1 Introduction

Let \(\mathbb {D}\) denote the unit disk of the complex plane. Let \({\text {d}}A(z)=\pi ^{-1} {\text {d}}x{\text {d}}y\) denote the Lebesgue normalized area measure. The Bergman space \(\mathcal {A}^2(\mathbb {D}) \) of the unit disk is the space of functions f analytic on \(\mathbb {D}\) for which the norm

$$\begin{aligned} \Vert f \Vert _ {\mathcal {A}^2}^2= \int _ {\mathbb {D}} \vert f( z ) \vert ^2 {\text {d}}A(z) \end{aligned}$$

is finite. We also consider the classical Hardy space \(\mathcal {H}^2(\mathbb {D})\), which consists of those analytic functions on \(\mathbb {D}\) whose Taylor coefficients are square summable. It is well known that the functions on \(\mathcal {H}^2(\mathbb {D})\) have boundary values (non-tangential limits) almost everywhere, see [7], Theorem 17.11, and, indeed, the norm

$$\begin{aligned} \Vert f \Vert _{\mathcal {H}^2}^2 = {1\over 2 \pi }\int _{-\pi }^\pi \vert f(e^{i\theta } )\vert ^2 \, {\text {d}}\theta \end{aligned}$$

is the natural one on \(\mathcal {H}^2(\mathbb {D})\).

We recall that the Bergman space \(\mathcal {A}^2(\mathbb {D})\) as well as the Hardy space \(\mathcal {H}^2(\mathbb {D})\) are Hilbert spaces. Next, consider an automorphism \(\varphi \) of the unit disk. The obvious change of variables shows that the composition operator \(C_\varphi f = f\circ \varphi \) acts boundedly on \(\mathcal {A}^2(\mathbb {D})\) as well as on \(\mathcal {H}^2(\mathbb {D})\), see [3] for instance. In addition, we can consider the Toeplitz operator \(T_{\varphi ^\prime }f = \varphi ^\prime f \), which is also bounded when \(\varphi \) is an automorphism of \(\mathbb {D}\). The obvious change of variables again shows that the operator \(U_\varphi f =T_{\varphi ^\prime } C_\varphi \) is a unitary operator on \(\mathcal {A}^2(\mathbb {D})\), that is, \(U_\varphi ^{-1}\) coincides with its adjoint \(U_\varphi ^\star \). Similarly, the operator \(U_\varphi = T_{{ {\varphi ^\prime }^{1/2}}}C_\varphi \) is a unitary operator acting on \(\mathcal {H}^2(\mathbb {D})\). Unicity results appear frequently in Analysis, see for instance [1, 2] and [4,5,6], where unicity theorems are obtained for the Klein–Gordon equation. On the other hand, it is a very basic fact that if the derivatives of any order at a point of an analytic function are zero, then the function itself is the zero function. Thus, the question whether it is possible or not to have a non-zero even (odd) function f such that \(U_\varphi f\) is odd (even) makes sense, that is, the function f has zero derivatives of odd (even) order and \(U_\varphi f\) has zero derivatives of even (odd) order. Somehow we are requiring that the zero Taylor coefficients are split between f and \(U_\varphi f\). We will show that in such a case f must be the zero function. We will also be considering a third space. The Dirichlet space of the unit disk \(\mathcal {D}^2(\mathbb {D})\), which is the space of analytic functions on \(\mathbb {D}\) for which

$$\begin{aligned} \Vert f \Vert _{\mathcal {D}_0}^2 =\int _{\mathbb {D}} \vert f ^\prime (z) \vert ^2 \, {\text {d}} A(z) \end{aligned}$$

is finite. Actually, to have a true norm we must take the space \(\mathcal {D}_0^2(\mathbb {D})\), which is equal to \(\mathcal {D}^2(\mathbb {D})\) modulo the constant functions. Alternatively, we can add \(\vert f(0)\vert ^2\) on the right-hand side of the display above to have a norm \(\mathcal {D}^2(\mathbb {D})\). In the latter case, for \(\varphi \) an automorphism of \(\mathbb {D}\), the operator \(C_\varphi \) itself acting on \(\mathcal {D}^2(\mathbb {D})\) is not unitary, but it is unitary when acting on \(\mathcal {D}_0^2(\mathbb {D})\).

