The Polynomial Numerical Index of L p (µ)

KYUNGPOOK Math. J. 53(2013), 117-124 http://dx.doi.org/10.5666/kmj.2013.53.1.117 The Polynomial Numerical Index of L p (µ) Sung Guen Kim Department of Mathematics, Kyungpook National University, Daegu 702-701, Republic of Korea e-mail : sgk317@knu.ac.kr Abstract. We show that for 1 < p <, k, m N, n (k) (l p) = inf{n (k) (l m p ) : m N} and that for any positive measure µ, n (k) (L p (µ)) n (k) (l p ). We also prove that for every Q P( k l p : l p) (1 < p < ), if v(q) = 0, then Q = 0. 1. Introduction Given a complex or real Banach space E we write B E for the closed unit ball and S E for the unit sphere of E. The dual space of E is denoted by E. For k N, a mapping P : E E is called a (continuous) k-homogeneous polynomial if there is a k-multilinear (continuous) mapping A : E E E such that P (x) = A(x,..., x) for every x E. P( k E : E) denotes the Banach space of all k-homogeneous continuous polynomials from E into itself with the norm P = sup x BE P (x). We refer to [6] for background of polynomials on a Banach space. Let Π(E) := {(x, x ) : x S E, x S E, x (x) = 1}. The numerical radius of P is defined [3] by v(p ) := sup{ x (P x) : (x, x ) Π(E)}. The polynomial numerical index of order k of E is defined [4] by n (k) (E) := inf{v(p ) : P P( k E : E), P = 1} = sup{m 0 : P M v(p ) for all P P( k E : E)}. Of course, n (1) (E) is the classical numerical index of E. Note that 0 n (k) (E) 1, and n (k) (E) > 0 if and only if v( ) is a norm on P( k E : E) equivalent to the usual Received July 11, 2012; accepted August 23, 2012. 2010 Mathematics Subject Classification: 46A22, 46G20, 46G25. Key words and phrases: Homogeneous polynomials, polynomial numerical index. This research was supported by the Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education, Science and Technology (2010-0009854). 117

118 Sung Guen Kim norm. It is obvious that if E 1, E 2 are isometrically isomorphic Banach spaces, then n (k) (E 1 ) = n (k) (E 2 ). The concept of the classical numerical index (in our terminology, the polynomial numerical index of order 1) was first suggested by G. Lumer [12]. In [4] the authors proved n (k) (C(K)) = 1 when C(K) is the complex spaces and some inequality n (k) (E) n (k 1) 1 (E) k(k+ k 1 ) n (k) (E) for every Banach space E. It was (k 1) k 1 shown that n (k) (E ) n (k) (E). The authors [10] found a lower bound for the polynomial numerical index of real lush spaces. They used this bound to compute the polynomial numerical index of order 2 of the real spaces c 0, l 1 and l. In fact, they showed that for the real spaces X = c 0, l 1, l, n (2) (X) = 1/2. They also presented an example of a real Banach space X whose polynomial numerical indices are positive while the ones of its bidual are zero. We refer to ([1 5, 7 12]) for some results about the polynomial numerical index. For general information and background on numerical ranges, we refer to [1 2]. In this paper, we show that for 1 < p <, k, m N, n (k) (l p ) = inf{n (k) (l m p ) : m N} and that for any positive measure µ, n (k) (L p (µ)) n (k) (l p ). We also prove that for every Q P( k l p : l p ) (1 < p < ), if v(q) = 0, then Q = 0. 2. Results For 1 < p < and m N, lp m denotes K m endowed with the usual p-norm, where K = R or C. We may consider lp m as a subspace of l p. Let {e n } N be the canonical basis of l p and {e n} n N the biorthogonal functionals associated to {e n } n N. Note that in general if X is a Banach space and Y is a subspace of X there is no comparison between n (k) (X) and n (k) (Y ) for k N. Theorem 2.1. Let 1 < p < and k N be fixed. Then n (k) (l p ) = inf{n (k) (l m p ) : m N} and the sequence {n (k) (l m p )} m N is decreasing. Proof. We proceed by steps. Let m N. We define P {1,,m} : l p lp m by P {1,,m} ( λ je j ) = m λ je j. Step 1: The sequence {n (k) (l m p )} m N is decreasing. Proof of Step 1. Let Q S P(k lp m:lm p ). We define Q P( k lp m+1 : lp m+1 ) by Q(x) = (Q P {1,,m} (x), 0) for x lp m+1. It is obvious that Q S P ( k lp m+1 : lp m+1 ). Claim A: v(q) = v( Q) Let (x, x ) Π(lp m ). Then ( (x, 0), (x, 0) ) Π(lp m+1 ) and ( ) x (Q(x)) = (x, 0)( Q( (x, 0) )) v( Q). By taking supremum in the left side of ( ) over (x, x ) Π(l m p ), we have v(q) v( Q). For the reverse inequality let ϵ > 0. Then there exists z 0 := m+1 a je j

