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Chapter VIII Banach Spaces of Analytic Functions
Chapter VIII
Banach Spaces of Analytic Functions
First, we recall our notation : D is the open disk {A E C ; 1A1 < l}, D(a,r) is the open disk with center a and radius r , D is the closed disk, C the unit circle { A E C ; 1x1 = 11, and II is the Torus,IR mod. 2 ~ . 1. Harmonic and subharmonic functions.
Let R be an open set in the plane. A function u , defined in R , with values i n E , is said to be subharmonic if : a) -m 5 u ( z ) < +m, for all z E R , b) u is upper semi-continuous on R , that is : if zn + z, u(z) 2 timsup,,, u ( z , ) . Equivalently, this property can be described by the fact that for every a EN,the set {z ; u(z) < a} is open. c) If the closed disk D(a,r) is contained in R ,
d) No integral in (1) 1s * -m.
remarks.
-
1) If u is continuous, a), b), d) are satisfied.
2) Properties a) and b) imply that u is bounded from above on every compact K C R . Indeed, if we set K , = { z E K ; u ( z ) 2 n}, K n is compact, and, by a), one of the Kn's must be empty. So, every integral in (1) is < +m. 3) The meaning of Condition c) is that the value of u at the center of a circle is smaller than the mean of the values on this circle.
A function u is said to be harmonic if the equality holds in (l), for all a , I, such that D ( a ,r) c R .
Lemma 1.1. - If u is subharmonic in R and 4 monotone, increasing, convex onIR then 4 o u is subharmonic. Proof. - Since
r$
is increasing, convex, it is continuous on IR. Therefore, 4 o u
Chapter VIII
164
is upper semi-continuous. Moreover, if
# 0 .(a)
I
#(I
D(a,r) C
+r
--T
u(a
R , we get :
+ rei8)-)2deu
#ou(a+reie)-
de 2R
,
since # is convex and d e / 2 a is a positive measure on [-x,x],of total mass 1.
Proposition 1.2. - Let f be holomorphic in R . Then continuous in n, for 1 < p < 00.
lflP
is upper semi-
Proof. - By virtue of Lemma 1.1, all we have to do is to show the result for p = 1. But :
In the sequel, %f and Sf are respectively the real and imaginary parts of the complex function f .
Proposition 1.3. - If f is holomorphic in Proof.
-
n, Qf and Sf are harmonic.
We write once more the equality :
and take real and imaginary parts.
Proposition 1.4. - Let u be a continuous subharmonic function in R , K a compact contained in n, h a real continuous function on K , harmonic in KO, such that u(z) 5 h ( z ) , if z E d K . Then u(z) 5 h ( z ) , z E K . Proof. - Put u1 = u - h. Then u1 is subharmonic in KO. Let's assume that for some z E KO, u1(z) > 0. Since u1 is continuous on K , it attains its maximum, m, on K , and since u1 I 0 on a K , the set : A = { z E K ; u~(z)= m} is a compact subset of KO. Let zo be a boundary point of A . Since zo E KO, one can find an r > 0 such that the closed disk D(zo,r) is also contained in KO. But some arc of do(zo,r) is contained in AC. Therefore,
and this contradicts the fact that ul is subharmonic in K"
A nalytic Functions
165
L e m m a 1.5. - Let C, be the circle of center 0, radius r , 0 < r < 1 , and f a continuous function on C,, with real values. There exists a function g, harmonic in D ( O , t ) ,with g = on C,.
Proof. - We define the Poisson Kernel : This is the family of functions Pr(d),
oir
and one checks immediately that if t = reie, eit
+z .
P,(d - t) = X(-) z
We recall moreover that the family (Pr), is an appmn'rnate unit in the space L,(lT, d 8 / 2 a ) , when 1 5 p < 00. This means that if f E L p ( l l ,dd/Zm),
f*P,
+
f ,
r+1-,
(3)
in L , ( n , d o / % ) . This property is an easy consequence of the following elementary facts about the Poisson Kernel :
The reader may find a proof of these facts in any book dealing with Harmonic Analysis (see for instance Hoffman (11, Katznelson [l],etc). However, in Proposition 2.2 below, we will prove a stronger fact, namely that the convergence in (3) holds almost everywhere (in short : a.e.). Now, to prove our Lemma, we put, for r'
< r,
z = r'ei',
Chapter VIII
166
Since f is in L 2 , the series Crmcneine is convergent, and, taking r' = r , we get g(reie) = j(rei8), so = g on C,. Moreover, by the properties of the Poisson Kernel,
and g is the real part of an holomorphic function, and so is harmonic in D ( 0 ,r) . Proposition 1.6. - If u is a continuous, subharmonic function in D , and if:
is the mean value of u on
C,, then
m(r) is an increasing function of r , 0
I
r < l .
Proof. - Let rl < r 2 . Let h , given by Lemma 1.5, the continuous function on D ( 0 ,r z ) , equal to u on C,, , and harmonic in D(0,r z ) . Then, by Proposition 1.4, u 5 h in D ( 0 , r z ) . Therefore,
and our Proposition is proved. 2. Basic facts about Fourier Series.
In this paragraph, we recall a few basic facts about Fourier series. Any function f E Ll(ll,d8/2a) (and, a fortiori, in Lp(ll,d8/2x) has a Fourier series :
- Cc,eine, 00
j
where :
The coefficients
c,
tend to 0 when n 4 f o o (Riemann-Lebesgue). We put
167
Analytic Functions
and :
We define the partial sums of the Fourier series :
-n
and the Cesaro aoerages of these partial sums : n-1
The following Theorem can be found for instance in Hoffman [l]:
Theorem 2.1. - a) The function f is in L,(II), 1 < p the sequence (un),,>0 is bounded in L,.
< 00, if and only if
b) The function f is in L l ( I I ) if and only if the sequence verges in L1 . C) The function f is continuous if and only if the sequence verges uniformly.
(Un)n>o
con-
(an)n>o
con-
d) The function f is in Lm(II) if and only if the sequence (Un)n>o converges to f in the topology u(Loo,Ll).This last fact implies that (un),,20 is bounded in L , . For j ( e i e ) E L l ( I I , d 8 / 2 ~ )we , define :
Then P r o p o s i t i o n 2.2. - For almost every 0 , f(reie)
+ j(e")
Proof. - Fix B E II. We have :
Let
,
when r
+
1-
.
Chapter VIII
168 /(reie) - / ( e i e ) = -
Let A =
-2T1[ q e
1 2R
-[P.(o - t ) ( g ( t )- t j ( e i e ) ) ] ~ ~ ~ r
1-1
P:(d
-"t)
- t/(ei8))
dt
g-
- t ) ( g ( t )- t j ( e i e ) ) ] : = ~ ,
+
1
= -(pr(e- lr)(g(n)- n f ( e i e ) ) - Pr(e n ) ( g ( - r ) 2x 1 = -(P.(e - R ) g ( n ) )- Pr(e- R)f(e") ,
+nf(eie)))
2R
and this last quantity tends to 0 when r -+ I - , except if B = T. (We recall that for every 8 # 0 , Pr(e)-+ 0 , when r + 1-.) Let now dt For 6
> 0 , we set
:
B: If
171
=
/
It-8156
dt 2n
P:(B- t ) ( g ( t ) - t / ( e i e ) ) -
> 6 > 0 , a direct computation shows that
.
:
I - . Therefore, BL
and this quantity tends to 0 when r For B t , we write :
-+
+ 0,
But P,!(r) is an odd function, since
P, is even. Therefore,
when r
+
1-
.
169
Anaiytic Functions
when E
7
4
0 , for almost every 8, since
> 0 , if 6 is small enough, we have
and then we let r
+
f E Ll(n,d8/2x).Therefore, for every
:
I - . This proves our Proposition.
The convolution with the Poisson Kernel can be extended to measures : P r o p o s i t i o n 2.3. - Let p be a finite Baire measure on the unit circle, and define :
f(reie) =
iL
P,(8 - t ) d p ( t ) .
Then f is harmonic in the open disk and the measures :
d8
dpr = f(reie) 2x
converge to p in the weak-* topology of measures, when r
+
1-
.
The proof of Harmonicity is as in Lemma 1.5 above, and if g is a continuous function, one checks immediately (using the fact that the Poisson Kernel is an approximate unit) that :
/-:
when r 3.
+
gdpr
-+
I_:
gdp
3
I-.
HP Spaces. For p , 1 5 p 5
00, we denote by H p ( I l ) the space of (equivalence classes) of functions f(eie) in L P ( n , d 8 / 2 x ) ,whose Fourier coefficients c n ( f ) are 0 for n < 0 . In other words. cneine .
-C
For f ( e i e ) E H P , we define :
Since (cnln2O D.
+0
, this series defines an analytic function inside the open disk
Chapter VIII
1 70
The reader will observe that we keep the same notation ( f ) for the function defined on C and its analytic continuation inside the disk. For p , 1 5 p 5
M, we have : HP C Lp(n) c L i ( n ) .
