Hermitian composition operators on Hardy-Smirnov spaces

Abstract Let Ω be an open simply connected proper subset of the complex plane and φ an analytic self map of Ω. If f is in the Hardy-Smirnov space defined on Ω, then the operator that takes f to f º φ is a composition operator. We show that for any Ω, analytic self maps that induce bounded Hermitian composition operators are of the form Φ(w) = aw + b where a is a real number. For ceratin Ω, we completely describe values of a and b that induce bounded Hermitian composition operators.


Introduction
Let be an open simply connected proper subset of the complex plane and be a Riemann map from the open unit disc onto . The Hardy-Smirnov space H 2 . / is the Hilbert space of functions F analytic on such that the integrals of jF j 2 over the images of the circles jzj D r, 0 < r < 1, under are uniformly bounded; see [5,Chapter 10] or [10, p. 63]. We will refer to the map as the underlying Riemann map of H 2 . /.
If f 2 H 2 . /; and is an analytic self map of , then the composition operator induced by on H 2 . /, denoted by C , is the linear operator defined by C .f / D f ı : Such operators on H 2 . / are studied in [10]. Also, composition operators on a Hardy space of a half-plane are studied in [7], [8], and [9].
In this paper we study bounded Hermitian composition operators on H 2 . /. In Theorem 3.3 we show that if C is bounded and Hermitian, then .w/ D aw C b where a 2 R: However, the self map having the form aw C b; a 2 R; is not always sufficient for C to be Hermitian. Some sufficient conditions are given in Theorem 3.3 in order for C to be Hermitian. We also show that if a constant map induces a Hermitian composition operator, then is a disc.
The sufficient condition in Theorem 3.3 requires the computation of 1 ı ı which could be complicated for some . Therefore in section 4, for some domains, we provide sufficient conditions that do not require the computation of 1 ı ı . In Lemma 4.7 it is proved that if .w/ D w C b is a self map of , then C is bounded and Hermitian on H 2 . /. In Theorem 4.10 we show that if the boundary of has infinite length, has a fixed point in and a 2 ¤ 1; then C is not Hermitian on H 2 . /.
Some concrete examples are provided in section 5. First we consider the strip D fz W 1 < I m.z/ < 1g; where I m.z/ denotes the imaginary part of z: It is proved in Theorem 5.2 that the composition operator C is bounded, non-trivial and Hermitian on H 2 . / if and only if .w/ D w C b where b 2 R: Next we consider D fz W 1 < I m.z/ < 1g [ fz W 1 < Re.z/ < 1g; where the real part of z is denoted by Re.z/: In Theorem 5.3 it is proved that the only non-trivial bounded Hermitian composition operator C on H 2 . / is induced by .w/ D w:

Background material
Throughout this paper will represent an open simply connected proper subset of the complex plane. The underlying Riemann map associated with H 2 . / is denoted by .

Notation
-The set of real numbers is denoted by R.
-The complex plane is denoted by C.
-The open unit disc fz W jzj < 1g is denoted by D.
-The real part of the complex number z is denoted by Re.z/: -The imaginary part of the complex number z is denoted by I m.z/:

Hardy-Smirnov spaces
Let be a Riemann map that takes D onto . For 0 < r < 1; let r be the curve in defined by r D .fjzj D rg/: where h ; i is the inner product in H 2 : Since V is an isometry it follows that hf; f i D kV .f /k 2 H 2 D kf k 2 : Thus H 2 . / is a Hilbert space with respect to the inner product h ; i : If g 2 H 2 ; then it is easy to see that Suppose that C is a composition operator on H 2 . /: If g 2 H 2 and z 2 D; then where C ;' .g/ D g ı '. Such an operator is called a weighted composition operator. From the discussion above, C on H 2 . / is isometrically similar to the weighted composition operator C ;' on H 2 (see [10, p. 66]).
Next we cite Theorem 2.1 of [3] which characterizes maps and ' when C ;' is Hermitian on H 2 . The following theorem can also be deduced from Theorem 6 of [4]. Conversely, suppose that a 0 2 D and c; a 1 2 R: If '.z/ D a 0 C a 1 z=.1 a 0 z/ maps the unit disc into itself and .z/ D c=.1 a 0 z/; then the weighted composition operator C ;' is Hermitian and bounded on H 2 .
We cite the following theorem from [6] that will be used later.
Theorem C. Let be an analytic self map of . The composition operator C is unitary on H 2 . / if and only if is an automorphism of that takes the form .w/ D e iÂ w C for some real number Â and a complex number :

