The Drury-Arveson Hardy space B

1 2

2

(B

ⁿ

)

The above theorem just misses capturing the Drury-Arveson Hardy space B

1 2

2 (Bⁿ).

If we take p = 2, σ = ¹₂ and η = 1 in the above proof, then the case η = 1 of (23) is weakened to the inequality

Re 1

1− z · z = Re (1− z · z )

|1 − z · z |² ≥ c + c22d(α∧α )−d([α]∧[α ]), z∈ K^α, z ∈ K^α, (24) which does not lead to the tree condition. We will however modify the proof so as to give a characterization in Theorem 7 below of the Carleson measures for B

1 2

2 (Bⁿ) in terms of the simple condition (31) and “split” tree condition (52) given below. We will proceed initially via the following three steps:

Step 1 If a positive measure µ is B

1 2

2 (Bⁿ)-Carleson, then the bilinear inequality

Bn Bn

1

1− z · z f (z ) dµ (z ) g (z) dµ (z) ≤ C f L²(µ) g _L2(µ), (25) holds for all f, g ∈ L²(µ).

Step 2 If a positive measure µ satisfies the Dirichlet tree condition, i.e. (8) for p = 2 and σ = 0;

β≥α

I^∗µ (β)² ≤ CI^∗µ (α) <∞, α∈ Tⁿ, (26)

then µ is B

1 2

2 (Bⁿ)-Carleson if and only if the bilinear inequality

Bn Bn

Re 1

1− z · z f (z ) dµ (z ) g (z) dµ (z)≤ C f L²(µ) g _L2(µ), (27) holds for all f, g ≥ 0.

Step 3 Suppose a positive measure µ satisfies (26) as well as the following growth condition on Bergman balls: for each R ≥ 1, there is a positive constant CR such that

µ (Bβ(w, R))≤ C^Rµ (Bβ(w, 1)) , w∈ Bⁿ, (28) where Bβ(w, t) = {z ∈ Bⁿ : β (z, w) < t} is the Bergman ball of radius t about w. Then the bilinear inequality (27) is equivalent to the discrete inequality,

α∈Tⁿ

|T^µg (α)|²µ (α)≤ C

α∈Tⁿ

|g (α)|²µ (α) , g ≥ 0, (29) where Tµ is the positive linear operator on the tree given by,

Tµg (α) =

β∈Tⁿ

22d(α∧β)−d([α]∧[β])g (β) µ (β) , α∈ Tⁿ, (30) and where Tⁿ ranges over all unitary rotations of a fixed Bergman tree (and [α] ∈ Rⁿwhich ranges over the corresponding unitary rotations of the associated ring tree).

Using the above three steps we can characterize Carleson measures for the Drury-Arveson Hardy space B

1 2

2 (Bⁿ)by the Dirichlet tree conditon (26) together with either (27) or (29). As an alternative to the Dirichlet tree conditon (26), we can also use the the simple condition

2^d(α)I^∗µ (α)≤ C, α∈ Tⁿ, (31) (recall that pσ = 1 and that θ = ^{ln 2}₂ so that 1 − |w|² ≈ 2^−d(α) = 2^−pσd(α) for w ∈ K^α by (20)), which is far stronger than the Dirichlet tree condition (26) in general:

β≥α

I^∗µ (β)²=

β≥α γ,γ ≥β

µ (γ) µ (γ ) =

γ,γ ≥α

[1 + d (α, γ∧ γ )] µ (γ) µ (γ ) (32)

=

γ≥α

µ (γ)

γ ≥α

[1 + d (α, γ∧ γ )] µ (γ )

≤

γ≥α

µ (γ)

d(α,γ)

j=0

(1 + d (α, γ)− j) I^∗µ A^jγ

≤

γ≥α

µ (γ)

d(α,γ)

j=0

(1 + d (α, γ)− j) C2^j−d(γ)

≤ C2^−d(α)I^∗µ (α)≤ CI^∗µ (α) . Recall that µ (α) = µ (α) = _K

αdµ for α ∈ Tⁿ.

Theorem 6 Let µ be a positive measure on the ball Bⁿ. Then the following conditions are equivalent:

1. µ is a Carleson measure on the Drury-Arveson space B

1 2

2 (Bⁿ), 2. µ satisfies both (31) and (27),

3. µ satisfies both (31) and (29), 4. µ satisfies both (26) and (27), 5. µ satisfies both (26) and (29).

