Fredholmness of multipliers on Hardy–Sobolev spaces

J. Math. Anal. Appl. 418 (2014) 1–10

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Fredholmness of multipliers on Hardy–Sobolev spaces Guangfu Cao 1 , Li He ∗ Department of Mathematics, Guangzhou University, Guangzhou, China

a r t i c l e

i n f o

Article history: Received 13 June 2013 Available online 18 March 2014 Submitted by Richard M. Aron Keywords: Hardy–Sobolev space Multipliers Fredholm operator

a b s t r a c t This paper discusses the Fredholmness of multipliers on Hardy–Sobolev Spaces and obtains an index formula for the multipliers with some special symbols. Our results show that Hardy–Sobolev spaces have richer properties than classical holomorphic function spaces, and the behavior of the operators on these spaces is complex. Some methods of Hardy or Bergman spaces fail in the case of the Hardy–Sobolev space. © 2014 Elsevier Inc. All rights reserved.

1. Introduction Let D be the unit disk in C, and T its boundary. Denote by dθ the Lebesgue measure on T. For β ∈ R and 0 < p < +∞, the Hardy–Sobolev space Hβp (D) consists of analytic functions f in D such that Rβ f ∈ H p , ∞ ∞ where f (z) = k=0 ak z k is the Taylor expansion of f and Rβ f = k=0 (1 + k)β ak z k . For p = ∞, we define the space Hβ∞ as    Hβ∞ = f  Rβ f ∈ H ∞ . The Hardy–Sobolev space is a general analytic function space which contains many classical function spaces. For example, it is easy to check that H 21 = D, the Dirichlet space; H02 = H 2 , the Hardy space; 2

2 2 H− 1 = La , the Bergman space. This means that the Hardy–Sobolev space has a complicated structure. In 2 recent years, a series of papers and books discussed these spaces and the integral operators on these spaces (for instance, [1,3–8]). In this paper, we study the Fredholmness of the multipliers with polynomial symbols on these spaces and obtain an index formula for these multipliers.

* Corresponding author. 1

E-mail addresses: [email protected] (G. Cao), [email protected] (L. He). Guangfu Cao is supported by NSF of China (grant No. 11271092) and NSF of Guangdong province (grant No. S2011010005367).

http://dx.doi.org/10.1016/j.jmaa.2014.03.024 0022-247X/© 2014 Elsevier Inc. All rights reserved.

G. Cao, L. He / J. Math. Anal. Appl. 418 (2014) 1–10

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2. Fredholmness of multipliers For ϕ ∈ Hβ2 (D), define Tϕβ as the multiplier with symbol ϕ on Hβ2 (D), that is Tϕβ f = ϕf for any f ∈ Hβ2 (D). Set    Mβ = ϕ ∈ Hβ2 (D)  Tϕβ is bounded on Hβ2 (D) . In [2], we show that an Hβ∞ -function may induce an unbounded multiplier on Hβ2 (D), but we don’t know whether Mβ is contained in Hβ∞ (D). For convenience, we simply denote Hβp (D) as Hβp for all 0 < p  ∞, and  · β is the norm in Hβ2 . β Lemma 2.1. Assume that λ ∈ D, β ∈ R. Then, Tz−λ is lower bounded on Hβ2 .

Proof. For any f (z) =

∞ k=0

ak z k ∈ Hβ2 , we have β Tz−λ f (z) =

∞ 

(ak−1 − λak )z k

k=0

where a−1 = 0, thus ∞ 2   β  = T f |ak−1 − λak |2 (1 + k)2β . z−λ β k=0

Note limk→∞ ak = 0, we claim that lim inf k→∞ | subsequence {akj } such that

ak −1 | ajk | j

ak−1 ak |

 1. Otherwise, we may find an  ∈ (0, 1) and a

< , i.e. |akj −1 | < |akj | which implies limj→∞ akj = 0, and this makes

a contradiction. Hence, there are an  > 0 and a K0 ∈ N such that | that ∞ 

|ak−1 − λak |2 (1 + k)2β =

k=0

∞  k=0

C

ak−1 ak

− λ|   for k  K0 . This means

 2  ak−1  |ak |2  − λ (1 + k)2β ak

∞ 

|ak |2 (1 + k)2β = Cf 2β

k=0 a

0 −2 where C = min{| aa−1 − λ|2 , | aa01 − λ|2 , . . . , | aK − λ|2 , 2 } is a positive constant, and the desired result holds 0 K0 −1 consequently. 2

