−∞. r→∞ r φ′ (r) lim
(3.12)
Then φ satisfies the condition Kp and there exists r1 > 0 such that ψp (r) ≍
1 φ′ (r)
for r ≥ r1 ,
where the involved constants might depend on p > 0. Proof. By hypothesis there is α < 1 and r2 ≥ r0 such that r ′ (3.13) ≤ αr on [r2 , +∞). pφ′ (r)
INTEGRAL OPERATORS ON WEIGHTED FOCK SPACES
17
So an integration by parts on (r2 , r] gives Z r Z r −s −pφ(s) ′ −pφ(s) se ds = −pφ (s)e ds pφ′ (s) r2 r2 Z r −re−pφ(r) r2 e−pφ(r2 ) s ′ −pφ(s) = e ds + + ′ pφ′ (r) pφ′ (r2 ) r2 pφ (s) Z r −re−pφ(r) r2 e−pφ(r2 ) + +α se−pφ(s) ds, ≤ pφ′ (r) pφ′ (r2 ) r2 that is Rr
−pφ(s) ds ≤ r2 se
1 1−α
−re−pφ(r) pφ′ (r)
+
r2 e−pφ(r2 ) pφ′ (r2 )
,
so taking the limit as r → ∞ and bearing in mind (3.12), we can assert that there is C1 = C1 (p, φ) > 0 such that for any r ≥ r2 Z ∞ re−pφ(r) (3.14) . se−pφ(s) ≤ C1 ′ φ (r) r R∞ In particular, 0 se−pφ(s) ds < ∞. We note that φ satisfies the Kp condition if ! R R ∞ −pφ(s) ′ (r) ∞ se−pφ(s) ds pφ se ds r r (3.15) lim sup − ≤ K < ∞. r 2 e−pφ(r) re−pφ(r) r→∞ Now, by (3.13) r = ′ φ (r)
Z r r2 s ′ pr 2 r2 ds + ≤ + ′ , ′ ′ φ (r2 ) 2 φ (r2 ) r2 φ (s)
which together with (3.14) implies that R ∞ −pφ(s) ds 1 p r2 r se . . + < ∞, ′ 2 rφ (r) 2 r φ′ (r2 ) r 2 e−pφ(r)
so the first addend in (3.15) is bounded. On the other hand, a straight-forward application of L’Hospital’s rule gives R∞ p r se−pφ(s) ds 1 ≥ lim inf > −∞, lim inf −pφ(r) 1 r ′ re r→∞ 1 − r→∞ ′ pr ( φ′ (r) ) φ (r)
consequently (3.15) holds. Finally, another application of L’Hospital’s rule gives lim inf r→∞
1 p−
1 r ′ r ( φ′ (r) )
≤ lim inf r→∞
which proves the lemma.
ψp (r) ψp (r) 1 , ≤ lim sup ′ ≤ lim sup 1 r ′ −1 −1 (φ (r)) r→∞ (φ (r)) r→∞ p − r ( φ′ (r) )′
Lemma 18. Assume φ : [0, ∞) → R+ is a twice continuously differentiable function such that ∆φ > 0 , (∆φ(z))−1/2 ≍ τ (z), where τ (z) is a radial positive
´ ANGEL ´ ´ OLIVIA CONSTANTIN AND JOSE PELAEZ
18
function that decreases to zero as |z| → ∞ and limr→∞ τ ′ (r) = 0. Then (a)
φ′ (r) = ∞. r→∞ r
(b)
lim τ (r)φ′ (r) = ∞, or, equivalently, lim
lim
φ′′ (r) = 0. r→∞ (φ′ (r))2
r→∞
(c) For any p > 0 we have ψp (r) ≍ (d) lim inf r→∞
1 φ′ (r) +
1
, for r ≥ 0.
r∆φ(r) ≥ C > 0. φ′ (r)
Proof. A simple calculation shows that 1 (rφ′ (r))′ ∆φ(r) = φ′′ (r) + φ′ (r) = . r r By L’Hospital’s rule we obtain (3.16)
∆φ(r) (rφ′ )′ rφ′ (r) = lim = ∞, = lim 2 r→∞ r→∞ 2r r→∞ r 2 lim
which proves (a). Let us now prove (b). Taking into account (3.16) we obtain Z Z τ (r) r s τ (r) r ′ ds. s∆φ(s) ds & τ (r)φ (r) = 2 r r 0 0 τ (s)
Again by L’Hospital’s rule we get
rτ −1 (r) τ (r) − rτ ′ (r) lim R r s = 0, = lim r→∞ r→∞ r 0 τ 2 (s) ds
which implies limr→∞ τ (r)φ′ (r) = ∞. By (a) and relation (3.16) this last fact is ′′ (r) equivalent to limr→∞ (φφ′ (r)) 2 = 0. Taking into account (a)−(b) it is straight-forward to check that the hypotheses in Lemma 17 are satisfied, indeed 1 r ′ = 0. lim r→∞ r φ′ (r)
and hence ψp (r) ≍ φ′1(r) for r ≥ r0 . Since φ′ ≥ 0 and limr→∞ φ′ (r) = ∞, we obtain (c). We now turn to (d). By (3.16) r∆φ(r) rτ −2 (r) r 2 τ −2 (r) R , ≍ ≍ r s φ′ (r) φ′ (r) 0 τ 2 (s) ds
so L’Hospital’s rule implies
r 2 τ −2 (r) rτ ′ (r) lim inf R r s ≥ 2, ≥ lim inf 2 1 − r→∞ r→∞ τ (r) 0 τ 2 (s) ds
and we are done. The previous considerations lead us to the following Littlewood-Paley formula.
INTEGRAL OPERATORS ON WEIGHTED FOCK SPACES
19
R+
Theorem 19. Assume that 0 < p < ∞ and φ : [0, ∞) → is twice continuously differentiable satisfying the hypotheses of Lemma 17. Then we have Z p p |f ′ (z)|p (ψp,φ (z))p e−pφ(z) dm(z), ||f || φ ≍ |f (0)| + Fp
C
for any entire function f , where the distortion function ψp,φ satisfies ψp,φ (z) ≍
1 φ′ (z)
for |z| ≥ r0 ,
for some r0 > 0. In particular, if φ ∈ I, the following holds Z e−pφ(z) dm(z), |f ′ (z)|p ||f ||p φ ≍ |f (0)|p + Fp (1 + φ′ (z))p C
for any entire function f .
