Ukrainian Mathematical Journal, Vol. 61, No. 3, 2009

ON THE COMPLETENESS OF ALGEBRAIC POLYNOMIALS IN THE SPACES Lp (R, d μ ) A. G. Bakan

UDC 517.538 0

We prove that the theorem on the incompleteness of polynomials in the space Cw established by de Branges in 1959 is not true for the space L p ( R, d μ ) if the support of the measure μ is sufficiently dense.

1. Preliminary Information and Main Result 1.1. De Branges Theorem. Let M+ (R ) be the cone of finite Borel measures on the real axis R , let M∗ (R ) be the set of measures μ ∈M+ (R ) with all finite moments sn (μ ) : =

∫x

n

d μ( x ) ,

n ∈ N0 : = {0, 1, 2, … } ,

R

and unbounded support supp μ : = { x ∈ R ∀ ε > 0 : μ ( x − ε, x + ε) > 0} , let M∗ (R + ) be the set of measures μ ∈M∗ (R ) for which supp μ ⊂ R + : = [ 0, + ∞ ), let C(R ) be the linear space of all functions real-valued and continuous on R, let P be the set of all algebraic polynomials with real coefficients, let P s [ D] be the set of polynomials from P all roots of which are simple and belong to the set D ⊂ R, let B(R ) be the family of Borel subsets of R, let P be the collection of linear topological spaces of real-valued functions on R for which P is a dense subset, and let W ∗ (R ) be the set of upper-semicontinuous functions w : R → R + such that f

xn

w

w

< ∞ for all n ∈ N0 , where

: = sup w( x ) f ( x ) . x ∈R

For w ∈W ∗ (R ) , the space Cw0 is defined as the linear set of all functions f ∈C (R ) for which lim w( x ) f ( x ) = 0

x →∞

Institute of Mathematics, Ukrainian National Academy of Sciences, Kiev, Ukraine. Translated from Ukrains’kyi Matematychnyi Zhurnal, Vol. 61, No. 3, pp. 291–301, March, 2009. Original article submitted June 24, 2008. 0041–5995/09/6103–0347

© 2009

Springer Science+Business Media, Inc.

347

348

A. G. B AKAN

equipped with seminorm ⋅ w . In this sense, the functions w ∈W ∗ (R ) are also called weights. If μ ∈M+ (R ) and D ∈B(R ) , then the measure μD defined on the Borel σ-algebra B(R ) by the relation μD ( A) : = μ ( A ∩ D), A ∈B(R ) , is called the restriction of the measure μ to the set D. For an entire function f, we define M f (r ) : = sup f ( z ) , z =r

r ≥ 0.

Let Λ f denote the set of all zeros of f . We say that f is of minimal exponential type if lim sup r −1 log M f (r ) = 0. r →+∞

Let E0S [ D] denote the set of entire transcendental functions f of minimal exponential type that are real on the real axis and all roots of which are simple and belong to the set D ⊂ R. Let E0H [ D] be the set of f ∈E0S [ D] such that

lim

λ →∞, λ ∈Λ f

λm = 0 f ′(λ )

for any m ∈ N0 .

In 1924, Bernstein [1] formulated the problem of finding conditions for a weight w ∈W ∗ (R ) under which algebraic polynomials are dense in the space Cw0 . This problem was solved in 1959 by de Branges [2] (see also [3 – 5]). In 1996, Sodin and Yuditskii [6] modified this solution, which can be formulated as the following theorem: Theorem A. Suppose that the set S w : = { x ∈ R w( x ) > 0} is unbounded for w ∈W ∗ (R ) . The algebraic polynomials P are dense in

Cw0 if and only if the following relation holds for any function

B ∈E0H [ Sw ] :

∑

λ ∈Λ B

1 = + ∞. w ( λ ) B ′( λ )

Consider condition (1) in more detail. If f ∈E 0S [ S w ] \ E 0H [ S w ], then there exists m0 ∈N0 such that

lim sup

λ →∞, λ ∈Λ f

λ m0 > 0. f ′(λ )