Hermitian operators
We begin our work by studying bounded Hermitian composition operators induced by constant maps. Notice that it is elementary to prove that a Riemann map from the unit disc onto another disc is a linear fractional transformation.
Lemma 3.1. Let be a constant self map of . If C is a bounded Hermitian operator on H 2 . /, then is a disc. Conversely, let be an open disc and .z/ D .az C b/=.cz C d /; where a; b; c; d 2 C; be the underlying Riemann map of H 2 . /. If is a constant map that takes the value .bd ac/=.jd j 2 jcj 2 /; then C is bounded and Hermitian.
Proof. First assume that is constant and C is a bounded Hermitian operator. For z 2 D let Since V is a linear isometry C ;' is bounded and Hermitian on H 2 . Therefore from Theorem B it follows that '.z/ D a 0 C a 1 z 1 a 0 z and .z/ D k 1 a 0 z where k; a 1 2 R. Since is a constant map ' is a constant self map of D. Hence a 1 D 0 and ja 0 j < 1: Then '.z/ D a 0 and from (4) it follows that 0 .z/ 0 .a 0 / D k 2 .1 a 0 z/ 2 : Since D .D/ and is an open set cannot be a constant map. Thus, k ¤ 0: Next we consider the two cases a 0 D 0 and a 0 ¤ 0: First assume that a 0 D 0: Then, 0 .z/ D k 2 0 .0/: Thus .z/ D k 2 0 .0/z C j for some j 2 C: Since k 2 0 .0/ is nonzero .D/ is a disc.
Next assume that a 0 ¤ 0: Then, 0 .z/ D k 2 0 .a 0 /=.1 a 0 z/ 2 and it easily follows that .z/ D k 2 0 .a 0 / a 0 .1 a 0 z/ C j for some j 2 C: Therefore is a linear fractional map without any poles on the closed unit disc. Thus .D/ is a disc. Next we prove the converse; From direct computations we get 1 ı ı .z/ D c=d and Theorem B it follows that C ;' is bounded and Hermitian on H 2 . Since C D V 1 C ;' V and V is an isometry C is bounded and Hermitian on H 2 . /.
The domain is said to be symmetric about the point p if . p/ D p: Let be an analytic self map of which is neither a constant nor the identity. Suppose that C is bounded and Hermitian on H 2 . / and 1 ı ı .0/ D 0: Then the following are true.
Since V is a linear isometry C ;' is bounded and Hermitian on H 2 . Therefore from Theorem B it follows that '.z/ D a 0 C a 1 z 1 a 0 z and .z/ D c 1 a 0 z where c; a 1 2 R. Since a 0 D '.0/; and ' D 1 ı ı it is easy to see that '.z/ D a 1 z; .z/ D c. Therefore By letting z D 0 in the equation above it can be readily seen that c 2 D 1: Notice that a 1 D . 1 ı ı / 0 .0/: (1) If a 1 D 1; from equation (6) it easily follows that .z/ D . z/ C 2 .0/: it can be easily seen that is symmetric about .0/: (2) Let .n/ denote the derivative of order n of . It follows from equation (6) that .n/ .0/ D a n 1 1 .n/ .0/ where n 2: Since is not the identity map ' is also not the identity thus, a 1 ¤ 1: Since a 1 is real and ja 1 j ¤ 1; it easily follows that .n/ .0/ D 0 for n 2: Since is a nonempty open set cannot be a constant thus is a polynomial of degree 1. Therefore .D/ is a disc.
with c; a 1 2 R; a 0 2 D; and 1 C ja 0 j 2 Ä a 1 Ä .1 ja 0 j/ 2 ; then C is a bounded Hermitian operator on (3) it follows that V C V 1 is the weighted composition operator C ;' . Since V is a linear isometry C ;' is bounded and Hermitian on H 2 . Therefore from Theorem B it follows that Since is an open set cannot be a constant map. Thus c ¤ 0: The map is nonconstant therefore ' is a nonconstant map. Hence a 1 ¤ 0: Thus, Proof of (2): we get that ' maps the open unit disc into itself. Then from Theorem B it follows that C ;' is bounded and Hermitian on H 2 . Since V C V 1 D C ;' and V is an isometry, C is bounded and Hermitian on H 2 . /.
Let … denote the upper half plane and take .z/ D i.1 C z/=.1 z/ to be the underlying Riemann map. The following result, which is also in [9] can be obtained using the theorem above.
The composition operator C is bounded and Hermitian on H 2 .…/ if and only if .w/ D w C i k; k 0:

