Composition Operators on the Fock Space

Transcription

1 Composition Operators on the Fock Space Brent Carswell Barbara D. MacCluer Alex Schuster Abstract We determine the holomorphic mappings of C n that induce bounded composition operators on the Fock space in C n. Furthermore, we determine which of these composition operators are compact, and we compute the operator norm of all bounded composition operators in this setting. We also consider extensions of these results in various generalizations of the Fock space. 1 Introduction Composition operators on spaces of analytic functions have been studied in many settings. Much has been written about the properties of these operators on the Hardy, Bergman, and Bloch spaces on the unit disk in the complex plane, or unit ball in C n (see, for example [3], [10] or [7]). Our purpose here is to study composition operators on the Fock space in C n. We will determine which composition operators are bounded, and which are compact. We will also compute the norm of every bounded composition operator in this setting, an interesting result since finding the norm of such an operator on the Hardy and Bergman spaces is still largely an open problem (see [1], [] or [4] for recent norm results in the Hardy space setting). We remark that while we will be doing our analysis in the several variable setting, we will make every effort to interpret our results and methods in the case n = 1 in order to clarify our ideas. The Fock space Fn is the Hilbert space of all holomorphic functions on C n with inner product f, g 1 (π) n f(z)g(z)e C z dν(z). n 1 1

2 Here ν denotes Lebesgue measure on C n. To simplify notation we will often use F instead of F n, and we will denote by f the norm of f. The reproducing kernel functions for the Fock space are given by k w (z) = e z,w /, where z, w = n 1 z jw j. Note that the substitution f = k w into the reproducing formula f, k w = f(w), which holds for all f F and w C n, leads to the identity k w = exp( w /4). For a given holomorphic mapping ϕ : C n C n, the composition operator C ϕ : F F is given by C ϕ (f) = f ϕ. Our first main result (see Theorem 1) will show that if the operator C ϕ is bounded, then ϕ must be of the form ϕ(z) = Az + B, where A is an n n matrix and B is an n 1 vector. Furthermore, it will follow that A 1 for bounded C ϕ, and that B will be restricted by the condition that Aζ, B = 0 for any ζ in C n with Aζ = ζ. Theorem 1 will also show that if C ϕ is compact, then A < 1 with no restriction on B. Our second result will be Theorem, which gives the converse to Theorem 1. In our third main result, Theorem 4, we determine the operator norm of C ϕ (where ϕ(z) = Az + B is chosen so that C ϕ is bounded) in terms of A and B. Finally, in the last section, we will outline some ideas about how to extend our results to various generalizations of the Fock space. We thank Christopher Hammond for several helpful remarks on the proof of Theorem 4. Main results Our first result concerns the boundedness and compactness of composition operators on F. Theorem 1. Suppose ϕ : C n C n is a holomorphic mapping. (a) If C ϕ is bounded on F, then ϕ(z) = Az +B, where A is an n n matrix and B is an n 1 vector. Furthermore, A 1, and if Aζ = ζ for some ζ in C n, then Aζ, B = 0. (b) If C ϕ is compact on F, then ϕ(z) = Az + B, where A < 1.

3 Note that when n = 1, Theorem 1(a) simply says that if C ϕ is bounded on F, then ϕ(z) = az + b, where a 1, and if a = 1, then b = 0. In the proof of Theorem 1 we will use the fact that every entire function on C n has a multi-index power series expansion of the form α c(α)zα, where α denotes a multi-index (α 1, α,, α n ) of non-negative integers, and z α = z α 1 1 z α zn αn. We use the standard notation α = α α n and α! = α 1! α n!. The quantity α is referred to as the total order of the multi-index α. Proof of Theorem 1. If C ϕ is bounded on F, then Cϕ(k w ) sup w C n k w k ϕ(w) = sup w C n k w ( 1 = w C sup ( exp ϕ(w) w )) <, (1) n 4 where the first equality follows from the easily verified property C ϕ(k w ) = k ϕ(w). From (1) it follows that lim sup w ϕ(w) w 1. () For each ζ B n and 1 j n, we define the analytic function ϕ j ζ by the equation ϕ j ζ (λ) = ϕ j(λζ) for λ C, where ϕ j is the j th coordinate function of ϕ. By () we must have lim sup λ ϕ j ζ (λ) λ 1. If ϕ j (z) = α c(α)zα has homogeneous expansion s=0 F s(z) (defined by F s (z) = α =s c(α)zα ), then since ϕ j ζ (λ) = λ s F s (ζ), we must have F s (ζ) = 0 for all s and ζ B n ; that is F s 0 for s and each coordinate function ϕ j is linear. This proves that ϕ(z) = Az+B as desired. If Aζ > ζ for some ζ of norm 1, then setting w = tζ, t > 0 in () and letting t we obtain a contradiction. Thus we must have A 1. Next we show that if Aζ = ζ, then Aζ, B = 0. As a special case of this, suppose Aζ = λζ where λ is a complex number of modulus 1. If Aζ, B 0, we may choose ρ C, ρ = 1 so that ρ Aζ, B > 0. Considering w = tρζ as t, we obtain a contradiction to (1). Now suppose Aζ = η, where ζ = η = 1. Let U be a unitary map of C n such that Uη = ζ. Then for τ(z) ϕ U(z) = A U(z) + B, we have C τ bounded on F and, since 3

