ON CLASSES OF EVERYWHERE DIVERGENT POWER SERIES

Transcription

1 ON CLASSES OF EVERYWHERE DIVERGENT POWER SERIES G. A. KARAGULYAN Abstract. We prove the everywhere divergence of series a n e iρn e inx, and 1 [ρn] a n cos nx, for sequences a n and ρ n satisfying some extremal conditions. These results generalize some well known examples of everywhere divergent power and trigonometric series. 1 We consider power series 1. Introduction c n e inx with complex coefficients c n. In 1912 Luzin [8] constructed an example of everywhere divergent series 1 with c n 0. It was an answer to Fatou s problem [4], asking whether power series converge almost everywhere, if c n 0. Then in 1912 Steinhaus [11] proved that the trigonometric series 2 n=2 cos nx + log log n log n is everywhere divergent. In 1916 Hardy and Littlewood [6] proved that the power series 3 n ρ e iαn log n e inx, 0 < ρ 1 2, α > Mathematics Subject Classification. 42B25. Key words and phrases. everywhere divergent series, power series, trigonometric series. 1

2 2 G. A. KARAGULYAN diverges for any x R. Many authors were interested in the problem of possible upper bound for the coefficients of everywhere divergent power series. In 1921 Neder [9] proved that if 4 a n 0, a 2 n =, then there exists a power series 1, which is everywhere divergent and c n = Oa n. Note that the second condition in 4 is necessary for almost everywhere divergence of series 1 according to Carleson s well known theorem [2]. Generalizing Neder s theorem Stechkin [10] constructed an example of series 1, which real an imaginary parts are everywhere divergent and c n = Oa n, where a n satisfies 4. Then Herzog constructed an example of series a n cos nx with a n 0, which is everywhere divergent. Dvoretzky and Erdos [3] proved that if a n a n+1, a n 2 =, then there exists a sequence of numbers ε n = 0 or 1, such that the series ε n a n e inx is everywhere divergent. In 1985 Galstyan [5] proved that if a sequence a n satisfies the conditions 4, then there exists a sequence ε n = 0 or 1, such that the series ε n a n cos nx diverges for any x R. Ul yanov [12] in We denote by It was the solution of a problem posed by x n = x n+1 x n, 2 x n = x n+1 x n. the first and second order differences of a given sequence of numbers x n, n = 1, 2,....

3 ON CLASSES OF EVERYWHERE DIVERGENT POWER SERIES 3 Theorem 1. If a sequence ρ n, n = 1, 2,..., satisfies the conditions 5 2 ρ n 0, 2 ρ n =, and 6 a n c 2 ρ n, n = 1, 2,..., c > 0, then the real and imaginary parts of the series 7 a n e iρn e inx, diverge for any x R. Theorem 2. If ρ n, n = 1, 2,..., satisfies the conditions 5 and 6 and a n 0, then the series 8 ε n a n cos nx, ε n = 1[ρn] + 1, 2 diverges for any x R. Remark 1. For a given number α R we shall use notation { α } α/2π = 2π, 2π where {x} denotes the fractional part of x. Note that in 5 and 6 we may everywhere replace 2 ρ n with 2 ρ n /2π. Remark 2. Notice that the values of ε n in 8 are 0 or 1. Without a significant change in the proof of Theorem 2 one can deduce also the everywhere divergence of series 1 [ρn] a n cos nx. Remark 3. For given coefficients a n, satisfying 4, we consider the sequence n 2 9 ρ 1 = 0, ρ n = n k 1a 2 k, n = 2, 3,.... It is easy to check that k=0 2 ρ n = a 2 n, n = 1, 2,... and hence we have 5 and 6. Thus we conclude that for a given sequence of coefficients 4, defining ρ n by 9, we get series 7 and 8 diverging at any point x R. So the theorems provide concrete