Geometry of
Throughout Section 4 the symbol will represent an open, simply connected proper subset of the complex plane. The underlying Riemann map of H 2 . / from D onto is denoted by . There are domains with simple geometric descriptions whose Riemann maps are complicated. For such domains computing 1 ı ı and 0 = 0 ı ' as required by Theorem 3.3 could be difficult. Thus in this section we find some conditions for C to be Hermitian that does not involve the computation of 1 ı ı and 0 = 0 ı ': The following lemma describes some geometric properties of the linear fractional transformation ' when C ;' is Hermitian on H 2 .
where 0 < ja 0 j < 1 and a 1 2 R: Then the following are true. (3) See the first paragraph on page 5778 of [3].
To prove the rest fixed points of ' must be investigated. Solutions of '.z/ D z are the fixed points of '. It is easy to see that the solutions of '.z/ D z are the roots of the quadratic equation (4) If r 1 ; r 2 are the roots of the equation (7), then r 1 r 2 D a 0 =a 0 : (5) If 1 C ja 0 j 2 D a 1 ; then a routine computation yields that '.z/ D a 0 z 1 a 0 z : It is easy to see that r 1 D .1 p 1 ja 0 j 2 /=a 0 is a root of the equation (7) and jr 1 j < 1: From part (4) it follows that the other fixed point must lie outside the closed unit disc.
(6) If a 1 D .1 ja 0 j/ 2 , then the equation (7) has only one root and it is readily seen that the root is ja 0 j=a 0 : A routine computation shows that ' 0 .z/ D .1 ja 0 j/ 2 =.1 a 0 z/ 2 ; hence ' 0 .ja 0 j=a 0 / D 1: The following lemma describes how the different values of '.0/ and ' 0 .0/ affect the operator theoretic properties of the Hermitian operator C ;' . For a proof of the lemma above see the third paragraph on page 5778 of [3].
To prove (2) let z D p in (8). Proof of (3); since C is bounded and Hermitian ' is continuous on the closed unit disc (see Theorem B). Since 0 exist at p we let z D p in (8). Then r D ' 0 .p/: From part (6) of Lemma 4.1 it follows that ' 0 .p/ D 1:

Fixed points
Suppose that .w/ D aw C b; a 2 R n f1g; b 2 C is a self map of . It is easy to see that has a natural extension Q to the whole complex plane. Since Q .b=.1 a// D b=.1 a/; and Q clearly does not have more than one fixed point in C it can be readily seen that has a fixed point in if and only if b=.1 a/ 2 : Since does not possess a fixed point in it follows that ' does not fix any points in D. But, ' is an analytic self map of the unit disc, therefore it has a fixed point on the unit circle (see page 59 of [2]). Since '.D/ is a disc properly contained in D it follows that ' has exactly one fixed point on the unit circle. Let be the fixed point of '.
To prove the remaining part suppose that lim z! .z/ exist finitely. Then, there is a disc U centered at such that .D \ U / is a bounded set. The closure of '.D/ n U is contained in D. Therefore, the set .'.D/ n U / is bounded. Since Note that since ' is a nonconstant a is nonzero. Since .'.D// is a bounded set, from the equation above it follows that the set .D/ is bounded. But is an unbounded set, hence our assumption that lim z! .z/ exist finitely must be false.