The Polynomial Numerical Index of L p (µ) 119 S l m+1 p such that (z 0, m+1 sign(a j) a j p 1 e j ) Π(lm+1 p ) and v( Q) m+1 ϵ < sign(a j ) a j p 1 e j ( Q(z 0 )) = sign(a j ) a j p 1 e j (Q( a j e j )) = C k+p 1 sign(a j ) a j C p 1 e j (Q( (where C := ( a j p ) 1 p 1) sign(a j ) a j C p 1 e j (Q( a j v(q), because ( C e j, which show v( Q) v(q). Thus v(q) = v( Q). It follows that a j C e j)) n (k) (l m p ) = inf Q S P( k l m p :l m p ) v(q) = inf Q S P( k l m p :l m p ) v( Q) inf v(r) R S P( k m+1 l p :l m+1 p ) = n (k) (l m+1 p ). Step 2: n (k) (l p ) n (k) (l m p ) for every m N a j C e j)) sign(a j ) a j C p 1 e j ) Π(l m p ), Proof of Step 2. Let Q S P(k l m p :lm p ). We define Q P( k l p : l p ) by Q(z) = (Q P {1,,m} (z), 0, 0, ) for z l p. It is obvious that Q S P(k l p :l p ). By the same argument as in Step 1, we have v( Q) v(q). Thus it follows. Step 3: lim m n (k) (l m p ) = n (k) (l p ) Proof of Step 3. Let Q S P( k l p:l p). For each m N, we define Q m P( k l m p : l m p ) by Q m (x) = P {1,,m} Q(x, 0, 0, ) for x l m p. It is obvious that Q m 1, Q m Q m+1 and v(q m ) v(q). For each m N, we define Q m P( k l p : l p ) by Q m (z) = (Q m P {1,,m} (z), 0, 0, ) for z l p. By the argument in Step 1, v( Q m ) = v(q m ). Claim B: lim m Q m = 1

120 Sung Guen Kim Let ϵ > 0. Choose x 0 S lp such that Q(x 0 ) > 1 ϵ. By continuity of Q at x 0 it follows that Q m P {1,,m} (x 0 ) Q(x 0 ) = P {1,,m} Q P {1,,m} (x 0 ) P {1,,m} Q(x 0 ) + P {1,,m} Q(x 0 ) Q(x 0 ) Q P {1,,m} (x 0 ) Q(x 0 ) + P {1,,m} Q(x 0 ) Q(x 0 ) 0 as m. Choose N 0 N such that Q m P {1,,m} (x 0 ) Q(x 0 ) < ϵ for all m N 0. Then for all m N 0, 1 Q m Q m P {1,,m} (x 0 ) > 1 2ϵ, which shows Claim B. Claim C: lim m v(q m ) = v(q) There exists (y 0, y ) Π(l p ) such that y (Q(y 0 )) > v(q) ϵ. Let y 0 := b je j. Then y = sign(b j) b j p 1 e j. For m N, we define y (m) 0 := m 1 b je j + ( j=m b j p ) 1 p em and ym := m 1 sign(b j) b j p 1 e j + ( j=m b j p ) p 1 p e m. It is obvious that (y (m) 0, ym) Π(l p ) and lim m y 0 y (m) 0 = 0 = lim m y ym. Note that lim m ym(q(y (m) 0 )) = y (Q(y 0 )). Indeed, y m(q(y (m) 0 )) y (Q(y 0 )) y m(q(y (m) 0 )) y (Q(y (m) 0 )) + y (Q(y (m) 0 )) y (Q(y 0 )) y m y Q(y (m) 0 ) + Q(y (m) 0 ) Q(y 0 ) y m y + Q(y (m) 0 ) Q(y 0 ) 0 as m. Choose N 1 N such that ym(q(y (m) 0 )) > v(q) ϵ for all m N 1. It is easy to show that for all m N 1, yn 1 ( Q m (y (N1) 0 )) = yn 1 (Q(y (N1) 0 )). It follows that for all m N 1, v(q) ϵ < y N 1 (Q(y (N 1) 0 )) = y N 1 ( Q m (y (N 1) 0 )) v( Q m ) = v(q m ) v(q), which show lim m v(q m ) = v(q). Thus we have ( ) v(q) = lim m v(q m) = lim sup [v( Q m m Q m ) Q m ] = lim sup v( Q m m Q m ) lim Q m m = lim sup v( Q m ) (by claim B) m Q m lim sup n (k) (lp m ) m

