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Necessary and sufficient conditions for solvability of the Hartman-Wintner problem for diff


NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY OF THE HARTMAN-WINTNER PROBLEM FOR DIFFERENCE EQUATIONS
N.A. CHERNYAVSKAYA DEPARTMENT OF MATHEMATICS AND COMPUTER SCIENCE BEN-GURION UNIVERSITY OF THE NEGEV P.O.B. 653, BEER-SHEVA, 84105, ISRAEL

arXiv:0705.3014v1 [math.CA] 21 May 2007

L.A. SHUSTER DEPARTMENT OF MATHEMATICS BAR-ILAN UNIVERSITY, 52900 RAMAT GAN, ISRAEL Abstract. The equation ?(rn?1 ?yn?1 ) = (qn + σn )yn , is viewed as a perturbation of the equation ?(rn?1 ?zn?1 ) = qn zn , n≥0 (1) (2) {qn }∞ n=0

which does not oscillate at in?nity. The sequences are assumed real, rn > 0 for all n ≥ 0, the sequences {σn }∞ may be complex-valued. We study the Hartmann=0 Wintner problem on asymptotic “integration” of (1) for large n in terms of solutions of (2) and the perturbation {σn }∞ n=0 .

{rn }∞ n=0 ,

n≥0

1. Introduction In the present paper we consider di?erence equations ?(rn?1 ?yn?1 ) = (qn + σn )yn , ?(rn?1 ?zn?1 ) = qn zn , n = 0, 1 , 2 , . . . (1.1)

n = 0, 1 , 2 , . . .

(1.2)

∞ where the sequences {rn }∞ n=0 , {qn }n=0 are assumed real, rn > 0 for all n ≥ 0, and the

∞ that in this case it has solutions {un }∞ n=0 (the principal, or recessive, solution) and {vn }n=0

{an }∞ n=0 . We assume that equation (1.2) does not oscillate at in?nity. It is known [1, Ch.VI]

sequence {σn }∞ n=0 may be complex-valued; ?an = an+1 ? an , n ≥ 0, for every sequence

(the non-principal, or dominant, solution), and there is an integer n0 such that the following

2000 Mathematics Subject Classi?cation. 39A11.
1

2

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

relations hold: u n > 0, vn > 0 for n ≥ n0


rn (vn+1 un ? un+1 vn ) = 1 for n ≥ 0 un lim = 0, n→∞ vn
n=n0

1 = ∞, rn un un+1

∞ n=n0

1 < ∞. rn vn vn+1

(1.3)

there exists a fundamental system of solutions (FSS) {u ?n , v ?n }∞ n=0 of equation (1.1) such that u ?n v ?n lim = lim =1 (1.4) n→∞ un n→∞ vn un+1 1 u ?n+1 = +o for n → ∞ (1.5) u ?n un rn un vn vn+1 1 v ?n+1 for n → ∞. (1.6) = +o v ?n vn rn un vn Here we assume that a FSS {un , vn }∞ n=0 of equation (1.2) with properties (1.3) is known. A similar question for di?erential equations was ?rst studied by P. Hartman and A. Wintner, and therefore we relate the above problem to their names (and denote it throughout as problem (1.4)–(1.6)); see [2],[6],[7], [8],[10] for generalities on problems (1.4)–(1.6); see [3] for a summary of results on the Hartman-Wintner problem. Note that a problem close to (1.4)–(1.6) was studied in [11], and the main result of [11] can be interpeted as a solution of problem (1.4)–(1.6). Therefore below we present statements of that paper (“projecting” them on to the Hartman-Wintner problem) and emphasize that the problem considered there is di?erent from problem (1.4)–(1.6). Here and throughout the sequel we say that problem loss of generality, we assume n0 equal to zero. Assertion 1.1. ([11]) If the series J =
def ∞ n=0

Our goal is to study the following problem: To ?nd conditions on {σn }∞ n=0 under which

(1.4)–(1.6) is solvable if (1.1) has a FSS {u ?n , v ?n }∞ n=0 satisfying (1.4)–(1.6); in (1.3), without

σn un vn

(1.7)

converges (at least conditionally), then the sequence Cn =
def

vn un

∞ k =n

σk u2 k,

n = 0, 1 , 2 , . . .

(1.8)

is well de?ned, and the following inequalities hold: |Cn | ≤ 2An , n = 0, 1 , . . . ; An = sup |Jm |,
m≥n def

Jm =

def

∞ k =m

σk uk vk .

(1.9)

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

3

Theorem 1.1. ([11]) Suppose that the series J (see (1.7)) converges (at least, conditionally), and, in addition,
∞ n=0

An+1 |Cn+1| <∞ rn un vn+1 An+1 |Cn+1 | 1 =?< . rn un vn+1 2 k =n


(1.10) (1.11)

1 lim sup m→∞ n≥m An

Then the Hartman-Wintner problem is solvable. Corollary 1.1.1. If the series J (see (1.7)) absolutely converges, then problem (1.4)–(1.6) is solvable. We prove Assertion 1.1 in §5. Theorem 1.1 is proved in §8. Note that assumption (1.11)

in its statement turned out to be super?uous. solvability of problem (1.4)–(1.6).

Let us now formulate our results. The following theorem contains necessary conditions for

Theorem 1.2. If the Hartman-Wintner problem is solvable, then the series σ : σ=
def ∞ n=0

σn u2 n

(1.12)

addition,

converges (at least conditionally), the sequence {Cn }∞ n=0 (see (1.8)) is well de?ned, and, in
n→∞

lim Cn = 0.

(1.13)

For the reader’s convenience, let us outline the plan of the paper. Below we follow the general approach to the study of problem (1.4)–(1.6) which was proposed in [3] for its di?erential analogue. This means that we restrict the initial problem in order to obtain a criterion for solvability of the narrow problem (see De?nition 1.1 and Theorem 1.3 below), and then to the narrow one (see Theorem 1.4 below). Finally, we analyze why the conditions for solvability of the Hartman-Wintner problems for di?erential and di?erence equations are not completely analogous whereas the statements of these problems are completely analogous over to precise statements. (see [3]). We also present examples to all the main results of the paper (see §9). We now go ?nd precise a priori requirements to {σn }∞ n=0 under which the initial problem is equivalent

De?nition 1.1. Problem (1.4)–(1.6) with the additional requirement
∞ n=0

rn un vn+1

u ?n+1 u ?n ? un+1 un

2

<∞

(1.?)

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N.A. CHERNYAVSKAYA AND L.A. SHUSTER

is called the narrow Hartman-Wintner problem (and is referred to below as problem (1.4)– (1.?)). To give a criterion for solvability of problem (1.4)–(1.?) and its consequences, we need the G= L=
def def ∞ n=0 ∞ n=0

following series:

|Cn+1|2 rn un vn+1

(see (1.8)) (see (1.9))

(1.14) (1.15) (1.16)

Re(Jn+1 C n+1 ) rn un vn+1
def ∞ n=0

P =

Re(σn C n )un vn .

Lemma 1.1. Suppose that the series J (see (1.7)) converges (at least, conditionally). Then the series L converges (at least, conditionally) if and only if the series G converges, and the series P converges (at least, conditionally) if and only if the series G (see (1.14) and B both converge, where B =
def ∞ n=0

|σn un vn |2 .

(1.17)

The next theorem contains the main result of the paper. Theorem 1.3. The narrow Hartman-Wintner problem is solvable if and only if the series J converge (at least, conditionally; see (1.7)) and either the series G or L converge (L at least, conditionally). Corollary 1.3.1. The narrow Hartman-Wintner problem is solvable provided the series J (see (1.7)) converges (at least, conditionally) and any of the following conditions I )–IV ) holds: I) at least one of the following inequalities holds (see (1.9)):
∞ n=0 ∞ n=0 ∞ n=0

|Jn+1Cn+1 | < ∞, rn un vn+1

An+1 |Cn+1 | < ∞, rn vn+1 un

|Jn+1 |2 < ∞. rn un vn+1

(1.18)

II) the series P converges (at least, conditionally); III) the following inequality holds (see (1.9)):
∞ n=0

|σn |An un vn < ∞

(1.19)

IV) the series J (see (1.7)) absolutely converges.

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

5

(1.4)–(1.6)). Therefore one can pose the problem: to ?nd requirements to the perturbation {σn }∞ n=0 under which the Hartman-Wintner and its narrow analogue are indistinguishable (equivalent). Clearly, if problem (1.4)–(1.?) is solvable, then problem (1.4)–(1.6) is also

Theorem 1.3 contains a criterion for solvability of problem (1.4)–(1.?) (but not of problem

solvable. Therefore the above question can be reduced to the following: to ?nd precise a FSS {u ?n , v ?n }∞ n=0 of equation (1.1) satisfying relations (1.4)–(1.6)) also satis?es condition priori requirements to {σn }∞ n=0 under which any solution of problem (1.4)–(1.6) (i.e., any

(1.?). In the last case, the Hartman-Wintner problem is completely reduced to the narrow Hartman-Wintner problem, and we say that the two problems are equivalent. A criterion for the equivalence of problem (1.4)–(1.6) and (1.4)–(1.?) is given in Theorem 1.4. FSS of (1.1) satisfying (1.4)–(1.6). Then inequality (1.?) holds if and only if G < ∞ (see Note that Theorem 1.4 often (but not always, see examples in §9) allows one to reduce Theorem 1.4. Suppose that the Hartman Wintner problem is solvable and {u ?n , v ?n }∞ n=0 is a

(1.14)).

of Theorem 1.3. More precisely, the set of equations (1.1) for which such a reduction is impossible is not larger than the set of equations (1.1) for which the sequence {Cn }∞ n=0 is de?ned, Cn → 0, as n → ∞, and G = ∞ (see (1.14)). Let us now analyze what is common and what is di?erent in the statement on solvability

the study of problem (1.4)–(1.6) to the study of problem (1.4)–(1.?), i.e., to the application

of the Hartman-Wintner problems for di?erential and di?erence equations. For brevity, we shall do that by conditional comparison of results, as follows. We assume that the criteria for solvability of the narrow Hartman-Wintner problems for di?erential and di?erence equations are completely analogous. Under this assumption, we obtain the following assertions (we put them in quotation marks because they are false): “Lemma 1.2”. Suppose that the series J (see (1.7)) converges (at least, conditionally). Then the series L and P (see (1.15)–(1.16)) converge (at least, conditionally) if and only if the series G (see (1.14)) converges. “Theorem 1.5”. The narrow Hartman-Wintner problem is solvable if and only if condition α) and any of conditions β ), γ ), or θ) hold: α) the series J (see (1.7)) converges (at least, conditionally); β ) the series G (see (1.14) converges;

6

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

γ ) the series L (see (1.15) converges (at least, conditionally); θ) the series P (see (1.18)) converges (at least, conditionally). Let us now compare “Theorem 1.5” with Theorem 1.3. We see that in the parts α)–β ) and α)–γ ), “Theorem 1.5” is true (and thus completely analogous to the corresponding theorem for di?erential equations). Here we see the complete analogy between the narrow HartmanWintner problems for di?erential and di?erence equations. In the part α)–θ), “Theorem 1.5” is no longer true, and here we see the di?erence between these problems. According to Corollary 1.3.1, in the part α)–θ) “Theorem 1.5” (di?erence case) is only a su?cient (but the narrow Hartman-Wintner problem shows that “Theorem 1.5” is no more precise in the part α)–θ). The reason can be explained by comparing Lemma 1.1 with “Lemma 1.2”. The convergence (at least, conditional) of both the series J and P (see (1.7), (1.16) implies not only the convergence of the series G (which already gives a criterion for solvability of the narrow Hartman-Wintner problem) but also the convergence of the series B (see (1.17)) which is, in general, is not obligatory and can be viewed as some “additional load” on the parameters of the problem. Example 1.1. Consider the Hartman-Wintner problem for the equations ?(nα ?yn ) = (?1)n yn+1 , (n + 1)β n≥1 (1.20) (1.21) not a necessary) condition for solvability of problem (1.4)–(1.?)). Thus, “discretization” of

?(nα ?zn ) = 0 · zn+1 , for α ∈ [0, 1), β ∈ 1 ? α,
3 2

n≥1

series P (see (1.16)) also diverges, which contradicts the assertion α)–θ) of “Theorem 1.5”.

(1.?)) is solvable but the series B (see (1.17)) diverges and, according to Lemma 1.1, the

? α . In this case (see Section 9, Example 9.3), problem (1.4)–

To conclude our analysis, we now have to ?nd a reason which explains the di?erence between Lemmas 1.1 and “1.2”. In this case such a di?erence arises because the integral calculus and the ?nite di?erence calculus are not always completely analogous. More precisely, in the case of the Hartman-Wintner problem for di?erential equations the proof of the criterion for solvability of type α)–θ) (see “Theorem 1.5”) relies, after all, upon the following obvious statement (see [3]): if x(t) is a continuously di?erentiable function on [1, ∞) such that x(t) → 0 as t → ∞, then the integral
1 ∞

x′ (t)x(t)dt

(1.22)

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

7

converges. The di?erence analogue of this integral (with step 1) is of the form
∞ n=1

(?xn )xn ,

xn = x(n),

n = 1, 2 , . . . ;

xn → 0 as n → ∞.

