Abstract. Weighted spaces of harmonic and holomorphic functions on the unit disc are studied. We show that for all radial weights which are not decreasing too ...
On weighted spaces of harmonic and holomorphic functions Wolfgang Lusky Abstract Weighted spaces of harmonic and holomorphic functions on the unit disc are studied. We show that for all radial weights which are not decreasing too fast the space of harmonic functions is isomorphic to c0 . For the weights that we consider we completely characterize those spaces of holomorphic functions which are isomorphic to c0 . Moreover, we determine when the Riesz projection, mapping the weighted space of harmonic functions onto the corresponding space of holomorphic functions, is bounded.
1
Introduction
This paper is a continuation of [6]. It is concerned with the classification of weighted spaces of harmonic and holomorphic functions on the unit disc. Put D = { z ∈ C : |z| < 1} and let v : [0, 1] → [0, ∞[ be a continuous, non-increasing function with v(1) = 0 and v(r) > 0 for all r 6= 1. For f : D → C put ||f ||v = sup |f (z)|v(|z|) |z| 0 define n
2 X
n+1 2X
2n+1 − k Rn (αk z + βk z¯ ) = (αk z + βk z¯ ) + (αk z k + βk z¯k ) n 2 k≥0 k=0 k=2n +1 X
k
k
k
k
(Convolution with a de-la-Vallee-Poussin kernel). It is well known that the Rn are uniformly bounded with respect to || · ||v , (see Lemma 3.1. below). Moreover put X X R (αk z k + βk z¯k ) = β0 + αk z k k≥0
k≥0
2
R is called the Riesz projection. It will turn out that R is not always bounded. We consider the following condition (⋆)
inf n
v(1 − 2−n−1 ) >0 v(1 − 2−n )
(see also condition (⋆⋆) of Lemma 3.9 below). For two norms || · || and ||| · ||| on a vector space X we write ||x|| ∼ |||x||| if there are constants a, b > 0 such that a||x|| ≤ |||x||| ≤ b||x|| for all x ∈ X. Observe that, for any weight v, by monotonicity, we have v(1 − 2−n−k ) lim sup ≤ 1, v(1 − 2−n ) n→∞
k = 1, 2, . . .
Theorem: Assume (⋆). −n−k ) (i) Let inf k lim supn→∞ v(1−2 = 1. v(1−2−n ) a) Then hv0 ∼ c0 b) Hv0 is not isomorphic to c0 and the Riesz projection R, as an operator on hv0 , is unbounded with respect to k · kv c) Hv0 has a shrinking basis d) There are m1 < m2 < . . . such that, for f ∈ hv0 and rn = 1 − 21−mn+1 , ||f ||v ∼ sup sup |(Rmn+1 − Rmn )f (z)|v(rn ) n
|z|=rn
) (ii) Let inf k lim supn→∞ v(1−2 < 1. v(1−2−n ) a) Then Hv0 ∼ c0 ∼ hv0 b) The Riesz projection R : hv0 → Hv0 is bounded c) For f ∈ hv0 , −n−k
||f ||v ∼ sup n
|(Rn+1 − Rn )f (z)|v(1 − 2−n )
sup |z|=1−2−n
Remark: (i) d) yields ||f ||v ∼ sup sup |f (z)|v(rn ) n
|z|=rn
A similar relation is true in case (ii) c). It follows from the construction in section 4 that for the sequence (mn ) in (i) we can take any subsequence 3
of the indices such that mn+1 is the smallest integer greater than mn with v(1−2−mn+1 ) ≤ 12 . We shall see that then supn (mn+1 − mn ) = ∞. v(1−2−mn ) Examples: v1 (r) = (1 − r)α for some α > 0 v2 (r) = (1 − log(1 − r))β for some β < 0 v3 (r) = logγ (1 − log(r)), r ≥ r0 , for some γ > 0 and some r0 > 0 (see Lemma 3.2. below). In each case (⋆) is fulfilled. v1 and v3 satisfy −n−1 ) lim supn→∞ v(1−2 < 1. Hence in these cases Hv0 ∼ c0 . v2 satisfies v(1−2−n ) inf k lim supn→∞
v(1−2−n−k ) v(1−2−n )
= 1. Here Hv0 is not isomorphic to c0 . More gen-
erally, (⋆) holds if, for some α > 0,
v(r) (1−r)α v(r) (1−r)β
↑ ∞. Hence in this case hv0 ∼ c0 .