4 A(U(η)) = η, by the special case just considered, we have AUη, B = 0 or Aζ, B = 0 as desired. This completes the proof of (a). To prove (b) we need only show that if ϕ(z) = Az + B induces a compact composition operator on F, then A < 1. Since the normalized reproducing kernel functions k w / k w tend to 0 weakly as w, we have C ϕ e lim sup exp w ( 1 4 ( ϕ(w) w ) where for an operator T, T e = inf{ T Q : Q is compact} is the essential norm of T. (c.f. Proposition 3.13 in [3] for the proof of a similar fact in other spaces). Now suppose that A = 1 and that ζ 0 satisfies Aζ = ζ. Let w = λζ, where λ C and λ Aζ, B 0. Letting λ, we see that C ϕ e 0, and so A is not compact. Thus A < 1. The converse to Theorem 1 holds as well. Theorem. Suppose ϕ(z) = Az + B, where A is an n n matrix with A 1, and B is an n 1 vector. ), (a) If Aζ, B = 0 whenever Aζ = ζ, then C ϕ is bounded on F. (b) If A < 1, then C ϕ is compact on F. Before we give the proof of this theorem, we need several preliminary results that will be of use throughout the rest of this section. First we recall the notion of the singular value decomposition of an n n matrix. Theorem 3. If A is an n n matrix of rank k, then A can be written as A = V ΣW, where V, W are n n unitary matrices, and Σ is a diagonal matrix (σ ij ) with σ 11 σ σ kk > σ k+1,k+1 = = σ nn = 0. The σ ii are the non-negative square roots of the eigenvalues of AA ; if we require that they be listed in decreasing order, then Σ is uniquely determined from A. A proof can be found in [6]. For brevity of notation we will write σ i for σ ii, the i th diagonal entry of Σ. Note that if A 1, then the σ i will all be less than or equal to 1, and at least one will equal 1 if A = 1. We make two notational definitions that will be in force for the rest of this section. Set j = max{r : σ r = 1}, (3) 4

5 (where j = 0 in the case that no σ r = 1), and k = max{r : σ r > 0}, (4) so that k = rank A. The singular value decomposition will allow us to perform a useful normalization in proving Theorem. This normalization will also play a role later in this section when we compute the norm of C ϕ on F. We remark that in the one variable setting this normalization is not needed, though its presence there does no harm. Proposition 1. Suppose ϕ(z) = Az +B, where A and B are as described in (a) of Theorem 1. Let ψ(z) = Σz+B, where the singular value decomposition of A is V ΣW, and B = V B. Then on F, C ϕ = C W C ψ C V, where C W and C V are the unitary operators given by C W (f) = f W and C V (f) = f V. Proof. We have C W C ψ C V (f)(z) = (f V ψ W )(z) = f(v ΣW (z) + V B ) = f(az + B) = C ϕ (f)(z) as desired, where we have used the relationship V B = B. Corollary 1. The operator C ϕ is bounded (resp., compact) on F if and only if C ψ is bounded (resp., compact) on F. Moreover, the norm of C ϕ is equal to the norm of C ψ. If ϕ(z) = Az + B and ψ(z) = Σz + B, where Σ and B are as given in Proposition 1, we call ψ a normalization of ϕ. The following lemma gives the reformulation of the hypothesis Aζ, B = 0 whenever Aζ = ζ from Theorem as applied to the normalization ψ. Lemma 1. Suppose that ϕ(z) = Az + B satisfies the hypothesis of (a) of Theorem, and suppose ψ = Σz + B is a normalization of ϕ. Then the first j coordinates of B are 0. Proof. Let A have singular value decomposition A = V ΣW as in Theorem 3. Since Aζ = ζ implies Aζ, B = 0, we have 0 = ΣW ζ, V B = ΣW ζ, B (5) 5