Wenowset,for 1 5 p < o o , O < r < l :
and from Proposition 1.2 and Proposition 1.6 follows that for given p , Mp(f,r) is an increasing function of r , 0 5 r < 1 .
f,
Moreover, if z = rei8,
and we have seen that f
* P,
-+
Finally, we can write, for 1 5 p
f in L,, when r
< 00
-+
1-
. This implies
:
:
This is also true for p = M , with :
by the maximum modulus principle for analytic functions. On H P , we put the norm induced by L,. Clearly, HP is a closed linear subspace of L,, since for each n, the mapping J -+ c,(f) is continuous (ICn(f)l
5 Ilfllp).
We also observe that for p
1 1, H m c HP c H’ c L1.
Formula (2) above allows us to extend f(eie) as an analytic function f ( 2 ) inside the unit disk, Conversely, one can recapture f(eie) from f(z) :
Analytic Functions
Proposition 3.1.
~
171
Let f E HP. For almost every 8, f(eie)
-+
,
when r
-t
1-
.
This is indeed a trivial consequence of Proposition 2.1, stated for functions in L1. This convergence is called radial, since reie moves along a radius, when r -+ I - . In fact, one can show that this convergence holds non-tangentially : in any angular sector centered at 8 , which does not contain the tangent at 8 to the circle (see for instance W. Rudin [l],p. 243). Before continuing our study, we list a few easy facts about H'
, H 2, H m .
Proposition 3.2. - I f f E H' and f ( e i e ) EIR a.e., then f is constant : there is a a EIR such that f ( e i B ) = a ax. Proof.
-
By definition, for n > 0 ,
Therefore, f has all Fourier coefficients equal to 0, except co(f) : this shows that f = co(f) a.e.. One could also say that f(z) is analytic and real-valued, therefore constant.
Corollary 3.3. - If both f and
f
are in H I , f is constant.
Indeed, (f + n / 2 and (f i n / 2 i are both in H1 , and are real-valued. Therefore, they are constant, by the previous Proposition, and their s u m is (1 - i ) f / 2 . ~
Proposition 3.4. - The space H 2 is a Hilbert space. If f , g are in H 2 , and :
Proof. - This is obvious on the definitions.
Chapter VIIl
172
Notation : To avoid ambiguity, we distinguish between : P+ : the set of polynomials N anz" , P : the set of trigonometric polynomials ~ f = a n z-n .
En=*
~
Proposition 3.5. - Let 4 in Lm(II). The space H Z is invariant by the multiplication by 4 (denoted by M 4 ) if and only if 4 is in H" . Proof. - a) Assume that M $ ( H 2 )c HZ.Then 4.1 E H z , so 4 E H Zn L , = H". b) Conversely, if E H - , M+(P+) c H 2 : indeed, if p(z) = C," aktk is a polynomial, dd e-*n'+(e'')p(e'') - = 0 ,
I_:
for n
2n
< 0. But P+ is dense in H Z for its norm, so MbHz c H 2 .
Corollary 3.6. - H" Indeed, if
is a Banach algebra.
4 , $J are in H" ,
so $t+hE H".
As a Corollary to Proposition 3.1 above, we give the following :
Theorem 3.7 (F.and M.Riesz). - If p is a complex mewure on II, and if : einedp(8) = 0 , for n = 1,2,. . . , J --*
then j~ is absolutely continuous with respect to Lebesgue measure : there is a function f in H' such that : P =
dd
fg.
Proof. - For z E D, we define :
-"
Analytic Functions
179
(1)
So f is in H I . But then we have :
and the conclusion now follows if we compare (1) and (2) and apply Proposition 2.3. 4. Jensen's Formula, Jensen's Inequality.
The following Proposition gives a useful example of an harmonic function : Proposition 4.1. - Let fl be a simply connected domain, and assume that f is analytic in fl and does not vanish in this region. Then there is a function F analytic in fl, such that eF = f . As a corollary, log If (.)I is harmonic in fl. Proof. - Fix
LO
in f l and put :
f(c)& ,
F(z) =
where I' is any path from zo to t,contained in fl. Then F is analytic in fl, and F'(z) = f ( z ) , z E f l . Since f does not vanish in fl, f'/f is analytic in fl, and we can apply this to f'/ f instead of f : we get a function F , such that F' = f'/ f , We can add a constant to F , so e F ( r o ) = f ( z 0 ) . Put h = f e - F : then h' = 0, and h(zo) = 1, so h is identically equal to I , and f = e F . Take now g = ERF. Then g is harmonic, I = eu, and log I f 1 = g.
Proposition 4.2. -
s_", log (1- e"(d8/2r
If
= 0.
Proof. - Let n = { z ; Rz < 1). Since fl is simply connected, and 1 - z # 0 in fl, there is a F analytic in fl, such that eF(') = 1- z , and F ( 0 ) = 0. Since R ( l - t ) > O in fl,wehave,for z E f l ,
% F ( z ) = log11 - Z I For small 6 > 0, let
r
,
ISF(Z)I < ~ / .2
be the path :
r(t) =
eit
,
6 5 t 5 2a - 6,
Chapter VIII
1 74
and 7 the circular arc whose center is at 1, and passes from ei6 to e-i6 within the unit disk. Then :
by Cauchy’s Theorem, and since F ( 0 ) = 0. But the last integral is smaller than C6 log 1 / 6 , where C is a constant, since the length of 7 is smaller than s6. Letting 6 -t 0, this gives the result.
Theorem 4.3 (Jensen’s Formula). - Let fl = D ( 0 ,R ) , f be analytic in R , with f ( 0 ) # 0. Let 0 < r < R , and al,..., a be ~ the zeros of f in o ( O , r ) , listed according t o their multiplicity. Then :
Proof. - We order the roots so that al,... ,a, are in * - - l c r ~=I r . P u t
D(O,r), la,+l I
+
=
so g(z) is analytic on some disk D(0,r E ) , e > 0, and has no zero in this disk, Therefore, by Proposition 4 . 1 , log Ig(z)l is harmonic in this disk, which implies :
dB g
log I!7(rei8)1
J_:
= log IS(0)l
ny k.To obtain the formula, we observe finally that :
But g(0) = f(0)
Indeed, each term r2
-
d,r
r(an - 2)
,
has modulus 1, and for n > m, each term
satisfies : by Lemma 4.2.
nsm,
175
Analytic Functions
Corollary 4.4 (Jensen's Inequality). - If f E H 1 and f(0)# 0 ,
Proof. - We cannot take immediately r = 1 in the previous theorem, since f is not analytic in a larger disk D(0,R ) , with R > 1. For any r, 0
Fix
E
< r < 1, by Jensen's Formula :
> 0 . A fortiori,
Now, when
E +
0, the sequence Iog(lf(eie)I
+E)
is decreasing, and thus :
Corollary 4.5. - For every function f in HI,not identically 0, the function log I f ( e i e ) l is integrable.
Proof. - First, we observe that :
and so all we have to show is that :
If f(0) # 0, this follows from Corollary z " g ( z ) , with g ( 0 )
#
4.4. If f(0) = 0 , we write f(z) =
0, and apply the Theorem to g.
176
Chapter VIII
We observe that, as stated here, Jensen’s inequality is a discontinuous estimate, in the following sense : assume that we have a sequence of functions (fn)n>O, with f n ( 0 ) + 0, and f n + f (say : uniformly on l?). Then, for each n, we apply Jensen’s inequality to fn, thus getting estimates which become worse and worse, when n + 00, since log Ifn(0)l+ -00, though a t the limit we apply Jensen’s inequality to g = f / z k . This phenomenon is partly due to the fact that Jensen’s inequality takes into account only the first coefficient of the Taylor expansion of f . A stronger version, taking into account a prescribed number k of coefficients in this Taylor expansion has been obtained by P. Enflo and the author (see the “Complements”, at the end of this Chapter, in which the general problem of continuity of the functional :
will also be studied). Corollary 4.0. - If a function f in H’ is not identically 0, it cannot vanish identically on a set of positive measure.
Indeed, since loglf) is integrable, the measure of the set where it takes the value -00 must be 0. From Jensen’s inequality, we deduce the following one, which can be interpreted as a similar inequality, involving the measure P,(t - 8)d8/25~instead of the measure dd/2x :
Proposition 4.7. - For any function f E H’ , for 0 5 T < 1 , t E
n,
Proof. - Fix zo , zo = r e i * , and apply Jensen’s inequality to the function :
We get :
Let A be the left-hand side of the above equation. Observing that ( eie + z ) / ( 1 t
177
Analytic Funct iom lei')
has modulus 1, we get, after a change of variables :
as we announced. 5 . Factorization of HP functions : Inner and Outer functions.
We recall our terminology : f ( e i e ) is the value of the function on the unit circle, obtained as a radial limit a.e., and f(z) is the value inside the unit disk. An i n n e r function is a function rn in H M ,satisfying Irn(cie)I = 1 a.e., and thus Irn(z)I 5 1 for every z E D , since the numbers M,(rn,r) are increasing with r (see Section 1). For instance, rn(eie) = ein8 ( n 2 0 ) is an inner function. We will see other examples later, with Blaschke products and singular functions. An outer function F is a function in H' such that :
F ( z ) = n ezp
L
+
'('1
eiti t e
dt 9
where n is a complex number of modulus 1, and k ( t ) a real integrable function. The meaning of formula (1) is not clear at this point. The following Proposition clarifies the links between k and F :
Proposition 5.1. - For a function F in H'
,
k ( e ) = IoglF(eie)J ,
ax.