Bergman space of the unit disc A 2
The set of analytic functions on the open unit disc for which where dA is the Lebesgue area measure is known as the Bergman space of the unit disc. This space is denoted by A 2 . Let w 2 D: In the following result we look at whose derivative is in the Bergman space. Proof. We prove this by contradiction. Assume that C is bounded and Hermitian on H 2 . /. Let ' D 1 ı ı (3)). Since V is an isometry C ;' is bounded and Hermitian on H 2 , hence '.z/ D a 0 C a 1 z 1 a 0 z and .z/ D c 1 a 0 z for some a 1 ; c 2 R (see Theorem B). Clearly .'.z// D . .z//: Hence Notice that ' 0 .z/ D a 1 =.1 a 0 z/ 2 is a scalar multiple of the reproducing kernel function at a 0 in A 2 : Thus, from Theorem 6 of [4] it follows that the weighted composition operator C ' 0 ;' is Hermitian on A 2 : Equation (9) yields that Therefore, 0 is an eigenvector of C ' 0 ;' : Since ' is a self map of the unit disc 1 C ja 0 j 2 Ä a 1 Ä .1 ja 0 j/ 2 ; (see part (2) of Lemma 4.1). Hence one of the following 3 cases must be true; (1) 1 C ja 0 j 2 < a 1 < .
Since the map does not have any fixed points in the map ' does not have any fixed points in D. If the case (1) or case (2) above is true, then ' has a fixed point in the open unit disc (see parts (3) and (5) of Lemma 4.1). Therefore, a 1 D .1 ja 0 j/ 2 : Now, from corollaries 2 and 15 of [4] it follows that C ' 0 ;' does not have any eigenvectors. This is the desired contradiction.
If C ' is the identity operator on the Hardy space of the unit disc H 2 , then it is very easy to see that ' is the identity function on the unit disc. Next we prove for that for any , if C is the identity on H 2 . /, then is the identity map on .
Lemma 4.6. Let be an analytic self map of . If the composition operator C is the identity operator on H 2 . /, then .w/ D w for all w 2 : Proof. If ' D 1 ı ı and D . 0 = 0 ı '/ 1=2 ; from (2) it follows that Therefore, if C is the identity operator on H 2 . / it easily follows that C ;' is the identity operator on H 2 . If Substitute the constant function f .z/ D 1 in (10) and it follows immediately that .z/ D 1 for all z 2 D: Thus, for all f 2 H 2 ; Next substitute the function f .z/ D z in (11) and it yields '.z/ D z for z 2 D: Since ' D 1 ı ı ; it follows that .w/ D w for all w 2 : For self maps .w/ D aw C b of with a 2 R; b 2 C; next we look at a D 1; a D 1 and jaj ¤ 1 separately.
Lemma 4.7. If .w/ D w C r; r 2 C maps into itself, then C is bounded and Hermitian on H 2 . /.