6 whenever Aζ = ζ. If ζ is chosen so that W ζ = (0,, 1,, 0) t (where the 1 appears in the m th position, 1 m j), then Aζ = ζ and ΣW ζ = W ζ, so that (5) implies that the m th coordinate of B is 0. Proof of Theorem. By Corollary 1 and the remarks following it, to prove (a) we need only show that if ψ(z) = Σz + B, where Σ is as described in Proposition 1 and the first j coordinates of B are 0 (where j is defined by (3)), then C ψ is bounded on F. If Σ is invertible, this will follow from a change of variables argument. We have ( f ψ(z) C e 1 z dν(z) = c f(w) C exp 1 Σ 1 (w B ) ) dν n n ( = c C f(w) exp 1 ) ( Σ 1 w w Re Σ 1 w, Σ 1 B ) e 1 w dν n where c = (det Σ) and c = c exp( Σ 1 B /). It suffices to show that Σ 1 w w Re Σ 1 w, Σ 1 B (6) is bounded away from as w ranges over C n. By Lemma 1 we see that the expression in (6) attains its minimum at points w C n satisfying w m = b m 1 σ m and w m b m 0 for j + 1 m n. This gives the desired result. If Σ is not invertible, then we have σ m = 0 for k < m n, where k is as defined in (4). Let Σ 1 = Σ and Σ be the diagonal matrix with Σ 1 = Σ except in the last n k positions along the diagonal, where Σ 1 has entries equal to 0 and Σ has entries equal to 1. By the argument just finished for the invertible case, C ψ is bounded on F, where ψ (z) = Σ z + B. Thus if f = c(α)z α is in F, so is f ψ. Notice we can write f ψ as g + h, where g is c(α)z α 1 1 z α j (σ j+1 z j+1 + b j+1) α j+1 (σ k z k + b k) α k (b k+1) α k+1 (b n) αn α j and h = f ψ g. The terms of h involve at least one of the variables z k+1,, z n, but the terms of g do not. The orthogonality of the monomials 6

7 in F then implies that f ψ g. But g = f ψ, and so f ψ F. By the closed graph theorem we conclude that C ψ is bounded on F. Finally we prove (b). Again first suppose that Σ is invertible. To show C ψ is compact, it suffices to show that a sequence {f n } that is bounded in F and converges to 0 uniformly on compact subsets of C n has its image under C ψ converge to 0 in norm. Changing variables as above, we have f n ψ = c f C n (w) e 1 Σ 1 w Σ 1 B e 1 w e 1 w dν(w). n Since σ i < 1 for all 1 i n, a calculation shows that e 1 Σ 1 w Σ 1 B e 1 w is as small as desired off a compact subset of C n. Since f n converges to 0 uniformly on compact sets, and f n is bounded, this guarantees that f n ψ 0. In the case Σ is not invertible, we again compare C ψ to C ψ, as defined above. Then f n ψ 0, since f n ψ 0 and f n ψ f n ψ. We turn now to the issue of norm calculations. Theorem 4. Suppose ϕ(z) = Az + B, where either A < 1 and B is arbitrary, or A = 1 and Aζ, B = 0 whenever Aζ = ζ. Then on F we have ( 1 ( C ϕ = exp w0 Aw 0 + B )), (7) 4 where w 0 is any solution to (I A A)w = A B. Before turning to the proof of Theorem 4, we make several remarks. First note that Theorem 4 is trivial in the case that A is unitary, so henceforth we will assume A is not unitary. Next note that in one variable, the equation (1 aa)w = ab has the unique solution w = ab/(1 a ) if a < 1, so that ( ) 1 b C ϕ = exp. 4 1 a 7

