (x)} converges uniformly on I2 = [−2, 2]. Repeating the process, we obtain two sequences B1 ⊃ B2 ⊃ · · · , I1 ⊂ I2 ⊂ · · · . (k)
For every natural number k, Bk = {n (t)} is a convergent sequence uniformly on close (m) interval [−k, k]=Ik . An application of the diagonal method gives a sequence {m (t)}=B0 , which converges uniformly on every Ik = [−k, k]. Now we turn to study the limit properties (m) of {m (t)}: (m) (1) The limit of {m (t)} defines a continuous function (t) in (−∞, +∞), and (0)=1. In fact, ∀t0 ∈ (−∞, +∞) there exists a natural number k, such that t0 ∈ [−k, k]. From (m) above, {m (t0 )} converges on [−k, k]. Denote the value of limit by (t0 ), and ∀t0 ∈ (−∞, +∞) arbitrarily, therefore we have lim (m) m (t) = (t),
t ∈ (−∞, +∞).
(m)
(m)
As m (t) is a characteristic function, m (0) = 1 for every natural number m, (0) = 1. (2) (t) is continuous in (−∞, +∞). In fact, ∀t0 ∈ (−∞, +∞), there exists a natural (m) (m) number k, such that t0 ∈ [−k, k]. As m (t) is continuous, and {m (t)} converges uniformly on [−k, k], therefore (t) should be a continuous function on [−k, k], it shows that (t) is continuous at t = t0 . (3) (t) is positive definite in (−∞, +∞). Given arbitrarily a natural number n, real numbers t1 , t2 , . . . , tn , complex numbers 1 , 2 , . . . , n , d = max |ti − tj |; i,j
K=
n n
|i j |.
i=1 j =1
Obviously, d is a real number, there exists a natural number k, such that d ∈ [−k, k]. Note (m) that {m (t)} is convergent to (t) uniformly on [−k, k]. ∀ > 0, ∃M > 0, when m > M, we have |(m) m (ti − tj ) − (ti − tj )| <
K
S. Cao / Nonlinear Analysis 63 (2005) e2347 – e2351
and
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n n n n (m) (ti − tj )i j m (ti − tj )i j − i=1 j =1 i=1 j =1
n n i=1 j =1
|(m) m (ti − tj ) − (ti − tj )| · |i j | < .
It means that as m → ∞ n n i=1 j =1
(m) m (ti − tj )i j →
n n
(ti − tj )i j .
i=1 j =1
(m)
Because m (t) is a characteristic function, therefore, from Lemma 2, we have n n i=1 j =1
(m) m (ti − tj )i j 0,
where “ 0” hints to us that the sum is a real number, and not a complex number. According to the completeness of the system of real numbers, the above sum should have the value of limit n n
(ti − tj )i j 0.
i=1 j =1
It shows that (t) is a characteristic function. (4) Finally, we give a definite and close interval [T1 , T2 ] arbitrarily, there exists a natural (m) number k, such that [T1 , T2 ] ⊂ [−k, k], sequence {m (t)} converges to (t) uniformly on [−k, k] and [T1 , T2 ]. (m) (m) Let F (x) be a distribution function of (t), Fm be a distribution function of m (t), by Lemma 3, we have lim Fm(m) = F (x) (as m → ∞) for every continuous point x of F (x). Till now, the main theorem has been proved.
References [1] P. Billiugsley, Probability and Measure [M], Wiley, New York, 1986.