The Polynomial Numerical Index of L p (µ) 121 Taking the infimum in the left side of ( ) over Q S P(k l p :l p ), we have n (k) (l p ) lim sup m n (k) (lp m ). By Step 2, we have n (k) (l p ) lim inf m n (k) (lp m ). Thus lim m n (k) (lp m ) = n (k) (l p ). Therefore, we complete the proof. Theorem 2.2. Let 1 < p <. Let Q P( k l p : l p ). Then v(q) = 0 if and only if Q = 0. Proof. It is enough to show that if v(q) = 0, then Q = 0. We will show that Q m := P {1,,m} Q span {e 1,,e m} : l m p l m p is the zero polynomial for every m N. Write Q m ( x k e k ) = k=1 k 1 + +k m =m,0 k 1,,k m m m! k 1! k m! xk 1 1 xk 2 2 xkm m A m (e k1,, e km ), where A m is the corresponding symmetric k-linear mapping to the k-homogeneous polynomial Q m. Let a k1 k m := A m (e k1,, e km ) l m p. Let p 1 := 0 and p n be the n-th prime (n 2). Let 0 t 1 be fixed and q R with 1/p + 1/q = 1. Put and y := t p 1 e 1 + t p 2 e 2 + + t p m e m (1 + t p p 2 + + t p p m) 1/p y := t(p 1) p1 e 1 + t (p 1) p 2 e 2 + + t (p 1) p m e m (1 + t p p 2 + + t p p m). 1/q Then (y, y ) Π(lp m ). Claim: a k1 k m = 0 for every k 1,, k m It follows that for every 0 t 1, so 0 = y (Q m (y)) = 1 (1 + t p p 2 + + t p p m) 1/q+k/p (t (p 1) p 1 e 1 + t (p 1) p 2 e 2 + + t (p 1) p m e m) (Q m (t p 1 e 1 + t p 2 e 2 + + t p m e m )), 0 = (t (p 1) p 1 e 1 + t (p 1) p 2 e 2 + + t (p 1) p m e m) = (Q m (t p 1 e 1 + t p 2 e 2 + + t p m e m )) + k 1 + +k m =m,0 k 1,,k m m [ 2 j m k 1 + +k m =m,0 k 1,,k m m t p k2 + + p k m m! k 1! k m! e 1(a k1 k m ) t p k2 + + p km +(p 1) p kj m! k 1! k m! e j (a k1 k m )].

