when is the inner factor of f a an interpolating blaschke
TRANSCRIPT
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When Is the Inner Factor of f � f (a) an
Interpolating Blaschke Product for All a?
Carl Cowen
IUPUI
Eva Gallardo Gutierrez
Universidad Complutense de Madrid
and
Pamela Gorkin
Bucknell University
UM Dearborn, Analysis Day, 29 April 2017
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When Is the Inner Factor of f � f (a) an
Interpolating Blaschke Product for All a?
Carl Cowen
Eva Gallardo Gutierrez
and
Pamela Gorkin
Thanks to: Simons Foundation Collaboration Grant 358080
Plan Nacional I+D MTM2013-42105-P
and
Simons Foundation Collaboration Grant 243653.
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Dedicated to the Memory of Donald E. Sarason
January 26, 1933 – April 8, 2017
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Some terminology:
H1(D) = H1 is Banach space of bounded analytic functions on unit disk D
Buerling: Every function in H1 has an inner-outer factorization f = Ig
where g is outer – polynomials in z times g are dense in Hp for p <1
and I is inner –�
�limr!1� I(rei✓)�
� = 1 almost all ✓
The inner function can be factored as I = BS with both B and S inner and
for |�| = 1 and integer N � 0, the Blaschke product B with zeros {zk} is
B(z) = �zNY
k
|zk|zk
✓
zk � z
1� zkz
◆
and, for µ a singular measure on @D, the singular inner factor S is
Sµ(z) = exp
✓
�Z 2⇡
0
eit + z
eit � zdµ(t)
◆
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In particular, if f = Ig = BSg is the inner-outer factorization of f ,
the Blaschke factor, B, tells us where, in the disk, the value of f is 0,
and the singular inner factor S, tells us where,
near the unit circle, the value of f is nearly 0.
In this talk, we’ll be interested in the di↵erent factorizations, for a in D:
f � f (a) = Iaga = BaSaga
and, in this factorization, the Blaschke factor tells us
where the value of f is the same as its value at a.
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It turns out that the singular factor,
and repeated zeros in the Blaschke factor, are rarely there!
Rudin’s Theorem (1967):
For f in H1, the inner factor of f � f (a) is a Blaschke product
with distinct zeros, except for a in a set of capacity zero in D.
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We know that a function analytic on D can take the value 0
only on a countable set with no accumulation points in D.
For an H1 function f , the set where f takes the value 0
are the zeros of the Blaschke product B in the factorization f = BSg
The infinite product
B(z) = �zNY
k
|zk|zk
✓
zk � z
1� zkz
◆
converges if and only ifX
k
(1� |zk|) <1 ({zk} is a Blaschke sequence)
so H1 functions vanish on a smaller set than general analytic functions!
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If {zk} is any sequence in D, the map E from H1 into `1
given by E : f 7! (f (zk))1k=1 is clearly continuous.
Definition: We say {zk} is an interpolating sequence if the map given by
E : f 7! (f (zk))1k=1 maps H1onto `1.
Definition: We say B is an interpolating Blaschke product
if B is a Blaschke product
B(z) = �Y
k
|zk|zk
✓
zk � z
1� zkz
◆
and the zero set {zk} is an interpolating sequence.
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Definition: We say {zk} is an interpolating sequence if the map given by
E : f 7! (f (zk))1k=1 maps H1onto `1.
Interpolating sequences have come up in many di↵erent ways in problems
related to analytic functions on the disk and spaces of such analytic functions:
• Identifying the algebras between H1(D) and L1(@D)
• Understanding the topology of H1 in the corona
• Constructions of Toeplitz or composition operators with special properties
• Creating tools to study Toeplitz operators and composition operators
• Characterizing commutants of analytic Toeplitz operators
• Building universal operators to study the invariant subspace problem
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In 1958, Carleson proved:
Theorem
The set of points {zj}j�1 in the disk is an interpolating sequence
if and only if there exists � > 0 such that
infk
Y
j�1,j 6=k
�
�
�
�
zj � zk1� zjzk
�
�
�
�
� �
Because ⇢(z, w) =
�
�
�
�
z � w
1� zw
�
�
�
�
is the ‘pseudo-hyperbolic distance’ between
z and w, Carleson’s result says the points in an interpolating sequence are,
in a very strong way, uniformly far apart.
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In functional Hilbert spaces,
functional f 7! f (↵) for ↵ in D is continuous, so there is vector K↵,
called the kernel function for ↵, so that f (↵) = hf,K↵i.
For H2, the kernel functions are
K↵(z) =1
1� ↵z
One tool, in studying operators on Hilbert spaces, is choosing a good basis.