It is worth to point out that the class of functions satisfying the hypotheses of Lemma 17 is quite large, containing functions whose growth ranges from logar rithmic (e.g. φ(r) = a log(1 + r), ap > 2) to highly exponential (e.g. φ(r) = ee ). The following result which is deduced from Lemma 5 shows that the distortion function is ”almost constant” on sufficiently small discs whose radii depend on ∆φ. Lemma 20. Assume that φ : [0, ∞) → R+ is a twice continuously differentiable function such that ∆φ > 0 , (∆φ(z))−1/2 ≍ τ (z), where τ (z) is a radial positive function that decreases to zero as |z| → ∞ and limr→∞ τ ′ (r) = 0. Then, there exists r0 > 0 such that φ′ (a) ≍ φ′ (z),
(3.17)
z ∈ D(a, δτ (a)),
for all a ∈ C with |a| > r0 . Moreover,
1 + φ′ (a) ≍ 1 + φ′ (z),
z ∈ D(a, δτ (a)),
for all a ∈ C. Proof. Since φ is radial, it is enough to show that there exists r0 > 0 such that for a ≥ r0 φ′ (a) ≍ φ′ (r),
(3.18) Recall that
r ∈ (a − δτ (a), a + δτ (a)).
1 φ (r) = r Our assumptions on τ imply ′
a ≍ r,
Z
r
s∆φ(s) ds,
r > 0.
0
r ∈ (a − δτ (a), a + δτ (a)),
for a > 2δ maxs∈[0,∞) τ (s). Hence, proving (3.18) reduces to showing that Z a Z r s∆φ(s) ds, r ∈ (a − δτ (a), a + δτ (a)). s∆φ(s) ds ≍ 0
0
´ ANGEL ´ ´ OLIVIA CONSTANTIN AND JOSE PELAEZ
20
We have R a−δτ (a) 0
(3.19)
Ra
s
τ 2 (s)
s 0 τ 2 (s)
ds
ds
R a+δτ (a) s Rr ds 2 s∆φ(s) ds 0 R a s τ (s) . . R0a 0 s∆φ(s) ds 0 τ 2 (s) ds
By L’Hospital’s rule and Lemma 5 we deduce
lim sup a→∞
R a+δτ (a) 0
Ra
s τ 2 (s)
s 0 τ 2 (s)
ds
ds
. lim sup
(1 + δτ ′ (a))(a + δτ (a))τ −2 (a + δτ (a)) aτ −2 (a)
∼ lim sup
(1 + δτ ′ (a))(a + δτ (a)) = 1. a
a→∞
a→∞
Analogously one can show that
lim inf a→∞
R a−δτ (a) 0
Ra
s τ 2 (s)
s 0 τ 2 (s)
ds
ds
≥ c > 0.
and (3.17) follows in view of (3.19). Since limr→∞ φ′ (r) = ∞ last assertion is now straight-forward.
4. Proof of Theorem 1 At the end ofnthis section we provide a proof of Corollary 2. √ o 2 Fix R > max 100, √p , 2 p , δ ∈ (0, mτ ) and consider the covering {D(δτ (zj )} given by Lemma 6 for t(z) = δτ (z).
4.1. Proof of (I): boundedness. Suppose first that Id : Fpφ → Lq (µ) is bounded. For a ∈ C with |a| ≥ η(R), consider the function Fa,R obtained in Proposition A. By Corollary 8, we have kFa,R kp φ ≍ τ (a)2 . Then, using (2.3) we get Fp
Z
eqφ(z) dµ(z) .
Z
D(δτ (a))
D(δτ (a))
≤
Z
C
|Fa,R (z)|q dµ(z)
|Fa,R (z)|q dµ(z) . kId kq φ
which implies that Kµ,φ ≤ CkId kqAp (w)→Lq (µ) .
Fp
2q
→Lq (µ)
τ (a) p ,
INTEGRAL OPERATORS ON WEIGHTED FOCK SPACES
21
Conversely, suppose that (1.4) holds. The idea of the proof goes back to [17]. Bearing in mind Lemma 6, Lemma 7 and Lemma 5, it follows that Z XZ |f (z)|q dµ(z) |f (z)|q dµ(z) ≤ C
j
.
XZ j
(4.1) .
D(δτ (zj ))
D(δτ (zj ))
X Z j
1 τ (z)2
|f (ζ)|p e−pφ(ζ) dm(ζ)
D(δτ (z))
p −pφ(ζ)
|f (ζ)| e
D(3δτ (zj ))
j
!q Z p dm(ζ)
eqφ(z) dµ(z)
eqφ(z) dµ(z) D(δτ (zj ))
2q
τ (z) p
!q
p
X Z
. Kµ,φ
!q
p
Z
D(3δτ (zj ))
|f (ζ)|p e−pφ(ζ) dm(ζ)
Now, using Minkowski inequality and the finite multiplicity N of the covering D(3δ τ (zj )) (see Lemma 6), we have Z
C
XZ |f (z)|q dµ(z) . Kµ,φ j
D(3δτ (zj ))
. Kµ,φ N q/p kf kq φ ,
q/p
|f (ζ)|p e−pφ(ζ) dm(ζ)
Fp
proving that Id : Fpφ → Lq (µ) is continuous with kId kq φ
Fp →Lq (µ)
. Kµ,φ .
4.2. Proof of (I): compactness. Suppose that (1.5) holds and let {fn } be a bounded sequence in Fpφ . By Lemma 7, Montel’s theorem and Fatou’s lemma, we may extract a subsequence {fnk } converging uniformly on compact sets of C to some function f ∈ Fpφ . Given ε > 0, fix r0 ∈ (0, ∞) with Z 1 (4.2) eqφ(z) dµ(z) < ε. sup 2q/p τ (a) D(δτ (a)) |a|>r0 Observe that there is r0′ ∈ (0, ∞) with r0 ≤ r0′ such that if a point zk of the sequence {zj } belongs to {|z| ≤ r0 }, then D(δτ (zk )) ⊂ {|z| ≤ r0′ }. So, take nk big enough such that sup|z|≤r0′ |fnk (z) − f (z)| < ε. Then, setting gnk = fnk − f , and arguing as in (4.1), it follows that Z X Z q q |gnk (z)|q dµ(z) |gnk (z)| dµ(z) + kgnk kLq (µ) ≤ |z|≤r0′
|zj |>r0
≤ Cε + Ckgnk kq φ sup
Fp |z |>r 0 j
D(δτ (zj ))
1 τ (zj )2q/p
This proves that Id : Fpφ → Lq (µ) is compact.
Z
eqφ(z) dµ(z) < Cε. D(δτ (zj ))
´ ANGEL ´ ´ OLIVIA CONSTANTIN AND JOSE PELAEZ
22
Conversely, suppose that Id : Fpφ → Lq (µ) is compact. Take Fa,R (z) , τ (a)2/p
fa,R (z) =
η(R) ≤ |a|,
where η(R) and Fa,R are obtained from Proposition A. By Corollary 8, sup kfa,R kF φ ≤ C < ∞ p
|a|≥η(R)
which together with the compactness of the identity operator implies that {fa,R : η(R) ≤ |a|} is a compact set in Lq (µ). Thus Z (4.3) |fa,R (z)|q dµ(z) = 0 uniformly in a. lim r→∞ r 2 2 On the other hand, it follows from Lemma 7 that Z eqφ(z) |fn (ζ)|q e−qφ(ζ) dm(ζ). |fn (z)|q ≤ C τ (z)2 D( δ τ (z)) 2
Integrate with respect to dµ, apply Fubini’s theorem, use (4.6) and Lemma 5 to obtain Z (4.7) |fn (z)|q dµ(z) {z∈C: |z|>r}
≤C
Z
{ζ∈C: |ζ|> 2r }
|fn (ζ)|q e−qφ(ζ)
1 τ (ζ)2
Z
!
eqφ(z) dµ(z) dm(ζ), r ≥ r0′ .
