A NOTE ON MULTIPLIER ALGEBRAS ON ... - CiteSeerX

0 downloads 0 Views 114KB Size Report
Mar 28, 2008 - we might try to characterize when I({fj}n j=1) = M(HE), where I({fj}n ... whose diagonal entries are djj = aj for j = 1, ..., N2. Then our example is ...
PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 136, Number 8, August 2008, Pages 2835–2838 S 0002-9939(08)09383-0 Article electronically published on March 28, 2008

A NOTE ON MULTIPLIER ALGEBRAS ON REPRODUCING KERNEL HILBERT SPACES TAVAN T. TRENT (Communicated by Michael T. Lacey)

Abstract. We construct a simple reproducing kernel space whose multiplier algebra does not satisfy a “corona theorem”.

Let HE denote a reproducing kernel Hilbert space on a set E. That is, HE is a Hilbert space of functions on E, satisfying: (i) for every x ∈ E, there exists a Cx < ∞, so that |f (x)| ≤ Cx f HE , for all f ∈ HE , and (ii) If f ∈ HE and f (x) = 0 for all x ∈ E, then f HE = 0; i.e. f = 0 in HE . For g ∈ HE , define Mg (f )(x) = g(x)f (x) for x ∈ E and f ∈ HE . Let M(HE ) = {f ∈ HE : Mf ∈ B(HE )}, the multiplier algebra for HE . When can we solve linear equations in M(HE )? As a very preliminary step, we might try to characterize when I({fj }nj=1 ) = M(HE ), where I({fj }nj=1 ) is the ideal generated by {fj }nj=1 ⊂ M(HE ). The model characterization is the ∞ 2 “corona theorem” of Carleson [1] nfor H (D) (= M(H (D))). Namely, for {fj }nj=1 ⊂ H ∞ (D), if 0 < 2 ≤ j=1 |fj (z)|2 for all z ∈ D, then I({fj }nj=1 ) = H ∞ (D). We are interested in whether an abstract “corona theorem” can hold a general for n 2 algebra, M(HE ). That is, if {fj }nj=1 ⊂ M(HE ) and 0 < 2 ≤ j=1 |fj (x)| n for all x ∈ E, does there exist {gj }nj=1 ⊂ M(HE ) with j=1 fj gj = 1? Of course, by very deep results of Cole (see [2]) and Sibony [3], there are counterexamples to the corona problem of the form H ∞ (Ω) with Ω ⊂ C3 and Ω ⊂ C2 , respectively. Since these algebras can be viewed as multiplier algebras for appropriate Bergman spaces, the answer to the corona question for a general multiplier algebra is “no”. In this note, we give a simple counterexample to the “corona problem” for a multiplier algebra of functions of one variable. This example is strong enough to give that: Received by the editors February 8, 2007. 2000 Mathematics Subject Classification. Primary 46E22, 47B32. Key words and phrases. Corona, reproducing kernel. The author was partially supported by NSF Grant DMS-0400307. c 2008 American Mathematical Society Reverts to public domain 28 years from publication

2835

2836

TAVAN T. TRENT

there is a g ∈ M(HE ), so that g1 ∈ HE and |g(x)| ≥  > 0 for all x ∈ E, but 1 / M(HE ). g ∈ Clearly, such an example cannot occur in H ∞ (Ω), and, in fact, functions in our space M(HE ), will, unfortunately, have no smoothness properties in general. 2 For a = (a1 , ..., aN 2 ) ∈ lN let Da denote the N 2 × N 2 diagonal matrix, 2, whose diagonal entries are djj = aj for j = 1, ..., N 2 . Then our example is based on the following lemma: 2 AN ∈ Lemma 1. For N = 1, 2, ... there exists an aN ∈ lN 2 and an invertible MN 2 (C) satisfying:

   (i) AN DaN A−1 N B(l2 2 ) ≤ 2,   N  −1  (ii) AN Da−1 A ≥ N, N  N 2 B(lN 2 )

  1 (iii) AN DaN B(l2 ) ≤ 2 , N N2   1   (iv) AN Da−1  2 ≤ 2 , and N N B(lN 2 )     (v) Da−1  2 ≤ C0 , for a fixed C0 < ∞ independent of N. N B(lN 2 )

Assume that the lemma holds. We proceed to construct our counterexample. 2 def def Let HN = {x ∈ CN : with xHN = AN (x)l2 2 }. For each N = 1, 2, ..., N ∞  1 ijt 2 2 let EN = { e 2πN : j = 0, ..., N − 1} and E = EN ⊂ C. Note that any N =1 N   zN,j

pairwise disjoint EN ’s with sufficiently many points will work. For f : E → C and N = 1, 2, ..., let f N = ( f (zN,0 ), ..., f (zN,N 2 −1 )). Then  2 def ∞   HE  { f : E → C: f 2HE = < ∞}. Since HE is isomorphic N =1 f N  HN ∞  to HN , we see that HE is a Hilbert space of functions on E. Also, for N =1

zN,j ∈ E,    f (zN,j ) ≤  f N  2

lN 2

    (f ) A 2 N ) N lN 2 N2    −1    = AN B(l2 ) f N  2 HN  −1  N   ≤ AN B(l2 ) f HE .    ≤ A−1 N B(l2

N2

Clearly, if f : E → C vanishes on E, then f HE = 0. reproducing kernel Hilbert space on E.