Therefore, it follows from the definition of the class W ∗ (R ) that lim inf

λ →∞, λ ∈Λ f

w(λ ) f ′(λ ) = 0,

(1)

ON THE COMPLETENESS OF ALGEBRAIC POLYNOMIALS IN THE SPACES L p (R, d μ )

349

i.e., equality (1) is true for B = f . Thus, the condition B ∈E0H [ Sw ] in Theorem A can be replaced by the equivalent condition B ∈E0S [ Sw ]. Assume that the set Sw is discrete and there exists a function E ∈E0S [ R ] such that Sw = Λ E . Then it follows from relation (1) for B = E that, for any entire function G that satisfies the condition E / G ∈P , one has G ∈E0S [ R ] , and condition (1) is satisfied for B = G. Indeed, since E (z ) / G (z ) = P (z ) = p0 z n + … + pn , where n ∈ N , pk ∈ R , 0 ≤ k ≤ n, p0 ≠ 0, we have G (z ) = E (z ) / P (z ) and there exists a number R ≥ 1 such that P (z ) ≥ 2 −1 p0 z Therefore, log MG (r ) ≤ log M E (r ) − log p0 / 2 − n log r

n

for any z ∈C,

z ≥ R.

for r ≥ R,

and, hence, G is of minimal exponential type. Thus, G ∈E0S [ R ] . Furthermore, since there are finitely many zeros of the function E inside the disk UR : = { z ∈C z ≤ R } and Λ P ⊂ UR , we can conclude that ∞ =

∑

λ ∈Λ E \ U R

≤

1 = w(λ ) E ′(λ )

∑

λ ∈Λ G \ U R

2 ′ λ ∈Λ G \ U R w(λ ) G (λ ) p0 λ

∑

n

1 w(λ ) G ′(λ ) P (λ ) 2 p0

≤

∑

λ ∈Λ G \ U R

2 ≤ w(λ ) G ′(λ )

2 p0

∑

λ ∈Λ G

2 , w(λ ) G ′(λ )

i.e., condition (1) is satisfied for B = G. This means that, for C w0 ∈P , it is necessary to require that condition (1) be satisfied for all entire functions from the set D0 [ Λ E ] , where D0 [ D] : =

{ g ∈E0S [ D]

}

card ( D \ Λ g ) = ∞ ,

D ⊂ R,

(2)

and card A ∈ N0 ∪ {∞} denotes the number of elements of the set A. The set D0 [ Λ E ] possesses the special property g ∈D0 [ Λ E ] ⇒ P (z ) g(z ) ∈D0 [ Λ E ]

∀ P ∈P S [ ΛE \ Λg ] ,

(3)

and, furthermore, the degrees of polynomials from the set P S [ Λ E \ Λ g ] are not bounded from above. Therefore, it follows from relation (1) for B = g ∈D0 [ Λ E ] that 1

∑ m λ ∈Λ (1 + λ ) w(λ ) g

g′ ( λ )

= +∞

∀ m ∈ N0 ,

which, by virtue of the inequality

∑

λ ∈Λ g

1

(1 +

λ

)1+ ε

< ∞,

ε > 0,

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A. G. B AKAN

is equivalent to lim inf

λ ∈Λg , λ → ∞

λ

m

w(λ ) g′(λ ) = 0

∀ m ∈ N0 ,

or, which is the same, log (w(λ ) g′(λ ) λ →∞ log λ

lim inf

λ ∈Λg ,

)

= – ∞.