D(δτ (ζ))
By condition (c) for any fixed ε > 0, there is r0 ≥ r0′ ∈ (0, ∞), such that ! p Z Z p−q p 1 qφ(z) p−q . e dµ(z) dm(ζ) < ε 2 {ζ∈C: |ζ|> r0 } τ (ζ) D(δτ (ζ)) 2
Then (4.7) and an application of H¨older’s inequality yields Z (4.8) |fn (z)|q dµ(z) {z∈C: |z|>r0 }
Z ≤ Ckfn kq φ Fp
≤ Cε.
{ζ∈C: |ζ|> r20 }
1 τ (ζ)2
Z
qφ(z)
e
D(δτ (ζ))
!
dµ(z)
p p−q
p−q p
dm(ζ)
R Moreover, we have limn→∞ |z|≤r0 |fn (z)|q dµ(z) = 0, which together with (4.8), gives limn→∞ kfn kLq (dµ) = 0. This completes the proof of Theorem 1. Proof of Corollary 2. Fix p ∈ (0, ∞) and put µ(z) = e−qφ(z) dm(z) in Theorem 1. For q > p, taking into account that τ decreases to 0 as a → ∞, we deduce that for any δ > 0 Z 1 eqφ(z) dµ(z) ≍ τ (a)2(1−q/p) → ∞, if a → ∞, τ (a)2q/p D(δτ (a)) so by Theorem 1 (I), Fpφ 6⊂ Fqφ . On the other hand, if 0 < q < p, using Theorem p 1 (II) and the fact that f ≡ 1 ∈ / L p−q (C, dm), we deduce that Fpφ 6⊂ Fqφ .
INTEGRAL OPERATORS ON WEIGHTED FOCK SPACES
25
5. Proof of Theorem 3 Let us first notice that, by Theorem 19, we have Z q (5.1) |f (z)|q |g′ (z)|q (1 + φ′ (z))−q e−qφ(z) dm(z), kTg f k φ ≍ Fq
C
for q > 0 and for any entire function f . This relation shows that the boundedness (compactness) of the integration operator Tg : Fpφ → Fqφ is equivalent to the continuity (compactness) of the embedding Id : Fpφ → Lq (µg,φ ) with dµg,φ (z) = |g ′ (z)|q (1 + φ′ (z))−q e−qφ(z) dm(z). 5.1. Proof of (I). Assume 0 < p ≤ q < ∞ and let δ ∈ (0, mτ ). By Theorem 1, the embedding Id : Fpφ → Lq (µg,φ ) is continuous if and only if Z 1 sup (5.2) |g′ (z)|q (1 + φ′ (z))−q dm(z) < ∞. 2q/p a∈C τ (a) D(δτ (a)) Bearing in mind Lemma 5, it is clear that (1.6) implies (5.2). On the other hand, by Lemma 20 and the subharmonicity of |g′ |q , we deduce Z −2 qp 2−2 qp ′ −q ′ q ′ −q |g ′ (z)|q dm(z) . (1 + φ (a)) τ (a) |g (a)| (1 + φ (a)) τ (a) D(δτ (a)) Z −2 qp . τ (a) |g′ (z)|q (1 + φ′ (z))−q dm(z), D(δτ (a))
and the proof of part (a) of (I) is complete. Concerning the compactness part (b) of (I), note that, by part (I) of Theorem 1, the embedding Id : Fpφ → Lp (µg,φ ) is compact if and only if Z 1 lim |g′ (z)|q (1 + φ′ (z))−q dm(z) = 0. 2q |a|→∞ τ (a) p D(δτ (a)) Proceeding as in the boundedness part, we see that this is equivalent to 2−2 qp
lim |g ′ (a)|q (1 + φ′ (a))−q τ (a)
|a|→∞
= 0.
5.2. Proof of (II). The equivalence (a) ⇔ (b) follows from part (II) of Theorem 1. Let us prove that (b) ⇒ (c). From part (II) of Theorem 1 we deduce that (b) is equivalent to ! p Z Z p−q 1 ′ q ′ −q dm(z) < ∞. |g (ζ)| (1 + φ (ζ)) dm(ζ) τ (z)2 D(τ (z)) C
´ ANGEL ´ ´ OLIVIA CONSTANTIN AND JOSE PELAEZ
26
Now, by Lemma 20 and the subharmonicity of |g′ |q , Z |g′ (z)|r (1 + φ′ (z))−r dm(z) C
.
Z
1 τ (z)2 (1 + φ′ (z))q
C
.
Z
1 τ (z)2
C
Z
′
Z
′
!
q
p p−q
|g (ζ)| dm(ζ)
D(τ (z)) q
′
−q
|g (ζ)| (1 + φ (ζ))
dm(z) !
dm(ζ)
D(τ (z))
p p−q
dm(z) < ∞.
′
g r older’s inequality a gives (c) ⇒ (b). If 1+φ ′ ∈ L (C, dm), then (5.1) and H¨ q/r q/p Z Z q ′ r ′ −r p −pφ(z) |g (z)| (1 + φ (z)) dm(z) |f (z)| e dm(z) kTg f k φ . Fq
C
C
g′ .k kqLr (C, dm) kf kq φ . ′ Fp 1+φ
′
g Thus Tg : Fpφ → Fqφ is bounded with kTg k . 1+φ ′
Lr (C,dm)
. This finishes the
proof.
6. Schatten classes on F2φ
Given a separable Hilbert space H, the Schatten p-class of operators on H, Sp (H), consists of those compact operators T on H with its sequence of singular numbers λn belonging to ℓp , the p-summable sequence space. If {en } is an orthonormal basis of a Hilbert space H of analytic functions in C with reproducing kernel Kz , then X X (6.1) Kz (ζ) = hKz , en ien (ζ) = en (ζ) en (z) n
n
Also, by (6.1) we have
X ∂ e′n (z)en (ζ), Kz (ζ) = ∂ z¯ n
Thus Parseval’s identity gives X (6.2) kKz k2H = |en (z)|2 n
and
z, ζ ∈ C.
