Examples

Example 1 ([24]). Put x_n=p_n, the nth prime and denote X_n=

2 p_n, 3

p_n, . . . ,p_n−1 p_n ,pn

p_n

.

The sequence of blocks X_n is u.d. and therefore the ratio sequence p_m/p_n, m= 1,2, . . . , n,n= 1,2, . . . is u.d. in [0,1].This generalizes a result of A. S c h i n -z e l (cf. W. S i e r p i ´n s k i (1964, p. 155)). Note that from u.d. ofXn applying for theL² discrepancy ofXn we get the following interesting limit

n→∞lim 1 n²p_n

n i,j=1

|pi−pj|= 1 3.

Example 2 ([24, Ex. 11.1]). Let γ, δ, and a be given real numbers satisfying 1≤γ < δ≤a. Letx_nbe an increasing sequence of all integer points lying in the intervals

(γ, δ),(γa, δa), . . . ,(γa^k, δa^k), . . . ThenG(X_n) =

g_t(x);t∈[0,1]

, whereg_t(x) has constant values gt(x) = 1

aⁱ(1 +t(a−1)) for x∈ (δ, aγ)

aⁱ⁺¹(tδ+ (1−t)γ), i= 0,1,2, . . . and on the component intervals it has a constant derivative

g_t(x) = tδ+ (1−t)γ

(δ−γ)(_a−1¹ +t) for x∈ (γ, δ)

aⁱ⁺¹(tδ+ (1−t)γ), i= 0,1,2, . . . and x∈

γ

tδ+ (1−t)γ,1

, where

F(Xnk, x)→gt(x) fornk for which xnk =

a^kγ+ta^k(δ−γ)

. (53) Here we write (xz, yz) = (x, y)z and (x/z, y/z) = (x, y)/z. Then the set G(X_n) has the following properties:

1⁰. Everyg ∈G(Xn) is continuous.

2⁰. Every g ∈ G(X_n) has infinitely many intervals with constant values, i.e., with g(x) = 0, and in the infinitely many complement intervals it has a constant derivative g(x) = c, where ¹

d ≤ c ≤ _d¹ and for lower d and upper dasymptotic density ofx_n we have

d= (δ−γ)

γ(a−1), d= (δ−γ)a δ(a−1). 3⁰. The graph of everyg∈G(X_n) lies in the intervals

1 a,1

× 1

a,1

∪ 1

a²,1 a

× 1

a²,1 a

∪. . . Moreover, the graph g in ₁

a^k,_ak−1¹

×₁

a^k,_ak−1¹

is similar to the graph of g in ₁

a^k+1,_a¹k

× ₁

a^k+1,_a¹k

with coefficient ¹_a. Using the parametric expression, it can be written for all x∈ ₁

aⁱ⁺¹,_a¹i

that g_t(x) = ^gt(a_ai^ix), i= 0,1,2, . . .

4⁰. G(Xn) is connected and the upper distribution function g(x) = g₀(x) ∈ G(X_n) and the lower distribution function g(x) ∈/ G(X_n). The graph ofg(x) on₁

a,1

×₁

a,1

coincides with the graph of y(x) =

1 +1

d 1

x −1 ₋₁

on_γ

δ,1

, further, on₁

a,^γ_δ

we haveg(x) =_a¹. 5⁰. G(X_n) = ^g_g^0(xβ)

0(β);β∈₁

a,_aγ^δ ! . For the proofs of 1⁰.−5⁰. we only note:

Assume that x_n ∈ a^k(γ, δ), i, i+ 1, i+ 2,· · · ∈a^j(γ, δ) for somej < k, and let F(Xn, x) →g(x) for some sequence of n. Then g(x) has a constant derivative in the intervals containing _xⁱ_n,ⁱ⁺¹_xn,ⁱ⁺²_xn, . . ., since

1 n i+1

xn −_xⁱ_n = x_n n , and thus ^x_nⁿ must be convergent tog(x), so ¹

d ≤g(x)≤ ¹_d. For xn=

ta^kδ+ (1−t)a^kγ we can find

g(x) = lim

n→∞

x_n n = lim

k→∞

a^k(tδ+ (1−t)γ)

#_k−1

j=0a^j(δ−γ) +a^k(tδ+ (1−t)γ)−a^kγ

= tδ+ (1−t)γ (δ−γ) ₁

a−1+t.

