To each function $f \in E$, we associate the endomorphism $U$ of $E$. Let $\lambda \in \mathbb{R}^*$ be an eigenvalue of $U$ with eigenvector $f$, which is a solution of $(E_{1/\lambda})$. We assume that $f$ is developable as a power series on $\mathbb { R } _ { + } ^ { * }$, that is, there exists a power series $\sum _ { n \geqslant 0 } a _ { n } x ^ { n }$ of infinite radius of convergence such that $$\forall x \in \mathbb { R } _ { + } ^ { * } , \quad f ( x ) = \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n } .$$ Using the results of Part IV, show that the only possible eigenvalues of $U$ are of the form $\lambda = 1 / p$ with $p \in \mathbb { N } ^ { * }$.
To each function $f \in E$, we associate the endomorphism $U$ of $E$. Let $\lambda \in \mathbb{R}^*$ be an eigenvalue of $U$ with eigenvector $f$, which is a solution of $(E_{1/\lambda})$. We assume that $f$ is developable as a power series on $\mathbb { R } _ { + } ^ { * }$, that is, there exists a power series $\sum _ { n \geqslant 0 } a _ { n } x ^ { n }$ of infinite radius of convergence such that
$$\forall x \in \mathbb { R } _ { + } ^ { * } , \quad f ( x ) = \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n } .$$
Using the results of Part IV, show that the only possible eigenvalues of $U$ are of the form $\lambda = 1 / p$ with $p \in \mathbb { N } ^ { * }$.