2 X ∂ Kz = |e′n (z)|2 . ∂ z¯ H
n
Now, we are going to give the proof of Theorem 4 on the description of the Schatten classes Sp := Sp (F2φ ). First we consider the sufficiency part of the case 1 < p < ∞. For this we need the following two lemmas. The first one is a L∞ version of Theorem 12 (see also [20, Theorem 2.1]).
Lemma 21. Let 0 < p ≤ ∞ and ϕ : [1, ∞) → R be a twice differentiable, positive and increasing function such that limr→∞ ϕ(r) = ∞. If ϕ satisfies (3.4), then for each entire function f , the following conditions are equivalent,
INTEGRAL OPERATORS ON WEIGHTED FOCK SPACES
27
(i) Mp (r, f ) = O(ϕ(r)), r → ∞. (ii) Mp (r, f ′ ) = O(ϕ′ (r)), r → ∞. Proof. Let {rn } the sequence defined by ϕ(rn ) = en . (i) ⇒ (ii). By Lagrange’s theorem for each n ∈ N (6.3)
ϕ(rn+1 ) − ϕ(rn ) = ϕ′ (tn )(rn+1 − rn ),
tn ∈ (rn , rn+1 )
so by Lemma 13 and Lemma 16 Cp Cp ϕ′ (tn ) Mp (rn+1 , f ) Mp (rn+1 , f ) = rn+1 − rn ϕ(rn+1 ) − ϕ(rn ) Cp ϕ′ (tn )ϕ(rn+1 ) ≤ Cϕ′ (tn ) ≤ Cϕ′ (rn ). ≤ ϕ(rn+1 ) − ϕ(rn )
Mp (rn , f ′ ) ≤
Now, for r ∈ [r0 , ∞) choose n such that r ∈ [rn , rn+1 ). Then
Mp (r, f ′ ) ≤ Mp (rn+1 , f ′ ) ≤ Cϕ′ (rn+1 ) ≤ Cϕ′ (r),
where the last inequality follows from Lemma 16. (ii) ⇒ (i). We assume that 0 < p ≤ 1 (the proof for p > 1 is analogous). Bearing in mind Lemma 14, (6.3) and Lemma 16, we get Mpp (rj+1 , f ) − Mpp (rj , f ) ≤ Cp (rj+1 − rj )p Mpp (rj+1 , f ′ ) ≤ C(rj+1 − rj )p ϕ′ (rj+1 )p ≤ C(rj+1 − rj )p ϕ′ (tj )p ≤ Cejp,
so a summation gives Mpp (rn+1 , f )
≤C
n X j=0
ejp + Mpp (r0 , f ) ≤ Cenp = Cϕ(rn )p
and the proof follows.
Lemma 22. If φ ∈ I satisfies (1.7), then ∂ Kz = O kKz k φ φ′ (|z|) , ∂ z¯ φ F2 F 2
|z| → ∞.
Proof. Let {en } be the orthonormal basis of F2φ given by en (z) = z n δn−1 ,
where δn2 = 2π
R∞ 0
∞ X
n=0
n ∈ N,
r 2n+1 e−2φ(r) dr. By Corollary 8 we have that
r 2n δn−2 =
∞ X
n=0
|en (z)|2 = ||Kz ||2F φ ≍ 2
So, if we consider the entire function defined by f (z) =
∞ X
n=0
z n δn−1 ,
e2φ(r) , τ 2 (r)
|z| = r.
´ ANGEL ´ ´ OLIVIA CONSTANTIN AND JOSE PELAEZ
28
then M2 (r, f ) =
1 2π
1 Rπ f (reiθ ) 2 dθ 2 ≍ Φ(r), as r → ∞, where −π Φ(r) =
eφ(r) . τ (r)
By Lemma 18 Φ′ (r) ≍ Φ(r)φ′ (r),
as r → ∞.
Moreover, in view of (1.7) and Lemma 18, a calculation shows that lim sup r→∞
Φ′′ (r)Φ(r) < ∞. (Φ′ (r))2
Thus by Lemma 21 M2 (r, f ′ ) = O Φ′ (r) .
Finally, since for r = |z|, 2 ∞ ∞ X X ∂ ′ 2 Kz = n2 r 2n−2 δn−2 = M22 (r, f ′ ) |en (z)| = ∂ z¯ φ F n=1 n=0 2
we obtain ∂ Kz ∂ z¯
F2φ
= M2 (r, f ′ ) = O Φ′ (r) ≍ Φ(r)φ′ (r) ≍ kKz kF φ φ′ (r), 2
r → ∞.
Proposition 23. Let g ∈ H(C), 1 < p < ∞ and φ ∈ I satisfying (1.7). If φ g′ p 1+φ′ ∈ L (C, ∆φ dm), then Tg ∈ Sp (F2 ). Proof. By Theorem 19, the inner product Z f ′ (z)g′ (z) (1 + φ′ (z))−2 e−2φ(z) dm(z) hf, gi∗ = f (0)g(0) + C
F2φ
gives a norm on equivalent to the usual one. If 1 < p < ∞, the operator Tg belongs to the Schatten p-class Sp if and only if X |hTg en , en i∗ |p < ∞ n
for any orthonormal basis {en } (see [28, Theorem 1.27]). Let {en } be an orthonormal set of F2φ , h, i∗ . Next, applying Theorem 19 for p = 1 and φ˘ = 2φ we deduce Z ˘ 2 2 |en (z)e′n (z)|(1 + φ˘′ (z))−1 e−φ(|z|) dm(z) 1 = ken kF φ =ken k φ˘ & F 2 1 C Z |en (z)e′n (z)|(1 + φ′ (z))−1 e−2φ(|z|) dm(z) ≍ C
INTEGRAL OPERATORS ON WEIGHTED FOCK SPACES
29
This together with H¨older’s inequality yields X n
p
|hTg en , en i∗ | ≤ .
X Z n
XZ n
=
Z
C
′
C
|g
′
(z)en (z)e′n (z)|(1
−2 −2φ(|z|)
+ φ (z))
e
p dm(z)
|g′ (z)|p |en (z)e′n (z)|(1 + φ′ (z))−(p+1) e−2φ(|z|) dm(z) p
|g (z)|
C
′
X n
!
|en (z)e′n (z)|
(1 + φ′ (z))−(p+1) e−2φ(|z|) dm(z),
and since kKz k2 φ e−2φ(|z|) ≍ ∆φ(z) as |z| → ∞ (see Corollary 8), the result will F2
be proved if we are able to show that X
(6.4)
n
|en (z) e′n (z)| . kKz k2F φ (1 + φ′ (z)). 2
To prove (6.4), we use the Cauchy-Schwarz inequality to obtain
X n
|en (z) e′n (z)|
≤
X n
2
|en (z)|
!1/2
∂ = ||Kz ||F φ Kz 2 ∂ z¯
X n
F2φ
|e′n (z)|2
!1/2
.
