Approximation of Bandlimited Functions by Exponential Sums

APPROXIMATION OF BANDLIMITED FUNCTIONS BY EXPONENTIAL SUMS

S. Norvidas∗

Institute of Mathematics and Informatics, Akademijos 4, 08663 Vilnius, Lithuania (e-mail: [email protected])

Abstract. Let K be a compact set in Rn. For 1 ≤ p ≤ ∞, the Bernstein space p p n BK is the Banach space of all functions f ∈ L (R ) such that its Fourier transform in a p n distributional sense is supported on K. If f ∈ BK , then f is continuous on R and has an extension onto the complex space Cn to an entire function of exponential type K. We study p the approximation of functions in BK by ﬁnite τ-periodic exponential sums X 2πi(x,m)/τ cme , m in Lp(τTn) norm as τ → +∞, where Tn = [ − τ/2, τ/2]n. Keywords: Fourier transforms, bandlimited functions, entire functions of exponential type, Bernstein spaces, exponential sums, trigonometric polynomials, approximations by trigonometric polynomials. 2000 Mathematics Subject Classification: primary 30D15, secondary 42B35.

1 Introduction

The Fourier transform on L1(Rn) is deﬁned here by Z fˆ(x) = f(t)e−i(x,t)dt, Rn P∞ n where (x, t) = k=1 xktk denotes the inner product in R . We normalize the inverse Fourier transform Z ˇ 1 i(ξ,t) f(ξ) = n f(t)e dt (1.1) (2π) Rn so that the inversion formula (cfˇ) = f holds for suitable f. If f ∈ Lp(Rn) ———————————————————————

∗ Partially supported by CRC 701 Spectral Structures and Topological Methods in Mathematics, Bielefeld 1 and p > 2, then the Fourier transform fˆis understood as tempered distribution on Rn. n p Let K be a compact set in R . For 1 ≤ p ≤ ∞, the Bernstein space BK is a closed subspace of Lp(Rn) consists of all f ∈ Lp(Rn) such that fˆ is supported p on K. In other terminology, any f ∈ BK is bandlimited to K. The Banach p space BK is equipped with the norm

µZ ¶1/p p kfkp = |f(x)| dx , if 1 ≤ p < ∞ Rn and

kfk∞ = ess supRn |f(x)|, if p = ∞. p Throughout this article, we consider only the cases when BK is inﬁnite- dimensional Banach space. For this reason, we will assume that K is a body in Rn, i.e., K coincides with the closure of its inner point set int(K). By the p Paley-Wiener-Schwartz theorem (see [4, p. 181]), any f ∈ BK is continuous on Rn and has an extension onto the complex space Cn to an entire function of exponential type K; this means that

|f(z)| ≤ MeHK (−y), (1.2) for all z = x + iy, x, y ∈ Rn, where M is a positive constant and

HK (ξ) = max(ξ, t), t∈K ξ ∈ Rn, is the supporting function of K. Let Zn denote the lattice of integer vectors in Rn and let Tn = (−1/2, 1/2]n. p We write N for the set of natural numbers. Fix τ > 0. Since each f ∈ BK is a continuous function on Rn, we have that the Fourier coeﬃcients of f on τTn Z 1 −2πi(m,u)/τ cm(f, τ) = n f(u)e du, τ τTn n p m ∈ Z , are well-deﬁned. The Fourier series of f ∈ BK is formally given by X 2πi(m,x)/τ cm(f, τ)e . (1.3) m∈Zn We will take in (1.3) only those terms whose Fourier transforms are supported on K. More precisely, set X 2πi(m,x)/τ fτ (x) = cm(f, τ)e , (1.4) n m∈Z ∩nτ K

2 where nτ = [τ/2π]Z is the integer part of τ/2π and nτ K = {nτ u : u ∈ K} . n Since K is a compact set, fτ is a trigonometric polynomial on R . We also n consider the following (τT )-truncated version of fτ

