Fix $N>1$. Let $f\in C(\mathbb{R},\mathbb{R})$ be such that the composition operator via $$ \begin{aligned} C_f:C(\mathbb{R},\mathbb{R}) &\rightarrow C(\mathbb{R},\mathbb{R}) \\ g & \mapsto g \circ f, \end{aligned} $$ is a bounded operator.

Let $C^f$ denote the adjoint operator $C_f$. Is there a criterion on $f$ to verify when $C^f$ has at most $N$ linearly independent eigenvectors?

Fix $N>1$. Let $f\in C^{\infty}(\mathbb{R},\mathbb{R})$ be such that the composition operator via $$ \begin{aligned} C_f:C^{\infty}(\mathbb{R},\mathbb{R}) &\rightarrow C^{\infty}(\mathbb{R},\mathbb{R}) \\ g & \mapsto g \circ f, \end{aligned} $$ is a bounded operator. Here the topology on $C^{\infty}(\mathbb{R},\mathbb{R})$ is the Whitney topology.

Let $C^f$ denote the adjoint operator $C_f$. Is there a criterion on $f$ to verify when $C^f$ has at most $N$ linearly independent eigenvectors?

Fix $N>1$. Let $f\in C(\mathbb{R},\mathbb{R})$ be such that the composition operator via $$ \begin{aligned} C_f:C(\mathbb{R},\mathbb{R}) &\rightarrow C(\mathbb{R},\mathbb{R}) \\ g & \mapsto g \circ f, \end{aligned} $$ is a bounded operator.

Let $C^f$ denote the adjoint operator $C_f$. Is there a criteria on $f$ to verify when $C^f$ has at-most $N$ linearly independent eigenvectors?

Fix $N>1$. Let $f\in C(\mathbb{R},\mathbb{R})$ be such that the composition operator via $$ \begin{aligned} C_f:C(\mathbb{R},\mathbb{R}) &\rightarrow C(\mathbb{R},\mathbb{R}) \\ g & \mapsto g \circ f, \end{aligned} $$ is a bounded operator.

Let $C^f$ denote the adjoint operator $C_f$. Is there a criterion on $f$ to verify when $C^f$ has at most $N$ linearly independent eigenvectors?