8 In several variables it is natural to ask whether the equation (I A A)w = A B always has a solution under our hypothesis on ϕ, and if so, whether the solution uniquely determines the expression in (7). Indeed, since A B is orthogonal to the kernel of I A A, a solution will always exist. To verify that the expression given in Equation 7 is constant on the solution set of (I A A)w = A B, note that if (I A A)w 0 = A B = (I A A)w 1 then w 0 w 1 ker (I A A), and thus w 0 w 1, A B = 0, so that w 0 Aw 0 = w 0, w 0 A Aw 0 = w 0, A B = w 1, A B = w 1 Aw 1. In order to prove Theorem 4, we will use a normalization of ϕ as we did in the proof of Theorem. Indeed, let ψ = Σz + B where Σ = diag {σ i } with σ 1 = = σ j = 1 and σ j+1,, σ n < 1, and with B = (0,, 0, b j+1,, b n) t. We first observe that if (I A A)w 0 = A B and (I Σ Σ)w = Σ B, then w 0 Aw 0 + B = w Σw + B. This, together with the fact that C ϕ = C ψ, implies that in order to prove Theorem 4 it suffices to prove it for the normalization ψ(z) = Σz + B. We next observe that solutions to (I Σ Σ)w = Σ B are easily seen to be w m = Furthermore, for such w we have { σm b m/(1 σm) for m j + 1 arbitrary for m j. w Σw + B = n m=j+1 b m. 1 σm This implies that in order to prove Theorem 4 we need only show that ( ( 1 b j+1 )) C ψ = exp + + b n, (8) 4 1 σj+1 1 σn which we will turn to shortly. The proof of Theorem 4 will require a representation for the adjoint of C ϕ, which we obtain in the following result. 8

9 Lemma. If ϕ(z) = Az + B where A, B satisfy the hypothesis of Theorem 1(a), so that C ϕ is bounded on F, then C ϕ = M kb C τ, where τ(z) = A z and M kb is multiplication by the kernel function k B. Proof. The proof follows by checking that C ϕ(k w ) = exp( z, Aw + B /) = M kb C τ (k w ) for all w C n. Since the span of the kernel functions is dense in F, the result follows. The next result gives a lower bound for C ψ, where ψ is a normalization of ϕ. Lemma 3. Suppose ϕ(z) = Az + B satisfies the hypothesis of Theorem 4 and let ψ be a normalization. We have Cψ C ψ sup w C (k ( ( w) 1 b j+1 )) = exp + + b n. n k w 4 1 σj+1 1 σn Proof. We have Cψ (k w) = k ψ(w) k w k w ( 1 ( = exp ψ(w) w )). Since Σ = diag(σ i ) with σ i = 1 for i j, where j defined is as in Equation 3, it is easy to check that ψ(w) w attains its maximum at points w C n which satisfy, for j + 1 m n, and w m = σ m b m 1 σ m arg w m chosen so that w m b m 0. Hence the maximum value of ψ(w) w is then n B σ + m b m n b = m. 1 σm 1 σm This gives the desired result. m=j+1 m=j+1 For these normalized maps ψ we will compute C ψ by identifying a reducing subspace M for C ψ and computing C ψ M and C ψ N, where N is the orthogonal complement of M. The next lemma identifies this subspace M. 9

10 Lemma 4. Let ψ(z) = Σz + B be a normalization of ϕ, where ϕ satisfies the hypothesis of Theorem 4. Define j, possibly 0, by Equation (3). Let M = {f F : f depends on z j+1, z j+,, z n only}. Then M is a reducing subspace for C ψ and the restriction of C ψ to M is compact. Proof. If f is in M, then the power series expansion of f about 0 has the form c(α)z α where the sum is over multi-indices α = (α 1,, α n ) satisfying α 1 = = α j = 0. That C ψ (M) M is then immediate from the form of Σ. To see that Cψ (M) M, let f M and use Lemma to write C ψ(f)(z) = k B (z)f(σz) = exp( z, B /)f(σz). Since the first j coordinates of B are 0, this is a function in M. Thus M is a reducing subspace for C ψ. To see that C ψ restricted to M is compact, consider the map ψ obtained from ψ by replacing any 1 s along the diagonal of Σ by 1/ s. By Theorem (b), C ψ is compact on F. Since C ψ and C ψ agree on the subspace M, and the restriction of a compact operator to a closed subspace is compact, the result follows. Corollary. Let ψ and M be as in Lemma 4. Then ( ) 1 n b C ψ M exp m. 4 1 σm Proof. When w C N satisfies m=j+1 w m = 0 for 1 m j and w m = σ m b m 1 σ m for j + 1 m n w m b m 0 10

