Eigenfunctions of composition operators acting on the space of entire functions.

Question

Let $\lambda$ and $b$ be complex numbers with $\lambda$ nonzero. Let $\varphi(z)=\lambda z+b$ and define the composition operator $C_\varphi$ on the space of entire functions by $$C_\varphi f(z) = f\circ \phi(z) = f(\lambda z+b).$$ Is there a known description of the eigenfunctions of $C_\varphi$?

Answer 1

Special Case 1. If $b=0$ and $\lambda\ne0,1$ then the eigenvalues of $C_\phi$ are $1,\lambda,\lambda^2\dots$. The eigenvectors corresponding to the eigenvalue $\lambda^k$ are the monomials $cz^k$.

Proof: Say $f\ne0$, $C_\phi f=\alpha f$ and $f(z)=\sum c_nz^n$. Then $$\lambda^nc_n=\alpha c_n\quad(n=0,1,\dots).$$Now since $f\ne0$ there exists $k$ with $c_k\ne0$, and it follows that $\alpha=\lambda^k$ and also $\alpha\ne c^n$ for $n\ne k$, hence $c_n=0$ for $n\ne k$.

Now the general $\phi$ (for $\lambda\ne0,1$ is conjugate in $\text{Aut}(\Bbb C)$ to the transformation just treated: We have $$\lambda z+b=\lambda(z+a)-a$$if $a=\frac b{\lambda-1}$. Hence

Special Case 2. If $\lambda\ne0,1$ then the eigenvalues of $C_\phi$ are $1,\lambda,\lambda^2\dots$. The eigenvectors corresponding to the eigenvalue $\lambda^k$ are the translated monomials $c(z-a)^k$, where $a=b/(1-\lambda)$.

If $\lambda=1$ the eigenspace is much larger:

Special Case 3. If $b=0$, $\lambda=1$ then $1$ is the only eigenvalue; every non-zero entire function is an eigenvector.

Special Case 4. If $b\ne0$, $\lambda=1$ then the eigenvalues are precisely the non-zero complex numbers; if $\alpha\ne0$ the corresponding eigenvectors are the entire functions of the form $f(z)=e^{\beta z}g(e^{2\pi iz/b})$, where $g$ is an entire function and $e^{\beta b}=\alpha$.

Proof: $0$ is not an eigenvalue since $C_\phi$ is invertible.

Suppose $\alpha\ne0$ and choose $\beta$ with $e^{\beta b}=\alpha$. If $g$ is entire and $f(z)=e^{\beta z}g(e^{2\pi iz/b})$ then $f\circ \phi=\alpha f$, so $\alpha$ is an eigenvalue with eigenvector $f$.

Conversely, suppose $\alpha\ne0$ and $f$ is an entire function with $f\circ \phi=\alpha f$. Choose $\beta$ with $e^{\beta b}=\alpha$ and define $$h(z)=e^{-\beta z}f(z).$$Then $h$ is entire and $h\circ \phi=h$, so a standard exercise shows that $h(z)=g(e^{2\pi i z/b})$