RIESZ BASES AND UNCONDITIONAL BASES

Transcription

1 In this paper we give a brief introduction to adjoint operators on Hilbert spaces and a characterization of the dual space of a Hilbert space. We then introduce the notion of a Riesz basis and give some equivalent definitions. Afterwards we discuss unconditional bases and prove that the Haar system is an unconditional basis for L p (R), where < p < and that L [, ] does not contain an unconditional basis. We begin our discussion of adjoint operators by looking at an important theorem. Theorem (Riesz s Lemma). Let H be a Hilbert space. If y H then the equation φ y (x) = x, y defines a bounded linear functional φ y H ; and furthermore, φ y H = y H. Conversely, if φ H then there exists a unique element y H such that φ = φ y. Proof. Let y H and let φ y (x) = x, y. Then φ y is linear by the linearity of, y. Further, for x H we have that φ y (x) = x, y x y and so φ y y and so φ y is bounded and hence φ y H. Further, ( ) y φ y = y y y, y = y and so φ y = y. Now, suppose φ H where φ is not identically zero( else trivial). Let M = ker(φ). Since φ is continuous we have that M = ker(φ) = φ ({}) is closed. Also, since φ is linear and nonzero we have that M is a closed subspace of codimension and so there exists z H so that z M and z =. We want to find y H so that φ =, y so let y = αz for some α which we will determine. Then y M so for any x M we have that φ(x) = = φ y (x). Also, we want φ(z) = z, y = z, αz = α z 2 = α so we want to let α = φ(z). Then, for x M = span{z} we have that x = βz for some β C and φ(βz) = βφ(z) = β z, φ(z)z = βz, y = x, y Lastly, in order to show uniqueness suppose x, y = x, y 2 for all x H. Then x, y y 2 = for all x H and so and so y = y 2. y y 2 = y y 2, y y 2 = Proposition 2. Suppose H and K are Hilbert spaces and T L(H, K), where L(H, K) is the set of bounded linear operators from H to K. Then there exists S L(K, H) so that T x, y = x, Sy for all x H and all y K. Date: December 22.

2 Proof. Let y K and define ϕ : H C by ϕ(x) = y, T X for all x H. Clearly ϕ is linear. Further, if x H such that x then ϕ(x) = y, T x y T x y T and so ϕ is bounded and hence ϕ H. Then, by Riesz s Lemma, there exists a unique Sy H so that x, Sy = ϕ(x) = y, T X = T x, y for all x H. Clearly S is linear since, if y, y 2 K and α C then x, αsy + Sy 2 = α x, Sy + x, Sy 2 = α T x, y + T x, y 2 = T x, αy + y 2 for all x H but S(αy + y 2 ) is the unique element of H so that x, S(αy + y 2 ) = T x, αy + y 2 so we must have that S(αy + y 2 ) = αsy + Sy 2. Further, for y K such that y we have Sy = Sy ( ) ( ) Sy, Sy Sy = T, y Sy Sy T y T Sy So we have that S L(K, H). Definition. If H and K are Hilbert spaces and T L(H, K) then the operator S in Proposition 2 is called the adjoint of the operator T and is denoted by T. Definition 2. A sequence of vectors (x n ) n A in a Hilbert space H is called a Riesz sequence if there exist constants < c C such that ( ) /2 c a n 2 ( ) /2 a n x n C a n 2 n A n A n A for all sequences of scalars (a n ) n A. A Riesz sequence is called a Riesz basis if additionally span{x n : n A} = H. Theorem 3. (a) If (x n ) is a Riesz basis in a Hilbert space H then there exists a sequence (x n) H so that x m, x n = δ m,n where δ m,n is the Kronecker delta function. Further, the sequence (x n) is also a Riesz basis in H. (b) If (x n ) is a Riesz basis in H then there exist constants < c C such that for all x H. ( ) /2 c x x, x n 2 C x Proof. Let (x n ) be a Riesz basis in H and let (e n ) be the usual basis for l 2 (Z), i.e., e n is the nth unit vector in l 2 (Z). Further, let A : span{e n : n Z} H be the unique linear operator such that A e n = x n for all n Z. Now, let x l 2 (Z). Since span{e n : n Z} = l 2 (Z), there exists (y n ) span{e n : n Z} so that y n x. Since y n span{e n : n Z} we have that y n = k Z αk ne k where all but finitely many α k n 2