ϕf,τ = fτ · χτTn , (1.5)

n where χτTn is the characteristic (indicator) function of the set τT . 2 2 Let Bσ := B[−σ,σ], where σ > 0. In a recent paper, Martin [7] studied the 2 approximation of f ∈ Bσ by the sums of one variable X 2πimx/τ χτT(x) · cm(f, τ)e . (1.6)

|m|≤[τσ/2π]Z Recall that a function f is called strictly bandlimited to [−σ, σ], if it is ban- 2 dlimited to [−%, %] for some 0 ≤ % < σ. It was proved in [7] that if f ∈ Bσ and f is strictly bandlimited to [−σ, σ], then (1.6) converges to f in L2 norm as τ → ∞ on R and on any horizontal lines in C. Gr¨ochening [2] studied similar to (1.6) trigonometric polynomials and obtained a quantitative estimate of 2 2 approximation in L (τT) norm. In [10], we investigated (1.5) for f ∈ BK in the case when K is an arbitrary Jordan-measurable compact subset of Rn. In the present paper, we give another proof of such approximation theorems. p Furthermore, our approach extends to any BK for 1 < p < ∞. To formulate the main result, recall that by an n-polytope in Rn we mean any intersection P of a ﬁnite number of half-spaces L(a, ω) = {x ∈ Rn :(x, a) ≤ ω}, a ∈ Rn, ω ∈ R, such that P has non-empty interior int(P ). p Theorem 1.1. Suppose f ∈ BK for 1 < p < ∞. If K is an n-polytope, then

lim kf − ϕf,τ kLp(Rn) = 0. (1.7) τ→∞ Until recently, the approximation of bandlimited functions by trigonometric polynomials was most completely investigated in the case p = ∞. Lewitan [6] was the ﬁrst to deﬁne the trigonometric polynomials X sin2( x + k) f˜ (x) := f(x + kτ) τ , τ > 0. (1.8) τ ( x + k)2 k∈Z τ

∞ ˜ It was proved in [6] that if f ∈ Bσ , then fτ → f as τ → ∞ uniformly on compact subset of R. Krein [5] given a new proof of this statement, obtained the degree of approximation, and considered several applications. Hörmander [3] considered trigonometric polynomials which are more general than (1.8) p (for a more detailed history and references see [1], p.p. 146–152). For f ∈ Bσ 3 and 1 ≤ p < ∞, Schmeisser [12] proved the convergence theorems for Lewitan (or Lewitan-Krein-Hörmander)polynomials (1.8) with respect to Lp norms on lines in C parallel to R and obtained the convergence of (1.8) with respect to lp norms. The approximation by polynomials of several variables similar to ∞ (1.8) was also studied in the case f ∈ BK (for details see [9] and the literature cited there). For our final theorem, we return to the polynomials (1.5). As in the case of Lewitan-Krein-Hörmanderpolynomials, it turns out that (1.5) also converges uniformly on compact subsets of Rn. p Theorem 1.2. Suppose f ∈ BK , where 1 < p < ∞. If K is an n-polytope, n then limτ→∞(f − ϕf,τ ) = 0 uniformly on compact subsets of R .

2 Preliminaries

Let us start by introducing some terminology and notation. For x ∈ Rn, n we denote by |x|p the lp norm on R , i.e.,

³X ´1/p p |x|p = |xk| , if p ∈ [1, ∞) k and |x|∞ = maxk |xk|. Set

n n n n U = {x ∈ R : |x|2 ≤ 1} and Q = {x ∈ R : |x|∞ ≤ 1}.

If A and B are two sets in Rn, then we write A + B for the set of all sums a + b, a ∈ A, b ∈ B. In particular, if A = {a}, a ∈ Rn, then we write simply a + B. For δ ∈ R \{0} we denote by δA the set {δa : a ∈ A}. Let K be a compact subset of Rn. The Lebesgue measure (in Rn) of K will be denoted by Ωn(K). Set n o n rK = inf |x|2 : x ∈ R \ K (2.1) and n o RK = max |x|2 : x ∈ K . (2.2)

For θ > 0, we denote by Kθ the θ-envelope of K, i.e.

n Kθ = ∪x∈K (x + θU ).

We deﬁne the following “θ-kernel of K”

n n K(−θ) = R \ (R \ K)θ. (2.3)