) If in addition, for some β > 0, < 1. ↓ 0 then lim supn v(1−2 v(1−2−n ) Hence Hv0 ∼ c0 . This is the case of normal weights ([8],[10]). The first example of a Hv0 -space which is not isomorphic to c0 was given in [4]. (See also [2].) It would be nice to obtain an isomorphic classification of Hv0 in case (i) of the theorem. Theorem (i)(d) might be helpful here. In particular the question arises if there are infinitely many Hv0 −spaces which are non-isomorphic to each other. If f is harmonic on D we consider the trigonometric conjugate −n−1
f˜ = −iRf + i(id − R)f + ifˆ(0) (i.e. f + if˜ is holomorphic). Then 1 1 Rf = (f + if˜) + fˆ(0) 2 2 It is a classical problem to relate the growth condition of f to that of f˜, (see also [10]). Corollary 1: Let v satisfy (⋆). Then the following are equivalent a) If f is harmonic and sup |f (z)| = o( |z|=r
then
1 ) as r → 1 v(r)
1 ) as r → 1 sup |f˜(z)| = o( v(r) |z|=r 4
b) There is a positive integer k such that lim sup n→∞
v(1 − 2−n−k ) 1 such that for each finite dimensional subspace E ⊂ X there is another finite dimensional subspace dim F ) ≤ λ. F ⊃ E with d(F, l∞ In [6] it was shown that Hv0 is a L∞ -space if and only if Hv0 ∼ c0 . Hence we obtain Corollary 2: Assume (⋆). Then Hv0 is a L∞ -space if and only if lim sup n
3
v(1 − 2−n−k ) < 1 for some k v(1 − 2−n )
Some lemmas
In the following we frequently use, for 0 < r ≤ 1, ||f ||r = sup |f (z)| |z|=r
c, c1 , c2 , . . . are always constants independent of the given indices. Sometimes we use the same symbols for different constants if no confusion can arise. 3.1. At first we collect the well known properties of Rn which are needed in the following. Consider the Cesaro means m X X m−k k k (αk z k + βk z¯k ) Sm (αk z + βk z¯ ) = m k≥0 k=0
5
Then we have Rn = 2S2n+1 − S2n . It is well known that ||Sm f ||1 ≤ ||f ||1 for each m and each trigonometric polynomial f (see e.g.[3]). For fixed z we put fz (λ) = f (λz) where |λ| = 1. Then we have (Rn fz )(1) = (Rn f )(z) and (Sm fz )(1) = (Sm f )(z) Hence ||Sm f ||r ≤ ||f ||r and the Rn are also uniformly bounded with respect to all || · ||r . Moreover, for all n and r with 0 < r < 1, we obtain n+1
(R(Rn+1 − Rn )fz )(λ) = λ2
¯ 2n fz (λ)) ¯ 2n+1 fz (λ)) − 1 λ2n S2n (λ S2n+1 (λ 2
([6]). Using the definition of Rn we arrive at the following Lemma: (i) Rm Rn = Rmin(m,n) , if m 6= n, and RRn = Rn R for all n. (ii) There is a constant c > 0 such that, for all f ∈ hv0 and 0 < r < 1, ||Rn f ||r ≤ c||f ||r ,
||R(Rn+1 − Rn )f ||r ≤ c||f ||r and
||(Rn − Rm )f ||r ≤ c||(Rq − Rp )f ||r if p ≤ m ≤ n ≤ q