2 Main Theorem

For the classical spaces of analytic functions presented in the introduction, we have

Theorem 1

Let \(\varphi \) be an automorphism of \(\mathbb {D}\). Let f be in \(\mathcal {A}^2(\mathbb {D})\) and \(U_\varphi = T_{{\varphi ^\prime } }C_\varphi \). If f is even and \(U_\varphi f\) is odd, then f is the zero function and the same is true if f is odd and \(U_\varphi f\) is even. The analogous statements for \(U_\varphi =T_{{\varphi ^\prime }^{1/2}} C_\varphi \) acting on \(\mathcal {H}^2(\mathbb {D})\) hold. The same is true for the operator \(C_\varphi \) acting on the Dirichlet spaces \(\mathcal {D}_0^2(\mathbb {D})\) or \(\mathcal {D}^2(\mathbb {D})\).

Proof

First we prove the statement for the Bergman space \(\mathcal {A}^2(\mathbb {D})\). It is well known that an automorphism \(\varphi \) of \(\mathbb {D}\) can be expressed as

$$\begin{aligned} \varphi (z)=w {p-z \over 1-{\bar{p}} z} , \end{aligned}$$
(1)

where \(w = e^{i\theta }\) for some \(\theta \) with \(-\pi < \theta \le \pi \) and \(\vert p \vert < 1\). Thus, \(C_\varphi =C_{\alpha _p}C_{\phi _\omega }\), where

$$\begin{aligned} \alpha _p(z)={ p-z\over 1 - {\bar{p}} z} \quad \text { and } \quad \phi _w (z)= w z. \end{aligned}$$

It follows that

$$\begin{aligned} w^{-1}U_\varphi f = T_{{\alpha _p^\prime }}C_{\alpha _p}C_ {\phi _w}f. \end{aligned}$$

Assume, by now, that f is even, so that \(g=U_\varphi f\) is odd. If we set

$$\begin{aligned} \mathcal {E}= \overline{\text {span}} \, \{ z^n: n \text { is even} \}, \end{aligned}$$

then \(\mathcal {E}\) is invariant under \(C_{\phi _w}\) and the orthogonal space \(\mathcal {E} ^\perp \) of \(\mathcal {E}\)

$$\begin{aligned} \mathcal {E} ^\perp = \overline{\text {span}} \, \{ z^n: n \text { is odd } \} \end{aligned}$$

is invariant under multiplication by \(w^{-1} \). Thus we find that \(C_ {\phi _w}f \in \mathcal {E}\) and \(w^{-1}U_\varphi f \in \mathcal {E}^\perp \). Thus \(U_ {\alpha _p}\) takes the function \(f_1=C_{\phi _w}f \in \mathcal {E}\) onto the function \(g_1= w^{-1}U_\varphi f \in \mathcal {E}^\perp \) if and only if \(U_\varphi \) takes \(f \in \mathcal {E}\) onto \(g=U_\varphi f \in \mathcal {E}^\perp \). Thus, we may replace \(U _\varphi \) by \(U_{\alpha _p}\) and \(f_1\) and \(g_1\) by f and g. If \(p =0\), the result is trivial. So we assume that \(p=\rho e^{i\eta }\) with \(\rho =\vert p \vert >0\) and \(-\pi < \eta \le \pi \). Since \(\alpha _p\) is its own inverse, we immediately see that \(C_{\alpha _p}^2=I\), where I is the identity operator. In addition, an easy computation also shows that \(U_{\alpha _p}^2=I\). Thus setting \(g=U_{\alpha _p} f\), we see that

$$\begin{aligned} {\left\{ \begin{array}{ll} g=U_{\alpha _p} f , \\ f =U_{\alpha _p} g. &{} \\ \end{array}\right. } \end{aligned}$$
(2)

It is clear now that if f is odd and \(U_\varphi f\) is even, we arrive to the very same conclusion. Therefore, to prove the whole statement of the theorem for the Bergman space, it is just enough to keep the assumption that f is even and \(g=U_\varphi f\) is odd.