(1.23)

(It is this series that appears in the course of the analysis of the condition α)–θ) of “Theorem 1.5”, see §5). It is easy to see that the series (1.23) converges if and only if the series D= converges. Indeed,
∞ n=1 ∞ n=1

(?xn )2

(1.24)

(?xn )xn = =

∞ n=1 ∞ n=1

(xn+1 ? xn )xn =

∞ n=1

(xn xn+1 ? x2 n)
∞ n=1

2 x2 x2 1 1 n+1 ? xn ? (xn+1 ? xn )2 = ? 1 ? 2 2 2 2

(1.25) (?xn ) .
2

Thus, should the integral (1.22) and the series (1.23) (with the above-mentioned requirements to x(t)) converge (or diverge) together, “Theorem 1.5” would be true in the part α)–θ). But this asmumption is wrong: for x(t) =
cos √ πt , t

the integral (1.22) converges and

the series (1.24), as one can easily see, diverges. Thus the analogy between conditions for solvability of the Hartman-Wintner problems for di?erential and di?erence equations is not complete although the statements of these problems are completely analogous.

2. Preliminaries To prove the result from Section 1, we only need some generalities from the theory of non-oscillating di?erence equations of order 2. See [1] for a comprehensive exposition. For the reader’s convenience, below we present a standard summary of all the needed facts. Recall that equation (1.2) is said to be non-oscillating at in?nity if all its solutions do not oscillate, i.e., do not change sign beginning from a certain number. Therefore, for any ?xed for all n ≥ 0. Indeed, otherwise we would just have to change numeration in (1.2) and for each solution ?nd the corresponding constant factor τ = ±1. The last line of relations in (1.2). Here the principal solution {un }∞ n=0 is determined uniquely up to a costant factor. Let
∞ (1.3) completely characterizes the principal {un }∞ n=0 and non-principal {vn }n=0 solutions of

FSS {un , vn }∞ n=0 of equation (1.2) one can assume that the inequalities un > 0, vn > 0 hold

8

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

{vn }∞ n=0 be any non-principal solution. Then the equality un = vn
∞ k =n

1 , rk vk vk+1

n = 0, 1 , 2 , . . .

(2.1)

determines the principal solution. The following inequalities are immediate consequences of (1.2), and we will use them repeatedly: un un+1 un+1 vn > , 1> , vn vn+1 un vn+1 Lemma 2.1. The following equalities hold:
∞ n=0

rn un vn+1 > 1,

n = 0, 1 , 2 , . . .

(2.2)

1 = ∞, rn un vn+1


∞ n=0

1 = ∞. rn vn un+1

(2.3)

Moreover, if vn+1 ≥ vn for all n ≥ 0, then
n=0

1 = ∞. rn un vn

(2.4)

Remark 2.1. We do not use equalities (2.3)–(2.4) in the proofs. We present them here because they enable one to evaluate the requirements from the statements in Section 1 more precisely. Proof of Lemma 2.1. For n ≥ 0 from (2.1) it follows that (see also (1.3)) Rn =
def ∞ s =n

1 = rs us vs+1 1 rs vs vs+1

∞ s =n ∞

1 rs vs vs+1

∞ k =s ?1

1 rk vk vk+1


?1



∞ s =n

1 rk vk vk+1 k =n

=

1 rk vk vk+1 k =n

?1

·

∞ s =n

1 rs vs vs+1

= 1.

Since the remainder Rn of the ?rst series in (2.3) does not converge to zero as n → ∞, the implies that the second series in (2.3) and the series (2.4) diverge.

series diverges. Together with inequalities (2.2) and the comparison theorem for series, this

Remark 2.2. Throughout Sections 3–9 below, in the whole proof the letter τ stands for absolute positive constants which are not essential for exposition and may di?er even within a single chain of computations.

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

9

3. A Problem Equivalent to the Hartman-Wintner Problem Our study of problem (1.4)–(1.6) relies upon the following assertion. Lemma 3.1. Problem (1.4)–(1.6) is solvable if and only if the equation ?(rn?1 un?1 un ?βn?1 ) = σn u2 n βn , has a solution {βn }∞ n=0 such that
n→∞

n = 1, 2 , . . .

(3.1)

lim βn = 1,

n→∞

lim rn vn un+1 ?βn = 0.

(3.2)

Proof of Lemma 3.1. Necessity. Suppose that problem (1.4)–(1.6) is solvable. Set βn = u ?n , un n = 0, 1 , 2 , . . .

and show that {βn }∞ n=0 is a solution of the problem (3.1)–(3.2). From (1.4) it follows that βn → 1 as n → ∞. From (1.5) it follows that u ?n+1 u ?n εn εn βn u ?n = + , lim εn = 0 ? ?βn = βn+1 ? βn = n→∞ un+1 un rn vn un+1 un rn vn un+1 ? lim rn vn un+1 ?βn = lim εn βn = lim εn · lim βn = 0.
n→∞ n→∞ n→∞ n→∞

Thus, relations (3.2) are proved. Furthermore, the equalities ?(rn?1 ?? un?1 ) = (qn + σn )? un , imply equality (3.1): σn u2 ?n = un ?(rn?1 ?? u n?1 ) ? u ?n ?(rn?1 ?un?1 ) n βn = σn un u = r n (u n u ?n+1 ? u ?n un+1 ) + rn?1 (un u ? n?1 ? u ? n u n?1 ) = rn un un+1(βn+1 ? βn ) ? rn?1 un?1 un (βn ? βn?1 ) = ?(rn?1 un?1un ?βn?1 ). ?(rn?1 ?un?1 ) = qn un , n = 0, 1 , 2 , . . .

Proof of Lemma 3.1. Su?ciency. Suppose that problem (3.1)–(3.2) is solvable, and let {βn }∞ n=0 be its solution. Set u ?n = βn un , n = 0, 1 , 2 , . . . (3.3)

10

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

Then ?(rn?1 un?1 un ?βn?1 ) = ? rn?1 un?1un u ?n u ? n?1 = ? [rn?1 (? u n u n?1 ? u ?n?1un )] ? u n u n?1 = rn (? un+1 un ? u ?n un+1) ? rn?1 (? u n u n?1 ? u ? n?1 u n ) = un [rn (? un+1 ? u ?n ) ? rn?1 (? un ? u ?n?1)] ? u ?n [rn (un+1 ? un ) ? rn?1 (un ? un?1)] = un ?(rn?1 ?? u n?1 ) ? u ?n ?(rn?1 ?un?1 ) = un ?(rn?1 ?? un?1 ) ? qn un u ?n u ?n = σn u2 = σn un u ?n . n un Since un = 0 for n ≥ 0, the last equality implies ?(rn?1 ?? un?1 ) = (qn + σn )? un , Here u ?n = lim βn = 1. (3.4) n→∞ n→∞ un for n ≥ n0 ? 1. Then (see (3.3)) we have u ?n = 0 for n ≥ n0 . Set (see (2.1)) lim
n

n = 0, 1 , 2 , . . .

Let |βn | ≥

1 2

v ?n+1 = u ?n+1
k =n0

1 rk u ?k u ?k+1

for n ≥ n0 .

(3.5)

By (3.5) we get v ?n 1 v ?n+1 ? = u ?n+1 u ?n rn u ?n u ?n+1 ? rn (? vn+1 u ?n ? u ?n+1 v ?n ) = 1, for n ≥ n0 . (3.6)

From (3.6) we obtain the following chain of equalities: u ?n+1(rn+1 ?? vn+1 ) ? u ?n (rn ?? vn ) = v ?n+1 (rn+1 ?? un+1 ) ? v ?n (rn ?? un ) ? u ?n+1?(rn ?? vn ) + rn ?? un ?? vn = v ?n+1 ?(rn ?? un ) + rn ?? un ?? vn ? u ?n+1?(rn ?? vn ) = v ?n+1 ?(rn ?? un ) = v ?n+1 (qn+1 + σn+1 )? un+1 ? ?(rn ?? vn ) = (qn+1 + σn+1 )? vn+1 , n ≥ n0 . Thus {u ?n , v ?n }∞ n=n0 is a FSS of equation (1.1). The following obvious equality is a consequence
n

of (1.3) (see [1]):

vn+1 = τ un+1 + un+1
k =n0

1 , rk uk uk+1

n ≥ n0 .

(3.7)

Indeed, from (1.3) we get + rn un1un+1 + rn?1 u1 n?1 un ........................ vn0 +1 v 0 = un + rn un 1un +1 un +1 n
0 0 0 0 0

vn+1 vn =u un+1 n vn?1 vn =u un n?1

? ? ? ? ?

vn+1 vn 1 ? = 0 + ? (3.7) ? un+1 un0 k=n rk uk uk+1 ? ? 0 ?

n

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

11

In the following relations, we use (3.5), (3.7), (3.4) and Stolz’s theorem (see [4, vol. I]):
n

v ?n+1 lim = lim n→∞ vn+1 n→∞

u ?n+1
k =n0

1 rk u ?k u ?k+1 n 1 rk uk uk+1

τ un+1 + un+1
k =n0 n

(3.8)
1 rk u ?k u ?k+1 1 rk uk uk+1

= lim

u ?n+1 · lim n→∞ un+1 n→∞

k =n0 n

= lim

τ+
k =n0

un un+1 = 1. n→∞ u ?n u ?n+1

Let us now check equality (1.5): rn un vn un+1 u ?n+1 u ?n u ?n+1 un+1 = rn un vn ? ? u ?n un u ?n un+1 un 1 = (rn vn un+1 ?βn ) → 0 as n → ∞. βn

It remains to prove equality (1.6). Below we use (3.6), (1.3), (1.5), (3.4), and (3.8): v ?n+1 u ?n+1 1 u ?n+1 vn+1 un ? un+1vn = + = + v ?n u ?n rn u ?n v ?n u ?n u ?n v ?n u ?n+1 un vn vn+1 un+1 u ?n+1 un vn vn+1 un+1 = = + ? + ?1 ? u ?n u ?n v ?n vn un u ?n u ?n v ?n vn un vn+1 un+1 vn+1 vn+1 un vn 1 u ?n+1 un+1 δn + = ? = + ?1 + ? + , vn un vn u ?n v ?n rn un vn u ?n un vn rn un vn un vn ? 1 + εn , δn → 0 as n → ∞. δn = u ?n v ?n Here {εn }∞ n=0 is the sequence de?ned by equality (1.5): un+1 εn u ?n+1 = + , u ?n un rn un vn
n→∞

lim εn = 0.

Corollary 3.1.1. Problem (3.1)–(3.2) is solvable if and only if the equation rn?1 un?1un ?βn?1 = ? has a solution {βn }∞ n=0 such that
n→∞ ∞ k =n

σk u2 k βk ,

n = 1, 2 , . . .

(3.9)

lim βn = 1,

n→∞

lim rn vn un+1 ?βn = 0.

(3.10)

Moreover, the solutions of the problems (3.1)–(3.2) and (3.9)–(3.10) coincide if they exist.

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N.A. CHERNYAVSKAYA AND L.A. SHUSTER

Proof of Corollary 3.1.1. Necessity. Suppose that problem (3.1)–(3.2) is solvable, let {βn }∞ n=0 be its solution, and let m ≥ 1. Then
n+m k =n

?(rk?1 uk?1 uk ?βk?1 ) = rn+m un+mun+m+1 ?βn+m ? rn?1 un?1un ?βn?1
n+m

=
k =n

σk u2 k βk .

(3.11)

From (3.2) and (1.3) it follows that
n→∞

lim rn un un+1 ?βn = lim (rn vn un+1 ?βn ) ·
n→∞

un = 0. vn

(3.12)

Therefore, as m → ∞, (3.11) implies (3.9), and (3.2) coincides with (3.10). Proof of Corollary 3.1.1. Su?ciency. Suppose that problem (3.9)–(3.10) is solvable, and let {βn }∞ n=0 be its solution. Then ?(rn?1 un?1 un ?βn?1 ) = rn un un+1?βn ? rn?1 un?1 un ?βn?1 =? Moreover, (3.10) coincides with (3.2). 4. Proof of Necessary Conditions for Solvability of the Hartman-Wintner Problem In this section, we prove Theorem 1.2. Proof of Theorem 1.2. If problem (1.4)–(1.6) is solvable, then by Lemma 3.1 there exists a solution {βn }∞ n=0 of problem (3.1)–(3.2). Note that from (1.3) and (3.2) it follows that
n→∞ ∞ ∞

σk u2 k βk

+

2 σk u2 k βk = σn un βn .

k =n+1

k =n

lim rn un vn+1 ?βn = lim (1 + rn vn un+1 )?βn = lim ?βn + lim rn vn un+1 ?βn = 0. (4.1)
n→∞ n→∞ n→∞ 1 2

Let |βn | ≥

for n ≥ n0 and m2 ≥ m1 ≥ n0 + 1. Denote αn = rn un un+1 ?βn , τ=

u0 + 1. v0

Then from (3.1) it follows that Z (m1 , m2 ) =
def m2 m2

σn u2 n
n=m1 m2

=

αn = βn

+
m1 ?1

n=m1 1 m2 rn?1 un?1 un (?βn?1 )2 n=m1

2 ?αn?1 = βn n=m

m

αn αn?1 αn?1 ?βn?1 ? + βn βn?1 βn?1 βn .

(4.2)

βn?1 βn

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

13

Let ε ∈ (0, 1]. From (3.2) and (4.1) we conclude that there is n1 > n0 such that
n≥n1

sup rn un vn+1 |?βn | ≤

ε , 16τ

n≥n1

sup rn vn un+1|?βn | ≤

ε . 16τ

(4.3)

Using (4.2), (4.3), (2.2), (2.1), we now estimate Z (m1 , m2 ) for m2 ≥ m1 ≥ n1 + 1 :
m2

|Z (m1 , m2 )| ≤ 4 sup rn un un+1|?βn | +
n≥n1

n=m1

rn?1 un?1 un |?βn?1 |2 |βn?1 βn |
m

2 un (rn?1 un?1 vn |?βn?1 |)(rn?1 vn?1 un |?βn?1 |) = 4 sup (rn vn un+1 |?βn |) + 4 rn?1 vn?1 vn n≥n1 vn n=m 1

≤4

ε u0 ε +4 v0 16τ 256τ 2

2

∞ n=1

ε ε2 u 0 1 ≤ + < ε. rn?1 vn?1 vn 4 64τ 2 v0

This estimate implie that the series σ (see (1.12)) converges in view of Cauchy’s criterion. Hence the sequence {Cn }∞ n=0 (see (1.8)) is well-de?ned. It remains to verify (1.13). Denote an = sup max{rm?1 um?1 vm |?βm?1 |,
m≥n?1

rm?1 vm?1 um |?βm?1 |},

n≥1

bn = sup

m≥n

1 1 , |βm | |βm βm+1 |

,

n ≥ 0.