4 In the sequel, we assume that in K(−θ) the number θ is less than the diameter of K. This implies that K(−θ) is nonempty set. p Let f ∈ BK . For θ > 0, we say that f is θ-strictly bandlimited to K if b supp f ⊂ K(−θ). Throughout, the term ”f is strictly bandlimited to K” means that f is θ-strictly bandlimited to K for some θ > 0. p We need a few preliminary properties of BK . Let us start with some p n n Nikol’skii’s inequalities (see [8, Sec. 3]). Let f ∈ B n , where Q = {x ∈ R : Qσ σ |xk| ≤ σk < ∞, k = 1, . . . , n}. If 1 ≤ p ≤ r ≤ ∞, then n ³Y ´1/p−1/r n kfkLr(Rn) ≤ 2 σk kfkLp(Rn). (2.4) k=1 In the case 1 ≤ p < ∞ lim f(x) = 0. (2.5) x→∞ p Of course, (2.4) and (2.5) still hold for all f ∈ BK in the case if K is a compact body in Rn. More precisely, given a compact body K ⊂ Rn and 1 ≤ p ≤ r ≤ ∞, there exists a positive constant A = A(n, p, r, K) < ∞ depending on n, p, r, and K so that

kfkLr(Rn) ≤ A kfkLp(Rn) (2.6)

p for all f ∈ BK . In other words, the embedding mapping

p r BK ⊂ BK (2.7) is continuous. Let p be a 2π-periodic in each variable trigonometric polynomial on 2πTn. If 1 ≤ p ≤ r ≤ ∞, then there exists a constant c = c(n) such that n ³Y ´1/p−1/r kpkLr(2πTn) ≤ c mk kpkLp(2πTn), (2.8) k=1 where mk is the degree of p in the variable xk (see [8, Sec. 3], [13]). Recall that the Schwartz class of test functions S(Rn) consists of the complex-valued inﬁnitely diﬀerentiable functions ξ on Rn satisfying

s u sup |(1 + |x|2) D ξ(x)| < ∞ (2.9) x∈Rn,|u|≤k Pn for all k, s ∈ N, where u is a nonnegative multi-index and |u| = j=1 uj. The dual space S0(Rn) is the space of tempered distributions. If ξ ∈ S(Rn) and τ > 0, then the Poisson summation formula gives X 1 X ³2π ´ ξ(x + τl) = ξb m e2πi(m,x)/τ , (2.10) τ n τ l∈Zn m∈Zn 5 x ∈ Rn (see [4, p. 177, (7.2.1)]). n ◦ n n Given a compact set K in R , we denote by SK (R ) the subspace of S(R ) consisting of strictly bandlimited to K functions. We will need the following easily implicit auxiliary statement. For completeness, we give also its proof. Lemma 2.1. Let K be a convex compact body in Rn. If 1 ≤ p < ∞, then ◦ n p n p SK (R ) is an L (R )-dense subspace of BK . Proof. Suppose ﬁrst that 0 ∈ int K. Take any ζ ∈ S(Rn) such that ζ is non-negative on Rn, supp ζ ⊂ U n, and Z ζ(x) dx = 1. (2.11) Rn By the Bochner theorem, the function ζˇ is continuous and positive deﬁnite on Rn. This, conjugate with (2.11), implies that

sup |ζˇ(t)| ≤ ζˇ(0) = (2π)−n. (2.12) Rn p Let f ∈ BK . Given δ ∈ (0, 1), we set fδ(x) = f(δx). Let ³(1 − δ)r ´ F (x) = (2π)nf (x) · ζˇ K x , (2.13) δ δ 2 b 0 n where rK is deﬁned in (2.1). We consider f as element of S (R ) with compact b 0 n support. Then fδ ∗ ζ is well deﬁned in S (R ) and (1 − δ)r supp Fc = supp (fb ∗ ζ) ⊂ δK + K Un ⊂ K δ δ 2 −(θ) p for all θ ∈ (0, (1 − δ)rK /2). By (2.5), each f ∈ BK , p ∈ [1, ∞], is a bounded n ◦ n function on R . Hence Fδ ∈ SK (R ). Let us prove that

lim kf − FδkLp(Rn) = 0. (2.14) δ→1

To this end, we estimate kf − fδkLp(Rn) and kfδ − FδkLp(Rn). Set ε > 0. Then there exists aε > 0 such that

p n n kfkL (R \aεU ) < ε. (2.15)