A compact condition of sequence of distribution functions Suyuan Cao Institute of Applied Maths, Guizhou University, Guiyang, Guizhou 550003, China
Abstract In this note, we establish a condition to give a sequence of distribution functions being compact. 䉷 2005 Elsevier Ltd. All rights reserved. Keywords: Compact condition; Distribution functions
1. Introduction In the probability theorem, Helly’s selection theorem is a famous one which is stated as follows. Theorem (Helly). For every sequence {Fn } of distribution functions, there exists a subsequence {Fnk } and a nondecreasing left-continuous function F such that lim Fnk = F (x) (k → ∞) at continuous points x of F . However, F in Helly’s theorem above may not be a distribution function. In this case, we say that the {Fn } is weakly compact. If F is a distribution function, then {Fn } is called compact. Naturally, we shall ask what condition is added to the family {Fn } such that it is compact. In this paper, we suggest such a condition. 0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.03.006
e2348
S. Cao / Nonlinear Analysis 63 (2005) e2347 – e2351
Definition. Let {Fn } be a sequence of distribution functions. ∀ > 0, if there exists a absolute constant A > 0, A is only correlated to , such that |x|>A
dF (x) < ,
then we call the sequence {Fn } to be an equi-convergence. We shall prove the following main theorem in the paper. Main theorem. Suppose that the sequence {Fn } of distribution functions is equiconvergence, then the sequence is compact.
2. Some lemmas Before the proof of main theorem, we firstly introduce some lemmas. They are famous theorems in probability theory or analysis and thus their proofs are omitted. Lemma 1. Suppose that the sequence {Fn } of distribution functions is equi-convergence. {n (t)} is a sequence of characteristics functions corresponding to {Fn }, then {n (t)} is bounded uniformly and equi-continuous for t ∈ (−∞, +∞). Proof. Because {n (t)} is a characteristic function of {Fn }, n (0) = 1, and for every n, t ∈ (−∞, +∞), |(t)| 1. It shows that {n (t)} is bounded uniformly. On the other hand, for every n, ∀t1 , t2 ∈ (−∞, +∞), according to the definition of characteristic function, we have +∞ |n (t1 ) − n (t2 )| = (eixt 1 − eixt 2 ) dFn (x) −∞ +∞ |eix(t1 −t2 ) − 1| dFn (x) −∞ +A = |eix(t1 −t2 ) − 1| dFn (x) + |eix(t1 −t2 ) − 1| dFn (x). −A
|x|>A
Obviously for ∀ > 0, |eix(t1 −t2 ) − 1|2, there exists A > 0, not be correlated with any n, such that |eix(t1 −t2 ) | dFn (x)2 · = . 4 2 |x|>A Now fix A, consider the integral |x|>A
|eix(t1 −t2 ) − 1| dFn (x).
S. Cao / Nonlinear Analysis 63 (2005) e2347 – e2351
e2349
In this case, due to |x|A (bounded) and eixt(t1 −t2 ) is continuous on variable t, there exists > 0, such that (|t1 − t2 | < ), |eix(t1 −t2 ) − 1| < 2 where is not correlated with any n and n (t). Therefore dFn (x) = |eix(t1 −t2 ) − 1| dFn (x) < 2 2 |x| A |x|>A i.e |n (t1 ) − n (t2 )| <
(|t1 − t2 | < ).
In other words, ∀ > 0, there exist > 0, for each n and n (t), when |t1 − t2 | < , we always have |n (t1 ) − n (t2 )| < . This is equi-continuity.
Definition. (t) is a continuous function with realvariable t ∈ (−∞, +∞), for each natural number m, t1 , t2 , . . . , tm are real numbers, 1 , 2 , . . . , m are complex numbers arbitrarily given. If we always have m m
(tj − tk )j k 0
k=1 k=1
hold, then we say the (t) is a positive definite, where k is the conjugate of k . Lemma 2 (Bochner). Suppose that (t) is a continuous function, and (0)=1. A sufficient and necessary condition that (t) is characteristic function is that (t) should be positive definite Lemma 3 (Levy–Cramer). Suppose that {Fn (x)} and {n (x)} are sequences of distribution functions and its (see [1]) characteristic functions, respectively. F (x) and (x) are the distribution function and its characteristic function. A sufficient and necessary condition lim Fn (x) = F (x) (n → ∞) at continuous points x of F (x), is lim n (x) = (x)
(n → ∞, x ∈ (−∞. + ∞))
and the convergence of {n (x)} is uniform in each definite close interval [T1 , T2 ] (see [1]). Lemma 4 (Ascoli–Arzela). Suppose that A = {f (x)} is a family of continuous functions defined on a compact metric space. If the family of functions is bounded uniformly and is equi-continuous, then we can select a sequence {fn } in the family, such that {fn (x)} is convergent uniformly. Now we turn to the proof of main theorem.