Using Theorem 18 and [3, Ex. 3] we shall add the following properties more-over:

6⁰. By definition (5) of the local asymptotic densityd_g and by (53) forg(x) = g_t(x) we have

d_gt = lim

k→∞

nk

x_nk = lim

k→∞

#_k−1

i=0 aⁱ(δ−γ) +ta^k(δ−γ) a^kγ+ta^k(δ−γ)

= (δ−γ)(1 +t(a−1))

(a−1)(γ+t(δ−γ)) (54)

and for t= 0 we havedg₀ =dand for t= 1 we havedg₁ =dand we see g_t(x) = 1

dgt

(55) forxwith the constant derivative ofg_t(x).

7⁰. For the function h_1,g(x) defined in (26), putting g(x) =g_t(x), we have:

d

d_gt = γ+t(δ−γ)

γ(1 +t(a−1)),1−dgt

1−d = γ

γ+t(δ−γ), d

d_gt

1−d_gt

1−d = 1

1 +t(a−1).

Then

h_1,gt(x) =

xγ(1+t(a−1))^{γ+t(δ−γ)} forx∈

0,_{γ+t(δ−γ)}^γ , x_d¹

gt + 1− _d¹_gt, forx∈ _γ

γ+t(δ−γ),1

, (56)

see the following figure.

,, ,, ,, ,, ,, ,, ,, ,

g_t(x)

h_1,gt(x)

(1,1)

(¹_a,¹_a)

(_a¹₂,_a¹₂)

Figure:gt(x) andh_1,gt(x).

8⁰. In the proof of the upper bound (29) we have proved that 1−₁

0 h_1,g(x) dxis maximal fordg = min(√

d, d). Lett₀∈[0,1] be such thatdgt0 = min(√ d, d) andt₀ can be computed by inverse formula to (54)

t= d_gt(a−1)γ−(δ−γ)

(δ−γ)(a−1)(1−dgt). (57) 9⁰. LetP(t) be the area in₁

a,1

×₁

a,1

bounded by the graph ofgt(x). Then 1

0

g_t(x) dx=P(t) 1 1−_a¹2

+ 1

a+ 1

= 1 2+ 1

2. 1

(a+ 1). (γa−δ)

(1 +t(a−1))(γ+t(δ−γ)) + 1

2. t(δ−γa)

(1 +t(a−1))(γ+t(δ−γ)) (58) and sinceg₀(x) =g(x) we have that the max_t∈[0,1]1

0 gt(x) dx is attained att= 0. Using derivative ofP(t) it can be see that the min_t∈[0,1]1

0 gt(x) dx

is attained at t= 1. It also follows from the fact that for x_n+1 = xn+ 1 we have

1 n+ 1

n+1

i=1

xi

x_n+1 − 1 n

n i=1

xi

x_n

= 1

n+ 1− 1

x_n+ 1+ 1 n+ 1. 1

1 +_x¹_n 1

n n i=1

x_i x_n

>0 becausec₁(x)∈/G(X_n) and thus lim sup_n→∞n¹#n

i=1 xi

xn <1. Now, denot-ing the indexn_k forx_nk = [a^kδ], the lim sup of ¹_n#_n

i=1 xi

xn is attained over n = nk, k = 0,1,2, . . . and for such nk we have F(Xnk, x) → g₁(x) for x∈[0,1].

10⁰. Thus we have lim inf

n→∞

1 n

n i=1

x_i x_n = 1−

1 0

g₀(x) dx= 1 2− 1

2. 1 (a+ 1)

γa−δ γ

, (59)

lim sup

n→∞

1 n

n i=1

x_i x_n = 1−

1 0

g₁(x) dx= 1 2+ 1

2. 1 (a+ 1)

γa−δ δ

. (60)

The upper bound (29) coincides with the maximal value of 1−₁

0 h_1,g(x) dx attained for dg = min(√

d, d). Since 1−₁

0 g₁(x) dx is maximal for all 1−₁

0 gt(x) dx, t∈[0,1] and 1−₁

0 g₁(x) dx≤1−₁

0 h₁,g₁(x) dxthen the upper bound (60) satisfies (29).

11⁰. Using explicit formulas

d= (δ−γ)

γ(a−1), d= (δ−γ)a

δ(a−1) (61)

for asymptotic densities we see again that (59) and (60) satisfy (28) and (29), respectively, in Theorem 19.