Now, the inequality (6.4) follows from Lemma 22. This completes the proof of the Proposition. Now we turn to prove the necessity for 0 < p < ∞. Proposition 24. Let g ∈ H(C), 0 < p < ∞ and φ ∈ I. If Tg ∈ Sp (F2φ ), then g′ p 1+φ′ ∈ L (C, ∆φ dm). Proof. We split the proof in two cases. Case 2 ≤ p < ∞. Suppose that Tg is in Sp , and let {ek } be an orthonormal basis in F2φ and R > 100. Let {zk } be the sequence from Lemma 6 for t(z) = δτ (z), δ ∈ (0, mτ ), and consider the operator A taking ek (z) to fzk (z) = Fzk ,R (z)/τ (zk ). It follows from Proposition 9 that the operator A is bounded on F2φ . Then Tg A belongs to Sp (see [28, p.27]), and by [28, Theorem 1.33] X k
kTg (fzk )kp φ = F2
X k
kTg Aek kp φ < ∞. F2
´ ANGEL ´ ´ OLIVIA CONSTANTIN AND JOSE PELAEZ
30
This together with Proposition A and Theorem 19 gives !p/2 Z X 1 ′ 2 ′ −2 |g (z)| (1 + φ (z)) dm(z) τ (zk )p D(τ (zk )) k
≍ .
X Z
2
D(τ (zk ))
k
X k
′
2
′
−2 −2φ(|z|)
|fzk (z)| |g (z)| (1 + φ (z))
e
!p/2
dm(z)
kTg (fzk )kp φ < ∞. F2
On the other hand, if δ is sufficiently small, applying Lemma 7, Lemma 5, Lemma 20 and Lemma 6, it follows that Z |g′ (z)|p (1 + φ′ (z))−p ∆φ(z) dm(z) C
.
k
.
D(δτ (zk ))
X
1 τ (zk )p
X
1 τ (zk )p
k
.
1 τ 2 (z)
XZ
k
Z
D(δτ (zk ))
Z
Z
′
D(δτ (z))
(1 + φ′ (z))−p
|g (ζ)| dm(ζ)
Z
′
D(δτ (z))
D(3δτ (zk ))
!p/2
2
2
′
−2
|g (ζ)| (1 + φ (ζ))
!p/2
dm(ζ)
!p/2
|g′ (ζ)|2 (1 + φ′ (ζ))−2 dm(ζ)
dm(z) τ 2 (z) dm(z) τ 2 (z)
.
This together with the previous inequality concludes the proof. Case 0 < p < 2. If Tg ∈ Sp then the positive operator Tg∗ Tg belongs to Sp/2 . Without loss of generality we may assume that g′ 6= 0. Suppose X Tg∗ Tg f = λn hf, en i∗ en n
Tg∗ Tg .
is the canonical decomposition of Then a standard argument gives that {en } is an orthonormal basis. So relation (6.2) together with Corollary 8 and H¨older’s inequality yields Z |g′ (z)|p (1 + φ′ (z))−p ∆φ(z) dm(z) C Z |g′ (z)|p (1 + φ′ (z))−p kKz k2 e−2φ(|z|) dm(z) ≍ C XZ |g′ (z)|p (1 + φ′ (z))−p |en (z)|2 e−2φ(|z|) dm(z) = n
≤ .
C
X Z n
′
C
2
′
−2
|g (z)| (1 + φ (z))
2 −2φ(|z|)
|en (z)| e
X X p/2 p/2 ∗ λp/2 hTg∗ Tg en , en i∗ = n = kTg Tg kSp/2 . n
n
p/2 dm(z)
INTEGRAL OPERATORS ON WEIGHTED FOCK SPACES
This completes the proof. Finally, we shall prove the main result in this section.
31
Proof of Theorem 4. Part (a) follows directly from Proposition 23 and Proposition 24. Moreover, if 0 < p ≤ 1 and Tg ∈ Sp (F2φ ) ⊂ S1 (F2φ ), then Proposition 24 and Lemma 18 (d) imply that for some r0 ∈ (0, ∞) Z Z |g′ (z)| |g ′ (z)| dm(z) . ∆φ(z) dm(z) ′ r0
2p α−1 2p p+ α−1
. The following conditions are equivalent:
(a) Tg : Fpφ → Fqφ is compact; (b) Tg : Fpφ → Fqφ is bounded;
32
´ ANGEL ´ ´ OLIVIA CONSTANTIN AND JOSE PELAEZ
(c) g is a polynomial with deg(g) < α − 2r , where r =
pq p−q .
α Moreover, Theorem 4 shows that, Tg ∈ Sp if and only if p > α−1 and g is a 1 polynomial with deg(g) < α(1 − p ). The above example illustrates that there does not exist a value of β = q−p pq big enough such that (1.6) implies that g is constant for all φ ∈ I. Example 2: The functions φ(|z|) = eβ|z| , β > 0 belong to I with ∆φ ≍ eβ|z| , so we obtain the following as a byproduct of Theorem 3.
Corollary 26. Let 0 < p, q < ∞, g an entire function, φ(|z|) = eβ|z| , β > 0.
(I) (a) Let 0 < p ≤ q < ∞. Then, • If 1p − 1q > 1, then Tg : Fpφ → Fqφ is bounded if and only if g is constant. • If 1p − 1q = 1, then Tg : Fpφ → Fqφ is bounded if and only if g is a polynomial with deg(g) ≤ 1. • If 1p − 1q < 1, then Tg : Fpφ → Fqφ is bounded if and only if
′
supz∈C |g (z)|e
|z|
1 − 1q −1 p
< ∞.
(b) Let 0 < p ≤ q < ∞. Then, • If p1 − 1q ≥ 1, then Tg : Fpφ → Fqφ is compact if and only if g is constant. • If p1 − 1q < 1, then Tg is compact if the ”little oh” versions of the boundedness conditions in (a) hold. (II) Let 0 < q < p < ∞. The following conditions are equivalent: (a) Tg : Fpφ → Fqφ is compact; (b) Tg : Fpφ → Fqφ is bounded; R r pq (c) C |g′ (z)|e−β|z| dm(z) < ∞, where r = p−q .
Moreover, Theorem 4 shows that Tg ∈ Sp if and only if ′
−β 1− 1p |z|
g (z)e
∈ Lp (C, dm),
p > 1.
The above example shows that for any r > 0 in Theorem 3 (II) (c) g can grow exponentially. |z| |z| Example 3: The function φ(|z|) = ee belongs to I with ∆φ ≍ e2|z|+e , so Theorem 3 provides the following |z|
Corollary 27. Suppose g is an entire function, 0 < p, q < ∞ and φ(|z|) = ee .
(I) (a) Let 0 < p ≤ q < ∞. Then, • If 1p − 1q ≥ 1, then Tg : Fpφ → Fqφ is bounded if and only if g is constant. • If 1p − 1q < 1, then Tg : Fpφ → Fqφ is bounded if and only if ′
|z|+
supz∈C |g (z)|e
(2|z|+e|z| )
1 − q1 −1 p
< ∞.