3 are zero. Then, since y n x we have that (y n ) is Cauchy so y n y m as n, m. Then A y n A y m = k Z(α ( ) /2 n k αm)x k k C αn k αm k 2 = C y n y m as n, m. So the sequence (A y n ) is Cauchy in H and since H is complete, there exists Ax H such that A y n Ax so it seems we can extend A to A where A : l 2 (Z) H but we should first check that A is well-defined. To this end, suppose (y n ), (w n ) span{e n : n Z} such that y n x and w n x. Then z n = (y n w n ). Let z n = αk ne k for all n Z. Then A z n = ( ) /2 αnx k k C αn k 2 = C z n So A y n A w n and so A is well-defined. Further, let x l 2 (Z). Then x = a ne n and hence Ax = a ne n C x and so A is continuous. Also, if y H then y = a nx n for some (a n ) C. Let x = a ne n. Then ( ) /2 x = a n 2 c a n x n < and so x l 2 (Z). Further, Ax = a nx n = y so A is onto. Further, suppose x l 2 (Z) and Ax =. Let x = a ne n. Then ( ) /2 c a n 2 a n x n = Ax = and so x = and hence A is injective. Therefore A is well-defined and, by the open mapping theorem, A is bounded so, by Proposition 2, we can discuss(a ). Now, for all n Z define x n = (A ) e n. Then, x m, x n = Ae m, (A ) e n = A Ae m, e n = e m, e n = δ m,n Further, we know A is bejective and so ker((a ) ) = ran (A ) = {} and ran ((A ) ) = ker(a ) = {} which gives that (A ) is also bejective and so (x n) is a Riesz basis for H. Now, for part (b), let x H. Then, by above, there exists (a n ) C so that x = a nx n. Then x, x n 2 = 2 a n x n, x n = 2 a n x n, x n = a n 2 Further, since we know (x n) is a Riesz basis we have that x = ( ) /2 ( ) /2 a n x n C a n 2 = C x, x n 2 k Z k Z 3

4 and ( ) /2 ( ) /2 x, x n 2 = a n 2 c a n x n = c x Definition 3. A sequence of vectors in a normed vector space X is said to be complete if its linear span is dense in X. Definition 4. Two sequences (x n ) n and (x n) n in a Hilbert space H are said to be biorthogonal if x m, x n = δ m,n for every m, n N. It is worth noting that if we further assume H is separable and (x n ) n is a basis for H then (x n) n is also a basis for H and every vector x H has a unique representation in the form x = x, x n x n however, we will do so without proof. n= Theorem 4. Let H be a separable Hilbert space and let (x n ) n H. Then the following are equivalent: () The sequence (x n ) n is a Riesz basis in H. (2) There is an equivalent inner product on H for which (x n ) n becomes an orthonormal basis for H. (3) There exists an orthonormal basis (e n ) n H and T L(H) which is invertible so that T ({e n : n N}) = {x n : n N}. (4) The sequence (x n ) n is complete in H and the matrix ( x i, x j ) i,j= operator on l 2. generates a bounded invertible (5) The sequence (x n ) n is complete in H and possesses a complete biorthogonal sequence (x n) n such that n= x, x n 2 < and n= x, x n 2 < for all x H. Proof. (3) (2). By assumption there exists an orthonormal basis (e n ) n H and T L(H) which is invertible so that T ({e n : n N}) = {x n : n N}. Without lost of generality, suppose T e n = x n for all n N (otherwise we can just reorder). Define, N : H H C where x, y N = T x, T y. Claim:, N is an inner product. Clearly x, x N. If x, x = then T x = and so x = since T is injective. Further, and αx + y, z N = T (αx + y), T z = α T x, T z + T y, T z = α x, z N + y, z N x, y = T x, T y = T y, T x = y, x N So, N defines an inner product on H. Further, for n, m N, x n, x m N = T x n, T x m = e n, e m = δ m,n So (x n ) n is an orthonormal basis with respect to, N. To see that, N is equivalent to, let x H. Then x 2 N = x, x N = T x, T x T 2 x 2 4

