Behaviour of the approximation numbers of a Sobolev embedding in the one-dimensional case.

Transcription

1 Behaviour of the approximation numbers of a Sobolev embedding in the one-dimensional case. D.E.Edmunds a, J.Lang b, a Centre for Mathematical Analysis and its Applications, University of Sussex, Falmer, Brighton BN1 9QH, UK b The Ohio State University, Department of Mathematics, 100 Math Tower, 231 West 18th Avenue, Columbus, OH , USA Abstract We consider the Sobolev embeddings E 1 : W 1,p 0 (a, b) L p (a, b) or E 2 : L 1,p (a, b)/{1} L p (a, b)/{1}, with < a < b < and 1 < p <. We show that the approximation numbers a n (E i ) of E i have the property that lim na n(e i ) = c p (b a) (i = 1, 2) n where c p is a constant dependent only on p. Moreover we show the precies value of a n (E 1 ) and we study the unbounded Sobolev embedding E 3 : L 1,p (a, b) L p (a, b) and determine precisely how closely it may be approximated by n-dimensional linear maps. Key words: Approximation numbers, Sobolev Embedding, Hardy-type operators, Integral operators 1991 MSC: 47G10, 47B10 Corresponding author. addresses: D.E.Edmunds@sussex.ac.uk (D.E.Edmunds), lang@math.ohiostate.edu (J.Lang). URL: lang (J.Lang). Preprint submitted to OSU MRI Preprint Series 11 September 2002

2 1 Introduction. Let Ω be a bounded subset of R n with smooth boundary, let 1 < p < and consider the embedding E 1 : W 1,p 0 (Ω) L p (Ω) where W 1,p 0 (Ω) is the usual first-order Sobolev space of functions with zero trace. This space is a closed subspace of the Sobolev space W 1,p (Ω). It is wellknown that E 1 is compact. More precise information about E 1 is available via its approximation numbers, for there are positive constants c 1 and c 2, depending only on p and Ω, such that the m-th approximation number a m (E 1 ) of E 1 satisfies c 1 m a m(e 1 ) c 2 m, m N (1) Of course, this is a very special case of quite general results concerning the approximation numbers of embeddings between function spaces, for which we refer to (T) and (ET). When p = 2 it is possible to sharpen (1) by using the familiar relation a m (E 1 ) = 1 λ 1/2 m between the approximation numbers of E 1 and the eigenvalues λ m of the Dirichlet Laplacian. Since the behaviour of the eigenvalues is well-known, it follows that lim ma m (E 1 ) exists; and even sharper statements about the asymptotic behaviour of a m (E 1 ) can be made. It is natural to ask whether or not lim ma m (E 1 ) exists when p is not equal to 2. In (EHL) a new technique was given for the study of the approximation numbers of the Hardy-type operator T on a tree Γ: x (T f)(x) = v(x) 0 f(t)u(t)dt, x Γ. Using this it was shown that T : L p (Γ) L p (Γ) has approximation numbers a m (T ) for which lim ma m (T ) exists, when 1 p. This technique was improved and extended in (EKL), where in the case in which Γ is an interval and p = 2, remainder estimates were obtained. These results were extended in (L) to cover the cases 1 < p <. 2

3 In the present paper we obtain sharper information about a m (E 1 ) than was previously known. We deal only with the case in which n = 1 and Ω is a bounded interval in the line. The techniques of this paper are based on methods derived from (EHL), (EKL), (L), (Li2) and (DM). In more detail, for the Sobolev embeddings E 1 : W 1,p 0 (I) L p (I) E 2 : L 1,p (I)/{1} L p (I)/{1}, where I = (a, b), < a < b < and L 1,p (I) is the space of all u L p loc (I) with derivative u L p (I), we show that there is a positive constant α p such that lim ma m(e i ) = α p for i = 1 or 2. m Moreover, it turns out that for every m N, there is a linear map P m with rank P m = m such that E 2 P m = α p /m a m+1 (E 2 ) α p /(m + 1). For embedding E 1 we have that for every m N, there is a linear map B m with rank B m = m such that E 1 B m = α p /(m + 1) = a m+1 (E 1 ). We also study the best approximation of the unbounded Sobolev embedding E 3 : L 1,p (I) L p (I) by linear maps of finite rank. We show that for every m N, there is a linear map R m with rank R m = m such that E 3 R m = α p /(m) = inf{ E 3 P ; P linear map, rank P < m + 1}. We also show that α p = ( 1 λ n,i ) 1/p where λ n,i is the first eigenvalue of a p Laplacian eigenvalue problem. Our conclusion appears to be the first result of this kind in the literature, apart from the special case p = 2. It remains to be seen whether or not this can be extended to higher dimensions. 3