INTEGRAL OPERATORS ON WEIGHTED FOCK SPACES
33
(b) Let 0 < p ≤ q < ∞. Then Tg is compact if the ”little oh” versions of the boundedness conditions in (a) hold. (II) Let 0 < q < p < ∞. The following conditions are equivalent: (a) Tg : Fpφ → Fqφ is compact; (b) Tg : Fpφ → Fqφ is bounded; R |z| r pq dm(z) < ∞, where r = p−q (c) C |g ′ (z)|e−|z|−e .
Moreover, Theorem 4 shows that Tg ∈ Sp if and only if ′
g (z)e
|z| |z|− 1− p1 e
2 −1 p
∈ Lp (C, dm),
p > 1.
8. Invariant subspaces of the Volterra operator In the particular case g(z) = z, the operator Tg becomes the Volterra operator Z z f (ζ) dζ, z ∈ C, f ∈ Fpφ , p > 0. V f (z) = 0
Recall that a closed subspace M in Fpφ is called invariant for V if V M ⊆ M. In this section we aim to characterize the invariant subspaces of V . For this we begin by showing that polynomials are dense in Fpφ , by two different methods. The first one is based on some smooth polynomials and the second one follows by a dilation argument. 8.1. Density of polynomials. For the first proof, we need some background on certain smooth polynomials defined in terms of Hadamard products. Let T be P the boundary of the unit disc D = {z : |z| < 1}. If W (eiθ ) = k∈J⊂Z bk eikθ is a P trigonometric polynomial and f (eiθ ) = k∈Z ak eikθ ∈ Lp (T), then the Hadamard product X (W ∗ f )(eiθ ) = ak bk eikθ k∈J
is well defined. If Φ : R → C is a C ∞ -function with compact support supp(Φ), we set AΦ,m = max |Φ(s)| + max |Φ(m) (s)|, s∈R
s∈R
and we consider the polynomials WNΦ (eiθ )
=
X k∈Z
Φ
k N
eikθ ,
N ∈ N.
With this notation we can state the next result on smooth partial sums. Theorem B. Let Φ : R → C be a C ∞ -function with compact support supp(Φ). Then the following assertions hold: (i) There exists a constant C > 0 such that Φ iθ 1−m −m (m) |θ| max |Φ (s)| , WN (e ) ≤ C min N max |Φ(s)|, N s∈R
s∈R
´ ANGEL ´ ´ OLIVIA CONSTANTIN AND JOSE PELAEZ
34
for all m ∈ N ∪ {0}, N ∈ N and 0 < |θ| < π. (ii) If 0 < p ≤ 1 and m ∈ N with mp > 1, there exists a constant C(p) > 0 such that p Φ iθ sup (WN ∗ f )(e ) ≤ CAΦ,m M (|f |p )(eiθ ) N
for all f ∈ H p . Here M denotes the Hardy-Littlewood maximal-operator Z θ+h 1 M (|f |)(eiθ ) = sup |f (eit )| dt. 2h 0 0 such that kWNΦ ∗ f kH p ≤ CAΦ,m kf kH p
for all f ∈ H p . Moreover, if Φ(0) = 1, then (iv) limN →∞ kf − WNΦ ∗ f kH p = 0, for any f ∈ H p , 0 < p < ∞. (iv) limN →∞ WNΦ ∗ f (eiθ ) = f (eiθ ) a. e. eiθ ∈ T, for any f ∈ H p and 0 < p < ∞. Theorem B follows from the results and proofs in [22, p. 111-113]. With this tool in our hands, we are going to prove the following. Theorem 28. Assume that φ ∈ I . Then the polynomials are dense in Fpφ . Proof 1. Pick up a function Φ : R → C be a C ∞ -function with Φ(0) = 1 and compact support. For simplicity, we shall write Wn for WnΦ . Take f ∈ Fpφ and fix r > 0. Let us denote fr (eiθ ) = f (reiθ ) and let us consider the sequence of polynomials {un }, where un (reiθ ) = (Wn ∗ fr ) (eiθ ). By (iii) of Theorem B, there is a constant C which does not depend on r neither nor on n such that Mpp (r, un ) ≤ CMpp (r, f ) so for any 0 < R < ∞ (8.1) Z
p −pφ(|z|)
|z|>R
|un (z)| e
dm(z) ≤ C
Z
|z|>R
|f |p e−pφ(|z|) dm(z) ≤ ||f ||p φ < ∞. Fp
So supn ||un ||F φ < ∞, which together to Lemma 7 implies that {un } is uniformly p bounded on each compact subset of C, so by Montel’s theorem {un } is a normal family. Consequently there is a subsequence {unk } that converges to an entire function g on compact subsets of C. By (iv) of Theorem B g = f . Finally, joining this fact to (8.1), and standard argument finishes the proof. Proof 2. For f ∈ Fpφ and r ∈ (0, 1) put fr (z) = f (rz), z ∈ C. Then (8.2)
lim kfr − f kF φ = 0.
r→1
p
INTEGRAL OPERATORS ON WEIGHTED FOCK SPACES
35
This follows by a standard that we sketch for the sake of completeness. R argument p −pφ Let R > 0 be such that |z|>R |f | e dA < ε. Hence Z Z Z p −pφ p p −pφ |fr − f | e dm + 2 |f |p e−pφ dm |fr − f | e dm ≤ |z|≤R |z|>R C Z +2p |fr |p e−pφ dm. |z|>R
It is clear that the first term in the above sum goes to zero as r → 1− . Now using polar coordinates together the fact that the integral means Mp (ρ, f ) are increasing in ρ we deduce that Z Z |f |p e−pφ dm < ε, |fr |p e−pφ dm ≤ |z|>R
|z|>R
and now (8.2) follows. We now show that fr can be approximated by its Taylor polynomials in the Fpφ −norm. Suppose first that p ≥ 1. If an is the n−th coefficient in the Taylor expansion of f , then by the Cauchy formula and H¨older’s inequality we get Z 2π p np |f (reit )|p dt, r ≥ 0. |an | r ≤ cp 0
Multiplying both sides of the above inequality by re−pφ(r) and integrating on [0, ∞) we obtain |an |kz n kF φ ≤ cp kf kF φ . p
Using this we obtain
∞
X
n n a r z
n
Fpφ
n=N
≤
∞ X
n=N
p
|an |kz n kF φ r n ≤ cp kf kF φ p
p
∞ X
rn,
n=N
which shows that fr can be approximated by its Taylor polynomials in the Fpφ −norm. Assume now 0 < p < 1. We have (see [12, Theorem 6.4]) Z 2π np p 1−p r |an | ≤ Cp n |f (reit )|p dt, 0
and proceeding as in the case p ≥ 1 we deduce
|an |kz n kF φ ≤ Cp n1/p−1 kf kF φ . p
p
Thus k
∞ X
n=N
an r n z n kp φ ≤ Fp
∞ X
n=N
|an |p kz n kp φ r np ≤ Cp kf kF φ Fp
p
∞ X
r np n1−p ,
n=N
and the conclusion follows, since the last expression above is the tail of a convergent series.