Example 3 ( [9, Ex. 2]). Let xn and yn, n = 1,2, . . ., be two strictly in-creasing sequences of positive integers such that for the related block sequences X_n=_x1

xn, . . . ,^x_xⁿ

n

andY_n=_y1

yn, . . . ,^y_yⁿ

n

, we have singleton for bothG(X_n) = g₁(x)

and G(Y_n) = g₂(x)

. Furthermore, let n_k, k = 1,2, . . ., be an in-creasing sequence of positive integers such thatN_k=#k

i=1n_i satisfies _N^nk

k →1.

Denote by z_n the following increasing sequence of positive integers composed by blocks (here we use the notationa(b, c, d, . . .) = (ab, ac, ad, . . .))

(x₁, . . . , xn₁), xn₁(y₁, . . . , yn₂), xn₁yn₂(x₁, . . . , xn₃), xn₁yn₂xn₃(y₁, . . . , yn₄), . . .

Then the sequence of blocks Zn=_z1

zn, . . . ,^z_zⁿ_n

has the set of d.f.s G(Zn) =

g₁(x), g₂(x), c₀(x)

∪

g₁(xyn);n= 1,2, . . .

∪

g₂(xx_n);n= 1,2, . . .

∪

$ 1

1 +αc₀(x) + α

1 +αg₁(x);α∈[0,∞)

%

∪

$ 1

1 +αc₀(x) + α

1 +αg₂(x);α∈[0,∞)

% , whereg₁(xy_n) = 1 if xy_n≥1, similarly forg₂(xx_n).

P r o o f. For everyn= 1,2, . . . there exists an integerk such that N_k−1 < n≤N_k

(hereN₀= 0). Put n=n−N_k−1. For everyn we have z_n=

x_n1y_n2. . . x_nk−1y_n if kis even, x_n1y_n2. . . y_nk−1x_n if kis odd.

Firstly we assume thatk is even. ThenZ_nhas the form Z_n=

. . . ,xn₁yn₂. . . yn_k−2(x₁, . . . , xn_k−1)

x_n1y_n2. . . x_nk−1y_n ,xn₁yn₂. . . xn_k−1(y₁, . . . , yn) x_n1y_n2. . . x_nk−1y_n

=

. . . , 1 x_nk−1y_n

y₁

y_nk−2, . . . ,y_nk−2 y_nk−2

, 1

y_n

x₁

x_nk−1, . . . ,x_nk−1 x_nk−1

,

y₁

y_n, . . . ,y_n

y_n

and thus forx > _x ¹

nk−1 we have

F(Z_n, x) =Nk−2+nk−1F(Xn_k−1, xyn) +nF(Yn, x) N_k−1+n

= N_k−2 N_k−1+n +

nk−1

N_k−1

1 +_Nⁿ

k−1

F(Xn_k−1, xyn) + 1

1 +^Nk−1_n F(Yn, x).

Ifn→ ∞, then the first term tends to zero. IfF(Z_n, x)→g(x) for some sequence ofn, we can select a subsequence ofn’s such that _Nⁿ

k−1 →αfor someα∈[0,∞), or _Nⁿ

k−1 → ∞. For suchn we distinguish the following cases:

(a) If n= constant, then

nk−1

N_k−1

1 +_Nⁿ

k−1

F(X_nk−1, xy_n)→g₁(xy_n)( hereg₁(xy_n) = 1 forxy_n >1) 1

1 +^Nk−1_n F(Y_n, x)→0

and thusF(Z_n, x)→g₁(xy_n).

(b) If n→ ∞, then F(Xn_k−1, xyn)→1; precisely F(Xn_k−1, xyn)→c₀(x).

(b1) If _Nⁿ

k−1 →0, thenF(Zn, x)→c₀(x).

(b2) If _Nⁿ

k−1 →α∈(0,∞), then F(Zn, x)→ _1+α¹ c₀(x) +_1+α^α g₂(x).

(b3) If _Nⁿ

k−1 → ∞, then F(Z_n, x)→0 +g₂(x).

Fork-odd we use a similar computation.

Now, identify xn = yn and select xn such that g₁(x) = x (e.g., xn = n or xn =pn, thenth prime) and putnk = 2^k2 fork = 1,2, . . .. Then the set of all d.f.s

G(Zn) =

g₁(x), c₀(x)

∪

g₁(xxn);n= 1,2, . . .

∪

$ 1

1 +αc₀(x) + α

1 +αg₁(x);α∈[0,∞)

%

is disconnected, as it can be seen in the figure on the page 174.

Example 4. Letx_n,n= 1,2, . . ., be an increasing sequence of positive integers for which there exists a sequencen_k, k= 1,2, . . ., of positive integers such that (ask→ ∞)

(i) ⁿ_n^k−1

k →0, (ii) _x^nk

nk →0, (iii) ^xnk−1_x

nk →0, and

(iv) xnk−i =xnk−i fori= 0,1, . . . , nk−n_k−1−1.

Then the sequence of blocks X_n=

x₁ x_n, x₂

x_n, . . . ,xn

x_n

has

G(Xn) =

hα(x);α∈[0,1]

.

P r o o f. For givenθ∈[0,1] andn=nk−

θ(nk−n_k−1)

and by (iv) we have xn=xnk−

θ(nk−nk−1) .

Fori≤nwe distinguish two cases:x_i ∈(x_nk−1, x_n] andx_i ≤x_nk−1. (I) Forxi ∈(xn_k−1, xn] we have

xi

x_n ∈

&

xnk−(nk−n_k−1) + 1 x_nk−[θ(n_k−n_k−1)] ,1

'

→[1,1]

asn→ ∞and for anyθ∈[0,1]. The number of suchx_i’s is (nk−n_k−1)−

θ(nk−n_k−1)

= (1−θ)(nk−n_k−1) +O(1).

(II) For xi≤xn_k−1 we have xi

x_n ∈

&

0, xn_k−1

x_nk−[θ(n_k−n_k−1)]

'

→[0,0].

We thus get, for anyx∈(0,1) and any sufficiently largen, F(Xn, x) = n_k−1

n = n_k−1

n_k−1+ (1−θ)(n_k−n_k−1) +O(1). This gives:

(a) If θ≤ε₀<1, for some fixedε₀, then F(X_n, x)→c₁(x).

(b) If θ= 1, then

F(X_n, x)→c₀(x).

(c) For anyα∈(0,1) there exists a sequence θk→1, ask→ ∞, such that n_k−1

n_k−1+ (1−θ_k)(n_k−n_k−1) →α, and in this case

F(X_n, x)→h_α(x).

Note that the sequences nk= 2^k2 and xnk = 2^(k+1)2 satisfy the assumptions (i), (ii), (iii) and (iv). We also see thatG(X_n) is connected but

F(X_nk+1, x)→c₀(x), and F(Xnk, x)→c₁(x),

a.e. on [0,1] and thusρ(t_nk+1, t_nk)→1. Using the permutationπ:N→N 1,2, . . . , n₁, n₂, n₂−1, n₂−2, . . . , n₁+ 1, n₂+ 1, n₂+ 2, . . . n₃, n₄, n₄−1, n₄−2, . . . , n₃+ 1, n₄+ 1, n₄+ 2, . . . , n₅, n₆, n₆−1, n₆−2, . . . , n₅+ 1, . . . we haveρ(t_π(n+1), t_π(n))→0 asn→ ∞, because the “neighbouring” d.f. oft_π(n) satisfies the scheme

c₁(x), c₁(x), . . . , c₀(x), c₀(x), . . . , c₁(x), c₁(x), . . . , c₀(x), c₀(x), . . . , c₁(x), c₁(x), . . . , c₀(x), c₀(x), . . .

Example 5. In [8] is proved that _x^xn

n+1 → 1 does not imply that G(Xn) is a singleton. This is a negative answer to the Problem 1.9.2 in [20].

Let a_k, n_k, k = 1,2, . . ., and x_n, n = 1,2, . . . be three increasing integer sequences andh₁< h₂ be two positive integers. Assume that

(i) _n^nk

k+1 →0 fork→ ∞; (ii) _n^ak

k+1 →0 fork→ ∞; (iii) for oddk we have

a^h_k² ≤x_nk = (a_k−1+n_k−n_k−1)^h1 ≤(a_k+ 1)^h2 and x_i= (a_k+i−n_k)^h2 forn_k< i≤n_k+1;

(iv) for even k we have

a^h_k¹ ≤xnk = (a_k−1+nk−n_k−1)^h2 ≤(ak+ 1)^h1 and x_i= (a_k+i−n_k)^h1 forn_k< i≤n_k+1.