In 1961, H. S. Shapiro and A. L. Shields showed that
if {↵j}j>0 is an interpolating sequence, then the set
K↵j
kK↵jkj > 0
is (Banach) equivalent to an orthonormal basis of the subspace H2 BH2
where B is the Blaschke product with zero set {↵j}j>0.
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Using the equivalence ofn
K↵j
kK↵jk
o
j>0to an orthonormal basis, the following
characterization of certain commutants is obtained:
Theorem[C., 1978]
Let ⌦ be bounded domain in C, let ' be the covering map of D onto ⌦, and
let G be the group of linear fractional maps on D satisfying ' � I = '.
If S is a bounded operator with ST' = T'S, then for I in G, there is an
analytic function CI defined on D so that for each ↵ in D and each g in H2,
(Sg)(↵) =X
I2GCI(↵)g(I(↵)) (1)
where the series converges absolutely and uniformly on compact subsets of D.
Conversely, if bounded operator S has representation (1), then ST' = T'S.
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An easy example of an interpolating sequence is {zn} where zn = 1� 2�n
Theorem: If ' maps the unit disk into itself, '(a) = a for |a| = 1 and
'0(a) < 1, then for any point z0 in D, the sequence {'n(z0)} is an
interpolating sequence.
For '(z) = (1 + z)/2, which maps D into itself, '(1) = 1, and
'0(1) = 1/2 < 1.
Since 'n(0) = 1� 2�n, this sequence is an interpolating sequence.
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For another easy example, consider the singular inner function
S1(z) = exp
✓
�1 + z
1� z
◆
which is associated with the singular measure µ = �1, point mass at {1}.
Clearly, 0 is not in the range of S1, but every other point of D is in
the range andS1(@D \ {1}) = @D
Indeed, S1 is a covering map of D onto D \ {0}
For p 6= 0 in D, the set
S�11 ({p}) = {z : S1(z) = p}
is an interpolating sequence in D.
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In thinking about ‘covering maps’, we follow W. A. Veech’s book:
A Second Course in Complex Analysis, 1967.
Definition: Let ⌦ and ⌦1 be regions. A triple (f,⌦1,⌦) is a covering if
a) f : ⌦1 ! ⌦ is analytic and,
b) if z0 2 ⌦1 and w0 2 ⌦ satisfy f (z0) = w0 and if � is an arc from w0 in ⌦,
then f�1 can be continued along � with values in ⌦1 and initial value z0.
We say ⌦1 is a covering surface of ⌦ and f is a covering map.
If f is a covering map, then f 0(z) 6= 0 for z 2 ⌦1, and f (⌦1) = ⌦.
For S1 above, the triple (S1,D,D \ {0}) is a covering.
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If f is a covering map and ⌦ is simply connected, then f is also one-to-one.
We say that a region ⌦1 is a universal covering of a region ⌦
if ⌦1 is simply connected and there is a covering, (f,⌦1,⌦).
The name “universal covering” stems from the “universal property”:
Theorem Let (f1,⌦1,⌦) be a covering with ⌦1 simply connected.
If (f2,⌦2,⌦) is a covering,
then there is a covering (c,⌦1,⌦2) such that f1 = f2 � c.
Example: The triple (z2, D \ {0}, D \ {0}) is a covering.
The triple (S1, D, D \ {0}) is a universal covering.
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Theorem If ⌦ is analytically equivalent to a bounded region and z0 is a
prescribed point in ⌦, then there exists a unique covering f : D! ⌦ such
that f (0) = z0 and f 0(0) > 0.
We also have the following:
Proposition If ⌦ is analytically equivalent to a bounded region, not simply
connected, and f : D! ⌦ is a covering map, then the fiber
{z 2 D : f (z) = f (a)} is an infinite sequence for all a in D.
Theorem[C., 1978] Let c be a covering map from D onto bounded domain ⌦.
If ⌦ is not simply connected, then for every a 2 D, the inner factor of
c� c(a) is an interpolating Blaschke product.
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Theorem[C., Gallardo-Gutierrez, Gorkin, 2016]
Let f be non-constant analytic mapping of D onto bounded domain, f (D).
Then f 0(z) 6= 0 for every z in D
if and only if
there is an analytic, universal covering, c, mapping the simply connected
domain, U, not equal to C, onto the domain f (D)
and a univalent, analytic map, � of D into U so that f = c � �.
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Example: Let f (z) = (1 + .5z)8 and consider f (D).
The set f (D) is contained in the annulus .0039 < 2�8 < |w| < 1.58 < 26
and is not simply connected.