Fix any such aε. If 1/2 < δ < 1, then

n/p p n n p n n kFδkL (R \aεU ) ≤ kfδkL (R \aεU ) < 2 ε (2.16) for all δ ∈ (1/2, 1). n Now we consider the functions f, fδ, and Fδ on aεU . It is known (see, for p n example, [10]), that if g ∈ BK , 1 ≤ p ≤ ∞, x, y ∈ R , and maxt∈K |(x−y, t)| ≤ π, then ³1 ´ |g(x) − g(y)| ≤ 2 sin max |(t, x − y)| · kgkL∞(Rn). (2.17) 2 t∈K 6 n Therefore, if x ∈ aεU and 1 − δ is suﬃciently small, then by (2.17), we obtain that ¯ ³ ´¯ ¡ n n ¢1/p¯ 1 − δ ¯ kf − fδk p n ≤ 2 a Ωn(U ) ¯sin max |(x, t)| ¯ L (aεU ) ε n 2 t∈K,x∈aεU ¡ n+p n ¢1/p 1/2 ≤ aε Ωn(U ) (1 − δ)n RK < ε, (2.18)

n where RK is deﬁned in (2.2). In view of (2.12), we have for x ∈ aεU ¯ ³ ´¯ ¯ (1 − δ)rK ¯ |f (x) − F (x)| = (2π)n|f (x)|¯ζˇ(0) − ζˇ x ¯ δ δ δ 2 Z ¯ ¯ ¯ i(1−δ)rK (x,t)/2¯ ≤ kfkL∞(Rn) ¯1 − e ¯ζ(t) dt ¯ Rn ¯ ¯ i(1−δ)rK (x,t)/2¯ ≤ max ¯1 − e ¯ kfk ∞ n n n L (R ) t∈U ,x∈aεU

(1 − δ)rK kfkL∞(Rn) 1/2 ≤ max |(x, t)| √ ≤ (n/2) (1 − δ)rK aεkfk ∞ n . n n L (R ) t∈U ,x∈aεU 2 Hence

¡ n n ¢1/p 1/2 p n ∞ n kfδ(x) − Fδ(x)kL (aεU ) ≤ aε Ωn(U ) (n/2) (1 − δ)rK aεkfkL (R ) < ε for suﬃciently small 1 − δ. Since ε is an arbitrary positive number, combining (2.15), (2.16), (2.18), and (2.19), we have (2.14). p In the general case, we take an arbitrary x0 ∈ int K. Let g ∈ BK . If

f(x) = g(x)e−i(x,x0), then f ∈ Bp , where the body int (−x +K) contains origin. For δ ∈ (0, 1), −x0+K 0 i(x,x0) we set Gδ(x) = Fδ(x)e , where Fδ is deﬁned in (2.13). Then (2.14) becomes ◦ n limδ→1 kGδ − gk = 0. Finally, each Gδ belongs to SK (R ), and this complete the proof of the lemma.

3 Proofs

For f ∈ S(Rn), let X ³ ´ 1 b 2π 2πi(m,x)/τ Ef,τ (x) := n f m e , (3.1) τ n τ τ m∈Z ∩ 2π K x ∈ Rn. We will compare this polynomial with our main trigonometric polynomial (1.4). n ◦ n Lemma 3.1. Let K be a convex compact body in R . If f ∈ SK (R ) and p ∈ [1, ∞), then 7 lim kfτ − Ef,τ kLp(τTn) = 0. (3.2) τ→∞

Proof. First we claim that there exists an τ0 > 0 such that if τ ≥ τ0, then Ef,τ can be written as X ³ ´ 1 b 2π 2πi(m,x)/τ Ef,τ (x) = n f m e , (3.3) τ n τ m∈Z ∩nτ K b where nτ = [τ/2π]Z. To that end, it suﬃces to show that if f((2π/τ)m0) 6= 0 n for m0 ∈ Z , then m0 ∈ nτ K. Since f is θ-strictly bandlimited to K for some n θ > 0 and K is convex, we have (2π/τ)m0 + θU ⊂ K. Hence there exists

δ0 > 0 such that if δ ∈ (0, δ0], then 2π (1 + δ)m ∈ K. (3.4) τ 0

Suppose τ0 is deﬁned be means of the equality (1 + δ0)(1 − 2π/τ0) = 1. Since

1−2π/τ < 2πnτ /τ ≤ 1, we have that for any τ ∈ [τ0, ∞) there exists δ ∈ (0, δ0] such that 2πnτ (1 + δ)/τ = 1. Using (3.4), we obtain that m0 ∈ nτ K. This proofs our claim (3.3).