5 Similarly, x 2 = x, x = T T x, T T x T 2 T x 2 = T 2 x 2 N So the inner products are equivalent. (2) () Suppose there is an equivalent inner product, say, N, on H for which (x n ) n is an orthonormal basis for H. Let x H. Then x = c nx n for some (c n ) C. Then x N = ( ) /2 c n x n N = c n 2 but, N is equivalent to, and so there exist d, d 2 > so that d x N x c 2 x N and so ( ) /2 d c n 2 ( ) /2 c n x n d 2 c n 2. () (5). The implication here is an immediate result from Theorem 3. () (3). Let (e n ) n be an orthonormal basis for H and let T : H H be the unique linear operator where T e n = x n. Let x H. Then x = α ne n for some (α n ) n C. Then, since (x n ) n is a Riesz basis, T x 2 = α n x n 2 B α n 2 = B x 2 So T is bounded. Further, let S : H H be the unique linear operator such that Sx n = e n. Let x H. Then x = α nx n for some (α n ) n C. Then, since (x n ) n is a Riesz basis, Sx 2 = α n e n 2 = α n 2 A α n x n 2 = A x 2 So S is also bounded. Clearly ST = T S = I and so T is invertible. (3) (4). Let (e n ) n H be an orthonormal basis and let T : H H be a bounded invertible operator such that T e n = x n for all n N. Since T is bounded and invertible we know that T is bounded and invertible, so T T is bounded and invertible, so if T T = (a ij ) i,j= is the matrix representation of T T with respect to (e n ) n then (a ij ) i,j= is a bounded and invertible operator on l 2. Further, a ij = T T e j, e i = T e j, T e i = x j, x i So (a ij ) i,j= = ( x i, x j ) i,j= and therefore ( x i, x j ) i,j= is a bounded and invertible operator on l 2. (4) (). Suppose (x n ) n is complete in H and the matrix M = ( x i, x j ) i,j= is a bounded and invertible operator on l 2. Let (e n ) n be an orthonormal basis for H and let T : H H where T x = e i x i, x j c j i N j N for H x = i N c ie i. T is really nothing but the matrix M acting on H so T is also bounded, linear, and invertible. Also 5

6 ( ) T c i e i, c i e i = e i x i, x j c j, c k e k i N i N i N j N k N = x i, x j c j c k e i, e k i N k N j N = x k, x j c j c k k N j N = c j x j, c k x k k N j N = c j x j, c k x k j N k N = c j x j 2 j N So T x, x for all x H, i.e., T is a positive operator. So, we can define T which is a positive, bounded, linear operator so that T T = T. Note that positive operators are self-adjoint and so, from above, we have, for x = j N c jx j, x 2 = T x, x = T T x, x = T x, T x = T x 2 T 2 x 2 = T 2 c i 2 i N Note that above we have x 2 = T x 2 and so c i 2 = c i e i 2 = ( T ) c j x j 2 ( T ) 2 j Nc j x j 2 i N i N j N Therefore, ( c i 2 c j x j 2 T ) 2 i N j N (3) (5). We have already seen that (3) () and Theorem 3 shows that () (5). (5) (3). Let A : H l 2 where Ax = ( x, x n ) n. Claim: A is closed. Let (z k ) k H such that z k z and Az k y l 2. Since Az k y we have that z k, x n y n 2 n as k where y = (y n ) n. Then we have that z k, x n y n as k for all n N but z k z and so z k, x n z, x n as k and so z, x n = y n for all n N and so Az = y and therefore A is closed. Then, by the Closed Graph Theorem, A is bounded and so x, x n 2 = ( x, x n ) n 2 A 2 x 2 n for all x H. Similarly, if we define B : H l 2 such that Bx = x, x n then B is bounded and so x, x n 2 B 2 x 2 n 6