4 2 Preliminaries and technical results. Throughout the paper we shall assume that < a < b < and that I = (a, b). We also assume that 1 < p < and denote by. p or. p,i the usual norm on the Lebesgue space L p (I). By the Sobolev space W 1,p 0 (I) we understand, as usual, the space of all functions u L p (I) with finite norm u p,i and zero trace. We consider the embedding E 1 : W 1,p 0 (I) L p (I) (2) and define the norm of E 1 by E 1 = u p,i. (3) u p,i >0 u p,i Plainly E 1 < ; moreover, it is well known (see, for example, (EE), Theorem V.4.18) that E 1 is compact. We will consider in this paper also the approximation numbers for the embedding E 2 : L 1,p (I)/{1} L p (I)/{1}, where L 1,p (I) is the space of all functions u L p loc (I) with finite pseudonorm u 1,p which vanishes on the subspace of all constant functions. By L 1,p /{1} we mean the factorization of the space L 1,p (I) with respect to constant functions, equipped with the norm u p,i. Then we have f L 1,p /{1} if and only if f p,i = inf c R f c p,i. In a similar way L p (I)/{1} is defined. The norm of E 2 is defined by E 2 = u p>0 u p u p. It is obvious that E 2 = a 1 (E 2 ) < and also lim n a n (E 2 ) = 0. We will also consider the unbounded embedding E 3 : L 1,p (I) L p (I). Since L 1,p (I) is defined by the pseudonorm u 1,p and E 3 is unbounded, we 4

5 will study the best approximation of E 3 by linear maps of finite rank (a n (E 3 ) are not well defined). Definition 2.1 Let J = (c, d) I. We define A 0 (J) = inf u p,j >0 α R u α p,j u p,j. Since every function in W 1,p (J) is absolutely continuous, we can rewrite A 0 (J) as A 0 (J) = inf u p,j >0 α R x c u (t)dt + u(c) α p,j u p,j. From this we can see the connection between A 0 and the Hardy operator. Lemma 2.2 Let I n be a decreasing sequence of subintervals of I with I n 0 as n. Then {A 0 (I n )} is a decreasing sequence bounded above by A 0 (I) and with limit 0. Proof. In this proof we extend u W 1,p (I n+1 ) outside I n+1 by a constant, i.e. u = 0 outside I n+1. From the definition of A 0 we have for I i+1 I i, A p 0(I i+1 ) = inf u p,ii+1 >0 α R inf u p,ii+1 >0 α R inf u p,ii >0 α R x c u (t)dt α p p,i i+1 u p p,i i+1 x c u (t)dt α p p,i i u p p,i i x c u (t)dt α p p,i i u p p,i i = A p 0(I i ) and so A 0 (I i ) A 0 (I i+1 ). For A 0 (J) we have x c A 0 (J) u (t)dt p,j u p,j =1 u p,j x = u (t) dt u p,j =1 c p,j 1/p u p J 1/p = J 1/p. u p,j =1 J p,j 5

6 From this observation it follows that A 0 (I n ) 0 as I n 0. Lemma 2.3 Let J = (x, y) I. Then A 0 ((x, y)) is a continuous function of x and y. Proof. Let us pose that there are x, y I and ε > 0 such that A 0 (x, y + h n ) A 0 (x, y) > ε for some sequence {h n } with 0 < h n 0. Then we have that there is ε 1 > 0 such that A p 0(x, y + h n ) A p 0(x, y) > ε 1 for any n N. But for all h > 0, A p 0(x, y + h) A p 0(x, y) = inf u p,(x,y+h) >0 α R u p,(x,y+h) >0 u p,(x,y+h) >0 inf u p,(x,y) >0 α R t x u α p p,(x,y+h) u p p,(x,y+h) t x u α p p,(x,y) u p p,(x,y) t x ( inf u α p p,(x,y+h) α R u p p,(x,y+h) t x inf u α p p,(x,y) α R u p ) p,(x,y+h) t x ( inf u α p p,(x,y) α R u p p,(x,y+h) t x + inf u α p p,(y,h) t x α R u p inf u α p p,(x,y) α R p,(x,y+h) u p ) p,(x,y+h) inf u p,(x,y+h) >0 α R u p,(y,y+h) >0 t x u α p p,(x,y) u p p,(x,y+h) t y u p p,(y,y+h) u p p,(y,y+h) (y, y + h) p/p h p/p, and we have a contradiction. Hence A 0 (x, y+h) A 0 (x, y) as h 0. Similarly we find that A 0 (x + h, y) A 0 (x, y) as h 0 and the result follows. Lemma 2.4 Let J = (c, d) I. Then there is a function f W 1,p (J) such that A 0 (J) = f p,j f α p,j = inf. f p,j α R f p,j Proof: It is possible to find a sequence {f n } n=1 of functions in W 1,p (J) such 6

