Statement: Supposeφis univalent self map ofU. thenCφ is compact onH2iff lim
|z|−→1−
1− |φ(z)|
1− |z| =∞ (2)
6.3.1 Proof of sufficiency:
We are assumingφis univalent and satisfies the condition (2) i.e. φsatisfies lim|z|−→1− 1−|φ(z)|
1−|z| =∞.
To showCphiis compact in H2, we employ the the sequential charaterization of compactness.
It is sufficient to show that, if{fn}is sequence that is bounded inH2and converges to 0 uniformly on the compact subset ofU then||cφfn|| −→0.
So without loss of generality let ,||fn| ≤1∀n.
Let >0 be given then the condition (2) guarantees a number 0< r <1 s.t.
1− |z|2≤(1− |φ(z)|2) for r <|z|<1 (3)
We fix this this for the remainder of the proof. According to the area integral estimate (1) of the H2norm:
1
2||Cφfn−fn(φ(0))||2≤ Z
rU
+ Z
U−rU
|(fn◦φ)0(z)|2(1− |z|2)dA(z) Givenfn−→0 uniformly on compact subset of U.
i.e. fn◦φ−→0 uniformly on the compact subset ofU. i.e. Cφfn−→0 uniformly on every compact subset of U.
=⇒ (fn◦φ)0 −→ 0 uniformly on every compact subset of U. So the first integral of above converges to 0.
Now for the second integral,
||Cφfn−fn(φ(0))||2≤o(1) + Z
U−rU
|fn(φ(z))φ0(z)|2(1− |φ(z)|2)dA(z)
≤o(1) + Z
U
|fn0(φ(z))|2(1− |φ(z)|2)|φ0(z)|2dA(z)
≤o(1) + Z
U
|fn0(w)|2(1− |w|2)|dA(w) (puttingφ(z) =w)
≤o(1) + 2||fn−fn(0)||2 (by (1) again)
≤o(1) + 2,
where in the last line we used the fact that||fn−fn(0)|| ≤ ||fn|| ≤1 for each n.
Sincefn(φ(0))−→0 , so the above estimate shows that limsupn||Cφfn|| ≤2.
=⇒ ||Cφfn|| −→0 becausewas an arbitrary positive number
=⇒ Cφ is compact.
The more subtle part of the theorem is the proof that condition (2) is necessary for compactness.
There are several paths to this result, each of which requires a new idea. We choose one based on elementary operator theory.
6.3.2 The Adjoint Operator
The scene now shifts to an abstract (separable) Hilbert space H. Recall that the norm of each elementy∈H can be expressed in terms of the inner product by
||y||= sup
x∈B
|< x, y >| (4)
whereB is the unit ball inH, and the supremum is attained at the unit vectorx=y/||y||(assuming y6= 0). In particular, the linear functional induced onH byy:
x7→< x, y > (x∈H) is bounded linear functional onH of norm||y||.
Now the Reisz representation Theorem asserts that each bounded linear functional onH is induced in this way by some (necessarily unique) vectory∈H.
If T is a bounded linear operator onH, and y∈H, then the linear functional x7→< T x, y > (x∈H)
is bounded, so there is a unique vector in H, which we denote byT∗y, that represents this functional in equation (4). The operator T∗ so defined on H is called the adjoint of T; its definition can be summarized like this:
< x, T∗y >=< T x, y > (x, y∈H).
ClearlyT∗ is a linear transformation on H, and (4) implies that||T∗ ||=||T||. It is also routine to check that
(T1 +T2)∗=T∗+T∗ and (cT)∗=cT∗.
where the symbolT, with or without subscripts, denotes a bounded linear operator on H and c is a complex number. In short, we have proved the following result, whereL(H) denotes the space of bounded linear operators onH, taken in the operator norm.
Lemma: The mapT −→T∗is a conjugate-linear isometry onL(H).
The adjoint of a finite rank operator: Suppose T is a bounded operator of rank one onH.
This means that for somex, y∈ H,
T z=< z, y > x(z∈ H) Using equation (5) we easily compute for eachz, w∈ H,
< z, T∗w >=< T z, w >=<< z, y > x, w >
=< z, y >< x, w >=< z, < x, w >y >
=< z, < w, x > y >, from which it follows thatT∗w=< w, x > y.
It is customary to write the one dimensional operator T as N., so the result just proved can be succiently rephrased as follows.
Lemma: The adjoint of a rank one operator has rank one; in fact ifx, y∈ H,then (xN
y)∗=yN x.
Since every finite rank operator is a sum of rank one operators, the Lemma and the linear na- ture of the adjoint operation yield this:
Corollary: The adjoint of a finite rank bounded operator again has finite rank.
All the results developed in this section now combine to show that the adjoint operation preserves compactness.
Proposition: The adjoint of a compact operator is compact.
Proof: Suppose T is a compact operator onH.
By the approximation theorem we that there exists a sequenceFn of bounded finite rank operators such that ||T −Fn|| −→ O. Since the adjoint operation is additive and isometric in the operator norm,
limn ||T∗−Fn∗||= lim
n ||(T−Fn)∗||= 0 Since each of the operatorsFn∗ is of finite rank and bounded.
So by the approximation theorem shows thatT∗ is compact.
Adjoint composition operators and reproducing kernels: Our second computation involves the adjoint of a composition operator. Although there is no good description of the adjoint that works for all composition operators on all H2 functions, we can always compute its action on an important special family of functions inH2: the reproducing kernels.
For each pointp∈U, let
kp(z) =def 1 1−pz¯ =
∞
X
n=0
¯ pnzn.
So clearly kp ∈H2. It is called reproducing kernel for the point p, and it gets the name from the fact that for eachf ∈H2,
< f, kp>=
∞
X
n=0
fˆ(n)pn=f(p). (6)
Lemma: Cφ∗kp=kφ(p)for eachp∈U. Proof: For each f ∈H2 we have
< f, Cφ∗kp >=< Cφf, kp>=Cφf(p) =f(φ(p)) =< f, kφ(p)>
=⇒ Cφ∗kp=kφ(p).
6.3.3 Proof of Necessity
TheoremNecessary condition for compactness Suppose φ is a holomorphic self-map ofU and that Cφ is compact onH2. Then
lim
|z|−→1−
1− |φ(z)|
1− |z| =∞ Proof: For each p∈U, let
fp(z) = kp
||kp|| =
p1− |p|2 1−pz¯ , the normalized reproducing kernel for p.
We are going to show that,
||Cφ∗fp|| −→0 as|p| −→1−. (7) This will finish the proof, since from the last lemma we can say that
||C∗φfp||2= (1− |p|2)||kφ(p)||2= 1− |p|2 1− |φ(p)2|
. To prove (7), recall that the adjoint operatorCφ∗ inherits the compactness ofCφ.
Thus the collection of Cφ∗ images of normalized reproducing kernels is a relatively compact subset ofH2, so every sequence of these images has a convergent subsequence.
We need only show that, the zero function is the only possible limit of such a subsequence. Suppose
|pn| −→1−
mboxand C∗phifpn−→g in theH2 norm.
We’ll be finished if we can show thatg= 0.
To see this, lethbe any polynomial.
Then the continuity of the inner product gives,
< g, h >= lim
n < Cφ∗fpn, h >
= lim
n
p1− |pn|2< C∗kpn, h >
= lim
n
p1− |pn|2< kφ(pn), h >
= lim
n
p1− |pn|2 h(φ(pn))
= 0
where the third line follows from the Lemma at the end of the last section, and the last one from the fact that h is bounded on U.
Thus g is orthogonal to every polynomial.
Since the polynomials form a dense subset ofH2, it follows thatg is the zero function.
With this result, the proof of the Univalent Compactness Theorem is complete.