In Theorem 7 of the next subsubsection, we will complete the characterization of Carleson measures for the Drury-Arveson space by giving a necessary and sufficient condition, namely (52) below, for (29) in the presence of (31).

Proof. If µ is B yields the simple condition (31). Indeed (we drop the from µ when no confusion should arise):

Note that we may choose structural constants for the Bergman tree so that the successor sets S (α) are sufficiently narrow that Re (1 − z · w)^{n+1+α2 +12} ≈

|1 − z · w|^{n+1+α2 +12} for z ∈ S (α) and w ∈ ∪β≤αKβ.

Conversely, suppose that (26) holds. We now define an associated measure µ that will satisfy (28) as well as (26):

dµ (z) =

where A > 0 will be chosen large enough that the Dirichlet tree condition (26) holds for the measure µ . One way to see that such an A exists is to note that for each N ≥ 1, the measure

dµ_N(z) =

α∈Tⁿ





d(α,β)≤N

µ (Kβ)



 χ_Kα(z) dλn(z)

is B2(Bⁿ)-Carleson with norm at most 2^CnN times that of µ. Clearly µ satisfies (28) as well. Then if either (27) or (29) holds, we obtain that µ is B

1 2

2 (Bⁿ )-Carleson by applying step 2 or 3 respectively to the measure µ in place of µ.

Next, we note that (28) implies µ ≤ µ and so µ is B

1 2

2 (Bⁿ)-Carleson, and hence also µ is B

1 2

2 (Bⁿ)-Carleson.

Given the three steps, this completes the proof of Theorem 6.

We now turn our attention to proving some parts of the three steps.

Proof. (Step 1) We use the presentation of H_n² given by the kernel function k (w, z) = _1−w·z¹ on Bⁿ as given earlier. Now µ is H_n²-Carleson if and only if the inclusion map T is bounded from H_n² to L²(µ), hence if and only if the adjoint T^∗ is bounded from L²(µ) to H_n², i.e.

T^∗f, T^∗f _H2

n ≤ C f ²L²(µ), f ∈ L²(µ) . (33) We claim that

T^∗f (z) = kw(z) f (w) dµ (w) , z ∈ Bⁿ. (34) Indeed, if µ is the delta mass δζ, then

T^∗f (z) = T^∗f, k_{z H}2

n = f, T k_{z L}2(µ)= f (ζ) kz(ζ) = f (ζ) kζ(z) = kw(z) f (w) dµ (w) . This also holds when µ is a finite linear combination of points masses, and a

limiting argument shows it holds in general. From (34) we obtain T^∗f, T^∗f _H2

n= kwf (w) dµ (w) , kwf (w ) dµ (w )

H_n²

= kw, kw H_n²f (w) dµ (w) f (w ) dµ (w )

= kw(w ) f (w) dµ (w) f (w ) dµ (w )

= 1

1− w · w f (w) dµ (w) f (w ) dµ (w ) ,

which in particular proves (25) if we substitute this in (33). which is equivalent to (27). Thus we have also proved step 2 without recourse to (26), thus simplifying some of the arguments below.

Proof. (Step 2) If µ is B

1 2

2 (Bⁿ)-Carleson, then (25) and hence also (27) holds, and we showed above that (26) holds.

Conversely, suppose that (26) and (27) hold. Given g ∈ L²(µ), we compute that S

where we have denoted the integral in braces above by Kα(z, z ).

A calculation shows that

Suppose now that f, g ≥ 0. Then using the asymptotic estimate (35) with α = 0 or 1, together with the fact that S

by using (27) on the first term after the equal sign, and using the Dirichlet tree condition (26) on the second term. Indeed, using inequality (83) in Subsubsection

5.2.1 of [8] together with the argument used for (41) below, we see that the second term after the equal sign in (36) is dominated by

C

Un α,α ∈Tn

d (α∧ α ) g U⁻¹α µ U⁻¹α g U⁻¹α µ U⁻¹α dU

= C

Uⁿ α∈Tⁿ

I^∗gµ U⁻¹α ² dU.

We have thus proved (21), and hence that µ is a B

1 2

2 (Bⁿ)-Carleson measure.