(4)

If Sw does not coincide with the set of zeros of any function of the set E0S [ R ] , then the set Sw \ Λ B is infinite for any function B ∈ E0S [ Sw ] . Therefore, in this case, we have B ∈E0S [ Sw ] ⇒ B ∈D0 [ Sw ] ⇒ PB ∈D0 [ Sw ]

∀ P ∈ P S [ Sw \ Λ B ] ,

(5)

whence, as above, we can conclude that condition (1) can be replaced by the equivalent condition (4) with g = B . Thus, we have proved the following modified version of Theorem A: Theorem B. Suppose that the set Sw : = { x ∈ R w( x ) > 0} is unbounded for w ∈W ∗ (R ) . The algebraic polynomials P are dense in Cw0 if and only if the following conditions are satisfied: log (a) lim sup λ ∈ΛF λ →∞

(b)

∑

λ ∈Λ E

1 − log F ′(λ ) w(λ ) log λ

= +∞

∀ F ∈ D0 [ Sw ],

1 = ∞, provided that there exists a function E ∈ E0S [ R ] such that Sw = Λ E . ′ w(λ ) E (λ )

1.2. Main Result. For any weight w ∈W ∗ (R ) with unbounded Sw , by virtue of Theorem A we establish that Cw0 ∉P ⇔ ∃ B ∈E0S [ Sw ]:

Cw0 χ Λ ∉P , B

(6)

where χ D denotes the indicator function of a set D ⊂ R . In other words, the incompleteness of algebraic polynomials in the space Cw0 leads to their incompleteness on the restriction of this space to a sufficiently sparse set that is the set of all zeros of a certain entire function of minimal exponential type such that all its simple zeros belong to the set Sw . In 1998, Borichev and Sodin [7] established an analogous property for the spaces L p (R, d μ ) in the case where the measure μ is discrete and its support satisfies the condition ∃ β > 0 : card ( [− r, r ] ∩ supp μ ) = O (r β ),

r → + ∞.

Theorem A and Proposition A1.5 in [7, pp. 225, 255] yield the following corollary:

(7)

ON THE COMPLETENESS OF ALGEBRAIC POLYNOMIALS IN THE SPACES L p (R, d μ )

351

Corollary A. Suppose that 1 ≤ p < ∞, a measure μ ∈M∗ (R ) is discrete, and its support supp μ satisfies condition (7). The algebraic polynomials P are not dense in the space L p (R, d μ ) if and only if there exists an entire function E ∈E0H [supp μ ] such that the algebraic polynomials P are not dense in the space L p (R, d μΛ E ). Let us show how this result can be obtained from the following theorem proved in [8, p. 38] (Theorem 2.1): Theorem C. Let μ ∈M∗ (R ) and 1 ≤ p < ∞. The algebraic polynomials P are dense in the space L p (R, d μ ) if and only if there exist a measure ν ∈M+ (R ) and a weight

w ∈W ∗ (R ) such that Cw0 ∈P

and dμ = w p dν, i.e.,

∫ w( x )

μ ( A) =

p

d ν( x )

A

for any set A ∈B (R ) . Consider an arbitrary discrete measure μ ∈M∗ (R ) defined by the expression d μ( x ) =

∑

λ ∈Sμ

μλ δ( x − λ) ,

x ∈ R,

where the set Sμ : = supp μ is countable and unbounded. If 1 ≤ p < ∞ , then, by virtue of Theorem C, L p (R, d μ ) ∈P if and only if there exist a measure ν ∈M+ (R ) and a weight w ∈W ∗ (R ) such that Cw0 ∈P and

∑

d ν( x ) =

λ ∈Sμ

∑

λ ∈Sμ

νλ δ( x − λ ) ,

w( x ) : =

μ λ = wλp νλ ,

νλ < ∞ ,

∑

λ ∈Sμ

wλ χ{λ } ( x ) ,

wλ , μλ > 0,

x ∈ R,

λ ∈Sμ .

Applying Theorem B to the weight w, we get L p (R, d μ ) ∈P

log lim sup λ ∈ΛF λ →∞

∑

⇔

∃ {νλ }λ ∈Sμ := supp μ ⊂ ( 0, + ∞ ) :

1 1 − log − p log F ′(λ ) μλ νλ log λ 1/ p

νλ

1/ p λ ∈Sμ μ λ

E ′(λ )

= +∞

if

= +∞

∑

λ ∈Sμ

νλ < ∞ ,

∀ F ∈ D0 [ Sμ ] ,

∃E ∈E0S [ R ] : Sμ = Λ E .