7 that for each n in N, f n p,j f n α p,j + 1/n = inf f n p,j α R f n p,j + 1/n > A 0 (J) and f n W 1,p (J) = 1. Since E is compact, it follows that there exists a subsequence of {f n }, again denoted by {f n } for convenience, which converges weakly in W 1,p (J), to f, say, and this subsequence converges strongly to f in L p (J). By a standard compactness argument we get that f n converges strongly to f in W 1,p (J) and then A 0 (J) = f p,j f α p,j = inf. f p,j α R f p,j Lemma 2.5 Let J = (c, d) I and let f be as in the previous lemma. Then f(x) = 0 only for x = (c + d)/2, f is monotone and f (c + ) = f (d ) = 0. Proof: Let f be from the previous lemma. Let f + (x) = max{f(x), 0} and f (x) = max{ f(x), 0}; then f + p p,j = f p p,j, f = f + f and {x : f(x) = 0} = 0. Since we know that for any g W 1,p (J), g 0 we have g p,j (g ) p,(0, J ) (where g is the non-increasing rearrangament of the function g). Then we have that f + p p,(0, J ) + f p p,(0, J ) (f +) p p,(0, J ) + (f ) p p,(0, J ) = A p 0(J). Now define r = {x : f(x) > 0} J and g(x) = f +(c + r x) for c x c + r and g(x) = f (c + r + x) for c + r x d. Then g p,j g p,j = A 0 (J), and g + p p,j = g p p,j. From all this we can see that we have found a function g such that: g is monotone, g(c + r) = 0 where c < c + r < d and ( g p,j / g p,j ) = A 0 (J). Now we show that g((c + d)/2) = 0 (i.e. r = (c + d)/2). Put J 1 = (c, c + r) and J 2 = (c + r, d); then we have g p p,j 1 + g p p,j 2 g p p,j 1 + g p p,j 2 = A p 0(J). (4) 7

8 Since A 0 (J) = J A 0 ((0, 1)), we see that g p p,j 1 g p p,j 1 A p 0((0, 1)) J 1 p 2 p. For if not then we can define h(x) = g(x) on (c, c + r) and h(x) = g( x + 2(r+c)) on (c+r, c+2r) and we have that inf α R h α p,(c,c+2r) = h p,(c,c+2r) and h p p,(c,c+2r) h p p,(c,c+2r) > A p 0((c, c + 2r)), which is a contradiction with the definition of A 0. Similarly we have g p p,j 2 g p p,j 2 A p 0((0, 1)) J 2 p 2 p. Observe that (4) holds if and only if g p p,j 1 g p p,j 1 = g p p,j 2 g p p,j 2 = A p 0(J) (do not forget that g p p,j 1 = g p p,j 2 ). This means that c + r = (c + d)/2 and moreover we can pose that g(x) = g( x + (c + d)) (i.e. g(x) is odd with respect to (c + d)/2). Next we show that g (c) = g (d) = 0. Note that g(c) = g(d) 0. Suppose that g (c) = g (d) < 0; then there are a number z > 0 and a sequence of numbers { } n=1 such that > c, c and g(c) g( ) c < z < 0 (i.e. c g (t)dt < ( c)z). A similar procedure can be carried out in the neighbourhood of d. Then we have z ( c) < c also we have g (t) dt ( c g (t) p dt) 1/p ( c) 1/p. And A p 0(J) = g p + c g p g p + c g p g p + ( c) g(c) p g p + ( c) z p 8