Lemma 2.2. Suppose p(z) is a polynomial on D, and p(z) has no zero point on T. Then, Tpβ is lower bounded on Hβ2 . n Proof. Assume p(z) = a i=1 (z−λi ) is the decomposition of the polynomial p on D, where a is the coefficient β is invertible of the highest order term of p(z) and λi ∈ D or λi ∈ C\D. Clearly, if λi ∈ C\D, then Tz−λ i β 2 on Hβ . Without loss of generality, assume λi ∈ D, then we know that each Tz−λi is lower bounded on Hβ2 n β by Lemma 2.1. Therefore, Tpβ = a i=1 Tz−λ is also lower bounded on Hβ2 . 2 i Theorem 2.3. Suppose p(z) is a polynomial on D, and R(p)(T) is the essential range of p(z) on T. If β is a Fredholm operator. λ∈ / R(p)(T), then Tp−λ

G. Cao, L. He / J. Math. Anal. Appl. 418 (2014) 1–10

3

β Proof. If λ ∈ / R(p)(D), then Tp−λ is invertible by Theorem 3.3 in [2]. Without loss of generality, assume β is lower λ∈ / R(p)(T) but λ ∈ R(p)(D), then p − λ has no zero points on T. Lemma 2.2 gives that Tp−λ β bounded on Hβ2 , so to complete the proof we still need to show that ker(Tp−λ )∗ has finite dimension. If n p(z) − λ = a i=1 (z − λi ) is the decomposition of the polynomial p − λ, then



β Tp−λ

∗

=a

n  β ∗ Tz−λi . i=1

β Hence, we just need to prove that each ker(Tz−λ )∗ has finite dimension. We may assume λi ∈ D, then i β β ∗ β (Tz−λi ) Kλi = 0, where Kλi (w) is the reproducing kernel of Hβ2 satisfying

−β Kλβi (w) = R−β λ i Rw

1 , (1 − w, λi )2

β β for more details, see [2]. Therefore, dim[ker(Tz−λ )∗ ]  1. Moreover, we can show that dim[ker(Tz−λ )∗ ] = 1. i i ∞ β )∗ f = 0, then for any n ∈ N, we have In fact, for any f (z) = k=0 ak z k ∈ Hβ2 , if (Tz−λ i

 β ∗  Tz−λi f, z n β = 0. That is 

β Tz−λ i

∗

f, z n

 β

   β

 β = f, Tz−λ z n β = Rβ f, Rβ Tz−λ zn i i  ∞   β k β n+1 β n = ak (1 + k) z , (2 + n) z − λi (1 + n) z = 0. k=0

By a direct computation, we have  an+1 = λi an

1+n 2+n

2β ,

β β β and this implies that dim[ker(Tz−λ )∗ ] = 1 and Ind(Tz−λ ) = −1, where Ind(Tz−λ ) is the Fredholm index i i i n β β β β ∗ is a of Tz−λi . Thus (Tz−λi ) is a Fredholm operator for all 1  i  n. Consequently, Tp−λ = a i=1 Tz−λ i Fredholm operator. 2 β If λ ∈ C such that Tp−λ is a Fredholm operator, then λ ∈ / R(p)(T), thus we have R(p)(T) = σe (Tpβ ). β β In the proof of Theorem 2.3, the Fredholmness of each Tz−λ and ker[T n i

i=1 (z−λi )

 β dim ker T n

i=1 (z−λi )

∗ 

 β = −Ind T n



i=1 (z−λi )

=−

n 

β

Ind Tz−λ i

i=0

=−

n  

 β ∗  − dim ker Tz−λ = n. i

i=0 β Since {Kλβi }ni=1 is independent and Kλβi ∈ ker(Tz−λ )∗ , we get that i

β β ∗ = ker T ker Tp−λ n

i=1 (z−λi )

∗

=

n  

Kλβi

i=1

where

n

β i=1 {Kλi }

is the linear span of {Kλβi }ni=1 .