(8.1)

(8.2)

(8.3)

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A. G. B AKAN

Note that, in these conditions, we can require the existence of summable positive sequences in (8.2) and (8.3) separately because if two positive summable sequences {νλj }λ ∈Sμ satisfy conditions (8.2) and (8.3), then the sequence νλ : = max {νbλ , νλc } , λ ∈Sμ , satisfies conditions (8.1) – (8.3). Moreover, by virtue of the known properties of sequences from l p (see [9], Chap. 4, Sec. 4.4 and Chap. 7, Sec. 1, Theorem 1), we have

∑

λ ∈Sμ

∑

λ ∈Sμ

1/ p

νλ

1/ p

μλ

νλ < ∞ ⇔

E′(λ )

< ∞

∑

1/( p −1) λ ∈Sμ μ λ

∀{νλ }λ ∈Sμ ⊂ ( 0, + ∞ ) ,

1 E ′(λ )

p /( p −1)

< ∞

if 1 < p < ∞,

and lim inf μ λ E ′(λ ) λ ∈ΛE λ →∞

> 0

if p = 1,

and, hence, the left-hand side of condition (8.3) can be replaced by the equalities

∑

1/( p −1) λ ∈Sμ μ λ

1 E ′(λ )

p /( p −1)

lim inf μ λ E ′(λ ) λ ∈ΛE λ →∞

= ∞

= 0

if 1 < p < ∞,

if p = 1,

(8.4)

which are already independent of the sequence {νλ }λ ∈Sμ . Using conditions (8.1) and (8.2), we establish that lim

λ ∈Sμ , λ →∞

ν λ = 0.

We can now replace the condition νλ > 0 in (8.1) by the equivalent condition νλ ≥ γ λ with any fixed positive summable sequence {γ λ }λ ∈Sμ . If the measure μ satisfies condition (7), then we can set γ λ = (1 + λ

)−β −1.

Then, for sufficiently large λ ∈Λ F , we get 0 > – log

1 1 ≥ – log = – (1 + β) log (1 + λ ) . νλ γλ

Substituting these inequalities into condition (8.2), we establish that it does not depend on the choice of the sequence {νλ }λ ∈Sμ . We can now formulate a theorem that is a modification of Theorem A and Proposition A1.5 in [7, pp. 225, 255].

ON THE COMPLETENESS OF ALGEBRAIC POLYNOMIALS IN THE SPACES L p (R, d μ )

353

Theorem D. Suppose that 1 ≤ p < ∞, a measure μ ∈M∗ (R ) is discrete, and its support Sμ : = supp μ satisfies condition (7). The algebraic polynomials P are dense in the space L p (R, d μ ) if and only if log lim sup λ ∈ΛF λ →∞

1 − p log F ′(λ ) μλ log λ

= +∞

∀ F ∈ D0 [ Sμ ] ,

(9)

and the corresponding equality (8.4) holds if there exists an entire function E ∈E0S [ R ] such that Λ E = Sμ . Corollary A obviously follows from Theorem D. If the discrete measure μ ∈M∗ (R ) does not satisfy condition (7), then condition (8.3) is not required, and, hence, the criterion L p (R, d μ ) ∈P includes only conditions (8.1) and (8.2). The theorem presented below is the main result of this paper. According to this theorem, Corollary A, generally speaking, cannot be generalized to measures that do not satisfy condition (7). Therefore, Corollary A is no longer true in the spaces L p (R, d μ ) , 2 ≤ p < ∞, for discrete measures with denser support. To formulate the theorem, we introduce new notation. For a countable set A ⊂ ( 0, + ∞ ) and the function 1 32 π

ψ (x ) : =

∑

m ≥1

( 2 π )m e − m x

1/ 4 m

,

m m!

x ≥ 0,

(10)

x ∈ R,

(11)

we define a discrete measure μ A by the relation d μ A (x ) : =

∑

λ e − λ ψ (λ ) δ( x − λ ),

λ ∈A

and denote n A (r ) : = card { λ ∈ A λ ≤ r }, r ≥ 0. Theorem 1. Suppose that L : =

{ log k }k ≥ 2 ,

A ⊂ L , and a function ψ and a measure μ A are de-

fined by relations (10) and (11), respectively. Then μ A ∈M∗ (R + ) , and if there exist positive constants a and Ca such that n A (r ) ≤ Ca er − a 2 −m

the same time, L p (R, (1 + x )

r

for all r ≥ 0, then L p (R, d μ A ) ∈P for every 1 ≤ p < ∞ . A t

d μ L ) ∉ P for any 2 ≤ p < ∞ and m ∈ N .

It is obvious that the measure μ L no longer satisfies condition (7) because card ( [− r, r ] ∩ supp μ L ) ≥ er − 2

for

r ≥ log 2.

2. Auxiliary Results To prove Theorem 1, we recall several known results concerning the moment problem. Every measure μ ∈M∗ (R ) is associated with the set Vμ (Vμ+ ) of all measures ν ∈M∗ (R ) (M∗ (R + )) such that sn ( ν) =

354

A. G. B AKAN

sn (μ ) for all n ∈N0 . According to the Hamburger (Stieltjes) moment problem, it is necessary, for a sequence of real numbers {γ n }n ∈N0 , to find measures μ ∈M∗ (R ) (M∗ (R + )) such that sn (μ ) = γ n for all n ∈N0 . If a solution of this problem exists and is not unique, then the corresponding problem is said to be indeterminate. The measures μ that solve these problems are also called indeterminate. In other words, a measure μ ∈M∗ (R ) (M∗ (R + )) is called indeterminate in the Hamburger (Stieltjes) sense (or, briefly, μ ∈ indet H (indet S ) ) if Vμ \ {μ} ≠ ∅ (Vμ+ \ {μ} ≠ ∅ ) and determinate in the Hamburger (Stieltjes) sense (or, briefly,

μ ∈ det H (det S ) ) if Vμ (Vμ+ ) = {μ} . In 1923, Riesz [10] established a direct relationship between determinate measures in the Hamburger sense and the problem of polynomial density in the space L 2(R, (1 + x 2 ) d μ ) and proved that (see [11], Proposition 1.3) μ ∈ det H ⇔ L 2(R, (1 + x 2 ) d μ ) ∈ P .

(12)

In 1991, Berg and Thill [11] generalized property (12) as follows (Theorem 3.8): μ ∈ det S ⇔ L 2(R, (1 + x ) d μ ) ∈ P

L 2(R, x (1 + x ) d μ ) ∈ P .

and

(13)

In 1941, Widder [12] published the result of Boas concerning sufficient conditions for the indeterminateness of the Stieltjes moment problem. For a sequence of positive numbers {λ n }n ∈N0 , Boas introduced the conditions λ 0 ≥ 1,

λ1 ≥ λ 0 ,

λ 2 ≥ 4 (1 + λ1 )2 ,

λ n ≥ (n λ n −1 )n , n = 3, 4, 5, … ,

(14)

which are called the Boas conditions, and proved (see [12], p. 142, Chap. 3, Theorem 16) that, for any sequence that satisfies conditions (14), there exist at least two different solutions μ ∈M∗ (R + ) of the problem of Stieltjes moments sn (μ ) = λ 2n , n ∈N0 . Lemma 1. Let a function ψ be defined by relation (10). Then, for any b > 0, the function x b ψ( x ) is integrable on R + , the values ∞

∞

ψ (log k ) b σ (b ) : = ∑ log k , k k =1

γ (b) : =

∫ 1

ψ (log t ) b log t dt t

(15)

are finite, and γ (b) ≤ σ (b) ≤ e γ (b). e4

(16)

Proof. Assume that N ∈N and denote ψN ( x ) : =

1 32 π

N

∑

m =1

( 2 π )m e − m x m m!

1/ 4 m

.