7 Let (e n ) n be an orthonormal basis for H and define S : span{x n : n N} H by Sx = n k= c ke k where x = n k= c kx k. Since (x n) n is a biorthogonal sequence for (x n ) n, we know that c k = x, x k and so n n () Sx 2 = x, x k e k 2 = x, x k 2 B 2 x 2 Similarly, if we define T : span{x n : n N} H by T x = n k= c kx k then k= k= (2) T x A 2 x 2 Further, since (x n ) n and (x n) n are complete, by () and (2), S and T can be extended to bounded linear operators on H. Let x, y H. Since (x n ) n and (x n) n are complete, there exist (c n ) n, (d n ) n C so that x = k= c kx k and y = k= d kx k. Let n N. Then ( n ) ( n ) n n n S c k x k, T d k x k = c k e k, d k e k = c k d k And So, k= k= k= n n c k x k, d k x k = k= k= k= n k= j= k= k= n c k d k x k, x k = k= k= n c k d k k= ( n ) ( n ) n n S c k x k, T d k x k = c k x k, d k x k Then, by continuity, we have that Sx, T y = x, y and so x, S T y = x, y for all x, y H, hence x, (S T I)y = for all x, y H and therefore S T I =, i.e., S Y = I. So S T = I implies S is onto. Further, S is onto and {} = ran (S) = ker(s ) and so S is also injective and hence (S ) exists. Further, (S ) is bounded by the Open Mapping Theorem and hence is a bounded invertible operator. Therefore S is invertible and so S is invertible and therefore S is a bounded invertible operator so that S {e n : n N} = {x n : n N}. k= Definition 5. Let X be a Banach space and let (x n ) n X. We say the series x n converges unconditionally to x if and only if there exists C > so that we have β n x n C x for all finitely non-zero sequences (β n ) n where β j {, }. Definition 6. A basis {x n : n N} of a Banach space X is said to be an unconditional basis if and only if in the unique representation x = α nx n the series converges unconditionally. Definition 7. For x L p (R), where p <, define h(x) = if x 2, h(x) = if 2 < x and h(x) = otherwise. Further, let h j,k (x) = 2 j/2 h(2 j x k) for j, k Z. Then the system {h j,k : j, k Z} is referred to as the Haar system. 7

8 For p < it is a well-known fact that the Haar system is a basis for L p (R) so we will simply assume this without proof. Our next goal is to show the Haar basis is an unconditional basis for L p (R), where < p <. For this, we will first need to consider a couple of theorems. Calderon-Zygmund Lemma. Let f : R n C be non-negative and integrable and let α >. Then there exists sets F and Ω such that: () R n = F Ω, (2) f(x) α almost everywhere on F, and (3) Ω is a union of cubes; Ω = k Q k whose interiors are mutually disjoint, and so that, for each Q k, we have that α < f(x)dx 2 n α µ(q k ) Q k Proof. Since f L (R n ), we have that R n f(x)dx = c < so choose k Z so that, if Q is a cube in R n with edge lengths 2 k then µ(q) > c α. Then, f(x)dx f(x)dx α µ(q) Q µ(q) R c c = α n Now, divide R n into cubes with disjoint interiors whose side lengths are all 2 k and let Q be such a cube. Divide Q into 2 n congruent cubes by bisecting each of its sides. Let Q be one of our new subcubes. If µ(q ) f(x)dx > α then let Q be one of the Q Q k mentioned in the statement of the proof. Note that α < µ(q f(x)dx ) Q µ(q f(x)dx = 2 n f(x)dx 2 n α ) Q µ(q) Q If instead, we have that µ(q ) f(x)dx α then divide Q into 2 n congruent cubes and repeat the process. Q Let Ω = k Q k and F = Ω c. Let x F. Then, for any cube Q constructed in our process which contains x, we have that µ(q) f(x)dx α and so Q f(x) = lim f(y)dy α a.e. diam(q) µ(q) Q Where the Q s in the limit can be taken as a sequence of cubes from our construction containing x whose diameters are converging to zero. This completes the proof of our lemma. Calderon-Zygmund Decomposition. Let f : R n C be non-negative and integrable and let α >. Then there exists functions g and (b i ) i N so that f = g + i N b i where the b i s are supported on disjoint dyadic cubes Q i. Further, if b = i N b i and Ω = i N Q i then we also have (i) g(x) = f(x) α for all x / Ω. (ii) g(x) 2 n α for all x Ω. (iii) Ω α f. (iv) Q i b i (x)dx = for all i N. 8