From (3.2) and (4.1) we get
n→∞

lim an = 0,

n→∞

lim bn = 1.

(4.4)

In the following relations we use the above notation and (1.8), (3.12), and (2.1): | Cn | = = vn un vn un vn ?αm?1 = βm un m=n
∞ ∞ ∞ m=n

αm αm?1 αm ?βm ? + βm+1 βm βm βm+1

rn?1 un?1 vn |?βn?1 | αm ?βm αn?1 ≤ ? β β βn βn m=n m m+1 (rm um vm+1 |?βm |)(rm vm um+1 |?βm |) |βm βm+1 |rm vm vm+1 m=n 1 + an vn un 1 r v v m=n m m m+1
∞ ∞

vn + un

(4.5)

≤ an bn

= an bn (1 + an ).

From (4.5) and (4.4) we get (1.13). 5. Auxiliary Assertions In this section we present various technical assertions needed for the proofs of Theorems 1.3 and 1.4 and their corollaries. Since we use Assertion 1.1, below we present its proof for the sake of completeness.

14

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

Proof of Assertion 1.1. Let Rn =
∞ k =n

σk u2 k,

n = 0, 1 , 2 , . . .

(5.1)

(In particular, R0 = σ (see (1.12)). Below we use notations (1.8)–(1.9), and relations (1.3): uk Rn = σk uk vk = v k k =n = =
∞ k =n ∞ k =n ∞ ∞ k =n

(Jk ? Jk+1 )

uk vk uk+1 uk ? vk+1 vk Jk+1 un = Jn ? . vn k=n rk vk vk+1


Jk

uk uk+1 ? Jk+1 + Jk+1 vk vk+1

(5.2)

uk+1 Jk+1 uk ? Jk+1 ? Jk vk vk+1 rk vk vk+1
∞ ∞ k =n

The series in the right-hand side of (5.2) absolutely converges since by (2.1) we have |Jk+1| ≤ An rk vk vk+1 un 1 = An , rk vk vk+1 vn n ≥ 0. (5.3)

k =n

sequence {Cn }∞ n=0 is well-de?ned. From (5.2)–(5.3) we get vn vn | Cn | = | Rn | = Jn ? un un
∞ k =n

Hence the series Rn , n ≥ 0 converges (at least, conditionally), and therefore by (1.8) the Jk+1 ≤ 2An , rk vk vk+1

n ≥ 0.

Lemma 5.1. Suppose that the series J (see (1.7)) converges (at least, conditionally). Then the series Hn =
def

Ck+1 , rk uk vk+1 k =n Jk+1 , rk vk vk+1 k =n




n = 0, 1 , 2 , . . .

(5.4)

also converges (at least, conditionally), and the following equality holds: vn Hn = un n = 0, 1 , 2 , . . . (5.5)

Proof. According to (1.8), we get (see (5.1) and (1.3)): ?Cn = Cn+1 ? Cn = = 1 rn un un+1 vn+1 vn vn+1 vn Rn+1 ? Rn = ? un+1 un un+1 un Cn+1 Rn+1 ? σn un vn = ? σn un vn . rn un vn+1
∞ k =n ∞ k =n

Rn+1 +

vn (Rn+1 ? Rn ) un

Since the series J converges, (1.9) implies ? Cn = ?Ck = Ck+1 ? Jn rk uk vk+1 ? Jn ? C n = H n . (5.6)

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

15

On the other hand, from (5.2) and (1.8), we get vn Jn ? C n = un
∞ k =n

Jk+1 . rk vk vk+1

(5.7)

Thus the series Hn , n = 0, 1, 2, . . . converges because of (5.6); and (5.5) follows from (5.6) and (5.7). Proof of Lemma 1.1. Since the series J (see (1.7)) converges (at least, conditionally) by Assertion 1.1, the sequence {Cn }∞ n=0 is well-de?ned (see (1.8)). Denote ?n = Re Cn ,
m2

ηn = Im Cn ,

n = 0, 1 , 2 , . . .
m2

L(m1 , m2 ) =
n=m1 m2

Re(Jn+1 C n+1 ) , rn un vn+1 ?n+1 rn un vn+1 ηn+1 rn un vn+1


G(m1 , m2 ) =
n=m1

|Cn+1 |2 rn un vn+1

? (m1 , m2 ) = L
n=m1 m2

?k+1 , r k uk vk +1 k =n+1 ηk+1 , r k uk vk +1 k =n+1 γn =
∞ k =n ∞

m2 ≥ m1 ≥ 0 (5.8) m2 ≥ m1 ≥ 0 n ≥ 0.

? (m1 , m2 ) = L
n=m1 ∞

?k+1 αn = , r k uk vk +1 k =n

ηk+1 , rk uk vk+1

well-de?ned. Moreover, by (5.6) and (1.13) we have
n→∞ n→∞

∞ ? ? By Lemma 5.1, the sequences {αn }∞ n=0 , {γn }n=0 and the values L(m1 , m2 ), L(m1 , m2 ) are

lim αn = lim γn = 0.

(5.9)

Let us now begin the proof. From (5.6) and (5.4), it follows that Jn+1 = Cn+1 + Hn+1 = Cn+1 + In turn, from (5.10) we get for n ≥ 0 : Re(Jn+1 C n+1 ) Re(C n+1 Ck+1) |Cn+1 |2 1 = + . rn un vn+1 rn un vn+1 rn un vn+1 k=n+1 rk uk vk+1 The last equality implies for m2 ≥ m1 ≥ 0 :
m2 ∞

Ck+1 , rk uk vk+1 k =n+1



n ≥ 0.

(5.10)

(5.11)

L(m1 , m2 ) = G(m1 , m2 ) +
n=m1

1 rn un vn+1

Re(C n+1 Ck+1 ) rk uk vk+1 k =n+1



(5.12)

? (m1 , m2 ) + L ? (m1 , m2 ). = G(m1 , m2 ) + L

16

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

? (m1 , m2 ) can be written in another way (see (1.25)): The expression L
m2

? (m1 , m2 ) = L
n=m1

(αn ? αn+1 )αn+1 ? 2
2 αm 2 +1 m2

1 2 2 2 =? [α ? αn + (αn+1 ? αn )2 ] 2 n=m n+1
1

m

= Similarly, we get

2 αm 1

1 ? . 2 n=m (rn un vn+1 )2
1

?2 n+1

(5.13)

2 2 2 γm ? γm ηn 1 2 +1 1 2 +1 ? L(m1 , m2 ) = ? . 2 2 n=m (rn un vn+1 )2
1

m

(5.14)

From (5.13), (5.14), (5.12) and (5.4), we get |Cn+1|2 |Hm1 |2 ? |Hm2 +1 |2 1 2 ? . L(m1 , m2 ) = G(m1 , m2 ) + 2 2 n=m (rn un vn+1 )2
1

m

(5.15)

Denote κ (m1 , m2 ) = From (5.6) and (1.13) it follows that
m1 →∞ m

|Hm1 |2 ? |Hm2 +1 |2 . 2 (5.16)

lim κ (m1 , m2 ) = 0.

From (5.15) and (2.2) it now follows that 1 2 |Cn+1|2 1 |L(m1 , m2 ))| ≥ G(m1 , m2 ) ? ?|κ (m1 , m2 )| ≥ G(m1 , m2 )?|κ (m1 , m2 )|. 2 2 n=m (rn un vn+1 ) 2
1

Hence by (2.2) and the last inequality, we have G(m1 , m2 ) ≤ 2|L(m1 , m2 )| + 2|κ (m1 , m2 )| (5.17) 3 |L(m1 , m2 )| ≤ G(m1 , m2 ) + |κ (m1 , m2 )|. (5.18) 2 From (5.17)–(5.18), we conclude that the series G and L satisfy (or do not satisfy) together Cauchy’s convergence criterion, which was to be proved. Let us now prove the second statement of the lemma concerning the series P (see (1.18)). Denote Re σn = ?n , αn =
∞ k =n m2

Im σn = ηn , γn =
∞ k =n

n ≥ 0; n ≥ 0;

?k u 2 k,

ηk u2 k,

P (m1 , m2 ) =
n=m1 m2

Re(σn C n )un vn , vn un
2

m2 ≥ m1 ≥ 0; m2 ≥ m1 ≥ 0; (5.19)

? (m1 , m2 ) = P
n=m1

?n u 2 n

∞ k =n

?k u 2 k ,

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY
m2

17

? (m1 , m2 ) = P
n=m1

vn un

2

ηn u2 n

∞ k =n

ηk u2 k ,
m2

m2 ≥ m1 ≥ 0; m2 ≥ m1 ≥ 0.
∞ k =n

κ (m, m2 ) = |Cm1 | ? |Cm2 | ,

2

2

B (m1 , m2 ) =
n=m1

(|σn |un vn )2 ,
2

According to Assertion 1.1, all values in (5.19) are well-de?ned. From (1.8) it follows that vn Re(σn un vn C n ) = Re σn un vn un = vn un
2 ∞ k =n ∞

σ k u2 k

=

vn un
∞ k =n

Re ηk u2 k .

σn u2 n

σ k u2 k (5.20)

?n u 2 n

2 ?k u 2 k + ηn un

k =n

From (5.20) it immediately follows that ? (m1 , m2 ) + P ? (m1 , m2 ). P (m1 , m2 ) = P Note that αn = Re un Cn , vn γn = Im un Cn , vn n ≥ 0, (5.22) (5.21)

? (m1 , m2 ), P ? (m1 , m2 ) can be writtenin a form di?erent than (5.19). For and the values P example (see (1.25)):
m2

? (m1 , m2 ) = 2 2P
n=m1 m2

vn un

2

m2

[αn (αn ? αn+1 )] =
2

n=m1

vn un

2 2 2 2 [αn ? αn +1 + (αn ? αn+1 ) ] 2 2 vn vn +1 ? u2 u2 n+1 n

=
n=m1

vn αn un
2

?

vn+1 αn+1 un+1
m2 2

2 2 + αn +1 2 αn +1 rn un un+1

+ (?n un vn )2
m2

= (Re Cm1 ) ? (Re Cm2 +1 ) + = (Re Cm1 )2 ? (Re Cm2 +1 )2 + Similarly, we get

n=m1 m2

vn+1 vn + un+1 un 1+ vn un+1 vn+1 un

+ +

(?n un vn )2
n=m1 m2

n=m1

(Re Cn+1)2 rn un vn+1

(?n un vn )2 .
n=m1

m2

? (m1 , m2 ) = (Im Cm1 )2 ?(Im Cm2 +1 )2 + 2P Hence
m2

n=m1

(Im Cn+1 )2 rn un vn+1

2 un+1 vn (ηn un vn )2 . 1+ + un vn+1 n=m 1

m

2P (m1 , m2 ) = κ (m1 , m2 ) +
n=m1

|Cn+1 |2 rn un vn+1

1+

vn un+1 vn+1 un

+ B (m1 , m2 ).

The last equality and (2.2) imply |P (m1 , m2 )| ≤ |κ (m1 , m2 )| + G(m1 , m2 ) + B (m1 , m2 ); 2|P (m1 , m2 )| + |κ (m1 , m2 )| ≥ G(m1 , m2 ) + B (m1 , m2 ). (5.23) (5.24)

18

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

Since
m1 →∞

lim κ (m1 , m2 ) = 0,

(5.25)

the assertion of the lemma now follows from (5.23), (5.24), (5.25) and Cauchy’s convergence criterion. Lemma 5.2. Suppose that the series J (see (1.7)) converges (at least, conditionally). Then the series G (see (1.14)) converges if the series I= converges. Proof. We use equality (5.6) (see also (5.4)): Jn+1 = Cn+1 + Ck+1 = Cn+1 + Hn+1 . r k uk vk +1 k =n+1 ?k+1 + (Re Hn+1 )2 ; rk uk vk+1 k =n+1
∞ k =n+1 ∞ ∞ ∞ n=0

|Jn+1 |2 rn un vn+1

(5.26)

(5.27)

From (5.27) we get (using notations (5.8)): (Re Jn+1 )2 = (Re Cn+1 )2 + 2?n+1 (Im Jn+1 ) = (Im Cn+1 ) + 2ηn+1 Let m2 ≥ m1 ≥ 0. Set I (m1 , m2 ) =
n=m1 2 2

(5.28)

ηk+1 + (Im Hn+1 )2 . rk uk vk+1
m2

(5.29)

m2

|Jn+1 |2 , rn un vn+1

H (m1 , m2 ) =
n=m1

|Hn+1 |2 rn un vn+1

From (5.28)–(5.29) it immediately follows that ? (m1 , m2 ) + L ? (m1 , m2 ) I (m1 , m2 ) = G(m1 , m2 ) + H (m1 , m2 ) + L (see (5.8)). From the last inequality by (5.13), (5.14), (5.15), and (5.16), we get 1 2 |Cn+1|2 I (m1 , m2 ) = G(m1 , m2 ) ? + H (m1 , m2 ) + κ (m1 , m2 ). 2 n=m (rn un vn+1 )2
1

m

(5.30)

From (5.30), by (2.2), we get |Cn+1 |2 1 2 + H (m1 , m2 )| I (m1 , m2 ) + |κ (m1 , m2 )| ≥ |G(m1 , m2 ) ? 2 n=m (rn un vn+1 )2
1

m

= G(m1 , m2 ) ? 1 ≥ G(m1 , m2 ). 2

1 1 |Cn+1 | + H (m1 , m2 ) ≥ G(m1 , m2 ) + H (m1 , m2 ) 2 2 n=m (rn un vn+1 ) 2
1

m2

2

(5.31)

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

19

From (5.16), (5.31) and Cauchy’s convergence criterion we now conclude that if the series I converges (see (5.26)), then the series G also converges. 6. Proof of the Criterion for Equivalence of the Hartman-Wintner Problem to its Restriction In this section we prove Theorem 1.4. Proof of Theorem 1.4. Necessity. We need the following obvious assertions. Lemma 6.1. Problem (1.1)–(1.?) is solvable if and only if problem (3.9)–(3.10) has a solu∞ n=0

tion {βn }∞ n=0 such that

rn un vn+1 |?βn |2 < ∞.