4 4 G. A. KARAGULYAN examples of everywhere divergence trigonometric and power series with coefficients having extremal rate. Moreover, the examples of series 2, 3 and the series constructed in the papers [9, 7, 10, 5] may be immediately deduced from the Theorems 1 or Auxiliary lemmas Lemma 1. If b 1 b 2... b m 0, then 2 m 10 b j jb 2 j, n m n 2 1. Proof. Observe that b i b j b 2 j. j=i+1 j=i+1 Thus, we get n 1 b i b j = b j b i 1 i<j n i=1 j=i+1 j 1b 2 j j=2 and therefore 11 b 2 j + b i b j 1 i<j n jb 2 j. On the other hand we have b i b j 12 1 i<j n nn 1 b 2 n 2 n 2n 2 1 nn mm + 1 nn + 1 b 2 n 2 2 = n n mb 2 n m jb 2 j. j=n+1 b 2 n

5 ON CLASSES OF EVERYWHERE DIVERGENT POWER SERIES 5 Combining the formula 2 13 b j = b 2 j i<j n with the inequalities 11 and 12, we get 10. b i b j. Lemma 2. If b 1 b 2... b m 0, 15 n m, and m m 14 c 1 = jb 2 j, c 2 = j nb 2 j then 15 Proof. Denote 16 c = We have 17 b 2 n = 2b2 n nn + 1 j=n+1 jb 2 j c 1 c 2 2 9c 1. j jb 2 j. 2 nn + 1 jb 2 j 2c n 2. If c < c 1 2c, then 15 is trivial. So we may suppose c 1 > 2c. Since n 15 there exists an integer p such that c1 18 n n 2c p 11n c1 10 2c. Using the relations 14, 16, 17 and 18, we obtain m c 1 c 2 = c + n 19 c + j=n+1 b 2 j 2cp n n + n m j=p+1 2cp n + nc 1 p 11 2c1 c + 2c 1 c < 3 c 1 c, 10 where we assume m b 2 j = 0 j=p+1 if p m. After simple calculations from 19 we get 15. b 2 j

6 6 G. A. KARAGULYAN Lemma 3. If a sequence x n satisfies the conditions 5 and l 0 is an arbitrary integer, then there exist integers p and q, q > p > l, such that q 21 0 < x j /2π < π 50, p j q, j=p 2 x j > Proof. Define the sequence of integers n k satisfying 22 x nk < 2πk x nk +1. Since 2 x k 0, there exists a number k 0, such that 23 2 x nk 1 < π 10 7, n k 1 > l, k > k 0. For a fixed integer k > k 0 we consider the sequence k 1 24 γ n = j n x j, n k 1 < n < n k. j=n We also assume γ nk = 0. Observe that, if 25 γ n < 3π, n k 1 < n < n k 1000, then 26 γ n 1 γ n = k 1 j=n 1 2 x j x nk From 24 and 25 we also get 27 2 x nk k 1 j=n π γ n 10 3 < j n x j π γ n n k n 6π n k n < π 100n k n. Define the integers p, q, n k 1 < p < q < n k, satisfying γ q+1 + α < 2π γ q + α, 28 γ p + α 2π + π 100 <γ p 1 + α. where 29 α = x nk /2π.

7 ON CLASSES OF EVERYWHERE DIVERGENT POWER SERIES 7 From 26 and 28 we obtain γ q γ q+1 + π 300 < 2π α + π 300, γ p γ p 1 π 300 2π α + π 100 π 2π = 2π α Using these estimates and the inequalities 10 and 15, we get q x j > γ p γ q 2 > γ q j=p It is easy to verify, that for any numbers n < m we have 31 k 1 x n x nk = x j j=n k 1 = j n x j n k n x nk j=n = γ n n k n x nk. Since x nk +1 x nk = 2 x nk, from 22 it follows that 32 x nk = 2πk θ 2 x nk, 0 θ 1. Substituting this in 31, we obtain which implies x n x nk = γ n + θn k n 2 x nk 2πkn k n, 33 x n = α + γ n + θn k n 2 x nk mod 2π, where α is defined in 29. From 27, 33 and 28 it follows that and this completes the proof. Lemma 4. If x n /2π [0, π/50], p n q, 34 x π 2 mod 2π, then there exists an integer ν = νx > 10, such that 35 #{k N : n < k n + ν, cos kx sinπ/} 3ν 5 for any n = 1, 2,....