Proof.- For z = reie, we decompose : ,it +
,it
reie
- rei9
= Pr(B- 1 )
+ iQ,(B - t ) ,
where P, , Qr are real : P, is the Poisson Kernel and Qr the conjugate Kernel :
Chapter VHZ
178
Therefore, (1) can be written, for t = reie,
F(z) = So we get :
K
ezp(Pr * k
+ iQr * k).
IF(z)I = ezp(P, * k ) .
(2)
Since k E L1, P, * k + k a.e., when r -+ 1- (see Section 2 ) . Since also F ( t e i e ) + F ( e i e ) (Proposition 3.1), our Proposition is proved.
We observe that by formula ( I ) , F cannot vanish inside the open unit disk.
Proposition 5.2. - Let F be a function in HI,not identically 0. The following are equivalent : a) F is an outer function,
b) If f is a function in H' such that lj(eie)I = IF(eie)I a-e., then : IF(z)I L
If(4l
3
Vz E
D,
- 1) a) implies b). Let F be outer, and f in H' such that If(e")I = IF(eie)I on r[. Then by Proposition 5.1,
Proof.
By formula (2) above, with k replaced by log lf(eie)l,
Now, by Proposition 4.7, the right-hand side is larger than log i f ( r e i e ) l . This proves our claim. 2 ) b) implies a). We consider the outer function :
dt
log IF(e")I-
2n
.
Since F is integrable, G is in H ' , and : IG(eie)I = IF(eie)(,
a.e.
Therefore, ]G(z)I 2 IF(z)I in D ,by b). Using again the fact that G cannot vanish in D , we find that F I G is analytic in D ,has modulus 1 on C , and modulus 2 1 in D : this is possible only if F / G is constant, and this constant must be of modulus 1. This proves that F is an outer function.
Analytic Functions
179
3) a) implies c) is obvious on the definition of an outer function. 4) c) implies a). We define G as above. Then lFl/lGl
C , and lF(O)l/lG(O)I = 1. Therefore, FIG is constant.
5
1 in D, = 1 on
We now turn to a first decomposition of functions in H'
:
Proposition 5.3. - Let f be a function in H I , non identically 0. Then f can be factored into m.F, where m is inner and F an outer function. This factorization is unique, up to a constant of modulus 1. Proof.
-
We put :
This is an outer function. Put now :
one obtains a function analytic in the open disk, and Irn(eit)I = 1, by Proposition 5.1. Therefore, m is an inner function. If f = mlF1 is another decomposition, IF1 = IFlI, so IF(z)I = IF1(z)I in D by Proposition 5.2, and F = nF1, for some n E C , with In1 = 1. 6. Factorization of Inner Functions : Blaschke Droducts. Sinaular Functions.
In this paragraph, we study inner functions and get for them further factorizations.
Theorem 6.1. - Let ( a , ) , , ? ~be a sequence of complex numbers, with 0
<
la01
<
la11
< - - .,
and
c(1
- la,[)
<
00.
n
Then the infinite product :
converges uniformly on each disk IzI 5 t < 1. Each a, is a zero of B ( z ) , with multiplicity equal to the number of times it is repeated in the sequence, and B ( z ) h a s n o o t h e r zero in D. Finally, IB(z)I < 1 in D, and [B(eie)l= 1 a.e.
Proof. - P u t :
Chapter VIII
180
Fix r < 1. For IzI
< r,
which shows that B ( z ) converges uniformly in the disk of radius r , and that B ( z ) is analytic in D(0,r). Furthermore, each a, is a zero of B with correct multiplicity, and B ( z ) # 0 otherwise. That ]B(z)l< 1 if IzI < 1 is obvious, since it is true for partial products. The radial limit B(eie) therefore exists a.e., and IB(eie)I5 1. It remains t o show that JB(eie)I = 1 a.e. But for any function in H I , the monotonicity of Ml(r, f ) (Section 3, formula (4)) gives :
We apply this t o Bk = IB,(e")I = 1 a.e., thus :
E/Bnrand we observe that
1 ~ 6 ( e " ) 1 = IB(eie)l
,
I6k(eie)I
= 1 a.e., so
ax.
But Bn(2) + B ( z ) uniformly on (121 = r}, so Bk -+ 1 uniformly on this circle. This implies that j : " , IB(e*e)ld0/2x2 1, and since IS(eie)I 5 1, we get IB(eie)I = 1 a.e.
A function of the form :
c,(1
with la,[) < 0 0 , is called a Elaschkc product. M is a positive integer, the set of an's may be infinite, finite or empty. ~
Proposition 6.2. - Let rn be an inner function. Then the zeros a, of m inside the open disk D satisfy Cn(l- I Q , ~ ) < 00. Proof. Dividing if necessary by a power of z , we may assume m(0) # 0 . Let la01 5 la11 5 be the enumeration of the zeros of m inside the unit disk. Fix an integer N and choose r , with ( aI 5~ r < 1. By Jensen's Formula, . a -
181
Analytic Functiom which implies :
Letting
t + 1-
N
,we get :
with C = - log Irn(0)I. But for 0 < z < 1 , log&
2 x . So we obtain : N 0
and the series converges.
Theorem 0.3. - Every inner function m can be f ztor ed as rn = E . G , where B is a Blaschke product and G is an inner function with no zero in the open disk. Proof. - We may assume that rn has infinitely many zeros a,, otherwise this is trivial. Dividing by tM if necessary, we may also assume that m(0) # 0. Put, as before,
We know that this product converges, by Proposition 6.2 and Theorem 6.1. Put B, = f l , " b k , g, = rn/B,. For fixed n, E , IEn(t)I > 1 - E , if 121 is sufficiently close to 1. Therefore, 1
sup 1gn(rei8)I Il--E 3 8
and by monotonicity, this holds for all r < 1. So, letting E + 0 , we get :
But g, + G = m / B , uniformly on each circle has no zeros.
It( =
r . So G is in HDO and
Chapter VIIl
182
It can be shown (cf. Hoffman [I], Duren [l])that the function G , inner with no zeros inside the disk, can be written under the form :
where p is a positive measure, singular with respect to Lebesgue measure. Such a function is called a singular inner function The simplest example (obtained with the Dirac measure at 8 = 0) is the function : z+l G ( a ) = e r p -. r-1
7. The Disk Algebra A[D).. The Disk Algebra A(D) is the space of functions which are analytic inside the open disk D and continuous on the closed disk b : therefore, it is C(n) n Hm equipped with the norm :
This is of course a separable Banach space : the polynomials the norm.
F+
are dense for
This Banach space has an interesting property, connected with the extension of functions defined an a compact of 0 measure. This property will be used, in an Operator Theory context, in the next Part. Theorem 7.1 (Fatou). - Let K be a compact in n, with 0 Lebesgue measure. There exists a function # in A (D), such that t
K = (8 ; #(eie) = 0). Remark.
- By virtue of Corollary 4.6, K must be of measure 0 for such a
function to exist.
Proof. - Since K is compact, II\K is the reunion of a countable family (I,,),,?o of open intervals, pairwise disjoint. Let E , = P(In).For every n, let yn be a positive function, of class C' on I n l satisfying yn < l / e , yn = 0 at the end-points of I,, and :
Analytic Functions
183
We put y = 0 on K , y = y, on I,,. The function y has the following properties :
-0
IY Il/e,
- K = ( 8 ; y(8) = 0}, - y is of class C' on II\K and continuous on - logy is integrable.
n,
P u t now w = logy. Then w satisfia : a) w = -00 on K ,and if dist(8,K) 0, w ( 8 ) b) w 5 -1 on n , c) w is continuous on II and of class C' on n\K, d) w is integrable. -+
-+
-00,
We now set :
= (J'r
* w ) ( e )+ i(Qr * w ) ( d )
9
if z = reie, where Pr is the Poisson Kernel and Qr the conjugate Kernel (see Section 2). Since P r 2 0, from b) follows that RRh 5 -1 in 0.Set ur(0) = (Pr * w ) ( 8 ) ; then ur + w , when r + I - , uniformly on I l l since w is continuous. We now show that, since w is C' on n\K, the functions ur(8) = (Qr * w)(O) converge uniformly on every closed interval contained in n\K. This will imply that h(eie) is continuous on n\K. L e m m a 7.2. - If w is of class C' on a closed interval interval, tlr converges uniformly to :
Proof. - Take 80 = 0. The function
is integrable on I = { t ; 0 5 t I rnin(B + a,a - 8 ) ) , and :
- 801
5
a, on this
Chapter VIII
181
And one checks for the Kernel
the following properties : 0 < gr(t)
<
1, and g , ( t )
-+
0, r
-+
I - , except for
t = 0. The Lemma follows by a computation similar to the one made for the
Poisson Kernel (Proposition 3.1). We now come back to the proof of the Theorem. We set g = l / h . Then, since h does not vanish in D, g is analytic in D , continuous in 0 ,by Lemma 7.2, and the zeros of g on C are exactly the points of K. This proves our Theorem. We observe that Wg < 0 on B\K .