(6.1)

Proof of Lemma 6.1. Necessity. Suppose that problem (1.1)–(1.?) is solvable. Then, clearly,

problem (1.4)–(1.6) is also solvable, and the two problems have a common solution, a FSS of equation (1.1). Then by Lemma 3.1 and its Corollary 3.1.1, equation (3.9) has a solution {βn }∞ ?n = βn un , n = 0, 1, 2, . . . (see (3.3)). Therefore (1.?) n=0 with properties (3.10), and u implies ∞>
∞ n=0

rn un vn+1

u ?n+1 u ?n ? un+1 un

2

=

∞ n=0

rn un vn+1 |βn+1 ? βn | =

2

∞ n=0

rn un vn+1 |?βn |2 .

Proof of Lemma 6.1. Su?ciency. Suppose that problem (3.9)–(3.10) has a solution {βn }∞ n=0 solvable, and u ?n = βn un , n ≥ 0. Then the series (6.1) is none other than the series (1.?).

with property (6.1). Then by Lemma 3.1 and its Corollary 3.1.1, problem (1.4)–(1.6) is

Lemma 6.2. The Hartman-Wintner problem is solvable if and only if the series σ (see (1.12)) converges (at least, conditionally), and the equation rn un un+1 ?βn = ?βn+1 has a solution {βn }∞ n=0 such that
n→∞ ∞ k =n+1 ∞ k =n+1 ∞ m=k +1

σk u2 k

?

?βk

σm u2 m

(6.2)

lim βn = 1,

n→∞

lim rn vn un+1 ?βn = 0.

(6.3)

Moreover, the solutions of problem (3.1)–(3.2) and (6.2)–(6.3) coincide if they exist. In addition, the narrow Hartman-Wintner problem is solvable if and only if the series σ converges (at least, conditionally) and there exists a solution of problem (6.2)–(6.3) with property (6.1)

20

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

Proof of Lemma 6.2. Necessity. Suppose that problem (1.4)–(1.6) is solvable. Then by Theorem 1.2 the series σ converges (at least, conditionally), and by Lemma 3.1 and its Corollary 3.1.1 there exists a solution {βn }∞ n=0 of problem (3.9)–(3.10). Denote αn =


σk u2 k,

n = 0, 1 , 2 , . . .

k =n

By Theorem 1.2 the sequence {αn }∞ n=0 is well-de?ned. Let us transform (3.9): rn un un+1?βn = ? =?
∞ k =n+1 ∞

σk u2 k βk = ?



k =n+1

(αk ? αk+1 )βk


k =n+1

[βk αk ? βk+1 αk+1 + αk+1 ?βk ] = ?βn+1 αn+1 ?


(?βk )αk+1 (6.4)

k =n+1

= ?βn+1

k =n+1

σk u2 k ?

∞ k =n+1

?βk

∞ m=k +1

σm u2 m .

The rest of the assertions of the lemma either hold automatically, or follow from Lemma 6.1.

Proof of Lemma 6.2. Su?ciency. we have well-de?ned values Since the series σ converges and there exists a solutions {βn }∞ n=0 of problem (6.2)–(6.3), rn un un+1 ?βn , βn+1
∞ k =n+1

σk u2 k,

n = 0, 1 , 2 , . . .

Then by (6.2) we conclude that the series
∞ k =n+1

?βk

∞ m=k +1

σm u2 m

converges, and therefore the chain of computations in (6.4) can be reversed. Equality (3.9) with the solution of problem (6.2)–(6.3). It remains to consider Lemma 3.1, Corollary 3.1.1 and Lemma 6.1. Let us now return to Theorem 1.4. Suppose that problem (1.4)–(1.6) is solvable. Then by Theorem 1.2 the series σ converges. Since the inequality (1.?) also holds, problem (6.2)– Lemma 6.2). Let us rewrite equality (6.2) in terms of {Cn }∞ n=0 (see (1.8)). We get uk+1 vn+1 ?βk Ck+1 , rn un vn+1 ?βn = ?Cn+1 βn+1 ? un+1 k=n+1 vk+1


is thus proved. Thus there exists a solution {βn }∞ n=0 of problem (3.9)–(3.10), and it coincides

(6.3) is also solvable and its solution {βn }∞ n=0 satis?es inequality (6.1) (see Lemma 6.1 and

n ≥ 0.

(6.5)

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

21

Denote Sn =
(θ ) Gm ∞ k =n

rk uk vk+1 |?βk | ;

2

vn+1 uk+1 Fn = |?βk | |Ck+1|, un+1 k=n+1 vk+1 ξn =
∞ k =n



n≥0 (6.6) n ≥ 0.

|Cn+1 |2 = , m ≥ 0, θ ≥ 0; r u v n n n +1 n=m lim

m+θ

uk+1 |Ck+1 |2 , vk+1 rk uk vk+1

Note that the following equality holds: vn ξ n = 0. n→∞ un Indeed, from (2.1) and (2.2) it follows that 0 ≤ ξn = uk+1 vk |Ck+1 |2 un ≤ sup |Ck |2 . v u r v v v k +1 k k k k +1 n k ≥n k =n
1 2 ∞

(6.7)

(6.8)

From (6.8) and (1.13) we obtain (6.7). Let us now consider (6.5). Let |βn | ≥ for n ≥ n0 . For such n from (6.5) it follows that
∞ 2

uk+1 |Cn+1|2 |βn+1 |2 |Cn+1 |2 vn+1 |?βk | |Ck+1| ≤ ≤ (rn un vn+1 |?βn |)2 + 8 2 un+1 k=n+1 vk+1 Let m ≥ n0 , θ ∈ N. From (6.9) we get
2 2 1 (θ) Fn Fn ≤ Sm + . Gm ≤ Sm ? Sm+θ+1 + 8 r u v r u v n n n +1 n n n +1 n=m n=m m+θ m+θ

.

(6.9)

(6.10)

Let us estimate the second sum in (6.10). First note that from the de?nition of Fn , Sn and Schwarz’s inequality, we obtain vn+1 1/2 Fn ≤ S un+1 n+1
m+θ m+θ ∞ k =n+1 2

uk+1 vk+1


|Ck+1|2 rk uk vk+1
2

1/2

.

(6.11)

In the following relations, we use (6.11), (2.2), (1.3) and (6.8):
2 Sn+1 Fn ≤ r u v r u v n=m n n n+1 n=m n n n+1

vn+1 un+1
∞ k =n+1

2

k =n+1

uk+1 vk+1

|Ck+1 |2 rk uk vk+1

1 vn+1 ≤ Sm r u u un+1 n=m n n n+1
m+θ

m+θ

uk+1 vk+1


2

|Ck+1 |2 rk uk vk+1 (6.12)

≤ Sm ≤ Sm = Sm

uk+1 |Ck+1 |2 vn+1 un+2 1 r u u un+1 vn+2 k=n+1 vk+1 rk uk vk+1 n=m n n n+1
m+θ

n=m

vn+1 vn+1 vn vn vn ξn+1 = Sm ? ξn+1 ? ξn + (ξn ? ξn+1 ) un+1 vn un+1 un un n=m
m+θ +1

m+θ

vk ξk uk

+
m

vn un+1 |Cn+1 |2 u vn+1 rn un vn+1 n=m n

m+θ

(θ ) ≤ Sm sup |Ck |2 + Sm Gm . k ≥m+θ

22

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

(6.10) imply

Let us choose a bigger n0 (if necessary) so that Sm ≤ 16?1 for m ≥ n0 . Then (6.12) and
θ) 2 G( m ≤ 16Sm + 16Sm sup |Ck | , k ≥m+θ

m ≥ n0 .

(6.13)

Let us take the limit in (6.13) (as θ → ∞). Then by (1.13) we have |Cn+1 |2 θ) = lim G( m ≤ 16Sm < ∞. θ →∞ r u v n=m n n n+1 Proof of Theorem 1.2. Su?ciency. Suppose that problem (1.4)–(1.6) is solvable and G < ∞ (see (1.14)). Then by Theorem 1.2 the series σ (see (1.2)) converges (at least, conditionally),
∞ the sequence {Cn }∞ n=0 is well-de?ned, and by Lemma 6.2 there exists a solution of {βn }n=0 ∞

of problem (6.2)–(6.3) and it coincides with the solution of problem (3.1)–(3.2). Let us transform (6.5): ?n = rn vn un+1 ?βn , n = 0, 1 , 2 , . . . Ck+1?k , r k vk vk +1 k =n+1
∞ ∞

(6.14)

We get ?n = ? un+1 vn vn Cn+1 βn+1 ? un vn+1 un n ≥ 0. (6.15)

From (6.15), taking into account (3.10), we get for n ≥ 0 vn un+1 vn |?n | = rn vk un+1 |?βn | ≤ τ |Cn+1| + un vn+1 un

|Ck+1 | . rk vk vk+1 k =n+1

(6.16)

Let us estimate the sum in the right-hand side of (6.16). Below we use Schwarz’s inequality and (2.1): vn un
∞ k =n+1 1/2 1/2

vn |Ck+1| ≤ rk vk vk+1 un = vn un

∞ k =n+1

|Ck+1|2 rk vk vk+1


∞ k =n+1 2 1/2

1 rk vk vk+1 .

(6.17)

un+1 vn+1

|Ck+1 | rk vk vk+1 k =n+1

vn+1 1 Let us multiply the last inequality by unr . We get vn un+1 n ? ∞ ? |C | |Ck+1|2 1 √ n+1 +√ rn un vn+1 |?βn | ≤ τ √ ? rn un vn+1 rn un un+1 k=n+1 rk vk vk+1

From (6.16) and (6.17) it follows that ? ?u vn vn n+1 rn vn un+1 |?βn | ≤ τ |Cn+1 | + ? un vn+1 un

un+1 vn+1

|Ck+1| rk vk vk+1 k =n+1



2

1/2

? ? ?

.

1/2

? ? ?

.

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

23

Hence rn un vn+1 |?βn | ≤ τ
∞ 2

|Cn+1 |2 1 + rn un vn+1 rn un un+1


|Ck+1|2 rk vk vk+1 k =n+1




.

(6.18)

Let us sum up the inequalities (6.18) for n ≥ 0. Then (see (6.6) and (1.14)) S =
def n=0

rn un vn+1 |?βn |2 ≤ τ
∞ k =n

G+

n=0

|Ck+1|2 1 rn un un+1 k=n+1 rk vk vk+1

.

(6.19)

Let us estimate the sum in the right-hand side of (6.19). Denote Wn = |Ck+1|2 , rk vk vk+1 n ≥ 0. (6.20)

∞ The sequence {Wn }∞ n=0 is well-de?ned. Indeed, as shown above, the sequence {Cn }n=0 is

de?ned and it is bounded because of (1.13). Hence, according to (2.1), we have 0 ≤ Wn =
∞ k =n

|Ck+1 |2 ≤ rk vk vk+1

sup |Ck+1|
k ≥n

2

∞ k =n

1 un = sup |Ck+1|2 . rk vk vk+1 vn k≥n

(6.21)

From (6.21) and (1.13) we get vn Wn = 0. n→∞ un In the following relations, we use (1.3), (6.22): lim
∞ n=0

(6.22)

|Ck+1|2 1 = rn un un+1 k=n+1 rk vk vk+1 =
∞ n=0



∞ n=0

vn vn+1 ? un+1 un

Wn+1 (6.23)

vn+1 vn vn Wn+1 ? Wn ? (Wn+1 ? Wn ) un+1 un un


v0 v0 vn |Cn+1 |2 = ? W0 + = G ? W0 ≤ G. u0 u rn vn vn+1 u0 n=0 n From (6.18) and (6.23) it follows that S=
∞ n=0

rn un vn+1 |?βn |2 ≤ τ G < ∞.

7. Double Purpose Lemmas In this section we present some technical assertions which will be applied in the proof of Theorem 1.3, both in the “necessary” and the “su?cient” part. In the statements below we use assumptions of two types – condition A) and condition B): A) the narrow Hartman-Wintner problem (problem (1.4)–(1.?)) is solvable;

B) the series G and J converge (J at least, conditionally) (see (1.4), (1.7)).