Proof. (Step 3) We first use (26) and (28) to discretize the bilinear inequality (27) to the following discrete bilinear inequality taken over all unitary rotations U⁻¹Tⁿ of the Bergman tree Tⁿ:

α,α ∈U⁻¹Tⁿ

22d(α∧α )−d([α]∧[α ])f (α) µ (α) g (α ) µ (α ) (37)

≤ C

α∈U⁻¹Tn

f (α)²µ (α)

1 2

α ∈U⁻¹Tn

g (α )²µ (α )

1 2

,

for all f, g ≥ 0 on U⁻¹Tⁿ, and for all U ∈ Uⁿ. At a crucial point in the argument below, we need to identify the distance 1 − |z · z |² in terms of the tree structure, and this is what leads to the associated ring tree Rⁿand the quantity d ([α] ∧ [α ]).

Recall that a slice of the ball Bⁿ is the intersection of the ball with a complex line through the origin. In particular, every point z ∈ Bⁿ\ {0} lies in a unique slice

Sz = e^iθz1, ..., e^iθzn : θ ∈ [0, 2π) .

We define two elements α and α of the Bergman tree Tⁿ to be slice-related if α ∼ α where ∼ is the equivalence relation introduced earlier. Now given α, α ∈ Tⁿ, let

[o, α] ={o, α¹, ..., αm = α} and [o, α ] = {o, α1, ..., α_m = α}

be the geodesics from the root to α, α respectively. We then have from (4) that αk and α_k are slice-related if and only if k ≤ d ([α] ∧ [α ]).

It may help the reader to visualize d ([α] ∧ [α ]) in the following way. Imagine that each slice S is thickened to a slab S of width one in the Bergman metric.

Thus in the Euclidean metric, a slab S is a lens whose “thickness” at any point is roughly the square root of the distance the boundary of the ball ∂Bⁿ. Moreover, given z ∈ Bⁿ, we denote by S^z the slab corresponding to the slice Sz, but trun-cated by intersecting it with B (0, |z|). The slabs S^cα and S^cα associated with the unique slices Scα and Sc_α through cα and cα will intersect in a “disc” of radius

roughly d ([α] ∧ [α ]) in the Bergman metric. Note that from this picture that αd([α]∧[α ]) is the exit point Eαα of the geodesic [o, α] from the slab S^α associated to the slice Sα through cα, and similarly, αd([α]∧[α ]) is the exit point Eαα of the geodesic [o, α ] from the slab S^α. Both points have the same distance from the root. Note that we can also define Eαα as the intersection of the geodesic [o, α]

with the ring [α] ∧[α ], which we will denote by E[α]∧[α ]α. Finally, note that since d ([α]∧ [α ]) = d (E^αα) = d (Eαα )and α ∧α = α where = max {k : α^k = α_k}, we have that d ([α] ∧ [α ]) satisfies

d (α∧ α ) ≤ d ([α] ∧ [α ]) ≤ min {d (α) , d (α )} . (38) The key feature of the quantity d ([α] ∧ [α ]) is that 2−d([α]∧[α ]) is essentially 1− |z · z |² for z ∈ K^α, z ∈ K^α. More precisely, for each z, z ∈ Bⁿ, there is a subset Σ of the unitary group Uⁿ with Haar measure |Σ| ≥ c > 0 and satisfying

c2^−d([^U−1α(z)]∧[^{U−1α(z )}]) ≤ 1 − |z · z |² ≤ C2−d([α(z)]∧[α(z )]), U ∈ Σ. (39) The main inequalities used in establishing the equivalence of (27) and (37) are (24), i.e.

Re 1

1− z · z ≥ c + c22d(α∧α )−d([α]∧[α ]), z ∈ K^α, z ∈ K^α, (40) for all α, α ∈ U⁻¹Tⁿ, U ∈ Uⁿ, together with a converse obtained by averaging over all unitary rotations U⁻¹Tⁿ of the Bergman tree Tⁿ,

Re 1

1− z · z ≤ C + C

Un

22d(α(U z)∧α(Uz ))−d([α(Uz)]∧[α(Uz )])dU. (41) This latter inequality is analogous to similar inequalities in Euclidean space used to control an operator by translations of its dyadic version. These inequalities follow from

Re 1

1− z · z = Re (1− z · z )

|1 − z · z |² ≈ 1− |z · z |²

|1 − z · z |² + 1. (42) With all this, we can complete the proof of the equivalence of (27) and (37) in the presence of (26) and (28).

Now (37) can be rewritten as

α∈Tⁿ

f (α){T^µg (α)} µ (α) ≤ C f ²(µ) g 2(µ), (43) for all f, g ≥ 0 on Tⁿ, and where Tµ is given in (30):

Tµg (α) =

α ∈Tn

22d(α∧α )−d([α]∧[α ])g (α ) µ (α ) .