11 then k w M, and by the calculations in the proof of Lemma 3 we have C ψ M C ψ (k ( ) w) 1 n b = exp m. k w 4 1 σm m=j+1 In the next lemma we compute the norm of C ψ restricted to M. We remark that in one variable this lemma provides the proof of Theorem 4, since then M = F if a < 1. Lemma 5. Let T be the restriction of C ψ to M, for ψ and M as given in Lemma 4. Then the norm of T is given by ( ( 1 b j+1 )) T = exp + + b n. 4 1 σj+1 1 σn Proof. Since T T is a positive, compact, self-adjoint operator, T T = T is an eigenvalue for T T and there exists F M such that (T T )F = T F. By Lemma we have k B (Σz + B )F (Σ(Σz + B )) = T F (z). (9) Evaluate both sides of this equation at any point z 0 with (I Σ Σ)z 0 = Σ B. If F (z 0 ) 0 we obtain immediately that T = k B (Σz 0 + B ). Since the condition (I Σ Σ)z 0 = Σ B implies that (1 σi )(z 0 ) i = σ i b i for 1 i n and b i = 0 for 1 i j we obtain ( ) T 1 n b = exp m 1 σm m=j+1 as desired. Thus it just remains to handle the case that F (z 0 ) = 0. For motivation we first show how to handle this case in the one variable setting, with ψ(z) = σz + b. If F has a zero of order k at z 0, then consider the equation F (a(az + b)) k b (ψ(z)) F (z) = C ψ and take limits as z z 0. Using L Hopitals rule we get ( ) b exp σ k = C (1 σ ψ. ) 11

12 This cannot be correct, since by Lemma 3 ( ) b C ψ exp 4(1 σ ) and σ < 1. For F a function of several complex variables, we say F has a zero of order k at z = z 0 if in the power series expansion for F about z 0, F (z) = α c(α)(z z 0) α, there is a multi-index α of total order k such that c(α) 0, but for all multi-indices of order less than k, c(α) = 0. Write z 0 = (z 1 0, z 0,, z n 0 ) and consider z s of the form (z γζ 1, z 0 + γζ,, z n 0 + γζ n ) where (ζ 1, ζ,, ζ n ) is a fixed point of B n and γ varies over the complex plane. Substitute such z s into the equation k B (Σz + B ) F (Σ(Σz + B )) F (z) = T and let γ 0. L Hopital s rule (differentiating k times with respect to γ) yields k B (Σz 0 + B α =k ) c(α)σα 1 1 σ α n n ζ α 1 1 ζ α n n = T (10) α =k c(α)ζα 1 1 ζ α n n for any choice of (ζ 1,, ζ n ) for which the denominator in this expression is not 0. The quotient on the left side of Equation (10) must be some positive constant, say µ, since T and k B (Σz 0 + B ) are positive. Thinking of ζ 1,, ζ n as variables for the moment, this says the homogeneous polynomials c(α)σ α 1 1 σ α n n ζ α 1 1 ζ α n n and α =k µ α =k c(α)ζ α 1 1 ζ α n n must agree and hence have the same coefficients. Thus c(α)σ α 1 1 σ α n n = µc(α) for every multi-index α of total order k. Since for some such α, c(α) 0, we must have σ α 1 1 σn αn = µ; this forces µ to be at most 1. But µ < 1 1