Corollary 7.3. - Let K a compact contained in the unit circle, with 0 Lebesgue measure. There is a function 4 in A(D), with d ( z ) = 1 for every E E K ,and I4(z)I < 1 if t E D\K. Proof. - We take 4
=
d ,where g is given by the previous theorem.
Theorem 7.4 (Rudin, Carleson). - Let K a compact contained in the unit circle, with 0 Lebesgue measure, and let f be a continuous function on K , with complex values. There exists a function F in A(D),such that the restriction of F to K is f , and such that :
Proof. Let 4 be the function given by Corollary 7.3. For every h in A (D), we have : ~
Write @"' = h
+ $n,
where $,, is in A(D) and $, = 0 on K . We get :
inf{Ilh
+ $llw
; $ E AID), $J = 0 on K }
5 inf \lh+ $Ilw n
= inf ll$"hll n
= lim ll4"hll n
= sup 1h(t)l. 2E K
But conversely, for every CC, in A(D),with
$J
=0
on K :
Analytic Functions
185
and so : inf{llh
+ $llm
; $ E A (D), rl, = 0 on K} = sup (h(z)l. ZE K
(1)
Call A K the set of restrictions t o K of functions in A(D), equipped with the norm induced by C ( K ) ,and call S the subspace of A(D) consisting of $ : 0 on K . Finally, let R be the restriction operator, from functions ff (D) into C ( K ) . Formula (1) means exactly that for every h in A ( D ) ,
+,
IlhllA(D)/S
=
IIRhllC(K)
*
(2)
This means that R is an isometry from the Banach space A (D)/S, into C ( K ), and therefore has closed range in C ( K ). The restriction operator from A (D) into C(K)has obviously the same range. We now show that this range is dense in C ( K ) : the restriction operator from ff (D)into C(K)will therefore be surjective. Assume not. Since the range is a vector space, it is contained in a closed hyperplane : there exists a measure p on K such that : /R(fJdP = 0
Vf
9
f
A(D).
By Hahn-Banach Theorem, p may be extended to a measure on denoted by p . Therefore : / f l ~ d = p 0 ,
VfEA(D).
In particular taking f = einol n 2 0, we find
I
einO
1Kdp = 0 ,
C(n),also
:
Vn>0.
By the F. and M. Riesz Theorem (Theorem 3.7), this implies that 1 K d p is absolutely continuous with respect to Lebesgue measure. But since P ( K ) = 0, this implies p = 0, and the image of R is dense. So, R , from A ( D ) / S onto C(K)is a surjective isometry. This does not quite finish the proof of our theorem : given f in C ( K ), we want t o find F in with IlFllm = Ilfllco. We may of course assume that llflloo = 1. The fact that R is a surjective isometry implies that there are functions t$ E A ( D ) and $ E S , with : q5 = f on K
,
141 < 1 on a\K,
(1)
Chapter VIII
186
tcI
ll4+tcIllca I 2. Put g = 6 + $. We have 191 5 1 on K , 191 5 2 on C. Let = Oon
0 1
There is an such that : On
01,we
~1
=
> 0 such that
get :
(2 E
I#[
1 -16"1g1 2
1 2
o k
(2)
1
0 1
be the set :
c ; Ig(2)l < 1 + 4). 1
< 1 - 01 on O f , so we can find an
< 1<
-14"'gl
For k = 1 , 2 , . . ., let :
K
{ZE
41 ,
1 1 - ( 1 + -)
<
2
4
c
; Ig(2)I < 1
1 1-- . 4
+ F1 } .
Finally, the function :
mk
have
1 1--. 2k
< Ok
in IN
on 0;.
Then the Ok's are decreasing and contain K. Assume ml < ... < been chosen so that :
Using the fact that there is an a mk such that :
ml
> 0 such that 141 < 1 - &k on O i , we find
m
.
is in A(D) and satisfies : F = f
onK,
and llFllm5 1. This finishes the proof of the Theorem.
187
Analytic Functiohs Exercises on Chapter VIII.
Exercise 1. - Show that the function f ( z ) = En,' zn/n is not in H m Exercise 2.
-
Show that if f is harmonic and z f (z) is harmonic, f is analytic.
Exercise 3. - Let f in H I . Show that there exist 9 , h in f =gh.
H 2 such that
Exercise 4. - Let h be a positive integrable function on n. A necessary and sufficient condition in order that h = If 1 2 , for some function f in H 2 is that J l o g h > -00. Exercise 5. - Let f E L,, 1 < p < 00, and :
Is
g
in HP ?
Exercise 6 (Hardy). - Let f f H'
, f (z) =
anz". Show that :
(Hint : first assume that u, 2 0 for all n , and use the formula :
In the general case, write f = gh, with g, h in H Z (see exercise 3 ) , with :
where B is the Blaschke product of the zeros of f . ) Exercise 7. - Let F be an outer function in H 1. Assume that f is in H', and that f / F is in L1. Show that f / F is in H' . Show that this property characterizes outer functions. Exercise 8. - Let f in a'. Show that a) or b) imply that f is an outer function : a) I / f is in H', b) Xf(z) > 0 , in D . If f is inner, show that 1 f is outer.
+
Chapter
188
vzzz
Exercise 9. - Let Hh be the subspace of H' consisting of functions f with f(0) = 0. Show that H m can be identified t o the dual of & / H J . Show that P+ is dense in H m for o ( H w , L 1 / H ; ) . Exercise 10. - Let f be analytic in the open disk. Prove that f is a Blaschke product if and only if : a)
lf(z)I 5 1 in D ,
Exercise 11 (Maximal ideals of A ( D ) ) .- Let to(.) be the identity function : f o ( z ) = z . Let $ be a character on A (D). Let a = $(fo). Show that for every polynomial p , $(p) = ~ ( c z )Deduce . the same relation for every function f in A (D),Deduce that the maximal ideals are of the form Z, :
z,
= {f E A ( D ) ; f(a)= 0).
Notes and Comments.
Our presentation is a mixture between many authors. For further information the reader should consult one of the excellent books which deal with HP spaces : Hoffman 111, Garnett [l],Duren 111, Rudin 111. The exercises also come from these books, except Exercise 11, which was communicated t o us by Richard Aron.
A nalytic Functio M
I89
Complements on Chapter VIII :
One may study the continuity of the decomposition :
f
=
B*S*F
(1)
which we have found during the previous sections. The function f is in H’ , B is the Blaschke product made with its zeros, S is a singular function and F an outer function in H’ . So the question is : if a sequence (f,),,?o converges to f in H1 , what about the corresponding terms B, , S, F, ? Do they converge to the terms B , S, F respectively ? The answer is uno” in general. Indeed, if :
l+z
S ( z ) = ezp - , 1-2
this is a singular function, as we have already seen. But if we set S,(z) = S{r.z), 0 < r < 1 , we get a sequence of functions in H’ ,which converges to S in H1. These functions are outer (to see this, one checks easily that they give an equality in Jensen’s inequality, so we apply Proposition 5.2). The following theorem was obtained by L. Bonvalot [ l ]:
Theorem 1. - The decomposition (1) is continuous a t every function f which has no non-trivial singular factor S. Since this decomposition is obtained by means of Jensen’s Inequality, its continuity is related to that of the “Jensen’s Functional” :
(the quantity e z p J ( f ) is known as Mahlcr’s measure, and it is used in Number Theory.) The continuity of Jensen’s Functional has been studied by the same author, and the result is similar :
Theorem 2. - Jensen’s Functional is continuous on H1 a t every function f which has no non-trivial singular factor in the decomposition (1). When f is a polynomial of degree n , say f = p = C,“ajz’ , estimates on J(f)have been intensively studied by many authors. For instance, Kurt Mahler [I] proved that : l o g C l a j l 2 J(!) 2 logClajI -nlog2.
Chapter VIIZ
190
(see also Arestov [l],Beller [I], Beller-Newman [l] Kurt Mahler 121 for other aspects of these estimates.)
Recently, P. Enflo and the present author (B. Beauzamy- P. Enflo [ l ] ) studied Jensen's Inequality for polynomials (or H1 functions) satisfying :
We say that a polynomial satisfying ( 2 ) has concentration d at degree k. For such a polynomial, P. Enflo and the author proved that there is a constant C ( d , k ) such that :
Numerical estimates where computed by the present author [13] for C ( d , k ) : it is the largest value (for t > 1) of the function :
Let C ( d ,k ) be the best constant satisfying (3) ; the precise value of C(d, k ) is unknown. However, it was shown in [13] that, for d = 1/2,
C ( 1 / 2 , k ) I-2klog2, and that, asymptotically, when k
--+
(5)
im,
C(d,k) 2 -2k.
(6)
The precise value of C ( d ,k ) has been computed by A. K. Rigler, S. Y.Trimble, R. S. Varga [l]for the class of Hurwitz polynomials : these polynomials have real positive coefficients, and their roots are either real negative, or pairwise conjugate, with negative real part. For this class, the best constant they find (in the case d = 1 / 2 ) is -2klog2, the upper bound in ( 5 ) , and it is obtained for the polynomial ( z 1)2k/22k,The problem is also solved for other values of d and the extremal polynomials are given ; we refer the reader to the above mentioned paper for an exposition of the complete results.