] = {0} imply



G. Cao, L. He / J. Math. Anal. Appl. 418 (2014) 1–10

4

β ∗ Lemma 2.4. Assume that λ ∈ D, n ∈ N. Then, dim[ker(T(z−λ) n ) ] = n. β Proof. Since ker(T(z−λ) n ) = {0} and

 β

  β

∗   β  β

dim ker T(z−λ) − dim ker T(z−λ) = Ind T(z−λ) = nInd Tz−λ = −n, n n n β ∗ we have dim[ker(T(z−λ 2 n ) ] = n. i)

Denote by wind(p, λ) the winding number of p around the point λ. Since wind(p(z)|T , λ) = wind((p(z) − λ)|T , 0) = n, we obtain Theorem 2.5. Suppose p(z) is a polynomial on D. If λ ∈ / σe (Tpβ ), then β

Ind Tp−λ = −wind(p − λ). Write  Hβ∞+

=

f∈

Hβ∞

 ∞      k β 1 ak z , ak (1 + k) ∈ l ,  f (z) = k=0

∞ ∞ and define Rβ : Hβ2 → H 2 as Rβ f (z) = k=0 ak (1 + k)β z k for f (z) = k=0 ak z k ∈ Hβ2 . Then, R−β is the inverse of Rβ . For ϕ ∈ L∞ (T), we denote by Tϕ the classical Toeplitz operator with symbol ϕ on H 2 . We also write K(H 2 ) for the space of compact operators on H 2 . Theorem 2.6. Assume that β ∈ R, ϕ ∈ Mβ ∩Hβ∞+ . Then, [Rβ Tϕβ R−β , Tz ] ∈ K(H 2 ). Furthermore, Rβ Tϕβ R−β is an essential Toeplitz operator on H 2 , that is, there are a ϕ˜ ∈ L∞ and a K ∈ K(H 2 ) such that Rβ Tϕβ R−β = Tϕ˜ + K. Proof. Assume ϕ(z) =

∞

ak z k is the Taylor series of ϕ. Then, a direct calculation gives that

k=0

 ∞  β β −β  n  ak 1 + R Tϕ R , Tz z = k=0

k 1+n

β 

1 1+ 1+n

−β 

1 1+ 1+n+k

β

− 1 z n+k+1 .

Now suppose {fn } is a sequence in H 2 which is weakly convergent to zero, that is fn (z) = as n → ∞. Then  ∞ ∞    β β −β  (n) bi ak 1 + R Tϕ R , Tz fn (z) = i=0

=



k=0

∞ 

l 

l=0

k=0

 (n) bl−k ak

k 1+i

β 

k 1+ 1+l−k

1 1+ 1+i

β 

−β 

1 1+ 1+i+k

1 1+ 1+l−k

−β 

β

∞

(n) i w i=0 bi z →

− 1 z i+k+1

1 1+ 1+l

β

 − 1 z l+1 .

Hence,  l  β  −β  β

2 ∞   β β −β  2  k 1 1   (n)  R Tϕ R , Tz fn  = 1+ 1+ bl−k ak 1 + −1     1+l−k 1+l−k 1+l l=0 k=0  l   2 −β  β ∞   (n)    1 1 β   1+  bl−k |ak |(1 + k)  1 + − 1 . 1+l−k 1+l l=0

k=0

0

G. Cao, L. He / J. Math. Anal. Appl. 418 (2014) 1–10

5

For any  > 0, there is a K0 ∈ N such that   −β  β   1 1  1+ − 1 <  1+  1+l−k 1+l for all l − k  K0 or for all k  l − K0 . Thus ∞   β β −β    R Tϕ R , Tz fn 2  l=0

l−K   −β  β 0  (n)    1 1 b |ak |(1 + k)β  1 +  1 + − 1 l−k   1+l−k 1+l k=0

 2 −β  β  (n)    1 1 β b |ak |(1 + k)  1 + − 1 1+ l−k  1+l−k 1+l

l 

+

k=l−K0 +1



∞ 





l−K 0

l=0

 (n)  b |ak |(1 + k)β l−k

k=0

 2 −β  β  (n)    1 1 β b |ak |(1 + k)  1 + 1+ − 1 . l−k  1+l−k 1+l

l 

+

k=l−K0 +1

∞ (n) i w 2 Since fn (z) = i=0 bi z → 0 in H as n → ∞, then for k > l − K0 , there is an N0 ∈ N such that (n)  |bl−k |  K 3/2 for all n  N0 . Thus, there exists a positive constant C such that 0