8 8 G. A. KARAGULYAN Proof. If x/2π = 0 35 is immediate. Consider the case if x/2π is a rational number and 36 x = 2πq mod 2π, q < p, p where p, q N are coprime numbers. We denote { } 2πk 37 G p =, k = 1, 2,..., p, 38 It is well known that = p 9π, 11π 29π, 31π 39 {kx/2π : k = n + 1, n + 2,..., n + p} = G p, for any n N. We suppose νx = p. To prove 35 for x defined in 36, it is enough to show that 40 #G p 2p 5, p 4. The case p = 4 is excluded because of 34. Since G p = if p = 2, 3, 5, 6, then 40 holds for such p s. If p 7, then rough estimation gives 9π # G p < p + 1, # G p, 11π 29π, 31π. < p + 1, which implies 41 p #G p < < 2p 5, p 7. In the case of irrational x/2π we shall use an approximation of x/2π by rational numbers. It is known that there exist coprime numbers l and ν, with ν > 40, such that 42 x 2πl ν < 2π ν. 2 We consider the set 0 = 8π, 12π 28π, 32π. Similarly as in 41, we get ν 43 #a + G ν 0 < < 2ν 5, ν > 40,

9 ON CLASSES OF EVERYWHERE DIVERGENT POWER SERIES 9 for arbitrary a R, where { a + G ν = a + 2πlk } /2π, k = 1, 2,..., ν. ν From 42 it follows that 2πlk kx ν < 2πk 2π ν 2 ν < π, k = 1, 2,..., ν, and therefore we conclude, that the condition a + kx/2π, k = 1, 2,..., ν,, implies a + 2πlk /2π 0. ν Taking a = nx, thus we obtain 44 #{k N : n < k n + ν, kx/2π } and the lemma is proved. = #{k N : 1 k ν, a + kx/2π } = #{k N : 1 k ν, a + 2πlk /2π 0 } ν = #a + G ν 0 < 2p 5 Lemma 5. If a sequence x n, n = 1, 2,..., satisfies the conditions 5, then for any sequence of coefficients a n, with 6, the series 45 a n cosx n is divergent. Proof. Applying Lemma 3, we may find sequences of numbers p n, q n N, q n > p n > n, n = 1, 2,..., such that < x j /2π < π 50, p n j q n, q n j=p n From 6 and 47 we obtain q n 2 x j > j=p n a j c q n j=p n 2 x j > c 10 3,

10 10 G. A. KARAGULYAN then, combining this with 46, we get q n q n a j cosx j > cosπ/50 a j > c cosπ/ j=p n j=p n This implies the divergence of series Proof of theorems Proof of Theorem 1. The real part of series 7 is 48 a n cosnx + ρ n. Defining x n = nx + ρ n, we have 2 x n = 2 ρ n. So x n satisfies the same conditions 5 and 6 as ρ n. Therefore the divergence of series 48 immediately follows from Lemma 5. The imaginary part of 7 is the series a n sinnx + ρ n = a n cosnx + ρ n π and its divergence can be obtained similarly, discussing the sequence x n = nx + ρ n π. Proof of Theorem 2. We fix an arbitrary number 49 x π 2 mod 2π and denote 50 x n = πρ n nx π 2. Applying Lemma 3 we may define sequences q n > p n > n with conditions 46 and 47. Observe, that if 51 p n k q n and cos kx sinπ/, then 52 1 [ρ k] cos kx = cos kx. Indeed, from 46 and 51 we have x k /2π 0, π/50, 9π kx/2π =, 11π 29π, 31π,