9 Proof. Using the same F and Ω = k Q k from the Calderon-Zygmund Lemma define { f(x) if x F g(x) = µ(q j) Q j f(t)dt if x Q j and let b = f g and define b i = b Qi. Then everything is immediate. We will also need a theorem by Marcinkiewicz which we will not be proving. Marcinkiewicz Interpolation Theorem. Let p < p and assume that T : M(U, µ) M(V, ν) is a sublinear operator of weak type (p, p ) and (p, p ). Then, for each p < p < p, T is of strong type (p, p). Theorem 5. The Haar basis is an unconditional basis for L p (R), where < p <. Proof. To prove the Haar basis is an unconditional basis we need to show β j,k f, h j,k h j,k C f j Z k Z for some C > where β = (β j,k ) j,k Z {, } has finite support. In otherwords, if we define T β : L p (R) L p (R) by T β f = β j,k f, h j,k h j,k j Z k Z then our goal is to show T β is bounded. To this end, we first want to show T β is of weak type (,). Let f L p (R) and let α >. By the Calderon-Zygmund decomposition there exists functions g and (b i ) i N so that f = g + i N b i where the b i s are supported on disjoint dyadic intervals Q i. Further, if b = i N b i and Ω = i N Q i then we also have (i) g(x) = f(x) α for all x / Ω. (ii) g(x) 2α for all x Ω. (iii) Ω α f. (iv) Q i b i (x)dx = for all i N. Clearly {x R : T β f(x) > α} {x R : T β g(x) > α 2 } {x R : T βb(x) > α 2 } and so (3) {x R : T β f(x) > α} {x R : T β g(x) > α 2 } + {x R : T βb(x) > α 2 } 9

10 Since the Haar system is an orthonormal basis for L 2 (R), we know T β is bounded as an operator on L 2 (R). So, we have that {x R : T β g(x) > α 2 } 4 α 2 Claim: If x / Q i then T β b i (x) =. Well, 4 α 2 R R 4 α 2 ( 4 α 2 ( T β g(x) 2 dx by Chebyshev g(x) 2 dx since T β is strong (2,2) R Ω R g(x) 2 dx + α f(x) dx + 4 α 2 ( α f + 4α 2 Ω ) Ω Ω ) g(x) 2 dx ) 4α 2 dx = C α f by (iii) T β b i (x) = j Z k Z β j,k h j,k (x) b i (y)h j,k (y)dy Q i by (i) and (ii) Let j, k Z. If Q i and Q j,k are disjoint (where Q j,k is the support of h j,k ) then h j,k (y) = for all y Q i and so β j,k h j,k (x) b i (y)h j,k (y)dy = Q i If Q i and Q j,k are not disjoint then one must be a subset of the other since they are dyadic intervals. If Q j,k Q i then x / Q j,k and so h j,k (x) = which means β j,k h j,k (x) b i (y)h j,k (y)dy = Q i If Q i Q j,k and Q i Q j,k then h j,k is constant on Q i and since Q i b i (y)dy = we have that Q i h j,k (y)b i (y)dy = and therefore Hence, T β b i (x) = for all x / Q i. T β b(x) α 2 and so {x R : T βb(x) > α 2 } Ω and therefore β j,k h j,k (x) b i (y)h j,k (y)dy = Q i So, if x / Ω then x / Q i for all i N and so T β b(x) = and so So, from (3), we have that {x R : T β b(x) > α 2 } Ω α f {x R : T β f(x) > α} C α f + α f = c f and so T β is of weak type (,). Since we know T β is of strong type (2,2), by the Marcinkiewicz interpolation theorem, T β is bounded on L p (R) for < p 2. A simple duality argument then shows T β is bounded for p > 2.