Upon using the Cauchy-Schwartz inequality and taking the supremum over all f with f 2(µ) = 1 in (43), we obtain the equivalence of (43) and the discrete inequality (29), where Tⁿ ranges over all unitary rotations of a fixed Bergman tree.

1.4.1 A characterization of Carleson measures for B

1 2

2

The bilinear inequality associated with (29) is

α∈Tn We rewrite the left hand side as

A,B∈Rⁿα∈A

Define the projection PC from functions h = {h (α)}_α∈C on the ring A to functions PCh on the ring C (provided C ≤ A) by

We also define the “Poisson kernel” P^C at scale C to be the mapping taking functions h = {h (γ )}_{γ ∈C} on C to functions P^Ch ={P^Ch (γ)}_γ∈C on C given by notice that the Poisson kernel

P^C(γ, γ ) = 2^{2d(γ∧γ )} 2^d(C)

is a geometric sum of averaging operators A^kC with kernel A^kC(γ, γ ) = 2^d(C)−kχ{d(γ∧γ )=d(C)−k}, namely

P^C(γ, γ ) =

d(C)

k=0

2^−kA^kC(γ, γ ) . (46) We now consider the bilinear inequality with P^C replaced by A⁰C:

C∈RnA,B∈Rn

A∧B=C

A⁰C(PC(fAµ)) , PC(gBµ) _C ≤ C f ²(µ) g 2(µ). (47)

The left side of (47) is

C∈Rⁿ

2^d(C)

A,B∈Rⁿ A∧B=C

PC(fAµ) , PC(gBµ) _C

=

C∈Rn

2^d(C)

γ∈C A,B∈Rn

A∧B=C

I^∗(fAµ) (γ) I^∗(gBµ) (γ) .

For fixed γ ∈ C, we dominate the sum A,B∈Rⁿ A∧B=C

above by

A,B∈Rⁿ A∧B=C

I^∗(fAµ) (γ) I^∗(gBµ) (γ)≤ I^∗(f µ) (γ) (gµ) (γ) + (f µ) (γ) I^∗(gµ) (γ) +

δ,δ ∈C(γ) [δ]=[δ ]

I^∗(f µ) (δ) I^∗(gµ) (δ ) .

The first two terms easily satisfy the bilinear inequality using only the simple condition (31). Indeed,

C∈Rⁿ

2^d(C)

γ∈C

I^∗(f µ) (γ) (gµ) (γ) =

γ∈Tⁿ

2^d(γ)I^∗(f µ) (γ) (gµ) (γ)

=

γ∈Tⁿ

I 2^df µ (γ) g (γ) µ (γ)

≤ I 2^df µ 2(µ) g 2(µ). Now the inequality I 2^df µ 2(µ) ≤ C f ²(µ) can be rewritten

γ∈Tn

Ih (γ)²µ (γ) ≤ C

γ∈Tn

h (γ)²2^−2d(γ)µ (γ) ,

which is equivalent to the condition

γ≥α

I^∗µ (γ)²2^2d(γ)µ (γ) ≤ CI^∗µ (α) , α∈ Tⁿ,

which in turn is trivially implied by the simple condition I^∗µ (γ)²2^2d(γ) ≤ C². It remains then to consider the “split” bilinear inequality

γ∈Tn

2^d(γ)

δ,δ ∈C(γ) [δ]=[δ ]

I^∗(f µ) (δ) I^∗(gµ) (δ )≤ C f ²(µ) g 2(µ), (48)

or equivalently the corresponding quadratic inequality obtained by setting f = g:

γ∈Tn

2^d(γ)

δ,δ ∈C(γ) [δ]=[δ ]

I^∗(f µ) (δ) I^∗(f µ) (δ )≤ C

α∈Tn

f (α)²µ (α) . (49)

Note that we have the following necessary condition for (49) that is obtained by taking f = χ_S(α) in (49):

γ≥α

2^d(γ)

δ,δ ∈C(γ) [δ]=[δ ]

I^∗µ (δ) I^∗µ (δ )≤ CI^∗µ (α) , α∈ Tⁿ. (50)

We now show that (50) and (31) together imply (49). To see this write the left side of (49) as

γ∈Tn

2^d(γ)

δ,δ ∈C(γ) [δ]=[δ ]

I^∗µ (δ) I^∗µ (δ )I^∗(f µ) (δ) I^∗(f µ) (δ ) I^∗µ (δ) I^∗µ (δ ) ,

and using the symmetry in δ, δ we bound it by

γ∈Tn

2^d(γ)

δ,δ ∈C(γ) [δ]=[δ ]

I^∗µ (δ) I^∗µ (δ ) I^∗(f µ) (δ) I^∗µ (δ)

2

.