11 ON CLASSES OF EVERYWHERE DIVERGENT POWER SERIES 11 which together with 50 implies sign cos kx = sign coskx + x k = sign cosπρ k π/2 = sign sin πρ k = 1 [ρ k], and therefore we get 52. According to Lemma 4 there exists a number ν = νx, satisfying 35. Hence, from 52 we obtain 53 #{k N : m < k m + ν, 1 [ρ k] cos kx sinπ/} = #{k N : m < k m + ν, cos kx sinπ/} 3ν 5 provided we have p n m, m + ν q n. We let [ pn ] [ qn ] s n = + 1, t n =. ν ν We have 54 s n 1ν p n < s n ν, t n ν q n < t n + 1ν. Then, denoting G k = {j N : kν < j k + 1ν, 1 [ ρ j π ] cos jx sinπ/} from 53 we obtain Hence we conclude, if 55 νk+1 j=νk+1 1 [ρ j] a j cos jx #G k 3ν 5, s n k < t n. 1 [ρj] a j cos jx 2ν 5 sinπ/a νk j G k 3ν sinπ/ 5 a νk+1 2ν 5 a νk = ν sinπ/ 5 3aνk+1 2a νk, provided s n k < t n. From 47 and the relation 2 x n = 2 ρ n 0 it follows that the inequality 56 νt n j=νs n+1 a j c νt n j=νs n+1 2 x j c 10 4

12 12 G. A. KARAGULYAN holds for sufficiently large n > n 0. From 55 and 56 we obtain νt n j=νs n+1 1 [ρj] a j cos jx t n 1 νk+1 = 1 [ρj] a j cos jx k=s n j=νk+1 > ν sinπ/ t n = ν sinπ/ tn 1 5 sinπ/ 5 k=s n a νk+1 2 k=s n+1 νt n 1 j=νs n+1+1 j=νs n+1 t n 1 k=s n a νk a νk + 3a νtn 2a νsn a j + ν3a νtn 2a νsn Using 56 and 54, we obtain νt n lim sup 1 [ρj] a n j cos jx > 0. So the series 8 is divergent if 49 holds. If x = π/2, then we have 1 [ρn] a n cos nx = 1 [ρ2k] a 2k cos kπ The last series is divergent because ρ 2k again satisfies the same conditions 5 and the divergent of such series at x = π has been already proved. This completes the proof of theorem. k=0 References 1. N. Bary, A treatise on trigonometric series, Pergamon Press, L. Carleson, On the convergence and growth of partial sums of Fourier series, Acta Math., , A. Dvoretzky, P. Erdos, On power series which diverges everywhere on the unit circle of convergence, Michigan Math. Journal, , No 1, P. Fatou, Series trigonometriques et series de Taylor, Acta Math., , S. Sh. Galstyan, Everywhere divergent trigonometric series, Math. Notes, , No 2, G. H. Hardy, J. E. Littlewood, Some problems of diophantine approximation: A remarkable trigonometrical series, Nat. Acad. Sci. U.S.A., 21916,

13 ON CLASSES OF EVERYWHERE DIVERGENT POWER SERIES F. Herzog, A note on power series which diverge everywhere on the unit circle of convergence, Michigan Math. Journal, , 3, No 1, N. N. Luzin, On a case of Taylor series, Mat. Sbornik, , L. Neder, ZüTheorie der trigonometrischen. Reihen-Math. Ann., , S. B. Stechkin, On convergence and divergence of trigonometric series, Math. Survey, 61951, No 2, H. Steinhaus, Une trigonometrique partout divergente, C.R. de la Soc. Sci. de Varsovie, P. L. Ul janov, Solved and unsolved problems in the theory of trigonometric and orthogonal series, Math. Survey, , No 1115, 369. Institute of Mathematics of Armenian National Academy of Sciences, Baghramian Ave.- 24b, 0019, Yerevan, Armenia address: g.karagulyan@yahoo.com