24

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

Throughout the sequel by condition A) we mean the collection of all criteria already proved for solvability of problem (1.4)–(1.?). Thus the expression “condition A) holds” means that the following requirements are satis?ed: A.1) the series σ (see (1.12) converges (at least, conditionally) (see Theorem 1.2); A.2) the sequence {Cn }∞ n=0 (see (1.8)) is well-de?ned, and equality (1.13) holds (see Theorem 1.2); A.3) the inequality G < ∞ holds (see (1.14) and Theorem 1.4);

A.4) the series σ converges (at least, conditionally), and problem (6.2)–(6.3) has a solution {βn }∞ n=0 which satis?es inequality (6.1) and equality (6.15) (see Lemma 6.2). Similarly, the expression “condition B) holds” means that the following requirements are satis?es: B.1) the series J converges (at least, conditionally); B.2) Condition A.1) holds (see Assertion 1.1); B.3) Condition A.2) holds and inequalities (1.9) hold (see Assertion 1.1); B.4) Condition A.3) holds (see Theorem 1.4). Let us proceed to the exposition of the results. Let n0 denote a number and de?ne θ = {θk }∞ k =n0 , { g n (θ )} ∞ n=n0 :
def

θ = sup |θk |.
n≥n0

def

(7.1)

In the space of sequences (7.1) let us introduce a linear operator A and a sequence g (θ) = { (A θ )n } ∞ n=n0 , vn (A θ )n = un g n (θ ) = ? Ck+1 θk , r k vk vk +1 k =n+1


Aθ =

n ≥ n0 ; n ≥ n0 .

(7.2) (7.3)

g (θ ) = { g n (θ )} ∞ n=n0 , Here {Cn }∞ n=n0

un+1 vn Cn+1 θn+1 , un vn+1 is the sequence de?ned in (1.8).

Lemma 7.1. Suppose that either one of the conditions A) or B) holds. Then: g (θ ) ≤ | (A θ )n | ≤ sup |Ck |
k ≥n n≥n0

sup |Cn |

θ ; ≤ sup |Ck |
k ≥n

(7.4) θ ; (7.5) (7.6)

k ≥n+1

sup |θk |
n≥n0

A ≤ sup |Cn |.

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

25

Proof. In the following relations we use the hypotheses of the lemma, (2.2) and (2.1): g (θ) = sup vn | (A θ )n | ≤ un = sup |Ck |
k ≥n ∞ n≥n0

un+1 vn |Cn+1| |θn+1 | ≤ un vn+1 sup |Ck |
k ≥n

n≥n0

sup |Cn |

θ ; vn un 1 rk vk vk+1 k =n+1 ≤ sup |Ck |
k ≥n ∞

|Ck+1| |θk | ≤ r k vk vk +1 k =n+1
k ≥n+1

k ≥n+1

sup |θk |

sup |θk |
k ≥n0

vn un+1 ≤ un vn+1
k ≥n0

sup |Ck |
k ≥n

k ≥k +1

sup |θk |
k ≥k 0

θ ;

Aθ = sup |(Aθ)k | ≤ sup

m≥k

sup |Cm |

θ ≤

sup |Ck |

θ .

Lemma 7.2. Suppose that either one of the conditions A) or B) holds. Then there is n0 ? 1 such that for all n ≥ n0 and any vector θ with ?nite norm (7.1), the following equality holds:
∞ k =1

(?1) A g (θ)

k

k

=
n



k =1

(?1)k (Ak g (θ))n .

(7.7)

Proof. Let n ≥ n0 , and let Pn be the functional de?ned by the equality Pn θ = θn , θ = {θk }∞ k =n0 , θ < ∞.

Clearly, Pn is linear. Obviously, Pn = 1 since |Pn θ| = |θn | ≤ θ and, in addition, Pn θ0 = 1 = θ0 , Let us now verify that if θ
(0) 0 ∞ θ0 = {θk } k =n0 , 0 θk =

0, 1,

if k = n if k = n

=

∞ k =1

θ ,

(k )

∞ k =1

θ(k) < ∞,

then the following equality holds: Pn Indeed, let m ≥ 1. Then
m ∞ k =1

θ

(k )

=

∞ k =1

Pn θ(k) .

(7.8)

m

m

| Pn θ

(0)

?

Pn θ
k =1

(k )

| = Pn θ ≤
∞ k =m+1

(0)

?

θ
k =1

(k )

≤ Pn

θ

(0)

?

θ(k)
k =1

θ(k) → 0 as m → ∞; ? (7.8).

26

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

|Cn | ≤ 2?1 for all n ≥ n0 . Then A ≤ 2?1 (see (7.6)), and from (7.7) and Lemma 7.1 it follows that
∞ k =1

Furthermore, from the hypotheses of the lemma it follows that there is n0 ? 1 such that

(?1) A g (θ) ≤

k

k

∞ k =1

A

k

g (θ ) ≤

∞ k =1

1 2k+1

θ = τ θ < ∞.

It remains to apply (7.8). Denote (see (7.2)–(7.3)): 1 fn (θ) = rn vn un+1 Φn (θ) =
∞ k =n ∞

|fn (θ)|,

κn (θ) =
∞ k =n

k =1 ∞

(?1)k (Ak g (θ))n ,


n ≥ n0 ; n ≥ n0 ,

(7.9) (7.10) (7.11)

s=1 k =n s

|(As g (θ))k | < ∞, rk vk uk+1 s ≥ 1, n ≥ n0 .

(s) κn (θ ) =

|(A g (θ))k | , rk vk uk+1

Lemma 7.3. Suppose that one of the conditions A) or B) holds. Then there is n0 ? 1 such
(s) κn (θ )

that for all n ≥ n0 and any sequence θ with ?nite norm (7.1), the following inequalities hold: ≤ Gn sup |Ck |
k ≥n s ?1

θ ,

s ≥ 1, n ≥ n0 ,

(7.12) (7.13) (7.14)

κn (θ) ≤ 2Gn θ , Φn (θ) ≤ 2Gn θ , Here Gn =
def ∞ k =n

n ≥ n0 , n ≥ n0 . n ≥ 0.

|Ck+1|2 , rk uk vk+1

(7.15)

Proof. We ?rst prove inequality (7.12) for s = 1. From the hypothesis of the lemma, we we consecutively use (7.11), (7.2), (7.3), (6.6), (1.3), (6.7) and (2.2):
(1) κn (θ ) ∞ conclude that the sequences {Gn }∞ n=0 , {Cn }n=0 are well-de?ned. In the following relations, ∞ ∞ k =n ∞

|(Ag (θ))k | = = rk vk uk+1 k =n ≤ =
∞ k =n ∞ k =n

1 vk rk vk uk+1 uk

Cm+1 gm (θ) rm vm vm+1 m=k +1
∞ k =n

1 rk uk uk+1

∞ m=k +1

|Cm+1 |2 um+1 θ = rm um vm+1 vm+1
∞ k =n

ξk+1 θ rk uk uk+1 θ

vk+1 vk ? uk+1 uk
∞ k =n

ξk+1 θ =

vk+1 ξk+1 vk ξk vk ? + (ξk ? ξk+1) uk+1 uk uk θ ≤ Gn θ .

vn = ? ξn + un

vk uk+1 |Ck+1 |2 uk vk+1 rk uk vk+1

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

27

Let us consider the case s ≥ 2. Denote (see (7.2)):
s) ∞ As?1g (θ) = h(s) (θ) = {h( n (θ )} n = n 0 ∞ (s) |Cm+1 | |hm |(θ)

?

s) s ?1 h( g (θ))n , n = (A

n ≥ n0 ,

s ≥ 2,

γ k (θ ) =

(s)

m=k

rm vm vm+1

,

s ≥ 2, k ≥ n0 .

(7.16)

Let us verify that for θ < ∞ we have the equality vk (s) γk+1(θ) = 0. k →∞ uk lim Indeed, by (1.13), (2.1) and (2.2), we get vk vk (s) γk+1(θ) ≤ uk uk uk+1 vk = vk+1 uk sup |Cm |
s) sup (|h( m )(θ )| m≥k +1 s) sup (|h( m (θ )| ) m≥k +1

(7.17)

m≥k +1

1 rm vm vm+1 m=k +1
m≥k +1



(7.18)
s) |h( m (θ )|

m≥k +1

sup |Cm |



sup |Cm |

sup
m≥k +1

.

Furthermore, from (7.16), Lemma 7.1, (7.3), (1.13) and (2.2) it follows that
s) (s) s ?1 sup (|h( g (θ ) ≤ A m (θ )| ) ≤ h (θ ) = A s ?1

m≥k +1

g (θ ) ≤ τ θ .

The last estimate and (7.18) imply (7.17). In the following relations, we consecutively use (7.11), (7.16), (7.2), (1.3), (7.16), (7.17), and (2.2):
(s) κn (θ )

|(As g (θ))k | = = r k vk uk +1 k =n ≤ =
∞ k =n ∞ k =n



|(Ah(s) (θ))k | = r k vk uk +1 k =n |Cm+1 | |hm (θ)| = rm vm vm+1
(s)



1 vk rk vk uk+1 uk k =n
∞ k =n



Cm+1 hm (θ) rm vm vm+1 m=k +1 γk+1(θ)
(s)



(s)

1 rk uk uk+1

∞ m=k +1

vk+1 vk ? uk+1 uk

vk+1 (s) vk (s) vk+1 (s) (s) γk+2 (θ) ? γk+1 (θ) + (γ (θ) ? γk+2(θ)) uk+1 uk uk+1 k+1
∞ k =n

vn (s) = ? γn+1(θ) + un ≤ sup |Ck |
k ≥n ∞

vk+1 |Ck+2| |hk+1 (θ)| ≤ uk+1 rk+1 vk+1 vk+2
s ?1

(s)

∞ k =n+1

|Ck+1 | |hk (θ)| rk uk vk+1

(s)

|(A g (θ))k | = rk vk uk+1 k =n+1

sup |Ck | κn+1 (θ).
k ≥n

(s?1)

We thus get the estimate
(s) κn (θ ) ≤

sup |Ck | κn+1 (θ),
k ≥n

(s?1)

s ≥ 2, n ≥ n0 .

(7.19)

28

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

To ?nish the proof of (7.12), we consecutively use inequality (7.19) and the obtained estimate for κn (θ) :
(s) κn (θ ) ≤ (1) s ?1

sup |Ck | κn+1 (θ) ≤ · · · ≤
k ≥n

(s?1)

sup |Ck |
k ≥n

κn+s (θ)

(1)



sup |Ck |
k ≥n

s ?1

Gn θ .

From the hypothesis of the lemma, it follows that there is n0 ? 1 such that sup |Ck | ≤ 2?1 . Then (7.10), (7.11) and (7.12) imply (7.13): κn (θ) =
∞ s=1 (s) κn (θ ) ∞ s=1 k ≥n0 ∞ s=1 s ?1

≤ Gn · θ 1 2 s ?1

k ≥n0

sup |Ck |

≤ Gn θ

= 2 Gn θ .

Estimate (7.14) follows from (7.13), the well-known theorem on changing summation order for the series with positive terms (see [9, §1.6]) and (7.9): Φn (θ) = ≤
∞ k =n ∞ k =n

|fk (θ)| = 1 rk vk uk+1

∞ k =n ∞ s=1

1 rk vk uk+1
s

∞ s=1

(?1)s (As g (θ))k
∞ s=1

|(A g (θ))k | =

|(As g (θ))k | = κn (θ) ≤ 2Gn θ . r k vk uk +1 k =n



8. Proof of the Criterion for Solvability of the Narrow Hartman-Wintner Problem In this section we prove Theorem 1.3. Proof of Theorem 1.3. Necessity. Suppose that problem (1.4)–(1.?) is solvable. This means

that condition A) from Section 7 holds, and to prove the necessity of the hypotheses of Theorem 1.3 it remains to prove that the series J (see (1.7)) converges (at least, conditionally). Let us prove this. Throughout the sequel {βn }∞ n=0 denotes a solution of problem (6.2)–(6.3). vn un+1 vn Cn+1 βn+1 ? un vn+1 un n = 0, Ck+1 ?k , r k vk vk +1 k =n+1
n→∞ ∞

Recall that by (6.14)–(6.15) and (3.2) we have the following basic equalities: ?n = ? n ≥ 0.

(8.1) (8.2)

?n = rn vn un+1?βn ,

lim ?n = 0,

n→∞

lim βn = 1.

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

29

From (7.2)–(7.3) it follows that the system of equalities (8.1) and, more precisely, its part for n ≥ n0 , can be written in the form ? = g (β ) ? A?, ? = {? n }∞ n=n0 , β = {βn }∞ n=n0 . (8.3)

Let us now choose n0 ? 1 so that the following inequality holds: 1 A ≤ sup |Cn | ≤ 2 n≥n0 (see (7.6)). Then the operator A in (8.3) is compressing, and therefore we can ?nd ? : ? = (E + A) g (β ) = g (β ) +
?1 ∞ k =1

(8.4)

(?1)k Ak g (β ).

(8.5)

Here E is the identity operator. By (7.7), equality (8.5) can be written coordinatewise (for n ≥ n0 ) : ? n = g n (β ) +
∞ k =1 ∞ k =1

(?1) A g (β )

k

k

= g n (β ) +
n

(?1)k (Ak g (β ))n ,

n ≥ n0 .

(8.6)

From (8.6), (8.1) and (8.2), we get (for n ≥ n0 ): 1 Cn+1 βn+1 = ?βn + rn un vn+1 rn vn un+1

∞ s=1

(?1)s (As g (β ))n .

(8.7)

as a solution of problem (6.2)–(6.3)). For a given n ≥ n0 , let us introduce a sequence {Qn }∞ k =0 :
(k )

Our next goal is to analyze equation (8.7) in detail (note that its solution {βn }∞ n=n0 exists

k) Q( n =

? ? 1, ?

if k = 0 (1 + ηn+s ) ,
?1

k ?1 s=0

if k ≥ 1,
(k )

ηm =

def

Cm+1 , rm um vm+1

m ≥ n0 .