9 Since A 0 (J) > 0 and z > 0, plainly g(c) p < z p A p 0(J) + g(c) p and there exists n 1 N such that for any n > n 1 we have ( c) g(c) p < ( c) z p g p g p + ( c) g(c) z( c) p and then g p g p ) + ( c) g(c) p g p ) < d d d ( )( ( g p g p ) + ( c) z p g p ) d d d ( )( ( + ( c) g(c) z( c) p ( g p ) + z p g(c) z(c ) p (c ) 2. From this it follows that for any n > n 1, g p + ( c) g(c) p g p + ( c) z p < g p + ( c) g( ) p g p. But this means that for l n = χ (xn,d)g + χ (c,xn)g( ) we have: A p 0(J) < c l n p c l n p for any n > n 1. In view of the antisymmetry of g we define a function r n (x) = χ (c,d+c xn)g(x) + χ (d+c xn,d)g(d + c ), and have A p 0(J) < c r n p c r n p for any n > n 1. Finally we define k n (x) = χ (xn,d+c )g(x)+χ (d+c xn,d)g(d+c )+χ (c,xn)g( ). Then for n large enough we have k n c p,j A 0 (J) < inf. c R k n p,j 9

10 But this contradicts the definition of A 0 (J) : hence g (c) = g (d) = 0. Now we recall the p-laplacian eigenvalue problem, which is defined, for 1 < p <, λ > 0 and T > 0 by ( u p 2 u ) + λ u p 2 u = 0, on (0, T ), u (0) = 0, u (T ) = 0. The set of eigenvalues of this problem is given by ( ) p 2nπp 1 λ n := for each n N. T p pp 1 The corresponding eigenfunctions are u 0 (t) = c, c R \ {0} and u n (t) = T ( nπp sin p nπ p T (t T ) 2n ). Here for p > 1 we put p = p p 1 and π p = 2B( 1 p, 1 p ) = π/ sin(π/p), where B denotes the beta function. Moreover sin p (.) can be defined as the unique (global) solution to the initial value problem ( u p 2 u ) + 2p p p p 1 u p 2 u = 0 u(0) = 0, u (0) = 1. Also sin p can be expressed in terms of hypergeometric functions, see ((AS), p.263), arcsin p (s) = ps 1/p F ( 1 p, 1 p, p ; s), or arcsin p (s) = B( 1 p, 1 p, (2s p )p ) where F (a, b, c; s) denotes the hypergeometric function and B is the incomplete beta function B(1/q, 1/p, x) = x 0 z 1/q 1 (1 z) 1/p dz, 10

11 see (AS). Moreover, for s [0, p/2] we have arcsin p (s) = p 2 2s p 0 dt (1 t p ) 1/p, (note that this integral converges for all s [0, p/2] ). We note that in this paper we are using the definition of π p and sin p functions from the paper (DM) which is slightly different from the definition of π p and the sin p function used in (Li1) and (Li2). Note that as arcsin p : [0, p/2] [0, π p /2] is strictly increasing then its inverse function sin p : [0, π p /2] [0, p/2] is also strictly increasing. We extended sin p from [0, π p /2] to all R as a 2π p periodic function by the usual way as in the p = 2 case (i.e. from sin ). For later use let us define cos p (t) := sin p(t). We have that ( p 2 ) p cosp (t) p + sin p (t) p = 1 for all t R, and π p = π p. From (DM) we have T 0 sin p ( nπ p T t) p dt = T p p p 2 p (p + p) and T 0 d dt sin p( nπ p T t) p dt = n p π p pp T p 1 (p + p). See (Li2) for more information about sin p (.) and cos p (.) functions. Definition 2.6 Given J = [c, d] R we denote by u n,j (t) the n-th eigenfunction of the p-laplacian eigenvalue problem on J and by λ n,j the corresponding n-th eigenvalue. 11

12 Note that u 0,J = C, u n,j (t) = J ( ) nπp J sin p (t nπ p J 2n c), for n 1 and λ n,j = ( ) p 2nπp 1, for each n N, J p pp 1 where π p = π/ sin(π/p). It is simple to see that for any n N, {u i,j } n is a linearly independent set. Lemma 2.7 Let J = (c, d) I. Then A 0 (J) = u ( ) 1/p 1,J p,j u 1,J α p,j 1 = inf =. u 1,J p,j α R u 1,J p,j λ 1,J Proof: We can see that A 0 (J) = u 1,J p,j u K(J) u 1,J p,j where K(J) = {f; 0 < f p,j <, inf α f α p,j = f p,j }. After taking the Fréchet derivative of A p 0(J) we can see that this lemma follows from the previous observation about eigenfunction and eigenvalues for the p-laplacian problem with Neumann boundary value conditions together with Lemma 4 (more can be found in (DKN)) We recall that, given any m N, the m th approximation number a m (T ) of a bounded linear operator T : X Y, where X and Y are Banach spaces, is defined by a m (T ) := inf T F X Y, where the infimum is taken over all bounded linear maps F : X Y with rank less than m. A measure of non-compactness of T is given by β(t ) := inf T P X Y, 12