Proof: Everything follows from the preceding remarks except the third inequality of (ii). From (i) we infer (Rn − Rm )(Rq − Rp ) if p < m ≤ n < q Rn (Rq − Rp ) if p = m < n < q Rn − Rm = (id − Rm )(Rq − Rp ) if p < m < n = q
Hence the third inequality follows from the fact that the Rm are uniformly bounded with respect to || · ||r . Corollary: 0 0 and all r ≥ r0 . Then hv0 ∼ hw0 and Hv0 ∼ Hw0 . Proof: Let u(r) =
v(r0 ), r ≤ r0 v(r), r0 ≤ r ≤ 1
Note, by assumption, v(r0 ) = w(r0 ). By monotonicity we have ||f ||u ≤ ||f ||v ≤
v(0) ||f ||u v(r0 )
||f ||u ≤ ||f ||w ≤
w(0) ||f ||u w(r0 )
and
3.3.Lemma: Let f be a trigonometric polynomial of degree n. Fix 1 0 < r < 1 and 1 − 4n+1 ≤ α ≤ 1. Then 21 ||f ||r ≤ ||f ||αr . Proof: Fix z with |z| = r. Using the maximum principle, we obtain, Z 1 1 n(id − Sn )(f )(tz) dt| ≤ 2n log( )||f ||r |f (z) − f (αz)| = | t α α Hence
1 1 ||f ||r − ||f ||αr ≤ 2n log( )||f ||r ≤ ||f ||r α 2 This yields the lemma. Corollary: 2−5 ||f ||r ≤ ||f ||αr
if 1 −
7
1 ≤ α ≤ 1 and n ≥ 5 n
Proof: Apply the lemma successively 5 times and observe that (1 −
5 1 1 )5 ≤ 1 − ≤ 1 − if n ≥ 5 4n + 1 4n + 5 n
3.4.Lemma: For any sequence of integers 1 ≤ m1 < m2 < . . . we have (mn+1 − mn ) log 2 ≤
mn+1 2X
k=2mn
Moreover supn |
P2mn+1
1 k=2mn k |
1 ≤ (mn+1 − mn + 1) log 2 k
< ∞ if and only if supn (mn+1 − mn ) < ∞.
Proof: The second assertion is a consequence of the first one which follows from Z 2mn+1 1 1 1 dx ≤ mn + mn +1 + . . . + mn+1 (mn+1 − mn ) log 2 = x 2 2 2 2mn and 1 2mn
+
1 2mn +1
+ ... +
1 2mn+1
≤
Z
2mn+1 2−1+mn
dx ≤ (mn+1 − mn + 1) log 2 x
3.5.Lemma: Fix integers 1 ≤ m1 < m2 < . . . and put rn = 1 − 21−mn+1 . If sup(mn+1 − mn ) = ∞ then there are fk ∈ hv0 with sup sup ||(Rmn+1 − Rmn )fk ||rn v(rn ) < ∞ n
while
k
sup sup ||(Rmn+1 − Rmn )Rfk ||rn v(rn ) = ∞ n
k
Proof: Let f (eiθ ) = i(π − θ), 0 ≤ θ < 2π. f is an P L∞ -function on ∂D, 1 k its harmonic extension on D (called f again) is f (z) = ∞ ¯k ). Fix k=1 k (z − z k so large that mk < mk + 1 < mk+1 − 1 < mk+1 . Put (1)
fk = v(rk )−1 (Rmk+1 −1 − Rmk +1 )f 8
Then, by Lemma 3.1, (2) (Rmn+1 − Rmn )fk =
fk 0
if k = n else
In particular fk ∈ hv0 and sup ||(Rmn+1 − Rmn )fk ||rn v(rn ) ≤ ||fk ||rk v(rk ) ≤ 2 sup ||Rj || · ||f ||1 n
j
by the maximum principle. On the other hand, by (1), (Rfk )(rk )v(rk ) ≥
k+1 −2 2mX
j=2mk +2
1 1 (1 − 2−mk+1 +1 )j ≥ j 2
k+1 −2 2mX
j=2mk +2
1 j
Here we used the fact that (Rmk+1 − Rmk )(g) = g if g(z) = z j or g(z) = z¯j where mk +2 ≤ j ≤ mk+1 −2. This is a consequence of Lemma 3.1.(i). Hence using (2), in view of Lemma 3.4., we obtain sup sup ||(Rmn+1 − Rmn )Rfk ||rn v(rn ) = ∞ n
k
3.6. We consider the case when Hv0 is complemented in hv0 . Fix |λ| = 1 and define Tλ : hv0 → hv0 by (Tλ f )(z) = f (λz). Then by [7] Theorem 5.18 there is a bounded projection P : hv0 → Hv0 with P Tλ = Tλ P for all λ ∈ ∂D. In particular, for f (z) = z¯n , n a positive integer, we obtain ¯ n (P f )(z), (P f )(λz) = λ
z∈D
w ¯ n Taking λ = eiθ , z = r > 0, w = reiθ , we see that (P f )(w) = ( |w| ) (P f )(|w|). P f must be analytic, hence P f = 0. We conclude that P is the Riesz projection. This is the argument of [7], p.129. It was also used in [3] and [2]. We have proved
Lemma: If Hv0 is complemented in hv0 then the Riesz projection R : hv0 → Hv0 is bounded. 3.7.Lemma: There is a constant c > 0 such that, for any f ∈ hv0 and integers p, q with 1 ≤ p < q, ||(Rq − Rp )f ||r1 r1 p ≤ c( )2 (q − p) ||(Rq − Rp )f ||r2 r2 9
whenever 0 < r1 ≤ r2 . Proof: By definition there are αk , βk such that (1)
((Rq − Rp )f )(z) =
q+1 2 X
(αk z k + βk z¯k )
k=2p
By Lemma 3.1 there are constants c1 , c2 > 0 such that, for any r, ||R(Rq − Rp )f ||r ≤ c1
q−1 X
||R(Rk+1 − Rk )f ||r ≤ c2 (q − p)||(Rq − Rp )f ||r
k=p
We obtain a constant c > 0 with (2)
max{||R(Rq − Rp )f ||r , ||(id − R)(Rq − Rp )f ||r } ≤ c(q − p)||(Rq − Rp )f ||r
Put g(z) =
2q+1 −2p X
αk+2p z k ,
h(z) =
2q+1 −2p X
βk+2p z¯k ,
k=0
k=0
Then, by (1), p
R(Rq − Rp )f (z) = z 2 g(z) p (id − R)(Rq − Rp )f (z) = z¯2 h(z) Hence using the triangle inequality and (2), ||(Rq − Rp )f ||r1 ||(Rq − Rp )f ||r2 max{||R(Rq − Rp )f ||r1 , ||(id − R)(Rq − Rp )f ||r1 } ≤ 2c(q − p) max{||R(Rq − Rp )f ||r2 , ||(id − R)(Rq − Rp )f ||r2 } r1 p max{||g||r1 , ||h||r1 } = 2c(q − p)( )2 r2 max{||g||r2 , ||h||r2 } r1 2p ≤ 2c(q − p)( ) r2 10
where the last inequality is a consequence of the maximum principle. 3.8.Lemma: Consider 0 < r1 < r2 < . . . and integers 1 ≤ m1 < m2 < . . .. Assume that, for some constant c > 0, for all f ∈ hv0 , ||(Rmn+1 − Rmn )f ||v ≤ c||(Rmn+1 − Rmn )f ||rn v(rn ),
n = 1, 2, . . .
and ||f ||v ∼ sup ||(Rmn+1 − Rmn )f ||rn v(rn ). n
Then hv0 ∼ c0 . Proof: Let Fn be the closure of the trigonometric polynomials on the circle Γn = { z : |z| = rn } with respect to the norm ||f ||n = ||f ||rn v(rn ). Of course, Fn is a C(K)−space. Using a suitable partition of unity we find a finite dimensional subspace En ⊂ Fn with dim (Rmn+1 +1 − Rmn −1 )hv0 |Γn ⊂ En and sup d(En , l∞
En
) 0 with (1)
||(Rn+1 − Rn )f ||v ≤ c||(Rn+1 − Rn )f ||(1−2−n ) v(1 − 2−n ),
n = 1, 2 . . .. Indeed, by definition, the degree of (Rn+1 − Rn )f is less than or equal to 2n+2 . Using Lemma 3.3. and the monotonicity of v we obtain a constant c1 > 0 such that, for each r with 1 − 2−n ≤ r ≤ 1, ||(Rn+1 − Rn )f ||r v(r) ≤ c1 ||(Rn+1 − Rn )f ||(1−2−n ) v(1 − 2−n ) If 1 − 2−m ≤ r ≤ 1 − 2−m−1 ≤ 1 − 2−n then Lemma 3.7. and (⋆⋆), Lemma 3.9, yield constants c2 , c3 with ||(Rn+1 − Rn )f ||r v(r) v(r) r n )2 ||(Rn+1 − Rn )f ||(1−2−n ) v(1 − 2−n ) ≤ c2 ( −n −n 1−2 v(1 − 2 ) −m −m−1 ) 1−2 2n v(1 − 2 ) ||(Rn+1 − Rn )f ||(1−2−n ) v(1 − 2−n ) ≤ c2 ( −n −n 1−2 v(1 − 2 ) ≤ c3 ||(Rn+1 − Rn )f ||(1−2−n ) v(1 − 2−n ) This proves (1) since ||(Rn+1 − Rn )f ||v = sup ||(Rn+1 − Rn )f ||r v(r) 0 0, a constant 0 < ρ < 1 and an index n0 with v(1 − 2−n−k ) ≤ ρv(1 − 2−n ) for all n ≥ n0 . Using Lemma 3.2. we can assume without loss of generality that n0 = 1. For 1 − 2−n+1 ≤ r ≤ 1 − 2−n we obtain, using Lemma 3.3., ||Rn f ||r v(r) ≤
n−1 X
||(Rj+1 − Rj )f ||r v(r)
j=0
≤ c1
n−1 X
||(Rj+1 − Rj )f ||(1−2−j ) v(1 − 2−j )
≤ c1
n−1 X
||(Rj+1 − Rj )f ||(1−2−j ) v(1 − 2−j )ρ[n−1−j]
j=0
v(1 − 2−n+1 ) v(1 − 2−j )
j=0
where [l] is the largest integer of the form mk which is less than or equal to l. Hence (2) implies that ||f ||r v(r) ≤ c sup ||(Rj+1 − Rj )f ||(1−2−j ) v(1 − 2−j ) j
and therefore ||f ||v ≤ c sup ||(Rj+1 − Rj )f ||(1−2−j ) v(1 − 2−j ) j
14
This implies (ii) c). This together with (1) proves hv0 ∼ c0 according to Lemma 3.8. The representation of the norm || · ||v in c) also shows, in view of Lemma 3.1. (ii) and the Corollary, that the Riesz projection R : hv0 → Hv0 is bounded. Hence Hv0 ∼ c0 . Case (i): Use induction to find m1 < m2 < . . . such that mn+1 is the smallest index greater than mn with 1 v(1 − 2−mn+1 ) ≤ −m n ) v(1 − 2 2
(3)
Hence v(1 − 2−mn ) < 2v(1 − 21−mn+1 ) by the choice of mn+1 . If 1 − 2−k+1 ≤ r ≤ 1 − 2−k then we find n with mn ≤ k − 1 < k ≤ mn+1 . By Lemma 3.1. we have ||(Rk − Rk−1 )f ||r ≤ c1 ||(Rmn+1 − Rmn )f ||r and ||Rk f ||r ≤ c1 ||Rmn+1 f ||r We infer from Lemma 3.3., since degree Rmn+1 f ≤ 21+mn+1 , (4)
||(Rk − Rk−1 )f ||r ≤ c||(Rmn+1 − Rmn )f ||(1−21−mn+1 ) and ||Rk f ||r ≤ c||Rmn+1 f ||(1−21−mn+1 )
Put p = mn+1 − 1. Then by (3), for j < n, v(1 − 2−p ) v(1 − 2−mn ) ≤ 2 ≤ 21−n+j v(1 − 21−mj+1 ) v(1 − 2−mj ) Since 1 − 2−mn ≤ r ≤ 1 − 2−mn+1 we obtain v(r) ≤ v(1 − 2−mn ) ≤ 2v(1 − 2−p ) by the choice of mn+1 . With Lemmas 3.1., 3.3. and (4), (5)
||Rk f ||r v(r) ≤ c1 ||Rmn+1 f ||r v(r) ≤ c2 ||Rmn+1 f ||(1−2−p ) v(1 − 2−p ) n X ≤ c2 ||(Rmj+1 − Rmj )f ||(1−2−p ) v(1 − 2−p ) ≤ c3
j=0 n X
||(Rmj+1 − Rmj )f ||(1−21−mj+1 ) v(1 − 21−mj+1 )21−n+j
j=0
≤ c4 sup(||(Rmk+1 − Rmk )f ||(1−21−mk+1 ) v(1 − 21−mk+1 ) k
15
(2) together with (4), (5) implies ||f ||r v(r) ≤ c sup ||(Rmk+1 − Rmk )f ||(1−21−mk+1 ) v(1 − 21−mk+1 ) k
This yields (i)(d). To show a) we use Lemma 3.8. To this end we have to check, with rn = 1 − 21−mn+1 , ||(Rmn+1 − Rmn )f ||v ≤ c||(Rmn+1 − Rmn )f ||rn v(rn )
(6)
We exploit the identities (Rmk+1 − Rmk )(Rmn+1 − Rmn ) =
By (d)
Rmn+1 Rmn
0 2 − Rm n+1 2 − Rm n
if k > n + 1 or k < n if k = n + 1 if k = n − 1
||(Rmn+1 − Rmn )f ||v ≤
(7)
c1 max{||(Rmn+1 − Rmn )2 f ||rn v(rn ), 2 2 ||(Rmn+1 − Rm )f ||rn+1 v(rn+1 ), ||(Rmn − Rm )f ||rn−1 v(rn−1 )}. n n+1 2 )f = (id − Rmn+1 )(Rmn+1 − Rmn )f is less than The degree of (Rmn+1 − Rm n+1 mn+1 +1 or equal to 2 . Hence by Lemma 3.3. and monotonicity of v, 2 2 ||(Rmn+1 − Rm )f ||rn+1 v(rn+1 ) ≤ c2 ||(Rmn+1 − Rm )f ||rn v(rn ) n+1 n+1 ≤ c2 ||(id − Rmn+1 )(Rmn+1 − Rmn )f ||rn v(rn ) ≤ c3 ||(Rmn+1 − Rmn )f ||rn v(rn )
We have 2 (Rmn − Rm )f = (R1+mn − Rmn )Rmn f = Rmn (Rmn+1 − Rmn )f n
and v(1 − 2−mn ) ≤ 2v(rn ). Hence (1) yields 2 ||(Rmn − Rm )f ||rn−1 v(rn−1 ) ≤ c4 ||(R1+mn − Rmn )Rmn f ||(1−2−mn ) v(1 − 2−mn ) n ≤ c5 ||(Rmn+1 − Rmn )f ||rn v(rn )
For the last inequality we used the maximum principle. Thus (7) implies (6) and we obtain a). In view of a), as a consequence of [6], we have c). 16
It remains to prove b). Here we use Lemmas 3.5. and 3.6. If Hv0 were isomorphic to c0 then Hv0 would be complemented in hv0 and hence R would be bounded. According to d) it suffices to show supn (mn+1 −mn ) = ∞. Then, by Lemma 3.5., R is unbounded. To this end fix k arbitrarily large. Since, by assumption, −n−j ) = 1 for all j we find integers p < q such that q − p ≥ k lim supn→∞ v(1−2 v(1−2−n ) and v(1 − 2−q ) 1 v(1 − 2−q+1 ) 1 ≤ while > −p −p v(1 − 2 ) 2 v(1 − 2 ) 2 If, for some n, mn ≤ p < q ≤ mn+1 then mn+1 − mn ≥ k. Otherwise there is n with p ≤ mn < q. Then necessarily q ≤ mn+1 since otherwise we would have 1 v(1 − 2−mn+1 ) v(1 − 2−q+1 ) ≤ ≤ −p −m n ) v(1 − 2 ) v(1 − 2 2 Hence p ≤ mn ≤ q ≤ mn+1 and similarly mn−1 ≤ p. Thus in any case mn+1 − mn−1 ≥ k. Since k was arbitrarily large, supn (mn+1 − mn ) = ∞. Acknowledgement.This research is part of an ‘Accion Integrada HispanoAlemana’. The author is grateful for the support by DAAD.
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7. W.Rudin, Functional Analysis, Mac Graw-Hill, New York, 1973 8. A.L.Shields, D.L.Williams, Bounded projections, duality and multipliers in spaces of analytic functions, Trans. Amer. Math. Soc. 162 (1971), 287-302 9. A.L.Shields, D.L.Williams, Bounded projections, duality and multipliers in spaces of harmonic functions, J. Reine Angew. Math. 299/300 (1978), 259-279 10. A.L.Shields, D.L.Williams, Bounded projections, duality and the growth of harmonic conjugates in the unit disc, Mich. Math. J. 29 (1982), 3-25
Wolfgang Lusky Fachbereich 17 Universit¨at-Gesamthochschule Warburger Straße 100 D-33098 Paderborn Germany
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