Next, since g is odd and f is even, we see that \(C_{\phi _{-1}}f=f\) and \(C_{\phi _{-1}}g=-g\) and since \(C_{\phi _{-1}}\) is its own inverse, we see that the equations in (2) become

$$\begin{aligned} {\left\{ \begin{array}{ll} g=- C_{\phi _{-1}} U_{\alpha _p }f , &{} \\ f = C_{\phi _{-1}} U_{\alpha _p }g. &{} \\ \end{array}\right. } \end{aligned}$$
(3)

From the equations in (3), we obtain the eigenvalue equation

$$\begin{aligned} f=-(C_{\phi _{-1}} U_{\alpha _{p}})^2f . \end{aligned}$$
(4)

Equation (4) implies that i or \(-i\) is an eigenvalue of \(U_\psi \), where

$$\begin{aligned} \psi (z)= \alpha _p \circ \phi _{-1}(z)={p+z\over 1+ {\bar{p}} z} . \end{aligned}$$

Now, since \(U_{\phi _{e^{i\eta } } }U_\psi U_{\phi _{e^{-i\eta }}}=U_\sigma \), where

$$\begin{aligned} \sigma (z)={\rho +z \over 1+\rho z} \end{aligned}$$

and eigenvalues are preserved by similarities, it is enough to show that \(U_\sigma \) has not i or \(-i\) as an eigenvalue. We observe that \(\sigma \) is a hyperbolic automorphism which have \(-1\) and 1 as fixed points, being the latter the attractive fixed point. Suppose that \(U_\sigma h=\pm ih\) for some non-zero function \(h\in \mathcal {A}^2(\mathbb {D})\). Let \(\sigma _0\) denote the identity map on \(\mathbb {D}\) and \(\sigma _n=\sigma \circ \sigma _{n-1} \) for each positive integer n and, then, for negative integer n, we write \(\sigma _n= \sigma ^{-1}_{-n}\).

Next, fix any closed disk \( D_r \subset \mathbb {D}\) with radius \(0<r<1\), whose boundary is tangent to the unit circle at 1. Take then the set \(\Omega =D_r \setminus \sigma _1(D_r)\). It is straightforward to check that

$$\begin{aligned} \bigcup _{n\in \mathbb {Z}} \sigma _n(\Omega )= \mathbb {D}. \end{aligned}$$
(5)

and that \(\sigma _n(\Omega )\) are pairwise disjoint. Using the change of variables \(z= \sigma (w)\), we have

$$\begin{aligned} \int _{\sigma _{n}(\Omega ) }\vert h (z) \vert ^2 \, {\text {d}} A(z)= & {} \int _{\sigma _{n-1} (\Omega )} \vert h (\sigma (w)) \vert ^2 \vert \sigma ^\prime (w) \vert ^2 \, d A(w) \\= & {} \int _{\sigma _{n-1} (\Omega )} \vert U_\sigma h(w) \vert ^2 {\text {d}}A(w) \\= & {} \int _{\sigma _{n-1} (\Omega )} \vert \pm i h(w) \vert ^2 {\text {d}}A(w) \\= & {} \int _{\sigma _{n-1} (\Omega )} \vert h(w) \vert ^2 {\text {d}}A(w). \\ \end{aligned}$$

Therefore, it easily follows by induction that

$$\begin{aligned} \int _{ \Omega } \vert h(w) \vert ^2 {\text {d}}A(w)= \int _{\sigma _n ( \Omega )} \vert h(w) \vert ^2 {\text {d}}A(w), \quad \text { for each } n\in \mathbb {Z}. \end{aligned}$$

Now, since h is not the zero function the integrals above are different from zero and since \(h \in \mathcal {A}^2(\mathbb {D})\), we have the first inequality below and using (5), we may conclude as follows:

$$\begin{aligned} \infty > \int _{\mathbb {D}} \vert h(z) \vert ^2 \, {\text {d}}A(z) = \sum _{n \in \mathbb {Z}} \int _{\sigma _n(\Omega )} \vert h(z) \vert ^2 \, {\text {d}} A(z) = \sum _{n\in \mathbb {Z}} \int _{\Omega }\vert h (z) \vert ^2 \, {\text {d}} A(z) =\infty , \end{aligned}$$

which is a contradiction. Thus, h cannot be an eigenfunction for \(\pm i\) and the proof for the Bergman space is complete.

The proof for the Hardy space works with almost no change and it just needed modify the end of the proof for the Bergman space, in which it is proved there is no eigenfunctions for \(\pm i\). In the present situation, we fix any point \(e^{i \gamma }\), with \(0<\gamma <\pi \), thus in particular, \(e^{i \gamma } \not = \pm 1\). Then we take the semiopen arc s on \(\partial \mathbb {D}\) that goes in the counterclockwise sense from \(e^{i\gamma }\) to \(\sigma (e^{i\gamma })\). Then we set \(S=s \cup {\bar{s}}\), where the bar stands for conjugate and denote \(S_n= \sigma _n(S)\) with \(n \in \mathbb {Z}\). Clearly \(\{S_n\}\) are pairwise disjoint and

$$\begin{aligned} \partial \mathbb {D} {\setminus } \{1, -1 \} = \bigcup _{n\in \mathbb {Z}} \sigma _n (S). \end{aligned}$$
(6)

The fact that \(h \in \mathcal {H}^2(\mathbb {D})\) implies that both h and \(U_ \sigma h\) have boundary values almost everywhere and, therefore, we have the second and third equalities below. The change of variables \(z= \sigma (w )\) shows that

$$\begin{aligned} \int _{\sigma _{n}(S) }\vert h (z ) \vert ^2 \, {\text {d}} \vert z \vert= & {} \int _{\sigma _{n-1} (S)} \vert h (\sigma (w)) \vert ^2\vert \sigma ^\prime (w) \vert \, {\text {d}} \vert w \vert \\= & {} \int _{\sigma _{n-1} (S)} \vert U_\sigma h(w) \vert ^2 {\text {d}} \vert w \vert \\= & {} \int _{\sigma _{n-1} (S)} \vert \pm i h(w) \vert ^2 {\text {d}} \vert w \vert \\= & {} \int _{\sigma _{n-1} (S)} \vert h(w) \vert ^2 d \vert w \vert . \end{aligned}$$

Therefore, it easily follows by induction that

$$\begin{aligned} \int _{ S} \vert h(z) \vert ^2 {\text {d}} \vert z \vert = \int _{\sigma _n ( S)} \vert h(z) \vert ^2 {\text {d}}\vert z \vert , \quad \text { for each } n\in \mathbb {Z}. \end{aligned}$$

Now, since h is not the zero function, the function \(h(e^{i\theta }) \) is different from zero almost everywhere, see [7, Theorem 17.18]. Thus, each of the integrals above are different from zero. Therefore, since \(h \in \mathcal {H}^2(\mathbb {D})\), we have the first inequality below, and using (6), we may conclude as follows:

$$\begin{aligned} \infty > \int _{\partial \mathbb {D}} \vert h( w) \vert ^2 \, {\text {d}}\vert z \vert = \sum _{n \in \mathbb {Z}} \int _{\sigma _n(S)} \vert h(z) \vert ^2 \, {\text {d}} \vert z \vert = \sum _{n\in \mathbb {Z}} \int _{S}\vert h (z) \vert ^2 \, {\text {d}} \vert z \vert =\infty , \end{aligned}$$

which is a contradiction. Thus h cannot be an eigenfunction for \(\pm i\) and the proof is also complete for Hardy space.