(8.8)

The sequence

(k ) {Qn }∞ k =0

can be de?ned recurrently:
k +1) Q( = n

Q(0) n = 1,

Qn , 1 + ηn+k

k ≥ 0, n ≥ n0 .

(8.9)

Note that from (1.13) and (2.2) it follows that
m→∞

lim ηm = 0.

(8.10)
(k )

Hence n0 can be chosen so that |ηn | ≤ 2?1 for n ≥ n0 , and therefore the sequence {Qn }∞ k =0 is well-de?ned. Let us write the solution {βn }∞ n=n0 of equation (8.7) in the form
k) βn+k = dk Q( n ,

n ≥ n0 , k = 0, 1, 2, . . .

(8.11)

where the factors dk , k ≥ n0 are to be determined later. From (8.11) it follows that βn = βn+k
k =0

= d0 Q(0) n = d0

?

d0 = βn .

(8.12)

30

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

To ?nd dk for k > 0, let us plug (8.11) into (8.7) with number (n + k ), n ≥ n0 , k = 1, 2, . . . Here we use the notations (7.9) and (8.8) (see also (8.9)): Cn+k+1 k +1) k) (k +1) fn+k (β ) = ?βn+k + βn+k+1 = dk+1Q( ? dk Q( n n + ηn+k dk +1 Qn rn+k un+k vn+k+1
k +1) k) (k ) = Q( (1 + ηn+k )dk+1 ? dk Q( n n = (dk +1 ? dk )Qn

?

dk+1 = dk + d0 = βn

1 (k) fn+k (β ), Qn

k ≥ 0, n ≥ n0 (8.13)

From (8.13) it immediately follows that
k

dk+1 = βn +
?=0

1

Qn

f (β ), (?) n+?

k ≥ 0, n ≥ n0 .

(8.14)

Finally, from (8.14) and (8.11) we get
k k +1) k +1) βn+k+1 = dk+1 Q( = βn Q( + n n ?=0

Qn

(k +1) (?)

Qn

fn+? (β ),

k ≥ 0, n ≥ n0 .

(8.15)

Below we study the sequence

a more convenient form. To obtain it, we set k = m ? 1, m ≥ 1 in (8.15). Then
(m) βn+m = βn Qn + m?1 ?=0

(k ) {Qn }∞ k =0

in detail for n ≥ n0 . It turns out that (8.15) admits
(m) (?)

Qn

Qn

fn+? (β ),

m ≥ 1, n ≥ n0 .

(8.16)

Let us now check the equality
(m) ?) Qn = Q( n · Qn+? , (m??)

m ≥ 1, ? = 0, 1, . . . , m ? 1; n ≥ n0 .

(8.17)

We consider two separate cases: 1) ? = 0, m ≥ 1; Qn
(m)

2) ? = 1, . . . , m ? 1;

m ≥ 2.

1) If ? = 0 and m ≥ 1, then (8.8) implies Qn
(m) (?) Qn ?=0

=

(0) Qn

(m) = Qn = Qn+?

(m??)

?=0

.

2) If ? = 1, . . . , m ? 1; m ≥ 2 then (8.8) implies Qn
(m) (?) Qn

Let m ≥ 1. Denote

1 (1 + ηn ) . . . (1 + ηn+??1 ) = (1 + ηn ) . . . (1 + ηn+m?1 ) (1 + ηn+? ) . . . (1 + ηn+m?1 ) 1 (m??) = = Qn+? . (1 + ηn+? ) . . . (1 + ηn+?+(m???1 ) = Qn
(m)

(m) Mn = max

?∈[0,m?1]

(?) Qn

?1 ,

[0, m ? 1] = {0, . . . , m ? 1}.

def

(8.18)

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

31

Clearly, there is ?0 ∈ [0, m ? 1] such that (see (8.17)):
(m) Mn

=

Qm

(m)

(? ) Qn 0

? 1 = Qn+?0 0 ? 1 .

(m?? )

(8.19)

Then from (8.19) and (8.16) it follows that βn+m = βn+?0 +(m??0 ) = =
(m?? ) βn+?0 Qn+?0 0 (m?? ) βn+?0 Qn+?0 0 m??0 ?1 ?=0

+

m??0 ?1 ?=0

Qn+?0 0 Qn+?0
(?)

(m?? )

fn+?0 +? (β )

+

0 Qn+?0 + ? fn+?0 +? (β ).

(m?? ??)

The last inequality implies
(m?? ) βn+?0 [Qn+?0 0

? 1] = βn+m ? βn+?0 ? ?
m??0 ?1 ?=0

m??0 ?1 ?=0

0 [Qn+?0 + ?

(m?? ??)

? 1]fn+?0 +? (β ) (8.20)

fn+?0 +? (β ),

m ≥ 1, n ≥ n0 .

Furthermore, note that (8.17) implies max
(m??0 ??) |Qn+?0 + ?

Let us replace, if necessary, n0 by a bigger number so that |βn | ≥ 2?1 for all n ≥ n0 .
(m??) |Qn+?

?∈[0,m??0 ?1]

? 1| ≤ max

?∈[0,m?1]

? 1| =

s∈[0,m?1]

max

Qn

(m) (m) , ? 1 = Mn

(s) Qn

(8.21)

m ≥ 1, n ≥ n0 . Therefore from (8.20) and (8.21), we get
(m) Mn m??0 ?1 ?=0

= +

(m?? ) |Qn+?0 0 m??0 ?1 ?=0 ∞ k =n

? 1| ≤ 2|βn+m ? βn+?0 | + 2

0 |Qn+?0 + ?

(m?? ??)

? 1| |fn+?0 +? (β )|
(m??0 ?1) Qn+?0 + ? ∞ k =n

|fn+?0 +? (β )| ≤ 4 max |βn+k ? βn | + 2
k ∈[0,m] k ∈[0,m] (m) 2Mn

?∈[0,m??0 ?1] ∞

max

?1

∞ k =n

fk (β )|

+2

|fk (β )| ≤ 4 max |βn+k ? βn | +

k =n

|fk (β )| + 2

|fk (β )|.

Thus for all m ≥ 1 and n ≥ n0 , the following inequality holds:
(m) Mn

≤ 4 max |βn+k ? βn | +
k ∈[0,m]

(m) 2Mn



k =n

|fk (β )| + 2

∞ k =n

|fk (β )|.

(8.22)

From Lemma 7.3 we conclude that the estimate (8.22) can be continued as follows (for all m ≥ 1, n ≥ n0 ) :
(m) (m) Mn ≤ 4 max |βn+k ? βn | + 4Mn Gn β + 4 Gn β . k ∈[0,m]

(8.23)

32

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

n0 (ε) ≥ n0 such that for all n ≥ n0 (ε), the following inequalities hold:
k ∈[0,m]

Let ε be a given positive number. From condition A) (see §7) we conclude that there is max |βn+k ? βn | ≤ ε , 16 4Gn β ≤ min 1 ε , 2 4 . (8.24)

Then from (8.23) and (8.24) for all n ≥ n0 (ε), m ≥ 1, we get
(m) Mn ≤

ε 1 (m) ε (m) + M + ? Mn ≤ ε. 4 2 n 4

(8.25)

By (8.8), (8.19) and (8.25), we now conclude that Qn
(m) (m) (m) ? 1 = |Qn ? 1| ≤ Mn ≤ε (0) Qn

for n ≥ n0 (ε), m ≥ 1.

Now by (8.8), (8.19) and (8.25), we get
(m) |Qn

? 1| =
m

Qn

(m) (n) ≤ ε, ? 1 ≤ Mn

(0) Qn

n ≥ n0 (ε),

m ≥ 1,

or, equivalently, 1 ?1 < ε 1 + ηn+s s=0 for n ≥ n0 (ε), m ≥ 0. (8.26)

for all n ≥ 0. The in?nite product

?1 Lemma 8.1. Let {an }∞ n=0 be a sequence of (maybe complex) numbers such that |an | ≤ 2 ∞

A=

(1 + an )

n=0

converges to a ?nite nonzero number if and only if for every ε > 0 there exists n0 (ε) ? 1 such that for all n ≥ n(ε) and all m ≥ 1, the following inequality holds:
m k =1

(1 + an+k ) ? 1 ≤ ε.

(8.27)

Proof of Lemma 8.1. Necessity. Suppose that the product A converges to a ?nite nonzero number. Denote
n m

An =

(1 + ak ),
k =1

(m) An =

(1 + an+k ),
k =1

m ≥ 1, n ≥ 0.

(8.28)

Then the sequence {An }∞ n=0 converges, and by Cauchy’s criterion for every ε > 0 there is n(ε) such that for all n ≥ n(ε) and m ≥ 1, the following inequality holds: |An+m ? An | ≤ ε |A| 2 ?
(m) |An | |An ? 1| ≤ ε

|A| . 2

(8.29)

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

33

inequality holds:

Let us replace, if necessary, n(ε) by a bigger number so that for all n ≥ n(ε), the following |An | ≥

|A| , n ≥ n(ε). 2 Then from (8.29) for n ≥ n(ε) and m ≥ 1, we get 1 |A| (m) (m) |A| |An ? 1| ≤ |An | |An ? 1| ≤ ε 2 2 ?
(m) |An ? 1| ≤ ε for n ≥ n(ε), m ≥ 1.

Proof of Lemma 8.1. Su?ciency. Suppose that (8.27) holds. Then there is a sequence of 1 2k for all n ≥ nk , m ≥ 1. Consider the value An . Suppose that
(m) |An ? 1| ≤

integers {nk }∞ k =1 such that

(8.30)

nk < n ≤ nk+1 , Then An

k = 1, 2 , . . .

n2 ?n1 ) n3 ?n2 ) n?nk ) . . . A( = An1 · A( · A( n1 n2 nk

? 1 2k ≤ τ < ∞, 1 > 0. τ

n2 ?n1 ) n3 ?n2 ) n?nk ) ? 1) + 1| . . . |(A( ? 1) + 1| |An1 | = |An1 | |(A( ? 1) + 1| |(A( n1 n2 nk

≤ |An1 | 1 +

1 2

1+

1 22

... 1 +

n2 ?n1 ) n3 ?n2 ) n?nk ) |An | = |An1 | |(A( ? 1) + 1| |(A( ? 1) + 1| . . . |(A( ? 1) + 1| n1 n2 nk

≥ |An1 | 1 ? Thus

1 2

1?

1 22

... 1 ?

1 2k



τ ?1 ≤ |An | ≤ τ, that the following inequalities hold:
(m) |An ? 1| ≤

n ≥ 1.

(8.31)

Let τ be the number from inequality (8.31). Let now ε > 0 be given, and choose n(ε) so ε τ

for n ≥ n(ε), m ≥ 1.

(8.32)

Let us verify that then |An+m ? An | ≤ ε We have
(m) |An+m ? An | = |An | |An ? 1| ≤ τ

for all n ≥ n(ε), m ≥ 1.

(8.33)

was to be proved.

Thus the sequence {An }∞ n=1

ε = ε, n ≥ n(ε), m ≥ 1. τ converges, and by (8.31) its limit is ?nite and nonzero, which

34

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

Let us now consider the in?nite product P =
∞ n=n0

(1 + ηn ),

ηn =

Cn+1 , rn un vn+1

n ≥ n0 .

(8.34)

A) it follows that the product P converges to a ?nite nonzero number. Suppose (see (8.8)) that Pn
(m)

Recall (see (8.10) and below) that |ηn | ≤ 2?1 for n ≥ n0 . Let us show that from condition is de?ned by the equality
(m) (m) Pn · Qn = 1,

n ≥ n0 , m ≥ 0.

(8.35)

following inequality holds:

According to (8.26), for any ε > 0 there is n(ε) such that for all n ≥ n(ε) and m ≥ 0, the 1
(m) Pn

?1 < ε

?

(m) (m) | Pn ? 1 | < ε | Pn |. 1 2

(8.36)

Let ε = 1 . Then there is n 2 (8.36)): 1
(m) Pn

1 2

such that for all n ≥ n 1
(m) Pn

and m ≥ 0, we have (in view of 1 1 = 2 2 ?
(m) | Pn |≤2

=

1
(m) Pn

?1+1 ≥1?

?1 ≥1? 1 2

for all n ≥ n n ≥ n(ε) and m ≥ 0

, m ≥ 0.
1 2

Let now ε > 0 be an arbitrary number, and choose m(ε) ≥ n 1
(m) Pn

≥ n0 so that for all (8.37)

?1 <

ε 2

?

ε (m) ε (m) | Pn ? 1 | ≤ | Pn | < · 2 = ε. 2 2

From (8.37) and Lemma 8.1, it follows that the product P (see (8.34)) converges to a ?nite nonzero number. This implies that the series Hn0 (see (5.4)) converges (at least, conditionally). To check that note that the series B=
∞ n=0 2 ηn ,

ηn =

Cn+1 , rn un vn+1

n≥0

absolutely converges. Indeed, from (2.2) and condition A) (see Section 7), we get
∞ n=n0

|ηk |2 =

∞ k =k 0

|Ck+1| rk uk vk+1

2

=

∞ k =n0

1 |Ck+1|2 ≤ G < ∞. rk uk vk+1 rk uk vk+1

(8.38)

Furthermore, since ηn → 0 as n → ∞, there is n0 ? 1 such that for all n ≥ n0 we have ln(1 + ηn ) = ηn + O (|ηn |2 ), n ≥ n0 . (8.39)

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

35

where the constant in O (·) is absolute. Denote
n

n

Sn =
k =n0

(1 + ηk ),

Bn =
k =n0

|ηk |2 ,

n ≥ n0 .

Then by (8.39) we get
n n n

ln Sn =
k =n0

ηk + O
k =n0

|ηk |

2

=
k =n0

ηk + O (Bn ),

n ≥ n0 .

∞ Since the sequences {ln Sn }∞ n=n0 , {Bn }n=n0 in the last equality have ?nite limits, the series

Hn0 (see (5.4)) converges (at least, conditionally). This almost immediately implies that the Cn+1 ? σn un vn , rn un vn+1
m

series J (see (1.7)) converges. Indeed, since Cn+1 ? Cn =
m

n≥0

(8.40)

(see the proof of Lemma 5.1), after adding up the equalities (8.40), we obtain Cm+1 ? Cn0 = Cn+1 τn un vn , ? rn un vn+1 n=n
0

n=n0

m ≥ n0 .

Condition A) (see §7) now implies: ?Cn0 = lim (Cn+1 ? Cn0 ) = Hn0 ? lim
m→∞ m→∞

m

σn un vn ,
n=n0

that is, the series J converges (at least, conditionally), which was to be proved. Proof of Theorem 1.3. Su?ciency. Below we need the following simple auxiliary assertion.

in?nite product P (see (8.34)) converges to a ?nite nonzero limit. Proof. It is known [9, §1.43, Example (1)], that if the series


Lemma 8.2. Suppose that condition B) holds (see §7). Then there is n0 ? 1 such that the

an ,



n=n0

n=n0

|an |2
∞ k =n0

(8.41) (1 + an ), n0 ? 1

converge (the ?rst one at least conditionally), then the in?nite product

B) it follows that the second series in (8.41) converges (see (8.38)), and Lemma 8.1 implies we conclude that |ηn | ≤ 2?1 for n0 ? 1.

converges to a ?nite nonzero limit. In our case, an = ηn , n ≥ 0 (see (8.8)). From condition

that the ?rst series in (8.41) converges. Finally, from the convergence of any series in (8.41)

36

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

Let us introdue an operator T (n0 ) acting in the space of sequences (7.1) by the rule T (n0 ) (θ) = {(T (n0 ) (θ))n }∞ n=n0 , Here (T Pn = and the values
∞ (n0 )

θ < ∞. n ≥ n0 . Ck+1 , rk uk vk+1 k≥n

(8.42)

(θ))n = ?

∞ k =n

Pn fk (θ), Pk ηk =

(8.43) (8.44)

(1 + ηk ),

hold (see (7.15)):

under condition B) there exists n0 ? 1 such that for all n ≥ n0 , the following inequalities |(T (n0 ) θ)n | ≤ τ Gn θ . (8.45)

k =n ∞ {fk (θ)}k=n0

n ≥ n0 ,

are de?ned by relations (7.9), (7.2) and (7.3). Let us verify that

|Pn | ≤ τ |Pk | for all k ≥ n ≥ n0 . Then (7.10) and Lemma 7.3 imply (8.45): | (T
(n0 )

Indeed, if n0 ? 1, by Lemma 8.2 the product P (see (8.34)) converges, and therefore (θ )n | ≤


k =n

Pn |fk (θ)| ≤ τ Pk



k =n

|fk (θ)| = τ Φn (θ) ≤ τ Gn θ .

if necessary, we get the estimate τ Gn0 ≤ 2?1 . Then we ?nally obtain 1 T (n0 ) ≤ . 2

From the proven estimate (8.45), we conclude that T (n0 ) ≤ τ Gn0 . Choosing a bigger n0 , (8.46)

Remark 8.1. Recall that according to the proof of Lemma 7.3 the validity of inequality (7.14) is guaranteed by a choice of n0 such that sup |Cn | ≤ 2?1 . Therefore below we assume n0 chosen big enough so that apart from the above requirements to n0 , the following inequalities hold together: 1 1 τ Gn 0 ≤ An0 ≤ , 4 2 (see Assertion 1.1). In (8.47) τ stands for the number from estimate (8.45). (8.47)
n≥n0

Consider an equation in the space of sequences (7.1) which is essential for that which follows: β = P + T (n0 ) β. (8.48)

operator T (n0 ) is compressing, and therefore equation (8.48) has a unique solution β in the space (7.1), and β ≤ τ P = τ < ∞.

Here n0 is chosen so that the estimate (8.46) holds, P = {Pn }∞ n=n0 (see (8.44)). Clearly, the

(8.49)

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

37

Let us study the properties of the solution β = {βn }∞ n=n0 of equation (8.48). From (8.46) and (8.48), we get β = (E ? T
(n0 ) ?1

) P =P+

∞ k =1

(T (n0 ) )k P

(8.50)

where E is the identity operator. Writing down (8.5) coordinatewise, we get (see Lemma 7.2): βn = Pn +
∞ k =1

[(T (n0 ) )k P ]n ,

n ≥ n0 .

(8.51)

Let us show that βn → 1 as n → ∞. Denote ω=
∞ k =1

(T (n0 ) )k?1P = (E ? T (n0 ) )?1 P

?

ω ≤ τ < ∞.

Then (8.50) takes the form β = P + T (n0 ) ω From (8.45) it follows that |(T (n0 ) ω )n | ≤ τ Gn ω ≤ τ Gn , and therefore
n→∞

?

βn = Pn + (T (n0 ) ω )n ,

n ≥ n0 .

(8.52)

lim βn = lim (Pn + O (Gn )) = lim Pn = 1.
n→∞ n→∞

equation (8.7). From (8.48) and the de?nition of the operator T (n0 ) , we get βn = Pn ? Pn fk (β ), Pk k =n


Let us now check that the solution β = {βn }∞ n=n0 of equation (8.48) is also a solution of

n ≥ n0 .

(8.53)

? = {d ?n }∞ by the equality Let us now de?ne a sequence d n=n0 ?n · Pn βn = d From (8.54) it follows that ?n+1Pn+1 ? d ?n Pn = d ?n+1 Pn+1 ? dn+1 ? fn (β ) Pn βn+1 ? βn = d Pn ?n+1Pn+1 ? d ?n+1 Pn + fn (β ) = d ?n+1 (Pn+1 ? Pn ) + fn (β ) =d ?n+1Pn+1 (1 ? 1 ? ηn ) + fn (β ) = ?ηn βn+1 + fn =d Cn+1 1 ? ?βn + βk+1 = rn un vn+1 rn vn un+1
∞ k =1

?

?n = 1 ? d

1 fk (β ), Pk k =n



n ≥ n0 .

(8.54)

(?1)k (Ak g (β ))n,

n ≥ n0 .

(8.55)

38

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

From (8.55) we immediately obtain vn un+1 rn vn un+1 ?βn = ? Cn+1 βn+1 + (?1)k (Ak g (β ))n , un vn+1 k =1 Then from (8.47), Assertion 1.1, (7.3), (7.5) and (2.2) we get rn vn un+1 |?βn | ≤ un+1 vn |Cn+1 | |βn+1 | + un vn+1
∞ k =1 ∞

n ≥ n0 .

(8.56)

|(Ak g (β ))n | ≤ τ An +

∞ k =1

(2An )k τ ≤ τ An ,

An → 0 as n → ∞, or, ?nally,
n→∞ ∞ k =1

lim rn vn un+1 ?βn = 0.

(8.57)

Let us write down the system of equalities (8.56) in the vector form ? = g (β ) + (?1)k Ak (g (β )) = (E + A)?1g (β ). (8.58)

the operator A is de?ned according to (7.3) and (7.2). From (8.58) it follows that ? = g (β ) ? A?. Equality (8.59) written in the coordinate form looks as follows: un+1 vn vn ?n = ? Cn+1 βn+1 ? un vn+1 un Ck+1?k , rk vk vk+1 k =n+1
∞ ∞

In (8.58), E is the identity operator, ? = {?n }∞ n=n0 , ?n = rn un vn+1 ?βn , n ≥ n0 , g (β ) and (8.59)

n ≥ 0.

(8.60)

Let us plug the values ?n , n ≥ n0 and Cn , n ≥ n0 (see (1.8)) into (8.60). We obtain (6.2): rn un un+1 ?βn = ?βn+1


σk u2 k

k =n+1

?



?βk

σm u2 m,

k =n+1

m=k +1

n ≥ n0 .

Thus, under the hypotheses of Theorem 1.3 there exists a solution of problem (6.2)–(6.3), and therefore problem (1.4)–(1.6) is solvable (see Lemma 6.2). Since the series G converges by assumption, from Theorem 1.4 it follows that the narrow Hartman-Wintner problem is also solvable. Proof of Corollary 1.3.1. I) Suppose that the series J (see (1.7)) converges (at least, conditionally). From (1.15) and (1.9) it follows that
∞ n=0 ∞ n=0 ∞ n=0

| Re(Jn+1 C n+1 )| ≤ rn un vn+1

|Jn+1 | |Cn+1 | ≤ rn un vn+1

An+1 |Cn+1 | . rn un vn+1

Therefore, if any of the ?rst two inequalities of (1.18) holds, the series L (see (1.15)) absolutely converges, and by Theorem 1.3 problem (1.4)–(1.?) is solvable. Furthermore, if the

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

39

third series in (1.18) converges, by Lemma 5.2 the series G also converges, and problem (1.4)–(1.?) is solvable in view of Theorem 1.3. Proof of Theorem 1.1. This statement immediately follows from the proven part I) of Corollary 3.1.1. Here assumption (1.11) is not used and is super?uous. II) If the series J (see (1.7)) and P (see (1.16)) converge (at least, conditionally), by Lemma 1.1 the seres G (see (1.14)) also converges, and then by Theorem 1.3 problem (1.4)– (1.?) is solvable. III) If the series J (see (1.7)) and the series (1.19) converge, then the series P absolutely IV) If the series J absolutely converges, inequality (1.19) holds, and problem (1.4)–(1.?)

converges, and problem (1.4)–(1.?) is solvable by part II) of this corollary. is solvable by part III) of this corollary.

9. Examples In this section we present three examples illustrating the power of Theorems 1.2, 1.3 and 1.4 and the simplest tools for their application. For the reader’s convenience and for brevity we analyze all the examples according to a common scheme consisting of steps I-IV. I. Study of the FSS {un , vn }∞ n=n0 of the basic equation (1.2). In the examples below, the FSS of equation (1.2) can be easily found in an explicit form.

To apply Theorems 1.2, 1.3 and 1.4 in these examples, it is enough to have sharp by order estimates of the FSS. Here such estimates can be obtained by standard methods without any we do not prove them. di?culties, using the classic Cauchy-Maclaurain Theorem [5, vol. II, §2.373], and therefore

II. Study of necessary conditions for the solvability of problem (1.4)–(1.6). In this step such conditions are obtained with the help of Theorem 1.2. More precisely, we ?nd conditions for the convergence of the series σ (see (1.12)), present a priori sharp by conditions guaranteeing equality (1.13). order estimates of the terms of the sequence {Cn }∞ n=n0 (see (1.8)), and, as a corollary, precise

III. Study of the condition for the coincidence of problems (1.4)–(1.6) and (1.4)–(1.?). the series G (see (1.14)). IV. Study of conditions for the solvability of problem (1.4)–(1.?).

In this step, following Theorem 1.4, we establish precise conditions for the convergence of

In this step, following Theorem 1.3, we ?nd precise conditions for the convergence of the

series J (see (1.7)); conditions for the convergence of the series G are given in step III.

40

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

Note that taking into account the above description, below we give no further comments on the problems arising at each step. The results obtained in each step of the above program and the corresponding conclusions are given in the course of the exposition. Finally, throughout the sequel the notation ?n ? ψn , n ≥ n0 means that the following inequalities hold: τ ?1 |?n | ≤ |ψn | ≤ τ |?n |, n ≥ n0 .

In the following example, we describe the situation where problem (1.4)–(1.6) and (1.4)– our theorems contains inevitable gaps. (1.?) do not coincide, and therefore the analysis of the Hartman-Wintner problem based on

Example 9.1. Consider problem (1.4)–(1.6) for the equations ?(e?n ?yn ) = (n + 1)γ e?(n+1) yn+1 , ?(e?n ?zn ) = 0 · zn+1 , n ≥ 0. n≥0 (9.1) (9.2)

In connection with Example 9.1, we have the following result. Assertion 9.1. For equations (9.1)–(9.2), problem (1.4)–(1.6) 2) is solvable for γ ∈ (?∞, ?1). 1) is not solvable for γ ≥ 0 and γ ∈ ?1, ?
1 2

;

, 0 , problem (1.4)–(1.6) is not equivalent to problem (1.4)–(1.?), Remark 9.1. For γ ∈ ? 1 2 and therefore the problem on solvability of problem (1.4)–(1.6) remains open in this case.

(1.3)). By formula (3.7), we ?nd a non-principal solution {vn }∞ n=1 of equation (9.2):
n

Proof of Assertion 9.1. I. The sequence un ≡ 1, n ≥ 0, is a principal solution of (9.2) (see vn+1 = un+1
k =0

1 = rk uk uk+1
∞ n=1

n

k =0

ek ? en+1 ,

n ≥ 0.

II. The series σ is of the form (see (1.12)): σ= σn u2 n =
∞ n=1

nγ e?n

and therefore converges for every γ ∈ R. Consider Cn , n ≥ 1 (see (1.8)) vn Cn = un


σn u2 n

k =n

?e

n



k =n

k γ e?k ? nγ

(9.3)

is not solvable.

Hence Cn → 0 as n → ∞ if and only if γ < 0. This implies that for γ ≥ 0 problem (1.4)–(1.6)

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

41

III. The study of the series G (see (1.14)) is based on the application of (9.3): G=
∞ n=1

|Cn+1|2 ? rn un vn+1

∞ n=1

n2γ .

Hence the series G converge if and only if 2γ < ?1. and for γ ∈ ?∞, ? We conclude that for γ ∈ ?
1 2 1 2

these problems coincide.

, 0 problems (1.4)–(1.6) and (1.4)–(1.?) are not equivalent,
∞ n=1

IV. By the results of Step I, we have (see (1.7)) J =

problem (1.4)–(1.?) is solvable.

problem (1.4)–(1.?) is not solvable ), and for γ < ?1 it is solvable (since the equivalent

We conclude that for γ ∈ ?1, 1 problem (1.4)–(1.6) is not solvable (since the equivalent 2

σn un vn ?

∞ n=1

nγ .

coincide provided at least one of them is solvable. Thus it turns out that problem (1.4)–(1.6) has been fully investigated because of the study of problem (1.4)–(1.?). Example 9.2. Consider the problem (1.4)-(1.6) for the equations 1 ?(nα ?yn ) = yn+1 , n≥1 (n + 1)β ?(nα ?zn ) = 0 · zn+1 , Here α ≥ 0, β ∈ R. n ≥ 1.

In the next example we consider a situation where problems (1.4)–(1.6) and (1.4)–(1.?)

(9.4) (9.5)

In connection with Example 9.2, we have the following result. Assertion 9.2. Problem (1.4)–(1.6) for equations (9.4)–(9.5) is solvable if and only if α + β > 2. (9.6)

Proof of Assertion 9.2. Throughout the sequel, each of the steps I–IV is subdivided into is necesary because the form of the FSS {un , vn }∞ n=1 of equation (9.5) changes with changing I.1) α ∈ [0, 1). three substeps according to the value of α : α ∈ [0, 1), α = 1, α ∈ (1, ∞). Such a subdivision

the parameter α ∈ [0, ∞).

vn , n = 1, 2, . . . is of the form

The sequence un ≡ 1, n ≥ 0 is the principal solution of (9.5). The non-principal solution
n

vn+1 = un+1
k =1

1 = rk uk uk+1

n

k =1

1 ? (n + 1)1?α . kα

I.2) α = 1.

42

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

vn , n = 1, 2, . . . is of the form

The sequence un ≡ 1, n ≥ 0 is the principal solution of (9.5). The non-principal solution
n

vn+1 = un+1
k =1

1 = rk uk uk+1

n

k =1

1 ? ln(n + 1). k

I.3) α > 1. un , n ≥ 1 is of the form (see (2.1)): The sequence vn ≡ 1, n ≥ 0 is the non-principal solution of (9.5), the principal solution 1 un = vn = r k vk vk +1 k =n II.1) α ∈ [0, 1). In this case the series σ is of the form σ=
∞ n=1 ∞

1 1 ? α?1 . α k n k =n



σn u2 n

=

∞ n=1

1 <∞ nβ

?

β > 1.

Consider the sequence Cn , n ≥ 1. We get vn Cn = un


σn u2 k

k =n

?n

1?α

∞ k =n

1 n1?α ? = n2?α?β . kβ nβ ?1

conditions α ∈ [0, 1), α + β > 2. II.2) α = 1.

obtained necessary conditions α ∈ [0, 1), β > 1, α + β > 2 is equivalent to the collection of

Therefore Cn → 0 as n → ∞

?

α + β > 2. The collection of the given and the

In this case the series σ is of the form σ=
∞ k =1

σk u2 k =

∞ k =1

1 <∞ kβ

?

β > 1.

Consider the sequence Cn , n ≥ 1. We get vn Cn = un


σk u2 k

k =n

?



k =n

1 ln n ? β ?1 → 0 β k n

as n → 0,

since β > 1. Thus the collection of the given and obtained neccessary conditions α = 1, β > 1 is equivalent to the collection of conditions α = 1, α + β > 2. II.3) α > 1. In this case the series σ is of the form σ=
∞ n=1 ∞ n=1

σn u2 n

?

1 1 <∞ β 2 n n α?2

?

β > 3 ? 2α.

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

43

Consider the sequence Cn , n ≥ 1. We get Cn = as n → ∞ ? vn un
∞ k =n α?1 σk u2 k ? n ∞ k =n

1 nα?1 1 ? ? n2?α?β → 0 k β k 2α?2 nβ +2α?3

collection of necessary conditions in this case: α > 1, α + β > 2. less of the value of α ∈ [0, ∞). III.1) α ∈ [0, 1). G= III.2) α = 1. In this case we get the series G in the form G=
∞ n=1 ∞ n=1

α + β > 2. Since for α > 1 we have β > 2 ? α > 3 ? 2α, we ?nally get the

We conclude that if problem (1.4)–(1.6) is solvable, then condition (9.6) must hold regard-

In this case we get the series G in the form |Cn+1|2 ? rn un vn+1
∞ n=1

n4?2α?2β = nα · 1 · n1?α

∞ n=1

1 n2α+2β ?3

<∞

?

α + β > 2.

|Cn+1 |2 ? rn un vn+1

∞ n=1

ln n nβ ?1

2

1 = n ln n

∞ n=1

ln n <∞ n2β ?1

?

β > 1.

Here the conditions α = 1, β > 1 are equivalent to the conditions α = 1, α + β > 2. III.3) α > 1. In this case we get the series G in the form G=
∞ n=1

|Cn+1|2 ? rn un vn+1

∞ n=1

1 n2β +2α?4

1 1 · = nβ nα?1

∞ n=1

1 n3α+3β ?5

<∞

?

α + β > 2.

solvable. Here condition (9.6) is a necessary condition for solvability of problem (1.4)–(1.?). IV.1) α ∈ [0, 1). In this case the series J is of the form ∞ ∞ n1?α J= σn un vn ? = nβ n=1 n=1

We conclude that problem (1.4)–(1.6) is solvable if and only if problem (1.4)–(1.?) is

∞ n=1

1 nα+β ?1

<∞

?

α + β > 2.

IV.2) α = 1. In this case the series J is of the form ∞ ∞ ln n σn un vn ? J= <∞ nβ n=1 n=1

?

β>1

?

α + β > 2.

44

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

IV.3) α > 1. In this case the series J is of the form ∞ ∞ 1 1 J= σn un vn ? · α?1 = β n n n=1 n=1 (9.6) holds.
∞ In the next example the perturbation {σn }∞ n=1 is oscillating, and its absolute value {σn }n=1 ∞ n=1

1 nα+β ?1

<∞

?

α + β > 2.

We conclude that the narrow Hartman-Wintner problem is solvable if and only if condition

coincides with the perturbation from Example 9.2. Thus we show that the exact account of the oscillation of the perturbation allows one to signi?cantly weaken requirements of solvability of problem (1.4)–(1.6). Example 9.3. Consider the problem (1.4)–(1.6) for the equations (?1)n ?(nα ?yn ) = yn+1 , n≥1 (n + 1)β ?(nα ?zn ) = 0 · zn+1 , n ≥ 1. Here α ∈ [0, 1), β ∈ R. In connection with Example 9.3 we have the following result.

(9.7) (9.8)

Assertion 9.3. The problem (1.4)–(1.6) for equations (9.7)–(9.8) is solvable if and only if α + β > 1, β > 0. (9.9)

is not specially mentioned. vn , n ≥ 1, is of the form

Proof of Assertion 9.3. Below in all the steps I–IV, we assume that α ∈ [0, 1) even when it

I. The sequence un ≡ 1, n ≥ 0, is the principal solution of (9.8). The non-principal solution
n

vn+1 = un+1
k =1

1 = rk uk uk+1

n

k =1

1 ? (n + 1)1?α . kα

(9.10)

II. In this case the series σ is of the form σ=
∞ n=1

σn u2 n

=

∞ n=1

(?1)n . (k + 1)β

From the necessary condition for convergence of the series and Leibnitz’s theorem, it follows that the series σ converges if and only if β > 0. Consider the sequence Cn , n ≥ 1. We get vn Cn = un


σk u2 k

k =n

?n

1?α

Rn (β ),

Rn (β ) =

def

(?1)k . β ( k + 1) k =n



NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

45

Below we need the following lemma. Lemma 9.1. For any β > 0 there exists n0 (β ) such that for all n ≥ n0 (β ), we have Rn (β ) ? n?β . Proof. Let m be any even number. Then (?1)n Rn (β ) = = 1 1 1 1 ? +···+ ? + (?1)n Rn+m (β ) β β β (n + 1) (n + 2) (n + m ? 1) (n + m)β
β

(9.11)

1 1 1? 1? β (n + 1) n+2

+···+

1 1 1? 1? β (n + m ? 1) n+m 1 k2

β

+ (?1)n Rn+m (β ).

By Taylor’s formula, for any β > 0 there exists n0 (β ) such that for all k ≥ n0 (β ), we have 1 1? k tion:
β

=1?

β +O k

,

k ≥ n0 (β ).

Here the constant O (·) is absolute and depends only on β. Let us now continue the calcula(?1)n Rn (β ) = β
m?1 k =1

1 +O (n + k )β (n + k + 1)

m?1 k =1

1 1 +(?1)n Rn+m (β ). (n + k )β (n + k + 1)2

All three summands in the last equality have ?nite limits as m → ∞. Hence (?1) Rn (β ) = β
n ∞ k =1

1 +O β (n + k ) (n + k + 1)



1 (n + k )β (n + k + 1)2

.

k =1

The obtained equality and Cauchy-Maclaurin’s theorem imply (see [5, vol.2, §2.373]) imply (9.11) Thus for n ≥ n0 (β ), we obtain for the value of Cn : Cn ? n1?α Rn (β ) ? n1?α?β → 0 as n → ∞ ? α + β > 1.

Thus if problem (1.4)–(1.6) is solvable, (9.9) holds. III. In this case, for a given β > 0, for n ≥ n0 (β ) (see Lemma 9.1) we get (see (7.15)): G n 0 (β ) = =
∞ n=n0 (β ) ∞ n=n0 (β )

|Cn+1 |2 ? rn un vn+1 1 n2α+2β ?1



1 n2α+2β ?2 nα

n=n0 (β )

1 · n1?α

<∞

?

α + β > 1.

and condition (9.9) is necessary for solvability of problem (1.4)–(1.6).

We conclude that problem (1.4)–(1.6) is solvable if and only if problem (1.4)–(1.?) is solvable,

46

N.A. CHERNYAVSKAYA AND L.A. SHUSTER

IV. In this case, for a given β > 0 and the number n0 (β ) (see Lemma 9.1) we get (see (1.9), (9.10)): J n 0 (β ) =
∞ n=n0 (β ) ∞ n=n0 (β )

σn un vn =

(?1)n (n + 1)β

n?1 k =1

1 . kα

Note that under condition (9.9) we have
n→∞

lim Rn (β )vn = 0.

Indeed, from (9.10) and (9.11) it follows (for n ≥ n0 (β )) that Rn (β )vn ? n1?α?β → 0 Let us now return to Jn0 (β ) : Jn0 (β ) =
∞ n=n0 (β )

as n → ∞
∞ n=n0

and

α + β > 1.

(Rn (β ) ? Rn+1 (β ))vn =
∞ n=n0 (β )

[Rn (β )vn ? Rn+1 (β )vn+1 + Rn+1 (β )(vn+1 ? vn )] (9.12)

= Rn0 (β ) vn0 (β ) +

Rn+1 (β ) . nα

But from (9.9) and (9.11) it follows that the series in the right-hand side of (9.12) absolutely converges:
∞ n=n0 (β )

|Rn+1 (β )| ? nα

∞ n=n0 (β )

1 nα+β

< ∞.

We conclude that problem (1.4)–(1.?) is solvable if (9.9) holds. 1?α < β ≤ We can now prove the assertion stated in Section 1 (see Example 1.1). Let α ∈ [0, 1) and
3 2

diverges. Indeed, from (9.10) it follows that B= for 2α + 2β ? 2 ≤ 1 ?
∞ n=1

3 (9.8) is solvable. Let us verify that if, in addition, α + β ≤ 2 , then the series B (see (1.17)) ∞ n=1

? α. Then β > 0, α + β > 1 and problem (1.4)–(1.6) for equations (9.7) and

[|σn |un vn ] ? α+β ≤
3 , 2

2

n2?2α = n2β

∞ n=1

1 n2β +2α?2

=∞

which was to be proved. References

[1] Ravi P. Agarwal, Di?erence Equations and Inequalities, Dekker, New York, 2000. [2] S. Chen, Asymptotic integration of nonoscillatory second order di?erential equations, Trans. Amer. Math. Soc. 327 (2) (1991). [3] N. Chernyavskaya and L. Shuster, Necessary and su?cient conditions for the solvability of a problem of Hartman and Wintner, Proc. Amer. Math. Soc. 125 (11) (1997), 3213-3228. [4] G.M. Fikhtengol’ts, A Course in Mathematical Analysis, vol. I, Moscow, Nauva, 1969 (Russian). [5] G.M. Fikhtengol’ts, A Course in Mathematical Analysis, vol. II, Moscow, Nauva, 1969 (Russian). [6] P. Hartman, Ordinary Di?erential Equations, Wiley, New York, 1964. [7] P. Hartman and A. Wintner, On non-oscillatory linear equations, Amer. J. Math. 75 (1953), 717-730.

NECESSARY AND SUFFICIENT CONDITIONS FOR SOLVABILITY

47

ˇ sa, Asymptotic integration of a second order ordinary di?erential equation, Proc. Amer. Math. [8] J. Simˇ Soc. 75 (1) (1987), 96-100. [9] C.C. Titchmarsh, The Theory of Functions, Oxford University Press, 1939. [10] W.F. Trench, Linear perturbations of a non-oscillatory second order equation, Proc. Amer. Math. Soc. 97 (3) (2001), 423-428. [11] W.F. Trench, Linear perturbations of a non-oscillatory second order di?erence equation, Jour. Math. Anal. Appl. 255 (2001), 627-635.



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