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S. Cao / Nonlinear Analysis 63 (2005) e2347 – e2351
3. The proof of main theorem First of all, we know from Lemma 1 that sequence B = {n (t)} satisfies the conditions of Lemma 4 on each close interval, therefore given the close interval I1 =[−1, 1]. We can select (1) (1) a subsequence B1 = {n (t)} in {n (t)}, such that B1 = {n (t)} converges uniformly on (2) I1 = [−1, 1]. For B1 and I2 = [−2, 2] ⊃ I1 , we can still select a subsequence B2 = {n (t)} (2) in B1 , such that B2 = {n (x)} converges uniformly on I2 = [−2, 2]. Repeating the process, we obtain two sequences B1 ⊃ B2 ⊃ · · · , I1 ⊂ I2 ⊂ · · · . (k)
For every natural number k, Bk = {n (t)} is a convergent sequence uniformly on close (m) interval [−k, k]=Ik . An application of the diagonal method gives a sequence {m (t)}=B0 , which converges uniformly on every Ik = [−k, k]. Now we turn to study the limit properties (m) of {m (t)}: (m) (1) The limit of {m (t)} defines a continuous function (t) in (−∞, +∞), and (0)=1. In fact, ∀t0 ∈ (−∞, +∞) there exists a natural number k, such that t0 ∈ [−k, k]. From (m) above, {m (t0 )} converges on [−k, k]. Denote the value of limit by (t0 ), and ∀t0 ∈ (−∞, +∞) arbitrarily, therefore we have lim (m) m (t) = (t),
t ∈ (−∞, +∞).
(m)
(m)
As m (t) is a characteristic function, m (0) = 1 for every natural number m, (0) = 1. (2) (t) is continuous in (−∞, +∞). In fact, ∀t0 ∈ (−∞, +∞), there exists a natural (m) (m) number k, such that t0 ∈ [−k, k]. As m (t) is continuous, and {m (t)} converges uniformly on [−k, k], therefore (t) should be a continuous function on [−k, k], it shows that (t) is continuous at t = t0 . (3) (t) is positive definite in (−∞, +∞). Given arbitrarily a natural number n, real numbers t1 , t2 , . . . , tn , complex numbers 1 , 2 , . . . , n , d = max |ti − tj |; i,j
K=
n n
|i j |.
i=1 j =1
Obviously, d is a real number, there exists a natural number k, such that d ∈ [−k, k]. Note (m) that {m (t)} is convergent to (t) uniformly on [−k, k]. ∀ > 0, ∃M > 0, when m > M, we have |(m) m (ti − tj ) − (ti − tj )| <
K
S. Cao / Nonlinear Analysis 63 (2005) e2347 – e2351
and
e2351
n n n n (m) (ti − tj )i j m (ti − tj )i j − i=1 j =1 i=1 j =1
n n i=1 j =1
|(m) m (ti − tj ) − (ti − tj )| · |i j | < .
It means that as m → ∞ n n i=1 j =1
(m) m (ti − tj )i j →
n n
(ti − tj )i j .
i=1 j =1
(m)
Because m (t) is a characteristic function, therefore, from Lemma 2, we have n n i=1 j =1
(m) m (ti − tj )i j 0,
where “ 0” hints to us that the sum is a real number, and not a complex number. According to the completeness of the system of real numbers, the above sum should have the value of limit n n
(ti − tj )i j 0.
i=1 j =1
It shows that (t) is a characteristic function. (4) Finally, we give a definite and close interval [T1 , T2 ] arbitrarily, there exists a natural (m) number k, such that [T1 , T2 ] ⊂ [−k, k], sequence {m (t)} converges to (t) uniformly on [−k, k] and [T1 , T2 ]. (m) (m) Let F (x) be a distribution function of (t), Fm be a distribution function of m (t), by Lemma 3, we have lim Fm(m) = F (x) (as m → ∞) for every continuous point x of F (x). Till now, the main theorem has been proved.
References [1] P. Billiugsley, Probability and Measure [M], Wiley, New York, 1986.