122 Sung Guen Kim Note that the elements of the set { p k2 + + p km, p k2 + + p km + (p 1) p kj : k 1 + + k m = m, 0 k 1,, k m m, 2 j m} are different. Thus e j (a k 1 k m ) = 0 for every 1 j m, which show a k1 k m = 0 for every k 1,, k m. Therefore, Q m = 0. Let x = k=1 x ke k l p be fixed. By continuity of Q at x, we have Q(x) = lim Q m(x) = 0. m Corollary 2.3. Let 1 < p <. Then for every k, m N, we have n (k) (l m p ) > 0. Proof. Assume that n (k) (lp m ) = 0 for some k, m N. Since the unit sphere of the finite dimensional space P( k lp m : lp m ) is compact, there exists some Q P( k lp m : lp m ) such that Q = 1 and v(q) = 0. Theorem 2.2 shows that Q = 0, which is impossible. Let (Ω, Σ) be a measurable space and µ a positive measure on Ω. We denote by P the collection of all partitions π of Ω into finitely many pairwise disjoint members of Σ with finite strictly positive measures. We order this collection by π 1 π 2 whenever each member of π 1 is the union of members of π 2. So P is a directed set. For each π = {E 1,, E m } P, we associate the subspace V π of L p (µ) defined by V π = { m i=1 a i1 Ei : a i K}. By P π we denote the projection of L p (µ) onto V π defined by 1 P π (f) = [ f(t)dt]1 Ei µ(e i ) E i i=1 for all f L p (µ). V denotes the union of all subspaces V π of L p (µ). We recall that V is a dense subspace of L p (µ), thus, for each f L p (µ), the sequence {P π (f)} π converges to f in L p (µ). We recall the following well-known result. Theorem 2.4 For 1 < p < and for every partition π = {E 1,, E m } P, the subspace V π is isometrically isomorphic to l m p. Thus n (k) (V π ) = n (k) (l m p ) for every k N. Theorem 2.5. Let 1 < p < and k N. Then for any positive measure µ, n (k) (L p (µ)) n (k) (l p ). Proof. Let Q S P( k L p(µ):l p(µ)). Let ϵ > 0. Choose x 0 S Lp(µ) such that Q(x 0 ) > 1 ϵ. By uniform continuity of Q on the closed unit ball of L p (µ), there exists some δ > 0 such that x, y B Lp(µ) with x y < δ implies that Q(x) Q(y) < ϵ. Choose π 0 P such that x 0 P π0 (x 0 ) < δ. Since P π0 (x 0 ) 1, we have Q(x 0 ) Q P π0 (x 0 ) < ϵ. Thus Q P π0 (x 0 ) > Q(x 0 ) ϵ > 1 2ϵ. Thus we can choose π 1 = {E 1,, E m } P such that

The Polynomial Numerical Index of L p (µ) 123 π 1 π 0 and P π1 Q P π0 (x 0 ) > 1 2ϵ. We define R P( k V π1 : V π1 ) by R(P π1 (x)) = P π1 Q P π1 (x) for x L p (µ). Obviously R 1. It follows that Thus ( ) R R( P π 0 (x 0 ) P π0 (x 0 ) ) = R(P π 0 (x 0 )) P π0 (x 0 ) k R(P π0 (x 0 )) P π1 Q P π0 (x 0 ) > 1 2ϵ. ( ) v(r) n (k) (V π1 ) R > n (k) (V π1 ) (1 2ϵ) (by ) = n (k) (l m p ) (1 2ϵ) (by Theorem 2.4) n (k) (l p ) (1 2ϵ) (by Theorem 2.1). Since V π1 is a finite dimensional space, there exists (y 0, y ) Π(V π1 ) such that v(r) = y (R(y 0 )). It follows that v(r) = y (R(y 0 )) = y (P π1 Q(y 0 )) = P π 1 y (Q(y 0 )) v(q), because (y 0, P π 1 y ) Π(V π1 ). By ( ), we have ( ) v(q) v(r) > n (k) (l p ) (1 2ϵ). By taking infimum in the left side of ( ) over Q S P(k L p (µ):l p (µ)), we conclude the proof. Acknowledgements The author would like to thank the referee for his comments and suggestions in improving this paper in both substance and presentation. References [1] F. F. Bonsall and J. Duncan, Numerical Ranges of Operators on Normed Spaces and of Elements of Normed Algebras, London Math. Soc. Lecture Note Ser. 2, Cambridge Univ. Press, 1971. [2] F. F. Bonsall and J. Duncan, Numerical Ranges II, London Math. Soc. Lecture Note Ser. 10, Cambridge Univ. Press, 1973. [3] Y. S. Choi and S. G. Kim, Norm or numerical radius attaining multilinear mappings and polynomials, J. London Math. Soc., 54 (1996), 135-147. [4] Y. S. Choi, D. Garcia, S. G. Kim and M. Maestre, The polynomial numerical index of a Banach space, Proc. Edinburgh Math. Soc., 49 (2006), 39-52.

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