Greedy bases are best for m-term approximation

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1 Greedy bases are best for m-term approximation Witold Bednorz Institute of Mathematics, Warsaw University Abstract We study the approximation of a subset K in a Banach space X by choosing first basis B and then using n-term approximation. Into the competition for best bases we enter all unconditional bases for X. We show that if the subset K X is well aligned with respect to a greedy basis B then, in certain sense, this basis is the best for this type of approximation. Our result extends the recent result of DeVore, Petrowa and Temlyakov. 1 Introduction Let X be a Banach space with a norm. We say that B = (e n ) n N is a natural biorthogonal system for X (which we call a basis), if e i = 1, span{e i } = X and there exists system of uniquely determined functionals e i such that e i (e i ) = 1 and e i (e j ) = 0 if i j. For each n 0 we denote Σ n (B) = {S = a i e i, Λ = n, a i R}, i Λ so Σ n is the set of all linear combinations of length n of elements (e n ) n N. For x X and n = 0, 1, 2,... σ n (B, x) = inf{x y : y Σ n }. Note in particular that σ 0 (B, x) = x. We define G n (B, x) = e i (x)e i, i Λ where Λ N is such that Λ = n and e i (x) e j(x) whenever i Λ and j Λ. We will say that basis B is greedy if there exists constant G such that x G n (B, x) Gσ n (x), for x X, n 0. 1

2 Throughout the paper we denote (ε i ) as a sequence of Rademacher variables i.e. independent, identically distributed and such that P(ε i = ±1) = 1. The basis B is 2 called unconditional if there exists constant U such that for each Λ N there holds i Λ ε i e i (x) Ux. If B is unconditional, there exists equivalent norm so that B is unconditional with constant U = 1. The basis B is called democratic if there exists constant D such that for any finite subsets Λ, Λ N, Λ = Λ we have e i D e i i Λ i Λ It is well known (Konyagin and Temlyakov [4]) that basis B is greedy if and only if it is unconditional and democratic. Set l n := e e n, the democracy implies D 1 l n e i Dl n, for Λ = n. (1) i Λ We do assume that numbers l n satisfy 2 condition, i.e. 2 k l 1 A2 n l 1 2 k 2n, for n 1. (2) k=1 Clearly if l n = n 1 p, p > 1 then (2) is satisfied. Lemma 1 If B is greedy and verifies (2), then AU 2 n a ie i 2 n l 2 n 2 n a i. Proof. We can assume that a i a i+1, since B is greedy, we have 2 n a ie i U 1 l 2 k a 2 k. Hence by (2) we obtain 2 n a i 2 k a 2 k U 2 k l 1 2 k 2 n a i e i AU2 n l2 1 2n a n i e i. In the sequel we will need some of the basic concepts of the Banach space theory. First let us recall the definition of type and cotype. Namely, if (ε i ) i=0 is a sequence of independent identically distributed random variables such that P(ε i = ±1) = 1 2 i.e. sequence of Rademacher variables, we say that X has type 2 if there exists a universal constant C 1 such that E ε i x i C 1 ( x j 2 ) 1 2, for n 1, xi X 2

3 and X is of cotype 2 if there exists a universal constant C 2 such that E ε i x i C 2 ( x j 2 ) 1 2, for n 1, xi X. In particular L p spaces has type 2 if p 2 and cotype 2 if 1 p 2, for more comprehensive information see e.g. book [6] (Chapter III A, 17). Since we work rather with bases we need a definition of type and cotype 2 in these settings. We say that basis B is Riesz, if a i e i A 1 ( a i 2 ) 1 2, and Bessel, if ( a i 2 ) 1 2 A2 a i e i, where A 1, A 2 are universal constants. Obviously if X has type or cotype 2 then B is Riesz or Bessel respectively, however it would be nice to have a characterization which uses only properties of l n. Lemma 2 If l n satisfy then B is Bessel basis and if then B is Riesz basis. n=1 l 2 2 n 2n L 2, (3) l2 2 n2 n L 1, (4) n=1 Proof. We can assume that a i a i+1. The unconditionality of B implies Hence by (3) a i 2 a i e i U 1 a 2 k 2 2 k U 2 2 k l 2 2 k a 2 k e i U 1 a 2 k l 2 k. a i e i 2 L 2 U 2 a i e i 2. Thus A 2 a ie i ( a i 2 ) 1 2, where A 2 = L U. Similarly assuming that a i a i+1, we have a i e i U k=1 2 k+1 a 2 k e i UD i=2 k a 2 k l 2 k, 3