Thus HE is a

MULTIPLIER ALGEBRAS ON REPRODUCING KERNEL HILBERT SPACES

2837

Suppose that f ∈ M(HE ); then a computation shows that, for h ∈ HE , 2

Mf (h)HE = = =



  AN ((f h)N )22 l

N2

N =1 ∞

 2   AN (Df N (h)N ) 2

lN 2

N =1 ∞

 2   )(A (h )) (AN Df N A−1 N N 2 . N lN 2

N =1

From this equality it follows that

    Mf B(HE ) = sup {AN Df N A−1 N 

2 ) B(lN 2

N ∈N

}.

Using the lemma to get the appropriate sequence of aN ’s, we define g : E → C, so that g N = aN , for N = 1, 2, ... . Then g ∈ HE , since by (iii) of the lemma 2

gHE = =

∞  

 2 g N  N =1 ∞

HN

=



2

AN (aN )l2 2 N

N =1

  AN Da (uN )22 N l

N2

N =1





N =1

N2 < ∞, N4

2

where uN denotes the vector in CN consisting of all 1’s. A similar argument, using (iv), shows that g1 ∈ HE . From (v), for all N = 1, 2, ... and j = 0, ..., N 2 − 1, we have g(zN,j ) ≥ C10 . So |g(x)| ≥ C10 , for all x ∈ E. By our construction,     } ≤ 2, by (i). Mg B(HE ) = sup{AN DgN A−1 N  2 N ∈N

B(lN 2 )

By a similar estimate, using (ii) we get that         = sup{AN Dg−1 A−1 M g1  N  B(HE )

N ∈N

N

2 ) B(lN 2

} = ∞.

This completes the construction. We finish with a proof of the lemma. Proof. For N = 1, 2, ... construct the N 2 × N 2 matrix ⎡ 1 −1 0 0 1+ 1 + 1 ⎢ N N3 1 ⎢ 0 −1 0 1 1+ N + N23 ⎢ ⎢ 1 0 0 − 1+ 1 + 3 1 −1 ⎢ N N3 ⎢ BN = ⎢ .. ⎢ . 0 0 0 ⎢ ⎢ . . . . .. .. .. .. ⎢ ⎣ 0 0 0 ··· Clearly, σ(BN ) = { 1+ 1 1+ N

k N3

···

0

···

0

···

0 .. .

.. ..

.

. 0

−1

1 2 1 1+ N +N N3

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎦

: 1 ≤ k ≤ N 2 } ⊂ D, the open unit disk.

2838

TAVAN T. TRENT

Since the spectral radius of BN is strictly less than 1, there exists a TN ∈ MN 2 (C) so that  −1  T BN TN  2 ≤ 1. N B(l ) N2

Now

 −1  B  2 N B(l

N2

)

N 2



( 1+

k=1

CN = TN−1 BN TN . Then

and

1 N

+

k N3 )

≥N

and

BN B(l2 2 ) ≤ 2.

Set

N

CN B(l2 2 ) ≤ 1,  N  TN CN T −1  2 ≤ 2, N B(l 2 )   N TN C −1 T −1  2 ≥ N. N N B(l ) N2

Since { 1+ 1 1+ k N N3

σ(CN ) = σ(BN ), σ(CN ) consists of the N 2 distinct eigenvalues : 1 ≤ k ≤ N 2 }. If we denote by DN the N 2 × N 2 matrix with

the same diagonal as BN , then CN is similar to DN , say CN = PN DN PN−1 . TN PN . Then the lemma follows with Set AN = N 2 T P  −1 ) N N B(l2 ) (DN B(l2 ) +DN  B(l2 ) 2 2 N

aN = (1 +

1 N

+

1 N3 , 1

+

1 N

N2

N

+

2 N 3 , ...,

1+

1 N

+

N2 N3 )

2

∈ CN .



References [1] L. Carleson, Interpolation by bounded analytic functions and the corona problem, Annals of Math. 76 (1962), 547-559. MR0141789 (25:5186) [2] T. W. Gamelin, Uniform Algebras and Jensen Measures, London Mathematical Society Lecture Note Series 32, Cambridge University Press, Cambridge, 1978. MR521440 (81a:46058) [3] N. Sibony, Un exemple de domain pseudoconvexe r´ egulier o` u l’´ equation ∂u = f n’admet pas de solution born´ ee pour f born´ ee, Invent. Math. 62 (1980/81), 235-242. MR595587 (82c:32020) Department of Mathematics, The University of Alabama, Box 870350, Tuscaloosa, Alabama 35487-0350 E-mail address: [email protected]