ON THE COMPLETENESS OF ALGEBRAIC POLYNOMIALS IN THE SPACES L p (R, d μ )

355

Then, for an arbitrary β ≥ 2, by using the integral representation of the gamma function and the multiplication formula for this function (see [13], Chap. 1, Sec. 1.2), we get ∞

β/4 −1

∫ ψ N (u) u 0

∞

du = 4 ∫ ψ N (u ) u 4

β −1

du =

0

1 2π

+∞ N

N

=

m 1 ( 2 π ) m Γ (mβ ) = ∑ m! 2 π mm β m =1

∫ ∑

1/ m

( 2 π )m e − mu m m!

0 m =1

uβ −1 du

m −1

N

r 1 ∑ m! ∏ Γ ⎛⎝ β + m⎞⎠ . r =0 m =1

Since the function Γ (2 + x ) increases for x ≥ 0 (see [14], Chap. 6), we conclude that the last sum does not exceed eΓ (β+1) − 1, and, according to the Levi theorem, for an arbitrary b > 0 we have the required summability of the function x b ψ( x ) on the positive semiaxis. Furthermore, the last equality yields β eΓ (β) − 1 ≤ γ ⎛⎝ − 1⎞⎠ ≤ eΓ (β+1) − 1, 4

β ≥ 2.

(17)

It is obvious that the function e − x ψ( x ) decreases on R + . Therefore, the following inequality holds on every segment of the form [ log k, log (k + 1)], k ≥ 1: e − x ψ( x ) ≤ Ck e − log(k +1) ψ (log(k + 1)) ,

x ∈[ log k, log (k + 1)],

where Ck =

e − log k ψ (log k ) , e − log(k +1) ψ (log(k + 1))

k ∈N .

Let us prove that C1 ≤ e 4 and Ck ≤ e for an arbitrary k ≥ 2, i.e., ψ (log x ) ψ (log 2 ) ≤ e4 , x 2

x ∈[1, 2] , (18)

ψ (log x ) ψ (log(k + 1)) ≤ e , x k +1

x ∈[k, k + 1] ,

Using the obvious relations

ψ (0 ) =

and

1 32 π

∑

m ≥1

( 2 π )m m m!

≤

e

2π

−1 32 π

k ≥ 2.

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A. G. B AKAN

1 32 π

ψ (1) =

∑

m −m

( 2π ) e

m m!

m ≥1

1 32 π

≥

2π e

e

−1− 2π e

2π e ,

we get

C1 =

ψ (0) e ψ (0) ≤ ≤ − log 2 ψ (1) e ψ (log 2)

2π

e

2π e

2π e

−1

−1−

2π e

< e4 ,

which means that the first inequality in (18) is true. We now estimate the constants Ck , k ≥ 2, from above. For m ≥ 1, we have

[

log (k + 1) − log k + m log1/4 m (k + 1) − log1/4 m k

]

1 ≤ log 1.5 + m ⎡⎣log1/ 4 m 3 − log1/ 4 m 2⎤⎦ = log 1.5 + 4

≤ log 1.5 +

log 1.5 4 log

1 − 1/ 4 m

2

≤ log 1.5 +

log 3

∫

log 2

1 t

1 − 1/ 4 m

dt

log 1.5 < 1, 4 log 2

whence − log k − m log1/4 m k ≤ 1 − log (k + 1) − m log1/4 m (k + 1) and, therefore, 1/ 4 m

e

−log k

ψ (log k ) =

1 32 π

∑

( 2 π )m e − log k − m log

∑

( 2 π )m e1 − log(k +1) − m log

k

m m!

m ≥1

1/ 4 m

≤

1 32 π

(k +1)

m m!

m ≥1

= ee − log(k +1) ψ (log(k + 1)) ,

which proves the second inequality in (18). Since, for any b > 0, the function logb (1 + x ) increases on R + , for an arbitrary natural N ≥ 3 we get N

ψ(log k ) ∑ k logb k ≤ k =2

N

ψ(log k ) ∑ k k =2

k +1

∫ k

logb x dx

(18)