11 Our next objective is to show L [, ] does not have an unconditional basis. To this end, we will show something even stronger, which is that L [, ] is not isomorphic to a subspace of a space with an unconditional basis. First we will need some lemmas and definitions. Definition 8. The space of all sequences (c n ) n C such that c n is a linear space and it is denoted by c. Definition 9. The sequence of Rademacher functions (r n ) is defined by r (t) = for all t [, ] and, for k, r k+ (t) = if and r k+ (t) = if t t 2 k l= 2 k l= [ ) 2l 2 2l, 2k+ 2 k+ [ ) 2l 2 k+, 2l 2 k+ Lemma 6. Let α [, ]. If f and g are two positive and measurable functions on [, ] we have ( /α ( /α ( ) /α f (w)dw) α + g (w)dw) α (f + g) α (w)dw Proof. Let p = α and let q = p. Then, from Holder s inequality we have Similarly, So, f α f α (w) (w)dw = (f(w) + g(w)) (f(w) + α/q g(w))α/q dw ( ) /p ( f(w) dw (f(w) + g(w)) α dw (f(w) + g(w)) /q ) /q ( ) /p ( g α g(w) ) /q (w)dw dw (f(w) + g(w)) α dw (f(w) + g(w)) /q ( /α ( ) /α f (w)dw) α + g α (w)dw ( ) /pα ( f(w) dw (f(w) + g(w)) α dw (f(w) + g(w)) /q ( ) /pα ( g(w) + dw (f(w) + g(w)) α dw (f(w) + g(w)) /q ( ) /qα (f(w) + g(w)) /q dw (f(w) + g(w)) α ( = (f(w) + g(w)) α dw ) /α ) /qα ) /qα The next theorem we need is Khintchine s Inequalities but this will be given without proof.

12 Khintchine s Inequalities. Let (r n ) be the sequence of Rademacher functions on [, ]. Then, for all p [, ), there exist constants A p, B p > such that, for all n N and a,..., a n R: ( n ) /2 ( ) /p ( n n ) /2 A p a 2 i a i r i (t) p dt B p a 2 i i= i= i= Lemma 7. The space c is not isomorphic to a subspace of L [, ]. Proof. Suppose T is an isomorphism from c into L [, ]. Let (e n ) n denote the unit vectors of c and let x n = T e n for all n N. By Khintchine s Inequalities, for a fixed w [, ], we have and so, ( n ) /2 A x i (w) 2 i= i= n x i (w)r i (t) dt A ( n ) /2 x i (w) 2 dw i= = = T = T n x i (w)r i (t) dtdw i= n x i ( )r i (t) dt i= ( n ) T e i r i (t) dt T i= dt n e i r i (t) dt i= Further, and so = e i 2 = T x i 2 T 2 x i 2 T 2 x i 2 2

13 Then, by Lemma 6, So we have that ( n ) /2 ( n ( ) 2 ) /2 x i (w) 2 dw (x i (w) 2 ) /2 dw i= i= ( n = x i 2 = i= ( n i= ( n + T 2 ) /2 ) /2 T 2 ) /2 ( ) /2 n + A T 2 T for all n N which is clearly a contradiction since T <. Theorem 8. L [, ] is not isomorphic to a subspace of a space with an unconditional basis. Proof. Let x L [, ] and let (r n ) n be the sequence of Rademacher functions. Claims: (i) r n x in the weak topology on L [, ], i.e., r n(t)x(t)h(t)dt for all h (L [, ]) = L [, ] (ii) x + r n x x Suppose x = A where A = [ 2l 2, 2l ) and let y = 2 k+ 2 k+ B where B = [ 2i 2 2, 2i j+ 2 ). Then, if n > k, j, we have j+ that r n(t)x(t)y(t)dt = so, if g = v m= a m Am where A m = [ 2im 2 2, 2im jm+ 2 ) then jm+ r n(t)x(t)g(t)dt. Let h L [, ]. Then h L [, ] and so, if ɛ >, there exists a linear combination of indicator functions of dyadic intervals g L [, ] so that h g < ɛ. Then, r n (t)x(t)h(t)dt r n (t)x(t)(h(t) g(t)) dt + r n (t)x(t)g(t)dt h(t) g(t) dt + r n (t)x(t)g(t)dt ɛ + r n (t)x(t)g(t)dt and, from above, r n(t)x(t)g(t)dt so r n(t)x(t)h(t)dt. Further, since the span of indicator functions of dyadic intervals is dense in L [, ] we have that r n x weakly for all x L [, ]. Further, if 3