Then _x^xn

n+1 →1 and the setG(X_n) of all distribution functions of the sequence of blocks XnisG(Xn) =G₁∪G₂∪G₃∪G₄, where

G₁ =

x^h12.t;t∈[0,1]

, G₂ =

x^h12(1−t) +t;t∈[0,1]

, G₃ =

max(0, x^h11 −(1−x^h11)u);u∈[0,∞) and G₄ =

min(1, x^h11.v);v∈[1,∞) .

In [24, Th. 5.2, p. 762 ] = Theorem 15, it is proved that the condition _x^xn

n+1 →1 implies the connectivity ofG(Xn)

P r o o f. 1.Firstly we prove that for anyh₁ < h₂ the sequencesak, nk, xn satis-fying (i)–(iv) exist:

Fori= 1, . . . , n₁ we putx_i=i^h1 and then we find a₁ such that a^h₁² ≤x_n1 ≤ (a₁+ 1)^h2. If we have selected, for an odd step k, alla_i, i= 1,2, . . . , k−1, x_i, i = 1,2, . . . , n_k, then we find a_k such that a^h_k² ≤ x_nk < (a_k+ 1)^h2, and then we putx_i= (a_k+i−n_k)^h2 forn_k < i≤n_k+1, where we choosen_k+1sufficiently large to satisfy the limits (i) and (ii). For an even step k we proceed similarly replacingh₂ byh₁.

2.In contrary to the independence of a_k and n_k+1 we have a_k

n

h1 h2

k

→1 for oddk → ∞, a_k n

h2 h1

k

→1 for evenk→ ∞. (62) This follows from (iii) and (iv), directly, e.g., from (iii) we have

a^h_k² n^h_k¹ <

a_k−1 nk

+ 1−n_k−1 nk

h₁

< (a_k+ 1)^h2 n^h_k¹ . As an application of (62) we have

ak

n_k →0 for odd k→ ∞, ak

n_k → ∞ for evenk→ ∞. (63) 3. Now we prove _x^xi

i+1 →1 as i → ∞. Let i ∈ (nk, n_k+1) and let, e.g., k be odd. Then by (iii)

x_i x_i+1 =

1− 1

a_k+i+ 1−n_k h₂

>

1− 1

a_k h₂

and for i=n_k again x_nk

xnk+1 > a^h_k² (ak+ 1)^h2 >

1− 1

ak

h₂

which implies the limit 1 as oddk→ ∞. Similarly for evenk.

4.LetN ∈[nk, n_k+1] be an integer sequence (we shall omit the index inNk) fork→ ∞. Forx∈(0,1) we have

F(XN, x) = #{1≤i≤n_k−1;_x^x_Nⁱ < x} N

+ #{n_k−1< i≤n_k;_x^xi

N < x}

N + #{n_k< i≤N;_x^xi

N < x} N

=o(1) + A N + B

N. (64)

To compute _N^A for oddk we use xi

x_N = (a_k−1+i−n_k−1)^h1

(a_k+N−n_k)^h2 < x⇐⇒i−n_k−1 < x^h11(a_k+N−n_k)

h2

h1 −a_k−1

and we have A

N =min(n_k−n_k−1,max(0,[x^h11(a_k+N−n_k)

h2

h1 −a_k−1]))

N . (65)

Similarly, for even k A

N =min(nk−nk−1,max(0,[x^h12(ak+N−nk)

h1

h2 −ak−1]))

N . (66)

For _N^B and odd k we use xi

xN

=

ak+i−nk

ak+N−nk

h2

< x⇐⇒i−n_k< x^h12(a_k+N−n_k)−a_k which gives

B

N = min(N−n_k,max(0,[x^h12(a_k+N−n_k)−a_k]))

N . (67)

Similarly, for even k we have B

N = min(N−nk,max(0,[x^h11(ak+N−nk)−ak]))

N . (68)

In the following we will distinguish three cases nk

N →t >0, nk

N →0 and N

n_k+1 →0, and N

n_k+1 →t >0.