13 yields a contradiction to Corollary, so we are left with µ = 1. k B (Σz 0 + B ) = T, as desired. Hence The final matter before we complete the proof of Theorem 4 in the several variable case is to estimate the norm of C ψ restricted to the orthogonal complement of our reducing subspace M. We address this issue next. Without loss of generality we may assume 1 j < n, where n is the dimension and j is defined by Equation (3). It is a straightforward calculation (see, e.g., Section of [9], where a slightly different normalization is used) to see that for a multi-index α, z α = α α!. In particular z k 1 1 z k j j zk j+1 j+1 zk n n = k 1+ +k j k 1! k j! z k j+1 j+1 zk n n. Lemma 6. If f F depends on z j+1,, z n only then z k 1 1 z k j j f = k 1+ +k j k 1! k j! f for any non-negative integers k 1,, k j. Proof. Write f = c(α)z α where the sum is over multi-indices α with α 1 = α = = α j = 0. Then z k 1 1 z k j j f = z k 1 1 z k j j c(α)zα α=(0,,0,α j+1,α n ) = c(α) z k 1 1 z k j j zα = k 1+ +k j k 1! k j! c(α) α α j+1! α n! = k 1+ +k j k 1! k j! f We next obtain an estimate on the norm of C ψ restricted to the orthogonal complement of the subspace M. To do this we will take a function g in M and write C ψ (g) as a sum of terms each of which is a function in C ψ multiplied by a monomial of the form z α 1 1 z α j j. The desired estimate will then follow by orthogonality and Lemma 6. Proposition. Suppose ψ(z) = Σz+B and M are as described in Lemma 4. Let g M. Then g ψ T g, where T denotes the restriction of C ψ to M. 13

14 Proof. Write g as a sum of pairwise orthogonal functions of the form z α g α where α = (α 1,, α j, 0 ) with α 1,, α j not all zero and g α is a function depending on z j+1,, z n only. This is accomplished by writing g is its power series expansion about 0 and grouping together all terms of the form c(α)z α 1 1 z α j j (z ) β, where α 1,, α j are fixed and not all zero and z denotes (z j+1,, z n ). Note that z α g α, z α g α = 0 if α α. Using Lemma 6 this orthogonality implies g = z α g α = α α 1! α j! g α where α = (α 1,, α j, 0 ). We have g ψ = z α g α ψ and, by the form of ψ, the terms of this sum are pairwise orthogonal as α ranges over multi-indices of the form α = (α 1,, α j, 0 ). Thus g ψ = z α g α ψ = α α 1! α j! g α ψ α α 1! α j! T g α = T g where the inequality holds since g α is in M. Finally, we combine the above results to obtain Theorem 4. Proof of Theorem 4. As previously discussed (see Equation (8), it suffices to show ( ( 1 b j+1 )) C ψ = exp + + b n (11) 4 1 σj+1 1 σn where ψ is a normalization of ϕ. By Lemma 5 we know that the expression on the right hand side of Equation (11) is the norm of the restriction of C ψ to the reducing subspace M, and by Proposition the norm of the restriction of C ψ to the orthogonal complement of M is bounded above by this same expression. This gives the result. 14

15 3 Concluding remarks In this final section we consider the possibility of extending the results of the previous section to various generalizations of F. For simplicity we restrict to the case n = 1. First consider the, in general, non-hilbert Fock space F p defined, for 0 < p <, to be the space of all entire functions f on C for which the norm f p p 1 f(z) p e 1 z da(z) < π C where da denotes Lebesgue measure in C. The boundedness and compactness of C ϕ can be determined by the Carleson properties of a certain measure associated with ϕ. In particular, recall that a finite measure µ on C is a (p, q)-carleson measure if the space F p is a subset of L q (µ). By the closed graph theorem, this is equivalent to the existence of a positive constant C such that f L q (µ) C f p (1) for all f F p. The inequality (1) is equivalent to the boundedness of the inclusion map i : F p L q (µ). If the map i is compact, we say that µ is a vanishing (p, q)-carleson measure. (A linear map is compact if the image of the closed unit ball has compact closure.) For a given entire function ϕ, the weighted pullback measure µ ϕ on C is given by µ ϕ (E) = e 1 z da(z) ϕ 1 (E) for every Borel subset E of C. Since one can easily see that C ϕ (f) q = f L q (µ ϕ ) for every f in F p, the operator C ϕ : F p F q is bounded (resp., compact) if and only if the pullback measure µ ϕ is (p, q)-carleson (resp., vanishing (p, q)-carleson). The next two results give a characterization of the Carleson and the vanishing Carleson measures. We will not give the proofs of these results since they are slight improvements of results contained in a paper of Ortega [8]. Here (ζ, r) denotes the Euclidean disk with center ζ and radius r. Theorem 5. Let 0 < p q <. The following statements are equivalent: (i) The measure µ is a (p, q)-carleson measure. 15