13 where the infimum is taken over all compact linear maps P : X Y. In our case we have X = W 1,p (I) and Y = L p (I). Then since L p (I) has the approximation property for 1 p, T is compact if and only if a m (T ) 0 as m, and β(t ) = lim n a n (T ). 3 The Main Theorem. Definition 3.1 Let ε > 0 and I = (a, b) R. We define N(ε, I) = inf{n; I = n I i, A(I i ) ε, I i I j = 0 for i j}. From our previous observation that A 0 (J) = ( ) 1/p 1 λ 1,J have: = (p p p 1 ) 1/p J 2π p we Observation 3.2 i) Given any ε > 0 we have N(ε, I) <. ii) Let ε > 0. Then there is a covering set of intervals (that is, a set of nonoverlapping intervals) {I i } N(ε) such that A 0 (I i ) = ε for i = 1,..., N(ε) and A 0 (I N(ε,I) ) ε. iii) For any n N there exist ε > 0, such that n = N(ε, I) and corresponding covering sets {I i } N(ε,I) for which A 0 (I i ) = ε for i = 1,...N(ε, I). Moreover we can see: Observation 3.3 Let n N and ε [ (p p p 1 ) 1/p, 2(n 1)π p (p p p 1 ) 1/p). Then N(ε, I) = n. From this observation we obtain the following two lemmas as in (EEH2). Lemma 3.4 Let n N. Then a n (E 1 ) (p p p 1 ) 1/p and a n+1 (E 2 ) (p p p 1 ) 1/p 13

14 and inf E 3 P n+1 (p p p 1 ) 1/p where the infimum is taken over all linear maps P n+1 : L 1,p (I) L p (I) with rank less than n + 1. Proof: Let {I i } n 1 be the partition from Observation 4 with ε = Set P f = n P i f where P i f(x) := χ Ii (x)(f((a i + b i )/2)), where I i = (a i, b i ). (p p p 1 ) 1/p. We can see that P i f is a linear map from L 1,p (I i ) into L p (I i ) (not necessarily bounded) and it is a bounded linear map from L 1,p (I i )/{1} into L p (I i ) with rank less or equal to 1. Then rank P n and P is a linear map from L 1,p (I) into L p (I) and it is a linear map from L 1,p (I)/{1} into L p (I). From (Li1) and u P Lemma 5 we have that A 0 (I i ) = i u p,ii u p,ii >0 u p,ii. Then we have: (E 3 P )f p p,i = (E 3 P )f p p,i i = (f(.) f((a i + b i )/2) p p,i i A p 0(I i ) f p p,i i (p p p 1 ) f p p,i 2nπ i p (p p p 1 ) f p 2nπ p,i. p (Note that f f((a + b)/2) p,i / f p,i < for any f L 1,p (I).) From this follows the third inequality for E 3. The proof of the inequality for E 2 is the same. For the first inequality for a n (E 1 ) we have to define a new partition of I. Let {I i } n 1 by the partition from Observation 4 with ε = (p p p 1 ) 1/p. Put J i = (a i + I i /2, b i + I i /2) for i = 1,..., n 1 and J 0 = (a, a + I 1 /2), J n = (a n + I n /2, b) where I i = (a i, b i ). Define {c i } n 0 and {d i } n 0 by J i = (c i, d i ). Set Gf = n i=0 G i f where G i f(x) := χ Ji (x)(f((c i +d i )/2)), for i = 1,..., n 1, G 0 f(x) := f(a) = 0 and G n f(x) := f(b) = 0 where I i = (a i, b i ). Then 14

15 rank G n 1 and G is a bounded linear map from W 1,p 0 (I) into L p (I). Since A p 0(I i ) = A p 0(J j ) then as before we have for f W 1,p 0 (I): (E 1 G)f p p,i = (E P )f p p,j i i=0 n 1 = (f(.) f((a i + b i )/2) p p,j i + f p p,j 0 + f p p,j n n 1 A p 0(J i ) f p p,j i + A p 0(I 1 ) f p p,i 0 + A p 0(I n ) f p p,j n (p p p 1 ) f p p,i i=0 2nπ i p (p p p 1 ) f p 2nπ p,i. p From this follows the first inequality for a n (E 1 ). From the proof of Lemma 6 we can see that for any n there exists K n, an n-dimensional linear subspace of L p, such that for any f L 1,p (I)/{1} (or from any f L 1,p (I)) we have inf f g p p g K n (p p p 1 ) f p p. Moreover, for any n there exists R n 1, an n 1 dimensional linear subspace of L p, such that for any f W 1,p 0 (I) we have Lemma 3.5 Let n N. Then inf f g p p g R n 1 a n (E 1 ) (p p p 1 ) f p p. (p p p 1 ) 1/p and a n (E 2 ) (p p p 1 ) 1/p. 15