5.Now, let ⁿ_N^k →t >0 as k→ ∞.

a) Assume that k is odd and compute the limit of _N^A by (65). We have

nk−nk−1

N →tand ift <1 we see x^h11

ak

N

h1 h2

+ N

N

h1 h2

1− nk

N ^hh21

−ak−1

N → ∞

since ^N

N

h1 h2

forh₁< h₂ is unbounded and by (62) a_k

N

h1 h2

= a_k n

h1 h2

k

n_k N

^h_h¹

2 →t

h1 h2

is bounded. Thus, for 0< t <1, we have A

N →t for odd k→ ∞. (69)

a1) Let for the momentt= 1. We have ^ak

n

h1 h2 k

→1 and

x^h11

a_k N

h1 h2

+ N−n_k N

h1 h2

^h_h²

1

− a_k−1

N →x^h11(1 +u)

h2 h1

assuming the limit ^N−nk

N

h1 h2

→u, whereu∈[0,∞) can be arbitrary. Putv= (1+u)

h2 h1. Thus fort= 1 and corresponding v∈[1,∞) we have

A

N →min(1, x^h11v) for odd k→ ∞. (70) If ^N−nk

N

h1 h2

→ ∞, then

A

N →1 for odd k→ ∞. (71)

b) Now, again 0< t≤1. For even k in (66) we have x^h12

a_k N

h2 h1

+ N

N

h2 h1

1− n_k

N ^hh12

− a_k−1

N →x^h12.t since by (62)

a_k N

h2 h1

= a_k n

h2 h1

k

n_k N

^h_h²

1→t

h2 h1. Thus

A

N →x^h12.t for even k→ ∞. (72)

c) For the limit ^B_N as odd k → ∞we compute (67) by using ^N−n_N ^k →1−t and

x^h12 a_k

N + 1− n_k N

−a_k

N →x^h12(1−t) since by (63) we have ^a_N^k = _n^ak_kⁿ_N^k →0. Thus

B

N →x^h12(1−t) for odd k→ ∞. (73) d) Again by (63), for evenkwe have ^a_N^k = ^a_n^k_kⁿ_N^k → ∞, then (assumingx <1)

x^h11 a_k

N + 1−n_k N

− a_k

N → −∞. Thus

B

N →0 for even k→ ∞. (74)

e) Summing up (69), (72), (73) and (74) we find, for everyx∈(0,1), F(X_N, x)→

x^h12(1−t) +t for oddk→ ∞,

x^h12.t for evenk→ ∞ (75)

for ⁿ_N^k →t, 0< t <1. For ⁿ_N^k →t= 1, ^N−nk

N

h1 h2

→uand v= (1 +u)

h2

h1 we have applying (70)

F(XN, x)→min(1, x^h11.v) for odd k→ ∞, (76) and for ^N−nk

N

h1 h2

→ ∞we have

F(XN, x)→c₀(x) for odd k→ ∞, (77) wherec₀(x) = 1 for x∈(0,1).

6. In the case ⁿ_N^k →0 and _n^N

k+1 →0 we have _N^A =o(1) and then it suffices to compute the limit _N^B by (67) or (68).

a) Assume that odd k → ∞. Since ^N−_N^nk → 1 and by (63) we have ^a_N^k =

ak

nk

nk

N →0 and thus x^h12

a_k

N + 1−n_k N

−a_k

N →x^h12. (78)

b) Assume that evenk→ ∞. In this case (by (62) and (ii)) we have ak

N = ak

n

h2 h1

k

n

h2 h1

k

N , ak

n

h2 h1

k

→1, ak

n_k+1 →0, then n

h2 h1

k

n_k+1 →0.

Thus, for anyu∈[0,∞) we can find a subsequence ofN such that n

h2 h1

k

N →u. (79)

Then

x^h11 a_k

N + 1−n_k N

− a_k

N →x^h11 −(1−x^h11)u. (80) c) Summing up (78) and (80) we find for everyx∈(0,1)

F(XN, x)→

x^h12 for oddk → ∞,

max(0, x^h11 −(1−x^h11)u) for evenk→ ∞ (81) for ⁿ_N^k →0, _n^N

k+1 →0 and foru∈(0,∞) satisfying (79) ifkis even. If ⁿ

h2 h1 k

N → ∞

then

F(X_N, x)→c₁(x) for even k→ ∞, (82) wherec₁(x) = 0 for x∈(0,1).