≤

N k +1

e∑

∫

k =2 k

ψ(log x ) b log x dx ≤ e γ ( b ) . x

ON THE COMPLETENESS OF ALGEBRAIC POLYNOMIALS IN THE SPACES L p (R, d μ )

357

This implies that σ ( b ) is finite for any b > 0 and the estimate σ ( b ) ≤ e γ ( b ) is true. To obtain the first inequality in (16), we again use the monotonicity of the function logb (1 + x ) for x ≥ 0 and inequality (18). As a result, we obtain ∞

k

ψ(log k ) logb x dx σ (b ) ≥ ∑ ∫ k k −1 k =2

(18)

≥

∞

k

ψ(log x ) b 1 1 log x dx = 4 γ (b). 4 ∑ ∫ x e k = 2 k −1 e

Lemma 1 is proved. Lemma 2. The sequence of numbers {σ (1 + n / 2 )}n ∈N0 satisfies the Boas conditions (14). Proof. For b ≥ 2 and n ∈ N, using estimates (17), we get γ ⎛⎝

b + 2n ⎞ − 1⎠ ≥ eΓ(b + 2 n) − 1 = (eΓ (b + 2 n −1) )b + 2 n −1 − 1 4 b + 2n − 2 ⎞ ≥ (eΓ (b + 2 n −1) − 1)b + 2 n −1 ≥ γ ⎛⎝ − 1⎠ 4

b + 2 n −1

,

i.e., b n b n − 1 ⎞ b + 2 n −1 γ ⎛⎝ + − 1⎞⎠ ≥ γ ⎛⎝ + − 1⎠ , 4 2 4 2

n ∈ N,

b ≥ 2.

(19)

Denote n γ n : = γ ⎛⎝ 1 + ⎞⎠ , 2

n σ n : = σ ⎛⎝1 + ⎞⎠ , 2

n ∈ N0 .

Using relation (17) with β = 8 + 2n and (19) with b = 8, we obtain eΓ(8 + 2 n) − 1 ≤ γ n ≤ eΓ(9 + 2 n) − 1,

n ∈ N0 ;

γ n ≥ γ 7n +−12 n ,

n ∈ N.

(20)

Using (20) and (16), we get σn ≥

1 1 7+2n 1 1 7+2n 4 γn ≥ 4 γ n −1 ≥ 4 7 + 2 n σ n −1 , e e e e

i.e., the sequence {σ n }n ∈N0 possesses the properties σ0 ≥

eΓ(8) − 1 , e4

σn ≥

1

11+ 2 n

e

σ n7 +−12 n ,

n ∈ N,

(21.1)

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A. G. B AKAN

eΓ(8 + 2 n) − 1 ≤ σ n ≤ e eΓ(9 + 2 n) − 1 , e4

(

)

n ∈ N0 .

(21.2)

We now prove that properties (21.1) and (21.2) yield relations (14) for the sequence {σ n }n ∈N0 . Using (21.1), we obtain σ 0 ≥ eΓ(8) − 5 = e 7!− 5 > e2 > 1.

(22.1)

Therefore, the first inequality in (14) is true. For n = 1, the second inequality in (21.1) yields σ1 ≥ σ 90 / e13. Using (22.1), we get σ 90 / e13 ≥ σ 0 . Therefore, σ1 ≥ σ 90 / e13 ≥ σ 0 , which proves the second inequality in (14). To verify the third inequality in (14), note that 4 (1 + σ1 )2 ≤ 16 σ12 < e 4 σ12 and, by virtue of the second 15 15 inequality in (21.1), σ 2 ≥ σ11 for n = 2. Using (21.2), we get σ1 ≥ eΓ(10) − 5 > e16 /9 , i.e., σ11 > 1 /e 1 /e 4 2 e σ1 , whence 15 σ 2 ≥ σ11 > e 4 σ12 > 4 (1 + σ1 )2 , 1 /e

which was to be proved. We now fix an arbitrary n ≥ 3. Then, by virtue of (21.1), the inequality required in (14), namely σ n ≥ (n σ n −1 )n ,

(22.2)

follows from the inequality σ 7n +−12 n / e11+ 2 n ≥ (n σ n −1 )n or, which is the same, from the inequality σ n−1 ≥ n n /( 7 + n) e(11+ 2 n)/( 7 + n) .