´ ANGEL ´ ´ OLIVIA CONSTANTIN AND JOSE PELAEZ
36
8.2. Invariant subspaces. Notice that, with respect to the standard orthonormal basis en (z) = z n /kz n kF φ , n ≥ 0, the operator V : F2φ → F2φ is a weighted 2 shift, i.e. V en = ωn en+1 with weight sequence (8.3)
ωn =
kz n+1 kF φ 2
(n + 1)kz n kF φ 2
,
n ≥ 0.
A simple argument using integration by parts and Lemma 18 shows that, for φ ∈ I, we have ωn → 0 as n → ∞. Theorem 29. Assume p > 0 and let φ ∈ I If the sequence (ωn )n≥1 given by (8.3) is eventually decreasing to zero, then a closed subspace M ⊂ Fpφ is a proper invariant subspace of V : Fpφ → Fpφ if and only if there exists a positive integer N such that Fpφ
M = {f ∈ Fpφ : f (k)(0) = 0 for 0 ≤ k ≤ N − 1} = Span{z k : k ≥ N }
.
Proof. Clearly, the sets ApN := {f ∈ Fpφ : f (k) (0) = 0 for 0 ≤ k ≤ N − 1} are invariant subspaces for V . Let us now prove that these are all the invariant subspaces of V . The case p = 2 follows directly from a result of Yakubovich on weighted shifts (see [27] Theorems 3-4). We are now going to show that the result for p 6= 2 can be obtained via the case p = 2 using the boundedness of Tg on Fock spaces Fpφ (see Theorem 3 above). In order to do this, let us first prove the following Claim: For each p > 0 there exist positive integers M1 = M1 (p), M2 = M2 (p) such that V M1 : F2φ → Fpφ and V M2 : Fpφ → F2φ are bounded. Let us first notice that by Theorem 3 and by (a), (b) of Lemma 18 we have V : Fpφ′ → Fqφ is bounded if F (q) ≤ p′ ≤ q,
(8.4)
V : Fpφ′ → Fqφ is bounded if F (p′ ) < q < p′ ,
(8.5) where F (s) =
2s s+2
for s > 0. From this it is easy to see that V : Fpφ → F2φ and
V : F2φ → Fpφ are bounded if p ≥ 2, and hence we can take M1 = M2 = 1 in this case. Let us now study the case 0 < p < 2. Consider sequence (xn ) given by x0 = 2, xn = F (xn−1 ), n ≥ 1 and notice that (xn ) decreases to zero as n → ∞. We aim to show that (8.6) (8.7)
V M : Fpφ → F2φ is bounded if
V M : F2φ → Fpφ is bounded if
xM −1 ≥ p ≥ xM ,
xM −1 ≥ p > xM , M ≥ 1.
Both assertions follow by induction. We only prove (8.7), since (8.6) is shown in a similar way. For M = 1 relation (8.7) follows directly from (8.4-8.5). Now assume that V M : F2φ → Fpφ is bounded if xM −1 ≥ p > xM . By (8.5) we deduce that
INTEGRAL OPERATORS ON WEIGHTED FOCK SPACES
Fqφ
FxφM +ε
→ V : we now obtain
37
is bounded if xM + ε > q > F (xM + ε). For 0 < ε < xM −1 − xM
V M +1 : F2φ → Fqφ is bounded if xM + ε > q > F (xM + ε). Let ε → 0 above to get
V M +1 : F2φ → Fqφ is bounded if xM ≥ q > F (xM ) = xM +1 ,
and with this the claim is proven. Now let M be an invariant subspace for V : Fpφ → Fpφ . Then let M1 , M2 ∈ N be such that V M1 : F2φ → Fpφ and V M2 : Fpφ → F2φ are bounded. Then V M2 M ⊂ F2φ F2φ
and V M2 M
is an invariant subspace for V on F2φ , so it is of the form F2φ
V M2 M
= A2N .
Now let f ∈ A2N . Then there exists a sequence (fn ) ⊂ M with V M2 fn → f in F2φ . But V M1 : F2φ → Fpφ is bounded, hence V M1 +M2 fn → V M1 f in Fpφ , which gives V M1 f ∈ M, since V M1 +M2 fn ∈ M. Put f = z N to deduce z N +M1 ∈ M and by applying V indefinitely to z N +M1 we get ApN +M1 ⊆ M.
If M 6= ApN +M1 , let N1 be the smallest nonnegative integer such that there exists f ∈ M with f (N1 ) (0) 6= 0 (clearly 0 ≤ N1 < N + M1 ). But, for this particular f , we have V N +M1 −N1 −1 f ∈ M, which in view of the above inclusion implies z N +M1 −1 ∈ M, and therefore ApN +M1 −1 ⊆ M. We repeat this procedure until we obtain ApN1 ⊆ M, and then the choice of N1 forces M = ApN1 , so that the proof is done. Corollary 30. For φ(z) = |z|α with α > 2, the proper invariant subspaces of V on Fpφ are precisely the spaces Fpφ
{f ∈ Fpφ : f (k) (0) = 0 for 0 ≤ k ≤ N − 1} = Span{z k : k ≥ N } Proof. By Theorem 29 it suffices to show that the sequence ωn =
kz n+1 kF φ 2
(n + 1)kz n kF φ 2
,
n≥0
is eventually decreasing to zero. By Stirling’s formula we deduce that ωn2 = 2−2/α
Γ( α2 (n + 2)) ≍ n2/α−2 , (n + 1)2 Γ( α2 (n + 1))
.