To deal with the Dirichlet spaces, we first prove the result for \(\mathcal {D}_0(\mathbb {D})\). Although we can provide a proof along the lines of the Bergman case, we choose to use similarities. The map D that to each function \(f \in \mathcal {D}_0(\mathbb {D})\) assigns \(f^\prime \in \mathcal {A}^2\) is clearly an isometric isomorphism from \(\mathcal {D}_0^2(\mathbb {D})\) onto \(\mathcal {A}^2(\mathbb {D})\), whose inverse is

$$\begin{aligned} (D^{-1}g) (z)= \int _0 ^z g(z) \, {\text {d}}z \quad \text{ for } \text{ each } g \in \mathcal {A}^2(\mathbb {D}) . \end{aligned}$$

The operator D as well as \(D^{-1}\) take even (odd) functions to odd (even) functions. On the other hand, the operator \(C_\varphi \) acting on \(\mathcal {D}_0(\mathbb {D})\) is, under D, isometrically similar to \(U_\varphi =T_{\varphi ^\prime } C_\varphi \) acting on \(\mathcal {A}^2(\mathbb {D})\). Therefore, if \(C_\varphi \) acting on \(\mathcal {D}^2_0(\mathbb {D})\) takes an even (odd) function to an odd (even) function, then \(U_\varphi \) acting on \(\mathcal {A}^2(\mathbb {D})\) takes an odd (even) function to an even (odd) function, a contradiction with the result for \(\mathcal {A}^2(\mathbb {D})\). Finally, to prove the result for the Dirichlet space \(\mathcal {D}^2(\mathbb {D})\) assume first that f is odd, and \(C_\varphi f\) is even, then \(C_\varphi f\) must be a constant c because the result for \(\mathcal {D}_0(\mathbb {D})\). It follows that \(c=0\), since \(C_\varphi ^{-1}=C_{\varphi _{-1}}\) takes the constant function c onto itself and is odd. If f is even and \(C_\varphi f\) is odd, then \(C_\varphi ^{-1}\) takes an odd function onto an even function, a contradiction with what is already proved. The proof is complete. \(\square \)

Remark 1

Just replacing \(\pm i\) by any point in the unit circle, we see from the proof of Theorem 1, that no point on the unit circle is an eigenvalue of \(U_\sigma \). A similar proof for parabolic automorphisms can of course be done. It is quite clear that the point spectrum plays a role in this kind of problems.

Remark 2

Observe that in general, we cannot use similarities, unless the similarities preserve even and odd functions or take even functions to odd functions and odd functions to even functions, which is the case of the similarity used for \(\mathcal {D}_0 (\mathbb {D})\) in the proof of Theorem 1.

Curiously enough, as a corollary of the proof of Theorem 1, we have the following Theorem.

Theorem 2

Let \(\varphi \) be an automorphism of \(\mathbb {D}\) and consider the operator \(U_\varphi = T_{{\varphi ^\prime } }C_\varphi \). Then there is an even function \(f \in \mathcal {A}^2(\mathbb {D})\) such that \(U_\varphi f\) is even if and only if \(\varphi (z)=w z\) for some \(w\in \partial \mathbb {D}\). The same is true if we replace even by odd. The analogous statements for \(U_\varphi =T_{{\varphi ^\prime }^{1/2}} C_\varphi \) acting on \(\mathcal {H}^2(\mathbb {D})\) hold. The same is true for the operator \(C_\varphi \) acting on the Dirichlet spaces \(\mathcal {D}_0^2(\mathbb {D})\).