(22.3)

To prove (22.3), note that n / (7 + n) ≤ 1, (11 + 2n) / (7 + n) ≤ 3, and, by virtue of (21.2), σ n−1 ≥ eΓ(6 + 2 n) − 5 . Therefore, relation (22.3) follows from the inequality eΓ(6 + 2 n) ≥ ne8 = e8+log n ,

(22.4)

which is true by virtue of the relations Γ(6 + 2n) = (5 + 2n) Γ (5 + 2n) ≥ (5 + 2n) Γ (11) ≥ (5 + 2 log (n + 1)) Γ (11) > 8 + log n. Thus, inequality (22.2) is true, and the sequence {σ n }n ∈N0 satisfies all Boas conditions (14). Lemma 2 is proved. 3. Proof of Theorem 1 Since, according to relation (11) [see also (15)], one has +∞

∫ 0

x n dμ L ( x ) =

∞

ψ (log k ) n +1 log k = σ ( n + 1 ) , k k =2

∑

n ∈ N0 ,

ON THE COMPLETENESS OF ALGEBRAIC POLYNOMIALS IN THE SPACES L p (R, d μ )

359

we conclude that μ A ∈M∗ (R + ) for any A ⊂ L . By virtue of Lemma 2 and the Boas theorem indicated above (see [12, p. 142], Chap. 3, Theorem 16), we have at least two different solutions ν ∈M∗ (R + ) of the Stieltjes moment problem +∞

∫

x n dν( x ) = σ ( n + 1 ) ,

n ∈ N0 .

0

Since μ L is a solution of this problem, we have μ L ∈ indet S . By virtue of relation (13) and the fact that 0 ∉ supp μ L , this means that L 2(R, x (1 + x ) d μ L ) ∉P . Since supp μ L is not the set of zeros of any function of the set E0S [ R ] , it follows from Proposition A1.2 in [7, p. 250] that L 2(R, (1 + x 2 )− m d μ L ) ∉P for any m ∈ N and, moreover, L p (R, (1 + x 2 )− m d μ L ) ∉P ,

m ∈ N , 2 ≤ p < ∞.

Now consider a nonempty set A ⊂ L such that n A (r ) ≤ Ca er − a Then, for an arbitrary n ∈ N0 , we get +∞

sn (μ A ) =

∫

n

x dμ A ( x ) =

0

∑

k ∈A

r

(23)

, r ≥ 0, for certain a and Ca > 0.

ψ (log k ) n +1 log n +1 k log k ≤ ψ(0) ∑ = ψ(0) k k k ∈A

+∞

∫

x n +1e − x dn A ( x ) .

0

However, +∞

0 <

∫

x

n +1 − x

e dn A ( x ) = x

n +1 − x

e nA (x )

0

+∞

=

∫

+∞ 0

+∞

−

∫ 0

[

x n n A ( x ) e − x [ x − (n + 1)] dx ≤

+∞

∫

x n +1 n A ( x ) e − x dx

0

0

+∞

≤ Ca

]

n A ( x ) (n + 1) x n e − x − x n +1e − x dx

∫x

n +1 − a x

e

dx =

0

2 Ca Γ(2n + 4 ) . a2 n + 4

Therefore, it follows from asymptotic relations for the gamma function (see [13, p. 62]) that the measure μ A satisfies the Cárleman condition in the sense of Definition 1 in [15, p. 222], i.e.,

∑ s2 n (μ A )−1/ 4 n

n ≥1

= + ∞.

By virtue of the well-known Berg–Christensen theorem [16] (see also [15, p. 222], Theorem A), this means that L p (R, d μ A ) ∈ P for any 1 ≤ p < ∞. Theorem 1 is proved.

360

A. G. B AKAN

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