´ ANGEL ´ ´ OLIVIA CONSTANTIN AND JOSE PELAEZ
38
and hence ωn → 0 as n → ∞. From the proof of Proposition 7 in [7] it follows that the function Γ(x + 2/α) f (x) = x2 Γ(x) is eventually decreasing for α > 1. Thus {ωn }∞ n=0 is eventually decreasing to zero. 9. Remarks on the Bergman space case In this section, we would like to illustrate that some of the methods we employed in the context of Fock spaces provide additional insight into the above mentioned results from [19, 21]. Given a positive radial weight w, the distortion function is defined as follows (see [19, 21]) Z 1 1 ψw (r) = w(u) du, 0 ≤ r < 1. w(r) r Analogues of Theorem 3 and Corollary 11 for the setting of Bergman spaces with rapidly decreasing weights were obtained in [19, 21]. In particular, the following holds. Theorem C. Suppose that w is a radial differentiable weight, and there is L > 0 such that Z w′ (r) 1 (9.1) w(x) dx ≤ L, sup 2 0 r0 . Suppose e−φ(r) =0 r→1 φ′ (r) φ′′ (r) lim inf ′ 2 > −1 r→1 φ (r) φ′′ (r) lim sup ′ 2 < ∞. r→1 φ (r) lim
Then (9.1) holds and there exists r1 ∈ [0, 1) such that (9.2)
ψw (r) ≍
1
φ′ (r)
for r ∈ [r1 , 1)
INTEGRAL OPERATORS ON WEIGHTED FOCK SPACES
39 φ′′ (r) φ′ (r)2
≥ α on Proof. By hypotheses there is α > −1 and r2 ≥ r0 such that [r2 , 1). So, an integration by parts on (r2 , r] ⊂ (r2 , 1), gives R r −φ(s) Rr R r ′′ (s) −φ(s) ′ (s)e−φ(s) −φ(r) −φ(r ) ds = r2 −φ −φ ds ds = −eφ′ (r) + eφ′ (r22) − r2 φφ′ (s) ′ (s) 2e r2 e R −φ(r) −φ(r ) r ≤ −eφ′ (r) + eφ′ (r22) − α r2 e−φ(s) ds, that is
Rr
−φ(s) ds ≤ r2 e
1 α+1
−e−φ(r) φ′ (r)
+
e−φ(r2 ) φ′ (r2 )
R1 so taking limits as r → 1− , we deduce that 0 e−φ(s) ds < ∞. Next, arguing as in the proof Lemma 17, (9.2) follows. These calculations also give (9.1). This finishes the proof. It is worth noticing that there are weights ω satisfying (9.1) but such that (9.2) α does not hold. The weight ω(r) = e−(1−r) ≍ 1, α > 0, gives a concrete example. Lemma 32. Assume φ : [0, ∞) → R+ is a twice continuously differentiable function such that ∆φ > 0 , (∆φ(z))−1/2 ≍ τ (z), where τ (z) is a radial positive function that decreases to zero as |z| → 1− and limr→1− τ ′ (r) = 0. Then (a) (b)
lim (1 − r)φ′ (r) = ∞.
r→1−
lim τ (r)φ′ (r) = ∞, or, equivalently, lim
r→1−
(c) ψω (r) ≍
r→1−
φ′′ (r) = 0. (φ′ (r))2
1
, for r ∈ [0, 1). +1 (d) There exists r0 ∈ [0, 1) such that for all a ∈ D with 1 > |a| > r0 , and any φ′ (r)
δ > 0 small enough we have
φ′ (a) ≍ φ′ (z),
z ∈ D(a, δτ (a)).
Proof. Since by L’Hospital’s rule 1−r −1 lim = lim ′ = +∞ r→1− τ (r) r→1− τ (r) bearing in mind (3.16) and using again L’Hospital’s rule, Rr s∆φ(s) ds ′ lim r(1 − r)φ (r) = lim 0 − − (1 − r)−1 r→1 r→1 Rr s r(1 − r)2 0 τ 2 (s) ds = +∞, ≥ C lim ≥ C lim −1 τ 2 (r) r→1− r→1− (1 − r)
which gives (a). Arguing as in the proof of Lemma 18 we obtain limr→1− τ (r)φ′ (r) = ∞. By ′′ (r) (a) and relation (3.16) this last fact is equivalent to limr→1− (φφ′ (r)) 2 = 0. Taking into account (a) − (b) it is clear that the hypotheses in Lemma 31 are satisfied, and hence ψω (r) ≍ φ′1(r) for r ≥ [r0 , 1). Since φ′ ≥ 0 and limr→1− φ′ (r) = ∞, we obtain (c).
40
´ ANGEL ´ ´ OLIVIA CONSTANTIN AND JOSE PELAEZ
Part (d) can be proved following the steps in the proof of Lemma 20. As a byproduct of the previous lemmas, we obtain the following improvements: • In view of (9.2) and [19, Theorem B] the more transparent LittlewoodPaley formula Z |f ′ (z)|p (9.3) ω(z)dA(z) ||f ||pAp ≍ |f (0)|p + ′ p ω D (1 + φ (z)) can be obtained for Bergman spaces with weights ω = e−φ belonging to the class I considered in [19]. • Part (d) of Lemma 32 shows that the hypothesis (6) on the distortion function in [19, Theorem 2] can be omitted. Going further, this observation allows us to extend the description of the boundedness and compactness of Tg to a wider class of weights. Especially, we admit for a considerably faster decay, including triple exponential weights of the form ω(z) = exp(−ee
1 1−|z|
).
References [1] S. Agmon, Sur une probl`eme de translations, C. R. Acad. Sci. Paris 229 (11) (1949) 540-542. [2] A. Aleman, A class of integral operators on spaces of analytic functions, Topics in complex analysis and operator theory, 3–30, Univ. Malaga, Malaga, 2007. [3] A. Aleman, J. A. Cima, An integral operator on H p and Hardy’s inequality, J. Anal. Math. 85 (2001), 157–176. [4] A. Aleman and B. Korenblum, Volterra invariant subspaces of H p , Bull. Sci. Math. 132 (2008), 510–528. [5] A. Aleman, A. G. Siskakis, Integration operators on Bergman spaces, Indiana Univ. Math. J. 46 (1997), 337–356. [6] A. Aleman, A. G. Siskakis, An integral operator on H p , Complex Variables Theory Appl. 28 (1995), 149–158. [7] A. Atzmon and B. Brive, Surjectivity and invariant subspaces of differential operators on weighted Bergman spaces of entire functions, Contemp. Math. 404 (2006), 27–38. [8] A. Borichev, R. Dhuez and K. Kellay Sampling and interpolation in large Bergman and Fock spaces, J. Funct. Anal. 242 (2007), 563–606. [9] B. Carswell, B. MacCluer, A, Schuster, Composition operators on the Fock space, Acta Sci. Math. (Szeged) 69 (2003), 871–887. [10] H. R. Cho and K. H. Zhu, Fock Sobolev spaces and their Carleson measures, J. Funct. Anal. 263 (2012), 2483–2506. [11] O. Constantin, A Volterra-type integration operator on Fock spaces, Proc. Amer. Math. Soc. 140 (2012), 4247–4257. [12] P. Duren, Theory of H p Spaces, Academic Press, New York-London 1970. [13] I. M. Gelfand, A problem, Uspehi Matem. Nauk 5 (1938), 233. [14] D. H. Luecking, Embedding theorems for spaces of analytic functions via Khinchine’s inequality, Michigan Math. J. 40 (1993), 333–358. [15] N. Marco, M. Massaneda and J. Ortega-Cerd` a, Interpolating and sampling sequences for entire functions, Geom. Funct. Anal. 13, (2003), 862–914. [16] M. Mateljevi´c and M. Pavlovi´c, Multipliers of H p and BMOA, Pac. J. Math. 146 n. 1 (1990), 71–84.
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