+
Banach Spaces of Analytic Functions
First, we recall our notation : D is the open disk {A E C ; 1A1 < l}, D(a,r) is the open disk with center a and radius r , D is the closed disk, C the unit circle { A E C ; 1x1 = 11, and II is the Torus,IR mod. 2 ~ . 1. Harmonic and subharmonic functions.
Let R be an open set in the plane. A function u , defined in R , with values i n E , is said to be subharmonic if : a) -m 5 u ( z ) < +m, for all z E R , b) u is upper semi-continuous on R , that is : if zn + z, u(z) 2 timsup,,, u ( z , ) . Equivalently, this property can be described by the fact that for every a EN,the set {z ; u(z) < a} is open. c) If the closed disk D(a,r) is contained in R ,
d) No integral in (1) 1s * -m.
remarks.
-
1) If u is continuous, a), b), d) are satisfied.
2) Properties a) and b) imply that u is bounded from above on every compact K C R . Indeed, if we set K , = { z E K ; u ( z ) 2 n}, K n is compact, and, by a), one of the Kn's must be empty. So, every integral in (1) is < +m. 3) The meaning of Condition c) is that the value of u at the center of a circle is smaller than the mean of the values on this circle.
A function u is said to be harmonic if the equality holds in (l), for all a , I, such that D ( a ,r) c R .
Lemma 1.1. - If u is subharmonic in R and 4 monotone, increasing, convex onIR then 4 o u is subharmonic. Proof. - Since
r$
is increasing, convex, it is continuous on IR. Therefore, 4 o u
Chapter VIII
164
is upper semi-continuous. Moreover, if
# 0 .(a)
I
#(I
D(a,r) C
+r
--T
u(a
R , we get :
+ rei8)-)2deu
#ou(a+reie)-
de 2R
,
since # is convex and d e / 2 a is a positive measure on [-x,x],of total mass 1.
Proposition 1.2. - Let f be holomorphic in R . Then continuous in n, for 1 < p < 00.
lflP
is upper semi-
Proof. - By virtue of Lemma 1.1, all we have to do is to show the result for p = 1. But :
In the sequel, %f and Sf are respectively the real and imaginary parts of the complex function f .
Proposition 1.3. - If f is holomorphic in Proof.
-
n, Qf and Sf are harmonic.
We write once more the equality :
and take real and imaginary parts.
Proposition 1.4. - Let u be a continuous subharmonic function in R , K a compact contained in n, h a real continuous function on K , harmonic in KO, such that u(z) 5 h ( z ) , if z E d K . Then u(z) 5 h ( z ) , z E K . Proof. - Put u1 = u - h. Then u1 is subharmonic in KO. Let's assume that for some z E KO, u1(z) > 0. Since u1 is continuous on K , it attains its maximum, m, on K , and since u1 I 0 on a K , the set : A = { z E K ; u~(z)= m} is a compact subset of KO. Let zo be a boundary point of A . Since zo E KO, one can find an r > 0 such that the closed disk D(zo,r) is also contained in KO. But some arc of do(zo,r) is contained in AC. Therefore,
and this contradicts the fact that ul is subharmonic in K"
A nalytic Functions
165
L e m m a 1.5. - Let C, be the circle of center 0, radius r , 0 < r < 1 , and f a continuous function on C,, with real values. There exists a function g, harmonic in D ( O , t ) ,with g = on C,.
Proof. - We define the Poisson Kernel : This is the family of functions Pr(d),
oir
and one checks immediately that if t = reie, eit
+z .
P,(d - t) = X(-) z
We recall moreover that the family (Pr), is an appmn'rnate unit in the space L,(lT, d 8 / 2 a ) , when 1 5 p < 00. This means that if f E L p ( l l ,dd/Zm),
f*P,
+
f ,
r+1-,
(3)
in L , ( n , d o / % ) . This property is an easy consequence of the following elementary facts about the Poisson Kernel :
The reader may find a proof of these facts in any book dealing with Harmonic Analysis (see for instance Hoffman (11, Katznelson [l],etc). However, in Proposition 2.2 below, we will prove a stronger fact, namely that the convergence in (3) holds almost everywhere (in short : a.e.). Now, to prove our Lemma, we put, for r'
< r,
z = r'ei',
Chapter VIII
166
Since f is in L 2 , the series Crmcneine is convergent, and, taking r' = r , we get g(reie) = j(rei8), so = g on C,. Moreover, by the properties of the Poisson Kernel,
and g is the real part of an holomorphic function, and so is harmonic in D ( 0 ,r) . Proposition 1.6. - If u is a continuous, subharmonic function in D , and if:
is the mean value of u on
C,, then
m(r) is an increasing function of r , 0
I
r < l .
Proof. - Let rl < r 2 . Let h , given by Lemma 1.5, the continuous function on D ( 0 ,r z ) , equal to u on C,, , and harmonic in D(0,r z ) . Then, by Proposition 1.4, u 5 h in D ( 0 , r z ) . Therefore,
and our Proposition is proved. 2. Basic facts about Fourier Series.
In this paragraph, we recall a few basic facts about Fourier series. Any function f E Ll(ll,d8/2a) (and, a fortiori, in Lp(ll,d8/2x) has a Fourier series :
- Cc,eine, 00
j
where :
The coefficients
c,
tend to 0 when n 4 f o o (Riemann-Lebesgue). We put
167
Analytic Functions
and :
We define the partial sums of the Fourier series :
-n
and the Cesaro aoerages of these partial sums : n-1
The following Theorem can be found for instance in Hoffman [l]:
Theorem 2.1. - a) The function f is in L,(II), 1 < p the sequence (un),,>0 is bounded in L,.
< 00, if and only if
b) The function f is in L l ( I I ) if and only if the sequence verges in L1 . C) The function f is continuous if and only if the sequence verges uniformly.
(Un)n>o
con-
(an)n>o
con-
d) The function f is in Lm(II) if and only if the sequence (Un)n>o converges to f in the topology u(Loo,Ll).This last fact implies that (un),,20 is bounded in L , . For j ( e i e ) E L l ( I I , d 8 / 2 ~ )we , define :
Then P r o p o s i t i o n 2.2. - For almost every 0 , f(reie)
+ j(e")
Proof. - Fix B E II. We have :
Let
,
when r
+
1-
.
Chapter VIII
168 /(reie) - / ( e i e ) = -
Let A =
-2T1[ q e
1 2R
-[P.(o - t ) ( g ( t )- t j ( e i e ) ) ] ~ ~ ~ r
1-1
P:(d
-"t)
- t/(ei8))
dt
g-
- t ) ( g ( t )- t j ( e i e ) ) ] : = ~ ,
+
1
= -(pr(e- lr)(g(n)- n f ( e i e ) ) - Pr(e n ) ( g ( - r ) 2x 1 = -(P.(e - R ) g ( n ) )- Pr(e- R)f(e") ,
+nf(eie)))
2R
and this last quantity tends to 0 when r -+ I - , except if B = T. (We recall that for every 8 # 0 , Pr(e)-+ 0 , when r + 1-.) Let now dt For 6
> 0 , we set
:
B: If
171
=
/
It-8156
dt 2n
P:(B- t ) ( g ( t ) - t / ( e i e ) ) -
> 6 > 0 , a direct computation shows that
.
:
I - . Therefore, BL
and this quantity tends to 0 when r For B t , we write :
-+
+ 0,
But P,!(r) is an odd function, since
P, is even. Therefore,
when r
+
1-
.
169
Anaiytic Functions
when E
7
4
0 , for almost every 8, since
> 0 , if 6 is small enough, we have
and then we let r
+
f E Ll(n,d8/2x).Therefore, for every
:
I - . This proves our Proposition.
The convolution with the Poisson Kernel can be extended to measures : P r o p o s i t i o n 2.3. - Let p be a finite Baire measure on the unit circle, and define :
f(reie) =
iL
P,(8 - t ) d p ( t ) .
Then f is harmonic in the open disk and the measures :
d8
dpr = f(reie) 2x
converge to p in the weak-* topology of measures, when r
+
1-
.
The proof of Harmonicity is as in Lemma 1.5 above, and if g is a continuous function, one checks immediately (using the fact that the Poisson Kernel is an approximate unit) that :
/-:
when r 3.
+
gdpr
-+
I_:
gdp
3
I-.
HP Spaces. For p , 1 5 p 5
00, we denote by H p ( I l ) the space of (equivalence classes) of functions f(eie) in L P ( n , d 8 / 2 x ) ,whose Fourier coefficients c n ( f ) are 0 for n < 0 . In other words. cneine .
-C
For f ( e i e ) E H P , we define :
Since (cnln2O D.
+0
, this series defines an analytic function inside the open disk
Chapter VIII
1 70
The reader will observe that we keep the same notation ( f ) for the function defined on C and its analytic continuation inside the disk. For p , 1 5 p 5
M, we have : HP C Lp(n) c L i ( n ) .