14 x = A where A is as above, then for n > k, we have x(t) + r n (t)x(t) dt = x(t) + r n (t) dt = + r n (t) dt A = 2dt + dt A D a D = 2 A 2 = x where D is the union of dyadic intervals as in the definition of the Rademacher functions. Further, by the density of the span of indicator functions of dyadic intervals, we have that x + r n x x for all x L [, ]. Suppose L [, ] is a subspace of Y where Y is a Banach space with an inconditional basis (e n ) n. Let n N and define P n : Y Y by ( ) P n a k e k = k= n a k e k Let x L [, ] such that x =. then P n x x so there exists N N so that x P N x 2. Note that dim(ran(p N )) < and so P N is compact. Further, r n x weakly and so, since P N is compact, we have that P N r n x in L [, ], hence there exists m N so that P N r k x 2 2 all k m. Further, since x + r n x x =, there exists m 2 N so that x + r k x 2 2 for all k m 2. Let k max{m, m 2 } and define x = r k x. Then there exists N N so that x P N x < 2 2 and we can do the same as above to find k 2 N so that P N (r k2 (x + x ) 2 3 and x + x + r k2 (x + x ) x + x 2 3 define inductively a sequence (k i ) i and a sequence (N i ) i so that and k= and let x 2 = r k2 (x + x ). We can continue in this fashion to P Ni (r ki (x + + x i )) 2 i+ x + + x i + r ki (x + + x i ) x + + x i 2 i+ x i = r ki (x + + x i ) x i P Ni x i 2 i+ Then we have that 2 x n x + + x n 2 Further, let u n = P Nn x n P Nn x n. Then, for all n N, u n x n = P Nn x n P Nn x n x n P Nn x n x n + P Nn x n 2 n+ + 2 n+ = 2 n for 4

15 So n= u n x n < and hence (u n ) n and (x n ) n are equivalent. Further, since (e n ) n is an unconditional basis so is (x n ) n. So there exists C > so that, for any A N and (a n ) n C, we have So, if (a n ) n is a finite sequence then Also, n A a n x n C n= n A a n x n C a n x n n= a n x n C sup a n x n 2C sup n N a n n= a n x n 2C sup a n n= So (x n ) n is equivalent to the unit vecot basis of c which contradicts Lemma 7, so we must have that L [, ] cannot be embedded into a space which possesses an unconditional basis. 5

16 References. E. Hernandez and G. Weiss (996), A First Course on Wavelets. New York: CRC Press. 2. Guerre-Delabriere, S. (992), Classical Sequences in Banach Spaces. New York: Marcel Dekker, Inc. 3. Stein, E.M. (993), Harmonic Analysis: Real-Variable Methods, Orthogonality and Oscillatory Integrals. Princeton: Princeton University Press. 4. Sunder, V.S. (998), Functional Analysis: Spectral Theory. Boston: Birkhauser Advanced Texts. 5. Wojtaszczyk, P. (99), Banach Spaces for Analysts. Cambridge: Campbridge University Press. 6. Wojtaszczyk, P. (997), A Mathematical Introduction to Wavelets. Cambridge: Campbridge University Press. 7. Young, R.M. (98), An Introduction to Nonharmonic Fourier Series. New York: Academic Press, Inc. 6