Proof

We use the same notation as in the proof of Theorem 1. It is clear that if \(p=0\) in (1), we obtain \(\varphi =\phi _w\), with \(w \in \partial \mathbb {D}\), then \(\varphi \) take even functions onto even functions and odd functions onto odd functions. If \(p\not =0\) in (1), and we assume that an even function is taken to an even function in the Bergman or Hardy space, one can follow the lines of the proof of Theorem 1 to arrive to the following eigenfunction equation

$$\begin{aligned} f=(C_{\phi _{-1}} U_{\alpha _{p}})^2f , \end{aligned}$$

We arrive to the same equation if f is odd and \(U_\varphi f\) is odd, which means that \(U_\sigma \) has \(\pm 1\) as an eigenvalue, which is impossible because the previous Remark 1. For the Dirichlet space \(\mathcal {D}_0(\mathbb {D})\), one can argue as in the proof of Theorem 1. The proof is complete. \(\square \)

Remark 1

Another form to state Theorem 2, when \(\varphi \not = \phi _w\), is that if f and Tf have vanishing derivatives of order odd (even), then f is the zero function, where T is any of the operators appearing in Theorem 2 acting in the corresponding space.

Remark 2

Observe that the Dirichlet space \(\mathcal {D}^2(\mathbb {D})\) has been excluded from the statement of Theorem 2. For the latter space, only the constant functions, which are even, are taken onto even functions, since constant functions are eigenfunctions for \(C_\varphi \) for the eigenvalue 1.

3 Concluding Remarks and Open Problems

Remark 1

A direct approach to prove Theorem 1 would be to prove that the equation \(PU_\sigma f=0\), where P is the projection onto even (or onto odd functions) has the zero function as the only solution. Theorem 1 then implies that the kernel of \(PU_\sigma \) is 0. This in particular implies that all the finite minors of the matrix of \(PU_\sigma \) are different from 0. The entries of the matrix of \(PU_\sigma \) are easily computed, which is not at all clear how to prove directly that the determinants of the minors are different from zero.

Remark 2

It seems that the unitariness of the operator does not play an essential role. For instance, it is quite obvious that one can produce a proof for invertible composition operators in the Hardy or Bergman space. Indeed, it also reduces to the eigenfunction equation for \(C_\sigma \) for the eigenvalues \(\pm i\). Now, the eigenfunctions of \(C_\sigma \) for \(\pm i\) are known, see [3], for instance, and one just need to check that there are no even nor odd eigenfunctions. What it seems to play a prominent role is the spectrum of the operators involved, specially the point spectrum. There are many problems related. One could try, for instance, to prove similar results for weighted composition operators and in particular composition operators. A reasonable condition for \(C_\varphi \) is to require that \(\varphi \) is univalent. For instance, \(\varphi (z)=z^2\) takes odd functions onto even functions. One could begin with non-unitary composition operators induced by linear fractional maps, or even invertible weighted composition operators. Another problem of some interest is to describe the maximum and minimum angle between \(U_\varphi f\) and \(\mathcal {E}\), and between \(U_\varphi f\) and \(\mathcal {E}^c\) when f runs the even (odd) functions \(\mathcal {E}^c\). For instance, if \(\varphi (z)=wz\), with \(\vert w \vert =1\), and \(f\not =0\) is even, the angle between \(U_\varphi f\) and \(\mathcal {E}\) is always 0 and the angle between \(U_\varphi f\) and \(\mathcal {E}^c\) is always \(\pi /2\).

Remark 3

If we write \(\mathbb {N}= E \cup E^c \), where E is a subset of the natural numbers \(\mathbb {N}=\{0,1,2, \ldots \}\) and denote by \(\mathcal {E}= \overline{\text {span}} \{z^n: n \in E \}\) and \(\mathcal {E}^c= \overline{\text {span}} \{z^n: n \in E^c \}\), we conjecture, under the conditions of Theorem 1, that if f is in \( \mathcal {E}\) and the image of f under the corresponding operator is in \(\mathcal {E}^c\), then f is also the zero function. Of course, one can ask the analogous question in the context of Theorem 2, in which \(\varphi (z)\not = w z\) with \(\vert w \vert =1\).