16 and inf E 3 P n+1 (p p p 1 ) 1/p where the infimum is taken over all linear maps P n+1 : L 1,p (I) L p (I) with rank less than n + 1. Proof: First we prove the second inequality for E 2. Let {I i } n 1 be the partition from Observation 4 with ε = (p p p 1 ) 1/p. From the definition of A 0 (I i ) we know that for i = 1,..., n there exists ϕ i W 1,p (I i ), ϕ i p,ii = 1 such that inf ϕ i α p,ii = A 0 (I i ) = ε. α R We extend each ϕ i to I by taking ϕ i = 0 outside I i and define φ i = ϕ i + c i where c i R is such that φ i L 1,p /{1}. Let P : L 1,p (I)/{1} L p (I)/{1} be a bounded linear operator with rank(p ) < n. Then there are constants λ 1,..., λ n, not all zero, such that P φ = 0, φ = λ i φ i. Note that φ L p (I)/{1}. Then, noting that the following summation is over λ i 0, E 2 φ P φ p p,i = E 2 φ p p,i = φ p p,i i inf φ α α p p,i i inf φ i α p α p,i i λ i p n ε p φ i p p,i i λ i p ε p φ p p,i. Then we have that E 2 P p,i ε, so that a n (E 2 ) ε. We prove the inequality for E 3 in the same way as for E 2. Let P : L 1,p (I) L p (I) be a linear operator with rank(p ) < n + 1. Let we have the system of functions {φ i } n considered previously and put φ n+1 = 1; then we have n + 1 linearly independent functions from L 1,p (I) (note that W 1,p (I)/{1} L 1,p (I)). 16

17 Then there are constants λ 1,..., λ n+1, not all zero, such that P φ = 0, φ = n+1 λ i φ i. Then, noting that the following summation is over λ i 0 we have E 3 φ P φ p p,i = E 3 φ p p,i = n+1 inf α n+1 ε p n+1 φ p p,i i φ α p p,i i n+1 inf α φ i p p,i i λ i p ε p φ p p,i. φ i α p p,i i λ i p Hence E 3 P p,i ε and then the third inequality for E 3 is satisfied. Now we prove the inequality for a n (E 1 ). Take u n,i the n-th eigenfunction of the p Laplacian eigenvalue problem on I with Neumann boundary condition. Let {I i } n 1 be the partition from Observation 4 with ε = (p p p 1 ) 1/p. Then we define φ i = u n,i χ Ii and φ i W 1,p 0 (I i ) and φ i p,i / φ i p,i = A 0 (I i ). Let P : L 1,p (I) L p (I) be a linear operator with rank(p ) < n. Then there are constants λ 1,..., λ n, not all zero, such that P φ = 0, φ = λ i φ i. Noting that the following summation is over λ i 0 we have E 1 φ P φ p p,i = E 1 φ p p,i = φ p p,i i φ p p,i i φ i p p,i i λ i p n+1 ε p φ i p p,i i λ i p ε p φ p p,i. Thus E 1 P p,i ε and so the third inequality for a n (E 1 ) is satisfied. 17

18 The previous two lemmas give us: Theorem 3.6 If <, then a n (E 1 ) = 2(n)π p (p p p 1 ) 1/p, (p p p 1 ) 1/p a n (E 2 ) (p p p 1 ) 1/p 2(n 1)π p and inf E 3 P n+1 = (p p p 1 ) 1/p where the infimum is taken over all linear maps P n+1 : L 1,p (I) L p (I) with rank less than n + 1. Then lim a n(e 1 )n = (p p p 1 ) 1/p, n 2π p and lim a n(e 2 )n = (p p p 1 ) 1/p, n 2π p where π p = π/ sin(π/p). Acknowledgement: Both authors thank the Leverhulme Foundation for its port; the second author is also indebted to the Ohio State Univeristy for an international travel grant. 18

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