16 (ii) For every r > 0 there is a constant C for which e αq z p dµ(z) C (13) (ζ,r) for all ζ C. (iii) There exists r > 0 and a constant C for which (13) holds for all ζ C. For vanishing Carleson measures, we have the following. Theorem 6. Let 0 < p q <. The following statements are equivalent: (i) The measure µ is a vanishing (p, q)-carleson measure. (ii) For every r > 0 e αq z p dµ(z) 0 (14) (ζ,r) as ζ. (iii) There exists r > 0 for which (14) holds. From the previous results, we can easily obtain Proposition 3 by replacing the measure µ with the measure µ ϕ. Proposition 3. Consider the operator C ϕ : F p F q for 0 < p q <. (a) If C ϕ is bounded, then ϕ(z) = az + b. Moreover, q a 1, and if p q a = 1, then b = 0. p (b) If C ϕ is compact, then q a < 1. p Conversely, suppose that ϕ(z) = az + b. q (c) If a = 1 and b = 0, then C p ϕ is bounded. (d) If q a < 1, then C p ϕ is compact. It is natural to ask whether one can compute the operator norm C ϕ : F p F q for 0 < p q < when at least one of p and q is different than. When p = q an obvious conjecture for this norm is ( ) 1 b exp q 1 a 16

17 when a < 1. When p < q a lower bound for the norm is ( ) 1/p 1/q ( ) 1 1 (q/p) b exp. π q 1 (q/p) a These results follow an argument similar to that in Lemma 3, using the point evaluation functionals on F p and F q. Returning to a Hilbert space setting, one might also consider generalizations to the Fock space obtained by defining F W to consist of those entire functions on C satisfying f(z) exp( W (z))da(z) < C for some positive weight function W (z) = W ( z ) satisfying appropriate hypotheses. To the extent that one can obtain good asymptotic information about k w, as w, (where, as usual, k w denotes the reproducing kernel function for evaluation at w in F W ) the methods of Theorem 1 will apply. For example, if W (z) = z c, c > 0, then a theorem of Folland and Rochberg (see, e.g. Theorem 4.6 in [?]) shows that k w = exp( w c ) w c c (1 + O(c w c ) γ ) for some γ > 0, as w. From this it follows, as in the proof of Theorem 1, that if C ϕ is bounded on F W, then lim sup z ϕ(z) z and hence that ϕ must be linear. The analogue of Theorem also holds for W (z) = exp( z c ), c > 0 with a similar proof, involving a change of variable argument, and elementary estimates on 1 w c (1 a c 1 b w c ) for w large. It seems unlikely that a precise calculation of C ϕ can be made in this setting, however it would be reasonable to expect that C ϕ FW Cϕ(k w ) = sup w C k w might continue to hold when W (z) = exp( z c ), c > 0. 17

18 References [1] M. Appel, P. Bourdon, and J. Thrall, Norms of composition operators on the Hardy space, Experiment. Math., 5 (1996), [] P. Bourdon and D. Retsek, Reproducing kernels and norms of composition operators, Acta Sci. Math. (Szeged), 67, (001), [3] C. Cowen and B. MacCluer, Composition Operators on Spaces of Analytic Functions, CRC Press, Boca Raton, FL, [4] C. Hammond, On the norm of a composition operator with linear fractional symbol, preprint. [5] F. Holland and R. Rochberg, Bergman kernels and Hankel forms on generalized Fock spaces, Contemp. Math. 3, (1999), [6] R. Horn and C. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, [7] F. Jafari, B. MacCluer, C. Cowen and D. Porter, eds. Studies on Composition Operators, Contemp. Math. 13, Amer. Math. Soc., [8] J. Ortega-Cerdà, Sampling Measures, Publ. Mat. 4 (1998), no., [9] W. Rudin, Function Theory in the Unit Ball of C n, Springer-Verlag, New York, [10] J. Shapiro, Composition Operators and Classical Function Theory, Springer, New York, The University of Michigan Department of Mathematics East Hall, 55 East University Ave Ann Arbor, MI carswell@umich.edu Department of Mathematics PO Box Kerchof Hall University of Virginia 18

19 Charlottesville, VA San Francisco State University Department of Mathematics San Francisco, CA