Wenowset,for 1 5 p < o o , O < r < l :
and from Proposition 1.2 and Proposition 1.6 follows that for given p , Mp(f,r) is an increasing function of r , 0 5 r < 1 .
f,
Moreover, if z = rei8,
and we have seen that f
* P,
-+
Finally, we can write, for 1 5 p
f in L,, when r
< 00
-+
1-
. This implies
:
:
This is also true for p = M , with :
by the maximum modulus principle for analytic functions. On H P , we put the norm induced by L,. Clearly, HP is a closed linear subspace of L,, since for each n, the mapping J -+ c,(f) is continuous (ICn(f)l
5 Ilfllp).
We also observe that for p
1 1, H m c HP c H’ c L1.
Formula (2) above allows us to extend f(eie) as an analytic function f ( 2 ) inside the unit disk, Conversely, one can recapture f(eie) from f(z) :
Analytic Functions
Proposition 3.1.
~
171
Let f E HP. For almost every 8, f(eie)
-+
,
when r
-t
1-
.
This is indeed a trivial consequence of Proposition 2.1, stated for functions in L1. This convergence is called radial, since reie moves along a radius, when r -+ I - . In fact, one can show that this convergence holds non-tangentially : in any angular sector centered at 8 , which does not contain the tangent at 8 to the circle (see for instance W. Rudin [l],p. 243). Before continuing our study, we list a few easy facts about H'
, H 2, H m .
Proposition 3.2. - I f f E H' and f ( e i e ) EIR a.e., then f is constant : there is a a EIR such that f ( e i B ) = a ax. Proof.
-
By definition, for n > 0 ,
Therefore, f has all Fourier coefficients equal to 0, except co(f) : this shows that f = co(f) a.e.. One could also say that f(z) is analytic and real-valued, therefore constant.
Corollary 3.3. - If both f and
f
are in H I , f is constant.
Indeed, (f + n / 2 and (f i n / 2 i are both in H1 , and are real-valued. Therefore, they are constant, by the previous Proposition, and their s u m is (1 - i ) f / 2 . ~
Proposition 3.4. - The space H 2 is a Hilbert space. If f , g are in H 2 , and :
Proof. - This is obvious on the definitions.
Chapter VIIl
172
Notation : To avoid ambiguity, we distinguish between : P+ : the set of polynomials N anz" , P : the set of trigonometric polynomials ~ f = a n z-n .
En=*
~
Proposition 3.5. - Let 4 in Lm(II). The space H Z is invariant by the multiplication by 4 (denoted by M 4 ) if and only if 4 is in H" . Proof. - a) Assume that M $ ( H 2 )c HZ.Then 4.1 E H z , so 4 E H Zn L , = H". b) Conversely, if E H - , M+(P+) c H 2 : indeed, if p(z) = C," aktk is a polynomial, dd e-*n'+(e'')p(e'') - = 0 ,
I_:
for n
2n
< 0. But P+ is dense in H Z for its norm, so MbHz c H 2 .
Corollary 3.6. - H" Indeed, if
is a Banach algebra.
4 , $J are in H" ,
so $t+hE H".
As a Corollary to Proposition 3.1 above, we give the following :
Theorem 3.7 (F.and M.Riesz). - If p is a complex mewure on II, and if : einedp(8) = 0 , for n = 1,2,. . . , J --*
then j~ is absolutely continuous with respect to Lebesgue measure : there is a function f in H' such that : P =
dd
fg.
Proof. - For z E D, we define :
-"
Analytic Functions
179
(1)
So f is in H I . But then we have :
and the conclusion now follows if we compare (1) and (2) and apply Proposition 2.3. 4. Jensen's Formula, Jensen's Inequality.
The following Proposition gives a useful example of an harmonic function : Proposition 4.1. - Let fl be a simply connected domain, and assume that f is analytic in fl and does not vanish in this region. Then there is a function F analytic in fl, such that eF = f . As a corollary, log If (.)I is harmonic in fl. Proof. - Fix
LO
in f l and put :
f(c)& ,
F(z) =
where I' is any path from zo to t,contained in fl. Then F is analytic in fl, and F'(z) = f ( z ) , z E f l . Since f does not vanish in fl, f'/f is analytic in fl, and we can apply this to f'/ f instead of f : we get a function F , such that F' = f'/ f , We can add a constant to F , so e F ( r o ) = f ( z 0 ) . Put h = f e - F : then h' = 0, and h(zo) = 1, so h is identically equal to I , and f = e F . Take now g = ERF. Then g is harmonic, I = eu, and log I f 1 = g.
Proposition 4.2. -
s_", log (1- e"(d8/2r
If
= 0.
Proof. - Let n = { z ; Rz < 1). Since fl is simply connected, and 1 - z # 0 in fl, there is a F analytic in fl, such that eF(') = 1- z , and F ( 0 ) = 0. Since R ( l - t ) > O in fl,wehave,for z E f l ,
% F ( z ) = log11 - Z I For small 6 > 0, let
r
,
ISF(Z)I < ~ / .2
be the path :
r(t) =
eit
,
6 5 t 5 2a - 6,
Chapter VIII
1 74
and 7 the circular arc whose center is at 1, and passes from ei6 to e-i6 within the unit disk. Then :
by Cauchy’s Theorem, and since F ( 0 ) = 0. But the last integral is smaller than C6 log 1 / 6 , where C is a constant, since the length of 7 is smaller than s6. Letting 6 -t 0, this gives the result.
Theorem 4.3 (Jensen’s Formula). - Let fl = D ( 0 ,R ) , f be analytic in R , with f ( 0 ) # 0. Let 0 < r < R , and al,..., a be ~ the zeros of f in o ( O , r ) , listed according t o their multiplicity. Then :
Proof. - We order the roots so that al,... ,a, are in * - - l c r ~=I r . P u t
D(O,r), la,+l I
+
=
so g(z) is analytic on some disk D(0,r E ) , e > 0, and has no zero in this disk, Therefore, by Proposition 4 . 1 , log Ig(z)l is harmonic in this disk, which implies :
dB g
log I!7(rei8)1
J_:
= log IS(0)l
ny k.To obtain the formula, we observe finally that :
But g(0) = f(0)
Indeed, each term r2
-
d,r
r(an - 2)
,
has modulus 1, and for n > m, each term
satisfies : by Lemma 4.2.
nsm,
175
Analytic Functions
Corollary 4.4 (Jensen's Inequality). - If f E H 1 and f(0)# 0 ,
Proof. - We cannot take immediately r = 1 in the previous theorem, since f is not analytic in a larger disk D(0,R ) , with R > 1. For any r, 0
Fix
E
< r < 1, by Jensen's Formula :
> 0 . A fortiori,
Now, when
E +
0, the sequence Iog(lf(eie)I
+E)
is decreasing, and thus :
Corollary 4.5. - For every function f in HI,not identically 0, the function log I f ( e i e ) l is integrable.
Proof. - First, we observe that :
and so all we have to show is that :
If f(0) # 0, this follows from Corollary z " g ( z ) , with g ( 0 )
#
4.4. If f(0) = 0 , we write f(z) =
0, and apply the Theorem to g.
176
Chapter VIII
We observe that, as stated here, Jensen’s inequality is a discontinuous estimate, in the following sense : assume that we have a sequence of functions (fn)n>O, with f n ( 0 ) + 0, and f n + f (say : uniformly on l?). Then, for each n, we apply Jensen’s inequality to fn, thus getting estimates which become worse and worse, when n + 00, since log Ifn(0)l+ -00, though a t the limit we apply Jensen’s inequality to g = f / z k . This phenomenon is partly due to the fact that Jensen’s inequality takes into account only the first coefficient of the Taylor expansion of f . A stronger version, taking into account a prescribed number k of coefficients in this Taylor expansion has been obtained by P. Enflo and the author (see the “Complements”, at the end of this Chapter, in which the general problem of continuity of the functional :
will also be studied). Corollary 4.0. - If a function f in H’ is not identically 0, it cannot vanish identically on a set of positive measure.
Indeed, since loglf) is integrable, the measure of the set where it takes the value -00 must be 0. From Jensen’s inequality, we deduce the following one, which can be interpreted as a similar inequality, involving the measure P,(t - 8)d8/25~instead of the measure dd/2x :
Proposition 4.7. - For any function f E H’ , for 0 5 T < 1 , t E
n,
Proof. - Fix zo , zo = r e i * , and apply Jensen’s inequality to the function :
We get :
Let A be the left-hand side of the above equation. Observing that ( eie + z ) / ( 1 t
177
Analytic Funct iom lei')
has modulus 1, we get, after a change of variables :
as we announced. 5 . Factorization of HP functions : Inner and Outer functions.
We recall our terminology : f ( e i e ) is the value of the function on the unit circle, obtained as a radial limit a.e., and f(z) is the value inside the unit disk. An i n n e r function is a function rn in H M ,satisfying Irn(cie)I = 1 a.e., and thus Irn(z)I 5 1 for every z E D , since the numbers M,(rn,r) are increasing with r (see Section 1). For instance, rn(eie) = ein8 ( n 2 0 ) is an inner function. We will see other examples later, with Blaschke products and singular functions. An outer function F is a function in H' such that :
F ( z ) = n ezp
L
+
'('1
eiti t e
dt 9
where n is a complex number of modulus 1, and k ( t ) a real integrable function. The meaning of formula (1) is not clear at this point. The following Proposition clarifies the links between k and F :
Proposition 5.1. - For a function F in H'
,
k ( e ) = IoglF(eie)J ,
ax.