By Theorem 3, this last term is dominated by the right side of (49) provided

γ≥α

2^d(γ)

δ,δ ∈C(γ) [δ]=[δ ]

I^∗µ (δ) I^∗µ (δ ) + 2^d(Aα)

δ ∈C(Aα) [α]=[δ ]

I^∗µ (α) I^∗µ (δ ) (51)

≤ CI^∗µ (α) , α∈ Tⁿ.

Now the necessary condition (50) shows that the first sum in (51) is at most CI^∗µ (α), while the simple condition (31) yields 2^d(Aα)I^∗µ (δ )≤ C, which shows

that the second sum in (51) is at most CI^∗µ (α). This completes the proof that (47) holds when both (50) and (31) hold.

The averaging operators A^kC for k > 0 are handled similarly, and we obtain the following theorem where

G^(k)(γ) = (η, η )∈ C^(k)(γ)× C^(k)(γ) : η∧ η = γ, [Aη] = [Aη ] , [η] = [η ] . Theorem 7 A positive measure µ on the ball Bⁿ is B

1 2

2 (Bⁿ)-Carleson if and only if µ satisfies the simple condition (31) and the following split tree condition,

k≥0 γ≥α

2^d(γ)−k

(δ,δ )∈G^(k)(γ)

I^∗µ (δ) I^∗µ (δ )≤ CI^∗µ (α) , α∈ Tⁿ, (52)

taken over all unitary rotations of the Bergman tree Tⁿ. 1.4.2 Related inequalities

Since

2d (α∧ α ) − d ([α] ∧ [α ]) = d (α) − d (α, E^αα ) ,

where Eαα denotes the exit point of the geodesic [0, α ] from the slab S^cα, we may rewrite the operator Tµ as the following variant of a fractional integral:

Tµg (α) =

α ∈Tn

2^{d(α)−d(α,Eαα )}g (α ) µ (α ) .

Inequality (38),

d (α∧ α ) ≤ d ([α] ∧ [α ]) ≤ min {d (α) , d (α )} ,

has the following interpretation relative to the kernel K (α, α ) = 22d(α∧α )−d([α]∧[α ]). If we replace d ([α] ∧ [α ]) by the lower bound d (α ∧ α ) in the kernel K (α, α ), then Tµ becomes

Tµg (α) =

α ∈Tn

2^{d(α∧α )}g (α ) µ (α ) , (53) whose boundedness on ²(µ) is equivalent to µ being a Carleson measure for B

1 2

2 (Tⁿ), which is in turn equivalent to the tree condition (8) with σ = ¹₂ and p = 2 (Alternatively, the above kernel is the discretization of the continuous kernel _1−z·z¹ , whose Carleson measures are characterized by the tree condition).

This observation provides another proof in the case of the Drury-Arveson space B

1 2

2 (Bⁿ), of a more general result that shows the tree condition characterizes Carleson measures supported on a tangential manifold (when d ([α] ∧ [α ]) = d (α∧ α ) for α, α in the support of the measure). In addition, we can see from this observation that the simple condition (31) is not sufficient for µ to be a

B

1 2

2 (Tⁿ)-Carleson measure. Indeed, let Y be any dyadic subtree of Tⁿ with the properties that the two children α+ and α₋ of each α ∈ Y are also children of α in Tⁿ, and such that no two tree elements in Y are equivalent. Now let µ be any measure supported on Y that satisfies the simple condition

2^d(α)I^∗µ (α)≤ C, α∈ Y, but not the tree condition

β∈Y:β≥α

2^σd(β)I^∗µ (β) ^p ≤ CI^∗µ (α) <∞, α∈ Y.

For α, α ∈ Y, we have d ([α] ∧ [α ]) = d (α ∧ α ), and so µ is a B

1 2

2 (Tⁿ)-Carleson measure if and only if the operator T in (53) is bounded on ²(µ) , which is equivalent to the above tree condition, which we have chosen to fail. Finally, to transplant this example to the ball Bⁿ, we take dµ (z) = _α∈Yµ (α) δcα(z) and show that the above tree condition fails on a positive proportion of the rotated trees U⁻¹Tⁿ, U ∈ Uⁿ.