Let τ ≥ τ0. Using (3.3) we can estimate kfτ − Ef,τ kLp(τTn) as follows:

kfτ − Ef,τ kLp(τTn) µZ ¶ ¯ X X ³ ´ ¯p 1/p 1 ¯ n 2πi(m,x)/τ b 2π 2πi(k,x)/τ ¯ = n ¯ τ cme − f k e ¯ dx τ n τ τT m∈Zn∩n K n τ τ k∈Z ∩ 2π K µZ ¶ ¯ X ³ ³ ´´¯p 1/p 1 ¯ n b 2π ¯ = n ¯ τ cm − f m ¯ dx n τ τT n τ m∈Z ∩nτ K X ¯ ³ ´¯ n/p−n ¯ n b 2π ¯ ≤ τ ¯τ cm − f m ¯. (3.5) n τ τ m∈Z ∩ 2π K Let s ∈ N. By (2.9) (with u = 0), there exists a constant c(f, s) < ∞ such that −s |f(x)| ≤ c(f, s)|x|2

8 If s > n, then via polar coordinates in Rn Z ¯ ³2π ´¯ 2πn/2 ∞ 2πn/2 ¯ n b ¯ ¡ ¢ n−s−1 ¡ ¢ n−s ¯τ cm − f m ¯ ≤ c(f, s) n % d% = c(f, s) n τ , τ Γ τ (s − n)Γ 2 2 2 where Γ is the usual Gamma function. Finally, if we choose in (2.9) u = 0 and s > 0 so that s > n(1 + 1/p), then (3.5) together with (3.6) proves (3.2) and the lemma. Let K be a compact subset of Rn. If k ∈ N, then X i(x,m) Dk,K (x) = e m∈Zn∩kK is called the Dirichlet kernel of order k related to K. Now the polynomial (1.4) can be presented as the value of the following integral operator Z 1 ³2π ¡ ¢´ fτ (x) = n f(t)Dnτ ,K x − t dt. (3.6) τ τTn τ Therefore, setting

Φτ (f) = ϕf,τ , p τ ∈ (0, ∞), we may assume that (1.4) deﬁnes on BK the one-parametric p p n family of bounded linear operators Φτ : BK → L (R ). In the next lemma we give some suﬃcient conditions under which the family (Φτ )τ>0 has uniformly bounded norms. Lemma 3.2. Let K be an n-polytope in Rn. If 1 < p < ∞, then there exists a constant $(K, p) < ∞ such that

kΦ (f)k p n ≤ $(K, p)kfk p (3.7) τ L (R ) BK

p for every f ∈ BK and all τ ∈ (0, ∞). p n n Proof Given f ∈ BK , we deﬁne on R the 2πT -periodic function gf n such that gf (v) = f((τ/2π)v) for v ∈ 2πT . Now, (3.7) implies

p p kΦτ (f)kLp(Rn) = kΦτ (f)kLp(τTn) Z ¯Z ³ ´ ¯ 1 ¯ 2π ¯p = np ¯ f(t)Dnτ ,K (x − t) dt¯ dx τ n n τ τT Z τT ¯Z ¯ τ n ¯ ¯p = n(p+1) ¯ gf (v)Dnτ ,K (y − v) dv¯ dy. (3.8) (2π) 2πTn 2πTn In [13] (theorem 2.4.5, p. 56), it is shown that under the conditions of this lemma, there exists a constant c(K) > 0, depending only on K, such that ³Z ¯Z ¯ ´ ³ ´ 1 ¯ ¯p 1/p 4p2 n p n n ¯ h(v)Ds,K (y −v) dv¯ dy ≤ c(K) 1+ khkL (2πT ) (2π) 2πTn 2πTn p − 1 9 for all s ∈ N and h ∈ Lp(2πTn). Combining this with (3.9), we get

³ 4p2 ń kΦ (f)k p n ≤ c(K) 1 + kfk p n . τ L (R ) p − 1 L (τT ) Thus we have (3.8) with $(K, p) = c(K)(1 + 4p2/(p − 1))n, and the lemma is proved. Proof of theorem 1.1. The Banach–Steinhaus theorem together with ◦ n Lemma 2.1 and Lemma 3.2 imply that it suffices to check (1.7) only on SK (R ). ◦ n Let f ∈ SK (R ). The Poisson summation formula (2.10) gives X f(x + τl) = Ef,τ (x), (3.9) l∈Zn where Ef,τ is defined in (3.1). Here the series converges absolutely and uniformly on compact subset of Rn. In particular, the series X f(x + τl) l∈Zn\{0}

n p n deﬁnes on τT an L (τT ) function. Since limτ→∞ kfkLp(Rn\τTn) = 0, we have from (3.2) and (3.10) that it remains to show that ° ° ° X ° lim ° f(x + τl)° = 0. (3.10) τ→∞ Lp(τTn) l∈Zn\{0}