Proof.- For z = reie, we decompose : ,it +
,it
reie
- rei9
= Pr(B- 1 )
+ iQ,(B - t ) ,
where P, , Qr are real : P, is the Poisson Kernel and Qr the conjugate Kernel :
Chapter VHZ
178
Therefore, (1) can be written, for t = reie,
F(z) = So we get :
K
ezp(Pr * k
+ iQr * k).
IF(z)I = ezp(P, * k ) .
(2)
Since k E L1, P, * k + k a.e., when r -+ 1- (see Section 2 ) . Since also F ( t e i e ) + F ( e i e ) (Proposition 3.1), our Proposition is proved.
We observe that by formula ( I ) , F cannot vanish inside the open unit disk.
Proposition 5.2. - Let F be a function in HI,not identically 0. The following are equivalent : a) F is an outer function,
b) If f is a function in H' such that lj(eie)I = IF(eie)I a-e., then : IF(z)I L
If(4l
3
Vz E
D,
- 1) a) implies b). Let F be outer, and f in H' such that If(e")I = IF(eie)I on r[. Then by Proposition 5.1,
Proof.
By formula (2) above, with k replaced by log lf(eie)l,
Now, by Proposition 4.7, the right-hand side is larger than log i f ( r e i e ) l . This proves our claim. 2 ) b) implies a). We consider the outer function :
dt
log IF(e")I-
2n
.
Since F is integrable, G is in H ' , and : IG(eie)I = IF(eie)(,
a.e.
Therefore, ]G(z)I 2 IF(z)I in D ,by b). Using again the fact that G cannot vanish in D , we find that F I G is analytic in D ,has modulus 1 on C , and modulus 2 1 in D : this is possible only if F / G is constant, and this constant must be of modulus 1. This proves that F is an outer function.
Analytic Functions
179
3) a) implies c) is obvious on the definition of an outer function. 4) c) implies a). We define G as above. Then lFl/lGl
C , and lF(O)l/lG(O)I = 1. Therefore, FIG is constant.
5
1 in D, = 1 on
We now turn to a first decomposition of functions in H'
:
Proposition 5.3. - Let f be a function in H I , non identically 0. Then f can be factored into m.F, where m is inner and F an outer function. This factorization is unique, up to a constant of modulus 1. Proof.
-
We put :
This is an outer function. Put now :
one obtains a function analytic in the open disk, and Irn(eit)I = 1, by Proposition 5.1. Therefore, m is an inner function. If f = mlF1 is another decomposition, IF1 = IFlI, so IF(z)I = IF1(z)I in D by Proposition 5.2, and F = nF1, for some n E C , with In1 = 1. 6. Factorization of Inner Functions : Blaschke Droducts. Sinaular Functions.
In this paragraph, we study inner functions and get for them further factorizations.
Theorem 6.1. - Let ( a , ) , , ? ~be a sequence of complex numbers, with 0
<
la01
<
la11
< - - .,
and
c(1
- la,[)
<
00.
n
Then the infinite product :
converges uniformly on each disk IzI 5 t < 1. Each a, is a zero of B ( z ) , with multiplicity equal to the number of times it is repeated in the sequence, and B ( z ) h a s n o o t h e r zero in D. Finally, IB(z)I < 1 in D, and [B(eie)l= 1 a.e.
Proof. - P u t :
Chapter VIII
180
Fix r < 1. For IzI
< r,
which shows that B ( z ) converges uniformly in the disk of radius r , and that B ( z ) is analytic in D(0,r). Furthermore, each a, is a zero of B with correct multiplicity, and B ( z ) # 0 otherwise. That ]B(z)l< 1 if IzI < 1 is obvious, since it is true for partial products. The radial limit B(eie) therefore exists a.e., and IB(eie)I5 1. It remains t o show that JB(eie)I = 1 a.e. But for any function in H I , the monotonicity of Ml(r, f ) (Section 3, formula (4)) gives :
We apply this t o Bk = IB,(e")I = 1 a.e., thus :
E/Bnrand we observe that
1 ~ 6 ( e " ) 1 = IB(eie)l
,
I6k(eie)I
= 1 a.e., so
ax.
But Bn(2) + B ( z ) uniformly on (121 = r}, so Bk -+ 1 uniformly on this circle. This implies that j : " , IB(e*e)ld0/2x2 1, and since IS(eie)I 5 1, we get IB(eie)I = 1 a.e.
A function of the form :
c,(1
with la,[) < 0 0 , is called a Elaschkc product. M is a positive integer, the set of an's may be infinite, finite or empty. ~
Proposition 6.2. - Let rn be an inner function. Then the zeros a, of m inside the open disk D satisfy Cn(l- I Q , ~ ) < 00. Proof. Dividing if necessary by a power of z , we may assume m(0) # 0 . Let la01 5 la11 5 be the enumeration of the zeros of m inside the unit disk. Fix an integer N and choose r , with ( aI 5~ r < 1. By Jensen's Formula, . a -
181
Analytic Functiom which implies :
Letting
t + 1-
N
,we get :
with C = - log Irn(0)I. But for 0 < z < 1 , log&
2 x . So we obtain : N 0
and the series converges.
Theorem 0.3. - Every inner function m can be f ztor ed as rn = E . G , where B is a Blaschke product and G is an inner function with no zero in the open disk. Proof. - We may assume that rn has infinitely many zeros a,, otherwise this is trivial. Dividing by tM if necessary, we may also assume that m(0) # 0. Put, as before,
We know that this product converges, by Proposition 6.2 and Theorem 6.1. Put B, = f l , " b k , g, = rn/B,. For fixed n, E , IEn(t)I > 1 - E , if 121 is sufficiently close to 1. Therefore, 1
sup 1gn(rei8)I Il--E 3 8
and by monotonicity, this holds for all r < 1. So, letting E + 0 , we get :
But g, + G = m / B , uniformly on each circle has no zeros.
It( =
r . So G is in HDO and
Chapter VIIl
182
It can be shown (cf. Hoffman [I], Duren [l])that the function G , inner with no zeros inside the disk, can be written under the form :
where p is a positive measure, singular with respect to Lebesgue measure. Such a function is called a singular inner function The simplest example (obtained with the Dirac measure at 8 = 0) is the function : z+l G ( a ) = e r p -. r-1
7. The Disk Algebra A[D).. The Disk Algebra A(D) is the space of functions which are analytic inside the open disk D and continuous on the closed disk b : therefore, it is C(n) n Hm equipped with the norm :
This is of course a separable Banach space : the polynomials the norm.
F+
are dense for
This Banach space has an interesting property, connected with the extension of functions defined an a compact of 0 measure. This property will be used, in an Operator Theory context, in the next Part. Theorem 7.1 (Fatou). - Let K be a compact in n, with 0 Lebesgue measure. There exists a function # in A (D), such that t
K = (8 ; #(eie) = 0). Remark.
- By virtue of Corollary 4.6, K must be of measure 0 for such a
function to exist.
Proof. - Since K is compact, II\K is the reunion of a countable family (I,,),,?o of open intervals, pairwise disjoint. Let E , = P(In).For every n, let yn be a positive function, of class C' on I n l satisfying yn < l / e , yn = 0 at the end-points of I,, and :
Analytic Functions
183
We put y = 0 on K , y = y, on I,,. The function y has the following properties :
-0
IY Il/e,
- K = ( 8 ; y(8) = 0}, - y is of class C' on II\K and continuous on - logy is integrable.
n,
P u t now w = logy. Then w satisfia : a) w = -00 on K ,and if dist(8,K) 0, w ( 8 ) b) w 5 -1 on n , c) w is continuous on II and of class C' on n\K, d) w is integrable. -+
-+
-00,
We now set :
= (J'r
* w ) ( e )+ i(Qr * w ) ( d )
9
if z = reie, where Pr is the Poisson Kernel and Qr the conjugate Kernel (see Section 2). Since P r 2 0, from b) follows that RRh 5 -1 in 0.Set ur(0) = (Pr * w ) ( 8 ) ; then ur + w , when r + I - , uniformly on I l l since w is continuous. We now show that, since w is C' on n\K, the functions ur(8) = (Qr * w)(O) converge uniformly on every closed interval contained in n\K. This will imply that h(eie) is continuous on n\K. L e m m a 7.2. - If w is of class C' on a closed interval interval, tlr converges uniformly to :
Proof. - Take 80 = 0. The function
is integrable on I = { t ; 0 5 t I rnin(B + a,a - 8 ) ) , and :
- 801
5
a, on this
Chapter VIII
181
And one checks for the Kernel
the following properties : 0 < gr(t)
<
1, and g , ( t )
-+
0, r
-+
I - , except for
t = 0. The Lemma follows by a computation similar to the one made for the
Poisson Kernel (Proposition 3.1). We now come back to the proof of the Theorem. We set g = l / h . Then, since h does not vanish in D, g is analytic in D , continuous in 0 ,by Lemma 7.2, and the zeros of g on C are exactly the points of K. This proves our Theorem. We observe that Wg < 0 on B\K .