  −β  β  (n)    1 1 b |ak |(1 + k)β  1 + 1+ − 1 l−k  1+l−k 1+l

l 

k=l−K0 +1

C

=



l 



K0 3/2 k=l−K0 +1 l 

C K0 3/2

K0

C = K0

|ak |(1 + k)β

k=l−K0 +1



C

|ak |(1 + k)β

 12

l 

|ak | (1 + k) 2

3/2

·

2β

 K0

k=l−K0 +1



 12

l 

|ak |2 (1 + k)2β

.

k=l−K0 +1

This implies that  β β −β    R Tϕ R , Tz fn 2 

∞ 



l=K0 +1



l−K 0 k=0

 (n)  b |ak |(1 + k)β + C l−k K0



l 

 12 2 |ak |2 (1 + k)2β

k=l−K0 +1

 l  2 −β  β K0      (n)  1 1 β   + − 1 . bl−k |ak |(1 + k)  1 + 1+ 1+l−k 1+l l=0

It is not difficult to see that

k=0

G. Cao, L. He / J. Math. Anal. Appl. 418 (2014) 1–10

6

 l−K 2  l 2 ∞ 0  (n)    (n)  β β b |ak |(1 + k) b |ak |(1 + k)  l−k l−k

∞  l=K0 +1

where Rβ (ϕ) ˆ =

∞ k=0

k=0

l=0

k=0

2  β  2   = Rβ (ϕ) ˆ f ˆ ∞ · fn 22 n 2  R (ϕ)

∞ (n) i |ak |(1 + k)β z k ∈ H ∞ and f n (z) = i=0 |bi |z . Also,

∞ 

l 

∞  



   β Sl R (ϕ) 2 − Sl−K Rβ (ϕ) 2 0 2 2

|ak |2 (1 + k)2β =

l=K0 +1 k=l−K0 +1

l=K0 +1

 2 = K0 Rβ (ϕ)2 −

 K −1  0   β

2 Si R (ϕ)  < ∞ 2 i=0

where l

 Sl Rβ (ϕ) = ak (1 + k)β z k . k=0

Moreover, there exists a positive constant C such that  l  2 −β  β K0    (n)    1 1 b |ak |(1 + k)β  1 + 1+ − 1 l−k  1+l−k 1+l l=0

k=0

 l 2 K0    (n)  b |ak |(1 + k)β → 0 C l−k l=0

k=0

(n)

as n → ∞, since bl−k → 0 as n → ∞. In other words, for arbitrary  > 0, there exists an N1 ∈ N such that  l  2 −β  β K0      (n)  1 1 β b |ak |(1 + k)  1 + 1+ − 1 <  l−k  1+l−k 1+l l=0

k=0

for all n  N1 . Consequently, there is a positive constant M such that   0 −1  β β −β  β 2  β 2 K  β  2

2  R Tϕ R , Tz fn   M  R (ϕ) Si R (ϕ)  + 1 ˆ ∞ · fn 2 + K0 R (ϕ)2 − 2 2 i=0

for all n  max{N0 , N1 }. This means that [Rβ Tϕβ R−β , Tz ]fn  → 0 as n → ∞, i.e. [Rβ Tϕβ R−β , Tz ] ∈ K(H 2 ). 2 Write  A+ β

=

 ∞    n ∞ ϕ  ϕ(z) = an z ∈ Hβ , an  0, n = 0, 1, 2, . . . , n=0

then we get the following property about A+ β. + Proposition 2.7. Suppose β > β  , then A+ β ⊆ Aβ  .

G. Cao, L. He / J. Math. Anal. Appl. 418 (2014) 1–10

7

∞ ∞ n β β n Proof. Assume ϕ(z) = ∈ A+ ∈ A. This means that n=0 an z n=0 (1 + n) an z β , so that R (ϕ) = ∞ β  (1 + n) a < +∞. Since β > β , we see that n n=0 ∞ 



(1 + n)β an =

n=0 

∞ 

(1 + n)β



−β

(1 + n)β an 

n=0

Furthermore, Rβ (ϕ) =

∞

n=0 (1

∞ 

(1 + n)β an < +∞.

n=0



+ n)β an z n ∈ A, that is ϕ ∈ A+ β .

2

+ β ∞ Theorem 2.8. Suppose β  0. Then A+ β ⊆ Mβ , and Tϕ   ϕHβ for ϕ ∈ Aβ .

∞ ∞ ∞ n 2 n Proof. Assume ϕ(z) = n=0 an z n ∈ A+ n=0 bn z ∈ Hβ , and write F (z) = n=0 |bn |z . Then, β , f (z) = F ∈ Hβ2 and Rβ (F )2 = Rβ (f )2 . Since R (ϕf ) = β

∞ 

 β

(1 + n)

n=0

n 

 ak bn−k z n

k=0

and Rβ (ϕF ) =

∞ 

 (1 + n)β

n=0

n 

 ak |bn−k | z n ,

k=0

it is obvious that  2 ∞ n   2   β   2β R (ϕf ) = (1 + n) a b   k n−k 2   n=0 k=0 2  n ∞   2β  (1 + n) ak |bn−k | n=0

 2 = Rβ (ϕF )2 .

k=0

Note that  n 2 ∞    β  β β R (ϕ)Rβ (F )2 = (1 + k) (1 + n − k) ak |bn−k | 2 n=0

k=0

and 1 + n  (1 + k)(1 + n − k) for all 0  k  n, hence,  β      R (ϕf )2  Rβ (ϕF )2  Rβ (ϕ)Rβ (F )2 2 2 2  β 2  β 2  β 2  2       R (ϕ) ∞ · R (F ) 2 = R (ϕ)∞ · Rβ (f )2 for all β  0. This shows that  β  β  Tϕ   R (ϕ) This completes the proof. 2

∞

= ϕHβ∞ .

G. Cao, L. He / J. Math. Anal. Appl. 418 (2014) 1–10

8

Moreover, we can find an example such that the equality Tϕβ  = ϕHβ∞ fails. β ∞ Example. There are a β ∈ R and a ϕ ∈ A+ β such that Tϕ  < ϕHβ .   ∞ ∞ 2 3 n 2 Let ϕ(z) = z + z , f (z) = n=0 bn z ∈ Hβ , and write F (z) = n=0 |bn |z n . Then, F ∈ Hβ2 . Since Rβ (ϕf )2  Rβ (ϕF )2 , we see that

 β T ϕ  =

  ∞    β   β n bn z , bn  0 . sup R (ϕf )2 = sup R (ϕF )2 : F Hβ2 = 1, F (z) =

f H 2 =1

n=0

β

Without loss of generality, assume bn  0 for f (z) = Rβ (ϕf ) =

∞ 

∞

n=0 bn z

n

with f Hβ2 = 1, then

(4 + n)β (bn+1 + bn )z n+3 + 3β b0 z 2 ,

n=0

which implies that ∞    β R (ϕf )2 = (4 + n)2β (bn+1 + bn )2 + 32β b20 2 n=0

and  β 

R (ϕ)2 = 3β + 4β 2 = 32β + 42β + 2 · 12β . ∞ A direct computation shows that 2β 2β ∞  ∞     β  4+n 4+n 2β 2 R (ϕf )2  2 (2 + n) b + 2 (1 + n)2β b2n + 32β b20 . n+1 2 2 + n 1 + n n=0 n=0 Note

4+n 1+n

and

4+n 2+n

are decreasing with

4+n 2+n

 2 for all n  0 and

4+n 1+n

 2 for all n  2, hence,

2β 2β ∞  ∞     β  4+n 4+n 2β 2 R (ϕf )2  2 (2 + n) b + 2 (1 + n)2β b2n + 32β b20 n+1 2 2 + n 1 + n n=0 n=0 =2

2β ∞   4+n 2+n

n=0

 2 · 22β

∞ 

(2 + n)2β b2n+1 + 2 · 42β b20 + 2 · 52β b21 + 2

n=2

(2 + n)2β b2n+1 + 2 · 42β b20 + 2 · 52β b21 + 2 · 22β

n=0

= 22 · 22β

∞ 

2β ∞   4+n

∞ 

1+n

(1 + n)2β b2n + 32β b20

n=2

(1 + n)2β b2n − 22 · 22β b20 − 2 · 24β b21 + 2 · 42β b20 + 2 · 52β b21 + 32β b20

n=0

= 2 · 2 f 2H 2 − 22 · 22β b20 − 2 · 24β b21 + 2 · 42β b20 + 2 · 52β b21 + 32β b20 β



2(β+1) =2 + 2 · 42β + 32β − 22 · 4β b20 + 2 · 52β − 2 · 42β b21 . 2

(1 + n)2β b2n + 32β b20

2β

Set β = 12 , then  β  R (ϕf )2  8 + 3b20 + 2b21  13 2

G. Cao, L. He / J. Math. Anal. Appl. 418 (2014) 1–10

9

and √   β R (ϕ)2 = 4 + 3 + 2 · 12 12 = 7 + 4 3. ∞ It is clear that  β 2 Tϕ  =

√  2  2 sup Rβ (ϕf )2  13 < 7 + 4 3 = Rβ (ϕ)∞

f H 2 =1 β

and this shows Tϕβ  < ϕHβ∞ . Theorem 2.9. Suppose ϕ ∈ Mβ . Then

0<δ<1

ϕ(D − Dδ ) ⊆ σe (Tϕβ ), where Dδ = {z ∈ D: |z| < δ}.

Proof. Without loss of generality, assume 0 ∈ 0<δ<1 ϕ(D − Dδ ). Then there is a sequence {zk } ⊆ D such that |zk | → 1 and ϕ(zk ) → 0 as k → ∞. Denote by kzβ the normalizing reproducing kernel of Hβ2 . There holds β ∗ β Tϕ kzk = ϕ(zk ) · kzβk , and   β ∗ β    Tϕ kz  = ϕ(zk ) → 0 k w

as k → ∞. However, kzβk → 0 as k → ∞ since |zk | → 1, so we see that Tϕβ is not Fredholm, that is 0 ∈ σe (Tϕβ ). 2 Write H(D) for the set of analytic functions on a neighborhood of D. It is easy to check that H(D) ⊂ Mβ for any β ∈ R1 . If λ ∈ / ϕ(D), then there exists an 0 > 0 such that |ϕ(z) − λ| > 0 > 0 for any z ∈ D. This shows that there exist a δ > 0 and an 1 > 0 such that |ϕ(z) − λ| > 1 > 0 for any z ∈ Dδ = {z | |z| < δ}, 1 1 hence ϕ−λ ∈ H(D). Furthermore, ϕ−λ ∈ Mβ , this means that λ ∈ / σ(Tϕβ ). Thus we have Theorem 2.10. Suppose ϕ ∈ H(D), then ϕ(D) = σ(Tϕβ ) for any β ∈ R1 . Theorem 2.11. Suppose ϕ ∈ H(D), then ϕ(T) = σe (Tϕβ ) for any β ∈ R1 . And if 0 ∈ / σe (Tϕβ ), then IndTϕβ = −wind(ϕ, 0). Proof. By Theorem 2.9, we only need to prove that σe (Tϕβ ) ⊂ ϕ(T). Without loss of generality, assume 0∈ / ϕ(T). We are able to prove 0 ∈ / σe (Tϕβ ). If 0 ∈ / ϕ(D), then Tϕβ is invertible by Theorem 2.10, hence we may assume 0 ∈ ϕ(D) − ϕ(T), thus ϕ(z) = 0 has only finitely many solutions in D. If {zi }ni=1 is the zero set of ϕ(z), then there is a ψ ∈ H(D) such that ϕ(z) =

n 

(z − zi )ψ(z)

(∗)

i=1 β and ψ(z) has no zero point in D. Noting that Tψβ is invertible and Tz−z is Fredholm, we get Tϕ = i β T Tψβ is also Fredholm, that is 0 ∈ / σe (Tϕβ ). Hence, ϕ(T) = σe (Tϕβ ). n i=1 (z−zi ) By (∗), we have that   n  wind(ϕ, 0) = wind m (z − zi ), 0 i=1

G. Cao, L. He / J. Math. Anal. Appl. 418 (2014) 1–10

10

and β IndTϕβ = IndT n

i=1 (z−zi )

.

Thus  IndTϕβ

=

β IndT n

i=1 (z−zi )

= −wind

n 

 (z − zi ), 0

= −wind(ϕ, 0) = −n.

2

i=1

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