7.Finally, let _n^N

k+1 →t >0. Then ^a_N^k →0, because (ii) _n^ak

k+1 →0. Computing the limit ^B_N by (67) or (68) we find

F(NN, x)→

x^h12 for oddk→ ∞,

x^h11 for evenk→ ∞. (83) 8.Now, assume that F(XN, x)→g(x) for some sequence of N ∈[nk, nk+1], i.e., g(x)∈G(Xn). Then we can find subsequence of N (denoting again asN) such that ⁿ_N^k, ^N−nk

N

h1 h2

, _n^N

k+1, and ⁿ

h2 h1 k

N converge. Consequently g(x) is contained in the collection of (75), (76), (77), (81), (82) and (83).

Thus the proof is finished.

L. M i ˇs´ık (2004, personal communication) found the following sequence x_n for which c₁(x)∈ G(X_n) andc₀(x) ∈/ G(X_n) and consequently the implication Q.7 in [9] does not hold.

Example 6. Letx_n,n= 1,2, . . ., be an increasing sequence of positive integers which satisfies the following conditions

(i) ifn_k= (k+ 1)(k−1)!2^k(k−1)2 fork= 1,2, . . ., thenx_nk = (k+ 1)n_k, (ii) ifn_k=k(k−2)!2^k(k−1)2 then x_nk =k²n_k,

(iii) ifn= 2ⁱn_k−1+j, 0≤j <2ⁱn_k−1 and 0≤i < k−1 fork= 1,2, . . ., then x_n=x_nk−1(i+ 1)2ⁱ+ (i+ 3)kj (i.e., n∈[n_k−1, n_k]),

(iv) ifn∈[n_k, n_k] fork= 1,2, . . ., then x_n=x_nk+n−n_k. Then for the sequence of blocks

Xn= x₁

x_n, x₂

x_n, . . . ,x_n x_n

we havec₁(x)∈G(Xn) but c₀(x)∈/G(Xn).⁵ P r o o f. We start with the following figure:

Here for n running through [2ⁱnk−1, 2ⁱ⁺¹nk−1], the xn is equi-distributed in [x₂ⁱ_nk−1, x₂i+1n_k−1] with difference Δi, wherei= 0,1, . . . , k−2.

5This and the Theorem 13 imply thatG(Xn)⊂ {cα(x);α∈[0,1]}.

1⁰.Using the definition of x_n we can see that ^x_x^nk

nk →1 and ⁿ_n^k

k →0 and thus we havec₁(x)∈G(X_n).

2⁰. On the contrary, assume that there exists increasing sequence m_l < m_l, l= 1,2, . . ., such thatm_l∈[n_k−1, n_k],k=k(l), (i) ^x_x^ml

ml →0 and (ii) ^m_m^l

l →1 as l→ ∞.

a) If [2^jn_k−1,2^j+1n_k−1]⊂[m_l, m_l] for some 0≤j ≤k−2, then m_l

ml ≤ 2^jn_k−1 2^j+1nk−1 = 1

2 which contradicts (ii).

b) If [m_l, ml]⊂[2^jnk−1,2^j+2nk−1], then x_ml

x_ml ≥ x₂^j_nk−1 x₂j+2nk−1

= (j+ 1)2^j (j+ 3)2^j+2 =

1− 2

j+ 3 1

4 which contradicts (i).

c) If [n_k, nk]⊂[m_l, ml], then m_l m_l ≤ n_k

n_k →0 which contradicts (ii).

d) Ifm_l ∈[2^k−2n_k−1, n_k] and ml ∈[n_k, nk], i.e.,ml=n_k+i, then (because n_k = 2^k−1m_k−1 andx_ml =x_nk+i)

xm_l

x_ml ≥ x₂k−2n_k−1

x_ml = x₂k−2n_k−1

x₂k−1n_k−1

· xn_k

x_ml =

k−1 k

· 1 2· 1

1 +_xⁱ

n k

, m_l

m_l ≤ n_k

m_l = 1 1 +_nⁱ

k

. Furthermore, (i) implies _nⁱ

k → 0 and (ii) implies _xⁱ

n k

= _k2ⁱ_n

k → ∞ which is impossible.

e) If [2nk,2²nk]⊂[m_l, ml] then m_l

m_l ≤ 2nk

2²n_k = 1 2 which contradicts (ii).

f) Finally, assume thatm_l ∈[n_k, nk] andml ∈[nk,2nk]. Sincex₂nk = 4xnk, we have

x_ml

x_ml ≥ x_nk

x_2nk = x_nk 4x_nk → 1

4

which contradicts (i).

In document DISTRIBUTION FUNCTIONS OF RATIO SEQUENCES. (Strani 35-50)