The derivative f 0(z) = 4(1 + .5z)7 6= 0 for z in the unit disk.
We can take U ⇢ {u : �8 log(2) < Re (u) < 8 log(1.5)} and c(u) = exp(u)
and �(D) ⇢ U, but |Im (�(z))| < 1.5⇡ (!!) so that f (z) = c(�(z)).
Indeed, �(z) = 8 log(1 + .5z) = log((1 + .5z)8)
so that f (z) = exp(log((1 + .5z)8) = (1 + .5z)8 !
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-10 -5 0 5 10 15 20 25 30-20
-15
-10
-5
0
5
10
15
20
Figure 1: The set f (D) for f (z) = (1 + .5 ⇤ z)8
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Sketch of proof: (Assume f 0(z) 6= 0 for z in D)
First, choose 0 as a base point in D and w0 = f (0) as a base point in f (D).
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Sketch of proof: (Assume f 0(z) 6= 0 for z in D)
First, choose 0 as a base point in D and w0 = f (0) as a base point in f (D).
Since f (D) is a bounded domain, there is a universal cover on any
domain U that is conformally equivalent to the unit disk. Choose u0 as a
base point in U; there is a unique universal cover, c, of U onto f (D) for
which c(u0) = w0 and c0(u0) > 0.
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Sketch of proof: (Assume f 0(z) 6= 0 for z in D)
First, choose 0 as a base point in D and w0 = f (0) as a base point in f (D).
Since f (D) is a bounded domain, there is a universal cover on any
domain U that is conformally equivalent to the unit disk. Choose u0 as a
base point in U; there is a unique universal cover, c, of U onto f (D) for
which c(u0) = w0 and c0(u0) > 0.
Since f 0(z) 6= 0, the function f is locally univalent at every point of D.
For any path, �, from 0 to z1 in D, f (�) is a path from w0 to w1 in f (D).
Because f is locally univalent, there is a unique lifting of this path in U from
u0 to u1. Now, define �(z1) = u1. Because D and U are simply connected,
all is well defined and � is univalent.
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Sketch of proof: (Assume f 0(z) 6= 0 for z in D)
First, choose 0 as a base point in D and w0 = f (0) as a base point in f (D).
Since f (D) is a bounded domain, there is a universal cover on any
domain U that is conformally equivalent to the unit disk. Choose u0 as a
base point in U; there is a unique universal cover, c, of U onto f (D) for
which c(u0) = w0 and c0(u0) > 0.
Since f 0(z) 6= 0, the function f is locally univalent at every point of D.
For any path, �, from 0 to z1 in D, f (�) is a path from w0 to w1 in f (D).
Because f is locally univalent, there is a unique lifting of this path in U from
u0 to u1. Now, define �(z1) = u1. Because D and U are simply connected,
all is well defined and � is univalent.
(Conversely, suppose f = c � �) f 0(z) = c0(�(z))�0(z); since c0 and �0 are
never 0, f 0 is never 0.
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Corollary[C., Gallardo-Gutierrez, Gorkin, 2016]
Let f be non-constant analytic mapping of D onto bounded domain, f (D).
Then f 0(z) 6= 0 for every z in D
if and only if
for each a in D, the zero sequence of f � f (a) is a single point,
or for each a in D, the zero sequence is an interpolating sequence.
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Sketch of proof: (Assume f 0(z) 6= 0 for z in D)
If f (D) is simply connected, then the universal covering map is univalent.
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Sketch of proof: (Assume f 0(z) 6= 0 for z in D)
If f (D) is simply connected, then the universal covering map is univalent.
If f (D) is not simply connected, there is a universal covering map so that
f = c � � and the zeros of c� c(u0) are all interpolating.
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Sketch of proof: (Assume f 0(z) 6= 0 for z in D)
If f (D) is simply connected, then the universal covering map is univalent.
If f (D) is not simply connected, there is a universal covering map so that
f = c � � and the zeros of c� c(u0) are all interpolating.
Conversely, suppose f 0(z0) = 0 for some z0 in D. In this case, for
g(z) = f (z)� f (z0), we have g(z0) = g0(z0) = 0, so the zero set of g, that is,
the zero sequence of f � f (z0), is not an interpolating sequence.
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Theorem[C., Gallardo-Gutierrez, Gorkin, 2016]
Let f be an H1 function.
Then f is an exactly n-to-1 map of D onto a bounded domain, where n > 1,
if and only if
f (D) is simply connected and there is a Riemann map c of D onto f (D)
such that f = c �B, where B is a finite Blaschke product of order n.
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