Let s ∈ N. If x ∈ τTn and l ∈ Zn \{0}, then (2.9) implies that there exists M(s) > 0 such that M(s) M(s) 2sM(s) 2sM(s) |f(x + τl)| ≤ s ≤ τ s ≤ s s ≤ s s . (1 + |x + τl|2) (1 + 2 |l|2) τ |l|2 τ |l|∞ We may take s ∈ N so that s > n. Then ¯ ¯ ∞ ¯ X ¯ 2sM(s) X 1 2sM(s) X 2n(m − 1)n−1 ¯ f(x + τl)¯ ≤ s ≤ s s . τ |l|∞ τ m l∈Zn\{0} l∈Zn\{0} m=1

Therefore exists M1(s) > 0 such that ° X ° ° ° n/p−s ° f(x + τl)° ≤ M1(s)τ . Lp(τTn) l∈Zn\{0}

This proves (3.11) and the proof of theorem 1.1 is complete. p Proof of theorem 1.2. Suppose f ∈ BK . We claim that

M(f) := sup kf − fτ kL∞(Rn) < ∞. (3.11) τ>0

10 In fact, by (2.4), it would suﬃce to show that

sup kfτ kL∞(Rn) < ∞. (3.12) τ>0

n N n To see this, let pτ (t) := fτ (τt/2π), t ∈ 2πT . Since K ⊂ RK U ⊂ RK Q , where RK is deﬁned in (2.2), the polynomial pτ is 2π-periodic and has degree at most nτ RK in each variable. Now, by (2.8), there exists an c < ∞ such that

n/p kpτ kL∞(2πTn) ≤ c (nτ RK ) kpτ kLp(2πTn).

Then

n/p kfτ kL∞(Rn) = kfτ kL∞(τTn) = kpτ kL∞(2πTn) ≤ c (nτ RK ) kpτ kLp(2πTn) = ³ ´n/p 2πnτ n/p c R kf k p n ≤ c R kf k p n . τ K τ L (τT ) K τ L (τT ) Thus, using lemma 3.2, we get

n/p n/p kfτ kL∞(Rn) ≤ c RK kΦτ (f)kLp(Rn) ≤ c RK $(K, p)kfkLp(Rn), (3.13) and this proves (3.13). In view of (2.5), to prove the theorem, it is enough to show that

lim kf − ϕf,τ kL∞(τTn) = 0. τ→∞

Assume, to the contrary, that there exist an a > 0 and a sequence τm → ∞

∞ n such that kf − ϕf,τm kL (τmT ) ≥ a for all m ∈ N. Then we may choose a ∞ n sequence (xm)1 , where xm ∈ τmT , so that

|(f − fτm )(xm)| = |(f − ϕf,τm )(xm)| ≥ a. (3.14)

We ﬁx δ ∈ (0, 1] and deﬁne

πδ n ∆m = xm + U , RK n m = 1, 2,... . Suppose y ∈ ∆m. Since y = xm + (πδ/RK )w, for some w ∈ U , we get πδ πδ max |(xm − y, t)| = max |(w, t)| ≤ max |t|2 ≤ πδ ≤ π. t∈K RK t∈K RK t∈K Now using (2.17) and (3.15), we obtain

∞ n ≥ a − πδkf − fτm kL (R ). (3.15) 11 Since a ≤ M(f) < ∞, we may take a δ = . 2πM(f) Then (3.16) implies Z Z p p |(f − fτm )(x)| dx ≥ |(f − fτm )(y)| dy ≥ n n τmT τmT ∩∆m ³a´p Ω (τ Tn ∩ ∆ ). (3.16) 2 n m m We may assume without of generality that 2π τm ≥ √ nRK for all m ∈ N. Then n n n πδ Ωn(U ) a Ωn(U ) Ωn(∆m ∩ τmT ) ≥ n = n+1 . (3.17) 2 RK 2 M(f)RK Then combining (3.17) and (3.18) we obtain

Z p+1 n p a Ωn(U ) |(f − fτm )(x)| dx ≥ n+p+1 n τmT 2 M(f)RK for m = 1, 2,... . This contradicts (1.7) and theorem 1.2 is proved.

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