If on the other hand, we replace d ([α] ∧ [α ]) by the upper bound min {d (α) , d (α )}

in the kernel K (α, α ), then Tµ becomes Tµg (α) =

α ∈Tⁿ

22d(α∧α )−min{d(α),d(α )}g (α ) µ (α ) , (54) whose boundedness on ²(µ) is shown in Theorem 8 below to be implied by the simple condition (31). Thus we see that the simple condition (31) characterizes Carleson measures supported on a slice (when d ([α] ∧ [α ]) = min {d (α) , d (α )}

for α, α in the support of the measure). In particular, this provides a new proof that the simple condition (31) characterizes Carleson measures for the Hardy space H²(D) = B

1 2

2 (D) in the unit disc.

Finally, when µ is invariant, the operator Tµ in (30) is bounded on ²(µ) if and only if µ is finite. To see this we need the “Poisson kernel” estimate

β∈B

2^2d(α∧β) ≈ 2d(B)+d([α]∧B), α∈ Tⁿ, B ∈ Rⁿ. (55) Using (55) we compute that Tµ1 is bounded (and hence a Schur function). Thus Tµ is bounded on ^∞(µ) with norm at most µ , and by duality also on ¹(µ).

Interpolation now yields that Tµ is bounded on ²(µ) with norm at most µ . Theorem 8 A positive measure µ satisfies the bilinear inequality

α,α ∈Tⁿ

22d(α∧α )−min{d(α),d(α )}f (α) µ (α) g (α ) µ (α )≤ C f ²_(Tn;µ) g 2(Tⁿ;µ), (56) if µ satisfies the simple condition (31).

Remark 2 Using the argument on pages 538-542 of [31], it can be shown that the bilinear inequality (56) holds if and only if the following pair of dual conditions hold:

β≥α

I2^d χ_S(α)µ (β) ²µ (β)≤ C

β≥α

µ (β) <∞, α ∈ Tⁿ, (57)

β≥α

2^dI χS(α)2^−dµ (β) ²µ (β)≤ C

β≥α

2^−2d(β)µ (β) , α∈ Tⁿ, where I is the fractional integral of order one on the Bergman tree given by,

Iν (α) =

β∈Tn

2^−d(α,β)ν (β) , α ∈ Tⁿ. (58)

The following simple sufficient condition of Schur type is used for the proof.

Recall that a measure space (Z, µ) is σ-finite if Z = ∪^∞N =1ZN where µ (ZN) <∞, and that a function k on Z × Z is σ-bounded if Z = ∪^∞N =1ZN where k is bounded on Zn× Zⁿ.

Lemma 9 (Vinogradov-Sen˘ı˘ckin Test, pg 151of [27]) Let (Z, µ) be a σ-finite mea-sure space and k a nonnegative σ-bounded function on Z × Z satisfying

Z×Z

k (s, t) k (s, x) dµ (s)≤ M k (t, x) + k (x, t)

2 for µ-a.e. (t, x) ∈ Z×Z.

(59) Then the linear map T defined by

T g(s) =

Z

k (s, t) g (t) dµ (t) is bounded on L²(µ) with norm at most M .

Proof. Let Z = ∪^∞N =1ZN where µ (ZN) < ∞ and k is bounded on Z^N × Z^N. The kernels

kN(s, t) = k (s, t) χ_ZN×ZN(s, t)

satisfy (59) uniformly in N , and the corresponding operators TNg(s) = _ZkN(s, t) g (t) dµ (t) are bounded on L²(µ) (with norms depending on µ (ZN) and the bound for k

on ZN × Z^N). However, (59) for kN implies that the integral kernel of the oper-ator T_N^∗TN is dominated pointwise by ^M₂ times that of T_N^∗ + TN, and this gives

TN

2 = T_N^∗TN ≤ ^M2 T_N^∗ + TN ≤ M T^N , and hence TN ≤ M. Now let N → ∞ and use the monotone convergence theorem to obtain T ≤ M.

Remark 3 If k (x, y) = k (y, x) is symmetric, then (59) ensures that for any choice of a, k(a, ·) can be used as a test function for Schur’s Lemma.

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Funct. Anal. 175 (2000), 111-124.

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在文檔中 NCTS 2005 Summer School on Harmonic Analysis : Lecture Note(IV) (頁 51-66)