Proof. If w 2 ; then it is easy to see that ı .w/ D . w C r/ C r D w: Thus is an automorphism of . From Theorem C it follows that C is unitary. Hence If f 2 H 2 . /, then C .f / D f ı and it follows that C C .f / D f ı ı : Hence C C is the identity operator on H 2 . /. Therefore C D C 1 and now it follows easily that C D C : Next we look at Hermitian C where .w/ D w C r: If is an automorphism of , then C is not a bounded Hermitian operator on H 2 . /.
Proof. Suppose that C is bounded and Hermitian. Since is an automorphism of , from Theorem C it follows that C is unitary. Now, C is both Hermitian and unitary therefore, C C is the identity operator on H 2 . /. Let f 2 H 2 . /: Since C .f / D f ı ; it is easy to see that C C .f / D f ı ı : Thus, Now, from Lemma 4.6 it follows that ı .w/ D w for w 2 : But, ı .w/ D w C 2b; and this leads to a contradiction since b ¤ 0: Next we look at domains with infinitely long boundaries.
Lemma 4.9. Suppose that one-dimensional Hausdorff measure of the boundary of is infinite. Let .w/ D aw C b; a 2 R; b 2 C be a self map of . If a 2 ¤ 1 and b=.1 a/ 2 ; then C is not a bounded Hermitian operator.
Proof. Suppose that C is bounded and Hermitian on H 2 . /. For z 2 D; let '.z/ D 1 ı ı .z/ and .z/ D . 0 .z/= 0 .'.z/// 1=2 : Then V C V 1 D C ;' (see (3)). Since V is an isometry C ;' is bounded and Hermitian on H 2 . Now from Theorem B it follows that '.z/ D a 0 C a 1 z=.1 a 0 z/ where a 1 2 R: Clearly ' is a self map of D, hence from Lemma 4.1 it follows that Therefore, one of the following 3 cases must be true; Next we will show that none of these cases are possible.
Case (1): if 1 C ja 0 j 2 < a 1 < .1 ja 0 j/ 2 ; then C ;' is compact (Lemma 4.2). Thus C is compact. Since the length of the boundary of is infinite, from Theorem 1.5 of [10] it follows that H 2 . / does not possess any compact composition operators. Thus, case (1) is not possible.
Case (2): if 1 C ja 0 j 2 D a 1 ; then C ;' is an isometry (Lemma 4.2). Since V is an isometry C is also an isometry. Thus, where I is the identity operator on H 2 . /. Since C is Hermitian we get C C D I: For g in H 2 . / it is readily seen that C C .g/ D g ı ı : Hence C C D C ı : Thus, C ı is the identity operator on H 2 . / and from Lemma 4.6 it follows that ı .w/ D w for w 2 : A direct computation shows that for w 2 ; ı .w/ D a 2 w C .ab C b/ and since a 2 ¤ 1 it is clear that ı .w/ ¤ w: Therefore case (2)  Since none of the three cases are possible it follows that our assumption, C is bounded and Hermitian is false.
The next theorem summarizes results above when boundary of is infinite and has a fixed point inside . The proof of the theorem above easily follows from Lemma 4.7 and Lemma 4.9.