Corollary 7.3. - Let K a compact contained in the unit circle, with 0 Lebesgue measure. There is a function 4 in A(D), with d ( z ) = 1 for every E E K ,and I4(z)I < 1 if t E D\K. Proof. - We take 4
=
d ,where g is given by the previous theorem.
Theorem 7.4 (Rudin, Carleson). - Let K a compact contained in the unit circle, with 0 Lebesgue measure, and let f be a continuous function on K , with complex values. There exists a function F in A(D),such that the restriction of F to K is f , and such that :
Proof. Let 4 be the function given by Corollary 7.3. For every h in A (D), we have : ~
Write @"' = h
+ $n,
where $,, is in A(D) and $, = 0 on K . We get :
inf{Ilh
+ $llw
; $ E AID), $J = 0 on K }
5 inf \lh+ $Ilw n
= inf ll$"hll n
= lim ll4"hll n
= sup 1h(t)l. 2E K
But conversely, for every CC, in A(D),with
$J
=0
on K :
Analytic Functions
185
and so : inf{llh
+ $llm
; $ E A (D), rl, = 0 on K} = sup (h(z)l. ZE K
(1)
Call A K the set of restrictions t o K of functions in A(D), equipped with the norm induced by C ( K ) ,and call S the subspace of A(D) consisting of $ : 0 on K . Finally, let R be the restriction operator, from functions ff (D) into C ( K ) . Formula (1) means exactly that for every h in A ( D ) ,
+,
IlhllA(D)/S
=
IIRhllC(K)
*
(2)
This means that R is an isometry from the Banach space A (D)/S, into C ( K ), and therefore has closed range in C ( K ). The restriction operator from A (D) into C(K)has obviously the same range. We now show that this range is dense in C ( K ) : the restriction operator from ff (D)into C(K)will therefore be surjective. Assume not. Since the range is a vector space, it is contained in a closed hyperplane : there exists a measure p on K such that : /R(fJdP = 0
Vf
9
f
A(D).
By Hahn-Banach Theorem, p may be extended to a measure on denoted by p . Therefore : / f l ~ d = p 0 ,
VfEA(D).
In particular taking f = einol n 2 0, we find
I
einO
1Kdp = 0 ,
C(n),also
:
Vn>0.
By the F. and M. Riesz Theorem (Theorem 3.7), this implies that 1 K d p is absolutely continuous with respect to Lebesgue measure. But since P ( K ) = 0, this implies p = 0, and the image of R is dense. So, R , from A ( D ) / S onto C(K)is a surjective isometry. This does not quite finish the proof of our theorem : given f in C ( K ), we want t o find F in with IlFllm = Ilfllco. We may of course assume that llflloo = 1. The fact that R is a surjective isometry implies that there are functions t$ E A ( D ) and $ E S , with : q5 = f on K
,
141 < 1 on a\K,
(1)
Chapter VIII
186
tcI
ll4+tcIllca I 2. Put g = 6 + $. We have 191 5 1 on K , 191 5 2 on C. Let = Oon
0 1
There is an such that : On
01,we
~1
=
> 0 such that
get :
(2 E
I#[
1 -16"1g1 2
1 2
o k
(2)
1
0 1
be the set :
c ; Ig(2)l < 1 + 4). 1
< 1 - 01 on O f , so we can find an
< 1<
-14"'gl
For k = 1 , 2 , . . ., let :
K
{ZE
41 ,
1 1 - ( 1 + -)
<
2
4
c
; Ig(2)I < 1
1 1-- . 4
+ F1 } .
Finally, the function :
mk
have
1 1--. 2k
< Ok
in IN
on 0;.
Then the Ok's are decreasing and contain K. Assume ml < ... < been chosen so that :
Using the fact that there is an a mk such that :
ml
> 0 such that 141 < 1 - &k on O i , we find
m
.
is in A(D) and satisfies : F = f
onK,
and llFllm5 1. This finishes the proof of the Theorem.
187
Analytic Functiohs Exercises on Chapter VIII.
Exercise 1. - Show that the function f ( z ) = En,' zn/n is not in H m Exercise 2.
-
Show that if f is harmonic and z f (z) is harmonic, f is analytic.
Exercise 3. - Let f in H I . Show that there exist 9 , h in f =gh.
H 2 such that
Exercise 4. - Let h be a positive integrable function on n. A necessary and sufficient condition in order that h = If 1 2 , for some function f in H 2 is that J l o g h > -00. Exercise 5. - Let f E L,, 1 < p < 00, and :
Is
g
in HP ?
Exercise 6 (Hardy). - Let f f H'
, f (z) =
anz". Show that :
(Hint : first assume that u, 2 0 for all n , and use the formula :
In the general case, write f = gh, with g, h in H Z (see exercise 3 ) , with :
where B is the Blaschke product of the zeros of f . ) Exercise 7. - Let F be an outer function in H 1. Assume that f is in H', and that f / F is in L1. Show that f / F is in H' . Show that this property characterizes outer functions. Exercise 8. - Let f in a'. Show that a) or b) imply that f is an outer function : a) I / f is in H', b) Xf(z) > 0 , in D . If f is inner, show that 1 f is outer.
+
Chapter
188
vzzz
Exercise 9. - Let Hh be the subspace of H' consisting of functions f with f(0) = 0. Show that H m can be identified t o the dual of & / H J . Show that P+ is dense in H m for o ( H w , L 1 / H ; ) . Exercise 10. - Let f be analytic in the open disk. Prove that f is a Blaschke product if and only if : a)
lf(z)I 5 1 in D ,
Exercise 11 (Maximal ideals of A ( D ) ) .- Let to(.) be the identity function : f o ( z ) = z . Let $ be a character on A (D). Let a = $(fo). Show that for every polynomial p , $(p) = ~ ( c z )Deduce . the same relation for every function f in A (D),Deduce that the maximal ideals are of the form Z, :
z,
= {f E A ( D ) ; f(a)= 0).
Notes and Comments.
Our presentation is a mixture between many authors. For further information the reader should consult one of the excellent books which deal with HP spaces : Hoffman 111, Garnett [l],Duren 111, Rudin 111. The exercises also come from these books, except Exercise 11, which was communicated t o us by Richard Aron.
A nalytic Functio M
I89
Complements on Chapter VIII :
One may study the continuity of the decomposition :
f
=
B*S*F
(1)
which we have found during the previous sections. The function f is in H’ , B is the Blaschke product made with its zeros, S is a singular function and F an outer function in H’ . So the question is : if a sequence (f,),,?o converges to f in H1 , what about the corresponding terms B, , S, F, ? Do they converge to the terms B , S, F respectively ? The answer is uno” in general. Indeed, if :
l+z
S ( z ) = ezp - , 1-2
this is a singular function, as we have already seen. But if we set S,(z) = S{r.z), 0 < r < 1 , we get a sequence of functions in H’ ,which converges to S in H1. These functions are outer (to see this, one checks easily that they give an equality in Jensen’s inequality, so we apply Proposition 5.2). The following theorem was obtained by L. Bonvalot [ l ]:
Theorem 1. - The decomposition (1) is continuous a t every function f which has no non-trivial singular factor S. Since this decomposition is obtained by means of Jensen’s Inequality, its continuity is related to that of the “Jensen’s Functional” :
(the quantity e z p J ( f ) is known as Mahlcr’s measure, and it is used in Number Theory.) The continuity of Jensen’s Functional has been studied by the same author, and the result is similar :
Theorem 2. - Jensen’s Functional is continuous on H1 a t every function f which has no non-trivial singular factor in the decomposition (1). When f is a polynomial of degree n , say f = p = C,“ajz’ , estimates on J(f)have been intensively studied by many authors. For instance, Kurt Mahler [I] proved that : l o g C l a j l 2 J(!) 2 logClajI -nlog2.
Chapter VIIZ
190
(see also Arestov [l],Beller [I], Beller-Newman [l] Kurt Mahler 121 for other aspects of these estimates.)
Recently, P. Enflo and the present author (B. Beauzamy- P. Enflo [ l ] ) studied Jensen's Inequality for polynomials (or H1 functions) satisfying :
We say that a polynomial satisfying ( 2 ) has concentration d at degree k. For such a polynomial, P. Enflo and the author proved that there is a constant C ( d , k ) such that :
Numerical estimates where computed by the present author [13] for C ( d , k ) : it is the largest value (for t > 1) of the function :
Let C ( d ,k ) be the best constant satisfying (3) ; the precise value of C(d, k ) is unknown. However, it was shown in [13] that, for d = 1/2,
C ( 1 / 2 , k ) I-2klog2, and that, asymptotically, when k
--+
(5)
im,
C(d,k) 2 -2k.
(6)
The precise value of C ( d ,k ) has been computed by A. K. Rigler, S. Y.Trimble, R. S. Varga [l]for the class of Hurwitz polynomials : these polynomials have real positive coefficients, and their roots are either real negative, or pairwise conjugate, with negative real part. For this class, the best constant they find (in the case d = 1 / 2 ) is -2klog2, the upper bound in ( 5 ) , and it is obtained for the polynomial ( z 1)2k/22k,The problem is also solved for other values of d and the extremal polynomials are given ; we refer the reader to the above mentioned paper for an exposition of the complete results.
+