4 so using the Schwartz inequality and (4) we get a i e i UD( a 2 k 2 2 k ) 1 2 ( l2k2 2 k ) DL 2 1 U( a i 2 ) 1 2. i=0 Thus a ie i A 1 ( i=0 a i 2 ) 1 2, where A 1 = 2DL U. Remark 1 Observe that if l n = n 1 p true. and p < 2 or p > 2 then respectively (3) or (4) holds Let K X, we define σ n (B, K) = sup σ n (B, x). x K We shall say that K X is aligned with B if for each a ie i K and b i a i we have b ie i uk, where u is a universal constant. The main result of the paper is the following one. Theorem 1 Let X be a Banach space and let B be a greedy basis, Riesz or Bessel which satisfies (2) ( 2 condition). Suppose that K is aligned with B and that B is an unconditional basis for X. There exist absolute constants: C > 0 and τ N such that σ 2 n(b, K) C τ σ 2 k(b, K), for n N, n τ. It says that in some sense greedy basis aligned with K X is the best among all unconditional basis. The first papers in this direction was by Kashin [3] who proved that if X is L 2 space then for each orthogonal basis B we have σ n (B, Lip α) C(α)n α, where 0 < α 1 and Lip α is a class of Lipschitz function according to the metric d(s, t) = s t α. Next step was due to Donoho [1], [2] who proved (also under the assumption that X is L 2 space) that if there exists an orthogonal basis B which is aligned with a subset K of X and such that lim sup n n β σ n (B, K) > 0 for some β > 0, then lim sup n n β σ n (B, K) > 0 for any other orthogonal basis B. In the recent paper by DeVore, Temlyakov and Petrowa [5] the result is extent from L 2 spaces to L p, i.e. supposing X is L p space where p > 1, B is greedy, aligned with K and such that lim sup n (log 2 n) α n β σ n (B, K) > 0, where α R, β > 0 implies that lim sup n (log 2 n) α+β n β σ n (B, K) > 0, for any other unconditional basis B. Theorem 1 gives a strengthening of the main result of [5]. 4

5 Remark 2 Under the same assumptions as in Theorem 1, if lim sup(log 2 n) α n β σ n (B, K) > 0, where β > 0, α R, n then lim sup n (log 2 n) α n β σ n (B, K) > 0. Proof. Assume that lim sup n n α 2 βn σ 2 n(b, K) = 0. That means for every ε > 0 σ 2 k(b, K) εk α 2 βk, for k N(ε). Thus for n N(ε) + τ we have A τ σ 2 k(b, K) εc τ k α 2 βk. Observe that k α 2 βk cn α 2 βn, where c is a universal constant (which depends on α, β, τ only). Theorem 1 implies that σ l (B, K) σ 2 n(b, K) εccn α 2 βn, for n N(ε) + τ, l 2 n, what is impossible since lim sup n (log 2 n) α n β σ n (B, K) > 0. In fact Theorem 1 is formulated in the language of general hypothesis proposed by Wojtaszczyk [7] (Problem 9), though it does not deal with the conjecture s full strength. 2 Basic results Remark 3 Let (ε i ) be a sequence of Rademacher variables and a i, a i,j R, i, j N. For each n 1 we have E a i ε i 2 = a i 2, E a i,j ε i ε j 2 2 a i,j 2. i, i, Proof. The first equality is classical and easy so we only prove the second one. Clearly E a i,j ε i ε j 2 = i, a i,j 2 + i,j a i,j a j,i 2 i,j a i,j 2, i,j where we used inequality 2ab a 2 + b 2. Suppose B = (e i ), B = (e i) are respectively greedy and unconditional basis for X. We define c i,j and d i,j by equalities e i = d i,je j, e i = c i,je j and denote by U the unconditionality constant for B. 5

6 Lemma 3 If B is Riesz or Bessel basis then 2 n 2 n c 2 l,jd 2 j,i c2 n, where c is a certain constant (not depending on n). Proof. Fix i 1. By the Bessel property of B we have 1 = e i = c i,j e j (U ) 1 ε j c i,j e j = = (U ) 1 c i,j d j,l ε j e l (U ) 1 A 1 2 ( Thus due to Remark 3 we obtain A 2 2(U ) 2 E c i,j d j,l ε j 2 c i,j d j,l ε j 2 ) 1 2 c i,j d j,l 2, and hence 2 n 2 n c i,jd j,l 2 (U ) 2 A 2 22 n. Now we fix l 1 and assume that B is Riesz. First observe that A 1 ( a i 2 ) 1 2 a i e i = a i c i,j e j = = ( a i ( c i,j d j,l ε j ))e l (U ) 1 a i ( c i,j d j,l ε j ). If we take a i = c i,jd j,l ε j, then A 2 1(U ) 2 E c i,j d j,l ε j 2 = It proves that 2 n 2 n c i,jd j,l 2 (U ) 2 A 2 12 n. c i,j d j,l 2. Proof of Theorem 1. Fix n 1, δ > 0. The definition of σ 2 n(b, K) implies that disregarding ε s there exists x K such that σ 2 n(b, K) (1 + δ)σ 2 n(b, x) = (1 + δ)x G 2 n(x). We can assume that e i (x) e i+1(x), i 1 and hence we can take G 2 n(x) = 2 n e i (x)e i. Since K is aligned to B, whenever a ie i K and b i a i, we have b ie i uk, where u is a universal constant. Consequently u 1 e 2 k (x) 6 ε i e i K.

7 It proves that setting y k := e 2 k (x), the cube K n,k := {u 1 y k i Λ k ε i e i, ε i { 1, 1}} is contained in K. Applying the triangle inequality we obtain x G 2 n(b, x) = i=2 n +1 e i (x)e i Thus due to the unconditionality and (1) we get 2 k+1 i=2 k +1 e i (x)e i U 2 k+1 i=2 k +1 2 k+1 i=2 k +1 e i (x)e i. e 2 k (x) e i DU e 2 k (x) l 2 k, hence σ 2 n(b, K) (1 + δ)x G 2 n(b, x) (1 + δ)du e 2 (x) l k 2 k = = (1 + δ)du y k l 2 k. (5) Fix k N. Let (ε i ) be a sequence of Rademacher variables. Without losing the generality we may assume that Λ k = {1,..., 2 k } and for simplicity denote d k j = (d j,1,..., d j,2 k), c k j = (c 1,j,..., c 2 k,j), c k j, ε = ε i c i,j, c k j, ε = ε i d j,i. Since x = u 1 y k i Λ k ε i e i K n,k K for any integer m we have σ 2 m(k, B ) σ 2 m(x, B ) and the definition gives σ 2 m(x, B ) = inf S Σ2 m(b ) u 1 y k 2 k ε ie i S, so σ 2 m(k, B ) inf S Σ 2 m(b ) u 1 y k Furthermore, by the unconditionality ε i e i S = inf Λ =2 m inf u 1 y k a Λ j c k R j, ε e j a Λ j e j. u 1 y k c k j, ε e j a Λ j e j (U ) 1 u 1 y k c k j, ε e j = j Λ = (uu ) 1 y k c k j, ε e j 2 k c k j, ε e j = (uu ) 1 y k ε i e i c k j, ε e j, 7

8 hence σ 2 m(k, B ) (uu ) 1 y k inf 2k ε i e i Λ =2 c k m j, ε e j. (6) Again using the unconditionality and e j = d j,ie i we get U 1 ε i e i c k j, ε e j = Due to Lemma 1 (ε i (ε i d j,i c k j, ε )e i i=2 k +1 d j,i c k j, ε e i d j,i c k j, ε )e i. (7) ε i e i c k j, ε e j A 1 U 1 2 k l 2 k ε i d j,i c k j, ε. Observe that ε i d j,i c k j, ε 1 ε i d j,i c k j, ε, hence by (6) and (7) we have AU 2 U σ 2 m(k, B ) y k l 2 k(1 2 k = y k l 2 k(1 2 k sup Λ =2 m The Schwartz inequality gives sup Λ =2 m ε i d j,i c k j, ε ) = ε i d j,i c k j, ε ) y k l 2 k(1 2 k sup d k j, ε c k j, ε ). Λ =2 m d k j, ε c k j, ε Λ 1 2 ( d k j, ε 2 c k j, ε 2 ) 1 m ( d k j, ε 2 c k j, ε 2 ) 1 2. Applying Remark 3 and Lemma 3, we get E( d k j, ε 2 c k j, ε 2 ) 1 2 ( E d k j, ε 2 c k j, ε 2 ) 1 2 Thus (2 d j,i 2 c j,l 2 ) 1 2 2c2 k 2. uau 2 U σ 2 m(k, B ) y k l 2 k(1 2c2 k 2 m 2 2 k 2 ) = yk l 2 k(1 2c2 m k 2 ). Taking m = k τ, and using (5) we get (1 + δ)uadu 3 U σ 2 k τ (K, B ) (1 + δ)(1 2c2 τ 2 )DU y k l 2 k (1 2c2 τ 2 )σ2 n(k, B). 8

9 We can find suitable τ such that 1 > 2c2 τ 2. Since δ > 0 is arbitrary we obtain σ 2 n(k, B) C σ 2 k(k, B ), τ where C = uadu 3 U (1 2c2 τ 2 ) 1. It ends the proof. Acknowledgment I would like to thank professor Przemyslaw Wojtaszczyk for involving me with the problem. References [1] Donoho, D. L. (1993) Unconditional Bases Are Optimal Bases for Data Compression and for Statistical Estimation. Appl. Comp. Harm. Anal. 1, pp [2] Donoho, D. L. (1996) Unconditional Bases and Bit-Level Compression. Appl. Comp. Harm. Anal. 3, [3] Kashin, B.S. (1985). Approximation properties of complete orthonormal systems. (Russian) Studies in the theory of functions of several real variables and the approximation of functions. Trudy Mat. Inst. Steklov. 172, [4] Konyagin, S. V. and Temlyakov V. N. (1999). A remark on greedy approximation in Banch spaces. J. Approx. 5, [5] DeVore, R., Petrowa, G. and Temlyakov V. (2003). Best basis selection for approximation in L p. J. of FoCM 3, pp [6] Wojtaszczyk, P. (1991). Banach spaces for analysts. Cambridge Studies in Advanced Mathematics. 25, xiv+382 pp. Cambridge University Press, Cambridge. [7] Wojtaszczyk, P. (2002). Greedy type bases in Banach spaces, Constructive Theory of Function Theory, Varna 2002 pp , Darba, Sofia 9