Strip
Throughout the subsection 5.1 the set fz W 1 < I m.z/ < 1g is denoted by and the underlying Riemann map of H 2 . / is denoted by .
Suppose that .w/ D aw C b; a 2 R and b 2 C is a self map of . It is easy to see that if jaj > 1; then . / 6 Â : Thus 1 Ä a Ä 1: Lemma 5.1. Let .w/ D aw C b; 1 < a < 1; b 2 C be a self map of . If C is a bounded Hermitian operator on H 2 . /, then has a fixed point in .
Proof. Recall that has a fixed point in if and only if b=.1 a/ is in . From Lemma 3.1 it follows that a ¤ 0: First consider 1 < a < 0: Since .0/ D b it follows that b 2 : Clearly 1 a > 1; and is a convex region that contains 0 therefore, b=.1 a/ 2 : Now consider 0 < a < 1: Let b D˛C iˇ: If w 2 ; then I m. .w// D aI m.w/ Cˇ: Thus 1 < aI m.w/ Cˇ< 1: Since I m.w/ can take any value in the interval . 1; 1/ it follows that a CˇÄ 1: We will next show that the assumption a CˇD 1; leads to a contradiction. If a CˇD 1; thenˇ=.1 a/ D 1: Since the imaginary part of b=.1 a/ isˇ=.1 a/ it follows that b=.1 a/ is on the boundary of . Let Since C is a bounded Hermitian operator '.z/ D a 0 C .a 1 z/=.1 a 0 z/ where a 1 2 R and a 0 2 D (see (3) and Theorem B). Clearly ' is a self map of D, hence from Lemma 4.1 it follows that 1 C ja 0 j 2 Ä a 1 Ä .1 ja 0 j/ 2 : Therefore, one of the following 3 cases must be true; If case (1) or (2) above is true, then from parts (3) and (5)  Then from part (6) of Lemma 4.1 it follows that ' has exactly one fixed point on the unit circle. The map ' is not an automorphism of the unit disc (see the second paragraph of page 5778 of [3]). Therefore cannot be an automorphism of . Let fw m g be a sequence in that converges to b=.1 a/: Clearly f 1 .w m /g has a subsequence f 1 .w m n /g that converges to some in the closed unit disc. From Theorem 2 of chapter 6 in [1] it follows that is on the unit circle. Let z n D 1 .w m n /: If z n ! ; from Proposition 4.4 it follows that f .z n /g diverges. Thus ¤ : Let Since ' is a linear fractional self map of D which is not an automorphism it is easy to see that y 0 2 D: The continuity of ' at yields that '.z n / ! y 0 : From equation (13) it follows that Letting n tend to infinity in equation (14) we get But .y 0 / 2 and b=.1 a/ 6 2 hence our assumption that a CˇD 1; cannot be true. Thus a Cˇ< 1: From inequality (12) it also follows that 1 Ä a Cˇ: Therefore, using a method similar to the one used above it can be proved that 1 < a Cˇ: Therefore 1 <ˇ=.1 a/ < 1 Since the imaginary part of b=.1 a/ isˇ=.1 a/ it follows that contains the point b=.1 a/: Hence has a fixed point in .
The next theorem characterizes bounded Hermitian operators on H 2 . / when is the strip fx C iy W 1 < y < 1g: Theorem 5.2. Let D fz W 1 < I m.z/ < 1g: The composition operator C is bounded, non-trivial and Hermitian on H 2 . / if and only if .w/ D w C b where b 2 R: Proof. First assume that C is bounded and Hermitian on H 2 . /. From Theorem 3.3 it follows that .w/ D awCb where a 2 R and b 2 C: Since is a self map of it is easy to see that 1 Ä a Ä 1: If 1 < a < 1; then from Lemma 5.1 it follows that has a fixed point in . Then, the part 1 of Theorem 4.10 says that C is not a bounded Hermitian operator. Thus a D˙1: Since maps to itself, if a D˙1, then it is clear that I m.b/ D 0: If a D 1; and b is a real number, then is an automorphism of . Now from Lemma 4.8 it follows that b D 0: Since C is not the identity operator it can now be concluded that a D 1: Thus, .w/ D w C b for some b 2 R: Now assume that .w/ D w C b where b is a real number. It is readily seen that is a self map of . Then from Lemma 4.7 it follows that C is bounded and Hermitian on H 2 . /.

Cross
Throughout the subsection 5.2 the set D fz W 1 < I m.z/ < 1g [ fz W 1 < Re.z/ < 1g is denoted by and the underlying Riemann map of H 2 . / is denoted by . The proof of Theorem 5.3 is similar to the proof of Theorem 5.2 therefore we will only provide an outline. The map .w/ D w is a self map of , therefore from Lemma 4.7 it follows that C is bounded and Hermitian on H 2 . /.
If C is bounded and Hermitian on H 2 . /, then .w/ D aw C b where a 2 R; b 2 C (see Theorem 3.3). If jaj > 1; then cannot contain . /: Thus jaj Ä 1: Using a proof similar to the proof of Lemma 5.1 it can be proved that a 6 2 . 1; 1/: Hence a D˙1: If .w/ D w C b maps into itself then it is clear that b D 0: Thus a D 1 results in the identity operator. If .w/ D w C b is a self map of it is not difficult to see that b D 0: Therefore .w/ D w: